Maths_2

3 About this book
Great Clarendon Street, Oxford OX2 6DP
Oxford University Press is a department of the University of Oxford. Endorsed by Edexcel, this book is designed to help you achieve your best
It furthers the University’s objective of excellence in research, scholarship,
and education by publishing worldwide in possible grade in Edexcel GCE Mathematics Core 3 and Core 4 units.
Oxford New York The material is separated into the two units, C3 and C4. You can use the
Auckland Cape Town Dar es Salaam Hong Kong Karachi tabs at the edge of the pages for quick reference.
Kuala Lumpur Madrid Melbourne Mexico City Nairobi
New Delhi Shanghai Taipei Toronto
Each chapter starts with a list of objectives and a ‘Before you start’
With offices in
section to check that you are fully prepared. Chapters are structured into
Argentina Austria Brazil Chile Czech Republic France Greece
Guatemala Hungary Italy Japan Poland Portugal Singapore manageable sections, and there are certain features to look out for
South Korea Switzerland Thailand Turkey Ukraine Vietnam
within each section:
© Oxford University Press 2009
The moral rights of the author have been asserted Key points are highlighted in a blue panel.
Database right Oxford University Press (maker)
First published 2009 Key words are highlighted in bold blue type.
All rights reserved. No part of this publication may be reproduced,
stored in a retrieval system, or transmitted, in any form or by any means, Worked examples demonstrate the key skills and techniques you need

EXAMPLE 3
Divide 3x2 + 5 by x2 + 1
without the prior permission in writing of Oxford University Press,
or as expressly permitted by law, or under terms agreed with the appropriate to develop. These are shown in boxes and include prompts to guide you ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Rewrite the numerator to involve a multiple of the denominator:
reprographics rights organization. Enquiries concerning reproduction
outside the scope of the above should be sent to the Rights Department,
through the solutions. 3x 2 + 5 = 3(x 2 + 1) + 2
x2 + 1 x2 + 1
Oxford University Press, at the address above Divide each term in the numerator:

=3+ 2
You must not circulate this book in any other binding or cover Derivations and additional information are shown in a panel. x2 + 1
and you must impose this same condition on any acquirer
British Library Cataloguing in Publication Data
Data available
Helpful hints are included as blue margin notes and sometimes as blue type
ISBN-13: 9780-19-911784 0
within the main text.
1 3 5 7 9 10 8 6 4 2
Misconceptions are shown in the right Investigational hints prompt you to
Printed in Great Britain by Ashford Colour Press, Ltd, Gosport.
margin to help you avoid making explore a concept further.
Paper used in the production of this book is a natural, recyclable product made from common mistakes.
wood grown in sustainable forests. The manufacturing process conforms to the
environmental regulations of the country of origin.
Each section includes an exercise with progressive questions, starting
Series Managing Editor Anna Cox
with basic practice and developing in difficulty.
Some exercises also include
Acknowledgements
‘stretch and challenge’ questions marked with a stretch symbol
The Publisher would like to thank the following for permission to reproduce and investigations to apply your knowledge in a variety of situations. INVESTIGATION
photographs: 10 Use computer software or a graphical calculator to check
P38 PA Archive/PA Photos; p74 Daniel Padavona/Shutterstock; p90 some of your answers to the questions in this exercise.
studiovanpascal/iStockphoto; p120 Image Source/Corbis; p138 evirgen/iStockphoto; At the end of each chapter there is a ‘Review’ section which includes You may have to use abs (absolute value) button to
input a modulus.
p158 servifoto/iStockphoto; p178 Sylvanie Thomas/Shutterstock; p192 Dijital
Film/iStockphoto; p206 Jenny Horne/Shutterstock; p212 Toni Räsänen/Dreamstime;
exam style questions as well as past exam paper questions. There are also
p260 pixonaut/iStockphoto; p267 Lawrence Freytag/iStockphoto; p274 Dmitry two ‘Revision’ sections per unit which contain questions spanning a
Nikolaev/Shutterstock; p296 Hal Bergman/iStockphoto.
range of topics to give you plenty of realistic exam practice.
The cover photograph is reproduced courtesy of Frans Lemmens/Photodisc/Getty The final page of each chapter gives a summary of the key points, fully
cross-referenced to aid revision. Also, a ‘Links’ feature provides an
The publishers would also like to thank Kathleen Austin, Judy Sadler and Charlie Bond for
their expert help in compiling this book.
engaging insight into how the mathematics you are studying is relevant
to real life.
At the end of the book you will find full solutions and an index.
The free CD-ROM contains a key word glossary and a list of essential
formulae as well as additional study and revision materials and the
student book pages.

Contents C4

6 Partial fractions 147–158 10.8 Integration by parts 240
C3 6.1 Separating fractions 148 10.9 A systematic approach
6.2 More partial fractions 152 to integration 246
1 Algebra and 3 Exponentials and Review 6 156 10.10 Volumes of revolution 248
functions 1–38 logarithms 79–90 Exit 6 158 Review 10 256
1.1 Combining algebraic fractions 2 3.1 The exponential function, e x
80
Exit 10 260
1.2 Algebraic division 4 3.2 The logarithmic function, ln x 84 7 Parametric
1.3 Mappings and functions 8 3.3 Equations involving e x and ln x 86 equations 159–178 11 Differential
1.4 Inverse functions 16 Review 3 88 7.1 Parametric equations and equations 261–274
1.5 The modulus function 22 Exit 3 90 curve sketching 160 11.1 First-order differential equations 262
1.6 Solving modulus equations 7.2 Points of intersection 166 11.2 Applications of differential
and inequalities 26 4 Differentiation 91–120 7.3 Differentiation 168 equations 268
1.7 Transformations of graphs 4.1 Trigonometric functions 92 7.4 Integration 172 Review 11 272
of functions 28 4.2 The exponential function, e x 98 Review 7 176 Exit 11 274
Review 1 34 4.3 The logarithmic function, ln x 102 Exit 7 178
Exit 1 38 4.4 The product rule 104
12 Vectors 275–296
4.5 The quotient rule 106 8 The binomial series 179–192 12.1 Basic definitions and notations 276
2 Trigonometry 39–74 4.6 The chain rule 110 8.1 The binomial series 180 12.2 Applications in geometry 282
2.1 Reciprocal trigonometric functions 40 4.7 Further applications 114 8.2 Using partial fractions 184 12.3 The scalar (dot) product 286
2.2 Trigonometric equations Review 4 118 8.3 Approximations 186 12.4 The vector equation of a
and identities 46 Exit 4 120 Review 8 190 straight line 290
2.3 Inverse trigonometric functions 50 Exit 8 192 Review 12 294
2.4 Compound angle formulae 54 5 Numerical methods 121–138 Exit 12 296
2.5 Double angle and half angle 5.1 Graphical methods 122 9 Differentiation 193–212
formulae 60 5.2 Iterative methods 130 9.1 Differentiating implicit functions 194 Revision 4 297–306
2.6 The equivalent forms for Review 5 136 9.2 Differentiating parametric functions 198
a cos + b sin 66 Exit 5 138 9.3 Growth and decay 200 Answers 307
Review 2 70 9.4 Rates of change 206 Index 345
Exit 2 74 Revision 2 139–146 Review 9 210
Exit 9 212
CD-ROM
Revision 1 75–78
Revision 3 213–216 The factor formulae 1–3
Formulae to learn
10 Integration 217–260 Formulae given in examination
10.1 The trapezium rule 218 Glossary
10.2 Integration as summation 222 C3 Practice paper
10.3 Integration using standard forms 224 C4 Practice paper
10.4 Further use of standard forms 226
10.5 Integration by substitution 228
10.6 Integration using trigonometric
identities 232
10.7 Integration using partial fractions 238

1
Algebra and functions
This chapter will show you how to
combine algebraic fractions
understand that some mappings are also functions
find and use composite functions and inverse functions
use the modulus function and sketch graphs involving it
transform graphs of functions using translations, reflections,
stretches and combinations of these.

Before you start
You should know how to: Check in:
1 Combine numerical fractions. 1 Calculate

e.g. Calculate 3 + 1 = 9 + 2 = 11 a 3+2

C3
4 6 12 12 12 8 3
3 1 3 1
× = =
4 6 24 8 b 3×2
8 3

2 Find important facts to help you sketch 2 Sketch the graphs of
graphs of functions. a y = x(x - 2)
e.g. The graph of f(x) = 1 passes through the
x−2 b y = x2 + x - 2
( )
point 0, − 1 , has a vertical asymptote, x = 2, and
2 c y = 1 +2
x
approaches the value 0 as x ® ± ¥

3 Translate, reflect and stretch the graph 3 Write the equation of the image of the
of y = f(x) graph of y = f(x) when
e.g. When the graph of y = x2 is translated a f(x) = x2 is reflected in the x-axis
+2 units parallel to the x-axis, its equation
b f(x) = x2 - 5x is reflected in the y-axis
becomes y = (x - 2)2
⎛ −5 ⎞
c f(x) = x2 + x is translated by ⎜ ⎟
⎝ 0⎠
d f(x) = x2 is stretched with scale factor 4
parallel to the y-axis.

1

1 Algebra and functions

Dividing by a fraction is equivalent to multiplying by its reciprocal.
1.1 Combining algebraic fractions

EXAMPLE 3
x 2 − 2x ÷ x 2 − 4
Addition and subtraction Simplify
x − 2x − 3
2 2x − 6
You can add or subtract algebraic fractions when they have a This method is similar to adding or ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

common denominator. subtracting numerical fractions.
x 2 − 2x ÷ x 2 − 4 = x 2 − 2x × 2x − 6 Multiply the first fraction by the
+ x − 2x − 3 2x − 6 x 2 − 2x − 3 x 2 − 4
2
reciprocal of the divisor.
E.g. b + y = by + by = ay by bx
a x ay bx
by is the common denominator.
x(x − 2) 2(x − 3)
= ×
You should find and use the lowest common denominator to (x + 1)(x − 3) (x − 2)(x + 2)
keep the calculation as simple as possible. 2x
=
(x + 1)(x + 2)
EXAMPLE 1

Evaluate a 3+1 b 2 + x−2
8 6 x2 + x x2 − 1
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Exercise 1.1
1 Simplify
a 3 + 1 = 9 + 4 = 13 The lowest common denominator
3p2q pq
8 6 24 24 24 6x 2 y
is 24. a b 8ab × 3ac
2 2 c ÷ 2
3xyz c 4b 2r 4r
b Factorise the algebraic expressions first:
( b ) × a b− a
2
2 + x−2 = 2 x−2 x × x +1 a 2
x + 3 × x −1
+ d e f
x 2 + x x 2 − 1 x(x + 1) (x + 1)(x − 1) x2 − 1 x2 2
x + x − 2 x2 + x − 6
2
C3

C3
2(x − 1) x(x − 2) The lowest common denominator x ÷ x 2 + 2x
2
h 1 × x2 − 9 ÷ x + 3
2
= + g
x(x + 1)(x − 1) x(x + 1)(x − 1) is x(x + 1)(x - 1). x 2 − 5x + 6 x −4 x x − 3x x

= 2x − 2 + x − 2x
2

x(x + 1)(x − 1) 2 Simplify and express as single fractions.
y
= x 2− 2
2
a x+ b a+b c 1 − 1
y z b a ax bx
x(x − 1)
d 1 + 1 e 1 +1 f 1 − 1
xy 2 x 2 y x +1 x a −1 a +1
Multiplication and division 1 1
g 2 + a +1 h 2+ 1 i 3− 2
You do not need to have a common denominator when you The method is similar to (a + 1) x x +1
multiply or divide algebraic fractions. multiplying or dividing
numerical fractions. 4 +2 2 + 1 3+ 2
j k l
y −2 3 x − 1 x2 − 1 x x2 + x
You can only cancel factors which are common to both the Remember when cancelling
numerator and denominator. brackets to cancel the whole 2 − y 1
bracket and not just part of it. m 1 − 2 n o + 2 2
z + 1 z2 − z − 2 y + 2 y + 3y + 2
2
x + 3x + 2 x + 4x + 3
2

3y y +1
EXAMPLE 2

p + q 2z − 2z r 2+ 1 − 1
Simplify x + 2 × x(x 2 − 1)
y2 − 4 y2 + y − 2 z + 2z − 3 z − 1
2
x 2 + 2x x 2 − 4
x2 − x x2 − x − 6
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

3 Simplify
x + 2 × x(x 2 − 1) = x + 2 × x(x − 1)(x + 1) Factorise as much as you can
x2 − x x 2 − x − 6 x(x − 1) (x + 2)(x − 3) before cancelling. 1+ 1
a 1− x b y c x3 + 1 d x2 − 1 × x + 1
3

= x +1 1− 1 x2 − 1 x +x x −1
Cancel the fraction down to its 1
1− 2
x−3 simplest form. x y

2 3


Some divisions result in a remainder.
1.2 Algebraic division

EXAMPLE 2
Divide a 4037 by 16
You can use long division to divide a polynomial of degree m The method is similar to dividing
by a polynomial of degree n, where m n. numbers using long division. b 4x3 + 3x + 7 by 2x - 1
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

The degree of the resulting polynomial is m - n. a 252
16 4 0 3 7 )
3 2↓↓
EXAMPLE 1

Work out (x3 - 2x2 - x + 2) ÷ (x + 1) 83↓
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
80↓
Set out as a long division: 37
32
)
x + 1 x 3 − 2x 2 − x + 2 Write each polynomial with the
highest power of x on the left. So 4037 ¸ 16 = 252 remainder 5
5 The remainder is 5.

The lead term of x + 1 is x. = 252 5
16
Divide the lead term x into x3 and write x2 in the answer space:
or 4037 = (252 ´ 16) + 5
Multiply this x2 by x + 1. x2 Write like terms in the

3 + x2 below and
x + 1 x 3 − 2x 2 − x + 2
Bring down the next
) same column. b 4x3 + 3x + 7 does not have an x2-term.
Insert a 0x2 to fill the place value for x2: The 0x2 is similar to the 0 in
Write x x3 + x2 ↓
subtract. term -x. the number 4037, which
−3x − x
2
2x 2 + x + 2
C3

C3
)
acts as place holder in the
2x − 1 4x 3 + 0x 2 + 3x + 7 hundreds column.
Repeat this cycle until the division is complete.
4x 3 − 2x 2 ↓ ↓
The next step is to divide the lead term x into -3x2 and write - 3x in the Compare this method with the 2x 2 + 3x ↓
numerical long division
answer space: 2x 2 − x ↓
4x + 7
x 2 − 3x + 2 243
4x − 2

x3 + x2 ↓ ↓
)
x + 1 x 3 − 2x 2 − x + 2 )
23 5 5 8 9
4 6↓ ↓
98↓
9 The remainder is 9.

−3x − x ↓
2 9 2↓ 4x 3 + 3x + 7 = 2x 2 + x + 2 remainder 9
So You can also write this result as
−3x − 3x ↓
2 69 2x − 1
Subtract: 69 4x3 + 3x + 7
2x + 2 0 quotient remainder = (2x - 1)(2x2 + x + 2) + 9
2x + 2 ¯ ¯
0 giving 5589 ÷ 23 = 243 You can expand these brackets
or 5589 = 243 ´ 23 = 2x + x + 2 + 9 2 to check your answer.
2x − 1
Since the remainder is 0, x + 1 is a factor of x3 - 2x2 - x + 2. Since the remainder is 0,
23 must be a factor of 5589. -
So (x3 - 2x2 - x + 2) ÷ (x + 1) = x2 - 3x + 2 divisor
You could also write this result as x3 - 2x2 - x + 2 = (x + 1) ´ (x2 - 3x + 2)
You can then expand these brackets to check your answer.

4 5

1 Algebra and functions 1 Algebra and functions

When the highest power of the numerator is the same as the 4 In each case find the quotient and remainder when the first expression is
highest power of the denominator you can use a quicker method. divided by the second expression.
a x3 + 5x2 - 5x + 1; x - 1 b x3 + 7x2 + 8x - 2; x + 2
EXAMPLE 3

Divide 3x2 + 5 by x2 + 1 Both the numerator and
c 2x3 - 13x2 + 17x + 10; x - 3 d x3 - 5x2 + x + 10; x - 2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
denominator have degree 2.
Rewrite the numerator to involve a multiple of the denominator: e 2n3 + 7n2 - 6; 2n + 3 f 9n3 - 22n + 6; 3n - 4
3x 2 + 5 = 3(x 2 + 1) + 2 Compare this example to the
x2 + 1 x2 + 1 numerical division g 3n3 - 11n2 + 2n + 1; 3n + 1 h x4 + 2x3 - 6x - 7; x2 - 3
23 (3 × 7) + 2 2
Divide each term in the numerator: = =3
7 7 7 i 2x4 + x3 + x; x2 + 1 j x3 - 2; x2 - 1
= 3 + 22
x +1
5 Find the functions f(x) which complete these equations.
a x3 - 5x2 + x + 10 = (x - 2) ´ f(x) b x3 - 7x2 - 10x - 2 = (x + 1) ´ f(x)
Exercise 1.2
1 In each case, divide the first expression by the second expression. c 2x3 - 11x2 + 15x - 50 = (x - 5) ´ f(x) d x4 - 3x3 + 3x - 1 = (x2 - 1) ´ f(x)

a x3 - 7x2 + 14x - 8; x - 1 b x3 - x2 - 26x - 24; x + 1 6 Find the functions and k values which complete these equations.
c x3 - 8x2 + 5x + 50; x + 2 d x3 - 10x2 + 31x - 30; x - 2 a x3 - 4x2 + 4x - 1 = (x - 3) ´ f(x) + k b x3 + 8x2 + 6x -1 = (x - 1) ´ f(x) + k
C3

C3
e 2x3 + 5x2 - 4x - 3; 2x + 1 f 9x3 + 5x - 2; 3x - 1 c 4x3 - 3x + 2 = (2x - 1) ´ f(x) + k d x4 + 3x - 4 = (x2 - x - 1) ´ f(x) + k In part d k is not
a constant.
g n3 - 7n + 6; n + 3 h 2n3 - 9n2 + 5n + 6; 2n - 3 7 Divide each polynomial by the given factor and hence factorise
it completely.
i 9y3 - 16y - 8; 3y + 2 j 2a4 - 5a3 - 10a + 3; a - 3
a x3 - 4x2 + x + 6 has a factor x - 2 b x3 - 8x2 + 19x - 12 has a factor x - 4
k 8a4 - 20a3 + 60a - 18; 2a + 3 l 4z3 - 17z2 + 19z - 5; z2 - 3z + 1
c 4x3 - 13x - 6 has a factor 2x + 3 d x4 - 13x2 + 36 has a factor x2 - 4
m 6x3 + 7x2 - 23x + 4; 2x2 + 5x -1 n x3 + 1; x + 1 e x3 + 8 has a factor x + 2 f x3 - 8 has a factor x - 2

2 Evaluate these divisions, giving each answer as an integer plus
an algebraic fraction. INVESTIGATION
x+6 x +1
a b c x−2 d 3x + 7 8 x2 - 1 factorises to give (x – 1)(x + 1) and
x+2 x+2 x+2 x+2
x3 - 1 factorises to give (x – 1)(x2 + x + 1).
h 3x2 + 1
2
e 2x + 1 4x + 3
g x2 − 1
2
f Factorise x4 - 1 and x5 - 1.
x−3 2x + 1 x +1 x −1
2−x x2 + x − 1 Can you make a deduction about x6 - 1?
i 1+ x
j
x2 − 2 Make a statement about xn - 1 and see if you can prove it.
3 In each case, divide the first expression by the second expression.
a 3x3 - 5x; x3 b 3x2 + 2; x2 - 3
c 4x2 + 1; 2x2 - 1 d 14x2 + 19; 2x2 + 3

6 7


There are four kinds of mapping.
1.3 Mappings and functions
One-to-one Many-to-one
Domain and range X Y
X Y
A relationship between two sets of numbers is called a mapping.
You can define a mapping as an equation.
E.g. Consider y = 3x - 1 X Y
1 2
Set X maps onto set Y y
2 5 y
under the mapping y = 3x - 1
3 8
You can also write this 5 y=x+2
4 11 5
as x ® 3x - 1 4
4
3 y = x2 – 4x + 5
You can also represent a mapping by a set of ordered pairs 3
2 2
(x, y) which you can show in a table or as a graph. 1 1
y x x
−2 −1 O 1 2 4 O 2 4 5 6
12
x 1 2 3 4 10 Each value of x maps onto one and Each value of x maps onto one and
y 2 5 8 11 8 only one value of y, and vice versa. only one value of y, but not vice versa.
6
4 A one-to-one mapping is a function. A many-to-one mapping is a function.
C3

C3
2

O x
1 2 3 4 One-to-many Many-to-many
X Y X Y
Set X is called the domain of the mapping. Its elements are
denoted by x, the independent variable.
Set Y is called the range of the mapping. Its elements are
denoted by y, the dependent variable.
y
y x2 + (y – 4)2 = 4
You say that ‘x maps onto y’ or that ‘y is the image of x’. 6
x = (y – 2)2
4 5
3 4
A mapping is a function only if each x-value maps onto one 2
3
and only one y-value. 2
1 1
O 2 3 4 x −2 −1O 1 2 x
In the example shown, the mapping is a function.
Each x-value maps onto more than one Many x-values map onto many y-values.
You can therefore write y = 3x - 1
y-value.
as either f(x) = 3x - 1 You say ‘f(x) equals 3x - 1’.
or f: x ® 3x - 1 You say ‘the function f that maps x onto 3x - 1’.
A one-to-many mapping is not A many-to-many mapping is not
a function. a function.

8 9


EXAMPLE 2
EXAMPLE 1
Find the range of these functions and state the type of the Find the range of the function f(x) = 4 - x2
function in each case. when the domain is x Î R means that the domain
a f(x) = 2x - 1, x = {0, 1, 2, 3} a xÎR b -1 < x 3 contains all real numbers.

b f(x) = x2 + 4, x Î R, -2 < x 3 ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Sketch the graph of the function for each of the given domains:
a Sketch the mapping diagram and graph: y y
y
4 4
5
X Y 3 3
0 −1 4
2 2
1 1 3 Both the domain and range are
discrete as x can take only the 1 1
2 3 2 The maximum and minimum
four given values. x x
3 5 1 −2 −1 O 1 2 −2 −1 O 1 2 3 y-values are not always at the
−1
end points of the domain.
O 1 2 3 4 x −2
−1
−3
−4
The range of f(x) is y = {-1, 1, 3, 5} −5
f(x) is a one-to-one function.
a The points on the graph for b The points on the graph
b Sketch the graph of f(x): x Î R have y-values from for -1 < x 3 have
C3

C3
- ¥ to 4. y-values from -5 to 4,
y
including -5 itself.
The range of f(x) is
12 y Î R, y 4 The range of f(x) is
10
y = x2 + 4 The domain is continuous as it
y Î R, -5 y 4
8
takes all values from -2 to 3
6
(including 3 but not -2). The range
4

EXAMPLE 3
is therefore also continuous.
8
2 The lowest value of the range is 4 a Find the range of f(x) = x Î R, x 1
which occurs when x = 0. (x − 2)2
−2 −1 O x
1 2 3
b Find the value of k such that f(k) = 4
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Refer to C1 for revision of
The graph of f(x) shows that the range of f(x) is Sketch the graph of f(x). sketching graphs of functions.
y Î R, 4 y 13
a From the graph, you can see that, for y
f(x) is a many-to-one function. the given domain, the range is
y Î R, y > 0 10
8 8
b When x = k, =4 8 y=
(x – 2)2
(k − 2)2
6
(k − 2)2 = 8 = 2
4 4
k−2 = ± 2
2
giving k = 2 ± 2
−2 O 2 4 6 x
The value x = 2 - 2 lies outside the domain, so the only
10 possible value of k is 2 + 2. 11


Composite functions

EXAMPLE 5
Given f(x) = 2x - 1 and g(x) = x2 + 1, prove that
A composite function is formed when you combine two
gf(x) ¹ fg(x) except for two values of x.
(or more) functions. The output from the first function
Find these two values.
becomes the input to the second function. ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

When f is applied first and g second, the composite function x f(x) gf(x) fg(x) = f(x2 + 1) = 2(x2 + 1) - 1 = 2x2 + 1
f g
gf(x) is formed. gf(x) = g(2x - 1) = (2x - 1)2 + 1 = 4x2 - 4x + 2

When g is applied first and f second, the composite function x g(x) fg(x) Equate fg(x) and gf(x) and solve to find the values of x:
g f
fg(x) is formed. 4x2 - 4x + 2 = 2x2 + 1
For a composite function to be formed, the range of the 2x2 - 4x + 1 = 0
first function must be all or part of the domain of the giving x = 4 ± 16 − 8 = 1 ± 1 2
second function. 4 2

Hence fg(x) ¹ fg(x) except when x = 1 ± 1 2
2
EXAMPLE 4

If f(x) = x2, g(x) = 3x - 1 and h(x) = 1
x
find Take care with the order of
a fg(x) b gf(x) the functions. Exercise 1.3
c fgh(x) d g2(x) 1 For each function, draw a mapping diagram or a graph, find
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
the range, and state if the function is one-to-one or many-to-one.
C3

C3
a fg(x) means that g(x) is the input of f(x). a f(x) = 4x - 3, x = {0, 2, 4, 6}
Draw a flow diagram: b f(x) = 4x - 3, x Î R, 1 < x < 6
2
x 3x – 1 (3x – 1)
g f c f(x) = 9 - x, x Î R, 1 < x < 4

d f(x) = x2 + 2, x Î R
fg(x) = f(3x - 1) = (3x - 1)2
e f(x) = + x, x = {0, 1, 4, 9}
b Draw a flow diagram:
f f(x) = - x, x = {0, 1, 4, 9}
x x2 3x2 – 1
g f(x) = x + 2 , x Î R, x ¹ 2
f g
x−2
gf(x) = g(x2) = 3x2 -1
⎧ x 2, x ∈ R, 0 x
⎪ 2
h f(x) = ⎨
c fgh(x) means that h(x) is the input of g(x) and then Draw a flow diagram if you
find it helps.
⎩4, x ∈ R, 2 < x
⎪ 6
the result of this composite function is the input
of f(x).
2 a Given g(x) = mx + c, g(2) = 8 and g(3) = 11, find m and c.
(x) ( ) (x ) (x )
2
fgh(x) = fg 1 = f 3 × 1 − 1 = f 3 − 1 = 3 − 1 b Given h(x) = ax2 + bx + c and h(0) = 1, h(1) = 0, h(2) = 1,
x
find a, b, and c.
d g2(x) = gg(x) = g(3x - 1) = 3(3x - 1) - 1 = 9x - 4 g²(x) means that g(x) is used
as the input of g(x).

12 13


3 Find the domain of each function. 9 When f: x ® 5 - x, g: x ® x and h: x → 1 ,
Say whether the function is one-to-one or many-to-one. x
find
a f: x ® 3x + 5 with the range y = {5, 20, 35, 50, 65} a fg b fh c gh
b f: x ® x2 with the range y Î R, 0 y 9 d fgh e hgf f f2
c f: x ® x − 3 with the range y Î R, 0 y 3 g g2 h h2
d f: x → 4 with the range y Î R, 1 y 4
x−2 10 a Given that f(x) = 2x - 1 and g(x) = x2, show that
fg(x) ¹ gf(x) for all values of x except x = a
4 Sketch each of these functions, given the domain x Î R. Find the value of a.
Find the range of each function.
b Solve the equations
a f(x) = (x - 2)2 + 3 b f(x) = (x + 4)2 - 3
i fg(x) = 49
c f(x) = x2 - 6x + 10 d f(x) = 1 + 2x - x2 ii gf(x) = 9
iii f 2(x) = 13
5 Explain why these mappings are not functions.
c Find the values of x which map onto themselves under
⎧3x, x ∈ ,0 the function fg.
b y = 6+ x,xÎR
x 2
a y =⎨
2−x
⎩6 − x, x ∈ , 2 x 8
11 Given that f(x) = 2x + 1 and g(x) = 3x + c,

6 Sketch the graph of the function a find c if fg(x) = gf(x) for all x
C3

C3
f(x) = x + 4 , x Î R, x ¹ 2 b find a if f 2(a) = a
x−2
Find the range of f(x) and the values of x which 12 Given f(x) = p + qx, g(x) = x2 - 4 and h(x) = 3x + 1,
are unchanged by this function. find p and q such that hgf(x) = 4(3x2 + 3x - 2)

7 If f(x) = x2 + 1 and g(x) = 2x - 1, find 13 a If f(x) = 2x + 3,
a f(3) b gf(3) c gf(x) find i f 2(x)
ii f 3(x)
d g(3) e fg(3) f fg(x)
iii f 4(x)
g f 2(3) h f 2(x) i g2(3)
b If f(x) = 2x + 3, find an expression in n for f n(x).
j g2(x)

8 Find fg(x), gf(x), f 2(x) and g2(x) when INVESTIGATION

a f(x) = x2 - 2, g(x) = 3x + 4 14 Explain why the composite function fg(x) is not allowed
when f(x) = 2x + 1, x Î R, -5 x 5
b f(x) = 1 , g(x) = 3x2 + 2, x ¹ 0 and g(x) = x2, x Î R, x 0
x
How would you change the domains so that the function
c f(x) = x + 3, g(x) = 4x - 2 fg(x) can exist?
2

d f(x) = 1 , g(x) = 2 - x, x ¹ 1
x −1

14 15


You can find an inverse function
1.4 Inverse functions
algebraically by x f f(x)
either using a flow
The inverse function f -1(x) of the function f(x) maps the range diagram in f –1(x) f –1 x
of f(x) back onto its domain. reverse
or rearranging the
function y = f(x) to give
E.g. If f(x) = 3x - 1 x ×3 –1 3x – 1
x in terms of y, then
interchanging x and
then f -1(x) = x + 1 x+1 ÷3 +1 x y and writing y as f -1(x).
3 3
graphically by y
The range of f(x) is the domain of f -1(x). The notation for the inverse
reflecting the graph of y=x
The range of f -1(x) is the domain of f(x). function uses -1.
Do not confuse it with the y = f(x) in the line y = x x
Both composite functions andff -1 f -1f reciprocal. (which is equivalent to
E.g. (sin x)-1 is the reciprocal interchanging x and y).
map x back onto itself,
1
so ff -1: x ® x and f -1f: x ® x , whereas sin-1 x is the
sin x
or ff -1(x) = x and f -1f(x) = x inverse function of sin x.

EXAMPLE 1
+ +
The inverse of a one-to-one function is a function. Find the inverse function f -1(x) when f(x) = 5x2 + 4, x Î 0 R0 means all positive real
Find the domain and range of the inverse function. numbers and zero.
C3

C3
E.g. For a one-to-one function,
f -1(x) = x + 1
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

If f(x) = 3x - 1 then f(x) = y leads to f -1(y) = x
3 with no ambiguity. Sketch a graph to help you to visualise the function and its inverse: y
and f(5) = 14 so f -1(14) = 5 f(x)
If x Î R, then the inverse of y = 5x2 + 4 would not be a y=x
The inverse of a many-to-one function is a one-to-many mapping so one-to-one mapping and therefore not a function.
+ f –1(x)
the inverse is not a function. In such a case, the inverse can be a By restricting the domain to 0 , the inverse is then a 4
function only if the domain of f(x) is restricted to certain values one-to-one mapping and a function.
to make the inverse a one-to-one mapping.
First method Second method O 4 x
E.g. Draw a flow diagram for f(x): Write y = 5x2 + 4
If f(x) = x2 then f -1 (x) = ± x x f(x) Rearrange to make x
square ×5 +4 the subject:
and f(3) = 9 so f -1 (9) = ±3 y
-1 Use the method that you feel most
So f (x) is not a function. y = x2
Draw the flow diagram in reverse: y−4 comfortable with. You can use a
9 x=+
For f -1 (9) to have a unique 5 different method to check your answer.
f –1(x) square x
value, the domain of f(x) must ÷5 –4 Interchange x and y and
be restricted to non-negative numbers. root
replace y by f -1(x):
In this case, f -1(x) = + x
This reverse flow diagram gives y = + x−4
5
−3 O
3 x f −1(x) = + x − 4 f −1(x) = + x − 4
5 5
+
The domain of f(x) is 0 (x 0) and the range of f(x) is y 4
So, the domain of f -1(x) is x 4 and the range of f -1(x) is y 0

16 17


EXAMPLE 2

EXAMPLE 3 (CONT.)
Find the inverse function and its domain and range if The domain has x ¹ 3 because c f(x) = f -1(x) gives the equation + x + 2 = x 2 − 2
f(x) = 4 for x Î R, x ¹ 3 when x = 3 the denominator, This is not easy to solve.
x−3 x - 3, is zero and division by
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• zero is not defined. However, the graphs of f(x) and f -1(x) intersect on the
Make x the subject of y = 4 : line y = x
x−3 So, at this point of intersection, f(x) = f -1(x) = x
y(x - 3) = 4
xy = 3y + 4 Solve the two equations x + 2 = x and x2 - 2 = x:
3y + 4 Both these equations give x2 - x - 2 = 0
x=
y (x - 2)(x + 1) = 0
x = -1 is outside the
Interchange x and y and replace y by f -1(x): So the only solution is x = 2 domain of f -1(x).
y = 3x + 4
x

f −1(x) = 3x + 4 Exercise 1.4
x
1 Find the inverse function f -1(x) of each of these functions, f(x),
f -1(x)exists for all real values of x except x = 0 where the domain of f(x) is R.
So, the domain of f -1(x) is x Î R, x ¹ 0
a f(x) = 3x - 2 b f(x) = 2x + 1 c f(x) = 2(x + 1)
The range of f -1(x) is the domain of f(x) giving y Î R, y ¹ 3
d f(x) = x + 3 e f(x) = 1 x + 3 f f(x) = (x + 1)2, x -1
2 2
C3

C3
EXAMPLE 3

a Find f -1(x) given that g f(x) = x2 + 1, x 0 h f(x) = 6 - x i f(x) = x − 3, x 3
f(x) = + x + 2, x Î R, x -2 f(x) only exists if x + 2 0 so
Sketch the graphs of f(x) and f -1(x). the domain of f(x) is limited to 2 Find the inverse function f -1(x) of each of these functions, f(x),
b Show algebraically that f -1f(x) = x x -2. Also, only the positive where the domain of f(x) is R.
value of the square root is
c Find x such that f(x) = f -1(x) In each case, sketch the graphs of f(x) and f -1(x) on the same axes.
allowed, so that f(x) is a function.
c f(x) = x + 4
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a f(x) = 2x + 1 b f(x) = 10 - 2x
a Make x the subject of y = x + 2: 2
y2 = x + 2 d f(x) = x2 + 2, x 0 e f(x) = (x - 2)2, x 2 f f(x) = x − 4, x 4
x = y2 - 2
1 ,x>2
Interchange x and y. Replace y by f -1(x): g f(x) = x + 3 , x -3 h f(x) = i f(x) = 1 + 2, x > 0
2−x x
y = x2 - 2
f -1(x) = x2 - 2 y 3 A function is self-inverse if the function and its inverse are identical.
The graph of the inverse function is half of the curve y = f –1(x) Determine which of these functions have an inverse function.
x2 translated 2 units downwards. y=x Find the inverse function where one exists and say whether the
The graph of f -1(x) is a reflection of f(x) in the line y = x f(x)
function is self-inverse or not.
From the graph, the domain of f -1(x) is x Î R, x 0 a f(x) = 8 - x, x Î R b f(x) = 12 , x Î R, x ¹ 0
and the range of f -1(x) is y Î R, y -2 O x
x
−2
c f(x) = 4 − x 2 , x Î R, 0 x 2 d f(x) = 4 − x 2 , x Î R, -2 x 2
b Substitute f(x) = + x + 2 into f -1f(x): –2

f −1f(x) = f −1 ( + x + 2 ) = ( + x + 2 ) − 2 = x + 2 − 2 = x
2
e f(x) = 8 - x, x Î R, 0 x 8 f f(x) = x , x Î R, x ¹ 1
x −1
So, f followed by f -1 leaves x unchanged.

18 19


4 Find the inverse of f(x) = 8 - x, x Î R, x 0 and explain why 10 The function g(x) is defined as g: x → 2 , x Î R, x ¹ 2
x−2
the function f(x) is not self-inverse.
Find the inverse function g -1(x) and its domain.
Solve the equation g(x) = g -1(x)
5 Prove that g(x) = x + 1, x Î R, x ¹ 1 is self-inverse.
x −1
Find the elements of the domain that map onto themselves 11 Find the inverse g-1(x) of the function g(x) = 3x + 1 , x Î R, Find the horizontal asymptote by
x−3 letting x ® ¥. Also find the
under g(x). x ¹ 3. And show that g(x) is self-inverse. vertical asymptote.
Sketch the graphs of g(x) and g -1(x).
6 The function f(x) is defined by f(x) = x2 - 6x + 14, x Î R, x 3 To complete the square, rewrite Find the values of x which map onto themselves.
By completing the square, draw the graph of f(x) and find in the form (x - a)2 + b
its range. See C1 for revision. 12 The function h(x) = cx + 1 , x Î R, x ¹ 2 is self-inverse.
x−2
Find the inverse function f -1(x), stating its domain and range. Find the value of c.
Sketch its graph on the same axes as f(x). Find the values of x which map onto themselves.

7 For each of these functions, find its range and its You can use the ‘completing 13 Prove that f(x) = a x + b , x ¹ a is self-inverse for all a and b.
the square’ method. x −a
inverse function. Find the values of x which map onto themselves.
State the domain and range of the inverse function.
Sketch the graphs of y = f(x) and y = f -1(x) on the same axes.
14 Sketch the graph of the function f(x) = x 2 + 1 , x ∈ R 0 , x ≠ 1
2
+
a f: x ® x2 - 4x + 5, x Î R, x 2 x −1
and its inverse f -1(x) on the same axes.
b f: x ® x2 - 8x + 21, x Î R, x 4
State the domain and range of the inverse function.
C3

C3
c f: x ® 4x - x2, x Î R, x > 2 Explain why the solution of the equation f(x) = f -1(x) is also
a solution of x3 - x2 - x - 1 = 0
d f: x ® x2 + 4x + 2, x Î R, x > -2
Show that an approximate solution is x = 1.84
8 For each of these functions, f(x), find f -1(x) and its domain.
Solve the equation f(x) = f -1(x)
Find the points where the graphs of y = f(x) and y = f -1(x) intersect. INVESTIGATION

a f(x) = 1 x + 4, x Î R 15 Which one of these four functions has an inverse function?
2
Explain why the other three functions do not have
b f(x) = x2, x Î R, x 0
inverse functions.
c f(x) = (x - 2)2, x Î R, x 2 f1(x) = x2 - 5, x Î R
d f(x) = x2 + 8x + 12, x Î R, x -4
f2(x) = 32 , x ∈
x
9 a Find a, b and c such that f -1(x) = x2 + ax + b, x Î R, x c
f3(x) = 5 – 2x, x Î R+
is the inverse function of f(x) = 1 + x + 1, x Î R, x -1
f4(x) = sin x, x Î R, 0 x p
b Find the values of
i f(8)
ii f -1(8)
iii x such that f(x) = f -1(x)

20 21


EXAMPLE 2
1.5 The modulus function Sketch the graph of y = 2|x| - 3
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

The modulus of a number |x| is its magnitude or absolute value. Sketch the graph of y = 2x - 3 for x 0 y
and reflect it in the y-axis:

In general, when x 0, |x| = x On your calculator, |x| may be
The reflection has the equation 3
y = -2x - 3 y = 2|x| – 3
when x < 0, |x| = -x written as ABS(x).
The function f(x) = 2|x| - 3
can be written as O x
E.g. 1.5
|2| = 2 and |-2| = 2
⎧2x − 3, x 0
|x| < 2 can also be written as -2 < x < 2 An empty circle shows that f(x) = ⎨
−2 2 the end value is not included. ⎩ −2x − 3, x < 0 –3

|x| 2 can also be written as x 2 or x -2 A filled circle on a number
−2 2 line shows that the end
value is included.
The graph y = |f(x)|

EXAMPLE 3
|f(x)| is always positive, so the graph y = |f(x)| always lies above
Sketch the graphs of
the x-axis.
a y = f(x) b y = |f(x)| c y = f(|x|)
when f(x) = x 2 - 2x - 8
EXAMPLE 1

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Sketch the graph of y = |2x - 3| y
a y = x2 - 2x - 8
C3

C3
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Differentiating or completing the
= (x - 4)(x + 2) y = x2 – 2x – 8
square shows that there is a
Sketch the graph of y = 2x - 3 Show the part below the x-axis minimum value at (1, -9).
When x = 0, y = -8
Where y < 0, reflect the graph in the x-axis: as a dashed line.
When y = 0, x = 4 or -2 You could use computer software
The reflected line has the equation y = -2x + 3 y
or a graphical calculator to check
The function f(x) = |2x - 3| can be written as these results.
−2 O 4 x
⎧2x − 3, x 1 1 3
⎪ 2 y = |2x – 3|
f(x) = ⎨
⎪−2x + 3, x < 1 1
⎩ 2
O 1.5 x –8
(1, –9)

–3
The graph of y = f(|x|) y
b For y = |f(x)|, reflect the
|x| is always positive. So, the two points on the graph with x = k part of the graph below the (1, 9)
and x = -k have the same value of f(|x|). The graph y = f(|x|) is x-axis in the x-axis. y = |x2 – 2x – 8|
8
therefore symmetrical about the y-axis.

−2 O 4 x

–8 The solution to part c is shown
on the next page.
22 23


EXAMPLE 3 (CONT.)
4 Sketch the graph of g(x) = |6 - 2x| for the domain 0 x 8
c For y = f(|x|), reflect the part of y
y= |x|2 – 2|x| – 8 Find the range of the function and the solution of the equation
the graph to the right of the
8 g(x) = 4
y-axis in the y-axis:

5 Sketch the graph of h(x) = |2x + 5| for the domain -4 x 1
Find the range of the function and solve the equation h(x) = 3

−4 O x
4 6 Sketch the graph of y = f(x), y = |f(x)| and y = f(|x|) for p x p
where
a y = |sin x|
–8
b y = sin |x|
(–1, –9) (1, –9)
c y = |cos x|
d y = cos |x|
Exercise 1.5
1 Sketch the graphs of these functions. 7 Find the point of intersection of the graphs of y = |5 - x|
Give the coordinates of any point where a graph meets the and y = |x + 1|
x-axis or y-axis.
a y = |x - 3| b y = |x| - 3 8 Find all points of intersection of the graphs of y = |x + 2|
and y = 2|x| - 4
C3

C3
c y = |2x - 1| d y = 2|x| - 1

e y = |4 - x| f y = 4 - |x| 9 How many solutions are there to the simultaneous equations
y = (|x| - 2)2 and y = |2x - 4|?
g y = |2x + 3| h y = -|2x + 3| Find them.
i y = |x2 - 4x + 3| j y = |x|2 - 4|x| + 3

k y = |3 - 2x + x2| l y = 3 - 2|x| + |x|2 INVESTIGATION
10 Use computer software or a graphical calculator to check
m y= 6 n y = -|x| some of your answers to the questions in this exercise.
x
You may have to use abs (absolute value) button to
o y = |log10 x | input a modulus.

2 Sketch the graphs of these functions.
a y = |x + 4| + |x - 1|
b y = |x - 2| + |x - 1|

3 On separate axes, sketch the graphs of y = f(x), y = |f(x)|
and y = f(|x|) where
a f(x) = x2 - 3x - 4
b f(x) = 4x - x2

24 25


Exercise 1.6
1.6 Solving modulus equations and inequalities
1 Use these two diagrams to help you solve
a b y
You can use a graphical or an algebraic method to find solutions y
y = |x – 4| y = |x| + 2
to equations and inequalities which involve modulus signs.
5
EXAMPLE 1

4 y=5 4 y = |x – 4|
Solve |2x + 1| = 5
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
2
Algebraic method Graphical method O 4 x –2 O 4 x
(2x + 1)2 = 52 Find the points of intersection y
4x 2 + 4x + 1 = 25 of y = |2x + 1| and y = 5: –4 –4
y = –2x – 1 y = 2x + 1
4x2 + 4x - 24 = 0 At P, 2x + 1 = 5 so x = 2
y=5
x2 + x - 6 = 0 The reflection of y = 2x + 1 in the Q P
i |x - 4| = 5 ii |x - 4| < 5 i |x - 4| = |x| + 2 ii |x - 4| > |x| + 2
(x + 3)(x - 2) = 0 x-axis is y = -2x - 1:
1
x = -3 or 2 At Q, -2x - 1 = 5
O x 2 Solve these equations and inequalities.
-2x = 6 so x = -3
The solutions are a i |x + 2| = 5 ii |x + 2| < 5
The solutions are
x = -3 and x = 2
x = -3 and x = 2 b i |2x + 3| = 7 ii |2x + 3| > 7
c i |3x - 1| = 8 ii |3x - 1| 8
EXAMPLE 2

Solve a |x| + 4 = 3x b |x| + 4 < 3x d i |6 - 2x| = 2 ii |6 - 2x| 2
C3

C3
The graph of y = |x| + 4 involves
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
a reflection in the y-axis.
e i |x - 2| = |x| ii |x - 2| < |x|
Sketch the graphs of y = |x| + 4 and y = 3x:
y
a The two graphs have only f i |3x - 2| = |x + 1| ii |3x - 2| < |x + 1|
b For |x| + 4 < 3x, you y = 3x
one point of intersection, P. need the graph of y = –x + 4

At P, x + 4 = 3x y=x+4 3 Solve these equations.
y = |x| + 4 to be below
4 = 2x the graph of y = 3x P a |x + 1| = x + 4 b |x| + 1 = x + 4 c |2x + 3| = 3x - 2
4
This occurs to the d |2x + 3| = 6 - x e 2|x| + 3 = 4x f |2x - 4| = |x| + 2
The solution is x = 2
right of P. O x
–4
The solution is x > 2 g |x2 - 4x| = 3x - 6 h |x|2 - 4|x| = x

4 Solve these inequalities.
EXAMPLE 3

Both graphs involve a reflection in a |2x - 1| > x + 1 b 2|x| - 1 < x + 1 c |3x - 6| < |x|
How many solutions has the equation |x2 - 3x| = |x - 2| ?
the x-axis.
Find the solution nearest to x = 3 d |2x - 2| |x| - 3 e |x2 - 3x| 2|x| - 4 f |x2 - 4| < |x| + 2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
y
y = |x2 – 3x|
Sketch the graphs of y = |x2 - 3x| and y = |x - 2|: x−2 < 4
5 Solve a |x2 - x - 6| = |x| + 2 b
The two graphs intersect four times. x+3
So, there are four solutions to the equation |x2 - 3x| = |x - 2| 2 y = |x – 2|
P
The solution nearest to x = 3 is given by the point P. O x INVESTIGATION
1 2 3
At P, the graphs of y = x - 2 and y = -(x2 - 3x) intersect. –2 6 If |x2 + bx + c| = x2 + bx + c for all x, find a condition
So x - 2 = -(x2 - 3x) which b and c must satisfy.
x2 - 2x - 2 = 0 giving x = 1 ± 3 Reflecting y = x2 - 3x in the x-axis
gives y = -(x2 - 3x) = -x2 + 3x
Reject the invalid value 1 - 3 which is negative.
The required solution is x = 1 + 3 = 2.73 (to 2 d.p.) Use the quadratic formula.
26 27
See C1 for revision.


EXAMPLE 2
1.7 Transformations of graphs of functions Describe the two transformations needed to transform
the graph of y = x2 into the graph of y = 4 - x2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Recall the rules of transforming a function: Refer to C1 for revision.
Work out the result of a reflection in y You can use a sketch to check
the x-axis: y = x2 your work.
y = f(x) ± a is the result of translating y = f(x) parallel to the
y-axis by ± a units If y = f(x) = x2, then after 4 In Example 2, the order matters:
reflection in the x-axis the reflection must be first and
y = f(x ± a) is the result of translating y = f(x) parallel to the
y = -f(x) = -x2 the translation must be second.
x-axis by a units If you did the translation first and
y = -f(x) is the result of reflecting y = f(x) in the x-axis Now work out the result of translating the reflection second, then the
y = f(-x) is the result of reflecting y = f(x) in the y-axis the new function, y = -x2, parallel to final graph would have the
−2 O 2 x
y = af(x) is the result of stretching y = f(x) parallel to the equation y = -(x2 + 4) = -x2 - 4
the y-axis 4 units upwards: y = 4 – x2
y-axis by a scale factor of a y
y = f(ax) is the result of stretching y = f(x) parallel to the After the translation, y = f(x) = -x2 y = –x2 y = x2 + 4

x-axis by a scale factor of 1 is transformed into y = f(x) + 4
a = 4 - x2
These are the two required transformations. y = x2
You can combine these transformations to give new functions. 4
2
E.g. When y = f(x) is transformed by:
−4 −2 O x
⎛a⎞ ⎛0⎞ −2
2 4
translations of ⎜ 0 ⎟ and then ⎜ ⎟ , the result is y = f(x - a) + b
C3

C3
⎝ ⎠ ⎝b⎠ −4
The order in which you do the
a stretch of scale factor 2 parallel to the x-axis and then a reflection transformations can sometimes
in the x-axis, the result is y = −f x ( 2) affect the final function.
y = –x2 – 4
EXAMPLE 1

The quadratic function f(x) = x2 - 3x is reflected in the y-axis

EXAMPLE 3
⎛2⎞ Sketch the graph of the function y = 3sin 2x, x Î R and find its range.
and then translated by the vector ⎜ ⎟ .
0
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

⎝ ⎠
Find the equation of the final function. If y = sin x, then y = sin 2x is the result of a stretch y
parallel to the x-axis with scale factor 1 .
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
2
The ‘stretch’ is really a ‘squash’. 3
After reflection in the y-axis, the function becomes The reflection in the y-axis
y = 3sin 2x
f(x) = (-x)2 - 3(-x) = x2 + 3x changes f(x) to f(-x).
y = sin 2x 2

After the translation, the function now becomes 1
The translation changes the new If y = sin 2x, then y = 3sin 2x is the
f(x) = (x - 2)2 + 3(x - 2) f(x) to f(x - 2). result of a stretch parallel to the y-axis −2p −p O x
= x2 - 4x + 4 + 3x - 6 p 2p
with scale factor 3. −1
= x2 - x - 2
You can check the answer using The range of y = 3sin 2x is -3 x 3 (or |x| 3). y = sin x
−2
The equation of the final function is f(x) = x2 - x - 2 a graph-plotter. Each point on y = sin 2x maps onto a point three
−3
times further away from the x-axis.
The stretch parallel to the x-axis has changed
the period from 2p to p (360° to 180°).

28 29


EXAMPLE 6
EXAMPLE 4

Sketch the graph of y = 1 + cos x − p , x Î R and find its range. ( ) This diagram shows the graph of y = f(x) This function is defined by its

(2 )
4 graph rather than by an equation.
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• On the same axes, sketch the graph of y = 3 − f 1 x
y
If y = cos x, then y = cos x − p is the
4 ( ) y

y = 1 + cos x – p
Find the coordinates of the images of points
O (0, 0) and P (2, 3).
result of a translation parallel to the 2 4 ( ) 3
P
y = f(x)
y = cos x
x-axis of + p units to the right. 1
4 y = cos x – p ( )
4
y = f(x - a) is the result of translating y = f(x) O O x
−2p −p p 2p x 2
parallel to the x-axis by +a units.
−1

( )
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

If y = cos x − p , then −2 The three transformations involved are, in this order:
4
y = 1 + cos x − p is the result of a
4 ( ) By not involving any stretches
y
y = f(x)
translation parallel to the y-axis of +1 unit upwards. in this problem, you have not 3
O#
P y=f 1x
2 ( )
changed the period of 2p.
y = f(x) + a is the result of translating y = f(x) parallel to the y-axis
by +a units. In this problem, the order in
y = 3 –f 1x (2 )
( 4)
which you do the two O x
2 4 P#
The range of y = 1 + cos x − p is 0 x 2. translations does not matter. –1
–2
–3 y = –f 1x ( )
C3

C3
2
–4
EXAMPLE 5

Sketch the graph of y = 2|x + 3| - 4, x Î R and find its range. –5
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

i a stretch parallel to the x-axis of scale factor 2
Three transformations of y = |x|
Firstly, there is a translation of y = |x| y ii a reflection in the x-axis
are involved.
parallel to the x-axis of -3 units to iii a translation parallel to the y-axis of +3 units upwards.
y = 2|x + 3| y = |x + 3|
the left. The image points are O¢(0, 3) and P¢(4, 0).
3
y = |x|
Secondly, there is a stretch parallel 2
to the y-axis of scale factor 2. 1
Exercise 1.7
Thirdly, there is a translation parallel −3 −2 −1 O 1 2 3 x
−1 1 This diagram gives the graph of y = x y
to the y-axis of -4 units downwards. On the same axes, sketch the graphs of
−2 y = √x
The range is y -4 y = 2|x + 3| – 4 −3 3
i y= x ii y = x + 3
−4
iii y = x − 3 iv y = x + 3 O x
3 6 9
v y= x -3 –3
Label each sketch with its equation.

2 On the same axes, sketch the graphs of
y = x2 y = x2 + 3 y = x2 - 3 y = (x + 3)2 y = (x - 3)2
Label each sketch with its equation.

30 31


3 a If f(x) = |x|, sketch on the same diagram the graphs of 8 Describe the transformations of the graph of y = cos x which
y = f(x), y = f(2x), y = 2f(x) result in a graph with the equation
a y = 1 + 2cos x
b If f(x) = |x|, sketch on the same diagram the graphs of
y = f(x) y = f(x + 2) y = f(x - 2) b y = 3 + cos x + p( 4 )
Label each graph with its equation.
c y = 3 - cos 2x
4 Sketch the graph of y = x2 - 3x
Sketch the graph and write the equation after the graph
(
d y = 4cos x − p
2 )
of y = x2 - 3x has been transformed by e y = 2cos 3x
a a reflection in the x-axis b a reflection in the y-axis f y = 1 cos(-x)
2
⎛2⎞ ⎛ 0⎞
c a translation of ⎜ ⎟ d a translation of ⎜ ⎟ .
⎝0⎠ ⎝ −4 ⎠ 9 On the same axes, sketch the graphs of y = tan x and

2
(
y = 1 + tan x − p for 0
2 ) x p
5 The graph of y = x is reflected in the x-axis and then translated
5 units upwards parallel to the y-axis. Sketch the final image of
10 The graph shows the function y = f(x)
y = x2 and write its equation.
y
6 Sketch the graph of y = 4 - x2
C3

C3
a Sketch the image of the graph of y = 4 - x2 after a reflection
x
in the x-axis followed by a stretch (scale factor 2) parallel (–5, 0) O (0, –1)
to the x-axis.
(–3,–4)
b Sketch the image of the graph of y = 4 - x2 after a
reflection in the y-axis followed by a stretch
a The graph of the function y = f(x) is transformed into the
(scale factor 2) parallel to the x-axis.
Write the equations for the resulting graphs. graph of y = f 1 x + 1 (2 )
i Describe the transformations which have taken place.
7 Sketch the graph of y = f(x) and its image when ii Sketch the graph of y = f(x) and its image on the same diagram.
a f(x) = x2 - 4x + 3 is stretched (scale factor 2) parallel to the b Repeat when f(x) is transformed into y = 2 - 3f(x - 1)
y-axis and then reflected in the x-axis
b f(x) = x2 - 3x is reflected in the x-axis and then translated
+1 units parallel to the x-axis INVESTIGATION

c f(x) = |x - 4| is stretched (scale factor 3) parallel to the x-axis 11 Which two transformations result in the graph of
and then translated -2 units parallel to the x-axis y = f(x) being
a enlarged with scale factor k and centre (0, 0)
d f(x) = sin x is reflected in the y-axis and translated +1 unit
parallel to the y-axis b rotated through 180° about the origin (0, 0)?
⎛π⎞ Write the equation of the image of y = f(x) in each case.
e f(x) = sin x is reflected in the x-axis and translated by the vector ⎜ 2 ⎟ .
⎜3 ⎟
⎝ ⎠
In each case, write the equation of the image.

32 33


Review 1 5 The function f is defined over the domain 0 x 4 by
f(x) = x 0 x<2

1 Express each of these expressions as a single fraction in its f(x) = 4 - x 2 x 4
simplest form. a Sketch the graph of f over its domain.
1 + 2 2 − 2x
a
x − 1 2x + 3
b
x +1 x +2 b Find all the values of x for which f(x) = x + 4 [(c) Edexcel Limited 2002]
4
2
3 + 3 2a3(a + b) × 6b
c (x + 1)(x − 4) (x − 1)(x − 4)
d 6 The function g is defined by g: x ® 4x2 + 9, x Î R, x 0
3b a2 − b2
a Find the inverse of g.
3m(m − 2) m2 − m − 2 4x 2 × x 2 − 3x + 2
e ÷ f
m −1 m2 (x − 1)2 8x 3(x − 2) b Find the value of g -1(18).

2a Express x + x + 12 as a single fraction in its
(x + 1)(x + 3) x 2 − 9 7 Prove that the function f(x) = x + 2 , x Î R, x ¹ 1 is self-inverse. Self-inverse means that the
x −1
simplest form. [(c) Edexcel Limited 2001] function and its inverse
Also prove that the only elements of the domain which map are identical.
onto themselves are x = 1 ± 3
b Hence, find the solution of x + x + 12 = −1 1
(x + 1)(x + 3) x 2 − 9 2
8 Given that f(x) = x2 - 5x + 6, sketch the graphs of
3a Find the quotient and remainder when a y = |f(x)|
C3

C3
i 4x3 + 4x2 - 5x - 3 is divided by 2x + 1 b y = f(|x|)
ii x3 - 7x + 8 is divided by x - 2
b Find the function f(x) and the constant k which satisfy the identity 9 This diagram shows the graph of the function
2x3 - 3x2 + 4 º (x -1) f(x) + k y = f(x), -1 x 6
y
c Divide f(x) = x3 - 2x2 - 9x + 18 by x - 2 and so factorise
f(x) completely. 2 y = f(x)

4 The functions f and g are defined by f(x) = x2 - 1, x Î R, x 0
and g(x) = 2x - 1, x Î R
a Find the values of
–1 O 3 6 x
i f(3) ii g(3) iii fg(3)
iv gf(3) v f -1(3) vi g -1(3)
b Find algebraic expressions for
Sketch, on separate diagrams, the graphs of
i fg(x) ii gf(x)
iii f -1(x) iv gf -1(x) a y = f(x) - 2

c Prove that there are no solutions to the equation b y = |f(x)|
fg(x) = gf(x) c y = f(|x|)
Give the coordinates of any turning points on your three diagrams.

34 35


10 The graphs of y = x - 2 and y = 1 are shown on this diagram. 15 The graph of the function f(x) = 3 - x, x 0 y

y is shown on this diagram.
y=x–2 a Give the range of f(x). 3
1 y=1 b Find the values of 2
y = 3 – √x
O 2 x i f -1f(6) ii f -1(-1) 1

O 2 4 6 8 10 12 14 16 18 x
–2
c Sketch the graph of y = |f(x)|
Find the values of the constant c if the equation
Sketch the graphs of y = |x - 2| and y = 1 on the same diagram |f(x)| = c has exactly two roots.
and find their points of intersection.
16 This diagram shows the graph of y = f(x)
11 Sketch the graphs of y
a y = |2x - 3| b y = 2|x| - 3 (2, 4)
4
c y = |4 - x| d y = 4 - |x| f(x)
3
e y = |x2 - 5x + 4| f y = |x|2 - 5|x| + 4 2
1
g y = |2x - x2| h y = 2|x| -|x|2
O 1 2 3 4 5 6 7 x
–1
12 Solve the equations
C3

C3
–2
a |3x - 2| = x + 2 b 3|x| - 2 = x + 2 –3
(5, –3)
c |2x - 3| = |4 - x| d 2|x| - 3 = |4 - x|
Sketch on separate diagrams the graphs of
e |6 - 2x| = x + 3 f |6 - 2x| = |x| + 3
a y = 1 f(x + 1)
2
b (2 )
y = f 1x +1
13 Solve the inequalities
17 The graph of y = x2 is transformed onto each of these graphs
a |2x - 3| > |4 - x| b 2|x| - 3 < |4 - x|
by successive transformations.
c |2x - 4| < x d |6 - 2x| > x + 3 Describe the transformations which have taken place in each case.
(|x| - 2)2 1
e 6 - 2|x| 1-x f x+4 a y = 3x2 - 5 b y = 2 (x - 2)2
c y = 1 - 2x2
14 Solve
a |x2 - 4| = x + 2
18 f(x) = 2x + 5 − 1 , x > −2
x+3 (x + 3)(x + 2)
b |x2 - 4| x+2
a Express f(x) as a single fraction in its simplest form.
c |x2 - 4| |x| + 2 1 , x > −2
b Hence show that f(x) = 2 −
x+2
c The curve y = 1 , x > 0, is mapped onto the curve y = f(x),
x
using three successive transformations T1, T2 and T3,
where T1 and T3 are translations.
Describe fully T1, T2 and T3. [(c) Edexcel Limited 2005]

36 37

1
Exit
2 Trigonometry
use reciprocal and inverse trigonometric functions
solve trigonometric equations and prove identities
Summary Refer to
use the compound angle formulae
Algebraic division follows the same procedure as numerical long division. 1.2
use the double angle and half angle formulae
A function maps each element of the domain onto one and only one
find equivalent forms for acos q + bsin q.
element of the range. 1.3
The composite function gf(x) means that the function f is applied
first and the function g is applied second. The function gf(x)
only exists if the range of f is part of the domain of g. 1.3
Before you start
To find the inverse function f -1 (x) of f(x) You should know how to: Check in:
either use a flow diagram in reverse
1 Use the special triangles. 1 Find the values of
or make x the subject of y = f(x) and then interchange x and y
or reflect the graph of y = f(x) in the line y = x. a sin 60° + cos 30°
A one-to-one function always has an inverse. A many-to-one function has 30º b tan 60° + tan 45°
an inverse only if the domain is restricted to make it a one-to-one function. 1.4
2
C3

C3
If |x| < a, then -a < x < a. If |x| > a, then x > a or x < -a. √3 c sin2 30° + sin2 60°
√2
1
To sketch the graph of y = |f(x)|, sketch y = f(x) and reflect
d sin2 45° - cos2 45°
any part below the x-axis in the x-axis. 60º 45°
To sketch the graph of y = f(|x|), sketch y = f(x) for x > 0 only 1 mm 1
and reflect it in the y-axis. 1.5, 1.6 1
e.g. sin 30° + tan 45° = 2 + 1 = 1.5
You can apply a combination of transformations to the graph of y = f(x).
The order of the transformations may affect the result. 1.7 8
2 Find cos q when q is acute and 2 a If q is acute and sin q = 17 , find the
tan q is known. values of cos q and tan q.
e.g. If tan q = 7 , 7
x
24 4
i b If q is obtuse and sin q = 5 , find the
Links then x = 7 + 24 = 25
2 2 24 values of cos q and tan q.
Many situations in real life involving several variables and 24
uncertainty can be modelled using mathematics. and cos q =
25

Computation using complex numerical methods is used in
forecasting weather, and mathematics is essential in 3 Find angles in all four quadrants. 3 Find q such that 0° q 360° when
understanding and predicting climate change. e.g. If cos q = - 1 , then q is in the second a sin q = 0.3 b tan q = -2
2
Even though, in reality, the variables involved are complicated, and third quadrants. 1
c cos q = 4
they can be simplified using a mathematical model which can Possible values of q = 180° ± 60° = 120° or 240°
then be used to understand the system being studied and to
make predictions. 4 Use identities involving sin q, cos q and tan q. 4 a If q is obtuse and sin q = 0.4, find cos q.
e.g. If sin q = 0.6 and q is acute, then sin2 q + cos2 q = 1 1 ≡ 1
b Prove the identity 1 +
gives cos q = 1 − (0.6) = 2
0.64 = 0.8 tan 2 q sin 2 q

38 39

2 Trigonometry

Similarly, you can sketch the graph of y = cosec q cosec q is not defined when
2.1 Reciprocal trigonometric functions sin q = 0
y
The reciprocal trigonometric functions, secant, cosecant and
cotangent, are defined as y = cosec i
3

Be careful not to confuse these. 2
cot q = tanq = cosq
1 1 1
sec q = cos q cosec q = sin q You would expect cosec to be
sin q
related to cos and sec to sin. 1
y = sin i

–360º–270º–180º–90º O 90º 180º 270º 360º i
EXAMPLE 1

Find, to 3 decimal places, the values of 162° is in the second quadrant y = sin q and y = cosec q have
2p where tan is negative. –1 periods of 360° (2p radians).
a sec 40° b cot 162° c cosec 5
–2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
When sin q = ±1, cosec q = ±1
1 1 When sin q = 0, the graph of
a sec 40° = cos 40° = 0.76604 = 1.305 –3
162º cosec q has a vertical asymptote.
O
1 1 1
b cot 162° = tan162° = - tan18° = -0.32492 = -3.077
The domains and ranges of sin q and cosec q are: When sin q is positive, cosec q is
also positive. When sin q is
1 1 Domain Range
c cosec 2p = sin 2p = 0.95106 = 1.051 2p
radians =
360°
= 72°
negative, cosec q is also negative.
5 5 5 5 y = sin q qÎR y Î R, -1 y 1
C3

C3
y = cosec q q Î R, q ¹ 0°, ±180°, ±360°, . . . y Î R, y 1, y -1

The graphs of sec q, cosec q and cot q The graph of cot q is shown here. cot q is not defined when
sec q is not defined when sin q = 0
Starting with the graph of y = cos q, you can sketch the graph of
cos q = 0 y
its reciprocal y = sec q
y 3
y = sec i
2
3

2 1

1 y = tan q and y = cot q have
y = cos i –360º –270º –180º –90º O 90º 180º 270º 360º i periods of 180° (p radians).
y = cos q and y = sec q both have
–540º –450º –360º –270º –180º –90º O 90º 180º 270º 360º 450º 540º i a period of 360° (2p radians).
–1
–1 When tan q = ±1, cot q = ±1
–2 When tan q = 0, the graph of cot q
–2 When cos q = ±1, sec q = ±1 y = tan i y = cot i
has vertical asymptotes.
When cos q = 0, sec q = ¥ and –3
–3 its graph has vertical asymptotes.

The domains and ranges of tan q and cot q are:
The domains and ranges of cos q and sec q are:
When cos q is positive, sec q is
also positive. When cos q is Domain Range
Domain Range You should be familiar with all of
negative, sec q is also negative. y = tan q q Î R, q ¹ ±90°, ±270°, . . . yÎR these graphs when q is in
y = cos q qÎR y Î R, -1 y 1
y = cot q q Î R, q ¹ 0°, ±180°, . . . yÎR radians as well as in degrees.
y = sec q q Î R, q ¹ ±90°, ±270°, . . . y Î R, y 1, y -1
40 41

2 Trigonometry 2 Trigonometry

Standard trigonometric results
1 1
The standard trigonometric results from these two special See C2 Chapter 12 for revision. Consider cot q = tanq = sinq = cosq
sinq
triangles are: cosq
Use Pythagoras’ Theorem.
(b ) (c )
sin 0° = 0 cos 0° = 1 tan 0° = 0 a
2 2
Divide a2 + b2 = c2 by b2: +1= b so tan2 q + 1 = sec2 q
1
sin 30° = 1 cos 30° = 3 tan 30° =
() ()
2 2 3 30°
b
2
c
2
√3 2 Divide a2 + b2 = c2 by a2: 1 + a = a so 1 + cot2 q = cosec2 q
1
sin 60° = 3 cos 60° = 2 tan 60° = 3 1
√2
2

sin 45° = 1 cos 45° = 1 tan 45° = 1 60° 45°
You have
2 2 1 1

You should also know these results
The same two special triangles also give you: in radians, where 180° = p radians. cot q = cosq You need to be able to recall
sin q all of these trigonometric
1
cosec 60° = 2 sec 60° = 2 cot 60° = tan q = cot (90° - q) values and formulae. They are
3 3 not provided in the formulae
cosec 30° = 2 sec 30° = 2 cot 30° = 3 sec2 q = 1 + tan2 q booklet in the examination.
3
cosec2 q = 1 + cot2 q
cosec 45° = 2 sec 45° = 2 cot 45° = 1

EXAMPLE 2
C3

C3
and from the trigonometric graphs you get: Find, giving each answer as a surd, the exact values of
sin 90° = 1 cos 90° = 0 tan 90° = ¥ a sec 330° b cosec 225° c cot 7p
6
sin 180° = 0 cos 180° = -1 tan 180° = 0 ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a 330° is in the fourth quadrant where cosine is positive.
The triangle shown gives See C2 for revision. 1 1 2 3
sec 330° = = = 1 = 2 =
a cos 330° cos 30° 3 3 3 330º
a = c 2
tan q = b O
b
c c
a
= sin q
cos q
i
By Pythagoras’ Theorem, a2 + b2 = c2 b b 225° is in the third quadrant where sine is negative.
Divide by c2: sin2 q + cos2 q =1 1 = 11 = − 2
cosec 225° = − sin 45° −
2
225º
O
You have the following results:

tan q = sin q sin2 q + cos2 q = 1
cosq
c 7p radians = 7 × 180° = 210° is in the third
sin q = cos (90° - q) cos q = sin (90° - q) 6 6
quadrant where tangent is positive.
210º
cot 7p = cot 210° = tan 210° = tan 30° = 1 =
1 1 1 3 O
6 3

42 43


EXAMPLE 3
5 Angle a is a reflex angle and cos a = 9
If cos q = − 5 and q is an obtuse angle, 41
13 Find the value of
find the exact value of cot q.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
a cosec a b cot a
Method 1 Method 2
6 Write each of these expressions as a power of sec b, cosec b or cot b.
Use 1 + tan2 q = sec2 q Draw a right-angled triangle for
with sec q = − 13 : 1
a first quadrant angle, f, where a b sec2b
5
cos f = 5 : tan 2 b cos b
13

( 13 ) − 1
2
cot 2 b sec 2 b
c 1 − cos b
2
tan2 q = − d
5 3
sin b sin 3 b
= 169 − 1 13 x
25
144 7 Find the value of cot q when
=
25
i z
a 2sin q = 3cos q b 4tan q = 1
tan q = ± 12 5
5 c cos q sin q = sin2 q d cos q = 9sin q tan q
But q is obtuse, so tan q Use Pythagoras’ theorem: e sin q = 3tan2 q cos q f cosec q = 2
is negative.
x = 169 − 25 = 12 and
8 Simplify these expressions.
So tan q = − 12
5 cot f = 5
1 12 a cot x tan x b cot x sin x tan x
and cot q = tanq
C3

C3
Angle q is obtuse, so tan q
=−5 and cot q are negative. c cot2 x sec x sin x d cosec x sec x sin2 x
12
cot q = -cot f = − 5 1 − sec 2 a
12 e sin x(cot x cos x + sin x) f
cos2 a cos ec 2a

Exercise 2.1 g 1 + cot2 a sin2 a h 1 − cot a
cosec 2a sin 2 a tan a
1 Use your calculator to find, to 3 significant figures, the values of
a sec 200° b cot 130° c cosec 340° tan a ⎞
sec a ⎛ 1
⎜ −
cos ec a ⎟
i
⎝ cos a ⎠
d sec 3p e cot 5p f cosec 2p
5 6 9

2 Find, in surd form where needed, the values of INVESTIGATION
a cot 135° b sec 120° c cosec 210° 9 What transformation maps the graph of y = sec x onto y = sec (x - 90°)?

d cot 4p e sec 7p f cosec 3p What two transformations are needed to map the graph of y = cot x
3 4 2 onto y = cot (90° - x)?

3 Find angle q such that -180° < q < 180° when Use the graphs of these Use computer software to compare the graphs of
functions to help you. y = sec (x - 90°) and y = cosec x and to compare the graphs of
a sec q = 1.25 b cot q = 2.5 c cosec q = 3.0
y = cot (90° - x) and y = tan x.
d sec q = -1.25 e cot q = -3.5 f cosec q = -2.0
What can you deduce?
Is this investigation sufficient as proof of your results?
4 Given that angle q is obtuse and tan q = − 8 , find the value of
15
a sec q b cot q
44 45

2 Trigonometry

EXAMPLE 3
2.2 Trigonometric equations and identities Solve the equation 1 + tan x = sec2 x for 0 x 2p
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Substitute 1 + tan2 x for sec2 x: 1 + tan x = 1 + tan2 x tan is positive in the first and
An equation is true for some, but not all, values of the third quadrants.
variables involved. 0 = tan2 x - tan x
0 = tan x (tan x - 1)
An identity is true for all values of the variables involved. so tan x = 0 or tan x = 1
There are two strategies for proving identities. tan x = 0 gives x = 0 or 2p 5p
You prove: p
4
either that the LHS of the identity is equal to the RHS It is usual to start with the more or tan x = 1 gives x = p or 5p 4
4 4 O
(or vice versa). complicated side of the identity.
or that both sides of the identity are equal to a common The solutions are x = 0, p , 5p , 2p
third expression. 4 4
EXAMPLE 1

Prove that tan x sin x + cos x º sec x Exercise 2.2
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• 1 Solve these equations for -180° x 180° See C2 for revision of
Manipulate the LHS of the identity: Choose one side of the translations of trignometric
a sec (x - 10°) = 3
identity to work on. functions.
sin x
LHS = tan x sin x + cos x = cos x ´ sin x + cos x b cosec (x + 20°) = -4

= sin x + cos x
2 2
c cot (x + 30°) = -2
C3

C3
cos x
1
1 d cot 2x = 2
= cos x = sec x = RHS
e sec (2x + 40°) = -2
So the identity is proved.
f cosec 1 x - 10° = - 3 (2 ) 2
g 3cos x = sec x
EXAMPLE 2

3
cos x tan x cos x
Prove that º h 4cot x = 3tan x
cosec 2 x cot 3 x
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• i 3cos x - cot x = 0
Show that both sides are equal to a common expression:
j 4sin x = 3tan x
sin x
LHS = cos x ´ cos x ´ sin 2 x = sin3 x Remember cosec2 x = 2
1
k 2cot x = cosec x
sin x

sin 3 x 1 cos 3 x l cot x = tan x
RHS = cos3x ´ 3 = sin3 x cot 3 x = 3
=
cos x tan x sin3 x
2 Solve these equations for -p q p
Hence, LHS º RHS and the identity is proved.
a sec2 q =2
b cot2 q = 3
c 4sin q = 3cosec q
d tan q = 4sin q cos q

46 47


3 Solve these equations for 0 x 360° 6 Find the points of intersection of these pairs of curves for
-360° x 360°, giving answers to 2 significant figures
a 2 + sec2 x = 4tan x
where necessary.
b 2cot x = tan x + 1 a y = 1 + cos x, y = 2sin2 x
c 3sin x - 2cosec x = 1 b y = 1 + 2tan x, y = 1 + sec x
2
d 2sec x - 1 = tan x c y = sec x, y = 1 + cos x
e 4cos x - 3sec x = 1 d y = tan x - sin2 x, y = (cos x - sec x)2
f cosec2 x = 4cot x - 3
7 Prove these identities.
g cos x + sec x = 2
a tan q sin q + cos q º sec q
h tan q + cot q = 2
b tan q + cot q º sec q cosec q
i tan q + 3cot q = 5sec q
1
c sec q - tan q º secq + tanq
4 Rewrite each pair of equations as one equation in terms of x and y.
d 1 − tan 2q ≡ 1 − 2 sin2q
2
a x = 4sec a, y = 2tan a
1 + tan q
b x = 3cosec a, y = 2cot a
e sinq ≡ cosq
c x = 4cos a, y = 3tan a 1 + tanq 1 + cotq
C3

C3
d x = 1 - sin a, y = 1 + cos a f (1 + sec q)(1 - cos q) º tan q sin q
e x = 3cos a, y = 4 + tan a sinq + 1 + cosq ≡ 2
g
1 + cosq sinq sinq
f x = asin a, y = bsec a

5 Describe the transformations which map the graph of the h 1 ≡ 1 − cosq
cot q + cosec q sin q
first equation onto the graph of the second equation.
a y = sec x, y = sec 1 x (2 ) i tan 2q + cos2q ≡ secq − sinq
sinq + secq
b y = sec x, y = 1 sec 2x
2

c y = cosec x, y = 2cosec (x + 90°) INVESTIGATION
8 Use a graphical package on a computer to check your
d y = cot x, y = -cot(-x) answers to all the questions in this exercise.
For example, you can check question 1 part a by plotting
the graph of y = sec (x - 10°) and finding the points
where y = 3.
For question 3 part a, you can plot the graphs of
y = 2 + sec2 x and y = 4tan x and then find their points
of intersection.

48 49

2 Trigonometry

The graphs show their domains and ranges as:
2.3 Inverse trigonometric functions
Domain Range
f(x) = cos x 0 x p -1 y 1
If f(x) = sin x, then the inverse function is Do not confuse sin-1 x
f -1(x) = arcsin x or sin-1 x. with (sin x)-1. -1
f (x) = arccos x -1 x 1 0 y p

p p
E.g. sin = 0.5, so arcsin 0.5 =
6 6 The principal values of arccos x are in the range 0 y p
arcsin 0.5 is the angle whose sine is 0.5.
Domain Range
f(x) = sin x is a many-to-one function. See Chapter 1 for revision of
p for it f(x) = tan x −p p yÎR
Its domain must be restricted to − p x inverse functions.
2
x
2
2 2 p
to have an inverse. f -1(x) = arctan x xÎR −p y
2 2
The graph of y = arcsin x is a reflection of y = sin x
in the line y = x The principal values of arctan x are in the range − p y p
2 2
y
y = arcsin x
3

EXAMPLE 1
2
Find the values of
1 y = sin x
a arccos 3 For arccos ⎛ 3 ⎞ , you need to find
⎜ 2 ⎟
For the reflection to work, the 2 ⎝ ⎠
x
–3 –2 –1 0 1 2 3 scales on the two axes must be b arctan (-1) the angle whose cosine is 3 .
–1 2
the same, with angles measured
C3

C3
–2
in radians. c arcsin (-0.3)
–3
y=x ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Use the special triangles:
The graphs show their domains and ranges as:

Domain Range
30°
f(x) = sin x −p x p -1 y 1
2 2 √3 2
−p p √2
-1(x) = arcsin x
f -1 x 1 y
2 2 1
60° 45°
1 1
The principal value of arcsin x is the unique value of arcsin x This is shown as the continuous
within the allowed range − p y p blue line on the graph.
2 2
a arccos 3 = p (or 30°)
2 6
The principal values of arctan are
Similarly, provided that domains are restricted, The alternative notations are in the first and fourth quadrants.
cos-1 x and tan-1 x.
b arctan (-1) = − p (or -45°)
y = cos x and y = tan x have inverse functions 4 So, an angle with a negative
y = arccos x and y = arctan x tangent (in this case, -1) is in the
fourth quadrant.
y y c Use a calculator:
y = tan x arcsin (-0.3) = -0.305 (or -17.5°) A negative principal value of
3 3
arcsin is a fourth quadrant angle.
2 2
1 1
y = cos x y = arctan x
x x
–3 –2 –1 0 1 2 3 –3 –2 –1 O 1 2 3
–1 –1
y = arccos x
–2 –2
–3 –3
50 y=x y=x 51


EXAMPLE 2 2 Find, as a surd where necessary, for angles between 0° and 180°
Find tan arcsin 3 .
4 ( ) a sin x, given that x = arccos 1
3
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

You need to find tan q given that q = arcsin 3 ;
4 (
b tan arccos 3
4 )
that is, find tan q given that sin q = 3
4 c cos (arcsin 1)
Take angle q to be an acute angle. Sine is positive, so angle q is in

Method 1 Method 2
the first or second quadrant.
(
d sin arcsin 5
8 )
tan q = sinq Draw a triangle to show sin q = 3 : e cos [arctan (-1)]
4
cosq

You know sin q = 3 and so f sin [arccos (-0.5)]
4
you need cos q.
3 4
g tan éarccos
ê
ë ( 2 )ùûú
3
-
Use sin2 q + cos2 q = 1:

(4)
2
3 Find the values of
cos q = 1 − 3

= 7
x
i
a arccos 3 + arcsin
2 ( 1)
−
2
4 Use Pythagoras’ Theorem:
Take +ve square root for cosine in b arctan 1 - arctan (-1)
x2 + 32 = 42
C3

C3
the first quadrant.
3 x = 16 − 9 = 7 4 a If q = arcsin x, find in terms of x
Hence, tan q = 4 So tan q = 3
i cos q ii tan q
7 7
4 b Express sec (arccos x) as an algebraic expression in x.
= 3
7
and tan arcsin 3 = 3
4 7
( ) 5 a Given that a = arctan x, express sin a + cos a in
terms of x.
b Given that x = tan q, find arccot x in terms of q.
Exercise 2.3
1 Giving answers in terms of p, find
a arcsin ⎛ 1 ⎞
⎜ ⎟ b arctan ( 3 ) a arctan x º p - arccot x
⎝ 2⎠ 2

c arccos 1 d arctan (-1) b arcsin x + arccos x º p
2

e arccos ⎜ 3 ⎞
⎛
⎟
⎝ 2 ⎠
f arcsin 0 c arctan x + arctan 1 º p (x) 2

g arccos ⎛ −
1 ⎞
⎜ ⎟ h arctan ⎛ − 1 ⎞
⎜ ⎟ INVESTIGATION
⎝ 2⎠ 3 ⎝ ⎠
7 Explore how to input the inverse trigonometric functions
using a computer’s graphical package.

52 53

2 Trigonometry

Similarly, you can derive formulae for cos (A ± B):
2.4 Compound angle formulae
cos (A + B) = cos Acos B - sin Asin B Try this for yourself.
You can find an expression for sin (A + B) in terms of the and cos (A - B) = cos Acos B + sin Asin B
trigonometric ratios of angle A and angle B.
It is not true that sin (A + B) = sin A + sin B You can derive expressions for tan (A + B) and tan (A - B)
You can prove this statement using a counter-example: using tanq = sinq :
cosq

Let A = B = 45° sin(A + B) sin A cos B + cos A sin B
tan (A + B) = = Divide all terms by cos A cos B.
cos(A + B) cos A cos B − sin A sin B
sin (A + B) = sin (45° + 45°) = sin 90° = 1
sin A cos B cos A sin B
and sin A + sin B = 1 + 1 = 2 = 2 ≠ 1 +
2 2 2 cos A cos B cos A cos B tan A + tan B
= = Cancel as shown and use
cos A cos B sin A sin B 1 − tan A tan B
So, sin (A + B) ¹ sin A + sin B − tan q = sin q
cos A cos B cos A sin B cos q

In fact, sin (A + B) = sin Acos B + cos Asin B By replacing B by -B, you can derive the formula
tan A − tan B
tan (A - B) =
1 + tan A tan B
Consider angles A and B in the R
C3

C3
right-angled triangles OPQ and OQR:
A These compound angle formulae can be summarised as:
Triangles OMN and RMQ are similar
so ÐMRQ = ÐMON = angle A L Q sin (A ± B) = sin Acos B ± cos Asin B Take care with the signs:
M sin (A ± B) uses only the ± sign
cos (A ± B) = cos Acos B sin Asin B cos (A ± B) uses only the sign
B tan A ± tan B tan (A ± B) uses both ± and .
A tan (A ± B) =
O P 1 tan A tan B
N

NR NL + LR PQ LR
In triangle ORN, sin (A + B) = OR = OR = OR + OR

EXAMPLE 1
NL = PQ
Find cos 15° as a surd.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

PQ OQ LR RQ Use the special triangles.
= OQ × OR + RQ × OR Use the formula cos (A - B) = cos Acos B + sin Asin B:

= sin Acos B + cos Asin B cos 15° = cos (45° - 30°)
30°
Although proved here for A and B as acute angles, = cos 45°cos 30° + sin 45°sin 30° √3 2
this formula is true for all values of A and B. √2
1 × 3 + 1 ×1 1
=
2 2 2 2
60° 45°
Replacing B by -B and using 3 +1
1 1
sin (-B) = -sin B and cos (-B) = cos B Angle -B is in the fourth =
2 2
gives quadrant where sine is negative 2
Multiply by to rationalise
= 1 2 ( 3 + 1)
sin (A - B) = sin Acos (-B) + cos Asin (-B) and cosine is positive. 2
or sin (A - B) = sin Acos B - cos Asin B 4 the denominator.
You could also have used
cos 15° = cos (60° - 45°)
54 55


EXAMPLE 5
EXAMPLE 2
Find the value of cos (a + b) when a is acute, b is obtuse, Solve the equation arctan (1 + x) + arctan (1 - x) = arctan 2
cos a = 3 and sin b = 5
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

5 13 Let a = arctan (1 + x): tan a = 1 + x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Let b = arctan (1 - x): tan b = 1 - x
From the triangles, you have
5 4
sin a = 4 and cos b = − 12 The equation to solve is now a + b = arctan 2
5 13 or tan (a + b) = 2
so cos (a + b) = cos acos b - sin asin b
a
Consider tan (a + b) = tana + tan b = (1 + x) + (1 − x)
= 3 × − 12 − 4 × 5
5 13 5 13 ( ) 3 1 − tana tan b 1 − (1 + x)(1 − x)
2 2
= 2 = 2
−36 − 20 56 13 1 − (1 − x ) x
= 5 × 13 = − 65 b
5
2
Hence =2
12 x2

The solution is x = ±1
EXAMPLE 3

Solve the equation sin (q - 30°) = 3cos q for 0° < q < 360°
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Exercise 2.4
Expand the brackets: sin qcos 30° - cos qsin 30° = 3cos q 1 Write as a single trigonometric ratio and so find the exact value of
a sin 35°cos 10° + cos 35°sin 10° b sin 70°cos 10° - cos 70°sin 10°
sin q ´ 3 - cos q ´ 1 = 3cos q
2 2
c cos 40°cos 10° + sin 40°sin 10° d cos 80°cos 40° - sin 80°sin 40°
C3

C3
3
sin q = 7 cos q
2 2 tan 70° − tan 45° tan100° + tan 35°
e f
1 + tan 70° tan 45° 1 − tan100° tan 35°
tan q = 7 tan q is positive, so q is in the
3 first and third quadrants.
2 Simplify
For 0° < q < 360°, q = 76.1° or 256.1°
a sin 2Acos A + cos 2Asin A b cos 3acos 2a - sin 3a sin 2a

c tan 2x + tan x d 1 + tan 3x tan x
1 − tan 2x tan x tan 3x − tan x
EXAMPLE 4

Prove the identity sin(A + B) º tan A + tan B
cos A cos B 3 Write each expression as a single trigonometric ratio.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

1 1
Show that the LHS is equivalent to the RHS: a cos x - sin x b 3
cos x + 1 sin x
2 2 2 2

LHS = sin A cos B + cos A sin B
cos A cos B c 3 + tan x d 1 + tan x
sin A cos B cos A sin B 1 − 3 tan x 1 − tan x
= +
cos A cos B cos A cos B e sin (90° - x)cos x + cos (90° - x)sin x f cos2 x - sin2 x
= tan A + tan B
Using tan q = sinq 4 By expanding sin (A - B), show that
= RHS cos q
a sin (90° - A) = cos A b sin (180° - A) = sin A
Hence the identity is proved.
5 Find the exact value of
a sin 15° b cos 75° c tan 75°
d tan 15° e tan 105° f sec 75°
56 57


6 Given acute angles a and b such that sin a = 12 and tan b = 3 , find 13 Find
13 4 i the greatest value ii the least value
a sin (a + b) b tan (a + b) c sec (a + b)
that each of these expressions can have, given that q varies with 0° < q < 360°
4 8 a sin q cos 40° + cos q sin 40°
7 If sin q = 5 and sin f = 17 where q is acute and f is obtuse, find
b cos q cos 20° - sin q sin 20°
a cos (q - f) b sin (q - f) c cot (q - f)
Give the values of q at which the greatest and least values occur in each case.
8 Angles A and B are obtuse and acute respectively, such that
tan A = -2 and tan B = 5. Find the value of ( 3 )
14 If tan q + p = 1 , show that tan q = 2 - 5 3
3 3
a cot (A + B) b sin (A - B)

9 a If tan (a + b) = 4 and tan a = 3, find tan b. 2 () 3 ()
15 Prove that arcsin 1 + arcsin 1 ≡ arcsin ⎛ 2 2 + 3 ⎟
⎜
⎞
⎝ 6 ⎠
b If sin (a + b) = cos b and sin a = 3 , find tan b.
5 16 Solve these equations.
c If tan (a - b) = 5, find tan a in terms of tan b. a arctan x = arctan 7 - arctan 2

10 If sin (q + f) = cos f, show that tan q + tan f = sec q ( 13 )
b arcsin x + arccos 12 = arcsin 4 (5)
11 Solve these equations for 0 < q < 360° c x = arcsin k + arcsin ( 1 − k ) 2

a 3sin q = sin (q + 45°) b 2cos q = cos (q + 30°)
C3

C3
c 2sin q + sin (q + 60°) = 0 d tan(q - 45°) = 3cot q
e sin(q - 60°) = 3cos (q - 30°) f sin(q + 90°) = tan q (x)
a arctan x + arctan 1 º p
2

g tan(60° - q) = tan(q - 45°) h sin q + cos q = 1 b arcsin x + arccos x º p
2 2

a sin (a + 30°) + sin (a - 30°) º 3sin a INVESTIGATION

b (sin a + cos a)(sin b + cos b) º sin (a + b) + cos (a - b) 18 Use graphical software on a computer to check your
answers to any equations in this exercise and to confirm
c sin(a + b ) º tan a + tan b any identities that you have proved.
cosa cos b For example, for question 11 a, draw the graphs of

(4 ) (4 )
d sin p + a + sin p − a ≡ 2 cos a
y = 3sin q and y = sin (q + 45°)

e cos (a - b) - cos (a + b) º 2sin asin b
cos(a + b )
f sina sin b
º cot acot b - 1

g cos (a + b)cos (a - b) º cos2 a - sin2 b

( 4 ) ( )
h sin2 q + p + cos2 q − p ≡ 1 + 2sinq cosq
4

i cot(a + b) ≡ cot a cot b − 1
cot a + cot b
58 59

2 Trigonometry

2.5 Double angle and half angle formulae cos2 A = 1 (1 + cos 2A) Learn these formulae. They are not
2 in the formulae booklet.
1
sin2 A = 2 (1 - cos 2A)
Double angle formulae
A + A = 2A is called a
Let A = B in the expansions of sin (A + B), cos (A + B) and tan (A + B): double angle.

EXAMPLE 1
Find the solutions of cos 2q + 3sin q = 2 for 0 < q < p
sin 2A = 2sin Acos A ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

2 2
cos 2A = cos A - sin A Choose an identity which gives the equation in terms of sin q or cos q only:
2 tan A cos 2q + 3sin q = 2
tan 2A =
1 − tan 2 A
Substitute cos 2q = 1 - 2sin2 q :
These are the double angle formulae. 1 - 2sin2 q + 3sin q = 2

There are two other forms of the formula for cos 2A: Rearrange to equate to 0: This is a quadratic equation
2sin2 q - 3sin q + 1 = 0 in sin q.

Use cos2 A + sin2 A = 1 Factorise:
(2sin q - 1)(sin q - 1) = 0
Substitute cos2 A = 1 - sin2 A into the formula for cos 2A:
sin q = 1 gives a first or second
cos 2A = cos2 A
- sin2 A Solve for sin q : sin q = + 1 or +1 2
2 quadrant angle.
= (1 - sin2 A) - sin2 A
C3

C3
= 1 - 2sin2 A For 0 < q < p,
sin q = + 1
5p
gives q = p or p - p = 5p
Similarly, substituting sin2 A = 1 - cos2 A Work through this proof on your own. 2 6 6 6 6
p
gives cos 2A = 2cos2 A - 1 and sin q = +1 gives q = p 6
2 O

The three versions of the formula for cos 2A are The solutions are q = p , p and 5p .
6 2 6

⎧cos2 A − sin 2 A

EXAMPLE 2
⎪
⎪ Prove the identity cot a - tan a º 2cot 2a
cos 2A = ⎨2 cos2 A − 1
⎪
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

⎩1 − 2 sin A
2
⎪ Show that the LHS of the identity is equivalent to the RHS:
Choose one side of the identity
LHS = cosa − sina to work with.
You can rearrange the identities for cos 2A to give expressions for sina cosa
= cos a − sin a
2 2
cos2 A and sin2 A:
sina cosa
= cos 2a
Rearrange cos 2A = 2cos2 A - 1: sina cosa
2cos2 A = 1 + cos 2A = 2 cos 2a
2 sina cosa
Divide through by 2:
= 2 cos 2a
sin 2a
cos2 A = 1 (1 + cos 2A) = 2cot 2a
2
Similarly, sin2 A = 1 (1 - cos 2A) Work through this proof on your own. = RHS
2
The identity is proved.
60 61


You can also find expressions for cos2 A and sin2 A .
EXAMPLE 3
Find y in terms of x given that x = sec q and y = cos 2q 2 2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Express x in terms of cos q : cos2 A = 1 (1 + cos A)
2 2
x = sec q = 1 A 1
Try to prove these results yourself.
cosq
sin2 = (1 – cos A)
Rearrange to make cos q the subject: 2 2

cos q = 1
x

EXAMPLE 5
Use the double angle formula to express y in terms of cos q : Choose the double angle Given that angle q is acute and sin q = 0.6
formula which converts
y = cos 2q = 2cos2 q - 1 = 2 ⎛ 12 ⎞ − 1 = 22 − 1
⎜ ⎟ cos 2q to cos q only.
find the value of cos q .
2
⎝x ⎠ x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

So, y = 22 − 1 Firstly find the value of cos q:
x Using sin2 q + cos2 q = 1
cos2 q = 1 - sin2 q
Substitute sin q = 0.6:
EXAMPLE 4

Given that 2arctan 3 = arccot x, find x. cos2 q = 1 - (0.6)2
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• = 0.64
Let arctan 3 = a so that tan a = 3 Hence cos q = ± 0.64 = ±0.8 = 0.8 as q is acute.
Let arccot x = b so that cot b = x and tan b = 1
x Now use cos2 q = 1 (1 + cos q) and substitute cos q = 0.8:
2 2
Substitute a and b into the equation:
cos2 q = 1 (1 + 0.8) = 0.9
C3

C3
2arctan 3 = arccot x 2 2
2a = b q = ±0.949 to 3 s.f.
Hence cos
tan 2a = 2 tan a = tan b
2
So 2
1 − tan a q is acute, so cos q = 0.949 to 3 s.f.
2
2×3 = 1
1 − 32 x

x = 1 - 9 = -1 1

EXAMPLE 6
The solution is
6 3 Solve the equation sin a = cos a for 0 a p
2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Change LHS into half angles:
Half angle formulae
You can also change between single angles and half angles. 2sin a cos a = cos a
2 2 2
You can find the half angle formulae by replacing 2A by A and 2sin a cos a - cos a = 0
2 2 2
replacing A by 1 A in the double angle formulae. Factorise by taking out cos
a
as a common factor:
2
2
cos a (2sin a - 1) = 0
2 2
sin A = 2 sin A cos A a
2 2 either cos = 0 or sin a = 1 a a
sin 2 is positive, so 2 is in
2 2 2
ìcos2 A - sin 2 A the first or second quadrant.
ï 2 2 a = p , 5p , . . .
a = p , 3p , . . .
ï
ï 2A
2 2 2 2 6 6
cos A = í2 cos - 1
ï
2 a = p, 3p, . . . a = p , 5p , . . .
ï1 - 2 sin 2 A 3 3
ï
î 2 For 0 a p, the solutions are a = p or p
A 3
2 tan
tan A = 2
1 − tan 2 A
62 2 63


Exercise 2.5 8 Solve the equations for 0 q 360°
1 Write each expression as a single trigonometric ratio.
a 2sin q cos q = 1 b cos2 q - sin2 q = 3
2 42° 2 tan 70° 2 2
a 2sin 23°cos 23° b cos - sin2 42° c
1 − tan 2 70°
c sin q cos q = 1 d cos 2q = sin q
d 2cos2 50° -1 e 2sin 3q cos 3q f 1- 2sin2 4q 2 2 8
1 e cos 2q + 3sin q + 1 = 0 f sin 2q = cos q
g 2 (1 + cos 40°) h 1 + cos 2q i sin q cos q
g cos 2q - cos q + 1 = 0 h tan 2q = 3tan q
2 tan 3q 1 - tan 4q
2
j 1 - tan 2 3q
k cos2 p - sin2 p l i cos q - 2 = 3cos 2q j sin 2q - 1 = cos 2q
5 5 2 tan 4q
k sin 2q + sin q = tan q l 4sin q = sin q
m 1 + cos q n sec q cosec q o cot q - tan q 2
m tan q = 6tan q n 3cos q = 2 + cos q
2 Find the exact value of each expression. 2 2
Give each answer as a surd where necessary. o sin q = cot q p cos q = 5sin q + 3
2 2
a 2sin p cos p b 2cos2 p - 1 c 1 - 2sin2 p
12 12 8 8 q cos 2q = tan 2q r sin q - 2 = 3cos q
2
sin p t 4sin2 q + 5sin 2q cos q = 4
° s tan q tan 2q = 2
1 - tan 22 1
2
8
d 2 e 1 - sin2 75° f
tan 22 1
°
sec p u sin q + 2cos q = 1
2 8
C3

C3
3 Find the values of sin 2a, cos 2a and tan 2a when 9 Prove these identities.

a cos a = 3 b sin a = − 1 c tan a = − 5 a 1 − cos 2A ≡ tan A b sin 2A º 2 tan A
5 3 12 sin 2A 1 + tan A
2

cos A + sin A
c sec 2A + tan 2A º cos A − sin A d cot A - tan A º 2cot 2A
4 Find the values of cos q and sin q when
a cos q = 3 b cos q = 1 e tan A + cot A º 2cosec 2A f cosec 2A + cot 2A º cot A
2 4 2 3
g 2cosec 2A º sec Acosec A h cosec A - cot A º tan A
2
5 Find the values of sin x, cos x and tan x when x is acute and
a cos 2x = 17 b sin 2x = 4 5 c tan 2x = 3 i tan Asec A º 2sin A sec A
25 9 4 2 2

6 Find the values of sin a , cos a and tan a when a is acute and 10 Prove that 2arctan 2 + arctan 3 = arccot 3
2 2 2
3 4
a cos a = 1 b sin a = 5 c tan a = 3
9 11 Use 3A = 2A + A to prove that
sin 3A = 3sin A - 4sin3 A
7 Find y in terms of x given that and cos 3A = 4cos3 A - 3cos A
a x = sin a, y = cos 2a Find an expression for tan 3A in terms of tan A.
b x = 3tan a, y = tan 2a
c x = 3sec a, y = cos 2a

64 65

2 Trigonometry

2.6 The equivalent forms for a cos q + b sin q In general, acos q ± bsin q is equivalent to rsin (q ± a) or
rcos (q ± a), where r is positive and a is an acute angle.
Consider y = 3cos q + 4sin q
Compare the graphs of y = 3cos q, y = 4sin q and
Let acos q + bsin q = rsin (q + a)
y = 3cos q + 4sin q
= rsin q cos a + rcos q sin a
y
For each value of x, you can add Equate the coefficients of sin q and cos q :
y = 3cos i + 4sin i the two y-values on the two blue a = rsin a
4 curves to give the y-value on the b = rcos a
black curve as the arrows show.
y = 3cos i Divide these two equations:
2 a = r sina = tan a
Cancel through by r.
b r cosa

O 90º 180º 270º 360º i
a = arctan a
b ()
Square a and b and add:
–2
a2 + b2 = r2(sin2 a + cos2 a) = r2 sin2 a + cos2 a = 1

–4 y = 4sin i So r = a2 + b2 r is positive.
y Hence, acos q + bsin q = rsin (q + a)
C3

C3
The graph of y = 3cos q + 4sin q is a transformation of the 5 where r = a2 + b2 and tana = a
4 y = 5sin (i + 37º)
b
basic sine curve y = sin q under a stretch (scale factor 5)
3
parallel to the y-axis and a translation of about 37°
2
to the left. You can also find r and a for each of rsin (q ± a) and rcos (q ± a).
1 y = sin i Try this yourself.
This diagram shows y = sin q transformed into y = 5sin (q + 37°) O 90º 180º 270º 360º i
–1

EXAMPLE 1
The 37° is only approximate.
–2 If 4sin q + 3cos q = rsin (q + a),
You will find a more accurate value later. –3 find r and a such that r > 0 and a is acute.
–4 ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

–5 Expand r sin (q + a): 4sin q + 3cos q = rsin qcos a + rcos q sin a Both sides contain + signs.
Compare coefficients of sin q and cos q :
4 = rcos a [1]
y 3 = rsin a [2]
Alternatively, the graph of y = 3cos q + 4sin q is a
transformation of the basic cosine curve y = cos q under 5 Divide equation [2] by equation [1]: 3 = r sin a = tan a
y = 5cos (i – 53º) a is acute, so sin a, cos a and
a stretch (scale factor 5) parallel to the y-axis and a 4 r cos a
4 tan a are all positive.
translation of about 53° to the right. 3 a = arctan 3 = 36.9°
4
2
This diagram shows y = cos q transformed into y = 5cos (q - 53°) 1 y = cos i Square equations [1] and [2] and add: 32 + 42 = r2(sin2 a + cos2 a)
The 53° is only approximate. O 90º 180º 270º 360º i = r2
–1 r = 25 = 5 r is positive so ignore r = -5
You will find a more accurate value later. –2
You can check your answer by
–3 Hence, 3cos q + 4sin q = 5sin (q + 36.9°) using a graphical package on
–4
a computer.
–5
66 67


Exercise 2.6 9 a Show that 5sin q -12cos q º rsin (q - a) for r > 0 and a acute.
1 In each equation find the values of r and a, where r > 0 and a is acute. Give the values of r and a.
Give r as a surd where appropriate and a to the nearest 0.1°.
b Find the maximum and minimum values of 5sin q - 12cos q
a 4cos q + 3sin q = rcos (q - a) b 5sin q + 12cos q = rsin (q + a) and the smallest positive values of a at which they occur.
c cos q - 2sin q = rcos (q + a) d 4sin q - 2cos q = rsin (q - a) Find the required stationary values of these expressions and,
e 3sin q - 4cos q = rsin (q + a) f 8cos q + 15sin q = rcos (q + a) in each case, give the smallest positive value of q at which
they occur.
2 Solve these equations, given that 0° q 360° c maximum of 5sin q - 12cos q + 20
a 5cos q + 12sin q = 6 b 2cos q - 3sin q = 1 d maximum of 20 - (6cos q + 8sin q)
c 8sin q + 15cos q = 10 d 3sin q - 5cos q = 4 20
e minimum of
6 cosq + 8 sinq
3 Prove that
( 4) ( 4)
f minimum of 15
a cos q + sin q = 2 sin q + p b cos q - sin q = 2 cos q + p 5 sinq − 12 cosq + 2

4 Prove that 10 a Express 3sin q - 2cos q in the form rsin (q - a) such that
a (
3cos q + sin q = 2 sin q + p
3 ) b cos q - 3sin q = 2cos q + p
3 ( ) r > 0 and a is an acute angle.
b Sketch the graph of y = 3sin q - 2cos q for -360° < q < 360°,
5 Solve these equations, given that -p q p labelling all points at which the graph crosses the axes
1
C3

C3
a cos q + 3 sin q = 2 b cos q + sin q = within this interval.
2
c 3cos q + sin q = 1 d cos q + 2sin q = 2 c Describe the transformations which the graph of y = sin q
undergoes to become the graph of y = 3sin q - 2cos q
6 Solve these equations, given that -180° q 180°
11 Two alternating electrical currents are combined so that the
a cos 2q + 2sin 2q = 1 b 2cos 3q - 6sin 3q = 5 resultant current I is given by I = 2cos wt - 4sin wt where the
constant w = 4 and t is the time (t > 0).
c 6cos q + 8sin q = 3 d sin q - 4cos q = 1
2 2 2 2 Find the maximum value of I and the smallest value of t at
which it occurs.
7 a Show that 2 cos q + 3 sin q can be written in the form rcos (q - a),
where r > 0 and a is acute. Find the values of r and a.
INVESTIGATION
b Hence, find the maximum value of 2 cos q + 3sin q and
the smallest positive value of q at which a maximum occurs. 12 Let 4cos q – 3sin q º r1cos (q + a1) º r2cos (q - a2)
1 º r3sin (q - a3) º r4sin (q + a4)
c Find the minimum value of and the
2 cosq + 3 sinq Find the values of r1, r2, r3, r4 and a1, a2, a3, a4.
smallest positive value of q at which a minimum occurs.
Use computer software to draw five sinusoidal graphs
to confirm your results.
8 a Express 8cos 2q - 6sin 2q in the form rcos (2q + a) where
r > 0 and a is acute. State the values of r and a.
b Find the minimum value of 8cos 2q - 6sin 2q and the smallest
positive value of q at which a minimum value occurs.
c Find the maximum value of 8cos 2q - 6sin 2q and the smallest
positive value of q at which a maximum value occurs.
68 69

2 Trigonometry

7a Given that sin2 q + cos2 q º 1, show that 1 + tan2 q º sec2 q
Review 2
b Solve, for 0 q < 360°, the equation 2tan2 q + sec q = 1
giving your answers to 1 decimal place. [(c) Edexcel Limited 2005]
1a If q is acute and tan q = 2 , find in surd form
i cot q ii sec q iii cosec q 8a Sketch, on the same diagram, the graphs of
b If q is obtuse and sin q = 5 , find as fractions y = cos x and y = sec x for -270° x 270°
13
i cosec q ii cot q iii sec q b On the same diagram, also sketch the graph of y = sec (x - 90°),
stating the transformation which maps sec x onto sec (x - 90°).
2 Solve these equations where 0° q 180°
15 3
a cosec q = 4 b cot (q + 20°) = 4 9 Given that sin A = 17 and sin B = 5 where A and B are both acute angles, find
a sin (A + B)
c cot (2q - 30°) = 3 d sec (3q - 80°) = 3
b cos (A + B)
e sin2 2q = 1
2
f (2 )
sec2 1 q = 4
c tan (A - B)
10 Use the expansions of sin (A ± B) and cos (A ± B) to find
a (sin2 q - 2cos2 q)sec2 q º sec2 q - 3
a sin 15° as a surd by substituting A = 60° and B = 45°
tan q + 1 cot q + 1
b ≡ b sin (A + B) where A is obtuse, sin A = 5 and B = 15°
sin q cos q 13
C3

C3
cosec q + sec q
c º sin q + cos q
cot q + tan q 11 Solve these equations for 0° q 360°
1 + cos q a cos (q + 30°) = 2sin q b cos 2q + cos q + 1 = 0
d (cot q + cosec q)2 ≡
1 − cos q
c cos 2q = sin q d tan2 q = 2sec q - 1
4 Solve these equations where -180° q 180°
e 1 + sin q = 3cos q f tan 2q = 3tan q
2
a 2cot q + tan q = 3 b 6sin q = 1 + cosec q
g 4tan q + 3tan q = 0 h 3sin q = 2 + cosec q
c cot q - 3cos q = 0 d tan2 q - 7 = 2sec q 2
e 3cos2 q = 2sin q cos q f tan q = 3 - 2cot q i sin q + cos q = 1

5a Describe the successive transformations which map the 12 Prove these identities.
graph of y = sin q onto cos 2q
a º cos q - sin q
i (
y = 2sin q + p
2 ) ii y = 3 - sin 2q (
iii y = 1 + 2sin q − p
4 ) cosq + sinq
cota cot b − 1
b cot (a + b) º
b Sketch, on one diagram, the graphs of y = cos q and y = 3 - cos 2q, cota + cot b
for -180° x 180°, giving all points of intersection with the coordinate axes. c sin 3q + sin q º 2sin 2qcos q
d tan a + cot a º 2cosec 2a
6a Find, in terms of p, the principal value of
i arcsin 1(2) ii arctan ( 3 ) iii arccos (-1)

b Find, as a surd, the value of

i
⎛
⎝
⎞
cos ⎜ arcsin 3 ⎟
2 ⎠
ii sin (arctan 1) (
iii tan arccos 1
2 )
70 71


Prove the identity 1 − tan 2 x º cos 2x
2
13 Solve these equations for -180° q 180° 21 a
1 + tan x
a cos q + cos 2q = 2
b Hence, prove that tan p = 7 − 4 3
12
b sin 2q = 1 sin2 q
2
c 4tan 2q tan q = 1 22 Find the values of r and a, where r > 0 and angle a is acute, when
a 12sin q + 5cos q º r sin (q + a)
14 Rewrite each pair of equations as an equation in terms of x and y.
b 8sin q - 15cos q º r sin (q - a)
a x = sin a, y = cos 2a
c 2cos q + sin q º r cos (q - a)
b x = 1 sec a, y = cos 2a
2 d cos q - sin q º r cos (q + a)
c x = tan 2a, y = tan a
23 a Find the values of r and a such that 3sin q + 4cos q º r sin (q + a),
15 a Given that sin x = 3 , use an appropriate double angle where r > 0 and a is an acute angle.
5
formula to find the exact value of sec 2x. b Write down the maximum value of 3sin q + 4cos q.
b Prove that c Solve the equation 3sin q + 4cos q = 2 for 0 < q < 360°
cot 2x + cosec 2x º cot x ( x ≠ np , n ∈ Z
2 ) [(c) Edexcel Limited 2004]
24 a Find the maximum and minimum values of each of these
16 Find the values of expressions and the smallest positive values of q at which

( 2)
they occur.
a arcsin ⎜ 3 ⎞
⎛
C3

C3
⎟ b arccos − 1 i 3sin q + cos q
⎝ 2 ⎠
ii cos q - 2sin q
c ( ( 3 ))
sin arctan 2 d tan ( arcsin ( 2 ) )
3 b Solve these equations for 0° q < 360°
i 3sin q + cos q = 2
(3) (5)
17 If a = arcsin 1 and b = arccos 3 , find the values of sin(a + b). ii cos q - 2sin q = 1

25 f(x) = 12cos x - 4sin x
2 ()
18 Given that a + b = arctan ( 5 3 + 8 ) and b = arctan 1 , use the expansion a Given that f(x) = Rcos (x + a), where R 0 and 0 a 90°,
of tan(a + b) to find the acute angle a. find the value of R and the value of a.

19 a Prove that, for all values of x, b Hence, solve the equation
cos x - cos(x + 60°) º cos(x - 60°) 12cos x - 4sin x = 7
for 0 x 360°, giving your answers to one decimal place.
b Use the fact that 36° = 120° - 84° to find the exact value of
a given that sin a = sin 84° - sin 36° c i Write down the minimum value of 12cos x - 4sin x.
ii Find, to 2 decimal places, the smallest positive value of x
c For 0 x 360°, solve the equation for which this minimum value occurs. [(c) Edexcel Limited 2006]
sin (60° + 2x) - 4sin 2x = 1 + sin(60° - 2x)
giving your answer in degrees correct to 1 decimal place.

20 a If sin(x + 30) = 2sin(x - 30°), prove that tan x = 2 3
3
b Solve the equation 2 - 2cos 2q = sin 2q,
for 0 q 360°, giving answers correct to 0.1° where necessary.

72 73

Revision 1

2Exit
1 Simplify 3 − 3 − 2
1 + x 2 + x (2 + x)2
and express your answer as a single fraction.

2 Simplify as far as possible
x 3 + x 2 − 2x
a
Summary Refer to x2 − 1
x + 2
sec q = 1
cosec q = 1 cot q = 1 b
cos q sin q tan q x +1 x +2
tan q = cot (90° - q) sec2 q = 1 + tan2 q cosec2 q =1 + cot2 q 2.1, 2.2
The inverse trigonometric functions are arcsin x, arccos x and arctan x. 3 Find the quotient and remainder when x4 + 4x3 + 2x2 + x - 5
Their principal values are unique values within the allowed range. is divided by x2 + x + 1.
q = arcsin x exists within the allowed range - p q p
2 2 4 Given that x4 - 3x3 + 7x2 - 8x + 5 º (x2 - 2x + 1) ´ Q(x) + R(x)
q = arcos x exists within the allowed range 0 q p
find the two function Q(x) and R(x).
q = arctan x exists within the allowed range - p q p 2.3
2 2
The compound angle formulae are 5a Find the quotient and remainder when x2 - 3x + 2
sin (A ± B) = sin Acos B ± cos Asin B tan (A ± B) = tan A ± tan B is divided into 2x3 - x2 - 9x + 6.
1 tan A tan B
cos (A ± B) = cos Acos B sin Asin B 2.4
b Hence, or otherwise, find the values of the constants l and m
C3

C3
The double angle formulae are
⎧cos2 A − sin 2 A so that there is no remainder when 2x3 - x2 + lx + m is divided
⎪
⎪ by x2 - 3x + 2.
sin 2A = 2sin Acos A cos 2A = ⎨2 cos2 A − 1 tan 2A = 2 tan A 2.5
1 − tan 2 A
⎪
⎩1 − 2 sin A
2 6a Express 4sin q + 3cos q in the form Rsin (q + a), where R > 0
⎪
The half angle formulae are and a is an acute angle.
sin A = 2sin A cos A ⎧cos2 A − sin 2 A b Hence, solve the equation 4sin q + 3cos q = 5 for 0° < q < 360°
2 2 ⎪ 2 2 2
⎪
⎪
2 tan A cos A = ⎨2 cos2 A − 1 cos2 A = 1 (1 + cos A)
tan A = 2 2 2 2 7a Express cos q + 4sin q in the form Rcos (q - a) where a is an
⎪
1 − tan 2 A ⎪1 − 2 sin 2 A A 1
sin2 = (1 - cos A) acute angle. Give the exact value of R and the value of a correct
2 ⎪
2.5
⎩ 2 2 2 to the nearest degree.
acos q ± bsin q can take the equivalent forms rcos (q ± a) or rsin (q ± a),
where r is positive and a is an angle. 2.6
b Hence, or otherwise solve the equation cos q + 4sin q = 3 for 0 < q < 360°

8 By writing 5sin q - 12cos q in the form Rsin (q - a) where R > 0 and
Links 0°< a < 90°, find
Trigonometry is behind the technology of modern a the greatest possible value of 5sin q - 12cos q
Satellite Navigation (Sat Nav) systems.
b the smallest possible value of q for which the greatest value occurs.
Sat Nav uses the Global Positioning System (GPS) which relies
on a collection of satellites, orbiting the earth and transmitting 9 Given that f(x) = 9 - (x + 2)2, x Î R
data. Information about how the satellites orbit, and their
position at a particular time, allows a GPS receiver to calculate a find the range of f(x)
its position on the surface of the Earth using basic trigonometry. b state whether f -1(x) exists or not
Combined with some maps and planning software, this is the
c find the value of ff(-4).
basis of in-car Sat Nav technology.
74 75

Revision 1 Revision 1

10 a Express x2 - 4x + 1 in the form (x - a)2 + b. 16 a Solve these equations.
b Given that the function f is defined by f: x ® x2 - 4x + 1, x Î R, x 2, i |3x - 1| = 8 ii |x| + 3 = 2x iii |x + 3| = 2x
find i the range of f ii the inverse function f -1. b Solve these inequalities.
i |3x - 1| > 8 ii |x| + 2 1x iii |x - 1| < 2x + 1
c Sketch the graphs of the functions f and f -1 on the same axes. 2

11 a Describe the transformations which are needed to transform
17 This sketch shows the curve y = f(x), x Î R. Point (1, 3) is a y
the graph of y = x3 into the graph of y = 1 - 2x3
turning point on the curve. The x-axis and the
Indicate the order in which the transformations occur. P
line x = 3 are both asymptotes to the curve. 3
b The graph of y = x2 - 2x + 5 is reflected in the y-axis and then
Sketch, on three separate diagrams, the graphs of
⎛ −1 ⎞
translated by ⎜ ⎟ . Find the equation of the final image in its
⎝ 0⎠ a y = |f(x)| b y = f(|x|) c y = f(x - 3) x
O 1 2 3
simplest form. showing any asymptotes and the coordinates of any
maximum or minimum points.
12 The functions f and g are defined by
f: x ® x2 - 5x + 4, x Î R, 2 x 5 18 a Sketch the graphs of y = |3x - 2| and y = 1 on the same diagram.
x
g: x ® kx - 2, x Î R, where k is a constant.
b Use your graphs to say why there is only one solution of the
a Find the range of the function f. equation x|3x - 2| - 1 = 0
b If gf(5) = 2, find the value of k. c Use algebra to find the solution of the equation x|3x - 2| - 1 = 0
C3

C3
13 a The function f is defined by f: x ® 5x, x Î R. 19 This diagram shows a sketch of the curve y = f(x), x -1 y
Write down f -1(x) and state the domain of f -1. The curve passes through the origin O, has a maximum value
at the point P(3, 2) and has the x-axis as an asymptote. P
b The function g is defined by g: x ® 3x2 - 2, x Î R. 2
Find gf -1(x) and state the range of gf -1. On separate diagrams, draw sketches of the curves with
these equations –1 O 3 x
14 This sketch shows the curve with equation y = f (x), x Î R, 0 x a y

The curve meets the coordinate axes at the points (a, 0) and (0, b). b
a y = |f(x)| b y = f(|x|) c y = f(x + 3)
a Sketch, on two separate diagrams, the curves On each sketch, indicate the coordinates of points at which
y = f(x)
i y = f -1(x) the curves have turning points and the coordinates of points

(2)
where the curves meet the x-axis.
ii y = 4f x
O a x
marking the coordinates of all points where these two curves 20 This figure shows part of the graph of y = f(x), x Î R. y

meet the coordinates axes. The graph consists of two line segments that meet at the point
(1, a), a < 0. One line meets the x-axis at (3, 0).
b If f defined by f(x) = (x - 3)2, x Î R, 0 x 3, find the value of The other line meets the x-axis at (-1, 0) and the
a and b. State the range of f. y-axis at (0, b), b < 0. –1 O 3 x
–b
c The function g is defined by g(x) = 1 + x , x Î R, x > 0 On separate diagrams, sketch the graphs with equations
Find gf(x), giving your answer in its simplest form. a y = f(x + 1) b y = f(|x|) (1, –a)

Indicate clearly on each sketch the coordinates of any points
15 a Sketch the graphs of y = |2x + 1| and y = |x - 1| on the same axes. of intersection with the axes.
b Solve the equation |2x + 1| = |x - 1| c Given that f(x) = |x - 1| -2, find
i the value of a and the value of b
ii the value of x for which f(x) = 5x [(c) Edexcel Limited 2005]
76 77

Revision 1

a

c
sin 2q ≡ cotq
1 − cos 2q
cosq + 1
secq 1 + cot 2q
≡1
b cot q - tan q º 2cot 2q

d tan 2q + 1 ≡ sec 2q
1 − tan 2q
3 Exponentials and logarithms
22 Solve these equations where 0° q 180° This chapter will show you how to
discover the value of the irrational number e
a 2sin2 q = 3(1 - cos q) b sec2 q = 2(2tan q - 1) use natural (or Napierian) logarithms
c 2cos q + cos (q + 60°) = 0 d cos 2q + cos q + 1 = 0 use the exponential function y = e x and its inverse function y = ln x
draw graphs of functions which involve e x and ln x
6 cos 2q solve equations which involve e x and ln x
23 a Prove the identity 6 - 3 sec2 q º 1 + cos 2q
use exponential and logarithmic functions to solve real-life problems.

b Solve the equation 6 cos 2q = 13 - 11tan q for 0° < q < 360°
1 + cos 2q

24 a Prove the cos 3a º 4cos3 a - 3cos a by substituting Before you start
(2a + a) for 3a. You should know how to: Check in:
b Solve the equation sec 2a cos 6a + 1 = 0 for 0° < a < 90° ax
1 Calculate and loga x for different 1 Calculate the value of
values of a and x. a ax for a = -3, x = -2
25 a Find, in radians, the values of e.g. If a = 1 and x = -3,
C3

C3
2 b a2x+1 for a = 1 , x = − 1
i arcsin ⎛ 1 ⎞ 9 4
⎜ ⎟
( 2)
−3
⎝ 2 ⎠ then ax = 1 = 23 = 8 c loga (x2 + 2x + 1) for a = 10, x = 9
ii sin (arctan 3 )
b Find the value of sin [arctan (-1)] 2 Use the laws of logarithms. 2 Find x, y and z given that log3 x = 2,
e.g. If log10 y = 3, then y = 103 = 1000 log10( y + 1) = 1 and z = log216
c If sin a = 3 and sin b = 8 where a is acute and b is obtuse, 2
5 17
find the value of cos (a - b). 3 Find the inverse function of f(x). 3 Find the inverse function, stating its
e.g. If f(x) =3x2+ 1, x Î R, then undoing domain and range, when
26 a i Express (12cos q - 5sin q) in the form Rcos (q + a), where the operations in reverse order gives
R > 0 and 0 < a < 90° a f(x) = 3x + 5
f −1(x) = + x − 1 , x Î R, x
b f(x) = x − 3
ii Hence solve the equation 1
3
2
12cos q - 5sin q = 4
for 0 < q < 90°, giving your answer to 1 decimal place. c f(x) = + 2x − 1

b Solve 8cot q - 3tan q = 2 4 Reflect, stretch or translate a graph and 4 Find the equation of the resulting curve
for 0 < q < 90°, giving your answer to 1 decimal place. [(c) Edexcel Limited 2004] find its new equation. when y = x2 + 2 is
e.g. When the graph of y = x2 - x - 1 is a stretched (scale factor 3) parallel to the
translated by ⎛ ⎞ , the equation of the new
3
⎜0⎟ y-axis and then reflected in the y-axis
⎝ ⎠
graph is y = (x - 3)2 - (x - 3) - 1 ⎛ 0⎞
giving y = x2 - 7x + 11
b translated by ⎜ ⎟ and then stretched
⎝ −5 ⎠
(scale factor 2) parallel to the x-axis.

78 79


Altering the values in the Altering the values in the
3.1 The exponential function, ex spreadsheet gives: spreadsheet again gives:

( ad x − 1 ) ( ad x − 1 )
ax is an exponential function for all values of a. a
dx
a
dx

You can use a table to draw the graph of a particular 2.60 0.9556 2.70 0.9933
exponential function, for example y = 2x: 2.65 0.9746 2.71 0.9970
2.70 0.9933 2.72 1.0007
x -2 -1 0 1 2 2.75 1.0117 2.73 1.0044

y 1 1 20 = 1 21 = 2 22 = 4 2.80 1.0297 2.74 1.0080
2-2 = 4 2-1 = 2
y
The value you want is between The value you want is slightly
You can also draw graphs of other members of the a = 2.70 and a = 2.75 less than 2.72.
family of curves y = ax y = 4x x
y=3
y = 2x This value of a is known as the exponential function, e, e is an irrational number (like p).
All the curves pass through the point (0, 1) where e = 2.718 28 to 5 decimal places.
y = 1.5x
because y = a0 = 1 for all values of a.
1
The graph of y = e x has a gradient of 1 at the point (0, 1). Sometimes e x is written as exp(x).
O x

EXAMPLE 1
Sketch the graphs, for x Î R, of
a y = e-x b y = 2e x - 3 c y = e 2x +1
C3

C3
To investigate the gradient of the curves at the point
y State the range of each function.
P(0, 1) from first principles, you calculate the gradient ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

of the chord PQ for small values of dx. y = ax
a The graph of y = e-x is the y
dy a −1 dx
Q (dx, adx ) reflection of y = e x in the 5
Gradient of chord PQ = = adx y = e–x y = ex
dx dx y-axis. Its range is y Î R, 4
adx – 1
In the limit, as dx ® 0, P
–1 y>0 3
the gradient of the chord PQ ® the gradient of the dx 2
tangent at P. O dx x
1

You need to find the limit of a − 1 as dx ® 0.
dx
–3 –2 –1 O 1 2 3 x
dx
Let dx take a very small value, say dx = 0.0001 b The graph of y = 2e x - 3 y
y = 2ex y = ex
( ad x − 1 ) is the result of a stretch
This spreadsheet shows the values of a − 1
dx
a
dx
dx (scale factor 2) of y = e x
for graphs with different values of a: 2.0 0.6932
parallel to the y-axis
2.2 0.7885
Between a = 2 and a = 3, there is a point on the curve y = ax followed by a translation
2.4 0.8755 2
of -3 downwards parallel y = 2ex – 3
where the gradient at P is exactly 1. This value is between 2.6 0.9556
a = 2.6 and a = 2.8 to the y-axis. 1
2.8 1.0297 Its range is y Î R, y > -3
3.0 1.0987 −3 −2 −1 O 1 2 3 x
You can get a more accurate approximation of this value by
−1
looking at the values between 2.6 and 2.8.
−2

−3
The solution to part c is shown
on the next page.
80 81

3 Exponentials and logarithms 3 Exponentials and logarithms

EXAMPLE 1 (CONT.) 4 The growth of algae in a polluted river is governed by the equation
c y = e 2x+1 = e 2x ´ e1 = e ´ e 2x y
N = N0eat,
The graph of y = e 2x+1 is the y = e2x

(
result of a stretch scale factor 1
2 ) e
y= ex
where N is the number of organisms per unit volume of river
water, t is the time in weeks from the start of the observation,
of y = e x parallel to the x-axis and N0 and a are constants.
followed by a stretch (scale y = e2x+1 a After 4 weeks, the number of organisms N is observed to be
1
factor e) parallel to the y-axis. double the initial number.
Its range is y Î R, y > 0 O x
Find the value of a to 4 significant figures.
b If N0 = 20, what is the value of N after 10 weeks of observation?
c How many weeks does it take for N to treble its initial value?
Exercise 3.1
d Give a reason why this model of pollution is unrealistic.
1 Sketch the graphs of these functions for the domain x Î R.
State the transformations of y = e x which are involved.
5 A radio-active substance decays such that its mass M at time t (hours)
a y = 1 + e -x b y = 1 - e-x is given by M = M0e-kt, where M0 and k are both constants.
c y = 3e x + 2 d y = 2 - 3e x a If k = 0.006 93 show that the half-life of this substance is 100 hours.
e y = 3e 2x f y = e x +1 b How long does the substance take to decay to a tenth of its
original mass?
g y = e x-2 h y = e -x +2
C3

C3
6 The number of cells N which are infected with a virus were
2 The population, P, of rats infesting a sewer grows
t observed to change with time t (hours) as given by
exponentially over time, t (weeks), according to P = Ae 20
N = 200 - 50e-2t
Find the value of A and copy and complete this table of values. a Construct a table of values of N for 0 t 15 and draw the
t 0 5 10 15 20
graph of N against t.
P 100 b How many infected cells were there initially?

Draw the graph of P against t. c What is the limiting value of N as time increases?
How long does it take for the population to double its initial size? d What series of transformations of N = e-2t result in the
given relationship?
3 A mass of M units of a radioactive substance decays t
−
exponentially over time t (seconds), where M = M 0e 10 7 State the transformation of the graph of y = e x which result in
Find the value of M0 and copy and complete this table of values. a y = e 2x+3 + 4

t 0 5 10 15 20
b y = e 2x-1 - 4
M 6 c y = e 2-x - 3

Draw the graph of M against t.
How long does it take for the mass of the substance to INVESTIGATION
reduce to 3 units? This time period is known as the 8 Find other examples of exponential growth and decay
half-life of the substance. which are governed by the equations
y = Aekt and y = Ae-kt
E.g. What is Newton’s law of cooling?
82 83


Exercise 3.2
3.2 The logarithmic function, ln x 1 a Find the values (to 3 significant figures) of
1
i e1.5 ii e-1.5 iii
The inverse of the exponential function y = e x You can show this by taking e1.5
is the logarithmic function y = loge x logarithms in any base n and
finding x in terms of y.
iv ln 1.5 v ln 1(1.5 ) vi 1
ln1.5
b Simplify
i ln (e5) ii ln (ex) iii eln 5
Logarithms with base e are called natural logarithms.
iv eln x v ln (2e5) vi e2ln 5
They are usually written as ln x rather than loge x.
2 Describe the transformations of y = ln x which result in each of
The function y = e x has the inverse function y = ln x Their graphs are reflections of these functions. Sketch the graphs of these functions, labelling
each other in the line y = x any points where the graphs intersect the coordinate axes.
State the domain and range of each function.
y y = e x has the x-axis as an
y = ex
asymptote a y = 2 + ln x b y = 3 - ln x c y = ln (x + 2) d y = ln (x - 2)
y=x y = ln x has the y-axis as an
asymptote e y = 1 + 3ln x f y = 1 + ln (2x) g y = 1 - ln x h y = 2ln (1 - x)
1 for y = e x, the domain is x Î R
and the range is y Î R, y > 0 3 a Describe the successive transformations of the graph
y = ln x for y = ln x, the domain is of y = ln x which produce the graph of y = 2ln (x + 3) + 1
O 1 x x Î R, x > 0 and the range
C3

C3
is y Î R b Find the points at which this graph cuts the x-axis and y-axis.
ex is positive for all values of x,
ln x does not exist for negative 4 Find the inverse function f -1(x) for each function f(x).
values of x. Sketch the graphs of f(x) and f -1(x) on the same diagram,
labelling any intersections with the coordinate axes.
EXAMPLE 1

Find the inverse function f -1(x) for the function f(x) = 1 + e-x In each case, give the domain and range of f -1(x).
Sketch the graphs of f(x) and f -1(x) on the same diagram, a y = 1 + 2ln x, x>0 b y = ln (x + 4), x > -4
labelling any intersections with the coordinate axes. 1x

Give the domain and range of f -1(x). c y = 3 + e-x, xÎR d y =2− e2 , xÎR
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

To find the inverse, let y = 1 + e-x: The graph of y = 1 + e-x is the 5 The functions f and g are defined by
y 1x
y - 1 = e-x graph of y = e x reflected in the f: x ® ln (6 - 2x), x Î R, x < 3 and g: x ® e 2 , x Î R.
y = – ln(x – 1)
y-axis and then translated
Take logarithms: upwards by 1 unit. a Find expressions for f -1(x) and g-1(x). State their domains and ranges.
-x y=x The graph of y = -ln (x - 1) is the
ln (y - 1) = ln (e ) 2
graph of y = ln x translated to the b Sketch the curves y = f(x) and y = f -1(x) on the same axes,
= -x ln e [ln e = 1] y=1+ e–x
right by 1 unit and then reflected stating the points where they cut the axes.
= -x 1
in the y-axis.
c Find an expression for gf(x) in its simplest form and
Interchange x and y: These transformations are the
inverses of each other. calculate the value of gf(-5).
ln (x - 1) = -y O 1 2
x
y = -ln (x - 1)
INVESTIGATION
6 What is semi-log and log-log graph paper?
Hence the inverse function is f -1(x) = -ln (x - 1) Explore why and how scientists and engineers use it to
draw graphs of y = Ae kx and other exponential curves.
The domain of f -1(x) is x Î R, x > 1 and its range is y Î R.
84 85


4 Find the three points on the graph of y = 2e3x+1 where
3.3 Equations involving e x and ln x
a y = 20 b y=2 c y=1
You can use the laws of logarithms and the techniques from 5 Find the point of intersection of the curves y = 2e2x - 1 and y = 3e x - 2
unit Core 2 to solve equations involving ex and ln x.
6 The graph of the function f(x) = ln (2x - 5), x Î R, x > k
EXAMPLE 1

Solve the equation e3x+2 = 25 has a vertical asymptote x = k
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• a Find the value of k. b Find x such that f(x) = -1
Take logarithms to base e of both sides:
7 a Find the equation of the asymptote to the curve y = 3e2x + 5
ln (e3x+2) = ln 25
(3x + 2)ln e = ln 25 loga (xn) = nloga x b Find the points of intersection with the curve y = 8e x
ln25 − 2 = 0.406 to 3 s.f.
x= loga a = 1 so ln e = 1
8 Find the point where the curve y = e x + 1 intersects the curve y = 6e-x
3

9 The population, P millions, of a country is growing exponentially
EXAMPLE 2

1 as given by P = 15e kt where t is the time in years.
Solve the equations a ln (3x + 1) = 2 b e2x = 2e x + 8 Given that P = 20 when t = 5, find the value of k.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Calculate the population when
a Use loga x = y Û x = ay:
a t=0 b t = 10
1
3x + 1 = e2 3x + 1 = e
10 The mass, m, of a radioactive material at a time t is given The half-life is the time taken
C3

C3
x= e −1
3
= 0.216 to 3 s.f. by m = m0e-kt where k and m0 are constants. for the material to decay to
9 half of its original mass.
If m = m0 when t = 10, find the value of k.
10
b Rearrange: e2x - 2e x - 8 = 0
Also find the half-life of the material.
Let y = ex: y2 - 2y - 8 = 0 y = ex is a useful substitution.
(y - 4)(y + 2) = 0 Remember it.
11 The isotope strontium-90 is present in radioactive fallout and
y = 4 or -2 has a half-life of 29 years. Its mass m at a time t (years) is given
Hence e x = 4 or -2 by m = m0e-kt where k and m0 are constants.
Find how many years elapse before any of this isotope in the
ex > 0 for all x, so the solution is x = ln 4 = 1.39 to 3 s.f.
atmosphere decays to a hundredth of its original mass.

Exercise 3.3 12 A hot liquid cools in such a way that the temperature difference, q,
1 Solve these equations. Give your answers to 3 s.f. between its actual temperature and the room temperature is given
a ex = 9 b e x = 7.39 c 4e x + 2 = 36 d e4x + 2 = 36 by q = ke-at where t is the time and k and a are constants. When the
room is at a constant temperature 20 °C and the initial temperature
e 4e-x + 2 = 36 f e-x = 7.39 g 7 - 2e-x = 4 h e x+2 = 20
of the liquid is 70 °C, the temperature falls to 60 °C after two minutes.
i 2e 3x-1 = 10 j e4x −1 = 1 k e3-x = 1 l e2x-3 × e x = 0.1 Find the temperature of the liquid after 10 minutes.
2

2 Solve these equations. INVESTIGATION
a ln x = 6 b ln (x + 1) = 2 c ln (2x - 1) = 0
13 The graph of y = Asin w t can represent an oscillation of The variable y can represent,
d ln (2x + 1) = 1 e ln (3 - x) = 2 f ln (5 - 2x) = 2.78 amplitude A over a time t, where A and w are constants. for example, the noise of a
If A = A0ekt, where A0 and k are constants, then the siren or an electrical current.
3 Solve these equations. amplitude A changes with time. Use computer software Start your exploration with
a e 2x - 3e x + 2 = 0 b e 2x + 12 = 7e x c e 2x = e x + 2 Use the substitution y = ex to draw the graphs of y = A0 sin wt and y = A0ekt sin wt A0 = 2, k = -0.2 and w = 1
d 2e 2x = e x + 10 e 4e 2x - 8e x + 3 = 0 f e2x - 16 = 0 for t > 0 for different values of A0 and k.
86 87


8 The population, P, of a certain organism grows exponentially
Review 3 t
over time t (days) according to P = Ae 20
Find the value of A and complete this table of values.
1 Solve these equations for x, giving answers to 2 significant figures. Draw the graph of P against t.
a 3e x = 10 b e1-2x = 0.5
t 0 5 10 15 20
c ln (2x - 1) = 3 d ln (1 - x) = -3 P 5

2 Sketch the graph of y = e x Calculate how long it takes for the population to double its
Sketch the graphs of these functions, stating the initial size.
transformations of y = e x which occur.
9 The temperature T °C in a boiler rises exponentially over
a y = e-x b y = 2e x + 3 t
24 hours so that, after a time t hours, T = T0e 20
c y = 3 - e 2x d y = e|x|
a Given that the temperature is 165 °C after 10 hours, find
3 Sketch the graph of y = ln x the value of T0.
Sketch the graphs of these functions, stating the
b What is the temperature after 24 hours?
transformations of y = ln x which occur.
a y = 2 - ln x b y = 2ln x + 1 10 Trees in a certain location are infected by a disease.
The number of unhealthy trees, N, was observed to change over
c 1 - 2ln x d y = |ln x| −t
C3

C3
time t (in years) as given by N = 200 - Ae 20

4 Solve the equations a If there are 91 unhealthy trees after 10 years, find the
a e2x + ex =6 b ex + e-x = 2.5 value of A.
b How many unhealthy trees were there initially?
5 Find the exact solutions of
c What is the limiting value of N as time increases?
a e2x+3 = 6 b ln (3x + 2) = 4 [(c) Edexcel Limited 2003]

11 Find, giving your answer to 3 significant figures where
6 Given that f(x) = ex and g(x) = 1 e x − 1, find appropriate, the value of x for which
2
a the values of f(2), f -1(2) and g -1(2), to 2 decimal places a 3x = 5

b the functions f -1(x) and g -1(x). b log2 (2x + 1) - log2 x = 2
c ln sin x = -ln sec x, in the interval 0 < x < 90° [(c) Edexcel Limited 2005]
7 Find the inverse function f -1(x) for each function f(x).
Sketch the graphs of f(x) and f -1(x) on the same diagram,
labelling any intersections with the coordinate axes.
In each case give the domain and range of f -1(x).
a f(x) = 3 + ln x b f(x) = ln (x - 2)
c f(x) = 2 + e-x d f(x) = 1 - e2x

88 89

3Exit
4Differentiation
This chapter will show you how to differentiate
the three basic trigonometric functions (sin x, cos x, tan x) and the
exponential and logarithmic functions (ex and ln x)
Summary Refer to
the sums, differences, products and quotients of these functions
f(x) = a x is an exponential function for all values of a.
composite functions formed by having functions within functions
f(x) = e x is the exponential function, where the irrational number
functions of the type x = f( y)
e has a value of 2.71 828 to 5 decimal places. 3.1
f(x) = e x has the inverse function f -1(x) = ln x,
where the natural logarithm ln x denotes loge x. Before you start
The graph of y = ln x is the reflection of the graph of y = e x in
the line y = x 3.2
You can solve the equation eax+b = q by taking natural 1 Find the gradient of the tangent at a 1 Draw the graph of y = x2 accurately.
logarithms of both sides. point P on a curve using the method The point P(1, 1) is fixed and point Q moves
You can solve the equation ln (ax + b) = q by rewriting it of small increments. from (3, 9) to (1, 1). Draw the chord PQ in
as ax + b = eq 3.3 See C2 Section 7.1. several positions and find its gradient each
Exponential growth occurs when a variable y changes with time. Compare these values with the
C3

C3
time t and y = Aekt where k > 0 gradient of the tangent at P.
Exponential decay occurs when a variable y changes with
time t and y = Ae-kt where k > 0 3.3 2 Find the gradient of a tangent to a curve. 2 Find the gradient of the tangent to these
dy curves at the point (1, 2).
e.g. If y = 3x2 - 4x + 1, then = 6x − 4
dx a y = x2 + 2x - 1 b y = x3 - x + 2
The gradient of the tangent at (1, 0) is 6 ´ 1 - 4 = 2
c y = (2 - x)(3 - x)
Links
3 Find the equation of a straight line if you 3 Find the equation of the straight line which
Exponentials and logarithms are used to understand
know its gradient and a point on the line. a has a gradient of 4 and passes through
and quantify many natural phenomena.
e.g. The point (1, 2) is on a straight line, gradient 3. the point (3, 2)
The decibel scale, used to measure sound, is based on Substitute in y = mx + c: 2 = 3 ´ 1 + c so c = -1 b passes through the points (3, 1) and (5, 5).
logarithms. The equation of the line is y = 3x - 1
The equation dB = 10log(I/I0)
is used to compute the intensity of a sound, where dB 4 Find a stationary value of a function and 4 Find the turning points on the curve
is a unit of sound in decibels, I is the intensity of the sound, decide if it is a maximum or minimum. y = x3 - 3x2 - 9x + 1
dy and determine whether they are maximums
and I0 is the softest sound that a human ear can detect. e.g. If y = x2 - 6x + 1, then = 2x − 6
dx or minimums.
dy
This equation can be used to calculate certain values, = 0 when x = 3 and y = 32 - 18 + 1 = -8
dx
such as the threshold for noise pollution.
d2 y
= 2 > 0, so -8 is a minimum point.
dx 2

5 Use the laws of logarithms. 5 Prove that log3 18 = 1 + 1 log3 2
2
e.g. log10 500 = log10 (5 ´ 102)
= log10 5 + log10 102
= log10 5 + 2log10 100
= log10 5 + 2
90 91

4 Differentiation

The derivatives of sin x, cos x and tan x.
4.1 Trigonometric functions
Consider the function y = sin x, where x is in radians.

You can find the value of the derivative of the function See C1 and C2 for revision
y = sin x graphically from the gradient of the tangent at of differentiation. Let P (x, y) and Q (x + dx, y + dy) be two points on y
a given point. When differentiating the graph of y = sin x
trigonometric functions you
p Q y = sin x
The tangent is parallel to the x-axis when x = must work in radians. For P, y = sin x y + dy
3p 2
and 2 and the gradient is 0. For Q, y + dy = sin (x + dx) dy
P R
QR y
The gradient of the chord PQ = dx
y PR

The tangent at the point sin(x + d x) − sin x O x + dx x
The gradient is also = x
where x = 0 rises at 45° 1
1 when x = 2p (x + d x) − x
to the x-axis and has a y = sin x
= sin x cos d x + cos x sin d x − sin x
Use the expansion of sin (A + B)
gradient of 1.
O p p 3p 2p x dx from Chapter 2.
2 2
−1 If P is fixed and Q approaches P, then dx ® 0 You can test these statements
cos dx ® 1 by ﬁnding the sine and cosine
and sin dx ® dx of small angles (in radians) on
By symmetry, the gradient your calculator.
is -1 when x = p and, in the limit, the chord PQ becomes the tangent at P.
For y = sin x
C3

C3
dy
The phrase ‘in the limit’ indicates that d x has reduced to zero and has
p 3p You can ﬁnd other values using dx
x 0 p 2p dy
2 2 computer software to draw the become an exact differential .
dx
dy curve and its tangents.
So, as dx ® 0, the gradient of PQ ® sin x × 1 + cos x × d x − sin x
1 0 -1 0 1
dx
dx

which, as a graph, gives these points: The values in the table and on the ® cos x × d x
dx
graph are the same as those for the ® cos x
function y = cos x
In the limit, when Q reaches P, the gradient of the tangent
y y at P is cos x.
dy
dx y = cos x
1 1

d(sin x) x is in radians in all of
= cos x
O p p 3p 2p x O p p 3p 2p x dx these results.
2 2 2 2
−1 −1
In a similar way, you can show that

This suggests that the derivative of d(cos x) −
= sin x The derivative of cos x
y = sin x is closely related to the dx has a negative sign.
function y = cos x
Derive the results for cos x and
d(tan x)
= sec 2 x tan x for yourself.
dx

92 93

4 Differentiation 4 Differentiation

EXAMPLE 3
EXAMPLE 1
Find the gradient of the curve y = x2 + cos x Find the values of q for which y = 3 + sin q has a
at the point where x = p maximum value.
2 ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

You have y = 3 + sin q
You have y = x2 + cos x
dy
Differentiate wrt x: Differentiate wrt q : = cos q = 0
wrt means ‘with respect to’. dq
dy
= 2x − sin x for stationary values when q = p , 3p , 5p , …
dx 2 2 2
Substitute x = p : d2 y
2 Differentiate again: = − sinq
dq 2
= 2 × p − sin p
dy
dx 2 2
d2 y
=p-1 At q = p = − sin p = −1 < 0
2 dq 2 2
d2 y
The required gradient is p - 1. which indicates a maximum value of y < 0 indicates a maximum
dq 2

d2 y 3p
2 = − sin 2 = +1 > 0
At q = 3p
dq
EXAMPLE 2

2
Find the equation of the normal to the curve d2 y
y = 1 + 2tan x at the point (0, 1). which indicates a minimum value of y > 0 indicates a minimum
dq 2
5p
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

5p d2 y
2 = sin = 2 = 1 < 0
At q = − −
C3

C3
You have y = 1 + 2tan x 2 dq
Differentiate wrt x: which indicates a maximum value of y
dy and so on.
= 2 sec 2 x
dx
Maximum values of y occur when
Substitute x = 0:
= 2sec2 0 sec x = 1 q = p , 5p , 9p , …
cos x 2 2 2
= 2
= p + 2np where n = 0, ±1, ±2, …
cos2 0 2
=2 =2
1

The gradient of the tangent at the point (0, 1) is 2. If the gradients of the tangent Exercise 4.1
and normal are m and m¢, then
The gradient of the normal at (0, 1) is − 1 . 1 Differentiate these functions with respect to x.
2 mm¢ = -1 or m¢ = − 1
m a x2 + sin x
The y-intercept of the normal is 1. See C1 Chapter 2 for revision of
equations of straight lines. b 6x - cos x
So, the equation of the normal is y = 1 x + 1 −
2 c sin x + tan x
d sin x + x + 1
e 3cos x - 4tan x
f 3x2 + 2x - 1 - 1 cosx
2

94 95


2 Find the gradient of the tangent to each of these curves at the 11 A pendulum bob swings such that the angle q that the pendulum makes
point with the given value of x. with the vertical is given by q = q0(sin t + cos t), where q0 is a constant.
a y = 2x + tan x when x = p a Find the value of q when
4
p p
b y = x2 - sin x when x = i t=0 ii t =
2 2
c y = 6x + 2cos x when x = p b Find the value of its angular velocity dq when
dt
p
d y = cos x + sin x when x = p i t=0 ii t =
2 2
e y = 3tan x - 2cos x when x = 0 c Explain why the time taken for one oscillation of the bob is 2p.

3 Find the equation of the tangent to the curve y = cos x at the 12 A particle moves in a straight line so that its distance, y metres,
point where x = p from a fixed point O is given by
4
y = 5cos t + 3sin t
4 Find the equation of the tangent to the curve y = sin x at the point
where x = p . At which point does this tangent cut the x-axis? where t is the time in seconds after it has begun to move.
4
a How far is the particle from O at the start of the motion?
5 Find the equation of the normal to the curve y = sin x + cos x at the
b Find the first two times that the particle is at rest.
(2 )
point p , 1 . At which point does this normal intersect the x-axis?
c What is the particle’s acceleration after one second of motion
6 Find the smallest positive value of q for which and when does the particle next have this acceleration?
1
C3

C3
a the gradient of the curve y = q + sin q has the value 2

b the gradient of the curve y = sin q - 3cos q has the value 0. INVESTIGATION
13 Construct a table of values with these headings for 0 x 0.5 in radians.
7 Find the smallest positive value of q for which these functions
have maximum values. x (radians) x (degrees) sin x cos x tan x

a y = sin q - 2 b y = sin q + 2cos q
a When x is in radians, for what range of values of x is
8 Find the smallest positive value of q for which these functions
i sin x » x ii tan x » x
have minimum values.
so that they differ by no more than 10% of the value of x?
a y = 5 + cos q b y = cos q - sin q
b The diagram shows the graph of y = 1 - kx2
9 Find the maximum and minimum values of the function y
f(x) = 3sin x + 4cos x for -p x p
Show that, if a and b are the smallest positive values of x at the 1
maximum and minimum values respectively, then a - b = p y = cos x

O x
10 An object moves in a straight line such that its distance y from –p p
2 2
a fixed point at a time t is given by y = A sin t, where A is a
positive constant. y = 1 – kx2

dy dy
a Find the velocity of the object and prove that dt = A2 − y 2 Find a value of k such that
dt
cos x » 1 - kx2 for small values of x (in radians).
d2 y
b Prove that +y=0
dt 2
96 97

4 Differentiation

You can use these values to draw the graph of y = e x: y
4.2 The exponential function, ex
y = ex
It is the exponential function and is one member of the family of
Suppose you want to find a function that stays unchanged when exponential functions y = ax which you ﬁrst met in unit C2.
it is differentiated.
In summary, 1
Start the search with the simple function y = 1 For y = 1, dy = 0
dx O x
Now work backwards. dy d (ex )
if y = e x, then = e x or = ex
dy
In each step, make the same as y from the previous step and integrate
dx dx
dx
to give a new y:
dy
=0

EXAMPLE 1
Þ y=1 Find the gradient of the tangent to the curve
dx
dy y = (e x + 1)2 - e2x at the point where x = 0
=1 Þ y=1+x ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

dx
2
You have y = (e x + 1)2 - e2 x
dy
=1+ x Þ y=1+x+ x = e2x + 2e x + 1 - e2x
dx 2
= 2e x + 1
dy x2 2 3
=1+ x + Þ y=1+x+ x + x dy
= 2e x
dx 2 2 2×3 Differentiate wrt x:
dx
Continue this process to produce a series with an inﬁnite number of terms: dy
At x = 0 = 2e0
2 3 4 n dx
C3

C3
y =1+ x + x + x + x + +x + This series differentiates to give =2´1=2
1× 2 1× 2 × 3 1× 2 × 3 × 4 n!
the same series.
The gradient of the tangent is 2 at the point where x = 0
By substituting different values of x, you can calculate the sum You need to also show that this
of the series and record the results in a table. series converges no matter what
value is given to x.
For x = 0 y=1

EXAMPLE 2
Find the point on the curve y = e x - x at which y has a
For x = 1 y = 1 + 1 + 1 + 1 + 1 + . . . = 2.718 282… stationary value.
2 6 24
Determine the nature of the stationary value.
For x = 2 y = 1 + 2 + 4 + 8 + 16 + . . . = 7.389056. . . and so on.
2 6 24 ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

dy
x 0 1 2 3 4
Differentiate y = ex - x wrt x: = ex − 1
dx
y 1 2.718 282. . . 7.389 056. . . 20.085 537. . . 54.598 150. . .
Recall from Chapter 3 that the At a stationary point, dy = 0 ex - 1 = 0
= 2.718 2821 = 2.718 2822 = 2.718 2823 = 2.718 2824 number e = 2.718282…
dx
ex = 1
Like 2 and p, the number e is irrational. x=0
It cannot be given an exact numerical value, but this series and y = e0 - 0 = 1
allows you to calculate the value of e to as many decimal places There is a stationary value at the point (0, 1).
as you wish.
d2 y
Differentiate again: = ex
You can also use negative values of x in the series. dx 2
For x = -1, y = 1 − 1 + 1 − 1 + 1 − ... = 1
= e−1 d2y d2 y
2 6 24 2.718282 At the point (0, 1) = e0 = 1 > 0 At a minimum point,
dx 2
is positive.
dx 2
x -1 0 1 2 3 4 ... n ... Try substituting other negative
Hence, the stationary value at the point (0, 1) is a minimum.
values of x into the series.
y e-1 e0 = 1 e e2 e3 e4 . . . en ...

98 99


Exercise 4.2 8 Find the values of x for which these functions have stationary values.
x x x Determine the nature of the stationary values.
1 a Draw the graphs of y = 2 , y = e and y = 3 accurately on the
same axes for -3 x 3 a f(x) = 1 - x + e x
b Calculate the gradient of y = e x when b f(x) = 1 + x - e x
i x=0 ii x = 1
c f(x) = 1 + 4x - 2e x
c Find the equation of the tangent to the curve y = ex when
i x=0 ii x = 1 9 Sketch the graph of the function y = x - e x
Indicate any stationary values.
2 Differentiate these functions with respect to x.
a 3x2 + e x b 4e x + 6x - 1 10 Consider the function y = e2x - 3e x + 2
a By letting t = e x, find the points where the graph of This substitution gives a
c 5sin x + 2e x d 3tan x − 1 e x quadratic equation in t.
2 y= e2x - 3e x + 2 cuts the x-axis.
e (e x + 1)(x - 1) - xe x f (e x + 1)(e-x - 1) - e-x
b Find the value of x at which y has a stationary value.
3 Find the gradient of the tangent to each of these curves at the Find whether the stationary value is a maximum or a minimum.
point with the given value of x.
Give each answer i in terms of e ii as a decimal to 3 significant figures. 11 Sketch the curve y = x2 + e x

a y = ex - x when x = 2
b y = 3x2 - 4e x when x = 1
C3

C3
INVESTIGATION
c y= ex - cos x when x = p 12 Use a spreadsheet with these headings, where
2
d y = x2 + 2x + 1 - e x when x = -1 f(x) = ax, f(x + dx) = ax+dx

e y = tan x + 2e x and the gradient g(x) = f(x + d x) − f(x)
when x = 0 dx

4 Find the equation of the tangent to each curve at the point a x dx f(x) f(x + dx) g(x)

where x = 0
a y = 3x + 2e x
b y = 3e x - sin x + 5 The function g(x) gives the gradient of the chord PQ which,
for small values of dx such as 0.0001, is approximately y
5 Find the equation of the normal to the curve y = 1 e x + 2x 2 equal to the gradient of the tangent at P.
( 2)
2 Q y = f(x)
at the point 0, 1 . Use the spreadsheet for a ¹ e (say, a = 4) and confirm f(x + dx)
that f(x) ¹ g(x) for any value of x. P
Repeat for other values of a ¹ e f(x)
6 Prove that the normal to the curve y = 1 - x + e x at the point (1, e)
passes through the point (e2, -1). Now let a = e
Confirm the result of Chapter 3 that the gradient at the O x x + dx x
7 Line L1 is the tangent to the curve y = 2e x - x at the point (0, 2). point (0, 1) is e.
Line L2 is the tangent to the curve y = sin x - x2 at the origin.
Show that, for other values of x, f(x) and g(x) have (almost)
Prove that the point of intersection of L1 and L2 is (2, 4).
the same value. You have now illustrated (but not proved)
that f(x) = e x stays unchanged when you differentiate it.

100 101

4 Differentiation

Exercise 4.3
4.3 The logarithmic function, ln x
1 Differentiate these functions with respect to x.
a 2ln x b ln x2 c ln (4x3)
y
y = ln x is the inverse of the function y = ex y = ex (2 )
d ln 1 x ( 2x )
e ln 1 f ln 5x 3

The graph of y = ln x is the reflection of the graph of y = ex ⎛ ⎞
g ln ⎜ 1 ⎟ h ln (e2x) i ln (xe x)
in the line y = x ⎝ x⎠
1
Consider the derivative of y = ln x j ln (x2 e2x)
O x
y = In x
2 Find the gradient of the tangent to each of these curves at the
point with the given value of x.
Rewrite the relationship: x = ey logab = c implies b = ac a y = 2x + ln x when x = 2

dx b y = x3 - ln x2 when x = 1
Differentiate with respect to y: = ey You differentiate x wrt y.
dy c y = ln x + ex when x = 1
The derivative of e y wrt y is e y.
( 3x )
dy 1
So = d y = ln 1 + 1 when x = -1
dx dx dy 1
The result dx = dx is shown 3x
dy
dy e y = ln xp + sin x when x = p
= 1y later in this chapter. 2
C3

C3
e
3 Find the points at which the graphs of these functions have zero gradient.
= 1
x Determine whether the functions have a maximum or a minimum
at these points.
a f(x) = x - ln x b f(x) = x2 - ln (x2)
dy d(ln x) 1
If y = ln x, then =1 or = x 4 Find the equation of the tangent to the curve y = x - ln x at the
dx x dx
point where x = 2. At which point does the tangent cut the x-axis?

5 Find the equation of the normal to the curve y = 2ln x3 at the point
EXAMPLE 1

Find the differential of where x = 1. Find the distance between the points at which the
b ln ⎛ ⎞ 3 normal cuts the x- and y-axes.
a ln (3x4) ⎜ ⎟
⎝ x⎠
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
6 Show that the point (1, e) lies on the curve y = e x + ln x
Find the area of the triangle bounded by the two coordinate
a Let y = ln (3x4) b Let y = ln ⎛ 3 ⎞ Where possible, simplify the
⎜ ⎟ axes and the tangent to the curve at this point.
= ln 3 + ln x4 ⎝ x⎠ logarithmic function to make it
1 easier to differentiate.
= ln 3 + 4 ln x = ln 3 - ln x 2
INVESTIGATION
()
1
dy
So =0+4 1
x = ln 3 − 1 lnx x = x2
7 Differentiate y = log10 x by ﬁrst changing the base of the
dx 2

()
logarithm to base e.
=4 dy 1
x So =0− 1 You can then differentiate y = ln x to ﬁnd d(log10 x).
dx 2 x dx
=−1 Try other bases. What do you notice?
2x

102 103

4 Differentiation

EXAMPLE 1 (CONT.)
4.4 The product rule b Let y = (x3 + 2)sin x
u = x3 + 2 v = sin x
Consider two functions u = f(x) and v = g(x) du = 3x 2 dv = cos x
which are multiplied so that y = uv y = f(x) ´ g(x) dx dx
dy
So = u dv + v du = (x 3 + 2)cos x + 3x 2 sin x
dx dx dx
Let dx be a small increase in x which produces small increases
dy d(tan x)
of du, dv and dy in u, v and y respectively. c When y = x2 tan x, = x 2 ´ sec2 x + tan x ´ 2x = sec 2 x
dx dx
You have y + dy = (u + du)(v + dv) = x2 sec2 x + 2x tan x
= uv + udv + vdu + dudv
Exercise 4.4
Subtract y = uv: dy = udv + vdu + dudv 1 Differentiate these functions with respect to x.
Divide each term by dx: dy
= ud v + vd u + d ud v (1) a xln x b x2sin x c e xtan x
dx dx dx dx
d (x2 - 3x + 1) e x e e x ln x f sin xln x
As dx ® 0 dy dy d v dv d u du
→ , → , → g x-2cos x h xe
1 x
i (x3 - 1)ln x
dx dx d x dx d x dx
d udn
and du ® 0, dv ® 0 As x ® 0, tends to zero.
dy dv du
dx j x tan x k (x2 - 1)(x3 - x2 + 1) l (x3 + 1)(x2 - 2x + 4)
In the limit from (1), dx = u dx + v dx
2 Find the gradient of the tangent to the curve y = x2sin x at the
C3

C3
point where x = p
2
In the limit, if y = uv then 3 Find the equation of the tangent to the curve y = x3 e x at the point where
dy dv du
=u +v a x=0 b x = -3
dx dx dx
d
or dx
[f(x) ´ g(x)] = f(x) ´ g¢(x) + g(x) ´ f ¢(x) 4 Find the equation of the normal to the curve
y = (x2 - 2x + 1)(x3 - 4x2 + l) at the point (1, 0).
Think of the product rule as:
(1st function ´ derivative of the 2nd) + (2nd function ´ derivative of the 1st) 5 a Prove that the equation of the normal to the curve y = sin x ´ ln x
at the point where x = 1 is given by y = 1.19(1 - x)
EXAMPLE 1

b Find the area of the triangle enclosed by the normal and the x- and y-axes.
Differentiate a x2 ln x b (x3 + 2)sin x c x2 tan x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

6 Find the turning points on the curve with the equation y = x2 (x + 1)8
a Let y = x2 ln x
Determine the nature of each turning point and sketch the curve.
u = x2 v = ln x
du dv 1 7 Find the stationary value of the function f(x) = x3ln x
= 2x = Differentiating both u and v wrt x.
dx dx x
and show that it is a minimum.
dy dv du
So =u +v
dx dx dx dy
8 Given that y = sin xcos x, prove that = 2cos2 x - 1
dx
= x × 1 + ln x × 2x
x
2

Find the stationary points on the graph of y = sin x cos x for 0 x p
= x + 2x ln x
and determine whether they are maxima or minima.

104 105

4 Differentiation

EXAMPLE 1
4.5 The quotient rule
Differentiate
sin x
ln x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Consider two functions u = f(x) and v = g(x) sin x
You have y =
which are divided to give y = u
v y = f(x) ln x
g(x) Take care not to mix up the
where u = sin x and v = ln x
functions u and v.
Let dx be a small increase in x which produces small increases du = cos x dv = 1
and dx x
of du, dv and dy in u, v and y respectively. dx

You have y + d y = u + du dy ln x × cos x − sin x × 1
x
v + dv So =
dx (ln x)2
Subtract y = u: d y = u + du − u
v v + dv v = x ln x cos x − sin x
2 Multiply both numerator and
x(ln x) denominator by x.
v (u + d u ) − u ( v + d v )
Combine the two fractions on the RHS: =
v (v + d v )

EXAMPLE 2
x 2 + 4x + 1
Expand and simplify: = vd2u − ud v Differentiate
cos x
v + vd v ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

y = x + 4x + 1
2

Divide by dx: dy v × du − u × dv (1) You have
= dx dx cos x
dx v 2 + vd v
dy cos x ´ (2x + 4) - (x 2 + 4x + 1) ´ (- sin x) d(cos x) −
= sin x
C3

C3
dy dy d u du d v dv
So = dx
As dx ® 0 → , → , → dx (cos x)2
dx dx d x dx d x dx
= (2x + 4)cos x + (x + 4x + 1)sin x
2
and du ® 0, dv ® 0 2
cos x
dy v du − u dv
= dx 2 dx = (2x + 4) + (x + 4x + 1)tan x
2
In the limit from (1),
dx v cos x

EXAMPLE 3
u Find the derivatives of a tan x b cot x These results are given on the
In the limit, if y = v then formulae list. You do not need
c cosec x d sec x
du dv to memorise them.
dy v× −u× ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

= dx dx
dx v2 a Let y = tan x = sin x
cos x
d ⎛ f(x) ⎞ g(x) × f ′(x) − f(x) × g′(x)
or
dx ⎜ g(x) ⎟
= dy cos x × cos x − sin x × (− sin x)
⎝ ⎠ (g(x))2 =
dx cos2 x

= cos x +2 sin x
2 2
Think of the quotient rule as: cos x
(bottom ´ derivative of top) - (top ´ derivative of bottom), all divided by (bottom)2
= 1
cos2 x
The solutions to parts b, c and d
= sec2 x are on the next page.

106 107


EXAMPLE 3 (CONT.) Exercise 4.5
cos x
b Let y = cot x = sin x 1 Differentiate these functions with respect to x.
x2
c x +1
2
dy sin x ´ (- sin x) - cos x ´ cos x x
d ln x
= a sin x b tan x
x x
dx sin 2 x

= (sin x + cos x)
2 2
− ex x +1 x cos x
2 e sin x f g h
sin x ex sin x x2
-1
= = -cosec2 x
k 2x 2 − 1
3
sin x 2
i 1− x j x2 + 1 l x2
1+ x x2 − 1 3x − 1 ln x
c Let y = cosec x = 1
sin x 2 Find the gradient of the tangent at the point where x = 0
dy sin x × 0 − 1 × cos x for the curve
=
dx sin 2 x
− a y = 3x − 2 b y = tan x
= cos x 2x + 1 x +1
sin 2 x

= -
1 ´ cos x
sin x sin x
3 The tangent to the curve y =
x3 + 1
1,
2 ( )
x at the point 1 cuts the

= -cosec xcot x x-axis at the point P. Find the coordinates of P.

d Let y = sec x = 1 4 Find the equation of the normal at the point (0, 0)
cos x
on the curve y = 2x
x
dy cos x ´ 0 - 1 ´ (- sin x) e
C3

C3
=
dx cos2 x 2
5 Find d y if y = 1 + x
= sin2x 2
dx x
cos x

= 1 ´ sin x = sec xtan x
cos x cos x 6 Prove that the maximum value of f(x) = ln3x is 1
x 3e

7 Use both the product rule and the quotient rule to differentiate
For some functions, you need to use both the product rule and
these functions.
the quotient rule. x
x sin x
a sin x cos x b cos x c xe
x ln x
EXAMPLE 4

2 x
Differentiate xe x 2
x sin x
sin x d e ln x e x ln x f
x x
e ln x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Firstly, consider the function f(x) = x2e x,
which is the product of x2 and e x.
INVESTIGATION
Differentiate f(x) using the product rule: f ¢(x) = x2e x + e x ´ 2x x
= x2e x + 2xe x 8 You can differentiate y = e sin x
2
2 x x
Let y = x e and use the quotient rule:
sin x either by ﬁrst using the product rule on ex sin x
dy sin x(x 2e x + 2xe x ) − x 2e x cos x followed by the quotient rule on the whole expression
= x
dx sin 2 x or by ﬁrst using the quotient rule on e 2 followed by
x
xe x(x sin x + 2 sin x − x cos x) the product rule on the whole expression.
=
sin 2 x
Show that these two methods give the same answer.
Is one method easier than the other?
108 109

4 Differentiation

EXAMPLE 2
4.6 The chain rule dy
Find when a y = cos3 x b y = x3 − 1
dx
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

A composite function is a ‘function of a function’. Examples of composite functions:
ln (x³ - 1) esin x a y = cos3 x, so if u = cos x then y = u3 dy
You can differentiate ‘functions of functions’ using the chain rule. = 3u2 = 3 cos 2 x
dy du
(x² - 2x - 1)4 2+ x 3
so = 3cos2 x ´ (-sin x)
dx du
Consider y = g(u) and u = f(x), so that y = g(f(x)) = − sin x
dx
= -3cos2 xsin x
1 1 1
− −
Let dx be a small increase in x which produces small increases b y = x 3 − 1, so if u = x3 - 1 then y = u = u2 dy
= 1 u 2 = 1 (x 3 − 1) 2
du 2 2
of du and dy in u and y respectively. dy 1 -
1
so = (x 3 - 1) 2 ´ 3x 2 du
= 3x 2
d y d y du dx 2 dx
Then = ×
d x du d x 3 x2
=
dy dy dy dy d u ® du 2 x3 − 1
As dx ® 0, ® , ® and (1)
dx dx du du dx dx
dy dy du
In the limit from (1), = ×
dx du dx

EXAMPLE 3
a Differentiate 2ln ( x x 2 + 1 )

dy dy du b Find the equation of the tangent to the curve
In the limit, if y = g(f(x)), then = ×
dx du dx y = 2ln ( x x 2 + 1 ) at the point where x = 1
C3

C3
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a Let y = 2ln ( x x 2 + 1 )
Think of the chain rule as:
Where possible, simplify functions
derivative of ‘outer’ function wrt u ´ derivative of ‘inner’ function wrt x 1 before you differentiate.
= 2ln x + 2ln (x 2 + 1)2
EXAMPLE 1

Differentiate a sin (x2 + 4x + 3) b ln (sin x) = 2ln x + ln (x2 + 1) ln(a ´ b) = ln a + ln b
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Differentiate, using the chain rule for ln (x2 + 1):
a Let y = sin (x2 + 4x + 3) u = x2 + 4x + 3 y = sin u
du = 2x + 4
dx
dy
du
= cos u Differentiating u wrt x and y wrt u.
dy
dx
1 ⎛ 1
= 2 x + 2x × ⎜ 2 ⎞ ⎟
⎝ x + 1⎠
() y = lnu and u = x2 + 1
so dy = 1 and du = 2x
du u dx
⎡ 2 1 2⎤
= 2⎢ x + 2 + x ⎥
dy dy du
Apply the chain rule: = ´
dx du dx ⎣ x (x + 1) ⎦
= cos u ´ (2x + 4) 2(2x 2 + 1)
=
Substitute u = x2 + 4x + 3: = (2x + 4)cos (x2 + 4x + 3) x (x 2 + 1)
1
b Let y = ln (sin x) u = sin x y = ln u b When x = 1 y = 2ln (1 × 2 ) = 2 ln 2 2 = ln 2
dy
du = cos x = 1 dy
=
2´3
=3
dx du u
dx 1 ´ 2

Apply the chain rule:
dy 1
= ´ cos x The equation of the tangent is y − ln2 = 3 y − y1
dx u x −1 Using =m
x − x1
y - ln 2 = 3(x - 1)
Substitute u = sin x: = 1 × cos x y = 3x + ln 2 - 3
sin x
= cos x cos x 1
= cot x = = cot x
sin x sin x tan x
110 111


Exercise 4.6 7 Find the gradient of the curve y = ln 1 + sin x at the point Simplify the logarithm before
1 Differentiate with respect to x where x = p you differentiate.
4
a sin (x2 + 1) b cos (x3 - 1) c tan (x2)

d ln (x2 + 2x + 3) e ln (x3 + 1) f ln (x3) Simpify part f before 8 Find the equation of the tangent to the curve y = x 2 + 16
you differentiate. at the point (3, 5).
g ln (sin x) h sin (ln x) i ln (ln x)
2 9 Differentiate with respect to x, where a and b are constants.
j esin x k e x+4 l ex 2
1
a a 2 + b 2x 2 b c (a2x 2 − b2)3
m (x2 + 1)5 n (2x + 6)7 o (x3 - 1)-2 ax 2 + b

1 Write part r as (x - 1)-1 ﬁrst.
p x2 − 1 q x3 + 1 r
x −1
10 Given that a and b are constants, find the only value of x for
You could use the quotient rule
which a has a stationary value.
3 instead of the chain rule but
2 1
the working is longer. ax 4 + b
s x 2 + 3x − 1 t u
2x + 1 x2 + 1
⎛ ⎞
11 a Prove that the curve y = ln ⎜ 1 + xe ⎟ has no stationary values.
x
1 2 ⎝ e ⎠
v ln x w x
e +1
x 3
2x − 1
b Show that the tangent to this curve at the point where
2 Find f ¢(q) when x = 0 has the equation x + 2y = ln a.
Find the value of a.
a f(q) = sin (q 2) b f(q) = sin2 q
C3

C3
c f(q ) = sin q d f(q) = tan (q 3) 12 a Prove that the curve y = 82 + 1
has only one turning point.
x (1 − x)2
1
(3 )
e f(q) = tan3 q f f(q) =
tan q b Show that the coordinates of the turning point are 2 , 27 .
2
dy c i Find an expression for d y .
3 Find when 2
dx dx
1 x
ii Hence, prove that the turning point is a minimum.
a y = sin x b y = ee
1
4 13 Find the position and nature of the stationary points of
c y= d y = 4e x
x 2 + 4x − 1 f(x) = sin2 x

e y = ln x − 1 f y = ln (1 −sin x x )
cos
Simplify the function in parts e
and f before you differentiate as in
question 1 part f. INVESTIGATION
4 Differentiate with respect to x.
14 Show that the function y = ln (x2 + 1) has only one stationary value.
a y = sin kx b y = cos kx Determine its nature and position.
c y = tan kx d y = ln kx The differential of tan kx is in the Sketch the graph of the function.
formulae list provided in the exam.
kx
e y=e None of the other results here are Check your answers by using computer software to draw the graph.
provided.
5 Find the equation of the tangent to the curve y = (2x - 4)3 Investigate the properties of ln (x2 + k) for different values of k.
at the point where x = 3

6 Find the point on the curve f(x) = (2x - 1)4 When the gradient is 1, f¢(x) = 1
where the gradient is 1.

112 113

4 Differentiation

EXAMPLE 3
4.7 Further applications Differentiate y = 3x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

You can combine the product, quotient and chain rules to Take the logarithms of both sides to remove the index:
differentiate more complicated functions. ln y = ln 3x = xln 3
Now, ln y is a function of y and y is a function of x.
So, ln y is a function of a function of x.
EXAMPLE 1

Differentiate y= ( x 2 + 1 ) sin x d(ln y) d(ln y) dy 1 dy
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• From the chain rule you have = × = × d(ln y) 1
=
dx dy dx y dx dy y
The function y = ( x 2 + 1 ) sin x is the product of Differentiate ln y = ln 3 ´ x with respect to x:
ln 3 is a numerical constant
x + 1 and sin x.
2
1 dy that is multiplying x.
× = ln 3 × 1
y dx
x 2 + 1 is a function of a function.
1 dy
You have u = x 2 + 1 = ( x 2 + 1) 2 and v = sin x = y ´ ln 3 = 3x ln 3
dx
Use the chain rule to differentiate u:
1
du 1 2 − dy
= (x + 1) 2 × 2x In general, if y = ax then = a x ln a
dx 2 dx
Use the product rule:
dy 1 −
1 Now consider a function of the type x = f(y) Examples are
= x 2 + 1 × cos x + sin x × (x 2 + 1) 2 × 2x
x = y 3 + 1 and
C3

C3
dx 2
x = ln (y2 - 3y + 1)
= ( x 2 + 1 ) cos x + x sin x Let dx be a small increase in x which produces a small
x2 + 1 increase of dy in y.
( x 2 + 1)cos x + x sin x Then dy
=
1
= dx dx
x2 + 1 dy
As dx ® 0 and dy ® 0

then d y → dy and d x → dx giving dy = 1
EXAMPLE 2

Differentiate y = ln sin (x2 + x + 1) dx dx dy dy dx dx
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• dy
You have y = ln u and u = sin v and v = x2 + x + 1
dy dy du dv
Use the chain rule: = ´ ´ There are three links in the dy 1
dx du dv dx
chain rule here. So, for x = f(y), =
dx dx
1 dy
= u × cos v × (2x + 1)

EXAMPLE 4
Substitute for u and v: = 1 × cos (x2 + x + 1) ´ (2x + 1) dy
sin v Find when x = y2 + 3
dx
(2x + 1)cos(x 2 + x + 1)
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

= dx
sin(x 2 + x + 1) Differentiate x = y2 + 3 with respect to y: = 2y
dy
= (2x + 1) cot(x2 + x + 1) dy
So = 1 = 1 = 1
dx dx 2y 2 x - 3 dy
Find in terms of y ﬁrst and
dy dx
then in terms of x.
114 115


Exercise 4.7 7 Find the value of x which gives a stationary point on the
1 Differentiate with respect to x. curve y = 52x - 4 ´ 5x
a (x2 - 1)3 sin x b (x3 + 1)2 tan x
dy
8 Find in terms of x when
c (3x + 2)4 ´ ex d ( x + 1) ´ ex dx
a x = 3y2 - 4 b x= y2 + 3
e sin2x f ( x − 1 ) ´ ln x
2
x2 c x = (y + 1)2
g x sin (x2) h x sin2 x
9 Find dy in terms of y when
i x cos 3x j x2 tan 2x dx
a x = 4y3 + y
k x e 2x+1 l x2 ln (x2 - 1)
b x = 3sin4 y
m sin x cos 2x n sin 4x cos x c x = 5tan3 (2y)
o sin x cos2 x p e x sin 3x d x = y ln y
e3x e x = e -y sin y
q e -2x cos x r x
sin x f x = 3sin 4y + 4cos 2y
s e t e -x sin 2x
x
(
g x = ln y 2 × y − 2 )
2 Differentiate with respect to x.
C3

C3
10 Find the equation of the tangent at the point (2, 1) on the
a ln sin (x2) b ln tan (x2) curve x = 2y 3
c sin2 3x d cos3 4x
11 Find the equations of the tangents to the curve x = 4y2 at the
Can you see an easier method for
e tan2 2x f tan3 (x + 4) points where x = 4
2
part h using inverse functions?
2 ln(x +1)
g esin x
h e
12 Prove that the curve x = y2 e y has no stationary points.
dy
3 Find when
dx
a y = 4x b y = 5x INVESTIGATIONS
Can you see a connection
c y = c x where c is a constant d y = 22x between parts a and d? 13 Find the turning points on the curve y = e x sin 2x for 0 x 2p
Sketch its graph over this range and check your answers using
e y = 52x f y = xx
computer software.
4 Find the value of x which gives a stationary point on the
14 a Find the turning points of the curve y = x + 1
x
curve y = x −1 ´
2
ex Sketch its graph.

d2y b Prove that the curve x = y + 1 has no turning points.
y
5 Find the value of 2
when x = 0 and y = e x sin 2x
dx Sketch its graph.

6 Find the gradient of the curve y = ln 1 + sin 2x at the c What is the connection between your two sketches?
point where x = p
2

116 117

4 Differentiation

9 The function y = sin (x) has the angle x measured in degrees.
Review 4
a Write the function so that the angle x is in radians.
d (sin x)
1 Use the product rule, quotient rule or chain rule to differentiate these functions with respect to x. b Find dx
tan x
a x3 tan x b tan3 x c tan (x3) d
x3
1 10 Find the derivative with respect to x of these functions.
e ex sin x f esin x g sin x h cos x ln x
e a f(x) = ln tan x b f(x) = ln (sin x cos x)
i sin 3x j tan 6x k ln (2x + 3) l e3x-1
c f(x) = (2x - 1)2(3x + 2)3 d f(x) = ln ((2x - 1)2(3x + 2)3)
m sinx x n ex (x2 - 3x + 4) o x
e x +12
11 Use the derivatives of cosec x and cot x to prove that
2 Differentiate these functions with respect to x, given that a and b are constants. d
[ln (cosec x + cot x)] = -cosec x [(c) Edexcel Limited 2004]
dx
ax
a y=e b y = eax+b c y = ef(x) d y = sin (ax)
d2 y
e y = sin (ax + b) f y = sin [f(x)] g y = tan (ax) h y = tan (ax + b) 12 Given that y = e x sin 2x, find the value of when x = 0
dx 2
i y = tan [f(x)] j y = ln (ax) k y = ln (ax + b) l y = ln [f(x)]
d2 y dy
13 If y = e-x cos x, show that + 2 + 2y = 0
dx 2 dx
3 Find the derivatives of these functions with respect to x.
a ex cos 3x b e3x tan x c e3x cos 2x d 4 tan 1 x (2 ) 14 Find
d2 y
dx 2
when y = 1 + x
1− x
C3

C3
e ln (x3 sin x) f ln ( x x + 1 ) g (3x + 1)2 h ln e x
(2x − 1)3 dy
i sin 5x
cos 5x
j (
x 2 sin 2x + p
2 ) k e3tan x l (1 + sin x )
ln 1 − sin x
15 Find
dx
in terms of x when

x a x = (y + 1)2 b x = y2 + 3
m xcot x n ln (tan x + sec x) o sec x
16 Differentiate with respect to x
4 Find the values of x for which these functions have stationary points for -p < x < p. 2x
a x3e3x b cos x c tan2 x
Determine the nature of the stationary points.
x dy
a y = xe-x b y = e ,x¹0 c y = e-x cos x d Given that x = cos y2 find dx in terms of y. [(c) Edexcel Limited 2002]
sin x

5 Find the equation of the tangent to the curve y = ln 1 + sin 2q at the point where q = p 17 a Differentiate with respect to x
2
3
3 i x2e3x+2 ii cos(2x )
6 The curve C has equation y = − ln(5x), where x > 0.
4x 2 3x
The tangent at the point on C, where x = 1, meets the x-axis at the point A. dy
b Given that x = 4sin (2y + 6) find in terms of x. [(c) Edexcel Limited 2006]
1 dx
Prove that the x-coordinate of A is 5 ln (5e). [(c) Edexcel Limited 2002]
dy 1
7 Differentiate with respect to x 18 Show that for the function y = 2 x + 3 +
dx 2x + 3
a y = 2x b y = ax c y = log10 x dy 2 ( x + 1)
is given by =
dx ( 2 x + 3 )3
8a Find the equation of the normal to the curve y = 1 + sin2 x
at the point where x = p
4 19 Find the turning points on the curve y = (x2 - 9)3 and determine
b Find the equation of the normal to the curve y = tan2 x whether each of them is a maximum or minimum.

118 where x = p and the point where the normal cuts the x-axis. 119
4

Exit
5 Numerical methods
use graphical methods to solve equations of the type f(x) = g(x)
use graphical methods to solve equations of the type f(x) = 0
Summary Refer to
find non-integer roots of the equation f(x) = 0
The derivatives of these functions f(x) are: 4.1
use iterative methods to solve equations
f(x) f¢(x) distinguish between a sequence of convergent and divergent iterations
sin x cos x represent iterative methods graphically.
cos x -sin x
tan x sec2 x
ex ex 4.2
1 Before you start
ln x 4.3
x
dy dv du
The product rule If y = uv, then =u +v 4.4 1 Rearrange an algebraic expression. 1 Find f(x) such that f(x) = 0 when
dx dx dx
e.g. x = 5x + 7
2
du dv a x= x3 + 1 b x= 3 x2 − 1
dy v dx − u dx 3x + 4
2x 2x
The quotient rule If y = u , then = 4.5 becomes 2x2 - 4x + 7 = 0
C3

C3
v dx v2
The chain rule If y = g(u) and u = f(x) so that y = g(f(x)), 4.6
2 Solve equations graphically. 2 a Plot the graph of y = x2 - 2x - 4
dy dy du for -2 x 4
then = × 4.7 e.g. Plot the graph of y = x3 - 4x2 + 3x
dx du dx Hence solve the equation x2 - 2x - 4 = 0
Hence find that the solutions of
d(sin ax) = a cos ax d(cos ax) = −a sin ax x3 - 4x2 + 3x = 0 are x = 0, 1 or 3.
b Plot the graph of y = x − 2x + 1
2
In particular, dx
and dx
x−2
for -2 x 4 and find the point where
dy dy 1
If x = f(y), then you can find from = 4.7 the graph touches the x-axis.
dx dx dx
dy
3 Solve simultaneous equations graphically. 3 Plot the graphs of y = 4 - x2 and y = 1 for
x
e.g. Plot the graphs of y = x2 - x + 1 and y = 2x - 1 |x| 3. Explain how the graphs can be
Hence find that the solutions of x2 - 3x + 2 = 0 are used to solve the equation x3 - 4x + 1 = 0
Links
x = 1 and x = 2 and find the solutions.
Mathematics is used extensively in the financial world,
which relies on accurate forecasts of the future.
4 Use a spreadsheet to find the values of 4 Create a spreadsheet to show the values of
In calculus, a derivative shows how a dependent variable f(x) for a range of values of x. f(x) = x2 + 2x - 3 and g(x) = x + 2 for
is affected as an independent variable changes. e.g. For f(x) = x3 - x + 2 integer values of x where 0 x 7.
x x3 - x + 2
This can be applied in a financial setting to, for example, 0 2
State the value of the integer a such
predict how the change in price of an underlying asset 1 2 that the solution of x2 + x - 5 = 0
will affect the related market value. Techniques such as 2 8 lies between a and a + 1.
the chain rule are usually used to calculate the 3 26
mathematical derivatives involved.

120 121

5 Numerical methods

Intersection with the x-axis
5.1 Graphical methods
Another graphical method for solving an equation of the type
f(x) = g(x) is to define a new function h(x) = f(x) - g(x)
Many equations, including those derived in real-life scenarios and then find the solution of the equation h(x) = 0
in science, engineering, economics and elsewhere, cannot be
solved algebraically. You are now searching for the x-values of the points at which the
graph of y = h(x) crosses (or touches) the x-axis.
Instead, you have to use numerical methods to find ‘Solutions’ are also called ‘roots’.
approximate solutions.

EXAMPLE 2
Find the solution of the equation cos x = x for 0 < x < p This equation is the same as in
Intersection of two graphs •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Example 1.
One method for finding an approximate solution to an equation Let h(x) = cos x - x y
of the type f(x) = g(x) is to draw the graphs of y = f(x) and y = g(x)
Sketch the graph of h(x) = cos x - x:
and find the x-values of any points of intersection.
1 y = cos x – x
The sketch gives an approximate value of the root as x = 0.7
EXAMPLE 1

Find the solution of the equation cos x = x for 0 < x < p x
The x-value where y = cos x - x crosses the x-axis is the same x-value –2 –1 0 1 2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

as that at the point of intersection in Example 1. –1
Sketch the graphs of y = cos x and y = x on the same axes: y
y=x Use a spreadsheet to record values with increments in x of 0.01: –2
There is only one point of intersection so the equation
1 For x 0.73, h(x) > 0 x cos x - x
cos x = x has only one solution. y = cos x
0.70 0.065 > 0 y
–3 –2 –1 O x For x 0.74, h(x) < 0
C3

C3
The sketch gives a rough approximation of the solution as x = 0.7 1 2 3 0.71 0.048 > 0 0.065
–1 0.72 0.032 > 0
This is an approximation. It is not an accurate answer. The root is between 0.73 and 0.74. 0.73 0.015 > 0
y = cos x – x
You can find a more accurate value using The root is nearer to 0.74 because
0.74 -0.002 < 0
either a graphical calculator or graphical computer software 0.75 -0.018 < 0
x is in radians. -0.002 is closer to 0 than is 0.015. 0.76 -0.035 < 0
or an ordinary calculator or computer spreadsheet.
y Reduce the increments in x to 0.001: O 0.7 x
Tabulate values of x and cos x with increments in x of 0.01: 0.76
0.77 For x 0.739, h(x) > 0 x cos x - x
For x 0.73, x < cos x x cos x y = cos x 0.737 0.0035 > 0
For x 0.74, x > cos x 0.70 < 0.765 y=x For x 0.740, h(x) < 0 0.738 0.0018 > 0 –0.035
0.71 < 0.758
0.739 0.0001 > 0
The solution is between 0.73 and 0.74 0.72 < 0.752 The root is between 0.739
0.73 < 0.745 0.740 -0.0015 < 0 y
and 0.740
0.74 and 0.738 are closer together than 0.74 > 0.738 0.741 -0.0032 < 0
0.75 > 0.732 0.0035
0.73 and 0.745, so the solution is between The root is closer to 0.739
0.76 > 0.725 0.7
0.735 and 0.740. than 0.740 because 0.0001 is
O x closer to 0 than is -0.0015. y = cos x – x
0.7 0.76
Reduce the increment in x to 0.001:
y
x < cos x for x 0.739 x cos x To 3 significant figures, the solution of the equation O 0.737 0.741
x
0.740
0.736 < 0.7412 y=x cos x = x is x = 0.739
x > cos x for x 0.740 y = cos x
0.737 < 0.7405
Check:
The solution is between 0.739 and 0.738 < 0.7398
h(0.7395) = cos(0.7395) - 0.7395 = -0.0007 is negative, so the root
0.740 but it is closer to 0.739. 0.739 < 0.7391 –0.0032
must lie between x = 0.739 and x = 0.7395
0.740 > 0.7385
To 3 significant figures, the solution 0.735
of the equation cos x = x is x = 0.739 O 0.735 x
0.740
122 123

5 Numerical methods 5 Numerical methods

Sometimes it is best to combine both approaches as shown in Example 3. Be aware that f(x) might cross the x-axis more than once in the [a, b] means the interval
interval [a, b]. a x b
EXAMPLE 3

Solve the equation x2 = sin x for − p < x < p Also be aware of the following three possibilities:
2 2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
y y y
y
y = f(x) y = f(x)
Sketch the graphs of y = x2 and y = sin x: y = f(x)
y = x2
There are two roots. y = sin x
x = 0 is an obvious root. f(b) f(a)
f(a) f(a)
From the sketch, the other root is approximately x = 0.8 –3 –2 –1 O 1 2 3 x f(b) b
O a b x O a b x O a x
Let h(x) = sin x - x2
f(b)
The solution of the equation x2 = sin x
occurs when h(x) = sin x - x2 = 0

y f(a) and f(b) are both positive, f(a) and f(b) are both positive f(a) and f(b) have different signs.
Record values for x from 0.85 to 0.90 in a table:
0.03 but there are two roots between but there is a single root when However there is no root between
h(x) changes from positive to negative y = sin x – x2
0.02 x = a and x = b the curve touches the x-axis. x = a and x = b because f(x) has
x sin x - x2
between x = 0.87 and x = 0.88 0.01
a discontinuity.
0.85 0.0288 >0
The solution is between x = 0.87 0.86 0.0182 >0
Sketching a graph will help you to visualise what is happening.
O 0.85 0.9 x
and 0.88. 0.87 0.0074 >0 –0.01
0.88 -0.0037 <0 When you have found an interval [a, b] which contains a root,
C3

C3
–0.02
h(0.875) = 0.0019 > 0, so the root is 0.89 -0.0150 <0 you can then reduce the width of the interval to find the solution
between 0.875 and 0.88. 0.90 -0.0267 <0 to the required accuracy.
h(0.875) = sin(0.875) - 0.8752
The solutions are x = 0 and x = 0.88 to 2 significant figures. = 0.001918. . .

EXAMPLE 4
Find all the solutions (to 2 d.p.) of the equation e x + x - 7 = 0
y
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

The location of roots on the x-axis y = ex
The graph of y = e x + x - 7 cannot be sketched quickly. 8
The roots of the equation f(x) = 0 are the x-values of the points
6
where the graph y = f(x) intersects the x-axis. Rearrange the equation e x + x - 7 = 0 as e x = 7 - x
4 y=7–x
Let f(x) be continuous between x = a and x = b 'Continuous' means that there is no and sketch the graphs of y = e x and y = 7 - x:
break or discontinuity in the graph. 2
y y The graphs intersect only once so there is only one root.
O 2 4 6 8 x
y = f(x) y = f(x) The sketch gives its approximate value as x = 1.5

Consider f(x) = e x + x - 7 y
f(b) f(a)
positive positive 1
Take values for x between 1.4 and 1.8 with steps of 0.1:
y = ex + x – 7
O f(a) a b x O a b f(b) x
negative negative f(1.6) < 0 and f(1.7) > 0 x ex + x - 7
1.4 -1.545 < 0 O x
The root is in the interval [1.6, 1.7] 1.4 1.8
1.5 -1.018 < 0
The root is nearer to 1.7 because 0.174
If f(a) and f(b) have different signs for distinct values a and b, 1.6 -0.447 < 0
is nearer to 0 than is -0.447. –1
then there must be at least one root of f(x) = 0 in the interval 1.7 0.174 > 0
from a to b. Example 4 is continued on the next page. 1.8 0.850 > 0

124 125


EXAMPLE 4 (CONT.) 3 a Sketch the graphs of y = x 3 and y = 9 - x on the same axes.
Explore within this interval in steps of x = 0.01: y
b Explain why there is only one root of the equation
f(1.67) < 0 and f(1.68) > 0 0.046
x ex + x - 7 y = ex + x – 7 x3 + x - 9 = 0
1.65 -0.143 < 0
The root is in the interval [1.67, 1.68] c Use your graph to state the interval [a, b] within which the
1.66 -0.081 < 0
0.014
As f(1.675) = 0.014 which is positive, 1.67 -0.018 < 0 root of the equation x 3 + x - 9 = 0 lies, where a and b are
O x consecutive integers.
the root is in the interval [1.67, 1.675] 1.68 0.046 > 0 1.67 1.675 1.68
1.69 0.109 > 0 –0.018
So, to 2 decimal place, the root d Find the root of the equation x 3 + x - 9 = 0
is x = 1.67 correct to 1 decimal place.

4 a Sketch the graphs of y = x 2 and y = e-x on the same axes.

Exercise 5.1 b Show that the only root of the equation x 2 - e-x = 0
1 Sketch the graphs of each pair of functions f(x) and g(x). lies in the interval [0.70, 0.71].
State how many roots there are to the equation f(x) - g(x) = 0 c Find the root correct to 3 decimal places.
a f(x) = x 2 g(x) = x + 2
1
b f(x) = x 3 g(x) = x + 2 5 a Sketch the graphs y = x and y = 5 - x 2 on the same axes.

c f(x) = x
1
g(x) = x + 2 b How many roots has the equation 1 + x2 - 5 = 0?
x
1 c Show that one root lies in the interval [2, 3] and find its
C3

C3
d f(x) = x g(x) = x 3
value correct to 2 decimal places.
e f(x) = x 2 - 4 g(x) = 1
x 6 Without drawing any graph, show that each of these equations
f f(x) = sin x g(x) = x + 2 In parts f and g, has a root within the given interval.
x is in radians.
a x 3 - 5x 2 + 6x - 1 = 0 [0, 1]
g f(x) = sin x g(x) = 1 x
2
b x 3 - 2x 2 + x - 3 = 0 [2, 3]
h f(x) = e x g(x) = 4 - x2
c x3 - 1 - 6 = 0 [1, 2]
x
2 By sketching appropriate graphs, find how many roots there d ex - x - 8 = 0 [2, 3]
are for each of these equations.
e ln x - x 2 + 5 = 0 [2.4, 2.5]
a x3 + x - 5 = 0
f sin x + x - 5 = 0 [5.6, 5.7]
b x3 - 1 + 1 = 0
x
c e - x2 + 1 = 0
x 7 Show that the equation xsin x + 2 = 0 has a root a such
that 3 < a < 4. Find the value of a correct to 1 decimal place.
d sin x + x - 1 = 0
8 The equation e-x - x + 1 = 0 has a root in the interval [a, a + 1]
e e-x + x 2 - 4 = 0
where a is a positive integer.
f x2 + 1 - sin x = 0 Find the value of a and the value of the root correct to 2 decimal places.

126 127


9 a Copy and complete this table for the function 13 a By sketching the graphs of y = 2 x and y = x 3 on the
f(x) = x 3 - 4x 2 - x + 5 same axes, find how many solutions there are to the
equation 2 x − x 3 = 0
x -2 -1 0 1 2 3 4 5
f(x) b Find the non-zero solution of the equation correct to
2 decimal places.
b Sketch the curve y = x 3 - 4x 2 - x + 5 for -2 x 5
c Write down the three intervals [a, b] where a, b are 14 a Sketch the graphs of y = ln x and y = 12 - x 2 on the same axes.
consecutive integers within which roots of the equation Explain why the equation ln x + x 2 - 12 = 0 has only one root.
x3 - 4x2 - x + 5 = 0 lie. b Find the root of the equation correct to 2 decimal places.
d Find the largest root of - x3 4x 2 -x+5=0
correct to 2 decimal places. 15 a Choose two functions f(x) and g(x) and sketch the graphs
of y = f(x) and y = g(x) to find the number of roots of the
10 a Copy and complete this table for the function equation cos x - x + 6 = 0
1 b Show that a root a exists such that 6 a 7
f(x) = 2 x 3 - 2x 2 - x + 1
c Find a correct to 3 decimal places.
x -2 -1 0 1 2 3 4 5
f(x)
16 Solve the equation 3x - x 3 = 0
b Write down the three intervals [a, a + 1], where a is integer, giving solutions correct to 3 decimal places where necessary.
within which roots of the equation 1 x 3 − 2x 2 − x + 1 = 0 lie.
C3

C3
2 17 Find all the roots of the equation xcos x + x = 0
c Find the smallest root of x 3 - 4x 2 - x + 5 = 0 Sketch the graph of y = xcos x + x
correct to 2 decimal places.
INVESTIGATION
11 a Find the interval [a, a + 1], where a is a positive integer,
1
such that the only root of the equation e x + 2 x − 10 = 0 18
P
lies within it.
Q
b Find the root correct to 2 decimal places. i
O

12 a Show that the function f(q ) = sin q − q + 2 for 0 q p
has a solution in the interval ⎡ 4 p , 8 p ⎤.
3 7
⎢ ⎣ ⎥ ⎦
In this diagram, the area of triangle OPQ is half the area
b Find the solution correct to 2 decimal places. of the sector OPQ.
Show that the angle, q, in radians, must be a solution of
the equation
2sin q - q = 0
Solve this equation to find q.

128 129

5 Numerical methods

Whether a sequence converges or diverges depends on the
5.2 Iterative methods gradient of f(x) at the point of intersection where x = f(x)

The root of the equation x - f(x) = 0 is at the point of intersection There are four possibilities:
of the graphs of y = x and y = f(x) where x = f(x) y At the point of intersection, x = a
y
You can show the steps of the iterative process of finding a root of An iterative method is a is the root of x = f(x) in each case.
y = f(x) y=x
the equation x - f(x) = 0 repetitive process which uses a y=x f(x1)
succession of approximations.
Numerical process Graphical process Each approximation builds f(x0) y = f(x) f(x0)
on the preceding approximation f(x1)
Choose a value x0 close to the Locate the value x0 on the x-axis f(x2)
root a until the required degree of
Calculate f(x0) Rise vertically at x0 to meet accuracy is achieved.
O a x0 x1 x2 x
the curve at a height of f(x0) O a x3 x2 x1 x0 x
Let f(x0) = x1 Go horizontally to the line y = x
where f(x0) = x1
y = f(x) is a rising curve and y = f(x) is a rising curve
Let x1 be the next Locate x1 on the x-axis
0 < f ¢(a) < 1 and f ¢(a) > 1
approximation to the root
The iterations converge. The iterations diverge.
Perform the next iteration and Go vertically to the curve and
calculate x2 then go horizontally to the line
to find x2 y
y y = f(x)
Perform more iterations. Go vertically to the curve and y = f(x)
y=x y=x
go horizontally to the line for
f(x0)
each iteration.
C3

C3
f(x2)

f(x1)
EXAMPLE 1

Find the root of the equation 3 x + 1 − x = 0 using an iterative
method, starting with x0 = 1 as the first approximation. O x2 x0 a x1 x3 x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
O x0 x2 a x3 x1 x

Rearrange the equation: y = f(x) is a falling curve y = f(x) is a falling curve
3
x= x +1 and -1 < f ¢(a) < 0 and f ¢(a) < -1
Define the iterative formula: The iterations converge. The iterations diverge.

xn+1 = 3 xn + 1 with x0 = 1 Overall, the iterative process converges provided that This condition can also be
Carry out the iterations: -1 < f ¢(a) < 1 where a is the root. written as |f¢(a)| < 1
x1 = 3 x0 + 1 = 3 1 + 1 = 3 2 = 1.259 921. . .
Try to choose a starting value x0 which is close to the root. A different iterative formula with
x2 = 3 x1 + 1 = 3 1.259921 + 1 = 1.312 293. . . the same starting value can lead
Start by looking for an interval which contains the root and then
The iterations seem to be converging. to a different root.
x3 = 3 x2 + 1 = 3 1.312293 + 1 = 1.322 353. . . choosing x0 from within that interval.
The same iterative formula with a
different starting value can lead
x4 = 3 x3 + 1 = 3 1.322353 + 1 = 1.324 268. . . and so on. n x 3
x+1 to a different root or even a
0 1 1.25992105 diverging sequence.
After four iterations, the root is correct to 1 1.25992105 1.31229384
2 decimal places and has a value of 1.32 2 1.31229384 1.32235382
3 1.32235382 1.32426874
More iterations will improve the accuracy.
4 1.32426874 1.32463263
You can do these calculations efficiently on a spreadsheet or 5 1.32463263 1.32470175
using the ‘Ans’ key on a scientific or graphical calculator. 6 1.32470175 1.32471488
7 1.32471488 1.32471737
After 8 iterations, the root is correct to 5 decimal places.
8 1.32471737 1.32471785
130 131


EXAMPLE 2 Exercise 5.2
Explore the roots of the equation x3 - 5x - 3 = 0 1 a Show, without drawing a graph, that the equation
using iterative methods. x2 - 5x + 2 = 0
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

has a root in the interval [4, 5].
First method
Rearrange x3 - 5x - 3 = 0 into the form x = f (x): b Show that x2 - 5x + 2 = 0 can be rewritten as x = 5 − 2
x
x3 = 5x + 3
c Using the iterative formula xn +1 = 5 − 2 and x0 = 4
x = 3 5x + 3 xn
2
Define the iterative formula: xn +1 = 3 5xn + 3 find a root of the equation x - 5x + 2 = 0
Let x0 = 2: Let x0 = -2:
x1 = 3 13 = 2.351 334 x1 = 3 −7 = −1.912 931 2 a Show that the equation x3 - 4x - 5 = 0
has a root in the interval [2, 3].
x2 = 2.452 803… x2 = -1.872 423…
x3 = 2.480 597… x3 = -1.852 964… b Show that the equation x3 - 4x - 5 = 0
x4 = 2.488 102… x4 = -1.843 470… can be rearranged as x = 3 4x + 5
x5 = 2.490 121… x5 = -1.838 803… c Use the iterative formula xn +1 = 3 4xn + 5 with x0 = 2
x6 = 2.490 664… x6 = -1.836 499… to find a root of the equation x3 - 4x - 5 = 0
x7 = -1.835 360…
d Show that the equation x3 - 4x - 5 = 0
The sequence is converging. The sequence is
C3

C3
xn − 5
3
The root is x = 2.49 converging on a can also produce the iterative formula xn +1 = 4
.
to 2 decimal places. different root. The root Take x0 = 2 and find whether this iterative formula
is x = -1.84
creates a converging or diverging sequence of
to 2 decimal places.
approximations to the root.
Second method
Rearrange x3 - 5x - 3 = 0 into the form x = f (x):
3 a Show that the equation x3 - x2 - 1 = 0
x = x −3
3
5x = x3 - 3 giving
5 has a root in the interval [1, 2].
xn − 3
3
Define the iterative formula: xn+1 = b Show that the equation x3 - x2 - 1 = 0
5
can produce the iterative formula xn +1 = 3 xn + 1
2
Let x0 = 2: Let x0 = 3:
Take x0 = 1 and find this root correct to 3 decimal places.
x1 = 3 − 3 = 4.8
3
x1 = 2 − 3 = 1
3

5 5
4 Show that the iterative formula xn +1 = 1 e n
−x

x2 = 4.8 − 3 = 21.51
3
x2 = 1 − 3 = −0.4
3
2
5 5 can be derived from the equation e-x - 2x = 0
x3 = -0.6128 x3 = 1992.18… Taking x0 = 1 find a root of this equation to 2 decimal places.
x4 = -0.64 602… This sequence is
x5 = -0.65 392… diverging very fast. 5 To find the value of 4 50, show that the equation x4 = 50 can be
No root is found.
x6 = -0.65 593… rearranged to give the iterative formula xn +1 = 3 50
xn
x7 = -0.65 644… 4
Use this formula with x0 = 3 to calculate 50
This series is converging on x3
The equation - 5x - 3 = 0 correct to 4 significant figures.
a third root which is is a cubic. This example shows
that it has three roots x = 2.49,
x = -0.66 to 2 decimal places.
-1.84 and -0.66 (to 2 d.p.).
132 133


6 a Find the values of l and m such that the iterative formula 11 a Show that the equation 2x - e x + 3 = 0 has a root between
-1 and -2.
xn +1 = l xn + m e xn + xn
xn b Show that the iterative formula xn +1 = −1
3
can be used to solve the equation x 3 - 5x2 - 7 = 0 can be used to solve this equation.

b Show that a root of the equation lies between x = 5 and x = 6 c Take x0 = -2 and find the root of 2x - e x + 3 = 0
c Take x0 = 5 and find the root correct to 3 significant figures.
d Hence find a root of the equation 2tan q - etanq + 3 = 0
7 a Evaluate f(1.1) and f(1.4) for the function f(q) = 6q - 5sin q - 3 in degrees correct to 1 decimal place.
where q is in radians.
12 a Show that the equation x 3 - 4x - 3 = 0 has a root in the
Explain why there is a root of the equation 6q - 5sin q - 3 = 0
interval [2, 3].
in the range 1.1 < q < 1.4
3(xn + xn + 1)
3
b Find the values a, b and c such that the iterative formula b Show that xn +1 = is an iterative formula for the
4xn − 1
2

q n +1 = a sin q n + 1 equation x 3 - 4x - 3 = 0
b c
c Take x0 = 3 and find a root of the equation correct to 3 decimal places.
can be used to solve the equation 6q - 5sin q - 3 = 0
d Hence, find a root of the equation 8y - 2y+2 = 3
c Taking q0 = 1.1 find a root of equation 6q - 5sin q - 3 = 0
correct to1 decimal place.
C3

C3
3
8 The iterative formula xn +1 = 6 − x is used to find the root INVESTIGATION
n
of the equation f(x) = 0 13 Create several of your own iterative formulae to solve the
Find the function f(x) and the root of the equation correct to cubic equation
3 decimal places, given that x0 = 2 ax3 + bx2 + cx + d = 0
for your choice of a, b, c and d.
9 Find the equation f(x) = 0 which each of these iterative formulae
can be used to solve. Use a computer spreadsheet to calculate the iterations
− 2 for different starting values x0.
a xn+1 = 5xn 2 7 b xn+1 = 3 5xn + 1 c xn+1 = 8 − x
xn See if your iterations converge or diverge.
2 Which one of your formulae converges most quickly to a root?
e xn+1 = 4x2 − 1
3 −x n
50 − xn
d xn+1 = 2 2
n f xn+1 = 6 − 2e
xn 3xn + 2 xn

10 a Find the constants a and b which would allow these two
iterative formulae to be used to solve the equation x4 = 20
2
xn +1 = a + xn xn +1 = b
2 2
and 3
xn
xn

b Taking x0 = 2 find which one of these two iterative formulae
converges faster to the root of the equation.

c Find the value of 4 20 correct to 5 decimal places.

134 135

5 Numerical methods

Review 5 6 f(x) = x 3 − 2 − 1 , x ≠ 0
x
a Show that the equation f(x) = 0 has a root between 1 and 2.
1 Sketch the graphs of each pair of functions f(x) and g(x). An approximation for this root is found using the iteration formula
State how many roots there are to the equation f(x) - g(x) = 0 1
= ⎛2 + 1 ⎞
3
xn +1 ⎜ with x0 = 1.5
a f(x) = x 2 g(x) = x + 2 b f(x) = e x g(x) = 4 - x ⎝ xn ⎟
⎠
b By calculating the values of x1, x2, x3 and x4, find an
2 By sketching appropriate graphs, find how many roots there
approximation to this root, giving your answer to 3 decimal places.
are for each of these equations.
a x3 + 2 − 1 = 0 b sin x - x 3 = 1 In part b, x is in radians. c By considering the change of sign of f(x) in a suitable
x interval, verify that your answer to part b is correct to
c ln x - x2 + 2 = 0 d |x + 2| - x 2 = 1 3 decimal places. [(c) Edexcel Limited 2004]

3 a Sketch the graphs of y = 3 - x 2 and y = e-x on the same axes. 7a Sketch, on the same set of axes, the graphs of
y = 2 - e-x and y = x
b Explain why there are only two roots of the equation
x 2 + e-x = 3 [It is not necessary to find the coordinates of any points of
Show that one root lies in the interval [1.6, 1.7] and find intersection with the axes.]
this root correct to 2 decimal places.
Given that f(x) = e −x + x − 2, x 0
c Find the other root correct to 2 decimal places.
b explain how your graphs show that the equation f(x) = 0
C3

C3
has only one solution
4 Without drawing any graph, show that each of these
equations has a root within the given interval. In each case, c show that the solution of f(x) = 0 lies between x = 3 and x = 4
find the root correct to 2 decimal places.
The iterative formula xn+1 = ( 2 − e −xn ) is used to solve the
2

a x − 12 − 2 = 0
3
[1.3, 1.4] b ln x - e-x =0 [1.0, 1.5]
x equation f(x) = 0
c cos x + 2x - 4 = 0 [2.3, 2.4] d e x + x2 - 8 = 0 [-3, -2] d Taking x0 = 4, write down the values of x1, x2, x3 and x4,
and hence find an approximation to the solution of
5 a Show that the equation x 3 - 3x - 4 = 0 has a root in the f(x) = 0, giving your answer to 3 decimal places. [(c) Edexcel Limited 2003]
interval [2, 3].
8 f(x) = x 3 + x 2 - 4x - 1
b Show that the equation x 3 - 3x - 4 = 0 can be rearranged
The equation f(x) = 0 has only one positive root, a.
as x = 3 3x + 4
c Use the iterative formula xn +1 = 3 3xn + 4 with x0 = 2 to find the a Show that f(x) = 0 can be rearranged as x = ( 4xx ++11), x ¹ -1
root of the equation x3 - 3x - 4 = 0 correct to 2 decimal places. ⎛ 4x + 1⎞
The iterative formula xn+1 = ⎜ n ⎟ is used to
d Show that the equation x3 - 3x - 4 = 0 can also produce the ⎝ xn + 1 ⎠
x3 − 4 find an approximation to a.
iterative formula xn +1 = n . Take x0 = 2 and find whether
3
b Taking x1 = 1, find, to 2 decimal places, the values
this iterative formula creates a converging or diverging
of x2, x3 and x4.
sequence of approximations to the root.
c By choosing values of x in a suitable interval, prove that
a = 1.70, correct to 2 decimal places.
d Write down a value of x1 for which the iteration formula
æ 4x + 1 ö
xn+1 = ç n ÷ does not produce a valid value for x2.
è xn + 1 ø
136 Justify your answer. [(c) Edexcel Limited 2004] 137

Revision 2

5Exit
1 Express as a single fraction in its simplest form.
3x 2 − x − 2 12
(3x − 1)(x + 2) x − 2x − 8

Simplify 3x − x − 14
2

Summary 2a
Refer to 2
x −4
Graphical methods
Hence, express 3x − x − 14 +
2
To solve an equation of the type f(x) = g(x), draw the graphs b 2 as simply as possible.
of y = f(x) and y = g(x) and find the x-values of any
2
x −4 x(x − 2)
points of intersection.
To solve an equation of the type f(x) = 0, draw the graph of 3a Express as a fraction in its simplest form
y = f(x) and find the x-values of the points at 2 + 1
x − 3 x 2 − 8x + 15
which the graph intersects (or touches) the x-axis.
If, between x = a and x = b, the graph of y = f(x) is continuous 2 + 1
b Hence, solve the equation =1
and f(x) changes its sign, then there is at least one root of the x − 3 x − 8x + 15
2

equation f(x) = 0 between x = a and x = b. 5.1

4 Given that 2x − 32 + x + 1 º (ax2 + bx + c) + dx + e ,
4
Iterative methods x2
(x − 1) 2
(x − 1)
To solve an equation of the form g(x) = 0
C3

C3
find the values of the constants a, b, c, d and e. [(c) Edexcel Limited 2008]
rearrange g(x) into the form x = f(x)
choose a value x0 which is close to the root 1
5a Express 2 - as a single fraction.
use the iterative formula xn+1 = f(xn) to generate the sequence x−4
x0, x1, x2, x3, …
1
decide if this sequence of x-values is converging or diverging b The function f is defined by f(x) = 2 - , x Î R, x ¹ 4
x−4
if it is converging, continue until you have the root to the Find an expression for the inverse function f -1(x).
required accuracy. 5.2
c Write down the domain of f -1.

6 The functions f and g are defined by
Links
f: x® 2x + ln 2, x Î R
In cases where a real-life problem cannot be solved
g: x ® e2x, xÎR
analytically, a numerical method is applied to find
an approximate soluton. a Prove that the composite function gf is gf: x ® 4e4x, x Î R
Applied problems can arise from diverse areas such b Sketch the curve with equation y = gf(x), and show the coordinates
as engineering, economics and biological sciences. of the point where the curve cuts the y-axis.
Solutions to such problems often require scientific c Write down the range of gf.
computation involving advanced iterative methods.
d Find the value of x for which d [gf(x)] = 3,
dx
giving your answer to 3 significant figures. [(c) Edexcel Limited 2006]

138 139


7 The function f is defined by 13 a Using sin2 q + cos2 q º 1, show that cosec2 q - cot2 q º 1
f: x ® |2x - a|, x Î R, where a is a positive constant.
b Hence, or otherwise, prove that cosec4 q - cot4 q º cosec2 q + cot2 q
a Sketch the graph of y = f(x), showing the coordinates of
the points where the graph cuts the axes. c Solve, for 90° < q < 180°, cosec4 q - cot4 q = 2 - cot q [(c) Edexcel Limited 2006]

b On a separate diagram, sketch the graph of y = f(2x), showing 14 a Show that
the coordinates of the points where the graph cuts the axes.
c
1
Given that a solution of the equation f(x) = 2 x is x = 4,
i cos 2x
cos x + sin x 4 ( )
º cos x - sin x, x ¹ n − 1 p, n Î Z
1
find the two possible values of a. [(c) Edexcel Limited 2002] ii 2 (cos 2x - sin 2x) º cos2 x - cos x sin x - 1
2
8 The functions f and g are defined by b Hence, or otherwise, show that the equation
1
f: x ® ln(2x - 1), x Î R, x>2 g: x ® 2 , x Î R, x ¹ 3 cos q æ cos 2q ö 1
x−3 ç ÷=
è cosq + sin q ø 2
a Find the exact value of fg(4).
can be written as sin 2q = cos 2q
b Find the inverse function f -1(x), stating its domain.
c Solve, for 0 q 2p, sin 2q = cos 2q
c Sketch the graph of y = |g(x)|. Indicate clearly the equation giving your answers in terms of p. [(c) Edexcel Limited 2006]
of the vertical asymptote and the coordinates of the point
at which the graph crosses the y-axis. 15 f(x) = 5cos x + 12sin x
2 Given that f(x) = Rcos(x - a), where R > 0 and 0 < a < p ,
d Find the exact values of x for which x − 3 = 3 [(c) Edexcel Limited 2008] 2
a find the value of R and the value of a to 3 decimal places
C3

C3
9 The functions f and g are defined by b hence, solve the equation 5cos x + 12sin x = 6 for 0 x 2p
f: x ® | x - a | + a, x Î R, g: x ® 4x + a, x Î R c i Write down the maximum value of 5cos x + 12sin x.
where a is a positive constant. ii Find the smallest positive value of x for which this
a On the same diagram, sketch the graphs of f and g, showing clearly maximum value occurs. [(c) Edexcel Limited 2008]
the coordinates of any points at which your graphs meet the axes.
16 a The graph of y = e x is transformed into the graph of each
b Use algebra to find, in terms of a, the coordinates of the point of the following equations.
at which the graphs of f and g intersect. Name the transformations involved in each case and give the order
c Find an expression for fg(x). in which they occur. Sketch the graph of y = e x and each of its images.
d Solve, for x in terms of a, the equation fg(x) = 3a [(c) Edexcel Limited 2003] i y = 1 + 2ex ii y = 2 + e -x iii y = 3e x-2
b Name the transformations (and the order in which they occur)
10 Prove that 1 − tan 2 x º cos 2x. Hence, prove that tan2 p = 7 - 4 3
2
that transform the graph of y = ln x into the graphs of each of these
1 + tan x 12
equations. Sketch the graph of y = ln x and each of its images.
1
11 a By writing sin 3q as sin(2q + q), show that sin 3q = 3sin q - 4sin3 q i y = 3 - ln x ii y = 1 + 2ln x iii y = 2 ln (x + 2)

b Given that sin q = 3 , find the exact value of sin 3q. [(c) Edexcel Limited 2007] 17 a A population P of individuals increases over a time t (days) from
4 t
an initial value of P0, according to the relation P = P0 e10
12 a Given that 2sin(q + 30)° = cos (q + 60)°, find the exact value of tan q °. How many days have to elapse for the population to double in size?
b i Using the identity cos (A + B) º cos Acos B - sin Asin B, b A number of cells are being infected with a virus. The number N
prove that cos 2A º 1 - 2 sin2 A of uninfected cells reduces with time t (hours) as given by
t
ii Hence solve, for 0 x 2p, cos 2x = sin x, N = 80 + 25e 2
−

giving your answers in terms of p. i How many uninfected cells were there initially?
iii Show that sin 2y tan y + cos 2y º 1 for 0 y < 1 p [(c) Edexcel Limited 2005]
ii What is the limiting value of N as time increases?
2
140 141


18 A savings account earns interest on the money invested in it 22 Differentiate with respect to x.
at a constant rate of 5% each year. An initial investment of ex
a e x sin x b x3 ln x c e -x (x 2 - 3) d
£1 thus has a value of £y after t years where y = 1.05t sin x
a Sketch the graph of y = 1.05t for t 0 ln x x3 − 1 3
e f g ln(tan x) h ex
x2 − 1 x3 + 1
b Find the total value of an initial investment of £500 after it has been
in the account for 6 years. Give your answer to the nearest £. 3
x 2 − 2x + 5
3 ex
i j k l cos2 x sin x
x3 − 1 e −1
x
c How many years does it take for any investment to double in value?
d What must be the interest rate (to 1 decimal place) if an 23 a Differentiate with respect to x.
investment is to double in value after 10 years? cos ( 2x 3 )
i x 2e 3x + 2 ii
3x
19 A particular species of orchid is being studied. The population dy
p at time t years after the study started is assumed to be b Given that x = 4sin (2y + 6), find in terms of x. [(c) Edexcel Limited 2006]
dx
0.2t
p = 2800ae0.2t where a is a constant.
1 + ae 24 a Differentiate with respect to x
Given that there were 300 orchids when the study started,
i e 3x(sin x + 2cos x) ii x3 ln(5x + 2)
a show that a = 0.12
Given that y = 3x + 6x 2− 7 ,
2
x ¹ -1,
b use the equation with a = 0.12 to predict the number of (x + 1)
years before the population of orchids reaches 1850. dy 20
b show that =
C3

C3
dx (x + 1)3
336
c Show that p = −
0.12 + e 0.2t d2y d2y 15
c Hence find 2 and the real values of x for which =− [(c) Edexcel Limited 2008]
d Hence show that the population cannot exceed 2800. [(c) Edexcel Limited 2005] dx dx 2 4

25 a The curve C has equation y = x
20 A heated metal ball is dropped into a liquid. As the ball cools,
9 + x2
its temperature, T °C, t minutes after it enters the liquid, is Use calculus to find the coordinates of the turning points of C.
given by
3
dy
T = 400 e -0.05t + 25, t 0 b Given that y = (1 + e2x)2, find the value of at x = 1 ln 3 [(c) Edexcel Limited 2007]
dx 2
a Find the temperature of the ball as it enters the liquid.
26 Find the stationary points on these curves and determine their nature.
b Find the value of t for which T = 300, giving your answer x
to 3 significant figures. a y=e b y = xe x
x
c Find the rate at which the temperature of the ball is y= x
c d y = e-2x sin x where -p x p
decreasing at the instant when t = 50. Give your answer x +1
2

in °C per minute to 3 significant figures.
27 A curve C has equation y = x2e x
d From the equation for temperature T in terms of t, given dy
above, explain why the temperature of the ball can never a Find , using the product rule for differentiation.
dx
fall to 20 °C. [(c) Edexcel Limited 2006] b Hence, find the coordinates of the turning points of C.
d2y
21 Solve the equations c Find
dx 2
a 6 + 3e-x = 8 b ln (x + 3)2 = 4 c e2x = 3ex - 2
d Determine the nature of each turning point of the curve C. [(c) Edexcel Limited 2008]

142 143


x and explain why
28 a Find dy for the curve C where y = 33 a Show that the equation 1 x4 - x - 3 = 0
dx x2 − 1 10
the graph of C is a falling curve for all values of x. has a root in the interval [2, 3].
b Find the equation of the tangent to C at the point where b Show that the equation 1 x4 - x - 3 = 0
x = 2. Also find the equation of the normal to C at the 10
point where x = -2. can produce the iterative formula xn + 1 = 4 10 xn + 30

c Find, to 2 decimal places, the coordinates of the point c Take x0 = 1 and find the root correct to 3 decimal places.
where this tangent and normal intersect.
34 Find the equation in the form f(x) = 0 which each of these
29 A curve C has equation y = e2x tan x, x ¹ (2n+1)p iterative formulae can be used to solve.
2
3 − xn
a Show that the turning points on C occur where tan x = -1 a xn +1 = b xn +1 = 3 5(xn - 2) c xn +1 = 3 + 2
2
xn xn
b Find the equation of the tangent to C at the point where x = 0 [(c) Edexcel Limited 2008]

x
30 a Given that y = loga x, x > 0, where a is a positive constant, 35 a Show that the iterative formula xn + 1 = a − n can be
xn b
3 2
i express x in terms of a and y used to solve the equation 2x + x - 24 = 0
ii deduce that ln x = y ln a Find the values of a and b in this case.
b Show that dy = 1 b Show that a root of the equation lies between x = 2 and x = 3
dx x ln a
c Take x0 = 2 and find the root correct to 3 significant figures.
C3

C3
The curve C has equation y = log10 x, x > 0
The point A on C has x-coordinate 10.
Using the result in part b, 36 f(x) = x3 - 2 - 1 , x ¹ 0
x
c find an equation for the tangent to C at A. a Show that the equation f(x) = 0 has a root between 1 and 2.
d The tangent to C at A crosses the x-axis at the point B. b An approximation for this root is found using the
Find the exact x-coordinate of B. [(c) Edexcel Limited 2004] iteration formula
1

31 a Sketch the graphs of y = x3 and y = 6 - x2 on the same axes. xn+1 = ⎛ 2 + 1 ⎟
⎜
⎞3 with x0 = 1.5
⎝ xn ⎠
b Explain why there is only one root of the equation By calculating the values of x1, x2, x3 and x4, find an
x3 + x2 - 6 = 0 approximation to this root, giving your answer to
c Use your graph to state the interval [a, b] within which the 3 decimal places.
root of the equation x3 + x2 - 6 = 0 lines, where a and b are c By considering the change of sign of f(x) in a suitable
consecutive integers. interval, verify that your answer to part b is correct to
d Find the root of the equation x3 + x2 - 6 = 0 3 decimal places. [(c) Edexcel Limited 2004]
correct to 1 decimal place.

32 a Find the interval [a, a + 1], where a is a positive integer,
such that the only root of the equation e x - x2 - 2 = 0
lies within this interval.
b Find the root correct to 2 decimal places.

144 145

Revision 2

37 This diagram shows part of the curve C with equation y
y = f(x), where f(x) = 0.5e x - x2
The curve C cuts the y-axis at A and there is a minimum at
the point B.
a Find an equation of the tangent to C at A.
A
C

6 Partial fractions
The x-coordinate of B is approximately 2.15. A more exact O x This chapter will show you how to
estimate is to be made of this coordinate using iterations B separate a fraction with different linear factors in its denominator
xn+1 = ln g(xn) into partial fractions
separate a fraction with a repeated linear factor in its denominator
b Show that a possible form for g(x) is g(x) = 4x
into partial fractions
c Using xn+1 = ln 4xn, with x0 = 2.15, calculate x1, x2 and x3. separate an improper fraction into partial fractions
Give the value of x3 to 4 decimal places. [(c) Edexcel Limited 2002] use the methods of equating coefficients and substitution, including
the cover-up rule.
38 f(x) = 1 - 1 + ln x , x > 0 y
2x 2
This diagram shows part of the curve with equation y = f(x). Before you start
The curve crosses the x-axis at the points A and B, and has a
minimum at the point C. You should know how to: Check in:
1
1 1 Substitute into formulae. 1 Find the value of y and z when x = 4
a Show that the x-coordinate of C is 2 .
e.g. Find z = 3x − 1 when x = 3
a y = 22x + 1
2
(x − 1)
2 4 if

C4
C3

b Find the y-coordinate of C in the form kln 2, O A B x x (1 − x)
9 −1
where k is a constant. z= 4 = 5 × 16 = 20
b z = (2 − x)
C 2
2
( )
2 4 1
3 −1
c Verify that the x-coordinate of B lies between 4.905 and 4.915 4 x (1 + 3x)

d Show that the equation 1 -1 + ln x = 0 can be rearranged into
2x 2 2 Factorise expressions. 2 Factorise
⎛ 1 ⎞
⎜ 1− 2x ⎟ e.g. 6x 2+ 11x - 10 a 4x 3 - 9x
the form x = 2e⎝ n ⎠
= (2x + 5)(3x - 2)
b x4 - 1
The x-coordinate of B is to be found using the iterative formula
⎛ 1− 1 ⎞
⎜ 2x ⎟
xn+1 = 2e⎝ n ⎠
with x0 = 5 3 Create and use identities. 3 Find A and B if
e.g. Find A and B if 3x + 8 º A(x + 2) + Bx a x(5x + 3) + 6x º x(Ax + B)
e Calculate, to 4 decimal places, the values of x1, x2 and x3. [(c) Edexcel Limited 2005]
Equate constants: 8 = 2A so A = 4
b (Ax + 3)(2x + B) º 8x 2 + 10x + 3
Equate coefficients of x:
39 f(x) = 3ex - 1 ln x - 2, x > 0 3 = A + B so B = -1
2
a Differentiate to find f ¢(x). 4 Add and subtract algebraic fractions. 4 Work out
3 + 4
e.g. 2 + 4 2 = 2(x + 1) + 4 (x + 3)
2
b The curve with equation y = f(x) has a turning point a
x + 3 (x + 1) (x + 3)(x + 1)2 x + 1 (x + 1)2
at P. The x-coordinate of P is a. Show that a = 1 e-a 2x 2 + 8x + 14
6 = b 1+ 2 + 3
The iterative formula xn+1 = 1 e −xn , x0 = 1, is used to find an (x + 3)(x + 1)2 x x +1 x −1
6
approximate value for a.
c Calculate the values of x1, x2, x3 and x4, giving your 5 Divide algebraic expressions. 5 Divide
answers to 4 decimal places. e.g. (x 2 + 4x + 7) ¸ (x + 3) a (x 3 - 3x 2 - x + 3) by (x + 1)
4
= x +1+ b (x 2 + 6x - 1) by (x - 2)
d By considering the change of sign of f ¢(x) in a suitable x+3
interval, prove that a = 0.1443 correct to 4 decimal places. [(c) Edexcel Limited 2005]
146 147

6 Partial fractions

You can use the method of substitution to separate a fraction into
6.1 Separating fractions partial fractions.

EXAMPLE 2
You can express a proper fraction of the type The initial fraction must be a 5x − 1
Express in partial fractions. This is the same as Example 1,
f(x) proper fraction. That is, the (x − 2)(x + 1) using a different method.
(ax + b)(cx + d )(ex + f ) numerator must be of a lower ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

degree than the denominator.
as partial fractions of the type A + B + C Let 5x - 1 º A + B º A(x + 1) + B(x - 2)
ax + b cx + d ex + f (x - 2)(x + 1) x - 2 x + 1 (x - 2)(x + 1)
where A, B and C are constants.
Equate the numerators: 5x - 1 º A(x + 1) + B(x - 2)
In particular, let x = -1 to eliminate A: An identity is true for all
You can use the method of equating coefficients to separate a values of x.
-5 - 1 = 0 + B(-1 - 2) so B=2
fraction into partial fractions.
Now let x = 2 to eliminate B:
10 - 1 = 3A so A=3
EXAMPLE 1

Express 5x − 1 in partial fractions.
(x − 2)(x + 1) So, 5x − 1 ≡ 3 + 2
Check your answer by adding
(x − 2)(x + 1) x − 2 x + 1 3
and 2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
x−2 x −1
Form an identity (which is true for all values of x):
Let 5x - 1 º A + B
See C 3 for revision of
You can use either of these methods or a mixture of the two
(x - 2)(x + 1) x - 2 x + 1 adding fractions.
to keep your working to a minimum.
º A(x + 1) + B(x - 2)
C4

C4
(x - 2)(x + 1)

EXAMPLE 3
º (A + B)x + A - 2B x 2 − 11x − 6
Express as partial fractions
(x - 2)(x + 1) (x + 2)(x − 2)(2x − 1)
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Equate the numerators: 5x - 1 º (A + B)x + A - 2B
Let x 2 − 11x − 6 ≡ A + B + C A, B and C are numerical
The two sides of this identity must be identical. (x + 2)(x − 2)(2x − 1) x + 2 x − 2 2x − 1 constants. The Core 4
specification does not extend
Equate the coefficients of x: 5 = A + B You now have two simultaneous A (x − 2)(2x − 1) + B (x + 2)(2x − 1) + C (x + 2)(x − 2)
≡ to algebraic numerators.
Equate the constants: -1 = A - 2B equations in A and B. (x + 2)(x − 2)(2x − 1)
Consider the numerators.
Subtract these two equations to eliminate A:
5 - (-1) = B - (-2B) Let x = 2 to eliminate A and C: 4 - 22 - 6 = 0 + B ´ 4 ´ 3 + 0 x2 - 11x - 6 = 4 - 22 - 6
6 = 3B so -24 = 12B and B = -2
B=2
Let x = -2 to eliminate B and C: 4 + 22 - 6 = A ´ (-4) ´ (-5) + 0 + 0
Substitute into 5 = A + B: A=3 so 20 = 20A and A = 1
Check your answer by adding
So, in partial fractions, 5x − 1 ≡ 3 + 2
3 and 2 Equate coefficients of x2: 1 = 2A + 2B + C To find the coefficients,
(x − 2)(x + 1) x − 2 x + 1 x−2 x −1 expand the brackets:
Substitute the values of A and B: C=3 A(x - 2)(2x - 1) = A(2x2 - 5x + 2)
B(x + 2)(2x - 1) = B(2x2 - 2x - 2)
So, x 2 − 11x − 6 ≡ 1 − 2 + 3 C(x + 2)(x - 2) = C(x2 - 4)
(x + 2)(x − 2)(2x − 1) x + 2 x − 2 2x − 1

148 149

6 Partial fractions 6 Partial fractions

You can use the cover-up rule when a fraction has only linear Exercise 6.1
factors. It is a shortened form of the method of substitution. 1 Use the method of equating coefficients to express these in partial fractions.
a 4x + 5 b x + 25 c 2
(x + 2)(x + 1) (x − 3)(x + 4) (x − 3)(x + 5)
EXAMPLE 4

Use the cover-up rule to express 3x 2 + 16x − 10
(x − 1)(x + 2)(2x − 1) 4x − 3 7x + 2 2x − 5
in partial fractions. d e f
x(x − 1) (x + 5)(2x − 1) x 2 − 6x + 8
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Let 3x 2 + 16x − 10 ≡ A + B + C 2 Use the method of substitution to express these in partial fractions.
(x − 1)(x + 2)(2x − 1) x − 1 x + 2 2x − 1
4x − 7 5x + 11 x+4
a b c
To find A, cover up (x - 1) and substitute x = 1 in the rest of the fraction: 3x2 + 16x − 10 (x − 3)(x − 2) (x + 1)(x + 4) x(x + 1)
(x − 1) (x + 2) (2x − 1)
A = 3 + 16 − 10 = 9 = 3 x − 11 3x x2 − x + 1
(1 + 2) × (2 − 1) 3×1 d e f
x − 7x + 6
2 (x − 1)(x − 2)(x − 3) (x 2 − 1)(2 − x)
To find B, cover up (x + 2) and substitute x = -2 in the rest of the fraction: 3x2 + 16x − 10
(x − 1) ( ) (2x − 1) 3 Use the cover-up rule to express these in partial fractions.
B = 3 × 4 + 16 × (−2) − 10 = −30
= −2 3x + 7
(−2 − 1) × (−4 − 1) (−3) × (−5) a 9
b (x − 1)(x + 4) c 1
(x − 2)(x + 1) (x − 5)(x − 3)
To find C, cover up (2x - 1) and substitute x = 1 in the rest of the fraction: 3x2 + 16x − 10
2 (x − 1) (x + 2) ( ) d x−3 e 4−x f 1
x(x − 2) x(x + 1)(x + 2) 4x 3 − x

C= 4 ()
3 1 + 16 − 10
2 =
−1
1
4 = 5×4 =1
( ) ( ) ( ) () 4 Express these in partial fractions.
C4

C4
1 −1 × 1 + 2 −
1 × 5 4 5
2 2 2 2 2x 3 6x
a b c
x 2 − 6x + 8 (x − 3)(x 2 + x − 2) (2x − 1)(3x − 2)
Hence 3x 2 + 16x − 10 ≡ 3 − 2 + 1
(x − 1)(x + 2)(2x − 1) x − 1 x + 2 2x − 1
d 3 − x2 e 2x + 3 f 2x + 5
1− x 4x 3 − x 9 − 4x 2

x2 + 4 2x 6
EXAMPLE 5

x +1
g h i
Express as partial fractions 2x − x 2 − x 3 (1 + 2x)(4 − x 2) x 4 − 5x 2 + 4
8x − 2x − 3
2

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Factorise the denominator: 8x 2 - 2x - 3 º (2x + 1)(4x - 3)
INVESTIGATION
x +1 ≡ A + B 5x − 1
Let
(2x + 1)(4x − 3) 2x + 1 4x − 3 5 a Examples 1 and 2 showed that ≡ 3 + 2
(x − 2)(x + 1) x − 2 x + 1
1 +1 − Use a graphical calculator or a graphical computer package
To find A, let x = - 1 : A= 2 =−1 Cover up (2x + 1) and substitute x = − 1 to draw the three graphs of
2 −2 − 3 10 2

y= 5x − 1 , y = 3 y = 2
3 +1 (x − 2)(x + 1) x − 2 and x +1
To find B, let x = 3 : B= 4 = 7 Cover up (4x - 3) and substitute x = 3
4 3 + 1 10 4
See how the graphs of the two partial fractions add
2
together to give the graph of the original fraction.
x +1 ≡− 1 + 7 Pay particular attention to the asymptotes.
So
8x 2 − 2x − 3 10 (2x + 1) 10(4x − 3)
b Repeat this graphical investigation using a fraction which
has a denominator with three factors,
3x
such as (x − 1)(x − 2)(x − 3)
150 151

6 Partial fractions

EXAMPLE 2
6.2 More partial fractions
Express as partial fractions 2x − 2x + 14
2
2
(x + 4)(x − 2)
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

A proper fraction of the type f(x) , where f(x) is a (ax + b)2 is a
Let 2x − 2x + 14 ≡
2
A + B + C
(ax + b)2 (1)
repeated linear factor. 2
(x + 4)(x − 2) x + 4 x − 2 (x − 2)2
polynomial in x, will produce two partial fractions of the type
A + B Use the cover-up rule.
ax + b (ax + b)2 To find A, cover up (x + 4) and let x = -4:
where A and B are constants.
A = 2x − 2x + 14 = 2 × (− 4) − 2 × (2 4) + 14 = 54 = 3
2 2
−
So 2
(x − 2) (− 4 − 2) 36 2
A repeated factor (ax + b)3 will produce partial fractions
A B C To find C, cover up (x - 2) 2 and let x = 2:
+ +
ax + b (ax + b)2 (ax + b)3
C = 2x − 2x + 14 = 2 × 2 − 2 × 2 + 14 = 18 = 3
2 2
So
(x + 4) 2+4 6
EXAMPLE 1

Express as partial fractions 4 − 7x To find B
(x + 3)(x − 2)2 either
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

2x 2 − 2x + 14 ≡ A(x − 2) + B(x + 4)(x − 2) + C (x + 4)
2
4 − 7x ≡ A + B + C
Let (x + 4)(x − 2) 2
(x + 4)(x − 2)2
(x + 3)(x − 2)2 x + 3 x − 2 (x − 2)2
A(x − 2)2 + B(x + 3)(x − 2) + C (x + 3) Equate coefficients of x²:
C4

C4
≡ 2=A+B
(x + 3)(x − 2)2
so B = 2 − A = 2 − 3 = 1
Equate the numerators: 4 - 7x º A(x - 2)2 + B(x + 3)(x - 2) + C(x + 3) 2 2
Let x = 2: 4 - 14 = 0 + 0 + C ´ (2 + 3) or
C = -2 Let x = 0 and substitute in (1): Any x-value would do,
Let x = -3: 4 + 21 = A ´ (-5)2 + 0 + 0 14 = 3 + B + 3 but x = 0 is the easiest.
A=1 4 × 4 2 × 4 −2 4

No other choice of x-value reduces a bracket to zero. Multiply by 8:
7 = 3 - 4B + 6
There are now two ways forward:
B= 1
2
either or
So 2x 2 − 2x + 14 ≡ 3 + 1 + 3
Use A = 1, C = -2 and Equate the coefficients
(x + 4)(x − 2)2 2(x + 4) 2(x − 2) (x − 2)2
let x = 1 in the numerators: of x2 in the numerators:
4 - 7 = 1 ´ (-1)2 + B ´ 4 ´ (-1) + (-2) ´ 4 0=A+B
-3 = 1 - 4B - 8 B = -A A fraction with a numerator of degree higher than (or equal to) x3 + 1
B = -1 B = -1 is an example of an
the degree of the denominator is an improper fraction. x2 − 1
improper fraction.
You could choose any value of x. Equating coefficients is usually the most
x = 1 and x = 0 are generally simple to use. efficient method at this point.
Before you can separate an improper fraction into
partial fractions, you must first change it to a mixed fraction, A numerical equivalent is
So 4 − 7x ≡ 1 − 1 − 2 consisting of a quotient and a remainder, by changing 9 to 2 1 .
(x + 3)(x − 2)2 x + 3 x − 2 (x − 2)2 either algebraic long division
4 4

or rearranging the numerator and finding the quotient
and remainder by inspection.

152 153

6 Partial fractions 6 Partial fractions

EXAMPLE 3 Exercise 6.2
Express 2 x + 7 as partial fractions.
2 2 1 These fractions have repeated factors in the denominators.
x −x−6 Express the fractions as partial fractions.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••v•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Numerator and denominator are both of degree 2, so it is an improper a x +1 b 16 c x2 − 7
fraction. You must change it to a mixed fraction. (x − 1)(x − 2)2 (x + 1)(x − 3)2 (2x − 1)(x + 1)2

either or d x e x − 10 f x2 − 1
Use long division: Rearrange the numerator: (x − 4)2 x 2(x − 2) x 3 − 2x 2
2 2x 2 + 7
x 2 - x - 6 2x 2 ) +7 x -x-62
The numerator is now
g 1
x(3x − 1)2
h x−4
x(x + 2)3
º 2(x - x2 - 6) + 2x + 19
2
2x 2 - 2x - 12 2 ´ denominator
2x + 19 x -x-6 + compensating terms
2 Express these improper fractions as partial fractions.
2x 2 + 7 ≡ 2 + 2x + 19 º 2 + 2 x + 19
2 The quotient is 2 and
a x b x2 c x2 − 2
So x -x-6 the remainder is 2 x + 19
2 x+2 x −1
2 (x − 1)(x + 3)
x −x−6
2
x2 − x − 6 x − x −6

2x + 19 x2 + 1 x2 x3
Let 2 x + 19 º
2
Factorise the denominator.
d
x(x + 1)
e
x −1
f
x −12
x - x - 6 (x + 2)(x - 3)
º A + B
g x 2− 3x + 5
3 2
h 9 + x2 x3
2
x+2 x-3 i
x + x−2 9−x 4x 2 − 1
Use the cover-up rule.
A = -4 + 19 = 15 = -3
C4

C4
3 Express x + 2 2 in partial fractions of the form A + B + C
3
Cover up (x + 2) and let x = -2: + D
-3 -5 -2
x(x - 2) x x−2 (x − 2)2
Cover up (x - 3) and let x = 3: B = 6 + 19 = 25 = 5
3+2 5
4 Express in partial fractions.
So 2x 2 + 7 ≡ 2 − 3 + 5
x −x−6
2 x+2 x−3 a 4 b 6 c 2x 2 + 1
(2x − 3)(x + 1) (x − 2)(x + 1)2 (x + 1)(2x − 1)

d 2x 2 e x2 + 5 f 2x + 1
EXAMPLE 4

x2 − 1 x + 2x − 3
2 (2x − 1)(3x + 1)
Express as partial fractions x 2 − 2x − 1
3 2

x − 3x + 2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
g x3 + 2 h x3 i x3 + 1
x(x + 1) x −x−2
2
x 2(x − 1)
This is an improper fraction.

Use long division: j 3 k x3 + 1 l (x 2 + 1)2
x (x + 1)2
2
x (x + 1)2 x 2(x 2 − 1)
So x 2 − 2x − 1 ≡ x + 1 + x−3
3 2
x +1
x − 3x + 2 x − 3x + 2
2
)
x 2 − 3x + 2 x 3 − 2x 2 −1
x−3 x−3 x 2 − 3x 2 + 2x
Let ≡ ≡ A + B x 2 − 2x − 1
INVESTIGATION
x − 3x + 2
2 (x − 1)(x − 2) x − 1 x − 2
x 2 − 3x + 2 5 Explore the graph of y = f(x) if f(x) is
x−3
Use the cover-up rule. a proper fraction with repeated linear factors in the denominator
Cover up (x - 1) and let x = 1: A = 1−3 = 2 an improper fraction.
1−2
2 − 3 = −1 Consider some of the fractions in this exercise, draw their
Cover up (x - 2) and let x = 2: B =
2 −1 graphs, and take particular notice of any asymptotes, both
vertical and horizontal.
Hence x 2 − 2x − 1 ≡ x + 1 +
3 2
2 − 1
x − 3x + 2 x −1 x − 2
154 155

6 Partial fractions

9a Express f(x) = 3x + 2 in partial fractions.
Review 6 (x + 4)(x − 1)
b Find f ¢(x) and deduce that the graph of y = f(x) has a
1 Use the method of equating coefficients to express these in partial fractions. negative gradient at all points on the curve.

a x+3 b 8−x c 6
(x − 2)(x − 1) x(x + 4) x2 − 9 10 The function f is given by
3(x + 1)
f(x) = , x ∈ R, x ≠ −2, x ≠ 1
2 Use the method of substitution to express these in partial fractions. (x + 2)(x − 1)

a x +1 b x2 - x + 5 c x 2 − 2x − 2 a Express f(x) in partial fractions.
(x − 1)(x + 3) x (x - 1)(x - 5) (x 2 − 1)(x + 2)
b Hence, or otherwise, prove that f ¢(x) < 0 for all values of
3 Use the cover-up rule to express these in partial fractions x in the domain. [(c) Edexcel Limited 2003]

a 8 b x2 + 1 c x2 - x - 4 The fraction in part b
(x − 1)(x + 3) (x − 2)(x + 1) x (x + 1)(x - 2) is improper. 1 − 3x
11 a Write (x − 2)(x + 3) in partial fractions.

4 Express these in partial fractions. b Find the gradient at the point where x = 1 on the graph of
4x 2 1 − 3x
a b y=
(2x + 1)(2x − 1) (3x − 2)(3x − 1) (x − 2)(x + 3)

c Explain why the graph is a curve which is always rising.
c x +1 d x+6
(x − 2)(x − 1)2 x 2(x − 3)
C4

C4
1
12 a Express in partial fractions.
9 x + 4x − 1
2 r (r + 1)
e f
x(2x − 3)2 (x 2 − 1)(x − 1) 1 + 1 + 1 + + 1 = n
b Deduce that
1´ 2 2 ´ 3 3´ 4 n (n + 1) n + 1

5 Show that x 2 + 4 can be expressed as A +
2
B + C r =∞
x −4 x−2 x+2 c Find the value of ∑ r(r 1 1)
+
Find the values of A, B and C. r =1

1
6 Express these in partial fractions. 13 a Express in partial fractions.
(r + 1)(r + 2)
a 3x 2 − 3 b 2x 2 − 3x − 24
r =n
(x − 1)(x + 2) x2 − x − 6
b Hence show that ∑ (r + 1)(r + 2) = 1 − n + 2
1 1
r =0
c x3 − 2 d x 3 + 2x 2 − 4
r =n
x 2(x + 1) x2 − 4
c Show that, as n ® ¥, ∑ (r + 1)(r + 2) converges.
1
r =0
1 State the sum to infinity.
7 Show that (x + 1)(x − k) can be expressed as partial fractions in the form
1 1
( − 1 )
a x − k x + 1 . Find a in terms of k.

8a Show that, if f(x) = x 3 - 2x 2 - x + 2, then f(2) = 0
Hence, factorise f(x) completely.
1
b Express f(x) in partial fractions.
1
c The line x = a is a vertical asymptote to the curve y = f(x)
156 State all possible values of a. 157

6Exit
7Parametric equations
sketch curves using their parametric equations
convert parametric equations to Cartesian equations
Summary Refer to
f(x) find points of intersection of curves and lines using parametric equations
For a proper fraction of the type (ax + b)(cx + d)(ex + f ) differentiate parametric equations to find equations of tangents and
where the factors of the denominator are all different and f(x) is a stationary values
integrate parametric equations to find areas under curves.
polynomial in x, the partial fractions are of the type A + B + C
ax + b cx + d ex + f
where A, B and C are constants. 6.1
For a proper fraction which has a repeated linear factor (ax + b)2 Before you start
in its denominator, there will be two partial fractions of
A + B You should know how to: Check in:
the type where A and B are constants.
1 If m = 2(x + 1) and n = 3x − 2
6.2
ax + b (ax + b)2 1 Substitute into formulae.
4
You can find the constants A, B, C, ¼ in the partial fractions by using e.g. If a = 2x + 3 and b = 1 - 4x, find y when
the method of equating coefficients find y when y = a2 - b
a y = 1 m + 2n
the method of substitution Substitute for a and b: 4
C4

C4
The cover-up rule is a shorter version of the method of substitution. 6.2
y = (2x + 3)2 - (1 - 4x)
= 4x2 + 12x + 9 - 1 + 4x b y = 2m2 + 16n2
You must change an improper fraction to a mixed fraction, consisting = 4x2 + 16x + 8
of a quotient and a remainder, before you can create partial fractions by
either algebraic long division 2 Solve simultaneous equations. 2 Solve these simultaneous equations.
or rearranging the numerator and finding the quotient and
e.g. Solve y = x + 1 and y + 5 = x2 a 2x + 3y = 1, y + 3x = 5
remainder by inspection. 6.2
Eliminate y: (x + 1) + 5 = x2
x2 - x - 6 = 0 b x2 + y2 = 3, x + 2y = 1
(x + 2)(x - 3) = 0
Links So, x = -2 or 3 and y = -1 or 4
Partial fractions allow you to express complicated The solutions are (-2, -1) and (3, 4).
fractions as the sum of simpler fractions. dy
These simplified expressions can be applied to find 3 Differentiate and integrate functions. 3 Find dx and y dx when
antiderivatives as well as inverses of transforms, dy 2
e.g. Find and y dx if y = ⎛1 + 12 ⎞
⎜ ⎟
such as the Laplace transform. dx ⎝ x ⎠ a y = x 2 + 1 + 12
x
Expand the brackets:
( )
Engineers use Laplace Transforms to simplify
problems by converting relationships that are y = 1 + 22 + 14 = 1 + 2x −2 + x − 4 b y = x(1 + x) 1 + 1
x x x
dependent on time t to a set of equations expressed
c y = x +1
2
dy −
in terms of the Laplace operator s. They can then Hence = 2(−2) x −3 + (− 4) x 5 2
dx x
use Inverse Laplace Transforms to return to the
= − 43 − 45
time domain. x x
−1
x −3
Laplace Transforms are particularly useful in and y dx = x + 2 x + +c
−1 −3
analyzing electronic circuits.
= x − − 13 + c
2
x 3x

158 159

7 Parametric equations

EXAMPLE 2
7.1 Parametric equations and curve sketching Sketch the curve given parametrically by the equations q is the parameter.
x = sin q, y = sin 2q for 0 q 2p
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

A Cartesian equation has the form y = f(x) e.g. y = x2 + 2x + 1
Construct a table of values for 0 q 2p:

Some relationships between x and y involve a third variable. p p 3p 5p 3p 7p
q 0 p 2p q is in radians.
This third variable is called a parameter. 4 2 4 4 2 4
x 0 0.707 1 0.707 0 -0.707 -1 -0.707 0
The equations x = f(t), y = g(t) are called parametric equations. t is the parameter. y 0 1 0 -1 0 1 0 -1 0

y
You can sketch a curve described by parametric equations by
finding points on the graph for a range of values of t.
Each point on the graph has a value of t associated with it. 5p p
i= i=
4 4
1
3p
i= i = 0, p , 2p
EXAMPLE 1

Sketch the graph of the curve with the parametric equations 2 As q increases from 0, the
–1 O 1 i=p x
curve traces the two
x = 2t - 1, y = t2 for -4 t 4 2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
loops of a figure-of-eight.
–1
7p 3p
Construct a table of values: i= i=
4 4
t -4 -3 -2 -1 0 1 2 3 4
x -9 -7 -5 -3 -1 1 3 5 7
C4

C4
When a curve is expressed using parametric equations, you can
y 16 9 4 1 0 1 4 9 16
find the Cartesian equation by eliminating the parameter t (or q).
y

EXAMPLE 3
t = –4 t=4 Find the Cartesian equation of the curves which have these
16
parametric equations.
It is useful to label each point a x = 2t - 1, y = 8t2 + 3 b x = 2sin q + 3, y = 2cos q – 5
with its value of t. You can ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

then see how the curve
a Substitute t from x = 2t - 1 into the equation for y:
takes shape as t varies.
( )
2
From x = 2t - 1, t = x + 1 , so y = 8 x + 1 − 1
2 2
t=3 = 2(x + 1)2 - 1
t = –3
= 2x2 + 4x + 1 2x2 + 4x + 1 is a quadratic
expression, which indicates
The Cartesian equation is y = 2x2 + 4x + 1 that the curve is a parabola.
b Find sin q and cos q in terms of x and y:
t = –2 t=2 From x = 2sin q + 3, sin q = x − 3
2
y+5
From y = 2cos q - 5, cos q =
2
t = –1 t=1
t=0 Substitute into sin2 q + cos2 q = 1:

( x 2 3) + ( y 2 5)
x 2 2
–9 –1 O 7 − +
=1
This equation represents a circle,
The curve is a parabola. (x - 3)2 + (y + 5)2 = 4 centre (3, -5) and radius 4 = 2.
See C2 for revision.
The Cartesian equation is (x - 3)2 + (y + 5)2 = 4
160 161

7 Parametric equations 7 Parametric equations

You can also find parametric equations of a curve represented 4 Find the Cartesian equation for each of the curves given
by a Cartesian equation. parametrically by these equations.
a x=t+4 y = 1 - 2t b x=3 y = 4t
t
EXAMPLE 4

Find parametric equations for the curve with the Cartesian c x=t+1 y = t2 - 2 d x = t2 y = t3
equation y = 6x 1 − x 2 e x = t2 - 1 y = t3 + 2 f x = t2 y= 2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• •••••••••••••••••••••••••••
t
•
g x= 1−t y= 1+t h x = 3cos q y = 4sin q
t t
You need to find a parameter which will simplify 1 − x 2
i x = sin q y = cos 2q j x = 3cos q y = 5cos 2q
Recall that 1 - sin2 q = cos2 q The parameter is q.
k x = 2sec q y = 3tan q l x=1+t y=2+t
Let x = sin q: 1−t 1−t
Letting x = cos q will also give
So y = 6 sin q 1 − sin q = 6sin q cos q
2
y = 3sin 2q. Try this yourself.
= 3(2sin q cos q) 5 Point P lies on the curve x = 2t - 4, y = t + 1
= 3sin 2q If the y-coordinate of P is 6, find its x-coordinate.

Hence, parametric equations for the curve are There may be more than one
x = sin q, y = 3sin 2q possible pair of parametric 6 Point Q lies on the curve x = 2 + t , y = 3 − 2t
equations for a given curve. 2−t 2−t
If the x-coordinate of Q is 4, find its y-coordinate.
Exercise 7.1
7 The point (4, k) lies on the curve x = t2 - 5, y = t - 1
1 A curve has the parametric equations x = 3t, y = t2 - 3
C4

C4
t -3 -2 -1 0 1 2 3 Find the possible values of k.
Copy and complete this table. x
Hence sketch the graph of the curve for -3 t 3 y 8 The variable point P (at, t2 - 1) meets the line y = 8 at the point (6, 8).
Find the possible values of a and the Cartesian equation of the
2 The parametric equations of a curve are x = 3t2, y = t3 t -2 -1 0 1 2
curve along which P moves.
Copy and complete this table. x
Hence sketch the graph of the curve for -2 t 2 9 Find the coordinates of the points where these curves meet the x-axis.
y
a x = t2 + 1 y=t-3 b x = 1 + t3 y=2-t
c x = 5t + 3 y = t2 - 4 d x = 3cos q y = sin q
3 Construct your own tables of values to sketch the graphs of
the curves with these parametric equations for the range of
10 Find the coordinates of the points where these curves meet the y-axis.
values given in each case.
a x=t-5 y = t2 - 2 b x = t 2 - 3t + 2 y=t+4
a x = t 2 - 4, y = 1t3 for -3 t 3
2 c x = t3 - t y = t2 d x = tan q y = sec q
b x= t3 - 2t + 4, y = t - 1 for -2 t 2
11 The curve x = at 2 - 3, y = a(t - 2) contains the point (17, 0).
c x = t 2, y=1 for -3 t 3
t Find the value of a.
d x = 4sin q, y = 4cos q for 0 q 2p
12 The point (20, 40) lies on the curve x = at 2, y = 4at
e x = 5cos q, y = 3sin q for 0 q 2p Find the value of a.
f x = sec q, y = tan q for 0 q 2p

162 163


13 The curve x = 2asin q, y = 1 + acos 2q intersects the y-axis at 21 A curve has parametric equations
the point (0, 4). x = t - 2sin t, y = 1 - 2cos t, 0 t 2p
Find the value of a. a Find the values of t, in terms of p, at the two points where
the curve crosses the x-axis.
14 Points A and B lie on the curve x = t 2 - 3, y = 2t + 3
where t = 2 and t = 3 respectively. b The curve crosses the y-axis at two points where t = a and t = b
Find Show that one of these points has a = 0. Find, by trial-and-improvement,
the value of b to 1 decimal place. Find the coordinates of these
a the distance between A and B
two points on the y-axis.
b the gradient of the chord AB
22 By substituting y = tx, find parametric equations for the
c the equation of the chord AB.
curves with these Cartesian equations.
15 Show that these two pairs of parametric equations represent a y3 = x2
the same straight line. Find the Cartesian equation of the line. b y = x2 - 2x
a x=1-t y = 3 - 2t
c x3 - y3 = x2
b x= 1 y=t +3
t +1 t +1 d x - y = xy

16 Find parametric equations of the curve with the Cartesian equation 23 A curve has the Cartesian equation x 3 + y 3 = 3xy
y = x 4 − x2 a Show, by substituting y = tx, that the curve can be represented
C4

C4
by the parametric equations
if q is the parameter such that x = 2cos q
x= 3t , y = 3t 2
17 Use the identity 1 + tan2 q = sec2 q to find parametric equations 1 + t3 1 + t3
for the curve with the Cartesian equation y = x
b i Find the points where t = 0 and t = ¥
1 + x2
ii Investigate the curve when t is close to -1.
c Hence, sketch the curve and find the equation of any asymptote.
18 The Cartesian equation of a curve is y = 3 1 − x
2

x
Find parametric equations for this curve if
INVESTIGATION
a x = sin q b x=1
t
24 a Use a computer’s graphical package to check your
answers to questions 1, 2 and 3.
19 The equation of a circle is x 2 + y 2 - 6x - 4y + 12 = 0 Refer to Example 3.
You can also check answers to other questions by
a The equation is written in the form (x - a)2 + (y - b)2 =1 drawing appropriate graphs.
Find the values of a and b.
b Investigate how changing the values of constants A, Your computer software may
b Hence, find parametric equations for the circle in terms B, m and n in these parametric equations alters the need to have q in degrees,
of the parameter q. graphs of the curves. that is, 0° q 360°
x = Asin mq, y = Bcos nq for 0 q 2p
20 A hyperbola has the equation 9x 2 - 4y 2 - 18x + 16y - 43 = 0
(x − a)2 (y − c)2
a The equation is written in the form − =1
b2 d2
Find the values of a, b, c and d.
b Hence, find parametric equations for the hyperbola in terms of q.

164 165


2 Find the points of intersection of each curve and the given line.
7.2 Points of intersection
a x = t 2 - 1, y = 2t + 1 y=x+2 b x = t 3, y = t 2 + 2t y = 2x + 1

You can use simultaneous equations to find the points of c x = t 2 - 1, y = t 2 + t + 1 2y - x - 3 = 0
intersection when a curve is expressed in parametric equations.
3 Find the points of intersection of the curve with parametric
equations x = 2t 2, y = 3t and the circle x 2 + y 2 - 6x - 1 = 0
EXAMPLE 1

Find the points of intersection of the curve with
parametric equations x = 2t - 1, y = t 2 4 Find the points of intersection of the parabola x + y2 = 9
and the straight line y = 3x - 2 and the curve x = (t - 3)2, y = 2t
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Solve the equations x = 2t - 1, y = t 2 ⎫ simultaneously. There are three unknowns, 5 Find the points where these curves cross the coordinate axes.
⎬
y = 3x - 2 ⎭ x, y and t.
a x=t-1 y=t-4 b x=t-2 y = t2 - 9 c x = t2 + 1 y = t - 3
Substitute x = 2t - 1, y = t2 into y = 3x - 2 to eliminate x and y:
t 2 = 3(2t - 1) - 2 d x = t3 - 1 y = t2 - 4 e x =1 − 1 y = t2 + 1 f x = p - 2t y = 1 - sin t
t t −4
t 2 - 6t + 5 = 0
(t - 1)(t - 5) = 0 so t = 1 or 5 6 The variable point P(t 2, 2t) moves along a locus.
Find the points where the locus crosses the straight line y = 2x - 4
When t = 1, x = 2(1) - 1 = 1 and y = 12 = 1
When t = 5, x = 2(5) - 1 = 9 and y = 52 = 25
7 The point P(t 2, 4t) moves as t varies. Q is the midpoint of OP
The points of intersection are (1, 1) and (9, 25). where O is the origin. Write down the coordinates of Q. Find the
C4

C4
Cartesian equation of the locus of Q.
EXAMPLE 2

Find the points A, B and C where the curve given parametrically 8 The point P(2t 2, 6t) lies on a curve. The foot of the perpendicular
by x = t 2 - 4, y = t - 1 intersects the two coordinate axes. from P to the x-axis is Q. The midpoint of PQ is M. Find
Hence, find the area of triangle ABC. a the coordinates of Q and M in terms of t
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

When y = 0, t - 1 = 0 giving t = 1 The curve meets the x-axis when y = 0 b the Cartesian equation of the locus of M as P moves.
When t = 1, x = 12 - 4 = -3,
giving the point of intersection A(-3, 0). 9 Find the points of intersection of the curve
x = 1 - 5t, y = t 3 + t 2 and the line x + y + 2 = 0
When x = 0, t 2 - 4 = 0 giving t = ±2 The curve meets the y-axis when x = 0
When t = 2, y = 2 - 1 = 1 y 10 The curve x = t + 1, y = t 2 - k intersects the x- and y-axes at
and, when t = -2, y = -2 - 1 = -3, points P and Q respectively.
giving the points of intersection B(0, 1) and C(0, -3). B Find the value of k (k ¹ 1) such that OP = 2OQ where O is
1
The area of triangle ABC is A the origin.
1 BC × OA = 1 × 4 × 3 = 6 square units –3 O x
2 2
INVESTIGATION
C
–3
11 Use a computer’s graphical software to draw graphs
using their parametric equations.
Exercise 7.2 Check your answers to the problems in this exercise
1 Find the points of intersection of the parabola x = t 2 y = 2t where you have found points of intersection.
and the straight lines
a x+y=3 b 4x + 2y = 15
166 167


EXAMPLE 2
7.3 Differentiation A curve is defined parametrically by x = 3 - t, y = 4 - t - t 2
A normal is drawn to the curve at the point A where t = 1
dy Find another point B at which this normal intersects the curve.
You can differentiate parametric equations to obtain .
dx ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Let t = 1:x = 3 - 1 = 2 and y = 4 - 1 - 12 = 2
If x = f(t) and y = g(t) dy
dy dy dx dy
So, the normal is drawn at the point A (2, 2).
dt
the chain rule gives dt = dx × dt or dx
=
dx Find
dy
and substitute t = 1:
dt dx
dy
dy −1 − 2t
dy
Once you know dx , you can use it to find equations of tangents = dt = −
dx dx 1
and normals to a curve and to find stationary values. dt
= 1 + 2t
EXAMPLE 1

A curve has parametric equations x = t3 + 2t + 4, y = t2 - 1 When t = 1,
Find a the equation of the tangent at the point where t = 2 the gradient of the tangent at the point A(2, 2) is 1 + 2(1) = 3 If the gradients of tangent
and normal are m and m¢,
b the nature of any stationary values and the points at which they occur. The gradient of the normal at the same point is - 1 . 1
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
3 then m¢ = - m
So, the equation of the normal at the point A(2, 2)
dx
a Differentiate x wrt t: = 3t 2 + 2 y−2
dt is = −1
x−2 3
dy
Differentiate y wrt t: = 2t 3y + x = 8
dt
C4

C4
dy
2t To find the intersections of the normal and curve, substitute the parametric
dy
So = dt = equations into 3y + x = 8:
dx dx 3t 2 + 2
dt 3(4 - t - t 2) + (3 - t) = 8
Substitute t = 2: = 2×2 = 2 2
is the gradient of the 3t 2 + 4t - 7 = 0 You know that the curve and
3 × 22 + 2 7 7
(t - 1)(3t + 7) = 0 normal intersect when t = 1
tangent when t = 2 7
When t = 2 x = 23 + 2 ´ 2 + 4 = 16 t = 1 or - 3
and y = 22 - 1 = 3
t = 1 gives the initial point A on the normal, y
So, the tangent passes through the point (16, 3) with a gradient of 2 .
7 Hence at B, t = - 7
y−3 y − y1
3
The equation of the tangent is =2 =m
x − 16 7 x − x1 A
Substitute t = − 7 into the parametric equations: 2
Rearrange: 7y = 2x - 11 See C1 for revision. 3
B

dy 2t x = 3 + 7 = 51 y = 4 + 7 − 49 = 8
b For stationary values, = =0 3 3 3 9 9 O 2 6 x
dx 3t 2 + 2
So the only stationary value occurs when t = 0
at the point where x = 03 + 2 ´ 0 + 4 = 4
The numerator 2t = 0 when t = 0 The required point of intersection is 5 1 , 8 . ( 3 9)
and y = 02 - 1 = -1 Investigate the gradient on either
y side of the point (4, -1):
t -1 0 1
Choose values of t either side of
t = 0 and make sure that the
x 1 4 7
x-values are either side of x = 4
dy −
2 2
dx 5 0 5 3

There is a minimum –1 O x
4 16
value at the point (4, -1).
168 169


Exercise 7.3 6 The tangent at point P(1, 1) to the curve x = 1 , y = t 2 intersects
1 Find the gradient of each curve with these parametric equations t
the curve at point Q.
at the point with the given value of t (or q).
Find the equation of the tangent at P and the coordinates of Q.
a x=t+1 y = 3t 2 when t = 2
b x = t2 - 2 y = t3 + 1 when t = 4 7 The point P lies on the curve x = 5cos q, y = 4sin q
A and B are the points (-3, 0) and (3, 0) respectively.
c x = t 3 - t2 y = (t + 3)2 when t = 1
a Find the distances AP and BP in terms of q.
d x = 3sin q y = 5cos q when q = p
4 b Show that the sum of the distances AP and BP is constant
e x=1+ t2 y = 1+ 1 when t = -2 for all points P.
t
f x = sin 2q y = qcos q when q = 0 ( t)
8 The line from the variable point P t, 1 to the origin O
intersects the line x = 1 at the point Q.
2 Find the equation of the tangent and the normal to the curves
with these equations at the point where t (or q) has the given value. y

x=1
a x = 2t 2 y = 4t when t = 1
2

b x = t2 + 1 y = t3 - 1 when t = 1

when q = p
P
c x = 2cos q y = cos 2q Q
4
C4

C4
O 1 x
y = 1 − t2
2
d x = 2t when t = 2
1+t 2
1+t

3 Find the stationary points on these curves.
a x=t y = t3 - t
b x = t2 y =t +1
t
c x = q - cos q y = sin q for 0 < q < 2p Find

d x = 3sin q + 2 y = 3cos q + 5 for 0 q 2p a the gradient and the equation of the line OP
b the coordinates of Q in terms of t
4 The curve x = 2t 2, y = 4t has a normal at point P(8, 8).
Find the equation of the normal. c the Cartesian equation of the locus of the midpoint of PQ.
Also find the point where the normal meets the curve a second time.

INVESTIGATION
5 Find the equation of the normal to the curve x = 6t, y = 6 at
t 9 Investigate how to draw tangents and normals to curves
the point where t = 2
using a computer’s graphical package. Hence, check some
Also find the point where the normal intersects the curve again.
of your answers to the problems in this exercise.

170 171


EXAMPLE 2
7.4 Integration This diagram shows the curve with parametric y

equations x = t 2 + 1, y = t 3 - 4t
t = –1
b y Find the values of t at the points A(1, 0) and B(5, 0).
t2
You can modify the expression f(x) dx for the area under a
a Find the area of the region enclosed by the loop of the curve. B
curve using the chain rule. t1 A t = ±2
O x
1 t=0 5
For parametric equations x = f(t), y = g(t),
the area under the curve between the points where
O x
t = t1 and t = t2 is given by a b t=1
t2 •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

∫ y dx dt The limits of this integral are t = t1 Find t at A and B:
t1 dt and t = t2, as the independent At points A and B, y = 0
variable is now t and not x. t 3 - 4t = 0
t(t - 2)(t + 2) = 0
EXAMPLE 1

This sketch shows the curve with parametric equation y t = 0 or ±2
4 t=3
x = t 2 + 1, y = t + 1 t=1 B Let t = 0: x = 02 + 1 = 1 and y = 0 - 0 = 0 Find x and y when t = 0
t 2
A So, t = 0 at the point A(1, 0).
Find the shaded area between the curve and the x-axis from
O x Let t = 2: x = 22 + 1 = 5 and y = 23 - 4 ´ 2 = 0 Find x and y when t = +2
x = 2 to x = 10 2 10
–2 Let t = -2: x = (-2)2 + 1 = 5 and y = (-2)3 - 4 ´ (-2) = 0 and t = -2.
C4

C4
–4 So, t = ±2 at the point B(5, 0).
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

You want the area under the curve from
Let x = 2: t 2 = 1, t = ±1 so y = 2 or -2. A(1, 0) where t = 0 to B(5, 0) where t = ±2
So, point A is (2, 2).
When t = 1, x = 2, y = 1 - 4 = -3, giving the point (2, -3) below the x-axis.
Let x = 10: 2
t = 9, t = ±3 so y = 3 1 or - 3 1 . So, integrating from t = 0 to t = 2 will give the area below the x-axis.
3 3
So, point B is 10, 3 1 . ( 3 ) Similarly, t = -1 gives the point (2, 3) and integrating from t = 0 to t = -2
gives the area above the x-axis.
You want the area under the curve from A(2, 2) where t = 1 You are calculating the area
area of loop = 2 ´ area enclosed by the curve above the x-axis The curve is symmetrical

3 (
to B 10, 3 1 where t = 3 ) above the x-axis only.

=2´
5
y dx = 2
−2

y
dx
dt where y = t3 - 4t and
dx
= 2t
about the x-axis.

dt dt
1 0
Integrate:
−2
10 3
dx =2 (t3 - 4t)(2t)dt
Required area = y dx = y dx dt where y = t + 1 and = 2t 0
dt t dt
2 1 Notice the change in the limits
−2
3 as the independent variable
=
1
( t)
t + 1 (2t) dt
changes from x to t. =4
0
(t4 - 4t2)dt

3 −
⎡ 5 3⎤ 2 You could find the answer directly
= 2 (t 2 + 1) dt = 4 ⎢ t − 4t ⎥ by integrating the whole way from
1
⎣5 3 ⎦0
t = -2 to t = 2 using

( )
3 −2
⎡ 3 ⎤ = 4 − 32 + 32 − 0
= 2 ⎢t + t ⎥ 5 3 2t4 - 8t2 dt
⎣3 ⎦1 2

= 17 1 square units Try this yourself.
(
= 2 9 + 3 − 1 − 1 = 21 1 square units
3 ) 3
15
172 173


Exercise 7.4 5 a This diagram shows the curve x = t 2 + 1 , y = 2t
2
y
1 Find, in each case, the area between the x-axis and the curve t
between the two points P and Q defined by the given values Find the coordinates of the points A, B and C on the C
of t or x. curve where t = 1 , 1 and 2 respectively.
y 2
a x = t + 2 y = 3t - 1 for t = 1 to t = 4
b Calculate the shaded area of the diagram bounded by
B
b x = 2t y = 4t 2 + 1 for t = 0 to t = 2 Q the curve and the line AC.
c x = t2 y = 1 + 12t - 3t 2 for t = 0 to t = 4 A
x
P O
d x = t - 1 y = 2t - 1 for x = 1 to x = 5
e x = 2t y = t2 + 2 for x = 2 to x = 4 O x

f x = 4t y=2 for x = 2 to x = 4
t

2 A and B are the points on the curve x = t 3, y = t2 y

where t = 0 and t = 2
Find the shaded area on the diagram. B 6 The curve x = t 2 - 3t + 2, y = 4t 2 + 1 is shown on y
R t=2
Also find the area of the region labelled R. this diagram.
A Find the points at which the curve cuts the y-axis.
O t=0 x
Also find the area bounded by the curve and
the line y = 5
C4

C4
3 a The curve x = t 2 - 1, y = 2t(2 - t) cuts the y O x
coordinate axes at the points A, B, C and D.
Find the positions of these points and their B
associated t-values. 7 a A curve is expressed parametrically by x = t 2 + 1, y = 2t
R
A
b Calculate the area of the region R in the C O x Another curve has the parametric equations x = 2s, y = 2
s
first quadrant enclosed by the curve and the Find any points of intersection.
two coordinate axes.
b Find the area enclosed by the two curves and the line x = 5
c Calculate the total shaded area.

8 a The straight line y = c - 2x touches the curve x = t, y = 1
t
D at point P.
Find the possible values of c and the coordinates of P.

b Find the area enclosed by the curve x = t, y = 1 and the
t
line y = 3 - 2x
4 This diagram shows the curve x = t 2, y = t(4 - t 2) y
Find the values of t at the two points where the curve
cuts the x-axis. INVESTIGATION
Hence, find the area enclosed by the loop.
9 Show that, when finding the area under a curve, you
O x get the same answer whether you use the curve’s
Cartesian equation or its parametric equations.
Consider some of the curves in questions 1 and 2 of
this exercise.
174 175


8 A curve has parametric equations x = t 2, y = 2t
Review 7 R is the point on the curve where t = r
a Show that the normal to the curve at point R has a gradient of -r.
1 Find the Cartesian equation for each of the curves given
parametrically by these equations. b If S is the point (s 2, 2s), find the gradient of the chord RS in terms of r and s.

a x = t - 1, y = t2 + 1 b x = t − 1, y = t +1 c If the chord RS is normal to the curve at R, show that r2 + rs + 2 = 0
t t
d At what point does the normal at the point (9, 6) meet the curve again?
c x = 4cos q, y = 3sin q d x = 2cos q, y = cos 2q
9 The trajectory of a cricket ball is given parametrically by the equations
2 Use the identity sin2 q + cos2 q º 1 to find parametric equations
for the curves with the Cartesian equation x = 10t, y = 2 + 10t - 5t 2

y= 5x where x and y are the horizontal and vertical distances travelled
a y = 3 x 1 − x2 b
4− x2 (in metres) after a time of t seconds from the ball being struck.
a Find the Cartesian equation of the trajectory.
3 The point (5, a) lies on the curve x = t 2 + 1, y = 1 (t − 1)
3 b Find the time taken before the ball hits the ground.
Find the possible values of a.
c What is the horizontal distance travelled by the ball before
4 Find the points where the curve given by these parametric equations it hits the ground?
a x = 3t + 1, y = t 2 - 1 intersects the straight line 2x + y = 6
10 This diagram shows a sketch of part of the curve C with y
C4

C4
b x = t3 - 4, y = t 2 - 4 intersects the x-axis parametric equations
c x = cos t, y = 5sin t intersects the circle x 2 + y 2 = 2 x = 1 + 3 , y = t 2 sin t, p t p
C P
t 2
R
⎛ 2⎞
5 Find the equation of the tangent and the normal to the curves with a The point P ⎜ a, p ⎟ lies on C. Find the value of a.
⎝ 4 ⎠
these parametric equations at the point where t has the given value. O a x
b Region R is enclosed by C, the x-axis and the line
a x = 2t - 1, y = t 3 + 1 when t = 1 b x = t 3 - 1, y = t 2 + t + 1 when t = 2
x = a as shown in the diagram. p
c x = 1 + t, y= 1 when t = 2 d x = 2sin t, y = sin 2t when t = p
1−t 6 Show that the area of region R is given by 3 p
sin t dt
2

6 Find, in each case, the area between the x-axis and the curve c Find the exact value of the area of R.
between the two points defined by the given values of t or x.
11 A curve has parametric equations
a x = t + 2 from t = 1 to t = 4 b x = 2t + 1 from x = 3 to x = 9
x = 2cot q, y = 2sin2 q, 0 < q p
y = 3t - 1 y= t 2
dy
c x = ln t from t = 2 to t = 3 d x = e-t from x = 1 to x = 2 a Find an expression for dx in terms of the parameter q.
y = tsin t y = e2t + 1
b Find an equation of the tangent to the curve at the point where q = p
4
7 The curve C is defined parametrically, for 0 q p, by the equations c Find a Cartesian equation of the curve in the form y = f(x)
x = 3cos q, y = 3cos 2q + 6 State the domain on which the curve is defined. [(c) Edexcel Limited 2005]
dy
a Find in terms of q.
dx
Explain why the gradient at any point on the curve C is never greater than 4.
b Find the Cartesian equation of C and sketch the graph of C.
176 177

7Exit
8 The binomial series
find the binomial expansion of (a + b)n and (1 + x)n when n is a positive integer
Summary Refer to
find the binomial expansion of (1 + x)n when n is a fraction or a negative number
You can sketch the graph of a curve given by the parametric equations x = f(t), and write the condition for which the expansion is valid
y = g(t) by using a table of values showing the values of x and y as t varies. 7.1
use partial fractions to express certain kinds of algebraic fraction as a
To convert the parametric equations x = f(t), y = g(t) of a curve into a binomial series
Cartesian equation, eliminate the parameter t from the two equations. 7.1
find numerical and algebraic approximations using binomial expansions.
You can find the points of intersection of two curves by solving
the equations of the curves simultaneously. Before you start
For example, you can substitute the parametric equations of one curve into
the Cartesian equation of the other curve. 7.2 You should know how to: Check in:
dy
To differentiate x = f(t), y = g(t) to find dx , 1 Use the laws of indices. 1 Write in the form (1 + x)n
−
2
dy 1 1 1
e.g. = = (1 + x) 3 a (1 + x)5 b
dy dt g ′(t) 3
(1 + x )2
2
(1 + x )3
the chain rule gives: dx = dx = f ′(t) 7.3 (1 + x )3
C4

C4
x =b
dt t =t 2
The area under a curve is given by
x =a
∫ y dx =
t = t1
y
dx
dt
dt 2 Manipulate surds. 2 Write in terms of the root of an integer
e.g. 0.99 = 9 × 11 = 3 11 a 1.21 b 4×3+ 4 c 1
where the parameter t has the value t1 at the point where x = a 100 10 9 1.2
and the value t2 at the point where x = b 7.4
⎛n⎞
3 Use Pascal’s triangle and calculate ⎜ ⎟ . 3 a Use Pascal’s triangle to expand (2 + x)3
r ⎝ ⎠
e.g. ⎛
10 ⎞ 10 10 × 9 × 8
⎜ ⎟ = C3 = = 120 b Find the value of 8C3 and expand (2 + x)8
Links ⎝ 3⎠ 1× 2× 3
The path of any projectile is the result of two independent
motions, horizontal and vertical, which can be expressed 4 Find terms when multiplying brackets. 4 In the expansion of
in terms of a parameter time t. e.g. Find the term in x2 in the expansion of
(1 − 3x − 5x 2
+ ) (1 + 1 x + 1 x 2
− )
You can model the path of a projectile at any time t by the
2
2 (
(1 + 2x + 3x + ) 1 − 1 x − 1 x 2 −
3 ) find
2 4

parametric equations
1
1 2
( )1
The term is (1) − x + (2x) − x + (3x 2)(1)
3 2 ( ) a the term in x
x(t) = (n0 cos q)t, y(t) = (n0 sin q) t − gt2
2
where q is the angle at which the projectile is launched at
1
( 5
= − − 1 + 3 x2 = x2
3 3 ) b the term in x2.

time t = 0, n0 is the initial velocity of the projectile, and g is
the acceleration due to gravity. 5 Express a fraction in partial fractions. 5 Express in partial fractions
7
e.g. Let ≡ A + B a 3
This model can be used to analyse the motion of a specific (1 − x)(5 + 2x) 1 − x 5 + 2x (1 − 2x)(1 + x)
projectile, which is useful in areas such as sports science The cover-up rule uses x = 1 to give A = 1
b 3
to study, for example, the flight of a golf ball. and uses x = −2 1 to give B = 2
2 (1 − 2x)(1 + x)2
7
So ≡ 1 + 2
(1 − x)(5 + 2x) 1 − x 5 + 2x

178 179


When n is negative or fractional
8.1 The binomial series the binomial series obtained is an infinite series – it does not
terminate after n + 1 terms
For n a positive integer, the binomial series is the series can be written as a series of ascending powers of x
Pascal’s triangle and nCr have no meaning when n is negative or
ænö ænö ænö ænö ænö n fractional and cannot be used to find coefficients
(a + b)n = ç ÷ an + ç ÷ an -1b + ç ÷ an -2b2 + + ç ÷ an -rbr + ç ÷b the condition -1 < x < 1 which restricts the range of values
è0 ø è1 ø è2 ø r è ø ènø
of x must always be stated
to expand (a + x)n as a series when n is not a positive integer,
( a)
n
This is a finite series with n + 1 terms. you must first rearrange (a + x)n as an 1 + x
ænö ænö
You can find the coefficients ç ÷ , ç ÷ ,
è 0 ø è1 ø

EXAMPLE 2
1
Find the first four terms in the expansion of
either or 1− x
from the nth row of State the range of values of x for which the expansion is valid.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Pascal’s triangle when n is small by using
1 −1
As = (1 − x ) 2 , let n = − 1 and replace x by -x in the expansion of (1 + x)n:
1 Row ⎛n⎞ n = n! 1− x 2
⎜ ⎟ = Cr (n − r)!r ! n! means factorial n
1 1 1st
⎝r ⎠ = n ´ (n - 1) ´ (n - 2). . . ´ 1
( 1 )( 2 ) ( x ) + ( 1 )( 2 )( 2 ) ( x )
−
3 −
3 5 − − −
( )
1
1 2 1 2nd 1 − 2 2
n(n − 1)(n − 2) (n − r + 1) There are r terms in both the = (1 − x ) 2 = 1 + − 1 ( − x ) + − 2 − 3 +
= 1− x 2 1× 2 1× 2 × 3
C4

C4
1 3 3 1 3rd
1× 2 × 3 × r numerator and denominator.
1 4 6 4 1 4th Write out the expansion in full with
= 1 + 1 x + 3 x2 + 5 x3 + brackets to avoid errors with the
2 8 16
n fractions and negative signs.
The special case when a = 1 and b = x gives the binomial series for (1 + x)
The expansion is valid for |-x| < 1 which gives |x| < 1

(1 + x)n = 1 + nx + n(n − 1) x 2 + n(n − 1)(n − 2) x 3 + + n(n − 1)(n − 2) (n − r + 1) x r + + xn
1× 2 1× 2 × 3 1× 2 × 3 ×r
This is valid for all x when n is a positive integer.

EXAMPLE 3
2
Find the first four terms in the expansion of (8 + 3x)3
Give the range of values of x for which the expansion is valid.
EXAMPLE 1

Find the first four terms in the expansion of (1 + 2x)20
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Rearrange so that the first term inside the bracket is 1.
⎛ n⎞
(1+ 2x )20 = 1 + 20(2x) + 20 × 2 (2x)2 + 20 × 19××318 ( 2x )3 +
1
× 19
1× 2
n is large, so use ⎜ ⎟ to find
⎝r ⎠
Then let n = 2 and replace x by 3 x in the expansion of (1 + x)n:
3 8

( )
the coefficients. Using 2
= 1 + 40x + 760x2 + 9120x3 + … 2 2
(8 + 3x)3 = 8 3 1 + 3x
3
Pascal¢s triangle is not an
8
which are the first four terms of the series. appropriate method in this case.
⎡ 2 2 −1
= 4 ⎢1 + 2 3x + 3 3 3x ( ) ( ) + 2 ( 2 − 1)( 2 − 2) ( 3x ) +
( 3 )( 8 ) 1 × 2 8 3 31 × 2 ×33 8
2 3
⎤
⎥ 83 = 4
2

⎢ ⎣ ⎥
⎦
The binomial expansion of (1 + is also valid for all x)n You can write -1 < x < 1 as ⎡ 2 3 ⎤ x2 x3
negative or fractional values of n provided that -1 < x < 1. |x| < 1. This restriction on = 4 ⎢1 + x − x + x − ⎥ = 4 + x − 16 + 96 −
As a check, substitute a small
⎣ 4 64 384 ⎦ value of x (such as x = 0.01)
the value of x ensures that
into the series and compare
the series converges. The expansion is valid for 3 x < 1 2
8
with the value of 8.03 3 from
which gives |x| < 8 or −2 2 < x < 2 2 your calculator.
3 3 3
180 181

8 The binomial series 8 The binomial series

EXAMPLE 4 4 Expand each expression as far as the term in x2.
Find the coefficient of x2 in the expansion of (1 + x )4 Find the values of x for which each expansion is valid.
31 − 3 x

For what values of x is the expansion valid? a (1 + x) 1 − x b 1 − 2x
1+ x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

(1 + x )4 −1
c 2+x d x+3
= (1 + x )4 (1 − 3 x ) 3
( ) x −1
2
31 − 3 x You can find the coefficients of
1− 1x
(1 + x)4 from Pascal’s triangle.
( )( )
2
⎛ 1 −1 − 1 ⎞
( )
−
⎜ ⎟
= (1 + 4x + 6x + 2
) ⎜1 + − 3
1 (−3x) + 3 3 ( −3x )2 + ⎟ 1+ x
2
⎝ 1× 2 ⎠ e f (3 − x) (1 + 2x ) 3
1− x
= (1 + 4x + 6x 2 + ) (1 + x + 2x 2 + ) g (2 + x)2 1 − 2x
The term in x2 is (1)(2x2) + (4x)(x) + (6x2)(1) = 12x2
The coefficient of the x2 term is 12. 5 Find the coefficient of x2 in the expansion of 1 + x + x 2
The expansion of (1 + x)4 is valid for all values of x because
the index 4 is a positive integer. 6 Find the first four terms in the binomial expansion of 1 + 2
x
The expansion of (1 − 3x )− 3 is valid for |−3x| < 1 or − 1 < x < 1
1
in descending powers of x.
3 3
For what values of x is the expansion valid?
The whole expansion is thus valid for − 1 < x < 1
3 3
7 The coefficient of x2 in the expansion of 1 + ax is -2.
C4

C4
Find two possible values of a and the first four terms of each
possible expansion.
Exercise 8.1
1 Expand as a series of ascending powers of x up to and including x3. 8 The first three terms in the expansion of (1 + ax)n are 1, +2x, and -2x 2.
State the range of values for which each series is valid. Find a, n and the coefficient of x3 in the expansion.
1 1
a 1 + 2x b 1+ x c 1 + 2x d
(1 − x)3
9 The second and fourth terms in the expansion of (1 + kx)n
e 3
1 + 3x f 1 g 1 h 4
1 + 2x are x and 5 x 3.
3
1− x (1 − 3x)2
Given that k, n > 0, find k and n. Also find the third term of the expansion.

( )
2
1 1 1− 1x
3
i j k (1 + 2x) 3
l
( )
2 2
1 − 2x 1− 1x
2
INVESTIGATION
1 1
2 a Expand i ii 10 Express f(x) = 1 + 1 as a series of descending powers of x.
1+ x 1− x x
4
as a series of ascending powers of x as far as x
Show that f(x) = 1 1 + x and so express f(x) as a series of
b Identify each series as a geometric progression and so find the x
sum of each series using the formula for a geometric progression. ascending powers of x. Use computer software to draw the
graphs of f(x) and these two series.
3 Find the first three terms in the binomial expansion of each expression. Notice the importance on the graphs of the values of x for
Give the values of x for which each expansion is valid. which the two expansions are valid.
1 1 1
a 4+x b c d
2+x (2 − 3x)2 9−x

182 183


Exercise 8.2
8.2 Using partial fractions
1 Write each expression in partial fractions and so expand each
expression as a series of ascending powers of x as far as x3.
You can express some fractions as a series of ascending powers Find the range of values of x for which each series is valid.
of x by firstly expressing them as partial fractions. 3 4 5
a b c
(1 − 2x)(1 − x) (1 + 3x)(1 − x)
(
1− )
1 x (1 + 2x)
2
EXAMPLE 1

Find the first four terms of the expansion of 1 + 5x
(1 − x)(2 + x) 3 6 2
d e f
as a series of ascending powers of x. (2 − x)(1 + x) (3 + x)(2 + x) (3 − 2x)(1 − 2x)
Find the range of values of x for which the expansion is valid. 4
g
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
(1 + x)(1 − x)2
Let 1 + 5x ≡ A + B
(1 − x)(2 + x) 1 − x 2 + x
2 a Express 2x + 32 in the form A + B
, where A and B are constants.
A(2 + x) + B(1 − x) (x − 4) x − 4 (x − 4)2
≡
(1 − x)(2 + x)
b Hence, or otherwise, express 2x + 32 as a binomial series up to and
Equate the numerators: (x − 4)
1 + 5x º A(2 + x) + B(1 - x) The cover-up rule gives including the term in x4. Give the range of values of x for which the
When x = -2, 1 - 10 = 0 + 3B the same results with series is valid.
B = -3 less written working.
When x = 1, 1 + 5 = 3A + 0 c Compare the value of 2x + 32 with the value of the series up to
(x − 4)
C4

C4
A=2 and including the term in x 4 when
1 + 5x i x=1 ii x = 1.5 iii x = 2
So = 2 − 3 = 2 − 3
(1 − x)(2 + x) 1 − x 2 + x 1 − x 2 1 + 1 x
2 ( ) What do you notice?

2( )
−1
= 2(1 − x)−1 − 3 1 + 1 x 3 f(x) = 1 + 14x , x <1
2 (1 − x)(1 + 2x) 2
a Express f(x) in partial fractions.
= 2 ⎢1 + (−1)(−x) + (−1)(−2) (−x)2 + (−1)(−2)(−3) (−x)3 + . . .⎤
⎡
⎥ 1
⎣ 1× 2 1× 2 × 3 ⎦ 3
b Hence find the exact value of 1
f(x) dx, giving your answer in the
3 ⎡1 + (−1) 1 x + (−1)(−2) 1 x
( ) ( ) + ( 11)( 22)(33) ( 1 x ) + . . .⎤⎥⎦
2 3 6
− − −
− form ln p, where p is rational.
2⎢⎣ 2 1× 2 2 × × 2
c Use the binomial theorem to expand f(x) in ascending powers
= 2 (1 + x + x 2 + x 3 +
2 2 4 (
) − 3 1 − 1 x + 1 x2 − 1 x3 + . . .
8 ) of x, up to and including the term in x 3, simplifying each term. [(c) Edexcel Limited 2003]

= 1 + 11 x + 13 x 2 + 35 x 3 +
2 4 8 16
INVESTIGATION
The expansion of (1 - x)-1 is valid for |x| < 1; that is -1 < x < 1
4 Use a computer’s graphical software to draw the graph of
( )
−1
The expansion of 1 + 1 x is valid for 1 x < 1; i.e. -2 < x < 2 -1 < x < 1 is a stricter condition y = f(x) where f(x) is one of the fractions in this exercise.
2 2 Also draw the graph of the equivalent binomial expansion.
than -2 < x < 2.
Both these conditions must apply, so the whole expansion is Explore the graphical significance of the range of valid
valid for -1< x < 1. values of x. Repeat for other functions f(x) selected
from this exercise, especially for Question 2.

184 185


You can compare the graph of y = f(x) with the graph of its
8.3 Approximations
binomial expansion.
2 3
The terms of the binomial expansion of (1 + x)n have ascending In example 2, you found that 1 + x = 1 + x − x + x − . . .
2 8 16
powers of x. for |x| < 1
If the value of x is small, such as 0.01, then successive terms in the
The graph of y = 1 + x and the graph of this infinite series are
expansion have smaller and smaller values.
identical for |x| < 1.

However, you can also compare the graph of y = 1 + x with
EXAMPLE 1

Find the value of 0.9998 correct to 6 significant figures by
the graphs of just parts of the series; for instance
letting x = 0.001 in the expansion of (1 - x)8.
2 2 3
y = 1+ x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

y = 1+ x − x and y = 1+ x − x + x
(1 - x)8 = 1 + 8(-x) + 8 ´ 7 (-x)2 + 8 ´ 7 ´ 6 (-x)3 +
2 2 8 2 8 16
+ x8
1´ 2 1´ 2 ´ 3
These are approximations to 1 + x and are valid when x is small
= 1 - 8x + 28x2 - 56x3 + … + x8
and high powers of x are negligible in size. That is, they are valid
Let x = 0.001 when x is close to x = 0 and the graph is in the neighbourhood of
so (1 - x)8 = 0.9998 the point (0, 1).
Then 0.9998 = 1 - 0.008 + 0.000 028 - 0.000 000 056 + . . . Keep your working to at least
» 0.992 027 944 7 significant figures if you The approximation y = 1 + x gives a straight line graph. The line y = 1 + 1 x is, in fact, the
2 2
= 0.992 028 correct to 6 significant figures. require accuracy to 6 s.f. It is a linear approximation. tangent to the curve y = 1 + x
at the point (0, 1).
C4

C4
2
You can check this value on a calculator. The approximation y = 1 + x − x is a quadratic approximation.
By including more terms of the series, the value could be calculated 2 8
beyond the accuracy of a calculator.
2 3
The approximation y = 1 + x − x + x is a cubic approximation.
2 8 16

This diagram shows that the more terms there are in the
EXAMPLE 2

Find the value of 2 correct to 5 decimal places by letting approximation, the closer the graph of the approximation
x = -0.02 in the expansion of 1 + x . is to the graph of y = 1 + x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

y

( ) ( )( ) ( )( )( )
1
1 −1 1 −1 − 3
1 + x = (1 + x) 2 =1+ 1 x + 2 2 x2 + 2 2 2 x3 + . . . linear
2 1× 2 1× 2 × 3 This expansion is valid for |x| < 1 approximation
2 3 and is thus valid for x = -0.02 6
= 1+ x − x + x −
2 8 16 4

Let x = -0.02, so 0.98 = 1 - 0.01 - 0.000 05 - 0.000 000 5 2 y= √1+x
= 0.989 949 5 to 7 decimal places x
–8 –6 –4 –2 0 2 4 6 8
–2
Now 0.98 = 49 × 2 = 7 × 2
100 10 –4
quadratic
So 2 = 10 ´ 0.989 949 5 = 1.41421 approximation
7 cubic –6
correct to five decimal places. approximation

186 187

8 The binomial series 8 The binomial series

EXAMPLE 3
1 7 a Find the first four terms in the binomial expansion of 1 − 1
Find linear and quadratic approximations to x
(1 + x)2 in descending powers of x.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
For what range of x-values is the expansion valid?
= (1 + x)−2 = 1 + (−2) x + (−2) × ( 3) x 2 −
1 − y
1× 2
b By letting x = 100, find the value of 99 correct to 6 decimal places.
(1 + x)2
5
= 1 − 2x + 3x 2 − for |x| < 1 c Choose a value of x and use the same series to find 101
4 quadratic correct to 6 decimal places.
1 ≈ 1 − 2x approximation
The linear approximation is (1 + x)2 linear 3
approximation 8 Show, by using a binomial expansion, that
1
2 ≈ 1 − 2x + 3x
2 2
The quadratic approximation is 1
(1 + x) a the linear approximation to is 1 - 3x
1 y=
1 (1 + x)3
(1 + x)2
b its quadratic approximation is 1- 3x + 6x2
–2 –1 0 1 2 x
–1

(1 + 2 )
3
9 Find the cubic approximation of the function x

Exercise 8.3 10 Find
1 Substitute x = 0.03 in the expansion of (1 - x)10 to find the You can check many of the
value of 0.9710 correct to 5 decimal places. answers in this exercise on a a quadratic approximation to the function 1 + 2 x
1 − 2x
a calculator. However, you
should not use a calculator
C4

C4
2 Expand (3 + x)6 and show, by substituting x = 0.02, that 5
b a cubic approximation to the function
to work out the answers. (1 − 2x)(2 + x)
3.026 = 758.650 correct to 3 decimal places.

3 By expanding 1 + x as a binomial series, find 11 A quadratic approximation to 1 is 1 − 3 x + Cx 2
(A + Bx)3 8 4
a the value of 5 correct to 5 decimal places by letting x = 1 Find the values of A, B and C.
4

b the value of 3 correct to 5 decimal places by letting x = − 1
4 INVESTIGATION
4 Expand 1 as a binomial series and, by substituting x = 0.1, 12 Use a computer’s graphical package to explore the graph
1− x 1
See question 8 in this exercise.
find the value of 10 correct to six significant figures. of y = and the linear, quadratic and cubic
(1 + x)3
approximations found from its binomial expansion.
5 a Find the first four terms of the binomial series for 3 1 + 3x
Explore other functions and their approximations from
b By substituting x = 1 find the value of 3
1003 questions in this exercise.
1000
correct to eight significant figures.

1
6 Expand as a series as far as the term in x3.
(1 + 2x)3
Substitute x = 0.001 and so find the value of 1.002-3

188 189


7 Given that 3 + 5x º A + B ,
Review 8 (1 + 3x)(1 - x) 1 + 3x 1 - x
a find the values of the constants A and B.
1 Expand as a series of ascending powers of x up to and including x3 b Hence, or otherwise, find the series expansion in ascending
State the range of values for which each expansion is valid. powers of x, up to and including the term in x2, of 3 + 5x
1 (1 + 3x)(1 − x)
a b 1 + 2x
1 − 3x c State, with a reason, whether your series expansion in
part b is valid for x = 1 [(c) Edexcel Limited 2004]
( )
-
2
2
1- 1x
3
c d 9 + 2x
2
8 Expand (2 + x)5 and show, by substituting x = 0.03, that
e 1 f 1 2.035 = 34.473 correct to 3 decimal places.
2−x 4−x
9 By expanding 1 + x as a binomial series, find the value of 10
g (1 − x) 1 + x h x+3
x −1 correct to 5 decimal places by letting x = 1
9

2 Find the constants A, B and C where 1 − x = A + Bx + Cx 2 +
2−x 10 Show, by using a binomal expansion, that
Find the range of values of x for which the expansion is valid. a the linear approximation to 1 is 1 - 3x
(1 + x)3
3 Find the coefficient of x2 in the expansion of 1 − x − x 2 b the quadratic approximation to 1 is 1 - 3x + 6x2
C4

C4
(1 + x)3

4 Find the first four terms in the binomial expansion of 1 − 2 3
x 11 The binomial expansion of (1 + 12x ) 4 in ascending powers of x up
in descending powers of x.
to and including the term in x3 is 1 + 9x + px 2 + qx 3, |12x| < 1
a Find the value of p and the value of q.
5 When (1 + ax)n is expanded as a series in ascending powers
b Use this expansion with your values of p and q together
of x, the coefficients of x and x2 are -6 and 27 respectively. 3
with an appropriate value of x to obtain an estimate of (1.6 ) 4 .
a Find the value of a and the value of n. 3
c Obtain (1.6 ) 4 from your calculator and hence make a comment
b Find the coefficient of x3. on the accuracy of the estimate you obtained in part b. [(c) Edexcel Limited 2003]
c State the set of values of x for which the expansion is valid. [(c) Edexcel Limited 2004]
3x 2 + 16 = A + B + C , |x| < 1
12 f(x) =
6 Write each expression in partial fractions and so expand it as a (1 − 3x)(2 + x)2 (1 − 3x) (2 + x) (2 + x)2 3
series of ascending powers of x as far as x3.
a Find the values of A and C and show that B = 0
Find the range of values of x for which each expansion is valid.
3 b Hence, or otherwise, find the series expansion of f(x), in
a
(1 − x)(1 + 2x) ascending powers of x, up to and including the term in x3.
Simplify each term. [(c) Edexcel Limited 2006]
b 3x + 1
(1 − x)(1 + x)2

190 191

8Exit
9 Differentiation
differentiate implicit functions and parametric functions
Summary Refer to use implicit and parametric functions in problems of coordinate geometry
When n is a positive integer, the binomial expansion of apply exponential functions to problems involving growth and decay
(a + b)n is a finite series, valid for all values of x, where find rates of change and explore how different rates of change
relate to each other in practical situations.
⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞ ⎛n⎞
(a + b)n = ⎜ ⎟ an + ⎜ ⎟ an −1b + ⎜ ⎟ an −2b2 + . . . + ⎜ ⎟ an −rbr + . . . ⎜ ⎟ bn 8.1
⎝0 ⎠ ⎝1 ⎠ ⎝2 ⎠ ⎝r ⎠ ⎝n⎠
⎛n⎞ Before you start
You can find the coefficients ⎜ ⎟ either from Pascal's triangle
r ⎝ ⎠
or by using n! 8.1
(n − r)!r! 1 Find the equation of a tangent and a 1 a Find the equation of the tangent to
When n is negative or fractional, the binomial expansion of normal to a curve. the curve y = 2x 2 − 12 at the point
(1 + x)n is an infinite series, valid only for |x| < 1, where e.g. Find the equation of the normal to the curve x
where x = 1.
(1 + x)n = 1 + nx + n(n − 1) x 2 + n(n − 1)(n − 2) x 3 + . . . +
1× 2 1× 2 × 3
n(n − 1)(n − 2). . .(n − r + 1) r
1 × 2 × 3. . . × r
x + ... 8.1
x (
y = x + 1 at 2, 2 1 .
2 ) b Find the equation of the normal
C4

C4
dy 1
When n is negative or fractional, you must rearrange (a + x)n =1- 2 to the curve y = sin x - cos x at the
dx x
( a)
point where x = p .
n
as an 1 + x to obtain its binomial expansion.
You can rewrite some algebraic fractions using partial fractions
8.1
(
At the point 2, 2 1 ,
2 )
dy
dx
1
=1− =
4 4
3 4

4
before expressing them as binomial series. 8.2 So, gradient of normal is −
3
You can use binomial expansions to find numerical and algebraic The equation of the normal is
approximations when x is small and terms containing high powers y − 21
2 =−4
of x are negligible in size and can be ignored. 8.3
x −2 3
which gives 6y + 8x = 31
Links
The binomial series is useful for approximations. There are 2 Use the chain rule and product rule. 2 Differentiate each expression with
many applications of this in physics. e.g. Differentiate y = sin2 x + sin x cos x respect to x.
dy
= (2sin x cos x) + (sin x ´ -sin x + cos x cos x) a y = tan3 x b y = x3 + 1
For example, in special relativity, which studies space and time, dx
the parameter g is defined as = 2sin x cos x + cos2 x - sin2 x c y = ex sin x d y = x3 ln x
1 = sin 2x + cos 2x
⎛ 2 ⎞− 2
g = ⎜1 − v2 ⎟
⎝ c ⎠
3 Manipulate logarithmic expressions. 3 Solve these equations,
where v is the velocity of a particle and c is the speed of light. e.g. Solve the equation 3x = 2x+4 giving answers to 2 s.f.
This is of the form (1 + x)n and, with v very much smaller than c, Take natural logarithms:
can be approximated by the first few terms of its series expansion a ln 3x = xln 2 + 1
xln 3 = (x + 4)ln 2
b e2x-3 = 20
() ()
2 4
g =1+ 1 v +3 v x(ln 3 - ln 2) = 4ln 2
c e1-0.02x = 1.5
2 c 8 c 4 ln 2 ln16
x= =
Using this approximation can make calculations easier and allow ln 3 − ln 2 ln1.5
related equations to be defined.
192 193

9 Differentiation

Implicit functions and coordinate geometry
9.1 Differentiating implicit functions
You can use the approaches from the Core 1 and Core 2 units Refer to C1 and C2 for revision.
with curves expressed by implicit functions.
Equations such as y2 + xy + y = 8, which are not easily rearranged
into the form y = f(x), are known as implicit functions.

EXAMPLE 2
Find the equation of the tangent to the curve x3 - y3ex + 8 = 0
You usually need to use the chain rule or the product rule See Chapter 4 for revision.
at the point (0, 2). Find the equation of the normal at this point.
(or both) to differentiate an implicit function. ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Differentiate with respect to x: The gradient of a tangent is given
dy

( )
n n by the value of at the point of
The chain rule gives d(y ) = d(y ) × dy = ny n −1 dy 3x 2 − y 3 × e x + e x × 3 y 2
dy
+0 =0 dx
dx dy dx dx dx contact with the curve.
d(xy) dy dx dy
and the product rule gives =x +y =x +y dy 3x 2 − y 3e x
dx dx dx dx so =
dx 3y 2e x

At the point (0, 2), dy 0 − 8×1 2
= =− e0 = 1
EXAMPLE 1

dy dx 3× 4 ×1 3
Find when a 2x3 + 4y3 = 3 b x2 + 3xy + y2 =6
dx
c x ln y = y2 + 1 d x3 y2 = sin(x - y) Find the equation of the tangent at (0, 2): Use y − y1 = m
x − x1
y −2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a Differentiate wrt x: 6x + 4 × 3y ×
2 dy
=0 2
Use the chain rule for 4y3 to give = −2 so the equation of the tangent is 3y + 2x = 6
dx x−0 3
The gradient of the normal at (0, 2) = + 3
3 1
d(4y ) dy
× = 4 × 3y 2 ×
dy The gradient of the normal, m¢ = -
2 dy dy x2 2
C4

C4
12y = -6x 2
and =− 2 dy dx dx m
dx dx 2y Find the equation of the normal at (0, 2):
y −2 3
b Differentiate wrt x: 2x + 3 x ( dy
dx
+ y × 1 + 2y
dy
dx
=0 ) Use the product rule for 3xy and
the chain rule for y2.
=
x−0 2
so the equation of the normal is 2y = 3x + 4

dy
(3x + 2y) = −(2x + 3y)
dx

EXAMPLE 3
dy 2x + 3y Find the stationary points on the curve x2 + y2 = 12x
=−
dx 3x + 2y ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Differentiate with respect to x:
x × ⎛ × ⎞ + ln y × 1 = 2y + 0
1 dy dy dy
c Differentiate wrt x: ⎜ y dx ⎟ dx
Use the product rule to 2x + 2y = 12
⎝ ⎠ differentiate x ln y wrt x. Within dx
dy x dy
ln y = 2y − the product rule, use the chain dy 12 − 2x 6 − x
dx y dx rule to differentiate ln y wrt x.
= =
dx 2y y
y ln y dy dy
= = 0 when x = 6 The gradient is zero at
2 y 2 − x dx dx
stationary points.
d Differentiate with respect to x: Use the chain rule to differentiate When x = 6, y2 = 12 ´ 6 - 62 = 36 and y = ±6
x3 ´ 2y
dy
dx (
+ y2 ´ 3x2 = cos (x - y) ´ 1 −
dy
dx ) ( ) sin(x - y) wrt x:
d(sin(x − y)) So, there are stationary points at (6, 6) and (6, -6).
dy dx
d( x − y )
(cos (x - y) + 2x3y) = cos (x - y) - 3x2y2 = cos (x - y) ´
dx dx
dy cos(x − y) − 3x 2 y 2 ⎛ dy ⎞ You can determine if a stationary point is a maximum or minimum by
= = cos (x - y) ´ ⎜ 1 − ⎟
dx cos(x − y) + 2x 3 y ⎝ dx ⎠ dy
either investigating the change in sign of dx near to the stationary point
d2 y
or finding whether 2 is negative or positive at the stationary point.
dx

194 195


Exercise 9.1 8 Tangents to the curve y2 = 4x at the points (1, 2) and (4, 4) meet
dy at the point P. Find the coordinates of P.
1 Find when
dx
a x2 - 3y2 = 5 b y3 - x3 = 2 c x2 + y2 - 3x + 4 = 0 9 The gradients of two tangents to the curve y2 = 16x are 2 and 1 .
3
Find the point at which the two tangents intersect.
d y3 + 3y2 = x e xy = 2 f x + xy + y = 1
g x2 + xy + y2 = 1 h x2y2 + x2 + y2 = 2 i 3x2 + 2y2 = x2y2 10 The curve y(x + y)2 + 15 = 3x3 has a tangent at the point (2, 1).
2 2 3 + 4 =1 Find the point where the tangent intersects the x-axis. Also find the
j x3 + 3x2y + 3xy2 + y3 = 1 k x − y =1 l angle that the tangent makes with the x-axis.
x2 y2
m cos x sin y = 1 n x + y = tan y
11 Find the two points of intersection of the curves y2 = x and x2 = 8y
dy Find the gradients of the curves at both points of intersection.
2 Find when Hence, find the angles at which the curves intersect.
dx
a y2 = ln x b yln x = ln y c x + y =1
12 Find the equations of the tangents to the curve y 2 = 3x + 12
2
x
d sin (x + y) = sin x e cos 3x sin 2y = 1 f ex+y = x which are parallel to the x-axis.
g xe y + yex = 1 h x = ex ln y
13 Find the x-coordinates of the turning points on the curve
x3 - y3 - 2x2 + 3y + x - 4 = 0
3 Find the gradient of the tangent to each of these curves at the given point.
a x2 - y2 = 7 (4, 3) b x2y = 12 (2, 3) 14 Find the points on each of these curves where the gradient of
C4

C4
c x2 + 3xy + y2 = 1 (3, -1) d cos x = sin y + 1 (0, 0) the curve is zero.
a x2 + 3y2 - 8x - 4y + 17 = 0 b 2x2 + 2y2 - 4x + 5y + 4 = 0
e cos (x − y) = x − p
2 ( )
p ,0
2
x
f xy + e y = 1 (0, 1)
2
y2 15 Find the turning points on each of these curves and determine
g y2 = 12x (3, 6) h x + = 1 (3, 0) whether they are maximum or minimum points.
9 16
i x3 + xy2 + y3 = 11 (2, 1) j xey = 2 (2, 0) a x2 - y2 + 10x - 5y + 19 = 0 b x2 - 2y2 + 6x - 3y + 18 = 0

4 Find the equations of the tangent and the normal to these curves at the given point. 16 Find the turning points on the curve x3 - 3xy2 - y3 + 3 = 0
a x(y - 3) = y2 (-4, 2) b 2sin x cos y = 1 ( p4 , p4 )
c ln (xy) = 2y - 1 ( 2, 1 )
2
d e2x + e2y = x + y + 2 (0, 0)
17 Find the maximum and minimum values of y when
a 3(x - 2)2 + 4(y - 1)2 = 16 b 3(y - 1)2 - 2(x + 1)2 = 12
5 Find the equation of the normal to the curve y2 = 8 at the point (1, 2).
1 + x2 18 a Find the values of x which give stationary values of y on the curve x3 + y3 = 3xy

6 The tangent to the curve y2 = x3 at the point (4, 8) y b Find the stationary values and determine whether they are maxima or minima.
intersects the x-axis at the point P.
The normal to the curve at the same point intersects 8
INVESTIGATION
the x-axis at point Q.
19 Show that the equation xy - x2 + 4y + x = 0 can also be written as y = x (x - 1)
P Q
Find the length PQ. O 4 x x+4
dy
Find in two ways: by using implicit differentiation of the first
7 a Find the equation of the normal to the curve dx
y2 + 3xy - 2x2 + 1 = 0 at the point A(2, 1). equation and by using the quotient rule in the second equation.
Show that your two answers are equivalent.
b The normal intersects the y-axis at the point B.
196 If O is the origin, find the area of triangle OAB. 197

9 Differentiation

3 The strophoid in this diagram has parametric equations y
9.2 Differentiating parametric functions
⎛ ⎞
x = t 2 − 1, y = t ⎜ t 2 − 1 ⎟
2 2

t +1 ⎝ t + 1⎠
You can find dy from the parametric equations x = f(t), y = g(t) The chain rule gives
dx dy dy dx Find the positions of its stationary points.
by using the chain rule to give = ×
dt dx dt O x

dy dy dx g′(t)
= ÷ =
dx dt dt f ′(t)

( )
4 The point P 4, 2 2 lies on the ellipse with parametric equations
5
EXAMPLE 1

The curve shown in the diagram y
x = 5cos q, y = 4sin q and S is the point (3, 0).
has parametric equations The normal to the ellipse at P meets the x-axis at Q.
x = 3cos q + cos 3q and p
Prove that the length QS = 3 PS
i=
y = 3sin q + sin 3q 6 5
P
Find the gradient of the 5 A curve has parametric equations
O x
tangent to the curve at the y
2 3
point P where q = p x= t , y = t
6 1 + t2 1 + t2
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
P

dx dy
The parameter t = 1 at the point P.
= -3sin q - 3sin 3q = 3cos q + 3cos 3q
dq dq Show that the tangent to the curve at P has the equation y = 2x − 1
C4

C4
2 O x
dy dy dx 3 cos q + 3 cos 3q cos q + cos 3q Find the point Q at which the tangent meets the curve again. Q
= ÷ = =−
dx dq d q −(3 sin q + 3 sin 3q) sin q + sin 3q
3
cos p + cos p +0
When q = p , dy = − 6 2 = −
2 = −
3
= −
1 INVESTIGATION
6 dx sin p + sin p 1 +1 3 3
6 2 2 6 When a curve (such as a circle) rolls along a fixed curve,
a point on the rolling curve traces out a locus called a roulette.
1
The gradient of the tangent at P is −
3 If a moveable circle rolls on the outside of a fixed circle,
a point on its circumference traces out a special roulette
called an epicycloid.
Exercise 9.2 If the moveable circle rolls on the inside of a fixed circle, the
1 A hyperbola has parametric equations x = sec q, y = tan q point traces out a hypocycloid.
Prove that dy = cos ec q d(sec q )
= sec q tanq Use a computer’s graphical software to investigate these
dx dq
Find the equations of the two tangents to the curve which special roulettes for different values of k.
are parallel to the y-axis. Their parametric equations are:
x = kcos q + cos kq, y = ksin q - sin kq for hypocycloids
2 The astroid in this diagram has parametric equations y
and x = kcos q - cos kq, y = ksin q - sin kq for epicycloids.
x = 4cos3 q, y = 4sin3 q
4
dy
Prove that = − tan q and find the equations of the four
dx
tangents to the astroid which are equally inclined at 45° to –4 O 4 x
both axes. Also find the area of the square that they enclose.
–4
198 199

9 Differentiation

EXAMPLE 3
9.3 Growth and decay Cells in a dish grow so that, at a time t (hours), the number
t
of cells, n, is given by n = 10 × 8 2
ax is an exponential function for all values of a. When a = e, the function ex is
Find
called the exponential function.
Consider y = ax where a > 0 a the number of cells when t = 1 and t = 5
b the average rate of increase over the period t = 1 to t = 5
Take natural logarithms of both sides: c the instantaneous rate of increase when t = 5.
ln y = ln ax = x ln a ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Differentiate with respect to x: Use the chain rule to
a When t = 1, n = 10 ´ 80.5 = 28 n is given to the nearest
1 dy differentiate ln y. When t = 5, n = 10 ´ 82.5 = 1810 whole number.
y dx = 1 × ln a ln a is a constant.
b The average rate of increase = 1810 − 28
dy 5 −1
dx = ln a × y = 445.5 cells per hour
c The instantaneous rate of increase is given by the gradient
dy dy of the curve at the instant when t = 5.
When y = ax, = ln a ´ ax When a = e, = ln e ´ ex = ex
dx dx t
dn t
1
For n = 10 × 8 2 , = 10 × (ln 8 × 8 2) ×
dt 2
EXAMPLE 1

= (5ln 8) ´ 82.5 when t = 5
If y = 2x, find the value of dy when x = 3
C4

C4
dx = 1882
Hence, find the equation of the tangent to the curve at the
point P where x = 3. The instantaneous rate of increase is 1882 cells per hour.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

y n
dy y = 2x
For y = 2 , = ln2 × 2 x x
dx The gradient of the chord PQ gives
= ln 2 ´ 23 = 8ln 2 when x = 3 the average rate of increase.
1810 Q
When x = 3, y = 23 = 8
P
Hence, the equation of the tangent at the point (3, 8) is
y −8
= 8 ln 2 The gradient of the tangent at the
x−3 point Q gives the rate of increase
y = (8ln 2)x - 24ln 2 + 8 at the instant when t = 5.
Substitute ln 2 = 0.693147… : y = 5.55x - 8.64 to 2 d.p. O 3 x
28 P
O 1 2 3 4 5 t
EXAMPLE 2

Find the value of f ¢(1) for the function f(x) = 32x+1 + 4 ´ 3x + 1
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Differentiate f(x) = 32x+1 + 4 ´ 3x + 1 with respect to x: Exponential functions of time of the type akt and a-kt are used in
Use the chain rule to real life to develop models of exponential growth and decay.
f ¢(x) = (ln 3 ´ 32x+1) ´ 2 + 4 ´ (ln 3 ´ 3x) + 0 differentiate 32x+1
= (2ln 3)32x+1 + (4ln 3)3x When a = e, the models use functions involving ekt for
So f ¢(1) = (2ln 3) ´ 33 + (4ln 3) ´ 3 exponential growth and e-kt for exponential decay, where k > 0
= 54ln 3 + 12ln 3
= 66ln 3
200 201


EXAMPLE 4

EXAMPLE 5
Sri Lanka had a population N0 in 2006 of 21 million people. Radioactive fallout in the atmosphere contains the isotope
Its population N in another t years is predicted to be strontium-90.
N = N0ert where r = 0.013 The mass m grams after a time t years is given by m = m0e-kt
where k = 0.024 and m0 is the mass at time t = 0.
Find
Find
a the predicted population of Sri Lanka in 2046
a the mass after 10 years if m0 = 5
b how long it will take to double the 2006 population
b the initial rate of decay and the rate of decay after 10 years
c the rate of growth of the population in 2026. if m0 = 5
m (grams)
c the time taken for m to reduce to a value of 1 m0
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a In 2046, t = 40 2
5
and the predicted population N = 21 ´ e0.013´40
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a When t = 10 and m0 = 5, m = 5 ´ e-0.024´10 3.93
= 21 ´ e0.52
= 35.3 million = 5 ´ 0.786… 2.5
= 3.93 grams (to 3 s.f.)
b When N = 2N0 N (million)
dm d(m0e −kt)
ert = 2 b The rate of decay = =
dt dt O 10 20 30 40 50 t (years)
rt = ln 2 = m0 ´ (-k) ´ e-kt
42 28.9
t = ln 2
35.3 = -km0e-kt
r dm
0.6931..
21 When t = 0, = −0.024 × 5 × e0
= dt
C4

C4
0.013 = -0.12 This value of -0.12 is the gradient
= 53.3 O 10 20 30 40 50 t (years) The initial rate of decay is 0.12 grams per year. of the curve when t = 0.
53.3 In the final answer, the negative
So, the population will have doubled in 53 years time; When t = 10, dm = −0.024 × 5 × e −0.024 × 10 sign is implied by the word ‘decay’.
that is, by 2059. dt
= -0.094
dN d(N 0ert)
c The rate of growth = The rate of decay after 10 years is 0.094 grams per year.
dt dt
= N 0rert
c When m = 1 m0 1 = e −0.024 × t
Take the reciprocal of both sides.
2 2
In 2026, t = 20 and
dN
= 21 × 0.013 × e0.013 × 20 2 = e0.024 ´ t
dt ln 2 = 0.024 ´ t
= 0.273 ´ e0.26 This value gives the gradient of the
curve when t = 20. t = 0.6931... = 28.9
= 0.354 0.024
This time of 29 years is called the
It takes almost 29 years for half the initial mass to decay. half-life of strontium-90.
So, according to this model, in 2026, the population of These calculations assume that
Sri Lanka will be growing at a rate of 0.354 million the mathematical model N = N0ert
= 354 000 people per year. will hold, but it is unlikely that r Exercise 9.3
will stay constant over many years.
1 Find dy when
dx
a y = 3x b y = 32x-1 c y = 4 ´ 35x+2
1
d y = 2x e y = 102x+5 f y = 3 ´ 51-2x
3

202 203


1 x +2 8 The number, N, of cells infected with a virus changes over t hours
2 Find f ¢(x) and f ¢(2) when f(x) = 2 2 + 3 (2 x) − 4
according to N = 200 - 50e-2t
dN
3 Find the equation of the tangent and the equation of the a Find the value of N and when t = 0.
dt
normal to the curve
a y = 2 ´ 3x when x = 1 b Find how many infected cells there are after 4 hours and
the rate of change in the number of infected cells at this time.
b y = 3 ´ 2x + 1 when x = 0.
9 A mass m grams of a substance decays exponentially over a time
4 A population P grows over a time t according to t hours where m = m0e-kt
a P = 5 ´ 23t b P = 20.4t + 2 a If m = 20 when t = 0, find m0.
1.5t 0.4t - 1
c P = 3e d P=e If m = 15 when t = 5, find k.
dP
In each case find the value of P and when t = 1. b Find the time taken for mass of the substance to decay to
dt
i half its original mass ii 10% of its original mass.
5 A population Q is in decline over time according to c Find the rate of decay when t = 0 and when t = 5.
a Q = 4 ´ 3-0.2t b Q = 32 - 0.4t
10 A hot liquid cools such that the difference, q, between its temperature
c Q = 2e-0.01t d Q = e6 - 2t and that of its surroundings at a time t minutes is given by q = ke-at
In each case, find the value of Q and dQ when t = 1. a If the liquid’s temperature is 70 °C after 1 minute, find the
dt
values of k and a. Room temperature is constant at 10 °C and
C4

C4
6 A number of cells, n, grows over a time t such that n = n0e0.2t the liquid’s initial temperature is 80 °C.
where n0 = 5. b Calculate the initial rate of cooling and the rate of cooling
Find after 5 minutes.
a the number of cells when t = 0 and when t = 10 c Write an expression for the temperature T of the liquid at time t.
b the average rate of growth over the period t = 0 to t = 10 d Calculate the time taken for the temperature of the liquid
c the instantaneous rate of growth when t = 5. to drop to 40 °C.

7 The population, P, of Manchester was 126 000 in 1821 and
236 000 in 1841. INVESTIGATION
a If t is the number of years after 1821, model the population
11 The population of London was 1 950 000 in 1841 and
as P = P0ekt and find the constants P0 and k.
2 800 000 in 1861.
dP Model the population in two ways as:
b Find and evaluate the rate of growth in 1831.
dt
P = P0at
c Estimate the population in 1851. Find the percentage error in this P = P0ekt
estimate compared to the actual population in 1851 of 303 000. where P0, a and k are constants.
The actual population in 1871 was 3 250 000.
Do the two models give the same estimates?
Explain your answer.

204 205

9 Differentiation

EXAMPLE 2
9.4 Rates of change A hot-air balloon is being blown up at a rate of
2 m3 per minute.
The rate at which water is flowing out of a tap affects the Assuming the balloon is spherical, find r

rate at which the depth of water in the bath increases. a the rate of increase in its radius r when r = 2.5 metres

This is an example of a rate of change in which a change in b the rate of increase in its surface area A when
one variable over a given time produces a change in the r = 2.5 metres.
other variable over that time.
2 m3 min–1
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

If V is the volume of water running into the bath,
dV a If the volume of the balloon is V m3, then dV = 2 m3 min-1
then the rate of change of V is . dt
dt
Similarly, if h is the depth of water in the bath, To find dr , use the chain rule dV = dV × dr
dt dt dr dt
dh
then the rate of change of h is .
dt To find dV , use the geometry of the sphere.
dr
These two rates of change are linked by the chain rule
V = 4 p r3
3
dV dV dh
= × dV 4
dt dh dt = p × 3r 2 = 4p r 2
dr 3
dV Substitute into the chain rule with r = 2.5:
The differential is independent of time t and can be derived
C4

C4
dh
dV dV dr
from some physical or geometrical relationship between V and h. = ×
dt dr dt
dr
2 = 4p ´ 2.52 ´ dt
EXAMPLE 1

A cylindrical water tank of radius 2 m holds water which 2m
dr 2
is being pumped in from the top of the tank at a rate of The rate of increase in the radius, =
dt 25p
3 m3 per hour.
3 m3 h–1 = 0.025 metres per minute to 2 s.f.
Find the rate at which the depth h of water is increasing.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
b The surface area of the balloon A = 4pr2
h V
If the volume of water in the tank is V, then dV = 3 m3 h-1 dA
= 8p r
dt dr
dV dV dh dh = 8p ´ 2.5 = 20p
The chain rule gives = × You have to find dt when r = 2.5
dt dh dt
dA dA dr
From the geometry of the cylinder, V = pr2h The chain rule gives = × where, from part a,
dt dr dt
dr 2
= 4ph for r = 2 =
dt 25p
dV dA 2
= 4p Substitution gives = 20p × = 1.6
dh dt 25p
dh
Substitute into the chain rule: 3 = 4p × When r = 2.5 m, the rate of increase in the surface area is
dt
1.6 m2 per minute.
dh 3
The rate of change of the depth of water, =
dt 4p
Make sure you use the correct
= 0.24 m h-1 units in your answer.

206 207


Exercise 9.4 9 A spillage of coffee on to a horizontal table forms a circular
1 A square sheet of molten glass is increasing in size such that stain with a radius increasing at a rate of 2 mm s-1.
the rate of increase of the length, x, of its sides is given by Find the rate at which the area of the stain is increasing
dx after 5 seconds.
= 0.2 m per minute.
dt
Find the rate of increase of its area, A, when x = 0.5 metres. 10 Sand falls onto horizontal ground at a constant rate of You will need to use the formulae
20 cm3 s-1 to form a pile in the shape of a circular-based for volume and height of a cone.
2 Oil is dripping onto a horizontal floor forming a circular pool cone with a semi-vertical angle of 45°.
of radius r cm. If the rate of increase of the radius is 0.5 cm per Calculate the rate at which the vertical height of the conical
second, find pile is increasing after 5 seconds.
a the rate of increase of the area of the pool when r = 15 cm
11 A boy 1.5 m tall runs directly away from a light which is fixed
b the rate of increase of the circumference of the pool. 2 m above a horizontal road. If he runs at a speed of 3 m s-1,
find the rate at which his shadow is lengthening.
3 A crystal forms in the shape of cube of edge length x mm.
If the rate of increase of its edges is 0.3 mm per minute, find 12 Robert Boyle (1627–1691) discovered that, for a fixed amount
a the rate of increase of its volume when x = 6 mm of gas at a constant temperature, its pressure varies inversely
b the rate of increase of its surface area when x = 6 mm. as its volume varies.
A quantity of gas has an initial volume of 0.25 m3 and an initial
4 An ice cube melts uniformly on all its faces. When its edges are pressure of 2 N m-2. Its volume is allowed to increase at a
2 cm long, the rate of decrease of its surface area is 4 cm2 per hour. constant rate of 0.05 m3 s-1. Find the rate at which its pressure
C4

C4
Find the rate of decrease in its volume at this instant. is changing at the instant when its volume is double its initial value.

5 A spherical bubble under water is rising to the surface and its
INVESTIGATION
radius r is increasing at a rate of 0.2 mm per second.
a When its radius is 4 mm, find the rate of increase in its volume. 13 As a piston moves into the cylinder of an engine, the
length L of the cylindrical space changes. r
b Find the rate of increase in its surface area when its volume is 36p mm3. This change in length causes a change in the volume V
of the gas in the cylinder, which causes the pressure P
6 Molten plastic is extruded from a nozzle at a speed of 20 cm s-1 of the gas to change. L V
and forms a cylindrical shape. The nozzle has a circular
cross-section of area 0.75 cm2. Find an expression for the rate of change of pressure
Find the rate of change in the volume of the extruded plastic. in terms of radius r, length L and time t.
Have you made any assumptions?
7 Air is leaking from a spherical balloon at a rate of 2 cm3 per second.
When its radius is 12 cm, find the rate of decrease of
a its radius
b its surface area.

8 A hollow cone, with a semi-vertical angle of 60°, is held vertex
downwards with its axis vertical. Water drips into the cone at
a constant rate of 4 cm3 per minute. Find the rate at which the 60º
depth of water is increasing when the water is 4 cm deep.

208 209

9 Differentiation

10 The parametric equations of a curve are
Review 9
x = 2cos q + cos 2q and y = 2sin q + sin 2q

dy
Show that stationary values occur on this curve when cos q = 1
1 Find when 2
dx Find two stationary points for 0 q 2p.
2 − 3 =4
a 3x2 + 4y2 = 9 b x2 + 3xy + y2 = 2 c
x2 y2 11 A curve is given by the parametric equations x = t2, y = 1
t
d cos 2x sin 3y = 1 e y3 = ln x f x2ey - y2ex = 2 a Find the equation of the tangent at the point A 9,( )
1 .
3
b The tangent at A intersects the curve at point B.
dy
2 Find for the curve x3 - 2x2y - 3xy2 + y3 = 9 Find the value of the parameter t at B.
dx
Find the gradient of the tangent to this curve at the point (2, -1).
12 The normal to the curve with parametric equations x = 1 , y = t2
t −1
3 Find the equations to the tangent to these curves at the given points. at the point P (1, 4) meets the curve again at points Q and R.
a y3 = 2x2 (2, 2) b x3 - 2x2y - y3 = 8 (3, 1) c ye2x = 3 (0, 3)
a Show that the gradient of the normal is 1 and find its equation.
4
4 The normal to the curve x2 y2
+ 3xy + = 11 at the point (2, 1) b Prove that the values of t at points Q and R are given by t = − 1 ± 2
meets the axes at the points P and Q. Given that O is the origin, show that 2
the area of triangle OPQ is 81 square units.
112 13 The curve C is given parametrically by x = t − 1 , y = t + 1 , t ¹ 0
t t
5 A curve has equation x3 - 2xy - 4x + y3 - 51 = 0
C4

C4
a Find the coordinates of the points on C at which the gradient is zero.
Find the equation of the normal to the curve at the point (4, 3).
b Find the equation of the normal to C at the point (0, 2).
Give your answer in the form ax + by + c = 0,
where a, b and c are integers. [(c) Edexcel Limited 2003]
dy
14 a Find when i y = 4x + 4 ii y = 42x + 2 ´ 4x + 1
dx
6 a Show that, for the curve x2 + 2y2 - 4x + 4y = 26
b Find the equation of the tangent to the graphs of both these
dy
= 2−x
dx 2 (1 + y) equations at the points where x = 0
b Find all the stationary points on the curve.
15 Bacteria grow so that, at a time t, the number n of bacteria
t
c Find the points on the curve where the tangents are parallel to the y-axis.
is given by n = 5 × 4 3
7 An ellipse is expressed by the parametric equations x = 4sin q, y = 3cos q Find
a the number of bacteria when t = 2 and t = 5
Find the equation of the tangent to the ellipse at the point where q = p
6
b the average rate of increase in the number of bacteria over the
8 A curve is expressed parametrically by the equations x = t2 + 1, y = t3 period from t = 2 to t = 5
a Find the equation of the tangent at the point where t = 2 c the instantaneous increase in the number of bacteria when t = 2
b Find the equation of the normal at the point (2, -1). d the value of t at which the value of n is double its initial value.

9 The curve C has parametric equation x = asec t, y = btan t, 0 < t < p , 16 The value £V of a car t years after the 1st January 2001 is given
2
where a and b are positive constants. by the formula V = 10 000 ´ (1.5)-t
Prove that dy = b cosec t a Find the value of the car on 1st January 2005
dx a
Find the equation in the form y = px + q of the tangent to dV
b Find the value of when t = 4
C at the point where t = p [(c) Edexcel Limited 2003] dt
210 4 c Explain what the answer to part b represents. [(c) Edexcel Limited 2005] 211

Revision 3

9Exit
1 Express these as partial fractions.
a x+4 b x+3 c 2x
(x − 2)(x + 1) x(x 2 − 1) x 2 − 5x + 6

d x2 + 2 e 2 f x+2
x ( x + 1)
2
x 2 ( 2x − 1) (1 − 2x ) ( x − 3 )2
Summary Refer to
You use implicit differentiation when the relation between
2 Show that x 2+ 3 can be expressed as A + B + C + D
3
x and y cannot be expressed explicitly by y = f(x) x ( x − 1) x x −1 x +1
d(y n) dy
The chain rule gives = ny n −1 Find the values of A, B, C and D.
dx dx

The product rule gives d ( xy ) = x dy + y dx = x dy + y 9.1 3 Express these as partial fractions.
dx dx dx dx x2 + 2 x 3 − 2x 2 + 2
a b x2 − 2 c
For parametric equations x = f(t), y = g(t), x2 − 1 ( x − 1) ( x + 2 ) x ( x − 1)
2
dy dy dx
the chain rule gives = × 9.2
dt dx dt
d x3 + 1 e 4x 3 + 1 f 2x 3 − 1
dy
x
If y = a , then = ln a × a x 9.3 x ( x − 2) x 2x − 1)
2(
( x + 1)2 (2x − 1)
dx
dy
When a = e, y = ex gives = ln e × e x = e x
C4

4 If x + 2x2 + 3x + 1 ≡ px 2 + qx + r + sx + t ,

C4
4 2
dx
You can model y x −2 2
x −2
y find the values of the constants p, q, r, s and t.
exponential growth by y = Aekt y = Aekt
exponential decay by y = Ae-kt
where k > 0 A 9.3 5a Point P lies on the curve with parametric equations x = t2 - 4, y = t + 1
A –kt
y = Ae If the y-coordinate of P is 6, find its x-coordinate.

O t t b The point (4, k) lies on the curve with parametric equations x = 1 − 5, y = t - 1
O t2
Find the possible values of k.
When changes in x effect changes in y over a period of time,
their rates of change are connected by the chain rule where
dy dy dx 6 The variable point P ( l t , 3t − 1) meets the line y = 11 at the point (6, 11).
= × 9.4
dt dx dt a Find the value of l.
b Find the Cartesian equation of the curve along which P moves in the form y = f(x)
Links
Differentiation is a versatile tool that can be 7 Find the points where the curve expressed parametrically
used in many fields, particularly within industry. by the equations x = 1 + t , y = 2t + 1 intersects
1−t
Derivatives can be used to express the rate of decay a the x-axis b the y-axis c the line y = x - 1
of a radioactive substance in a chemical power plant.
Engineers can use differentiation to calculate rates 8 A curve is defined parametrically by x = 2t + 1, y = t2 - 4t + 1
of change of variables when designing systems to A normal is drawn to the curve at the point where t = 4.
ensure efficiency. Find
a the equation of the normal
Managers can solve maximum and minimum problems to
determine how to maximize profit or minimize waste. b the point at which the normal intersects the curve a second time.
212 213


9a A curve has parametric equations x = t2 + 3, y = 1 + t y B 13 Expand as a series of ascending powers of x up to and including x3
Find the two points A and B on the curve at which State the range of values of x for which each expansion is valid.
A
x = 7 and x = 12. 1
a b 1 + 5x c 9 + 2x
1 + 3x
b Find the area between the curve, the x-axis and the
ordinates x = 7 and x = 12. x 1 1
d e f
4 + 3x 4−x
O 7 12 x 1− 1x
4

14 Find the first three terms in the binominal expansion of each expression.
10 The diagram shows the graph of the curve with parametric y Give the values of x for which each expansion is valid.
equations x = t2 - 12, y = t3 - 9t 1+ x
a (1 − 2x) 1 + x b
a Find the values of t at the points where the curve 1 + 3x
intersects the x-axis.
O x 15 Find the first four terms in the binomial expansion of 1− 3 in
b Find the shaded area on the diagram. x
descending powers of x. For what values of x is the expansion valid?
Hence, find the total area of the loop.
16 The coefficient of x2 in the expansion of 4 + ax is -1. Find two possible
values of a and the first three terms of each possible expansion.
11 The curve expressed parametrically by y

x = 4 - 2t, y = t 2 + 1 17 Write each expression in partial fractions and so expand it
C4

C4
2
t as a series of ascending powers of x as far as x3.
is shown on this diagram. The curve cuts the y-axis at P
Find the range of values of x for which each expansion is valid.
R
the point P and has a minimum value at the point Q.
a 2−x b 3 − 2x
a Find the coordinates of points P and Q. Q (1 − 2x)(1 + x) (2 + x)(1 − 3x)

b Find the shaded area on the diagram.
18 f(x) = 3x − 12 ,
O x
x <1
c Find the area of the region labelled R. (1 − 2x) 2
y
Given that, for +x ≠ 1 , 3x − 1 ≡ A + B
L 2 (1 − 2x)2 (1 − 2x) (1 − 2x)2
12 This diagram shows the curve C with parametric equations where A and B are constants,
p P
a find the values of A and B.
x = 8cos t, y = 4sin 2t, 0 t 2
C

The point P lies on C and has coordinates ( 4, 2 3 ). b Hence, or otherwise, find the series expansion of f(x), in
a Find the value of t at the point P. R ascending powers of x, up to and including the term in x3,
simplifying each term. [(c) Edexcel Limited 2006]
The line L is a normal to C at P. O 4 x
b Show that an equation for L is y = −x 3 + 6 3
dy
19 Find when
The finite region R is enclosed by the curve C, dx
the x-axis and the line x = 4, as shown shaded a x2 - 5xy + y2 = 1 b xey = y3 -1 c xy = 1 + x sin y
in the diagram. p
2
c Show that the area of R is given by the integral 64 sin2 t cos t dt d sin x cos y = 1 e y − x =2 f x2y = tan(x + y)
p
3

d Use this integral to find the area of R, giving your answer in 20 a Find the equation of the tangent to the curve exy + 1 = x at (2, 0).
the form a + b 3, where a and b are constants to be determined. [(c) Edexcel Limited 2008] b Find the equation of the normal to the curve ln (xy) = y2 - 1 at (1, 1).
214 215

Revision 3

21 A curve is expressed parametrically by x = t + 1 , y = t2 - 1 (t ¹ 0)
2

10
t
Show that there are tangents to the curve which are parallel to
the y-axis and that they meet the curve at the points (2, 0) and (-2, 0).

22 a
dy
Find dx for an ellipse with parametric equations x = 3sin q, y = 2cos q, 0 q p
Integration
b Prove that the equation of the tangent to the ellipse at the point P find the area under a curve to a specified accuracy using a
where q = a is given by 3ycos a + 2xsin a = 6 numerical method
c The tangent at P intersects the coordinate axes at the points A and B. find the exact area under a curve by integration
Find, in terms of a, the area of triangle OAB, where O is the origin. integrate a variety of functions by using standard integral forms,
substitution, trigonometric identities and partial fractions, and
d Find the value of a which gives the smallest possible value of the
integration by parts
area of triangle OAB. State this area.
be systematic in your approach to integration
use integration to find volumes of revolution.
23 The tangent and normal to the curve y = 5 ´ 2x, at the point where
x = 1, intersect the y-axis at the points P and Q respectively.
Find the distance PQ as a decimal, correct to 1 decimal place.
Before you start
24 The population N0 of a town in 2008 is 56 000. A model of its growth You should know how to: Check in:
predicts that its population N in t years after 2008 will be N = N0e0.008t 1 Differentiate various functions using the 1 Differentiate
Find product, quotient and chain rules. a x2 ln x b x3ex
C4

C4
a its population in 2018 e.g. Differentiate y = x2sin 3x c ex tan x d x x2 + 1
dy
b the rate of growth of its population in 2018 = 2x sin 3x + (3 cos 3x) x 2
dx e ln x f ln (sin x)
x
c how long it takes for its population to be double the 2008 figure. = 2x sin 3x + 3x cos 3x
2

25 A spherical balloon is being inflated at a constant rate of 0.2 m3 per minute. 2 Manipulate trigonometric identities. 2 Prove these identities.
e.g. Prove that sin 2A ≡ cot A a tan A + cot A º 2cosec 2A
Find 1 − cos 2A
Use the double-angle formulae: 2 tan A
a the rate of increase in its radius r when r = 0.5 metres b º sin 2A
sin 2A 2 sin A cos A
1 + tan 2 A
b the rate of increase in its surface area when r = 0.5 metres. ≡
1 − cos 2A 1 − (1 − 2 sin2 A)
c cos 3A º 4cos3 A – 3cos A
2 sin A cos A cos A
≡ ≡ ≡ cot A
26 This diagram shows a right circular cylindrical metal rod which 2 sin2 A sin A
is expanding as it is heated. After t seconds, the radius of the
rod is x cm and the length of the rod is 5x cm. x 3 Find partial fractions. 3 Express these in partial fractions.
The cross-sectional area of the rod is increasing at the constant 5x
x
e.g. Express 2 in partial fractions. a
rate of 0.032 cm2 s-1 x(x − 2) x 2 − 25
dx Let 2 A
≡ +
B 6
a Find when the radius of the rod is 2 cm, giving your x(x − 2) x x−2 b
x (x − 1)(x + 2)
dt
answer to 3 significant figures. A (x − 2) + Bx
≡ 1
x (x − 2) c
b Find the rate of increase of the volume of the rod when x = 2. [(c) Edexcel Limited 2008] x (x − 1)2
Hence 2 º A(x - 2) + Bx
Letting x = 0 gives A = -1 d x2 + 3
Equate coefficients of x: A + B = 0 so B = 1 x (x − 2)
2 1 1
So ≡ −
x (x − 2) x−2 x

216 217

10 Integration

10.1

EXAMPLE 1
The trapezium rule a Estimate (to 3 significant figures) the area under the curve
y = sec2 x from x = 0 to x = p by using the trapezium rule with
3
When the area enclosed by a graph, the x-axis and two ordinates For n equal intervals, i 4 equal strips ii 8 equal strips.
x = a and x = b is split into n equal intervals of width h, then the use (n + 1) x-values.
p
trapezium rule gives x-values are sometimes 3
called ‘ordinates’. b Calculate the exact value of sec2 x dx as a surd.
0

b Find the percentage error in each of the two estimated values of
f(x) dx ≈ 1 h [ y0 + yn + 2(y1 + y2 + ... + yn −1)] where h = b − a area found using the trapezium rule.
a
2 n ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

p −0
y
a i With 4 strips, the width of each strip, h = 3 = p
4 12
y = f(x) Record the values in a table:

x 0
p 2p 3p 4p Use radians on your calculator.
12 12 12 12
y = sec2 x 1 1.0718 1.3333 2 4
yn
Estimate of the area required = 1 × p × [1 + 4 + 2(1.0718 + 1.3333 + 2)]
2 12
y2
y1 1 × p × 1 + 4 + 8.8102
y0 =
2 12
[ ]
C4

C4
x
O a h b = 1.8078 = 1.81 units2 (to 3 s.f.)
8
p −0
Using more strips involves more calculation but gives a more ii With 8 strips, the width of each strip, h = 3 = p
8 24
accurate approximation.
p 2p 3p 4p 5p 6p 7p 8p = p
x 0 24 24 24 24 24 24 24 24 3
convex y 1 1.0173 1.0718 1.1716 1.3333 1.5888 2 2.6984 4

concave See for revision.
Estimate of the area required = 1 × p × [1 + 4 + 2 × 10.8812]
C1
All the ‘middle values’ add
2 24 together to give 10.8812
= 1 × p × 26.7624
The shaded areas are the errors 2 24
in each approximation. = 1.7516 = 1.75 units2 (to 3 s.f.)
t
If the graph is a concave curve, If the graph is convex, the
the trapezium rule overestimates trapezium rule underestimates p
3 p
the actual area. the actual area. b sec 2 x dx = ⎡tan x⎤0
⎣ ⎦
3 d (tan x)
= sec² x
0 dx
= 3 − 0= 3

i The percentage error with 4 strips = 1.8078 − 3 × 100
3
= 4.4% (to 2 s.f.)

ii The percentage error with 8 strips = 1.7516 − 3 × 100 The percentage error is smaller
3 when more strips are used (but
= 1.1% (to 2 s.f.) more calculation is involved).

218 219

10 Integration 10 Integration
1
Exercise 10.1 3 Find an approximate value for I = ex sin x dx using the
1 Use the trapezium rule to estimate the value of these integrals trapezium rule with 0
correct to 3 significant figures using the number of strips given.
a six ordinates
1
a 2x dx 5 strips b eleven ordinates.
0
Give reasons why one of these values is more accurate than
5 the other.
b ln(1 + x2) dx 4 strips
1
4 The semicircle y = + 36 − x 2 is split into twelve vertical strips.
p
2
c ( sinq ) dq 6 strips Find an estimate of the area of the semicircle (to 3 s.f.) using
0 the trapezium rule. Hence, obtain an approximate value of p.
2 2
d e −x dx 5 strips 5 Use the trapezium rule with seven ordinates to estimate the
3
1 2
value of I = 1 − x dx to 2 decimal places.
p 9
2 0 2
e cos2 q dq 6 strips x2 + y = 1
−
p Sketch the graph of the ellipse 9 4
2
and use your value of I to estimate its area.
4
8
1 + e-x dx
f 6 strips
6 a Estimate the value of the integral I = (3 x + 1) dx
2
C4

C4
2
0
using the trapezium rule with 8 strips.
p
2 b Calculate the exact value of I.
2 a Estimate the value of the integral I = cos x dx
0
p c Find the percentage error in the estimated value to 1 decimal place.
using the trapezium rule by dividing the interval from 0 to 2
into six strips.
b Find the exact value of I by integration. INVESTIGATION

c Calculate the percentage error in the estimated value of I. 7 Do some research to find a formula for the area of an ellipse.

d Explain, using a suitable diagram, why the answer to part a How does this formula also give you the area of a circle?
is an underestimate of the exact value of I. Use your answer to question 5 to find another estimate for
the value of p.

220 221

10 Integration

Exercise 10.2
10.2 Integration as summation 1 Find the values of these definite integrals.
p p 1 2
You can find the exact area under a curve by summing an 2 2 1 dx
See C2 for revision. a 2cos x dx b p
5sin x dx c ex dx d x
infinite number of infinitely thin rectangles. y 0 4 −1 1

b b
∑dA ≈ ∑ ydx
p 4 4 p
Area PQRS = y = f(x) 4
x + 1 dx (x + 1)2 dx 3
cos x + sec x dx
a a e sin x - cos x dx f g h
Q −p 1 x 1 x 0
cos x
4
Each strip of area dA is approximated by a rectangle of height y and (x, y)
width dx, so its area dA » ydx P 2 Find these indefinite integrals.
y
The Greek letter sigma, å, indicates the sum of many of these rectangles. 1 sec 2 q dq
S R a
2
b x − 2 dx c 5cos x - 3sin x dx
In the limit, as dx ® 0, O x x
a dx b
b
area PQRS = lim ∑ yd x As dx® 0, the number of 3 a Find the area bounded by the graph of y = ex, the two
d x →0 rectangles ® ¥.
a
coordinate axes and the line x = 2
b In the limit, the Greek letters dx
= y dx and å become the English letters b The region bounded by the graph of y = ex, the two coordinate axes
a
dx and respectively. and the line x = k has an area of 2 units2. Find the value of k.

is an elongated letter S (for Sum) 4 a Find the point of intersection of the graphs y
invented by Leibnitz in about 1680. y = cos x and y = sin x for 0 x p
C4

C4
2
Integration is also the reverse of differentiation, so you should 1
b Find the area, A, bounded by the graphs of y = cos x, y = cos x y = sin x
already recognise these basic results:
y = sin x and the x-axis as shown in this diagram. A
O p x
d(sin x) d(e x) = e x c Find the area bounded by the two graphs and the y-axis.
= cos x cos x dx = sin x + c x
e dx = e + c x 2
dx dx
d(cos x) = − sin x
sin x dx = -cos x + c d(ln x) = 1 1 dx = ln x + c 5 Find the area between the graphs of y = ex and y = 1 from x = 1 to x = 2.
dx dx x x x
c is the constant of integration.
d(tan x) = sec 2 x
sec2 x dx = tan x + c 6 If the region under the graph of y = 1 from x = 1 to x = a has
dx x
an area of 4 units2, find the value of a.
A sketch graph will help you to
EXAMPLE 1

Find the area enclosed by the graphs of y = cos x and y = ex visualise the problem.
INVESTIGATION
and the line x = 1 y
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
7 If a temperature is measured discretely n times with the
y = ex
From the sketch, the area required is the area under y = cos x results q1, q2, . . . , qn , then the average (mean) temperature
i
subtracted from the area under y = ex n
1 1 is 1 ∑q i
n
Area = ex dx - cos x dx 1
0 0 1
If the temperature varies with time t such that q = f(t) and
1
= ⎡e x ⎤ − [sin x ]0
1 O x 5
–1 1 it is measured continuously over a period T, then the
⎣ ⎦0 y = cos x mean
T
= (e − 1) − (sin 1 − 0) = 2.718 − 1 − 0.841 = 0.88 You could write this as one integral mean temperature is 1 f(q) dt
1 T 0 t
The area required is 0.88 units2 to 2 d.p. (ex - cos x) dx O p
T
0 Find the mean temperature if q = 5sin t from t = 0 to t = p
sin 1 means the sine of 1 radian
222 223
(not 1°).

10 Integration

Exercise 10.3
10.3 Integration using standard forms 1 Integrate with respect to x.
a cos 5x b sin 4x c sec2 3x
The Core 3 and 4 specifications list the derivatives and integrals
that you are expected to remember. d cos 1 x e cosec2 4x f e4x -3
2
The formulae booklet lists derivatives and integrals that are
g (3x + 2)4 h tan 3x i 1
provided for you in your examinations.
3x − 1
1
Here is a list of standard integral forms where a, b, c and Try to derive these results yourself. j k cot 2x l cos (2x + 3)
n are constants: (3x − 1)2 5
m sec2(4x + 1) n sec 4x tan 4x o sec 4x
1 x n +1 + c (n ≠ −1)
x n dx =
n +1
(ax + b)n dx = 1 × 1 (ax + b)n +1 + c p cosec 4x q e-2x r cos 3x + sin 1 x
a n +1 3
s cosec 2x cot 2x t (e x - e-x)2

cos x dx = sin x + c cos(ax + b) dx = 1 sin(ax + b) + c 2 Evaluate these integrals.
a
p p p
3 8
sin x dx = -cos x + c sin(ax + b) dx = − 1 cos(ax + b) + c a cos 3x dx b sec2 2x dx c sin x dx
a p 2
0 0
2

sec 2(ax + b) dx = 1 tan(ax + b) + c
p 1
sec2 x dx = tan x + c 3
2
sin 3x dx 1 dx
C4

C4
a d e e2x+1 dx f
4 −1 0
3x + 1
0

ex dx = ex + c eax +b dx = 1 eax +b + c
a
3 Integrate with respect to x.
1 + 1
a b 1+ 1
x 2 cos2 x x sin 2 x
1 You can always check
1 dx = 1 ln |ax + b | + c
x dx = ln x + c ax + b a your integration by
c e2x −1 + e1−2x d
sin 3x
differentiating your answer. cos2 3x

See the formulae booklet for other standard integrals. e cos 3x cosec2 3x
EXAMPLE 1

p
3
Evaluate (cos 3x + 4sin x) dx INVESTIGATION
0
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

4 An AC current i varies with time t such that AC means ‘alternating current’.
p
p ‘Rectified’ means that the current
3 i = 5sin wt where w = 3
(cos 3x + 4sin x) dx = ⎡ 1 sin 3x − 4cos x ⎤3 always flows in the same direction;
0 ⎢
⎣3 ⎥
⎦0 The current is rectified so that i = |5sin wt| that is, i > 0 at all times.

(3 3 ) (3 )
Draw the graph of the rectified current for 0 t 2p
= 1 sin p − 4cos p − 1 sin 0 − 4cos 0
Use the ideas of the investigation in Section 10.2 to find
=0−4 × 1 −0+4=2 the mean value of the rectified current for t = 0 to t = p
2 3

224 225

10 Integration

10.4 Further use of standard forms In general, f ′(x) dx = ln|f(x)| + c
f(x)

There are two special cases which use standard forms and the

EXAMPLE 3
chain rule in reverse.
x dx cos x dx
Consider y = (3x2 + 2x + 1)5 Integrate a b
3 sin x − 4
x2 + 1
dy The derivative of x2 + 1 is 2x.
Apply the chain rule: = 5(3x2 + 2x + 1)4 ´ (6x + 2) ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

dx Modify the integral so that you
x dx = 1 2x dx = 1 ln(x 2 + 1) + c
= 5(6x + 2)(3x2 + 2x + 1)4 a
x2 + 1 2 x2 + 1 2 have 2x ‘on the top’.
The derivative of 3sin x - 4
In reverse, (6x + 2)(3x 2 + 2x + 1)4 dx = 1 (3x 2 + 2x + 1)5 + c cos x dx = 1 3 cos x dx = 1 ln |3sin x - 4| + c is 3cos x. You need to have
5 b
3 sin x − 4 3 3 sin x − 4 3 3cos x ‘on the top’.

In general, you can perform an integration of the form Check your answers mentally
by differentiating them using Exercise 10.4
f ¢(x) ´ g[f(x)] dx by sight. the chain rule.
1 Integrate each expression with respect to x.
a 3x 2 b 3x2(x3 - 1)5 c 2x + 3
x3 − 1 x 2 + 3x − 1
EXAMPLE 1

Integrate x cos(x2 + 1) dx d (2x + 3)(x 2 + 3x - 1)4 e x−2 f (x - 2)(x2 - 4x + 1)3
x 2 − 4x + 1
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
The ‘inside’ function f(x) is x2 + 1
cos x
C4

C4
and its derivative is 2x. g h xcos (x2 + 1) i x x2 − 1
Introduce 1 ´ 2 to make a sin x + 1
x cos (x2 + 1) dx = 1 2xcos (x2 + 1) dx = 1 sin (x2 + 1) + c 2
2 2 2
2x on the ‘outside’.
j x k x l xe x
x2 − 1 x2 − 1
2
m xe
−x
n (x + 1) x 2 + 2x + 3 o x +1
EXAMPLE 2

1
x 2 dx x + 2x + 3
2
Evaluate 3
1
(x − 2)2 3
p cos xesin x q x 2e x r
0 x ln x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

1 1 1 p
x 2 dx = 1 3x 2 dx = 1 The ‘inside’ function f(x) is x3 - 2
2 By writing cot q º cos q , find cot q dq and show that
−2 2
3x (x − 2) dx
2 3
and its derivative is 3x2. cot q dq = 1 ln 2
(x 3 − 2)2 3 (x 3 − 2)2 3 sin q p 2
Introduce 1 ´ 3 to make a
0 0 0 4
3
1
⎡ 3 −1
⎤ 1 3x2 on the ‘outside’.
= 1 ⎢ (x − 2) ⎥ = − 1 ⎡ 3 1 ⎤
⎢ ⎥ 3 By writing tanq ≡ sin q , prove that tan q dq = ln |sec q | + c
3⎣ −1 ⎦0 3 ⎣ x − 2 ⎦0 cosq

3( 1 2)
p
= −1 1 − 1 = −1 × −1 = 1 Find the value of
4
tan q dq
− − 3 2 6
0

Now consider y = ln |3x2 + 2x + 1| INVESTIGATION
× (6x + 2) = 26x + 2
dy 1 x dx
Apply the chain rule: = 4 By writing x º x - 1 + 1, find
dx 3x 2 + 2x + 1 3x + 2x + 1 x −1
2 3 n
6x + 2 dx Use long division to rewrite x , x ,..., x
In reverse, = ln |3x2 + 2x + 1| + c x −1 x −1 x −1
3x 2 + 2x + 1 and so integrate each of them.

226 227

10 Integration

10.5 Integration by substitution To evaluate a definite integral, you can
either change back to the variable x and use the original limits of x
You can simplify an indefinite integral using a substitution. or stay with the new variable u, provided you change the limits
on the integral to the corresponding values of u.
EXAMPLE 1

Consider the indefinite integral (3x + 2)5 dx

EXAMPLE 3
p
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
2 cos x dx
Evaluate
Change the variable x by substituting u = 3x + 2: 4 + sin x
0
5
You now have u dx ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

du
You cannot integrate a function of u with respect to x. Let u = 4 + sin x, so dx = cos x and du = cos x dx
Find a substitution for the operator ‘dx’ in terms of u. Either or
When x = 0, u = 4 + sin 0 = 4
Differentiate u = 3x + 2 wrt x: du = 3 p u2
dx 2 cos x dx
= ∫ 1 du When x = p , u = 4 + sin p = 5
4 + sin x u1 u 2 2
Separate the operators du and dx: du = dx 0
3 p 5
2 cos x dx 1 du
The actual values u1 and u2 are not needed. =
You now have (3x + 2) dx = u × du = 1 5 5
u du 5 4 + sin x u
3 3 = [ lnu ]u
u
2 0 4

= [ lnu ]5
1

= 1 × 1 u6 + c Change back to x and use the limits of x: 4
C4

C4
3 6 p
= ⎡ ln(4 + sin x)⎤0
⎣ ⎦
2 Stay with u and use the limits of u.
= 1 u6 + c = ln 5 - ln 4
18
= ln (4 + 1) - ln (4 + 0)
= ln (1.25)
Substitute u = 3x + 2: (3x + 2)5 dx = 1 (3x + 2)6 + c You must give the final = ln (1.25)
18
answer in terms of x.

Exercise 10.5
EXAMPLE 2

Perform this integration x dx This substitution is chosen 1 Use the given substitutions to find these integrals.
x +1 to simplify the denominator
using the substitution u = x + 1 in the integral. a x2(1 + x3)4 dx u = 1 + x3 b x 2 1 + x 3 dx u = 1 + x3
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Rearrange u = x + 1: u2 = x + 1 You could also use the c cos x sin4 x dx u = sin x d sec2 x (1 + tan x) dx u = tan x
substitution u = x + 1. Show
Differentiate wrt x using the chain rule: 2u du = 1 that it gives the same answer.
dx 1 x dx
e dx u = 3x - 1 f u=x+5
Separate the operators: 2u du = dx (3x − 1)4 (x + 5)2

( x x− 1) dx
Rewrite the integral in terms of u: 2
g u=x-1 h ex dx u = ex + 2
x dx
= u − 1 × 2u du = 2 (u2 - 1) du
2
(e + 2)3 x

x +1 u

(3
= 2 1 u3 − u + c = 2 u3 − 2u + c ) 3
Substitute for u and return to x:
x dx 2 (x + 1)3 − 2 x + 1 + c
=
x +1 3

228 229


2 Evaluate each integral using the given substitution. 4 The x-axis is a tangent to the curve y = (x - 1)(x 2)4
at the point P.
a e x 1 + e x dx u = 1 + ex b sin x 1 − cos x dx u = 1 - cos x
y

x dx
c u2 = x - 1 d x 3 1 + x 4 dx u2 = 1 + x4
x −1

(ln x)2 dx 1 dx A
e u = ln x f u2 = x O P x
x 2+ x

1 x
g dx u2 = x + 1 h dx x = sin q
1+ x +1 1 − x2 a Write down the coordinates of P.
b Let u = x - 2 and find the area A enclosed between
1
i dx x = sin q j sec3 x tan x dx u = sec x the curve and the x-axis.
1 − x2

dx dx
5 Find the points where each of these curves meets the x-axis.
k u = ex l x = 2sin q For each curve find the area enclosed between the curve
e x − e −x x2 4 − x2
and the x-axis.
m x + 4 dx u2 = x + 1 a y= x 4− x
x x−2
b y = x (x - 2)2
C4

C4
3 Calculate the values of these definite integrals using the given substitutions. c y = 5cos x sin3 x for 0 x p
p
2 3
a 2cos x esin x dx u = sin x b (x - 1)(x - 2)3 dx u = x - 2 INVESTIGATION
0 2

6 All these integrations can be performed by choosing
1 3 appropriate substitutions.
x
c dx u2 = 1 + x2 d 2x 2x + 3 dx u2 = 2x + 3 Some of them, however, can be done immediately on
1 + x2 1
2
0
sight using f ¢(x) ´ g[f(x)] dx
p2 2 Decide which integrals can be written down on sight.
4 1 cos x dx x dx
e u2 = x f u = 2x - 1 Find all the integrals.
x 1
(2x − 1)4
0
a 3x 2(x 3 + 1)4 dx b 2x3(x2 + 1)2 dx
1 5
g xe x
2
−1
dx u = x2 - 1 h x 2 − 1 dx u2 = x - 1
0 x −1 c x x + 1dx d sec 2 x 1 + tan x dx
1

x x 2 dx
e dx f
x2 + 1 x +1

2
+1
g xe x dx h e x e x − 1 dx

230 231

10 Integration

EXAMPLE 3
10.6 Integration using trigonometric identities Integrate a cos4 x dx b cos5 x dx
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

You can integrate some trigonometric expressions after
( )
2
rearranging them into one of the standard integral forms. a cos4 x dx = 1 (1 + cos 2 x) dx cos2 x = 1 (1 + cos 2x)
2 2
1
= 4 (1 + 2cos 2x + cos2 2x) dx
EXAMPLE 1

cosec x dx
Integrate a b sec2 x sin x dx
sec x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
=4
1
(1 + 2cos 2x + 1 (1 + cos 4x)) dx
2

( 2 + 2cos 2x + 1 cos 4x ) dx
cosec x dx 1 × cos x dx cos x dx 1
a = = = cot x dx 3
sec x sin x 1 sin x The formula book gives =4
2
= ln |sin x| + c cot x dx = ln|sin x|

sec x sin x dx
and 4 2 (
= 1 3 x + sin 2x + 1 × 1 sin 4x + c
2 4 )
b sec2 x sin x dx = cos x
= sec x tan x dx d ( sec x )
dx
= sec x tan x = 3 x + 1 sin 2x + 1 sin 4x + c
= sec x + c 8 4 32

b Write cos5 x as
You could use the substitution
You can integrate powers of sine and cosine by first changing the cos x(cos4 x) = cos x (cos2 x)2 = cos x (1 - sin2 x)2 method with u = sin x. Try this
powers to multiples. method to show that it gives
cos5 x dx = cos x (1 - sin2 x)2 dx
C4

C4
the same result.
EXAMPLE 2

= cos x (1 - 2sin2 x + sin4 x) dx
Integrate a sin2 x dx b cos2 3x dx c (1 + tan x)2 dx
cos x sin2 x dx = 1 sin3 x is of the form
3
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
= (cos x - 2cos x sin2 x + cos x sin4 x) dx
a sin2 x dx = 1 (1 − cos 2x) dx = 1 x − 1 sin 2x + c = 1 x − 1 sin 2x + c
2 2 ( 2 ) 2 4
sin2 x = 1 (1 - cos 2x)
2 = sin x − 2 sin3 x + 1 sin5 x + c
f ¢(x) ´ g[f(x)] dx, as is

3 5
cos x sin4 x dx
b cos2 3x dx = 1 (1 + cos 6x) dx = 1 x + 1 sin 6x + c = 1 x + 1 sin 6x + c
2 2 ( 6 ) 2 12 The simplest examples of a product of sine and cosine involve the
cos 2 3x = 1 (1 + cos 6x) same multiple of the angle.
2

(1 + tan x)2 dx = (1 + 2tan x + tan2 x) dx = (2tan x + sec2 x) dx

EXAMPLE 4
c 1 + tan2 x = sec2 x
Find sin 3x cos 3x dx
tan x dx and sec2 x dx
= 2ln |sec x| + tan x + c
are both standard forms. ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Use the double-angle formula sin 2A = 2sin Acos A:

For higher powers of sine and cosine, the method you use depends sin 3x cos 3x = 1 × 2 sin 3x cos 3x = 1 sin 6x
2 2
on whether the power is even or odd.
Hence, sin 3x cos 3x dx = 1 sin 6x dx
2
For even powers, use the double-angle formulae as many times as is needed.
For odd powers, use sin2 A + cos2 A = 1 as shown in Example 3 part b. =1×
2 ( −
1 cos 6x + c = − 1 cos 6x + c
6 12 )

232 233


If the product involves different multiples of the angle, you can Exercise 10.6
write the product as the sum or difference of sines and cosines 1 Integrate with respect to x.
and then integrate. a cos 3x b sin 4x

(2 )
c cos 1 x (2)
d sin 3x
EXAMPLE 5

Find sin 5x cos 3x dx
e cos (2x + 1) f sin (3x - 2)
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Consider the expansions of sin (A ± B) where Refer to Section 2.7 g sec2 4x h sec2 (2x - 3)
A = 5x and B = 3x : on the CD-ROM .
2 Find these integrals by rearranging into standard forms.
sin (5x + 3x) = sin 5x cos 3x + cos 5x sin 3x
1 dx
and sin (5x - 3x) = sin 5x cos 3x - cos 5x sin 3x a b tan 3x cos 3x dx
cos2 3x
Add the expressions together:
1 dx
sin 8x + sin 2x = 2sin 5x cos 3x c d tan2 x cosec2 x dx
sin 2 4x
1
Hence, sin 5x cos 3x dx = 2 (sin 8x + sin 2x) dx sec 2 x dx
e cot 2x sec 2x dx f
cosec x
=1
2 ( −
1 cos 8x − 1 cos 2x + c
8 2 ) g tan x cosec x dx h (1 + sec x)2 dx
= − 1 cos 8x − 1 cos 2x + c
C4

C4
16 4
i sin 3x(1 + cot 3x) dx j (cosec x + 2)2 dx

(cos x + 1)2 dx
EXAMPLE 6

p k l cot 2x dx
Find
8
sin 6x sin 2x dx sin 2 x sin 4x
0
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

3 Find these integrals.
Consider the expansions of cos (A ± B) where
A = 6x and B = 2x : a cos2 x dx b sin2 3x dx

(2)
cos (6x - 2x) = cos 6x cos 2x + sin 6x sin 2x
and cos (6x + 2x) = cos 6x cos 2x - sin 6x sin 2x c cos2 x dx d sin2(3x + 1) dx

Subtract:
e (1 - tan x)2 dx f (1 + sin x)2 dx
cos 4x - cos 8x = 2sin 6x sin 2x
p p
8
1 8 g sec2 x tan4 x dx h cos3 x dx
Hence, sin 6x sin 2x dx = 2 (cos 4x - cos 8x)dx
0 0
p i sin3 x dx j sin4 x dx
= 1 ⎡ 1 sin 4x − 1 sin 8x ⎤
8
2⎣⎢4 8 ⎥0
⎦
k cos7 x dx l tan3 x dx
(
= 1 1 ×1 − 0 − 0 + 0 = 1
2 4 8 )

234 235


4 Integrate with respect to x. 11 Find the shaded areas in these diagrams where
a sin x cos x b sin 2x cos 2x a y = 2sin x cos3 x b y = 2cos3 x
c tan x cos2 x d sin2 3x cot 3x y
2 2 y
2 2
e tan 3x f cot 3x

g 1 − sin 2x
2
1
( )
2
h
sin 2x 1 − sin 2 1 x
2

p x O x
–p p
O
5 By expanding sin (A + B) and sin (A - B), show that 2 2 2

2sin Acos B = sin (A + B) + sin (A - B)
12 Find the area enclosed by the graphs of y = 2cos2 x and
Hence find sin 6x cos 2 x dx 1
y = 2 cos 3x and the y-axis, as shown in this diagram.
y
6 Find

a sin 4x cos x dx b cos 5x cos 4x dx c sin 3x sin 2x dx y = 2cos2 x

7 Evaluate these definite integrals. O x
y = 1 cos 3x
C4

C4
p p p 2
2 4 4
a sin 3x cos 2x dx b sin 4x sin 6x dx c cos 2x cos 3x dx
0 0 0

INVESTIGATION
8 Evaluate p
p p
p 13 Consider the integral sinn x dx where n is a positive integer.
2 4
2 5 −p
a 1 + sin x dx b cosec x tan x dx c sin x dx
−p −p −p
Use computer software to explore the graph of y = sinn x
2 4
for different values of n.
d Explain your answer to part c in terms of the graph of y = sin5 x
Give a reason why
p
p
2 2 2
a sinn x dx = 0 when n is odd
9 Find cosec x cot x dx using the substitution u = cot x −p
p
4
p p
b sinn x dx = 2 ´ sinn x dx when n is even.
10 Prove that the area between the graph of y = tan x and the −p 0

x-axis from x = 0 to x = p is ln 2.
3

236 237

10 Integration

Exercise 10.7
10.7 Integration using partial fractions 1 Find these integrals.
a 4 dx b x+5 dx c 4x + 1
x(x + 2)
dx
You can use partial fractions to help you to integrate certain algebraic fractions. (x + 1)(x − 3) (2x − 1)(x + 1)
EXAMPLE 1

2x 2 dx 8 dx
x+5 d dx e f
Find dx (x − 1)(x + 3) 4x 2 − 1 x(x 2 − 1)
(x − 1)(x + 2)
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

1 1 9
≡ A + B ≡ A(x + 2) + B(x − 1)
x+5 g dx h dx i dx
Let
(x − 1)(x + 2) x − 1 x + 2 (x + 2)(x − 1) x 2(x − 1) x(x − 1)2 (x − 2)(x + 1)2

Equate the numerators: x + 5 º A(x + 2) + B(x - 1)
2 Evaluate these integrals.
Let x = 1: 1 + 5 = A(1 + 2) + 0 so A = 2 4 3 4
Let x = -2: -2 + 5 = 0 + B(-2 - 1) so B = -1 a 3 dx b x + 2 dx c 1 + 2x dx
3
(x − 2)(x + 1) 2 (x − 1)2 3 (3 + x)(2 − x)
x+5 ≡ 2 − 1
So
(x − 1)(x + 2) x − 1 x + 2
x + 5 dx = 2 − 1 dx 3 Show that the area enclosed by the curve y = 4 , the
Hence, (x + 3)(x − 1)
(x − 1)(x + 2 x −1 x +2 When you work with
= 2ln |x - 1| - ln |x + 2| + c
partial fractions, you often lines x = -4 and x = -5 and the x-axis is ln 5 square units.
get an answer involving 3
logarithmic functions.
= ln (x − 1) + c
2

x+2 4 Find the area between the graph of y = 1
C4

C4
and the
x 3 − 3x 2 + 2x
x-axis from the ordinates x = 3 to x = 4.
EXAMPLE 2

1
x + 1 dx 5 Either rearrange the numerator or use long division before
Evaluate
0 (x − 2)2 integrating these expressions using partial fractions.
x 2 dx x 2 + 1 dx
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a b
Let x +1 ≡ A + B ≡ A(x − 2) + B x2 − 9 x2 − 1
(x − 2)2 x − 2 (x − 2)2 (x − 2)2
Equate numerators: x + 1 º A(x - 2) + B c (x + 2)(x − 1) dx d x 3 + x 2 + 1 dx
Let x = 2: 2 + 1 = 0 + B so B = 3 x(x + 1) x2 − x − 6
Equate coefficients of x: A=1

So x +1 ≡ 1 + 3
INVESTIGATION
(x − 2)2 x − 2 (x − 2)2
x dx
1 1 6 You can evaluate the integral
x + 1 dx = 1 + 3 dx x2 − 1
Hence, by several different methods.
(x − 2)2 0
x − 2 (x − 2)2
0
1 Integrate using
1 + 3(x − 2)−2 dx
= x−2 a a logarithmic standard form on sight
0
1 1 b partial fractions
= ⎡ ln | x − 2| + − (x − 2)−1 ⎤ = ⎡ ln | x − 2| − 3 ⎤
⎢
3
⎥ ⎢ ⎥
⎣ 1 ⎦0 ⎣ x − 2 ⎦0 c the substitution u = x2 - 1

= ln1 − 3 − ln 2 + 3 = 3 − ln 2 ln 1 = 0 d the substitution x = sec u
−1 −2 2
Show that your four answers are equivalent.

238 239

10 Integration

EXAMPLE 2
10.8 Integration by parts Find a x4 ln x dx b ln x dx
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

You can use integration by parts to integrate the product of
a Let u = ln x and dv = x 4
dv 1
two functions. = x 4 so v = x 5 and
dx dx 5
du 1
The method is based on reversing the product rule for differentiation. u = ln x so =
x4 ln x dx = 1 x 5 ln x − 1 x 5 × 1 dx
dx x
Then,
5 5 x
If u and v are both functions of x, then the product rule states:
d(uv) = u dv + v du
dx dx dx = 1 x 5 ln x − 1 x4 dx
5 5

Rearrange: u dv = d(uv) − v du = 1 x 5 ln x − 1 × 1 x 5 + c
dx dx dx 5 5 5

Integrate with respect to x:
= 1 x 5(5 ln x − 1) + c
25

b Let u = 1 and dv = ln x
u dv dx = uv − v du dx dx
dx dx
This example is important.
By thinking of ln x as the product 1 ´ ln x, You can differentiate ln x
du you can use ‘integration by parts’. by sight but you can not
The overall aim is to make sure that v is easier to integrate dx integrate it by sight.
than the u dv that you started with. Then, ln x dx = 1 ´ ln x dx
dx
C4

C4
Your first step in choosing which function is u and which is dv u must be simple to differentiate. dv
dx dv = xln x - x × 1 dx = 1 so v = x and
is crucial for success. dx
dx
must be simple to integrate. x
Keep u Integrate Integrate Differentiate u = ln x so du = 1
dx x
steady dv dv u wrt x = xln x - 1 dx
dx dx
= xln x - x + c

Eu dx dx = u v –Ev du dx
dv
dx

EXAMPLE 3
Use integration by parts to evaluate the definite integral
1
xex dx
EXAMPLE 1

Find x cos xdx 0
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
Let u = x and dv = e x
dx
Let u = x and dv = cos x 1 1
dx 1
Then, xex dx = ⎡ xe x ⎤ -
⎣ ⎦0 ex ´ 1 dx
Then, 0 0
1 1
Keep Integrate Integrate Differentiate Choosing u = x gives a simplified = ⎡ xe x ⎤ − ⎡e x ⎤
⎣ ⎦0 ⎣ ⎦0
x steady cos x cos x x wrt x
final integral because du = 1
dx
= (1 ´ e1 – 0) – (e1 – 1)
=1
Excos x dx = xsin x –Esin x × 1 dx
= xsin x + cos x + c

240 241


In some cases, integration by parts gives you an integral which is Exercise 10.8
still not simple enough to integrate. 1 Find
However, if you integrate the new integral by parts again, you can
a x sin x dx b xex dx
sometimes solve the problem in two stages.

c x ln x dx d xe2x dx
EXAMPLE 4

Find x2 e3x dx
e x sec2 x dx f xe-x dx
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Integrate by parts: ln x dx
g x sin 2x dx h
Let u = x and dv = e3x 2 dv
dx
= e3x so v = 1 e3x
3
x3
dx
and u = x 2 so du = 2x
dx i xe2x+1 dx j x2 ln x dx
x2 e3x dx = x 2 × 1 e3x − 1 e3x × 2x dx
3 3

= 1 x 2e3x − 2 xe3x dx
3 3
k x ln x dx l (
x cos x − p dx
4 )
Now integrate xe3x by parts: m (
x sin x + p dx
6 ) n x tan2 x dx
= 1 x 2e3x − 2 x × 1 e3x − 1 e3x × 1 dx
3 ( 3 33 ) Let u = x, so du = 1, and integrate
dx
e3x using a standard form. o (ln x)2 dx
= 1 x 2e3x − 2 ( 1 xe − 1 × 1 e 3x 3x
+c )
C4

C4
3 3 3 3 3
2 Find
= 1 x 2e3x − 2 xe3x + 2 e3x + c
3 9 27

=e
3
3x
( x2 − 2 x + 2 + c
3 9 ) a xcos nx dx b xenx dx

c xnln x dx d sin nx ln (sec nx) dx
EXAMPLE 5

Find ex cos x dx 3 Evaluate these definite integrals.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
p 2
2
Let u = e and dv = cos x x
You could also solve this problem
a xcos x dx b x3 ln x dx
dx 0 1
using u = cos x and dv = e x
dx
ex cos x dx = ex sin x - ex sin x dx Try it for yourself. 3 2
c ln x dx d x log10 x dx
2 1
Integrate by parts a second time: 4 p

{ }
2
e log10 x dx f xcot2 x dx
= e x sin x − e x(− cos x) − (− cos x) e x dx 2
p
4

1 p
= ex sin x + ex cos x - ex cos x dx 3 x2 2
g x e dx h ex sin x dx
0 0
So, 2 ex cos x dx = ex sin x + ex cos x Rearrange with both integrals on LHS.

x
ex cos x dx = e (sin x + cos x) + c
2
242 243


4 The graph of y = x sin x for 0 x 2p is shown here. 9 Evaluate these definite integrals.
p
y 2
a e2xcos x dx
0
y = xsin x
1
Q b 2x2e-2x dx
O x
p 2p 0
R
2
c x(ln x)2 dx
1

Find area Q and area R.
10 This diagram shows the graph of y = x2e2x
5 This diagram shows the graph of y = xe-x y
Find the position of the stationary value. y = x2e2x
Find the area between the curve, the x-axis and the ordinates
x = 0 and x = 5.
y

y = xe–x
O x
–2
C4

C4
x
O 5
Find
a the y value of the maximum stationary point
6 a Prove that the curve y = x ln x has a minimum and find this b the area enclosed by the curve, the x-axis and
minimum value. the ordinate x = -2.
b Find the area enclosed by the curve and the x-axis.

7 Integrate each of these functions either by parts or by INVESTIGATION
using a substitution of your choice.
11 Let In = sinn x dx
a x(1 + x)4 dx b (x + 1)2ex dx c x x − 1 dx
By writing sinn x as sin x sinn-1 x, use integration by parts
to show that nIn = -cos xsinn-1 x + (n - 1)In -2
8 Find
Use this formula to find sin3 x dx and sin4 x dx
a x2ex dx b x2sin x dx c x2e2x dx
p
Find the value of sin6 x dx
d x2e-3x dx e e2xcos 2x dx f x2cos 3x dx 0

g e3xsin 2x dx

244 245

10 Integration

9 2 sin2 3x 10 4sec2 x 11 5 cos 4x 12 xsin 2x
10.9 A systematic approach to integration
13 x2 ln (2x) 14 3x 15 x(x + 1)4 16 x(x2 + 2)5
When integrating a particular function, you should look to use: x2 + 7
standard integrals to see if the function can be integrated on sight
20 cot 2x
2
the two particular cases f ′(x) dx and f ¢(x)g[f(x)] dx 17 xe2x 18 xe2x 19 2 tan2 2x
cosec 2x
f(x)
the method of substitution
trigonometric identities
partial fractions
(2 ) (2 )
21 tan 1 x cot2 1 x 22 ln (2x) 23 ln (x 2) 24 x + 2
x +1
the method of integration by parts.
25 (x + 3)(x2 + 1) 26 cos 2x + p ( 2 ) 27 tan(2x - p) 28 e2xsin 2x
EXAMPLE 1

Find a x(x3 + 1)2 dx b x(x + 1)7 dx
29 sin 2x cos 2x 30 sin 2x cos 4x 31 5 32 1
(x + 1)(x − 4) x 2 − 2x
Let u = x + 1
c x2(x3 + 1)7 dx d x x + 1 dx
in part d.
33 x 34 e x e x + 1 35 e2x 36 cos x
4
e +1
2x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

x−2 sin x

x(x3 + 1)2 dx = x(x6 + 2x3 + 1) dx (x + 1)8 (x + 1)8
a b x(x + 1)7 dx = x ´ - ´ 1 dx 37 sin 5x cos 2x 38 cos x sin7 x 39 (x2 - 9)-1 40 sin x cos x
8 8
C4

C4
1 1 1
= 8 x (x + 1) − 8 × 9 (x + 1) + c
8 9
= (x7 + 2x4 + x) dx
41 Evaluate
p p
= 1 (x + 1)8(8x − 1) + c
p
= 1 x8 + 2 x5 + 1 x2 + c 72
4 2
1
4
tan x dx
12
8 5 2 a sin2 2x dx b dx c d ∫ 3sin 3x cos 3x dx
0 1
x(2x − 1) sin x 0
0

c x (x + 1) dx = 1 3x2(x3 + 1)7 dx
2 3 7
d x x + 1 dx = (u2 - 1) ´ u ´ 2u du
3
42 Find
= 1 × 1 (x 3 + 1)8 + c = 2 (u4 - u2) du sec 2 x dx 3
3 8 a b tan4 x dx c x 2e x dx d cos5 x dx
tan 3 x
2 2
= 1 (x 3 + 1)8 + c = 5 u5 − 3 u3 + c
24 1 dx sec x tan x dx
3 e sin4 x dx f g h 3x dx
= 2 (x + 1)2(3x − 2) + c x ln x 3 + sec x
15

You could also use integration INVESTIGATION
b
by parts in d.
Try it yourself to show that the two 43 What is a geometrical interpretation of the integral f(x) dx for b > a?
a
methods give the same answer.
Without evaluating these integrals, find whether they are positive,
Exercise 10.9 negative or zero.
1 p
Integrate these functions with respect to x. 1 4
1 dx 5 2
a x−2
b x (1 - x ) dx c sin3 x dx
−p
1 x2(x - 3) 2 x(x - 3)2 3 (x - 6)6 4 3x2(x3 - 2)7 0 0 4

x2
5 x2 x 3 + 1 6 7 x2 cos (2x3) 8 3cos2 x
246 x +1 3
247

10 Integration

EXAMPLE 1
10.10 Volumes of revolution The area enclosed by the curve y = x2 + 1, the x-axis and the
ordinates x = 1 and x = 2 is rotated about the x-axis
The area under a curve y = f(x) between the ordinates x = a through 360°.
and x = b is given by Find the volume of the solid of revolution.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

2 2 y = x2 + 1
b b The volume = py2 dx =p (x2 + 1)2 dx y p is a constant. You can take it
Area A = lim ∑ yd x = y dx 1 1 outside the integral sign.
d x →0 a
a
2
=p (x4 + 2x2 + 1) dx
1
If you rotate a rectangular strip of width dx about the x-axis, the strip
2
generates a thin circular disc. = p ⎡ 1 x5 + 2 x3 + x ⎤
⎢
⎣5 3 ⎥
⎦1
O 1 2 x
The disc is a cylinder, radius y
The volume of this thin disc is dV = py2dx.
y
and thickness d x. 32 16
(
1 2
=p 5 + 3 + 2 − 5 − 3 −1 )
y You can leave your answer as a
178
y = f(x) = 15 p cubic units multiple of p.
y = f(x)

(x, y) dx (x, y)

y y y

EXAMPLE 2
O x x O x Show that the volume of a sphere of radius r is given by 4 p r 3.
C4

C4
a dx b a b 3
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

dx The circle x2 + y2 = r2 has a radius r and a centre (0, 0).
y
Area A volume dV Volume V The shaded semicircle is rotated about the x-axis through
360° to make a sphere of radius r. (x, y)
If you rotate the whole of area A about the x-axis, you generate a Imagine a thin disc of thickness dx and radius y.
solid which is formed by summing an infinite number of thin discs. y
The volume of the disc, dV = py2dx = p(r2 - x2)dx
As the number of discs increases, dx ® 0 and, in the limit, the r –r O x r x
summation gives the exact value of the volume, V, of the solid. So, the volume of the sphere = p(r2 - x2)dx
−r
This is known as a volume of revolution.
r
= p ⎡r x − 3 x ⎤
2 1 3
⎢ ⎥ ⎣ ⎦ −r dx

Volume V = lim
d x →0
b b
∑ p y 2d x = a py2 dx (
= p r 3 − 1 r 3 − r 2(−r) + 1 (−r)3
3 3 )
a
= 4 p r3
3

248 249


EXAMPLE 4
EXAMPLE 3
y
A hollow bowl is formed by a solid of revolution when the The curve with parametric equations x = t 2 + 1, y = t + 1
area between the parabola y2 = 4x and the straight line t t=2
is shown in this diagram. t=1
y = 2 x is rotated 360° about the x-axis.
3 Find the volume of revolution when the shaded area
Find the volume of the bowl. O x
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
bounded by the curve from t = 1 to t = 2 is rotated
about the x-axis through an angle of 2p.
(3 )
2
The curve and line intersect when 2 x = 4x y
y = 2x
3
4x2 = 9 ´ 4x ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

( t)
6 y2 = 4x 2 2 2
4x(x – 9) = 0 dx
The volume V = py2 dx dt = p t + 1 × 2t dt dt
= 2t
x = 0 or 9 dt
1 1 The limits are values of t
y = 0 or 6
(not values of x).
( ) (t )
2 2
The points of intersection are at (0, 0) and (9, 6). 1
+ 2t + 1 dt
x
O 9 Expand the bracket t + t : = 2p 3
t
A solid of revolution is formed from 1

2
thin discs (of radius y = 4x ) with the = 2p ⎡ 1 t 4 + t 2 + ln t ⎤ = 27 p + p ln 4 cubic units
central parts of radius y = 2 x removed. ( 3 ) ⎢4
⎣ ⎥1
⎦ 2

(3 )
2
( )
2
The volume of the disc, dV = p 4x d x − p 2 x d x

EXAMPLE 5
= (p × 4x − p × 4 x )d x
A circle of radius r has parametric equations y
2
9
C4

C4
(x, y)
x = rcos q, y = rsin q
The volume of the solid of revolution
x =9 The shaded semicircle is rotated 360° about the x-axis. y
= lim
d x →0
∑dV Show that the volume V of the sphere generated is 4 p r 3 B A
x =0 3 –r O x r x

(p × 4x − p × 4 x ) dx
9
2
= A slightly quicker method is to
9
0 realise that the ‘hollow’ in the bowl
dx
is a cone of base radius 6 units

(4x − 4 x ) dx
9
2
and height 9 units. ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

=p
0
9 The volume of the cone is thus At points A and B, y = 0, sin q = 0, so q = 0 and p respectively.
1
9 3
p ´ 36 ´ 9 = 108p and you p p
dx
= p ⎡2x 2 − 4 x 3 ⎤
⎢
can subtract it immediately from V= py2 dq
dq = p(r sin q)2 ´ (-r sin q) dq dx = -r sin q
⎣ 27 ⎥0 ⎦ 0
dq
9 0
p
= p (162 – 108 – 0) p ´ 4x dx = 162p
0 = -pr 3 (1 - cos2q)sin q dq Using sin2 q + cos2 q = 1
= 54 cubic units 0
p

= -pr 3 ∫ (sin q - sin q cos2 q) dq
0
Curves with parametric equations p Either recognise the integral
If a curve is specified by parametric equations x = f(t), y = g(t), = −p r 3 ⎡ − cosq + 1 cos3 q ⎤ of -sin q cos2 q on sight or use
⎢
⎣ 3 ⎥0
⎦
then the volume V of the solid of revolution is given by the substitution u = cos q

b t2 The independent variable is t.
= −p r 3 ⎛ 1 − 1 −
⎜
⎝
( 3 ) ( 1 + 1 ) ⎟⎠⎞ =
3
− −
4 pr 3
3
The negative sign indicates that
you have integrated anticlockwise
2 dx
= ∫ p y dt dt
around the circle from A (where
V= py 2 dx The limits of this integral are The volume of the sphere = 4 p r 3 cubic units t = 0) to B (where t = p).
a t1 t = t1 and t = t2. 3

250 251


Exercise 10.10 5 The region R is defined as the area between the curve y = tan x,
1 Each of these three shaded areas is rotated 360° about the x-axis. the x-axis and the line x = p . If R is rotated 180° about the
Find the volumes of revolution. 4
x-axis, find the volume of the solid which is generated.
a b y c y
y
y = 3x2
y = x(3 – x) 6 a Find the volume generated when the area enclosed by the
y = √x – 1 curve y = x2, the x-axis and the line x = 3 is rotated 360°
about the x-axis.

x
b Find the points of intersection of the curve y = x2 and the
O 1 2 4
O 2 3 x line x + y = 12. Find the volume generated when the area
O x in the first quadrant enclosed by the curve, the line and
1 2
the y-axis is rotated 360° about the x-axis.
2 In each case, the region R is bounded by the curve y = f(x), the
7 a Find the points of intersection of the line y = 2x and the
x-axis and the given ordinates. Find the volume generated
curve xy = 8
when R is rotated through an angle of 2p about the x-axis.
b The area in the first quadrant bounded by this line and
a f(x) = x2 - 1, x = 1, x = 2 b f(x ) = 1 , x = 2, x = 3
x curve, the x-axis and the ordinate x = 4 is rotated through
2p about the x-axis. Find the volume generated.
c f(x) = x(x – 2), x = 0, x = 1 d f(x) = sin x, x = 0, x = p
2
8 The area between the two curves y = x3 and y2 = x is rotated
e f(x) = 1 ,
x = 0, x = p f f(x) = e x x , x = 1, x = 2
through 2p about the x-axis. Find the volume of the solid
cos x
C4

C4
4
which is generated.
g f(x) = x − 1, x = 2, x = 3 h f(x) = x ln x , x = 2, x = 3
x
9 The region R is defined as the area enclosed by the y-axis and y
3 The line y = mx passes through the point (h, r). y the two curves y = sin x and y = cos x. If R is rotated through
y = mx 2p about the x-axis, find the volume generated. y = sin x
The area bounded by the line, the x-axis and the
ordinate x = h is rotated 360° about the x-axis (h, r) O x
r y = cos x
to generate a cone of height h.
Find the value of m in terms of r and h.
Prove that the volume of the cone is 1 p r 2h. 10 A metal washer has the shape of a solid of revolution. y
3
O h x The shaded area in this diagram, enclosed between the
curve y = x2 - x + 4 and the line y = 4, is rotated through P Q
4 A solid of revolution is formed by rotating the curve an angle of 2p about the x-axis to form the washer.
y = x 4 − x 2 , where x > 0, a full turn about the x-axis. a Write down the coordinates of the points of
Sketch the graph of the curve and find the volume of the solid. intersection P and Q.
O x
b Find the volume of revolution.

252 253


11 If the area between the curve y = ln x, the x-axis and the 16 That part of the curve, defined parametrically by y
ordinate x = 3 is rotated through 180° about the x-axis, x = y = 1 , which lies between the ordinates of
t 2,
1+t
find the volume generated.
( )
R 1,
2 ( )
1 and S 4, 1 is rotated a full turn about the x-axis.
3 R S
12 A sphere of radius 13 cm is intersected by two parallel planes
O x
7 cm apart. Find the volume of revolution.
The larger intersection is 5 cm at its nearest point from the
centre of the sphere.
Find that volume of the sphere which lies between the two planes. 17 The shape of a rugby ball is called a spheroid. It is the solid
of revolution when an ellipse is rotated about one of its axes.
13 Find the volumes of revolution when the shaded areas on Find the volume of the spheroid which is generated by
these diagrams are rotated about the x-axis through 360°. rotating the ellipse with parametric equations x = acos q,
y = b sin q about the x-axis.
a b y c
y
t=2 18 This diagram shows the curve with parametric equations y

y t=1 x = sin q, y = sin 2q
1
Find the volume of revolution when one of the loops is
t=2 rotated a full turn about the x-axis.
t=1 O x
–1 1

O x x x –1
O O 2 4
C4

C4
19 Find the volume of the solid of revolution when the area
The parametric equations are bounded by the curve x = tan q, y = sin q, the x-axis and the
x = t3 x = t2 + 3 x = 4t ordinates x = 1 and x = 3 is rotated through an angle of 2p
y=t 2
y = 3t + 1 y=2 about the x-axis.
t

14 A curve has parametric equations x = 1, y = t 2 y INVESTIGATION
t
The area bounded by the curve, the x-axis and 1 2
x= ,y=t 20 y
t
the ordinates x = 1 and x = 2 is rotated through an angle
2
of 2p about the x-axis.
Find the volume of the solid of revolution. 2

O 1 2 x
2
O 2 x

15 The part of the curve, defined parametrically by x = t2 - 1, y This circle, with centre (2, 2) and radius 1 unit, is rotated
y = et, from the ordinate of P where t = 0 to the ordinate about the x-axis through an angle of 2p to generate a solid.
of Q where t = -1 is rotated about the x-axis through 360°.
P What name can be given to the solid?
Q
Find the volume of revolution.
O x Can you find an expression for the volume of the solid?

254 255

10 Integration

4 Use standard forms to integrate these expressions with respect to x.
Review 10
a cos 3x b sec2 5x c sin x d (6x + 1)4
4
1 Use the trapezium rule to estimate the value of these integrals e cot 2x f 1 g 1 h e4x
4x + 3 ( 4x + 3)2
correct to 3 significant figures using the number of strips given.

a
2
3x dx 5 strips b
1
e dxx2
6 strips
i (ex + 1)2 j
2( )
cosec 2 x + 1 k sec 2x l sec 2x tan 2x
0 0
5 Find
4 p
4 x dx
c ln(x2 - 1) dx 6 strips d tan q dq 5 strips a 2x(x2 + 3)5 dx b c x cos (x2 + 1) dx d cos x sin6 x dx
2 0 x2 + 3

x − 1 dx sec 2 x dx 2
p
4 e f
1 + tan x
g xe −x dx h x 2 1 + x 3 dx
2a Use the trapezium rule to estimate the value of I = sec2 x dx x 2 − 2x − 1
0
by dividing the interval from 0 to p into 5 strips. 6 Use the given substitutions to find these integrals.
4
b Find the exact value of I by integration. 1
a dx u=x+3 b (x - 1)(x - 4)3 dx u = x - 4
c Calculate the percentage error in the estimated value of I. ( x + 3)2
d Explain, using a suitable diagram, why the answer to x dx
c e x e x + 3 dx u = ex + 3 d u2 = x + 1
part a is an overestimate of the exact value of I. x +1
C4

C4
2

( )
3 This figure shows a sketch of the curve with equation y 2
7 Calculate the value of x dx using the substitution u = x + 2
y = (x - 1)ln x, x > 0 y = (x – 1) In x x+2
1
a Copy and complete the table with the values of 1
2
y corresponding to x = 1.5 and x = 2.5 1
8 Use the substitution x = sin q to find the exact value of 3 dx [(c) Edexcel Limited 2005]
0 (1 − x 2 ) 2
x 1 1.5 2 2.5 3
y 0 ln 2 2ln 3
9 Find these integrals, using appropriate trigonometric identities where necessary.
3 O 1 3 x
tan x dx
b Given that I = (x - 1)ln x dx, use the trapezium rule a cos x cosec x dx b tan2 x cosec2 x dx c
1 1 − cos2 x
i with values of y at x = 1, 2 and 3 to find an approximate
value for I to 4 significant figures d cos2 3x dx e sin2 3x dx f (2 + tan x)2 dx
ii with values of y at x = 1, 1.5, 2, 2.5 and 3 to find another
approximate value for I to 4 significant figures.
c Explain, with reference to the figure shown, why an increase
g (2)
cos4 x dx h (2)
cos3 x dx i sin5 x dx

in the number of values improves the accuracy of the approximation.
3
j sin4 2x dx k (tan4 x - sec4 x) dx l ⎜
⎝ 4 ⎠
⎟( )
⎛1 − sin 2 1 x ⎞ dx

d Show, by integration, that the exact value of (x - 1)ln x dx is 3 ln 3. [(c) Edexcel Limited 2006]
1
2
10 a By expanding cos (A + B) and cos (A - B), show that
2cos Acos B = cos (A - B) + cos (A + B)
Hence, find cos 6x cos 4 x dx
p
4
256 b Use a similar method with sin (A ± B) to find the value of sin 6x cos 4x dx 257
0


11 Integrate using partial fractions. 18 Integrate

a x+3
( x − 1) ( x − 3)
dx b 5x
(2x − 1) ( x + 2)
dx c 7
( x + 1) ( x − 3)2
dx a 1 − sin 2 x dx
1 − cos2 x
b sec 2 x − 1 dx
sin x
c cos3 (4x) dx d (2)
sin 4 x dx

12 Find the value of e (2 - sin2 x) dx f sin 7x sin 3x dx g exsin 3x dx h x2 cos 4x dx
3 2
a 1 b 9 dx
dx
2
( x − 1) (2x − 1) 1
x 2 (3 − x )
19 Evalute
p
2 3 3
sin x
5x + 8 a x(x - 1)5 dx b (x - 1)2ln (x - 1) dx c dx
13 g(x) = 1 + tan 2 x
(1 + 4x)(2 − x) 1 2 p
4
a Express g(x) in the form A + B , where A and B
(1 + 4x) (2 − x)
are constants to be found. 20 The region R is bounded by the curve y = f(x), the x-axis and
the given ordinates. In each case, find the volume generated
b The finite region R is bounded by the curve with equation when R is rotated through an angle of 360° about the x-axis.
y = g(x), the coordinate axes and the line x = 1
2 a f(x) = 4 - x2 x = 1, x = 2 b f(x) = sin x x = p , x = p
4 2
Find the area of R, giving your answer in the form
a ln 2 + b ln 3 [(c) Edexcel Limited 2003] y y
f(x) = 4 – x2
14 Use integration by parts to find
C4

C4
f(x) = sin x
a x cos x dx b x cos 3x dx c xe3x dx d x sin nx dx
R R

e x2e2x dx f e2xsin x dx g x3ln x dx O 1 2 x O p p x
4 2

15 Evaluate these integrals using integration by parts. 21 The area enclosed by the y-axis, the curve y = x2 + 4 and the straight
p p
2
line y = 5x is rotated about the x-axis through an angle of 2p.
4 1 3
a x sin x dx b x2ex dx c excos x dx d x4ln x dx Draw a sketch of the area and find the volume of the solid
0 0 0 1 which is generated.

16 a Use integration by parts to find x cos 2x dx 22 Part of a curve is defined by a range of values of the parameter t.
In each case, find the volume of revolution when the area between
the x-axis and the defined part of the curve is rotated about
b Hence, or otherwise, find x cos2 x dx [(c) Edexcel Limited 2005]
the x-axis through 360°.
a x = t2 + 1, y = t3 1 t 2 b x = et, y = t + 1 0 t 1
17 Choose an appropriate method of integration to find each of these integrals.

a x3(x2 - 1) dx b (x - 2)(x2 - 1) dx c (x - 5)6 dx d 2x(x2 - 2)5 dx 23 This diagram shows part of the curve with y
equation y = 1 + 1
2 x
e x x − 2 dx f x dx g 1 dx h 1 dx The shaded region R, bounded by the curve, the
x2 − 1 x2 − 1 x2 − x x-axis and the lines x = 1 and x = 4, is rotated through R
O 1 4 x
x2 x 360° about the x-axis. Using integration, show that
i xe 2 dx j xe 2 dx k x ln (3x) dx l x2 ln (3x) dx the volume of the solid generated is p 5 + 1 ln 2 ( 2 ) [(c) Edexcel Limited 2003]

258 259

10
Exit
11
Differential equations
solve first-order differential equations of the forms
dy dy dy
Summary Refer to = f(x), = f( y) and = f(x)g(y)
dx dx dx
The trapezium rule gives an estimate of the area between a curve and the x-axis. 10.1 use the method of ‘separating the variables’
The area A between a curve and the x-axis from x = a to x = b is given by use first-order differential equations to solve problems in practical contexts.
b
either A= y dx where the curve has the Cartesian equation y = f(x)
a
t2 Before you start
or A= y dx dt where the curve has parametric equations x = f(t), y = g(t) 10.2
t1
dt You should know how to: Check in:
The volume of revolution when the area between a curve and the 1 Solve simple problems involving 1 a The gradient function of the curve
x-axis from x = a to x = b is rotated 360° about the x-axis is given by gradients of curves. y = f(x) is f ¢(x) = x2 - 5
b
either V = p y2 dx where the curve has the Cartesian equation y = f(x) e.g. A curve passes through the point (1, 5) such If the curve passes through the point
dy
a that = 2x. Find the equation of the curve. (3, -4), find its Cartesian equation.
t2 dx
or V= p y 2 dx dt where the curve has parametric equations x = f(t), y = g(t) 10.10 If
dy
= 2x , then y = x2 + c dy
= (x + 1)2
C4

C4
t1
dt dx b The derivative of a curve is
To pass through (1, 5), 5 = + c 12 dx
To integrate a function, you may need to use If the curve passes through the point
so c = 4
standard integrals which can be integrated on sight 10.3 (2, 5), find its equation.
The equation of the curve is y = x2 + 4
the two particular cases f¢(x) dx and f ¢(x) ´ g[f(x)]dx 10.4
f(x)
the method of substitution 10.5 2 Integrate various functions. 2 Integrate
trigonometric identities 10.6 1 a (2x + 1)3
e.g. Find a 4e-3x dx b dx
partial fractions 10.7 x (x + 2)
b cos 2x + sec2 3x
the method of integration by parts, where u dv dx = uv – v du dx 10.8 a 4e −3x
−3x
dx = 4 × e + c = − 4 e−3x + c
dx dx −3 3 c 1
(x − 1)(x + 3)
Links
b 1
x ( x + 2)
dx = 1
2 ( 1x − x + 2 ) dx
1
d x cos x
Engineers use integration to determine the pressure exerted = 1 (ln x − ln ( x + 2)) + c x
on the vertical gates of a dam by the water. 2 e
x2 + 3
Pressure is defined as the force per unit area. In a fluid, the = 1 ln
2 ( x +x 2 ) + c f x
force exerted on a submerged object increases if either the x −1
density of the fluid, the depth of the object, or the exposed
area of the object increases. The force can also vary at 3 Use exponentials and logarithms. 3 Find x in terms of k when
different points on the object if its shape is not uniform, e.g. If ln (x + 1) - ln x = k, find x in terms of k. a ln (x - 2) = ln x + ln 2 + k
as is the case with many dam gates. (x)
ln(x + 1) − ln x = ln x + 1 = k b 2ln x = ln (x2 + 1) + k
x +1
The force exerted by the water on the gate can be modelled by So = ek c k2 = 1 + ke-x
x
b
F=w xy dy 1= x(ek - 1)
1
a x=
ek − 1
where w is the density of the water, (b - a) is the vertical length of the gate in the water and
x is the horizontal length of the gate at a point at depth y below the surface of the water.
260 261

11 Differential equations

EXAMPLE 1
11.1 First-order differential equations Find the particular solution of the differential equation
dy
x2 − 3x = 1 if y = 4 when x = 1.
dx
A differential equation is a relationship between two (or more) ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

variables and one (or more) of their derivatives.
dy
x2 − 3x = 1
dx
The order of a differential equation is given by the highest
dy
derivative in the equation. Rearrange: x2 = 1 + 3x
dx
E.g. dy 1 + 3x 1 3
= 2 = 2+
dy dx x x x
= x sin x is a first-order differential equation in x and y.
dx
d2 y
Integrate with respect to x:
− 6 dy + 9 = 0 is a second-order differential equation in x and y.
dx 2 dx
y= ( x1 + 3x )dx
2
This chapter considers only certain types of first-order
differential equations. = − 1 + 3 ln x + c This is the general solution.
x
When x = 1, y = 4, so 4 = -1 + 3ln1 + c
Consider the differential equation dy = 2x c=4+1-3´0
dx
=5
You have y = 2x dx To find the solution of the equation,
integrate with respect to x. The particular solution is y = − 1 + 3 ln x + 5
giving y = x2 + c where c is an arbitrary constant. x
C4

C4
This solution is called the general solution.

Different values of c give different solutions but the graphs If dy = f(y), then dx = 1 dx 1
y dx dy f(y) =
of these different solutions all have the same basic shape. dy dy
and integrating with respect to y gives the general solution
The graphs form a family of curves, in which each curve dx
can be formed from any other by a simple translation parallel x= 1 dy = g(y) + c
f(y)
to the y-axis.
(3, 14)
If you choose one member of the family, then that solution
is called a particular solution.

EXAMPLE 2
5 Find the particular solution of the differential equation
For example, consider the graph which passes through the dy
= cos2 y
point (3, 14). O x dx
Substitute x = 3 and y = 14 into y = x2 + c given that y = 0 when x = 1.
14 = 32 + c
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

giving c = 5
dx 1
The particular solution in this case is y = x2 + 5 Rearrange dy = cos 2 y : = = sec 2 y
dx dy cos2 y
Integrate with respect to y:
If dy = f(x), then the general solution is given by
dx The general solution is x = sec2 y dy = tan y + c
y = f(x) dx = g(x) + c
Given y = 0 when x = 1, 1 = tan 0 + c, so c = 1
The particular solution is x = 1 + tan y

262 263

11 Differential equations 11 Differential equations

EXAMPLE 5
EXAMPLE 3
dy dy y
Find the particular solution of (1 − y) + y 2 = 1, if y = 2 Find the particular solution of = 2 if y = 3 when x = 2.
dx dx x −1
when x = 0. ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• •••••••••••••••••••••••••••

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
•

Factorise and use partial fractions:
dy (1 + y)(1 − y) Separate the variables and integrate:
Rearrange: = = 1+ y 1
= 1
dx 1− y dy
= 2 dx
x 2 − 1 (x − 1)(x + 1)
dx 1
y x −1
So =
dy 1 + y

1 dy = ln(1 + y) + c
=1
2 ( x 1− 1 − x 1+ 1 ) dx
Integrate with respect to y: x= This is the general solution.
1+ y
ln y = 1 ( ln(x − 1) − ln(x + 1)) + c
2
When x = 0, y = 2 so 0 = ln (1 + 2) + c
c = -ln 3 ln y = ln x − 1 + ln A Define a new arbitrary constant A
x +1 so that c = ln A
The particular solution is x = ln (1 + y) - ln 3 ⎛ ⎞
= ln ⎜ A x − 1 ⎟
You can now incorporate A within
= ln 1 + y
3 ( ) ⎝ x +1⎠ the logarithm, using the fact that
ln p + ln q = ln (pq)
The general solution is y = A x − 1
ex = 1 + y x +1
3
y = 3e x - 1 Substitute for x and y:

3=A 1
3
C4

C4
You could have used this method so A = 3
dy
If = f(x) × g(y), then the general solution is given by dy
dx on = 1 + y in Example 3. The particular solution is y = 3 x − 1
dx x +1
1 dy = Try it yourself to show that it
f(x) dx
g(y) gives the same answer.
Exercise 11.1
This method is called separating the variables. x and y (with the operators dx 1 Find the general solutions of these differential equations.
dy dy
and dy) are separated onto the a = 3x 4 b = cos 2x
two sides of the equation. dx dx
EXAMPLE 4

dy dy dy
Find the general solution of xy 2 = x2 + 1 c (x 2 + 1) =x d (x + 1) =x
dx dx dx
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Separate the variables and integrate: 2 Find the general solutions of these differential equations.
x + 1 dx
2 dy dy
y 2 dy = a = 4y 4 b = e −2y
x dx dx

= ( x + 1x ) dx c
dy
dx
−y=3 d
dy
dx
= cot y
1 y 3 = 1 x 2 + ln x + c
3 2 3 Find the general solutions of these differential equations.
A new arbitrary constant c¢ has
y 3 = 3 x 2 + 3 ln x + c ′ dy x dy y dy
2 been introduced where c¢ = 3c a = b = c = xy
dx y dx x dx
dy dy dy
d x − y =1 e y = et f t = y + ty
dx dt dt
dx dx
g t2 = x +1 h t = cot x
264 dt dt 265


4 Find the particular solutions of these differential equations. 8 The gradient at the point (x, y) on a curve is given by
dy dy x
a = x 2 + x + 1, given y = 2 when x = 0 =
dx dx y sec x
If the curve passes through the point (0, 2),
dy 1
b = , given y = 2 when x = 1 find its Cartesian equation.
dx y −1

( )
2 dx
c
dy
= x, given y = 4 when x = 1 9 Find the particular solution of the equation (1 + cos 2q ) =2
given that x = 1 when q = p
dx dq
4
2 dy
d (1 + x) = xy , given y =1 when x = 0
2
dy
dx 10 Find the general solution of the equation (y 3 + 1) − xy = x
dx
, given that y = 0 when x = p
dy sec y
e =
dx sec x 2 dy
11 The differential equation xy = x 2 + y 2 can be solved
dy dx
f ex+ y = 1, given y = ln 2 when x = 0 using the substitution y = xz
dx
dy
a Find an expression for in terms of x and z.
dx
5 Use a mixture of methods to find the general solutions of these
differential equations. dy dz
dy
b Eliminate y from xy = x 2 + y 2 and show that xz =1
dy dx dx
a (x + 1) dx = 1 b = e −3y
dx
c Hence, prove that y2 = 2x2ln (ax) where a is a constant.
dy x+2 dy
= tan y
c = d
C4

C4
dx y−2 dx 12 Prove, by using the substitution y = tz, that the general solution
y
dy y (t + y)
e y
dy
−x =1 f x cos y
dy
= sin y of the equation = is Aty = e t , where A is a constant.
dx dx dt t (y − t)

dy dy
g 3x + x = x2 h 2y + y =1
dx dx INVESTIGATION
dy y dy x +x 2
13 Oscillations and waves in, for example, simple
i = j = 2
dx x(x + 1) dx y +y pendulums and electrical circuits, can be modelled
dy dy using differential equations. Architects and structural
k tan x = cot y l (x + 1) − xy = 0
dx dx engineers use differential equations when designing
buildings and bridges to take account of their natural
dy dy
m cos2 x = sin2 x n y = sec y frequencies when these structures sway in the wind. If
dx dx
the oscillations induced by the wind match the natural
dy dy oscillations of the structure, then resonance occurs.
o ex + y2 = 4 p (cos x − sin x) = 2 sin x
dx dx Use the Internet to investigate
the early discovery of resonance;
6 Find the equation of the parabolic curve which passes through the conditions under which resonance leads to instability;
dy
the point (-1, 2) and for which (y − 1) =4 the structural collapse of the Tacoma Narrows Bridge
dx
in the USA in 1940.
7 Show that the curve, which contains the point (3, 7) and for
which dy = 1 + y , has the equation y + 1 = k(x + 1)
dx 1 + x
Find the value of k.

266 267


EXAMPLE 3
11.2 Applications of differential equations The radioactive element strontium-90 has a half-life of In radioactive material, atoms
29 years. Find what percentage of the initial amount of disintegrate spontaneously. The
radioactive strontium is left after a hundred years. rate of disintegration at a given
You can solve problems involving rates of change by forming a time t is proportional to the
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

differential equation and then using integration. amount of radioactive material left
dM dM
You have ∝ M giving = −kM where k is positive. in the sample. The amount of
dt dt radioactive material M is
EXAMPLE 1

The acceleration, a m s-2, of a moving particle depends on the decreasing over time, so the rate
time, t seconds, which has elapsed since the start of the Separate the variables and integrate:
of disintegration dM is negative.
motion. When a = 6cos 2t dM dt
= -k dt
a find the velocity of the particle, v m s-1, in terms of the M
time t, given that the velocity after p seconds is 7 m s-1 ln M = -kt + c = -kt + ln A Let c = ln A, so that ln M - ln A
4
b find the initial velocity.
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
(A)
ln M = −kt giving M = Ae-kt
can be combined.

dv Acceleration is the rate of A half-life of 29 years means that, if the initial amount of
a Acceleration a= = 6 cos 2t change of velocity.
M
dt
strontium is M0, then 1 M0 remains after 29 years. M = MOe–kt
2
Integrate with respect to t: v = 6cos 2t dt = 3sin 2t + c This is the general solution. MO

When t = p , v = 7, so
4
7 = 3sin 2 × p + c ( 4 ) When t = 0, M0 = Ae0 so A = M0 and M = M0e kt
e −29k = 1
1M
c=7-3=4 When t = 29, 1 M 0 = M 0e −29k so 2 O
2 2
The velocity of the particle, v = 3sin 2t + 4 This is the particular solution.
e29k = 2 gives 29k = ln 2 and k = 1 ln 2 O 29 t
29
C4

C4
b When t = 0, v = 3 ´ 0 + 4 = 4
The initial velocity = 4 m s-1 Hence M = M 0e 29( )
− ln 2 t
This graph illustrates the half-life of
strontium as 29 years.
−100 ln 2 −
When t = 100, M = M 0e 29 = M 0e 2.39
EXAMPLE 2

At any given time t, the rate of increase of a population of = 0.0916M0 (to 3 s.f.)
bacteria is proportional to the size of the population, N. The
After 100 years, just over 9% of the strontium is radioactive.
initial population is 50. If the population has increased to 100
when t = 1, find the size of the population when t = 5.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Exercise 11.2
dN
The rate of increase of the population is . 1 As a train enters a tunnel with a velocity of 6 m s-1, its acceleration, a m s-2,
dt
So
dN
∝ N giving
dN
= kN where k > 0 is given by a = 1 t + 1 , where t is the time in seconds spent
dt dt 100 10
in the tunnel.
Separate the variables and integrate:
1 dN = k dt a Find an expression for the speed of the train in terms of t.
N
Let c = ln A, so that ln N - ln A
b If the train leaves the tunnel after 30 seconds, find its speed on exit.
ln N = kt + c
can be combined.
= kt + ln A
(A)
N = kt 2 The population of a certain kind of insect is growing at a rate
ln which is proportional to the number, N, of insects present at a
N = Ae kt given time t (in days), where d N = 0.1N
dt
You have N = 50 at t = 0 and also N = 100 at t = 1 There are two unknown constants,
The population is monitored over time and N = 200 when t = 0.
A and k. You need two items of
When t = 0, 50 = Ae0, so A = 50
information to find their values. a Find the size of the population one week after monitoring has begun.
When t = 1, 100 = 50e k, so e k = 2
Hence N = Aekt = A (e k)t = 50 ´ 2t This is the particular solution. b How long will it take for the initial population to increase tenfold?
When t = 5, N = 50 × = 1600 25
The size of the population when t = 5 is 1600.
268 269


3 Pine trees in a forest are dying due to a fungal disease. Initially there 8 Newton’s law of cooling states that, for an object at a temperature q °C,
were 2000 trees, but, with a rate of infection proportional to the the rate of decrease in its temperature is proportional to the difference
number of trees still unaffected by the fungus, the number unaffected between its temperature and the ambient temperature. An object is
after 2 years is 1600. initially at 70 °C in a room at a constant temperature of 10 °C.
Find how long it will take before half the trees are infected. During the first 10 minutes its temperature falls to 60 °C.

4 A crystal suspended in a chemical solution is increasing in size
a Prove that q = 10 + 60e-kt where k = 1 ln 6
10 (5)
over time. The rate of increase of volume is inversely proportional to b Find how much longer elapses before its temperature falls to 50 °C.
the square of its volume. Initially, the crystal had a volume of 3 cm3
c What will its temperature be one hour after it initially started to cool?
and, one day later, its volume was 4 cm3.
How long will it take for its volume to increase to 10 cm3?
9 The growth of a leaf on a plant depends on the water absorbed
from the plant and the water lost by evaporation from its surface.
5 When the power is switched off in an electrical circuit, it takes
If the width of a leaf is w cm, then the rate of increase of the width
time for the current to stop flowing. At the moment of switching
is equal to 2w - w 2.
off the time t = 0 and the current i = i0. 2t
If the circuit has a resistance R and an inductance L, then a If w = 1 when t = 0, show that w = 2e 2t
1+ e
L di + Ri = 0 where L and R are constants.
dt b Find the maximum width of the leaf.
a Find an expression for i as a function of t in terms of L and R.
1
b How long does it take the current to fall to a value of 100 th of 10 In a chemical reaction in a solution a new chemical is formed.
its initial value? At a time t seconds, y grams of this chemical are present in the solution
dy
C4

C4
Give your answer in terms of R and L. and the rate of increase of the chemical is given by = 2(3 - y)(1 - y)
dt

6 A oil-tank 4 metres tall has a leak. The depth of oil, h metres, in a Given that y = 0 when t = 0, find an expression for y in terms of t.
the tank is decreasing over time at a rate, in metres per hour, b Find the mass, y grams, which has formed 2 seconds after the
which is proportional to the square root of the depth. start of the reaction.
a Given that the tank is full initially and that the initial rate at which
c Find the limiting value of y as t increases.
the depth is decreasing is 16 cm per hour, find the depth of
oil in the tank after 10 hours.
INVESTIGATION
b How long does it take for the tank to empty?
11 Shake 100 drawing pins on to a table.
7 Radium is a radioactive element and it decays at a rate proportional Pins landing point down can be used to simulate radioactive
to the mass M of radium which exists in a sample at a time t. atoms which have decayed in the first second. Remove them.
a If the initial mass of radium in a sample is M0, show that the
Repeat over and over again, removing all the pins which land
mass of radium remaining after a time t is given by M = M0e-kt,
point downwards, until you have only half of the original
where k is a constant.
100 pins left.
b If the half-life of radium is 1620 years, show that the value of k is
The number of shakes that you have made gives you the
approximately 4.28 ´ 10-4
half-life in seconds.
c How long will it take for the initial mass of radium to decay by 99% ?
Construct a mathematical model and check its validity for
predicting the number of shakes required to leave 10 pins.

270 271


7a The acceleration, a m s-2, of a moving particle is inversely
Review 11 proportional to the velocity, v m s-1, at which it is travelling
after a time of t seconds.
1 Find the general solutions of these differential equations. i Find v in terms of t, given that its initial velocity is 10 m s-1
dy dy and, after 5 seconds, its velocity is 20 m s-1.
a x = x3 + 2 b (x 3 − 1) = 3x 2 ii Find its velocity after a further 5 seconds of motion.
dx dx
dy b If the acceleration is directly proportional to the velocity,
c sin x = cos x d dy = y 2
dx dx find its velocity after 10 seconds of motion, given that v = 10
dy dy when t = 0 and v = 20 when t = 5.
e = e −3y f + y =1
dx dx
8 Liquid is stored in a cylindrical tank of radius 5 metres. Algae
2 Find the particular solutions of these differential equations. on the surface of the liquid is growing at a rate proportional to
dy
the surface area, A, which is covered with algae. The algae was
a = 1 − 2x + 2x 2 given y = 4 when x = 3 first noticed when it covered 10% of the surface and, after a
dx
further 10 days, it covered 20% of the surface.
dy
b y = y2 − 1 given y = 2 when x = 1 a Write down a differential equation involving dA .
dx 2 dt

3 Separate the variables to find the general solutions of these Solve the equation to find the area of the surface covered with
differential equations. algae after another 10 days.
dy
b (x 2 + 1)
dy
= xy b How many days in total does it take for 75% of the surface
a y2 = 1 + ex
C4

C4
dx dx to be covered with algae?
dy dy
c sec x = cos2 y d 3y − y =1 9 Newton’s law of cooling states that an object at a temperature q °C
dx dx
cools in such a way that the rate of decrease in its temperature is
dy y dy
e = 2 f y = 2x sec y proportional to the difference between its temperature and room
dx x − 3x + 2 dx
temperature. An object is initially at 70 °C in a room at a constant
temperature of 20 °C. During the first 5 minutes, its temperature
4 Find the particular solutions of these different equations, given falls to 60 °C.
that y = 2 when x = 1.
a Show that q = 20 + 50e-kt and find the value of the constant k.
3 dy
a y =x 2
dx b How many more minutes elapse before its temperature falls to 50 °C?
dy
b x = y +1 10 Chemical A is converted into chemical B during a reaction. At any
dx
time during the reaction, the rate at which A is converted into B
dy y
c cosec(x − 1) = 2 is proportional to the quantity of A that remains at that time.
dx y −1
a The quantity of A at time t is x and, when t = 0, x = x0.
dy
5 A curve contains the point (0, p). Show that, if = x sec 2 y, then Write down a differential equation involving t, x and x0.
dx
the curve has the equation y + sin ycos y - x2 = k b In a particular experiment, the initial quantity of A is reduced by
Find the value of k. a half in 5 minutes. Find how many more minutes it takes for A
to reduce to only 10% of its initial quantity.
6 Given that y = 1 when x = p, solve the differential equation
2
dy
= y 2x cos x
dx
272 273

11
Exit
12Vectors
express vectors in different ways and use the components of a vector to
calculate its magnitude
Summary Refer to
investigate properties of vectors, including how to add and subtract them
dy
= f(x) has the general solution y = f(x)dx = g(x) + c 11.1 calculate the distance between two points and find their midpoint in
dx
dy 1 3-dimensional space
= f(y) has the general solution x = f(y) dy = g(y) + c 11.1 find the scalar product (or dot product) of two vectors
dx
dy 1 calculate the angle between two vectors and the intersection of two lines
= f(x) ´ g(y) has the general solution found from dy = f(x)dx 11.1
find the vector equation of a straight line.
dx g(y)
This method is known as separating the variables. 11.1
The general solution of a first-order differential equation has just Before you start
one arbitrary constant. 11.1
You can represent the general solution graphically by a family of curves. You should know how to: Check in:
You can represent the particular solution of a first-order differential equation 1 Describe a translation using a vector. 1 a Find the vector which maps the point
by just one curve selected from the family of curves. 11.1 e.g. The point (4, 1) maps onto the point (5, 3) (3, 4) onto these points.
C4

C4
⎛1 ⎞ i (4, 6) ii (4, -1) iii (2, 7)
under a translation given by the vector ⎜ ⎟ .
⎝ 2⎠ b Find the image of the square
(3, 3), (5, 5), (3, 5), (5, 3) under the
Links
⎛ 6⎞
An example of a differential equation with translation ⎜ −2 ⎟ .
⎝ ⎠
many applications is the logistic equation
dP 2 Use Pythagoras’ theorem. 2 a Find the distance from the point (0, 0)
= P(1 - P)
dt
where P is a variable dependent on t. e.g. If the two shorter sides of a right-angled to the point (4, 5).
triangle are 8 cm and 6 cm, then the longest b A right-angled triangle has two sides
This differential equation is often used to side is given by 64 + 36 = 10 cm of length 5 cm and 7 cm. Calculate
model population growth, where the rate two possible values of the length of
of reproduction is proportional to both its third side.
the existing population and the
supportability of that population. 3 Solve simultaneous equations in 3 Solve these simultaneous equations.
three unknowns. a 3x + 4y - z = 3
In this setting, the equation now takes the form
e.g. 2x + 3y + z = 8 (1) 2x - 2y + z = 7
dt (
dP = rP 1 − P
K ) 3x - y + z = 1
x + 4y + z = 9
(2)
(3)
x - 3y + 2z = 8
where the constant r defines the growth rate of the b x + y + 2z = 4
population, and K is the supportable population Subtract (2) from (1) to get
2x - 3y + z = 7
within the given environment. -x + 4y = 7 (4)
3x + 2y + 3z = 7
Subtract (3) from (1) to get
x - y = -1 (5)
Add (4) and (5): 3y = 6
y=2
Substitute into (5) and (1):
x = 1, z = 0
274 275

12 Vectors

Unit vectors parallel to the coordinate axes are denoted by ⎛4⎞
12.1 Basic definitions and notations The vector PQ = ⎜ ⎟ can be
⎛ 1⎞ ⎛0⎞ ⎝ 3⎠
i = ⎜ ⎟ , j = ⎜ ⎟ in two dimensions
⎝0⎠ ⎝ 1⎠ written as PQ = 4i + 3j
A vector is a quantity which has both a magnitude (size) and a
direction. For example, velocity and force are both vectors. ⎛ 1⎞ ⎛0⎞ ⎛0⎞
⎜ ⎟, ⎜ ⎟ and
and by i = ⎜ 0 ⎟ j = ⎜ 1 ⎟ k = ⎜ 0 ⎟ in three dimensions.
⎜ ⎟
A scalar is a quantity which needs only a magnitude to describe it ⎜0⎟ ⎜0⎟ ⎜ 1⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
fully. For example, speed and mass are both scalar quantities.

A displacement vector represents a movement from a point P to xi + y j
The unit vector parallel to xi + y j is
a point Q. x2 + y2
The vector is represented by a directed line segment or an arrow.
The length of the arrow represents the magnitude of the vector.
⎛4⎞
For the vector PQ = ⎜ ⎟ with |PQ | = 5, a unit vector in the Check: its magnitude
The vector in the diagram is printed as PQ , PQ or a. Q ⎝ 3⎠

(4) + (5)
2 2
When handwritten, this vector can be written as PQ , or a . same direction will have components one-fifth of those of PQ . = 3
= 16 + 9 = 1
a 5 25
⎛4⎞
The magnitude of the vector PQ (also called its modulus) can ⎜5⎟
The unit vector parallel to PQ is ⎜ ⎟ = 4 i + 3 j
⎜3⎟ 5 5
be printed or handwritten as PQ, |PQ |, |a| or just a. It can also P ⎝5⎠
be printed as a .
Two vectors are equal if they have the same magnitude and
The components of a vector are the movements parallel to the
the same direction.
C4

C4
coordinate axes when the vector is drawn on a Cartesian grid. Q
One vector is the negative of another vector if they have the
The components of this vector are 4 and 3. same magnitude but opposite directions.
3
⎛4⎞
The vector can be written as PQ = ⎜ ⎟ One vector is a scalar multiple of another vector if they have If k is positive, ka and a have the
⎝3⎠ P the same direction but different magnitudes. same direction.
Pythagoras’ theorem gives you 4 If k is negative, ka and a have
2
j In general, the vector ka is parallel to the vector a and has a
|PQ | = 42 + 32 = 25 and |PQ | = 5 i magnitude k times that of a. opposite directions.

⎛ 2⎞ ⎛ 4⎞ −2
In general, in two dimensions e.g. Consider a = ⎜ ⎟ , b = ⎜ ⎟ and c = ⎛ ⎟
⎜
⎞
z ⎝ −1 ⎠ ⎝ −2 ⎠ ⎝ 1⎠
⎛x⎞ You have b = 2a, c = -a and b = -2c b
the magnitude of PQ = ⎜ ⎟ is x2 + y2 Q
y ⎝ ⎠ P
and, in three dimensions, the magnitude of
a c
⎛x⎞
PQ = ⎜ y ⎟ is
⎜ ⎟ x2 + y2 + z 2 O
⎜z ⎟
⎝ ⎠ y
Adding and subtracting vectors
x You can represent two successive displacements by two
The zero vector O has zero magnitude and no direction. R
~ In three dimensions there are vectors PQ and QR.
three axes for x, y and z.
A unit vector is a vector with a magnitude of 1. They are equivalent to one displacement given by the b
r=a+b
The notation for a unit vector uses a ‘hat’ so that â or ~ is the
â vector PR .
The symbol ~ is known as ‘twiddle’. Q
unit vector in the direction of vector a.
a
So PR = PQ + QR P

276 or r=a+b 277

12 Vectors 12 Vectors

R

EXAMPLE 1
PR (or r) is called the resultant vector or, simply, the resultant. Find a unit vector which is parallel to 6i + 8j.
Using the components of the vectors, the resultant of the Find the angle between this vector and the direction
(6 )
5
(2 ) of the x-axis.
⎛6⎞ 3
vectors PQ and QR is ⎜ ⎟ . ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

5 ⎝ ⎠ Q
Sketch a diagram to help you to visualise the problem.
On diagrams, the resultant is
marked by a double arrow. ( )
4
2
If p = 6i + 8j
P then the magnitude of p, p = 62 + 82 = 10 p
8
p 6i + 8j 3
You can change the order of displacements without affecting and the unit vector p =
ˆ = = i + 4j
p 10 5 5
the result. i
So r = a + b = b + a b From the diagram, tan q = 4
r 3 6
When the two vectors are drawn tip-to-tail to make a triangle,
the addition process is called the triangle law of addition.
and the required angle, q = tan −1 4 = 53.1°
3 () x

a

EXAMPLE 2
⎛3⎞ ⎛n⎞
Find the values of k and n if q = kp where p = ⎜ ⎟ and q = ⎜ ⎟ .
⎝2⎠ ⎝8 ⎠
When the two vectors start from the same point to make ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a parallelogram, the addition process is called the ⎛n⎞ ⎛3⎞
parallelogram law of addition. You have ⎜ ⎟ = k ⎜ ⎟
⎝8 ⎠ ⎝2⎠
C4

C4
r
The triangle law and the parallelogram law give the The y-components give 8 = 2k, so k = 4
b
same resultant. The x-components give n = 3k, so n = 12
a

EXAMPLE 3
If p = 2i + j - 3k and q = 4i + 2k
find a |p + q| b |2p - q|
To add several vectors you need to apply the triangle law b ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

several times in succession.
If the vectors are drawn tip-to-tail, you can find the
c ⎛ 2⎞ ⎛4⎞ ⎛ 6⎞
resultant vector from the total displacement. a p + q = ⎜ 1 ⎟ + ⎜ 0 ⎟ = ⎜ 1 ⎟ and |p + q| = 36 + 1 + 1 = 38
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
a ⎜ −3 ⎟ ⎜2⎟ ⎜ −1 ⎟
This process is called the polygon law of addition. ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
As before, the order of the addition does not matter. d
⎛ 2⎞ ⎛4⎞ ⎛ 0⎞
In this diagram, r = a + b + c + d b 2p − q = 2 ⎜ 1 ⎟ − ⎜ 0 ⎟ = ⎜ 2 ⎟ and |2p - q| =
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 0 + 4 + 64
r
⎜ −3 ⎟ ⎜ 2 ⎟ ⎜ −8 ⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ = 68 = 2 17
Subtracting a vector is equivalent to adding the negative of
that vector.
EXAMPLE 4
b Find KL when LN = 2i - 3j + k and KM = 4i - 2j + 3k
That is a - b = a + (-b) given that M is the midpoint of KN.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a N
L
⎛4⎞ ⎛2⎞ ⎛4⎞ ⎛ −2 ⎞ ⎛ 2⎞ KL = KN + NL = 2KM - LN
In this diagram, r = a - b = ⎜ ⎟ − ⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟ = ⎜ ⎟ –b
2 ⎝ ⎠ 3 ⎝ ⎠ 2 ⎝ ⎠ −3 ⎝ −1⎠ ⎝ ⎠ = 2(4i - 2j + 3k) - (2i - 3j + k)
M
r=a–b = 6i - j + 5k

278 K 279


Exercise 12.1 10 If the vectors r and s are parallel, find l and m if
1 Find the magnitude of a u = 4i + 2j – 3k, v = li + 4j + mk
a 5i + 12j b 5i - 12j c i + 2j + 4k d 2i - 3j + k
b u = li + j + 2k, v = 12i + mj – 6k
s
2 If r = 2i - 3j + k, s = 4i - 2j + 2k and t = 2i + k
11 Use this diagram to write these vectors as column vectors. –s
find the magnitude of
a r b s
a r+s b r+s-t c 2s + t d 2r - 3s + 4t r
c r+s d r-s
3 Find a unit vector in the direction of
a 6i + 8j b 10i - 24j c i+j+k d 2i - j + 2k

4 Find a vector, writing it as a column vector, which
a has a magnitude of 5 and makes an angle of 30° with the 12 Using this diagram, write the resultants of these vector q r
positive direction of the x-axis additions as column vectors.
b has a magnitude of 2 and makes an angle of 120° with the a p+q b p+q+r
p
positive direction of the x-axis. c p+q+r+s d p+q+r+s+t
s

5 a Find a vector which has a magnitude 20 and is parallel to 4i + 3j.
b Find a vector which has a magnitude of 5 and is parallel to 2i - 2j - k.
C4

C4
t
6 Find the value of n, given that
a |3i - 3j + nk| = 22 b |ni + nj + 2k| = 4 13 AB = 5i + 7j, CB = 11j and CD = -4i + 6j

7 The vectors p and q are given by p = 3i + 2j and q = 2i + 5j Prove that the vectors AC and BD are perpendicular.
Find
a the angle that p makes with the positive direction of the x-axis 14 PQ = 6i + 3j - 4k and PR = 2i - j – 2k
b the angle between the directions of p and q M is the midpoint of QR. Find the vector PM .

c a unit vector in the direction of p - q.
INVESTIGATION
8 a If LM = 5i + 2j and MN = 4i – j, find LN .
15 Points A and B have position vectors a and b. M is the
midpoint of AB. P is the midpoint of AM. Q is the
b If PQ = 3i + 4j - 3k and RQ = i - 2j + k, find PR.
midpoint of MB. P, M and Q are the points of
c If GE = 2i - j - k and EF = 3i + 2j + 3k, find FG. quadrisection of AB.
a Find the position vectors for M, P and Q in terms of
9 a Find the values of l and m if p = 3i + 2j - 4k, q = 9i + lj +mk a and b.
and q = 3p
b Find the position vectors for the two points of
b Find the value of a if the vectors p = 2i + 5j and trisection of AB.
q = ai - 10j are parallel.
c Can you write down the position vectors for the four
points of quintisection of AB?

280 281

12 Vectors

The midpoint of a line
12.2 Applications in geometry
Points A and B have position vectors a and b.
Position vectors The midpoint M of the line AB has the position vector
The position vector of point A is the fixed vector to the The lower-case notation a is m = 1 (a + b)
2
point A from the origin O. potentially confusing as it can
be used either as a general
It is printed as OA or a and is written as OA or a. vector or as the position vector You can prove this result in two ways.
for point A.
E.g. In two dimensions, for the point A(3, 4),
⎛ 3⎞ First method Second method
the position vector OA = ⎜ ⎟ = 3i + 4j
⎝4⎠
A A C
In three dimensions, for the point B(2, 1, -3),
⎛ 2⎞ M M
⎜ ⎟
the position vector OB = ⎜ 1 ⎟ = 2i + j - 3k a a
⎜ −3 ⎟ m m
⎝ ⎠
B B
b b
The distance between two points O O
E.g. Points A(5, 8, 2) and B(8, 12, 14) have position vectors A (5, 8, 2) Let OACB be a parallelogram
a = 5i + 8j + 2k and b = 8i + 12j + 14k
whose diagonals intersect at M,
The displacement from A to B can be AB direct a the midpoint of both diagonals.
C4

C4
or in two stages via the origin AO + OB .
B (8, 12, 14) OM = OA + AM OM = 1 OC
2
b
You have AB = AO + OB = -a + b O
=b-a
= OA + 1 AB = 1 (OA + AC )
2 2
= (8i + 12j + 14k) - (5i + 8j + 2k)
= a + 1 (− a + b) = 1 (OA +
= 3i + 4j + 12k 2 2 OB )
=a+ 1b − 1a
The distance from A to B = AB = 9 + 16 + 144 2 2 = 1 (a + b)
2
= 169 = 1 (a + b)
2
= 13

EXAMPLE 1
In general, AB = b - a Prove that the points A(0, -1, 2), B(1, 1, 5) and C(3, 5, 11) Collinear points lie on the same
are collinear. straight line.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Using coordinates in three dimensions, the distance from
AB = b – a = (i + j + 5k) – (0i – j + 2k) From the coordinates of A, B and C,
A(x1, y1, z1) to B(x2, y2, z2) is
= i + 2j + 3k the position vectors are
AB = ( x2 − x1 )2 + (y2 − y1)2 + (z2 − z1)2 a = - j + 2k
BC = c – b = (3i + 5j + 11k) – (i + j + 5k) b = i + j + 5k
= 2i + 4j + 6k c = 3i + 5j + 11k
= 2(i + 2j + 3k)
So, BC = 2AB
Hence, BC and AB are parallel and have the point B in
common. So the points A, B and C are in the same
straight line.
282 283


Exercise 12.2 7 The parallelogram OACB has OA = a and OB = b
1 a Find the distance between points A(2, 4, 1) and B(3, 5, 3).
Point R lies on OA and point S on AC such that OR : OA = 1 : 4
b Given point C(4, 8, 1), find the length AC and show that and CS : CA = 1 : 4. Find the vector RS in terms of a and b.
triangle ABC is right-angled.
Deduce two facts relating RS and OC.
2 In triangle P(1, 4, 5), Q(3, -2, 1), R(5, 0, 3), M and N are the
midpoints of sides PQ and PR respectively. 8 The parallelogram OACB has OA = a, OB = b and OC = c
Find the lengths QR and MN. Point P lies on OB such that OP : PB = 1 : 2
Point Q lies on AB such that AQ : QB = 2 : 1
3 For each of the following sets of points A, P and B, show that
Prove that OC and PQ are parallel and find the ratio PQ : OC.
the points A, P and B are collinear.
Find the fraction such that AP = 1 AB
n (3 )
9 L and M are the points 1 , − 1, 2 and (1, 5, 6) respectively.
a A(2, 1, 4), P(3, 3, 5), B(5, 7, 7) The position vector of point P relative to the origin O
b A(-1, 1, -4), P(0, 1, -2), B(4, 1, 6) is i + 3j + 6k. Point R lies on OP such that OR : RP is k : 1
Find the value of k if R, L and M are collinear.
c A(1, -3, 0), P(3, -2, -1), B(9, 1, -4)
d A(3, 0, 1), P(7, 2, 5), B(9, 3, 7) 10 The skew quadrilateral OABC has OA = a, OB = b and OC = c
The midpoints of its four sides in order are P, Q, R and S.
4 M and N are the midpoints of sides OA and OB of triangle OAB. Find the vectors PQ and SR in terms of a, b and c.
C4

C4
B Prove that PQRS is a parallelogram.

N
n
INVESTIGATION
11 a The distance between points A(1, 2, t) and B(t, t, 0)
O m M A is x, where t is a variable.
Find x2 as a function of t and hence find the
Find the vectors MN and AB in terms of m and n. minimum distance between these two points.
Make a deduction about the lines MN and AB. b Use a similar method to find the minimum
distance between P(2, 0, t) and Q(t, t, 3).
5 In triangle OAB, point P lies on AB and points A, B and P
have position vectors a, b and p.
Find p in terms of a and b given that AP : PB is
a 1:2 b 1:3 c 2:3
d 3:5 e m:n

6 In triangle OPQ, OP = p and OQ = q. Point M lies on OP such that
OM : MP = 1 : k and point N lies on OQ such that ON : NQ = 1 : k
Find the vector MN in terms of p and q.
Deduce two facts about MN and PQ.

284 285

12 Vectors

12.3 The scalar (dot) product In component form in three dimensions, you have
a · b = a1b1 + a2b2 + a3b3

The scalar product (or dot product) of two vectors a and b is
a
For a = a1i + a2 j + a3k and b = b1i + b2 j + b3k
a · b = |a| |b| cos q i·j = i·k = j·k = 0
i a · b = (a1i + a2 j + a3k) · (b1i + b2 j + b3k) i · j = |i||j|cos 90° = 0
= a1b1i · i + a1b2i · j + a1b3i · k + a2b1j · i + a2b2j · j + a2b3j · k because cos 90° = 0
where q is the angle between the directions of vectors a and b. b
+ a3b1k · i + a3b2k · j + a3b3k · k
The scalar product is a scalar quantity. |a|, |b| and cos q are all scalar. i·i = j·j = k·k = 1
Angle q will lie between 0° and 180°. Therefore, the scalar product is = a1b1 + a2b2 + a3b3 i · i = |i||j|cos 0° = 1
positive when q is acute, zero when q is 90°, and negative when q is obtuse. because cos 0°= 1

Properties of the scalar product You can calculate the angle between two vectors using
a · b = |a||b|cos q = a1b1 + a2b2 + a3b3
If a and b are perpendicular, then a · b = 0

EXAMPLE 1
b Find the angle between the vectors a = 3i + 2j + k and b = i + 3j - 4k
When a and b are perpendicular, ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

q = 90°, cos 90° = 0 and a · b = |a||b|cos 90° = 0
⎛3⎞ ⎛ 1 ⎞
In particular, i · j = j · k = k · i = 0 a
a × b = ⎜ 2 ⎟ . ⎜ 3 ⎟ = (3 × 1) + (2 × 3) + (1 × −4) = 3 + 6 − 4 = 5
⎜ ⎟ ⎜ ⎟
⎜ 1 ⎟ ⎜ −4 ⎟ a · b = a1b1 + a2b2 + a3b3
⎝ ⎠ ⎝ ⎠
C4

C4
a · a = a2 (la) · b = a · (lb) = l(a · b) a ⋅ b = a b cosθ = 9 + 4 + 1 × 1 + 9 + 16 × cosθ
= 14 × 26 × cos q
As |a||b|cos q = a1b1 + a2b2 + a3b3
The angle between two equal vectors is 0° The length of la is l times the length of a.
cos q = 5 = 5
as they have the same direction. That is, |la| = l|a|. so q = 74.8° to nearest 0.1°
14 × 26 364
a · a = |a||a| cos 0° = a ´ a ´ 1 = a2 So, l(a · b) = l|a||b| cos q = |la||b| cos q
In particular, i · i = j · j = k · k = 1 = (la) · b

EXAMPLE 2
Similarly for the lengths of lb and b,
Given the four points P(2, 4, 1), Q(3, -2, 4), R(2, 0, 1) and S(-1, 2, 6), prove that
|lb| = l|b| and l(a · b) = a · (lb)
the lines PQ and RS are perpendicular.
a·b = b·a •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• •••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Q
PQ = q - p = (3i - 2j + 4k) - (2i + 4j + k) = i - 6j + 3k
By definition, a · b = |a||b| cos q = |b||a| cos q = b · a RS = s - r = (-i + 2j + 6k) - (2i + k) = -3i + 2j + 5k
q R
So PQ · RS = (i - 6j + 3k) · (-3i + 2j + 5k) = -3 - 12 + 15 = 0 P
p r
a · (b + c) = a · b + a · c As the scalar product is zero, PQ and RS are perpendicular.
O s S

V

EXAMPLE 3
By definition, a · (b + c) = |a||b + c| cos q b c
Prove the cosine rule c2 = a2 + b2 - 2abcos C for this triangle.
= OA ´ OV ´ cos q OV cos q = OU B W ••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

= OA ´ OU A
b
b+c
In this triangle, b = a + c and c = b - a.
= OA ´ (OT + BW)
Hence, c · c = (b - a) · (b - a) = b · b - b · a - a · b + a · a c
= OA ´ (OBcos a + BVcos b) ai b
O T a U A
= a · a + b · b - 2a · b
= |a||b|cos a + |a||c|cos b
which gives c = a2 + b2 - 2abcos C
2
= a·b + a·c OA ´ OB cos a = |a||b|cos a B
a
286 OA ´ BV cos b = |a||c|cos b C 287


Exercise 12.3 9 Find the acute angle between LM and LN when L, M and N are the points
1 Find p · q when L(2, 1, 4), M(4, -1, 2) and N(3, 0, 1).
a p = 3i + 2j - 4k q = 2i + 5j - 2k b p = 2i + j - 3k q = 5i - 2j + 2k
10 Find the size of angle EFG, given the three points E(1, -1, 3),
c p=i-j-k q = i + 3k F(2, 1, 3) and G(-1, 0, 4).

2 If p = 2i + 3j - k, q = 4i - j + 2k and r = i + j - k, find the values of 11 Show that triangle PQR is right-angled when its vertices are
a p·q b p·r c p · (q + r) a P(3, -3, 7), Q(4, 0, 2), R(1, -2, 3) b P(1, 0, -1), Q(2, 1, 0), R(3, -1, 1)
d p·q + p·r e p·q - p·r f p · (q - r)
12 Triangle PQR has vertices P(2, 5, 4), Q(1, 6, -2) and R(4, 6, 3). Find Do not round during
your working.
3 Find the acute angle between the vectors p and q when a angles P, Q and R to the nearest 0.1° b the lengths of the three sides
a p = 3i + 4j - 3k q = 2i - j - 2k b p = 2i - j - 2k q = 5i + 3j + k c the area of triangle PQR.
c p = i - j - 4k q = i + j - 2k d p = 4i - 3j + 3k q = i + 2j - k 13 a Show that the vectors u = 3i + j - k and v = i - 2k + j are perpendicular.
e p = 2i - j - k q = i + 3k f p=i+k q = 2j - k b Find a vector r that is perpendicular to both u and v. Use ai + bj + ck
⎛ 2⎞ ⎛0⎞ ⎛ −3 ⎞ ⎛4⎞
14 Simplify
g p = ⎜ 5⎟
⎜ ⎟ q = ⎜1 ⎟
⎜ ⎟ h p = ⎜ 0⎟
⎜ ⎟
q = ⎜5⎟
⎜ ⎟
⎜ −2 ⎟ ⎜2⎟ ⎜ 2⎟ ⎜6⎟ a (p + q) · r + (p - q) · r b p · (q + r) - p · (q - r)
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
c (p + r) · (p - r) d p · (p - r) - (p - r) · r
C4

C4
4 These pairs of vectors are perpendicular. Find the values of l.
15 A semicircle, centre O, contains the vectors a and b as shown. Y
a 3i + j + 4k and 3i - 5j + lk b 4i - lj + lk and i + 3j - 5k
Prove that the angle XYZ is a right angle.
5 For which values of m are these vectors perpendicular? b
⎛ 4⎞ ⎛2⎞ ⎛ 3⎞ ⎛ −3 ⎞
⎜ ⎟ and ⎜ ⎟ ⎜ ⎟ and ⎜ ⎟
b ⎜ m⎟ X a a Z
⎜ m⎟
a ⎜ −2 ⎟ O
⎜1 ⎟
⎜ m⎟ ⎜2⎟ ⎜ −1 ⎟ ⎜ 7⎟
⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 16 The rhombus PQRS has PQ = a and PS = b
Prove that the diagonals of the rhombus are perpendicular.
6 State whether these pairs of vectors are parallel, perpendicular or neither.
⎛ 2⎞ ⎛ 8⎞ ⎛ ⎞ 17 Find a vector which is perpendicular to both
⎜ ⎟ ⎜6 ⎟ ⎛ 3⎞ ⎛ 2⎞ ⎛6⎞
b ⎜ 2 ⎟ , ⎜ −4 ⎟
i + 2j + 3k and 2i - 3j - 8k
a ⎜ 1⎟, ⎜ 2⎟ c ⎜− ⎟ ⎜ ⎟
⎜ 4 ⎟ , ⎜3⎟
⎜ 2⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎜ −1 ⎟ ⎜ −4 ⎟ ⎜1⎟ ⎜ 1⎟
⎜ ⎟ ⎝ ⎠
⎜ 5⎟ ⎜0⎟
⎝ ⎠ ⎝ ⎠ ⎝2⎠
⎝ ⎠ ⎝ ⎠ 18 Find the angle between vectors a and b when
|a| = 1, |b| = 2 and |a - b| = 3
7 If p = 2i + 3j and q = li + 2j, find the value of l such that
19 Triangle OAB has vertices A(1, 2, -2), B(6, 8, 0) and the origin O.
a p and q are perpendicular b p and q are parallel
Find the value of cos AOB and prove that the area of triangle OAB = 2 26
c the angle between p and q is p .
4

8 a Find the angle between 2i + 3j + 12k and INVESTIGATION
i the x-axis ii the y-axis iii the z-axis.
20 The tetrahedron ABCD has two pairs of opposite edges perpendicular.
b Find the angle between i + j + k and each of the three coordinate axes. Prove that the third pair of opposite edges is also perpendicular.

288 289

12 Vectors

EXAMPLE 2
12.4 The vector equation of a straight line Find the vector equation of the line through the points
P(2, 3, 0) and Q(3, 5, 1).
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

Consider a straight line through point A parallel to vector b b
as in this diagram. R The direction of the line is
A Q
PQ = q - p = (3i + 5j + k) - (2i + 3j) = i + 2j + k
Point A has a position vector a with respect to the origin O. P
The line passes through point P, so the vector equation of
Any other point R on the straight line has a position vector r. a r
the line is
r = p + t(q - p) = (2i + 3j) + t(i + 2j + k) p q
Since the line AR is parallel to b, then the vector AR is a
multiple of b. Alternatively, as the line passes through Q, the vector
O
That is, AR = tb, where t is a scalar. equation could also be
r = q + s(q - p) = (3i + 5j + k) + s(i + 2j + k) O
A can be any point on the line and
The vector equation of the straight line through point A in b can be any parallel direction, so
the direction b is the equation r = a + tb is not In general, the vector equation of the line through points
r = OA + AR = a + tb unique. One line can have many
vector equations.
P and Q is r = p + t(q - p)

Compare this equation with the cartesian equation of a straight line, y = mx + c.

EXAMPLE 3
Show that the two vector equations r = i + 2j + 3k + t(2i - j + 2k)
Given A(a1, a2, a3), R(x, y, z) and b = b1i + b2j + b3k, then the and r = 7i - j + 9k + s(4i - 2j + 4k) both describe the same
vector equation of the line in component form is
C4

C4
straight line.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

⎛ x ⎞ ⎛ a1 ⎞ ⎛ b1 ⎞ ⎛ a1 + tb1 ⎞ The two equations give lines with directions 2i - j + 2k and
r = ⎜ y ⎟ = ⎜ a2 ⎟ + t ⎜ b2 ⎟ = ⎜ a2 + tb2 ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ 4i - 2j + 4k.
⎜z ⎟ ⎜a ⎟ ⎜ b ⎟ ⎜ a + tb ⎟
⎝ ⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝ 3 3⎠ ⎛ 4⎞ ⎛ 2⎞
and any point R on the line has coordinates b t=3
Because ⎜ −2 ⎟ = 2 ⎜ −1⎟ these two vectors are parallel
⎜ ⎟ ⎜ ⎟
t=1
t=2 ⎜ 4⎟ ⎜ 2⎟
(a1 + tb1, a2 + tb2, a3 + tb3). t=0 ⎝ ⎠ ⎝ ⎠
t = –1
t = –2 A and so the lines have the same direction.
As t varies, R moves along the line. The second line passes through the point (7, -1, 9) when s = 0.
a r = a + tb
When t = 0, R and A coincide. Points on the first line are given by (1 + 2t, 2 - t, 3 + 2t) and,
When t is positive, R is on one side of A and when t = 3, this gives the point (7, -1, 9).
when t is negative, R is on the other side of A.
O Hence, the two lines pass through the same point in the same
direction. The two equations represent the same line.
EXAMPLE 1

Find the vector equation of the straight line through the
point A(3, 2, -1) in the direction of 4i - j + 2k.
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• Intersection of straight lines
The vector equation is r = 3i + 2j - k + t(4i - j + 2k) r = a + tb In two dimensions, two straight lines can coincide (that is, they
are the same line), they can be parallel to each other, or they can
An alternative form of the equation is
intersect each other.
r = (3 + 4t)i + (2 - t)j + (-1 + 2t)k
In three dimensions, two straight lines can coincide, be parallel,
Using column vectors with R(x, y, z), you can also write
intersect or be skew.
⎛x⎞ ⎛ 3⎞ ⎛ 4⎞
the equation as ⎜ y ⎟ = ⎜ 2 ⎟ + t ⎜ −1 ⎟
⎜ ⎟ ⎜ ⎟
⎜ ⎟
When two lines are skew, one passes above the other without Skew lines in 3D
⎜ z ⎟ ⎜ −1 ⎟ ⎜ 2⎟ intersecting it. There is still an angle between them.
⎝ ⎠ ⎝ ⎠ ⎝ ⎠
290 291


EXAMPLE 4 2 Find a vector equation of the line through the points P and Q when
a Determine whether the lines given by
a P(2, 3, 1) Q(3, 0, 4) b P(2, -1, 1) Q(5, 0, 1)
r = 2i - j + 4k + t(i + j - k) and r = i - 2j + 3k + s(2i + 2j - k)
intersect or are skew. c P(1, 0, 0) Q(4, 5, -2) d P(0, 0, 0) Q(1, 2, 3)
If they intersect, find the point of intersection. e P(1, -2, -3) Q(2, -1, -2) f P(2, -3, -1) Q(5, 0, -1)
b Find the angle between the two lines, giving your answer
in degrees to 1 d.p. 3 Find the vector equation of the line through the point A(2, 1, -2) which is
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

a parallel to the x-axis b perpendicular to the xy-plane.
a The equations of the lines can be written as
r = (2 + t)i + (-1 + t)j + (4 - t)k 4 Show that the given point lies on the given line and find the
and r = (1 + 2s)i + (-2 + 2s)j + (3 - s)k corresponding value of t.
They will intersect if the coefficients of i, j and k are equal. a (4, 1, -1) and r = 2i - 3j + k + t(i + 2j - k)
You need to find values of s and t to make b (9, -4, 1) and r = -i + j + 6k + t(-2i + j + k)
2 + t = 1 + 2s (1)
-1 + t = -2 + 2s (2) c (3, 2, 0) and r = (3 - t)i + (2 + 3t)j - tk
4-t=3-s (3)
5 Determine whether these pairs of lines are parallel, intersect or are skew.
Add (2) and (3): 3=1+s so s=2 Find the acute angle between the lines. If they intersect, find their point of intersection.
Substitute into (2): -1 + t = -2 + 2 ´ 2 so t=3 The values s = 2 and t = 3 a r = 3i + 5j + 7k + t(i + 2j - 3k) r = 4i + j + 2k + s(3j + k)
satisfy all three equations. b r = 6i - 2j + 8k + t(i - 5j + 7k) r = i + 3j + 2k + s(3i + 5j - 8k)
Substitute in (1): LHS = 2 + 3 = 5 and RHS = 1 + 2 ´ 2 = 5
The two lines do intersect.
c r = 2i - 4j + k + t(2i - j - k) r = 3i + j - 6k + s(4i - 2j - 2k)
C4

C4
Substitute into one of the vector equations: If all three equations were not
r = 2i - j + 4k + 3(i + j - k) = 5i + 2j + k satisfied, then the lines would
d r = 2j - k + t(4i - 2j + 3k) r = 4i + j - 2k + s(i + 4j + 4k)
not intersect.
so the point of intersection is (5, 2, 1).
e r = 2i - k + t(6i + j + 8k) r = 12i - 15j - 5k + s(9i + 3j + 4k)
b The directions of the two lines are given by
6 a Find the vector equations of the two lines PQ and RS where the
b1 = i + j - k and b2 = 2i + 2j - k
four points are P(-1, 1, 3), Q(8, 7, 6), R(0, 5, 2) and S(-2, 7, 0).
The angle q between the lines is found from b1 · b2 = | b1|| b2| cosq
b Show that the two lines intersect and find the point of intersection.
1 × 2 + 1 × 2 + (−1) × (−1) = 1 + 1 + 1 × 4 + 4 + 1 × cos q
b1 = 12 + 12 + (−1)2 = 3 c Find angle PRS to the nearest 0.1°
5= 3 × 9 × cos q
cos q = 5 b2 = 22 + 22 + (−1)2 = 9 7 a Find the vector equation of the line through the points A(1, 2, 3) and B(2, -1, -1).
3 3 Show that the point R(1 + t, 2 - 3t, 3 - 4t) lies on this line for all t.
The acute angle between the two lines is 15.8°
b C is the point (5, -2, -6). Find the value of t that makes the vectors CR and AB perpendicular.
c Find the shortest distance from point C to line AB.
Exercise 12.4
1 Find the vector equation of the line through point A INVESTIGATION
which is parallel to vector b, when
8 Using the parameter l, find the general point R on the line joining
a A(1, 2, -3) b = 2i - 3j + k b A(4, -1, 3) b = -2i + j + 2k
points A(1, 2, 3) and B(0, 1, 2).
c A(3, 0, 1) b = 4i + 2k d A(1, -1, 0) b=i+j-k Repeat using the parameter m for the general point S on the line joining
⎛1 ⎞ ⎛2⎞ C(0, -2, 2) and D(1, 0, -1).
⎜ ⎟
e A(2, 1, 0) b = ⎜0⎟
⎜ ⎟ f A(0, 0, 1) b = ⎜0⎟ Find l and m so that the line RS is perpendicular to both AB and CD.
⎜3⎟ ⎜0⎟
⎝ ⎠ ⎝ ⎠ Find the shortest distance between lines AB and CD.

292 293

12 Vectors

8a Write down the vector equation of the line L1 which passes
Review 12
through point P(1, -2, 4) and is parallel to the vector 2i + 3j -k.
b Find the vector equation of the line L2 which contains point P
1 Find a the magnitude of the vector a = 3i + 12j + 4k and also point Q(0, 1, 3).
b the unit vector parallel to vector a c Find the angle between the two lines L1 and L2.
c a vector of magnitude 5 which is parallel to vector a Hence, or otherwise, find the shortest distance from Q to L1.

d the angle between a and the direction of the x-axis. 9a Show that the point A(7, 8, 1) lies on the line L with the vector equation
⎛ 3⎞ ⎛2⎞
2 The points A, B and C have coordinates (1, 4, 3), (3, 6, 6) and ⎜ ⎟
r = ⎜ 0 ⎟ + t ⎜ 4 ⎟ , but that the point B(1, -4, 0) does not lie on L.
⎜ ⎟
(7, 10, 12) respectively. Find the vector AB and show that the ⎜ −1 ⎟ ⎜1⎟
⎝ ⎠ ⎝ ⎠
points A, B and C are collinear.
b Find the acute angle between L and the line through A and B.
3 A triangle has vertices L(3, 1, 0), M(1, -2, 5) and N(2, 5, 2). c Find the shortest distance from B to L.
a Prove that the triangle is right-angled and find its area.
10 a Show that the two lines with these vector equations intersect.
b Find point K such that the quadrilateral KLMN is a parallelogram.
⎛ 3⎞ ⎛ 2⎞ ⎛ 1⎞ ⎛ 1⎞
c Show that the midpoint of KM coincides with the midpoint of LN. ⎜− ⎟ ⎜ ⎟ and r = ⎜ 4 ⎟ + m ⎜ −1 ⎟
r = ⎜ 1⎟ + l ⎜ 1 ⎟ ⎜ ⎟ ⎜ ⎟
⎜ 2⎟ ⎜ −1 ⎟ ⎜ −7 ⎟ ⎜ 2⎟
⎛ 1⎞ ⎛ 4⎞ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
C4

C4
4a ⎜ ⎟
Show that the vectors ⎜ 2 ⎟ and ⎜ −1⎟ are perpendicular.
⎜ ⎟
⎜ −2 ⎟ ⎜ 1⎟ b Find their point of intersection.
⎝ ⎠ ⎝ ⎠
b Find the acute angle between the two vectors i + 2j - 2k and 3i - j - k. c Find the acute angle between the two lines.

5a Given points A(-1, 0, 2), B(2, 4, 1) and C(0, 5, -3), find the 11 Two straight lines have the vector equations r = i - 2k + t(-2i + j + 2k)
and r = 3i - 4j + k + s(i - j + k)
vectors AB and BC . Hence, use the scalar product to find
angle ABC to the nearest degree. a Find the acute angle between the directions of the two lines.

b D is the point (1, 1, k). Find the value of k such that the lines b Do the two lines intersect or are they skew?
AB and BD are perpendicular.
12 a Find the values of t and s which show that the point
6 Find a unit vector ai + bj + ck which is perpendicular to the P(-2, 6, -1) lies on the line r = i - 4k + t(-i + 2j + k) and
two vectors 2i + 2j - k and 4i + 2k. that point Q(1, 7, 0) lies on the line r = 8j + 2k + s(-i + j + 2k)
b Prove that PQ is perpendicular to both lines.
7 Relative to a fixed origin O, the point A has position vector 3i + 2j - k,
the point B has position vector 5i + j + k, and the point C has c Find the shortest distance between the two lines.
position vector 7i - j. d Find the angle between the two lines.
a Find the cosine of angle ABC.
b Find the exact value of the area of triangle ABC.
c The point D has position vector 7i + 3k.
Show that AC is perpendicular to CD.
d Find the ratio AD : DB. [(c) Edexcel Limited 2003]

294 295

Revision 4

12
Exit
1 Express these as partial fractions.
6 3x − 1 x3
a b c
x(x + 2) (x + 1)2(x − 3) x −4
2

2 f(x) = 11 + 2x + x 2 =
2
A + B + C x <1
Summary Refer to (1 − 2x)(3 + x) 1 − 2x 3 + x (3 + x)2 2
Equal vectors have the same magnitude and direction. 12.1
a Find the values of A and C. Show that B = 0
The vector ka is parallel to vector a and has a magnitude k times that of a. 12.1
The vector p = xi + yj + zk has a magnitude (or modulus) |p| = x 2 + y 2 + z 2 12.1 b Hence, find an expansion of f(x), in ascending powers of x up to
and including the term in x3. Simplify your answer as far as possible.
p xi + y j + z k
The unit vector parallel to p is p =
ˆ = 12.1
p x2 + y2 + z 2
3 y
The position vector of point A(a1, a2, a3) relative to the origin O
is given by OA = a = a1i + a2 j + a3k 12.2
If points A and B have position vectors a and b, then AB = b - a 12.2
The midpoint M of AB has a position vector m = 1 (a + b) 12.2
2
The distance between points A(a1, a2, a3) and B(b1, b2, b3) is given by R
C4

C4
AB = ( a1 − b1 ) + ( a2 − b2 ) + ( a3 − b3 )
2 2 2
12.2 O x
2
The scalar (or dot) product a · b = |a||b| cos q = a1b1 + a2b2 + a3b3
where q is the angle between a and b. 12.3 The diagram shows part of the curve with equation y = f(x), where
The vector equation of the line through point A and parallel to
f(x) = x2 + 1 , 0 x<3
vector b is r = a + tb 12.4 (1 + x)(3 − x)
The vector equation of the line through points P and Q is r = p + t(q - p) 12.4 a Given that f(x) = A + B + C , find the values of the
1+ x 3− x
constants A, B and C.

Links b The finite region R, shown in the diagram, is bounded by the curve
To ensure safety during aircraft landing, the aviation industry with equation y = f(x), the x-axis, the y-axis and the line x = 2
uses vectors to guide the aircraft to the runway. Find the area of R, giving your answer in the form p + qln r,
where p, q and r are rational constants to be found. [(c) Edexcel Limited 2002]
As an aircraft approaches an airport, the Instrument Landing
System (ILS) uses a series of radars, which are sent up from
4 A curve has parametric equations x = tan 2 t y = sin t, 0 < t < p
the perimeter of the runway, to guide it to a safe landing. 2
dy
Vectors and the dot product are used to determine if the a Find an equation for in terms of t. You do not need to simplify your answer.
dx
aircraft has intercepted the area covered by the beams. b Find an equation of the tangent to the curve at the point where t = p
4
Suppose that the aircraft has position vector s and moves with Give your answer in the form y = ax + b, where a and b are constants to be determined.
velocity v, and that p is the position vector of a point P in the
area covered by the radar beams. Then, if the dot product c Find the Cartesian equation of the curve in the form y2 = f(x) [(c) Edexcel Limited 2007]
v·(s - p) is positive, the aircraft is moving towards P.
When this product is 0 the aircraft is passing P and when it is
negative it is moving away from P.

296 297


5 y 7 y
P
C
1
R
Q
–1 O 1 x O x

–1
The diagram shows a sketch of part of the curve C with parametric equations
x = t 2 + 1, y = 3(1 + t)
The curve shown in the diagram has parametric equations
x = cos t y = sin 2 t 0 t < 2p The normal to C at the point P(5, 9) cuts the x-axis at the point Q,
as shown in the diagram.
dy
a Find an expression for in terms of the parameter t.
dx a Find the x-coordinate of Q.
dy
b Find the values of the parameter t at the points where =0
dx b Find the area of the finite region R bounded by C, the line
c Hence, give the exact values of the coordinates of the points PQ and the x-axis. [(c) Edexcel Limited 2005]
on the curve where the tangents are parallel to the x-axis.
8 y
d Show that a Cartesian equation for the part of the curve
where 0 t < p is y = 2x 1 − x 2
1a A
e Write down a Cartesian equation for the part of the curve 2
C4

C4
B
where p t < 2p [(c) Edexcel Limited 2003] O a x

6 y (metres)
The curve shown in the diagram has parametric equations
x = a cos 3t, y = a sin t, 0 t p
C
6
R The curve meets the axes at points A and B as shown.
A The straight line shown is part of the tangent to the curve at the point A.
B x (metres)
Find, in terms of a,
This diagram shows a cross-section R of a dam. The line AC a an equation of the tangent at A
is the vertical face of the dam, AB is the horizontal base and the
curve BC is the profile. b an exact value for the area of the finite region between the curve,
Taking x and y to be the horizontal and vertical axes, then A, B and the tangent at A and the x-axis, shown shaded in the diagram. [(c) Edexcel Limited 2006]

C have coordinates (0, 0), (3p 2, 0) and (0, 30) respectively.
The area of the cross-section is to be calculated. 9 Expand as a series of ascending powers of x up to and including x3.
Initially the profile BC is approximated by a straight line. State the range of values for which each expansion is valid.

a Find an estimate for the area of the cross-section R using a 1 b 31+ 1x c (1 − x ) 1 + x
this approximation. 1 − 3x 2

The profile BC is actually described by the parametric equations 2+x 3
d e
x = 16t − p ,
2 2
y = 30 sin 2t, p
4
t p
2
4−x (
(1 − x) 1 + 1 x
2 )
b Find the exact area of the cross-section R.
c Calculate the percentage error in the estimate of the area of
the cross-section R that you found in part a. [(c) Edexcel Limited 2004]
298 299


10 a Write down the first four terms of the binomial expansion, 15 The volume of a spherical balloon of radius r cm is V cm3,
( )
n
in ascending powers of x, of 1 − 1 x , where n < 1. where V = 4 p r 3
3 3
State the range of values of x for which the expansion is valid.
a Find dV
dr
b x3
Given that the coefficient of in this expansion is four times The volume of the balloon increases with time t seconds according
the coefficient of x2, find
i the value of n to the formula dV = 1000 2 , t 0
dt (2t + 1)
ii the coefficient of x4 in the expansion.
b Using the chain rule, or otherwise, find an expression in terms
of r and t for dr
dt
11 a Prove that, when x = 1 , the value of 1 − 3x is exactly equal
12 c Given that V = 0 when t = 0, solve the differential equation
to cos 30° dV = 1000 , to obtain V in terms of t.
dt (2t + 1)2
b i Expand 1 − 3x , x < 1 , in ascending powers of x
3
up to and including the term in x3, writing your answer d Hence, at time of t = 5
in as simple a form as possible. i find the radius of the balloon, giving your answer
ii Use your expansion to find an approximation for cos 30° to 3 significant figures
ii show that the rate of increase of the radius of the
c Find the percentage error in your approximation for the balloon is approximately 2.90 ´ 10-2 cm s-1 [(c) Edexcel Limited 2006]
value of cos 30°
16 a Copy and complete this table by finding the three missing values of y,
dy given that y = sec x
12 Find , in terms of x and y, when
C4

C4
dx
p p 3p p
a y2 = xy + 2 x 0 16 8 16 4

b x3 - 3x2y + y 3 = 1 y 1.01959 1.08239

c 1 + 1 + 1 =1
b By using the trapezium rule with all five y-values in the table,
x 2 xy y 2 p
4
d x ln y = y ln x find an estimate for sec x dx
0
Show all your working, giving your answer to 4 decimal places.
13 A set of curves is given by the equation sin x + cos y = 0.5
dy c Use a standard integral to show that the exact value of
a Use implicit differentiation to find an expression for p
dx
ln(1 + 2 )
4
sec x dx is
b For -p < x < p and -p < y < p, find the coordinates of the 0
dy
points where =0 [(c) Edexcel Limited 2007]
dx d Find the percentage error in the estimate that you obtained
using the trapezium rule.
14 a Given that y = 2x, and using the result 2x = exln 2, or otherwise,
dy 17 Evaluate using an appropriate method.
show that = 2 x ln 2
dx
2
b Find the gradient of the curve with the equation y = 2(x ) at the a e x e x − 2 dx b (x − 1)(x + 2)7 dx c 2 cos2 x dx
point with the coordinates (2, 16). [(c) Edexcel Limited 2007]
d cos x sin 4 x dx e 9x 2 dx f (x + 3) e x dx
(x − 1)(x + 2)2

g ex cos 2x dx h x 2 dx i 1 tan x sin 2x dx
x +3
3
2
300 301


1
y
18 Use the substitution u = 2x to find the exact value of 2 x dx [(c) Edexcel Limited 2007]
23
0
(2 + 1)2
x

19 a By using the formulae for sin(A ± B), with A = 5x and B = 2x,
show that 2sin 2xcos 5x can be written as sin lx - sin mx, O x
where l and m are positive integers. State the values of l and m.
b Hence, or otherwise, find sin 2x cos 5x dx
3p
4 A table top, in the shape of a parallelogram, is made from two types
c Hence find the exact value of sin 2x cos 5x dx
p of wood. The design is shown in the diagram. The area inside the
4
ellipse is made from one type of wood, and the surrounding area
20 a Use the identity for cos (A + B) to prove that is made from a second type of wood.
cos 2A = 2cos2 A - 1 The ellipse has parametric equations
b Use the substitution x = 2 2 sin q to prove that x = 5cos q, y = 4sin q, 0 q < 2p
6
( 8 − x 2 ) dx = 1 (p + 3 3 − 6 ) The parallelogram consists of four line segments, which are tangents to
2 3 the ellipse at the points where q = a, q = -a, q = p - a, q = -p + a
c A curve is given by the parametric equations a Find an equation of the tangent to the ellipse at (5cos a, 4sin a),
x = sec q, y = ln(1 + cos 2q), 0 < q < p and show that it can be written in the form 5ysin a + 4xcos a = 20
2
C4

C4
Find an equation of the tangent to the curve at the point b Find by integration the area enclosed by the ellipse.
where q = p [(c) Edexcel Limited 2003]
c Hence show that the area enclosed between the ellipse and
3
80
the parallelogram is sin 2a − 20p
21 a Show, by using the substitution x = sin q, that, for |x| < 1,
d Given that 0 < a < p , find the value of a for which the areas
1 dx = x + c, where c is an arbitrary constant.
3 1
(1 − x 2)2 (1 − x 2)2 4
of the two types of wood are equal. [(c) Edexcel Limited 2002]

b Use integration by parts to show that the exact value of
4 y
24
x2 ln x dx can be written as 8 (p ln 2 - q),
2
9
where p and q are integers. 2
Find the values of p and q. R

O 2 x
22 a Use integration by parts to show that

( ) ( ) ( )
This diagram shows part of the curve with equation y = x2 + 2
xcosec x + p dx = −x cot x + p + ln ⎡sin x + p ⎤ + c,
2
⎢
−
p <x<p
6 6 ⎣ 6 ⎥
⎦ 6 3 The finite region R is bounded by the curve, the x-axis and the
lines x = 0 and x = 2.
b Solve the differential equation
a Use the trapezium rule with four strips of equal width to estimate
( 6 ) dx
sin2 x + p dy = 2xy (y + 1) the area of R.

to show that 1 ln
2
y
y +1 6 (⎢
⎣ )
= −x cot x + p + ln ⎡sin x + p ⎤ + c
6 ⎥
⎦ ( ) b State, with a reason, whether your answer in part a is an
under-estimate or over-estimate of the area of R.
c Given that y = 1 when x = 0, find the exact value of y when x = p [(c) Edexcel Limited 2005] c Using integration, find the volume of the solid generated when R is rotated
12
through 360° about the x-axis, giving your answer in terms of p. [(c) Edexcel Limited 2002]
302 303


25 y 28 In an experiment a scientist considered the loss of mass of a
collection of picked leaves. The mass, M grams, of a single leaf
was measured at times t days after the leaf was picked.
1
The scientist attempted to find a relationship between M and t.
O a b x
In a preliminary model she assumed that the rate of loss of mass
was proportional to the mass M grams of the leaf.
The curve shown in the diagram has the equation y = 1
2x + 1 a Write down a differential equation for the rate of change
The finite region bounded by the curve, the x-axis and the lines of mass of the leaf, using this model.
x = a and x = b is shown shaded. This region is rotated through b Show, by differentiation, that M = 10(0.98)t satisfies this
360° about the x-axis to generate a solid of revolution. differential equation.
Find the volume of the solid generated. Express your answer
as a single simplified fraction, in terms of a and b. [(c) Edexcel Limited 2008] Further studies implied that the mass, M grams, of a certain leaf
satisfied a modified differential equation
26 a The curve C1 in Figure 1 has parametric equations y 10 dM = −k (10M − 1) (1)
dt
x = 2 , y = 3t2 C1
where k is a positive constant and t 0
t
The area bounded by the curve, the x-axis and the
c Given that the mass of this leaf at time t = 0 is 10 grams, and that
ordinates x = 2 and x = 4 is rotated through an
its mass at time t = 10 is 8.5 grams, solve the modified differential
angle of 360° about the x-axis.
equation (1) to find the mass of this leaf at time t = 15. [(c) Edexcel Limited 2003]
Find i the values of t when x = 2 and x = 4 O 2 4 x
ii the volume of the solid of revolution. Figure 1
C4

C4
29 Fluid flows out of a cylindrical tank with constant cross-section.
At time t minutes, t 0, the volume of fluid remaining in the
b The curve C2 in Figure 2 is defined parametrically by y tank is V m3. The rate at which the fluid flows, in m3 min-1, is
x = t + 4t, y = 1
2 proportional to the square root of V.
2+t
The region between the x-axis and that part of C2 from C2
a Show that the depth, h metres, of fluid in the tank satisfies the

( 3) ( 4)
the point P 5, 1 to the point Q 12, 1 is rotated P
Q
differential equation dh = −k h, where k is a positive constant.
dt
a half-turn about the x-axis. b Show that the general solution of the differential equation may
O x
Show that the volume of the solid generated is p ln 4
5 12 be written as h = (A - Bt)2, where A and B are constants.
3 Figure 2
c Given that, at time t = 0, the depth of fluid in the tank is 1 m, and
27 a Find the general solutions of these differential equations. that 5 minutes later the depth of fluid has reduced to 0.5 m, find
dy dy the time, T minutes, which it takes for the tank to empty.
i = e 2y ii (x + 1) =y
dx dx
d Find the depth of water in the tank at time 0.5T minutes. [(c) Edexcel Limited 2003]
dy dy
iii + xy = x 2 y iv cos y = sec 2 x sin y
dx dx
30 Points P(2, 3, -1), Q(4, -1, 4) and R(2, -1, 3) are the vertices of a triangle.
b Find the particular solutions of these differential equations.
2
a Find the vectors PR and PQ and the angle between them, to
⎛ dy ⎞
⎜ dx ⎟ = yx , given that y = 4 when x = 2 the nearest tenth of a degree.
2
i
⎝ ⎠
dy b Find the lengths PR and PQ, giving your answers in surd form.
ii (x 2 + 1) − xy = 0, given that y = 10 when x = 1
dx c Find the area of triangle PQR, correct to 3 significant figures.

304 305

Revision 4

31 Relative to a fixed origin O, the point A has position vector 4i + 8j - k
and the point B has position vector 7i + 14j + 5k.
Answers
a Find the vector AB .
Before you start Answers 3x + 7 2(2 y 2 − 2 y − 1)
b Calculate the cosine of ÐOAB. o ( x + 1)( x + 2)( x + 3)
p ( y − 1)( y − 2)( y + 2)
Chapter 1
c Show that, for all values of l, the point P with position vector 2x3 − 8x − 2
1 a 11 b 1
q z
r
li + 2lj + (2l - 9)k lies on the line through A and B. 24 4 (z + 1)(z + 3) x ( x 2 − 4)
2 a y
d Find the value of l for which OP is perpendicular to AB. y = x(x – 2)
3 a
-x b
e Hence find the coordinates of the foot of the perpendicular y
from O to AB. [(c) Edexcel Limited 2002] y −1

c
32 Referred to an origin O, the points A, B and C have position vectors O
2
x x2 − x + 1
(9i - 2j + k), (6i + 2j + 6k) and (3i + pj + qk) respectively, where x − 21
d x + x +1
p and q are constants. b y y = x2 + x – 2 x

Exercise 1.2
a Find, in vector form, an equation of the line l which passes
through A and B. 1 a x2 - 6x + 8 b x2
–2
O x - 2x - 24
Given that C lies on l, c x2 - 10x + 25 d
–2
b find the value of p and the value of q 2
x - 8x + 15
c calculate, in degrees, the acute angle between OC and AB. c y e x2 + 2x - 3 f 3x2 +
C4

C3
1
y=x+2
The point D lies on AB and is such that OD is perpendicular to AB. x+2
2
d Find the position vector of D. [(c) Edexcel Limited 2002] g n2 - 3n + 2 h n2
x
- 3n - 2
1O
33 The points A and B have position vectors 2i + 6j - k and 3i + 4j + k. –
2 i 3y2 - 2y - 4 j 2a3 +
The line L1 passes through the points A and B. 2
a + 3a - 1
3 a y = -x2 b y = x2 + x
a Find the vector AB. c y = x2 + 11x + 30 d y = 4x2 k 4a3 - 16a2 + 24a - 6
l 4z - 5
b Find a vector equation for the line L1. Exercise 1.1
m 3x - 4 n x2 - x
2x 6a2
A second line L2 passes through the origin and is parallel to the vector i + k. 1 a b bc
c 6pr +1
z
The line L1 meets the line L2 at the point C. 4 1 4
d 1
e a
f 1 2 a 1+ b 1− c 1−
c Find the acute angle between L1 and L2. ( x − 1)x a −1 x2 −4 x+2 x+2 x+2

1 7 1
d Find the position vector of the point C. [(c) Edexcel Limited 2008] g 1 h 1 d 3+ e 2+ f 2+
x−3 x x+2 x−3 2x + 1

2 4 3
34 Lines L1 and L2 are given by the equations xz + y 2 a2 + b2 b−a g 1− h 3+ i −1 +
2 a yz
b ab
c abx x2 + 1 x2 − 1 x +1
L1: r = i - 2k + l(2i + 4j - k) j 1+ x +1
x+y 2x + 1 2
d e x( x + 1)
f 2 x −2
L2: r = 8i + 5j - 10k + m(-i + j + 2k), where l and m are parameters. x2 y2 a2 −1
a+2 3x + 1 3 a 3 − 52 b 3 + 211
2x + 1 x −3
a Show that L1 and L2 intersect and find the coordinates of the g h i x +1
x
(a + 1)2 x
point of intersection. c 2+ 3
d 7− 2
2y + 8 2x + 3 3x + 5 2x2 − 1 2x2 + 3
j 3( y − 2)
k l
b Show that L1 and L2 are mutually perpendicular. x2 − 1 x( x + 1)
4 a x2 + 6x + 1 rem 2 b x2 +
z −4 1 5x - 2 rem 2
c Show that the point P (3, 4, -3) lies on L1. m n
z2 − z − 2 y +1

d P is reflected in L2. Find the coordinates of the image of P.
306 307

Answers Answers

b y g y c y 10 a a = 1 b i ±5 ii 2, -1 iii 4
21
c −1, 1
2
10
11 a 2 b -1
1 12 p = ±1, q = ±2
13 a i 4x + 9 ii 8x + 21 iii 16x + 45
b 2nx + 3 (2n - 1)
−2 O 2 x 1
14 The range of g(x) is not a subset of the domain of
−1
O x f(x). Change domain of g(x) to 0 x 5
1 3
O1 x
6 Range y Î R, y 1 Exercise 1.4
Range y Î R, y ¹ 1 d y 1 a x+2 b x −1 x
−1
Range y Î R, 1 < y < 21 3 2
c
2
h y 2
d 2x - 3 e 2(x - 3) f x −1
c y 1
4 g x −1 h 6-x i 2
x +3
8 x
O 1
2 a f −1 (x ) = x − 1
5 2
y
Range y Î R, y 2
O x O 2 6 x 5 a x = 2 maps onto two values of y
1 4
Range y Î R, 0 y 4 b x = 2 does not map onto a value of y
Range y Î R, 5 < y < 8 6 y
One-to-one mappings: a, b, c, e, g
Many-to-one mappings: d, h 1 1
–2
2 a 3, 2 b 1, -2, 1
d y O x
3 a {0, 5, 10, 15, 20} one-to-one 1
C3

C3
1
b x Î R, -3 x 3 many-to-one 1
–2
c x Î R, 3 x 12 one-to-one
d x Î R, 3 x 6 one-to-one x
−4 O 2
4 a y 1
2 b f −1 (x ) = 5 − x
2
−2
O x y
7
Range y Î R, y ¹ 1 10
Range y Î R, y 2 y = f (x)
3 x = 4 and -1 are unchanged
7 a 10 b 19 c 2x2 + 1
d 5 e 26 f 4x2 - 4x + 2 5
e X Y O x
2 g 101 h x4 + 2x2 + 2 i 9
0 0 y = f –1(x)
1 1 Range y Î R, y 3 j 4x - 3
8 O 5 10 x
4 2 y
b
9 3 fg(x) gf(x) f2(x) g2(x)
13
Range y = {0, 1, 2, 3} c f -1(x) = 2x - 4
a 9x2 + 24x + 14 3x2 - 2 x4 - 4x2 + 2 9x + 16 y
1 3
b +2 x 27x4 + 36x2 + 14
f X Y 3x2 + 2 x2 y = f (x)
0 0
1
c 2x + 2 2x + 10 ( x + 18) 16x - 10
1 –1 4
2 y = f –1(x)
1 2x − 3 x −1
4 –2 d 1− x x −1 2− x
x
–4 O 2 x
−4 O x 1
9 –3 9 a 5− x b 5− 1 c d 5− 1 –4
−3 x x x
Range y = {0, -1, -2, -3} Range y Î R, y -3 1
1
e f x g x4 h x
5−x

308 309

Answers Answers

1 b y
d f −1 (x ) = x − 2, x 2 i f −1 (x ) = , x >2 11 x = 3 ± 10
x −2 y = f (x)
y y
y = f (x) y

y = f –1(x) 5
4
y = f (x) 3
2 2 y = f –1(x)

O 2 x y = f –1(x) O x
45 1 O 3 x
–
Range of f(x) is y Î R, y 5 3–1
O 2 x 3
e f −1 (x ) = x + 2, x 0 f -1(x) = 4 − x − 5
y 3 a f -1(x) = 8 - x, x Î R self-inverse Domain of f -1(x) is x Î R, x 5
Range of f -1(x) is y Î R, y 4 12 c = 2, x = 2 ± 5
b f −1 (x ) = 12 , x ∈ R, x ≠ 0 self-inverse
y = f –1(x) x c y 13 x = a ± a 2 + b
c f −1 (x ) = 4 − x 2 , x ∈ R, 0 x 2 self-inverse y = f –1(x) 14 y
2 y = f (x) 4 5
d No inverse function exists
O x
2 e f -1(x) = 8 - x, x Î R, 0 x 8 self-inverse 2

f f -1(x) = x2 + 4, x 0 f f (x ) = x , x ∈ R, x ≠ 1
−1
self-inverse y = f –1(x)
x −1 –1
y f (x) 1 y = f (x)
O 2 4 y = f (x) x
y = f –1(x) 4 f -1(x) = 8 - x, x Î R, x 8
f(x) and f -1(x) do not have the same domain. Range of f(x) is y Î R, y < 4 x
–5 –1 O 1 5
f -1(x) = 2 + 4 − x –1
5 1± 2
Domain of f -1(x) is x Î R, x < 4
C3

C3
4 y = f (x)
6 y Range of f -1(x) is y Î R, y > 2
y = f (x)
d y
O 4 x
y = f (x) –5 y = f (x)
y = f –1(x)
4
5 x +1
g f -1(x) = x2 - 3, x 0 f -1(x) = x − 1
3
y
2 Domain x Î R, x -1, x > 1
–1
y = f (x) y = f –1(x) Range y Î R, y 0, y ¹ 1
O 3 5 x
x
15 f3 has an inverse.
–2 O 2 4
y = f (x) All the others are many-to-one functions.
Range of f(x) is y Î R, y 5
f (x) = 3 + x − 5
-1 –2 Exercise 1.5
–3 O x
Domain of f -1(x) is x Î R, x 5 1 a y
Range of f(x) is y Î R, y > -2
–3 Range of f -1(x) is y Î R, y 3
f -1(x) = x + 2 − 2
7 a y Domain of f -1(x) is x Î R, x > -2
Range of f -1(x) is y Î R, y > -2 3
h f (x ) = 2 − 1 , x < 0
−1
x y = f (x)
8 a f -1(x) = 2x - 8, x Î R, x = 8 (8, 8)
y y = f –1(x) O x
3
b f (x ) =
−1
x, x ∈ , x 0 x = 0, 1 (0, 0), (1, 1)
y = f –1(x)
b y
2 c f −1 (x ) = 2 + x, x ∈ , x 0 x = 4 (4, 4)
1 d f −1(x ) = x + 4 − 4, x ∈ , x −4
2
O x x = −3, (−3, −3)
1 2
9 a a = -2, b = 0, c = 1 O x
O 2 x Range of f(x) is y Î R, y 1 –3 3
y = f (x) b i 4 ii 48 iii 3
f -1(x) = 2 + x − 1 –3
2
Domain of f -1(x) is x Î R, x 1 10 g −1 (x ) = 2 + , x ∈ R, x ≠ 0 solutions are x = 1 ± 3
x
Range of f -1(x) is y Î R, y 2

310 311

Answers Answers

c y j y 3 a y 4 y
3 10

6

–1 O 4 x
1 x
–3 –1 O 1 3
x O 3 8 x
O 1
–4
2 k y
Range 0 y 10
y y = f (x)
Solutions x = 1 or 5
d
3 5 y
y
2 7

4 5
1O 1 x O 1 x
–
2 2 3
l y
–1 –1 O 4 x
y = |f (x)|
3
y
e y 2 O x
1
–4 –2 1
2
Range 0 y 7
4 –1 O 1 x
Solutions x = -1 or -4
–4 O 4 x 6 a y
m y
O 4 x –4
C3

C3
f y –p O p x
y = f (|x|)
b y
4 b y
O x

n y –p O p x
–4 O 4 x
c y
O x O 4 x

g y
–p O p x
o y
d y
y = f (x)
3 y

–p O p x
1O x
–1 O 1 x
2 7 (2, 3)
2 a y 8 (-2, 0), (6, 8)
h y 9 4 solutions; (-6, 16), (0, 4), (2, 0), (4, 4)
O 4 x
1
–1 5 y = |f (x)| Exercise 1.6
2
O x y 1 a i -1, 9 ii -1 < x < 9
b i 1 ii x<1
–3 2 a i -7, 3 ii -7 < x < 3
–4 O 1 x b i -5, 2 ii x < -5, x > 2
b y c i −2
1
,3 ii x 21, x 3
i y –4 O 4 x 3 3
d i 2, 4 ii 2 x 4
3 3 e i 1 ii x > 1
y = f (|x|) f i 1 1
,1 ii 1 < x < 1 1
4 2 4 2
1

O 1 3 x O 1 2 x
312 313

Answers Answers

1 1 b y 5 y b y
3 a −2 b −1 c 5 d -9, 1
2 2 y = |x + 2| y = x2 – 3x
y = |x| 5
e 11 f 6, 2 g 3, 6 h -3, 0, 5 y = 5 – x2
2 3 y = |x – 2|
2 2
4 a x > 2, x < 0 b − <x<2 c 11 < x < 3
3 2
d xÎR e 2.56 x 4
–2 O 2 x O 1 3 4 x
f -3 < x < -1 and 1 < x < 3 O x
4 a y
y = x2 – 3x
5 a −2 2, 1 − 5, 2, 4 b x < −4 2 , x > −2
3
6 b2 4c y = –x2 + 5x – 4
6 a y
Exercise 1.7 y = x2 – 4 c y
1 y 4 y = |x – 4|
O 3 x
y = √x + 3 2
y= (2 ) – 4
x

3 y = √x + 3
4 y= | x + 2 – 4|
3
y = √x y = –x2 + 3x
y = √x – 3 –4 –2 O 2 4 x O x
b y 4 10
2 2
y = x + 3x y = x – 3x

–3 O 3 x d y
y = 1 – sin x
2
2
y=4–x 1
C3

C3
–3 y = sin x
–3 O 3 x
b y –p p O p p 3p 2p x
– −1
8 2 2 2
y = √x – 3
y = 2(4 – x2) e y y = 3 – sin x– p ( )
c y 2
2 y y = 3 + cos x
y = x2 – 3x 4
y = (x + 3)2 y = x2 + 3 y = x2 y = (x – 3)2 3
4
2
1 y = sin x

3 O x p 3p 2p x
2 3 5 –2p – 3p –p– p O p
2 2 −1 2 2
–2 O 2 x
y = x2 – 7x + 10 8 a First: Stretch parallel to y-axis with scale factor of 2
2
y=4–x
–3 O 3 x y ⎛0⎞
d Second: Translation of ⎜ ⎟
y = x2 – 3x ⎝1⎠
7 a y
y = x2 – 3 y = x2 – 4x + 3 ⎛ −p ⎞ ⎛0⎞
–3 b Translation of ⎜ 4 ⎟ and translation of ⎜ ⎟ in
⎜ ⎟ 3 ⎝ ⎠
3 ⎝ 0 ⎠
either order
3 a y x c First: A stretch parallel to x-axis with scale factor
–1 O 3 4
y = |2x| = 2|x|
of 1
2
O 1 2 3 4 5 x
y = |x| Second: A reflection in the x-axis
⎛0⎞
–4 Third: A translation of ⎜ ⎟
⎝ 3⎠
d A stretch parallel to the y-axis with scale factor of
O x y = x2 – 3x – 4
⎛p ⎞
4 and a translation of ⎜ 2 ⎟ , in either order.
2
y = –2x + 8x – 6 ⎜ ⎟
314
–6 ⎝0⎠ 315

Answers Answers

e A stretch parallel to the y-axis with scale factor 2 11 a Two stretches, both with scale factor k, one parallel 9 a y c y
and a stretch parallel to the x-axis with scale factor to the x-axis and the other parallel
1 to the y-axis
, in either order. O 3 x
(k)
3
y = f(x) y = kf x 4
f A reflection in the y-axis and a stretch parallel to
–2
1 b Two reflections, one in the x-axis and the other in
the y-axis with a scale factor of , in either order.
2 the y-axis O 4 x
9 y y = 1 + tan x – p
( ) y = f(x) y = -f(-x)
Turning point (3, 0)
y = tan x 2
b y d y

Review 1 4
1 4x + 1 2(2 − x 2 ) 6x
1 a b c 2
( x − 1)(2 x + 3) ( x + 1)( x + 2) ( x − 4)( x 2 − 1)
–4 O 4 x
4 a 3b 3m3 1
O p p x d a−b e 2 f
m −1 2 x( x − 1) O 3 x
2
–1 2( x + 2) 1 Turning point (3, 2) e
2 a b x = − or 1 y
( x + 1)( x − 3) 3
y = tan x c y
3 a i 2x2 + x - 3 rem 0 ii x2 + 2x - 3 rem 2
b (x - 1)(2x2 - x - 1) + 3 2 4
10 a i A stretch parallel to the x-axis of scale factor 2 c (x - 2)(x - 3)(x + 3)
⎛0⎞
and a translation of ⎜ ⎟ . 4 a i 8 ii 5 iii 24
⎝1⎠ –3 O 3 x O x
iv 15 v 2 vi 2 1 4
ii y
b i 4x (x - 1) ii 2x2 - 3 iii x +1 Turning points (±3, 2)
f y
y = f(x) iv 2 x + 1 − 1 10 y
C3

C3
5 a y
4
–1 O x 2
2
1
(–6, –3) –4 –1 1 4 x
(–3, –4) 1 O x
2 ( )
y=f x +1 O 2 4 x 2
g y
(1, 1), (3, 1)
⎛1⎞ b 11 , 2 2
b i A translation of ⎜ ⎟ , a stretch parallel to the 3 5
0 11 a y
⎝ ⎠
y-axis with scale factor 3, a reflection in the 6 a 1 x −9 b 1.5
2
⎛0⎞ 3
x-axis, and a translation of ⎜ ⎟ . y
⎝2⎠ 8 a O x
2
y = -3x3 + 18x2 - 33x + 20
image point (0, 20) h y
6
y O 1 x
ii 1
2
(–3, 14)
y = 2 – 3f(x – 1) –2 O 2 x
y
O x b
2 3

b y 12 a 0, 2 b -1, 2 c −1, 21
3
(1, 5) d −7, 21 e 1, 9 f 1, 9
3
6
1 O 1 x 13 a x < −1, x > 2 1 b −7 < x < 21
–1 1 3 3
2 2
–1 O x
c 11 < x < 4 d x < 1, x > 9
3
–3
(–3, –4) y = f(x)
–3 –2 O 2 3 x e −1 2 x 5 f -3 x 5
3

316 317

Answers Answers

14 a -2, 1, 3 b x 1, x 3 2 a -1 b -2 c -2 b (30°, 2.2), (150°, -0.15), (-210°, -0.15), (-330°, 2.2) j tan 6q k cos 2p l cot 8q
c (±52°, 1.6), (±310°, 1.6) 5
2cos2 q
c x -3, -1 x 1, x 3
( )( )
d 1 e 2 f -1 m n 2cosec 2q o 2cot 2q
3 d (0, 0), (±180°, 0), (±360°, 0), 45°, 1 , 225°, 1 , 2
2 2
( )( )
15 a y 3 b i 6 ii 16
1 , 1 1 1
3 a ±36.9° b 21.8°, -158.2° c 19.5°, 160.5° −135°, −315°, 2 a 1 b c
c y 2 2 2 2 2
d ±143.3° e -15.9°, 164.1° f -30°, -150°
0<c 3 Exercise 2.3 d 2 e 2− 3 f 1
3 17 15 2
4 a − b − 4 4
15 8 1 a p b p c 0 d −p
4 3 4
O 9 x
9 3 a 24 , − 7 , − 24 b ± 4 2 , 7, ±4 2
5 a − 41 b − e p f 0 g 3p h − p 25 25 7 9 9 7
40 40 6 4 6
16 a y
6 a cot2 b b sec3 b c ± 120 , 119 , ± 120
2 a 2 2 b 7
c 0 d 5 169 169 119
c cosec b d cosec5 b 3 3 8
7
y=
1
f(x + 1) 2 4 a 1, ± 3 7 b − , ±4 2
(1, 2) 1 3 9 9
2 2 7 a 3 b 4 c 1 e f g − 5 8 8
2 2 2
± 3 5 a 2 , 21 , 2 b 2, 5 , 2 1
, 3 ,1
b 90° or p
1 d ±3 e 3 f c
3 a 0 5 5 21 3 3 5 10 10 3
8 a 1 b sin x c cot x 2
–1 O 1 2 4 5 6 x d tan x e 1 f 1 4 a i 1 − x2 ii x b 1 6 a 2, 5 , 2 b 1
, 3 ,1 c 1
, 2 ,1
–1 x 3 3 5 10 10 3 5 2
g 1 h 1 i 1 1 − x2 5
(4, – 3 )
2 ⎛ 90° ⎞
5 a 1+ x b p −q
2 7 a 1 - 2x2 b 6x c 18 −2 x
2

9 A translation of ⎜ ⎟ 1 + x2 9 − x2 x
b y ⎝ 0 ⎠ 8 a 22.5°, 67.5°, 202.5°, 247.5°
⎛ 90° ⎞ Exercise 2.4 b 15°, 165°, 195°, 345°
(4, 5) A translation of ⎜ ⎟ followed by a reflection 1 3 3
⎝ 0 ⎠ 1 a b c c 45°, 135°
1 in the y-axis 2 2 2
2 ( )
y=f x +1
sec (x - 90°) = cosec x
d −1 e 1 f -1
d
e
30°, 150°, 270°
210°, 330°
cot (90° - x) = tan x 2
O
f 30°, 150°, 90°, 270°
6 14 x 2 a sin 3A b cos 5a c tan 3x d cot 2x
Exercise 2.2 g 60°, 90°, 270°, 300°
C3

C3
(10, –2) 3 a sin (45° - x) or cos (45° + x)
1 a 80.5°, -60.5° b -34.5°, 174.5° h 0°, 30°, 150°, 210°, 330°, 360°
b sin (60° + x) or cos (30° - x)
17 a A stretch (scale factor 2) parallel to the y-axis c -56.6°, 123.4° i 60°, 109.5°, 250.5°, 300°
c tan (60° + x) d tan (45° + x)
⎛ 0⎞ d 31.7°, -58.3°, 121.7°, -148.3° j 90°, 270°, 45°, 255°
e 1 f cos 2x
followed by a translation of ⎜ −5 ⎟ e 40°, -80°, 100°, -140° f -63.6° k 0°, 180°, 60°, 300°
⎝ ⎠ 3 −1 3 +1
g ±54.7°, ±125.3° h ±49.1°, ±130.9° 5 a b 3 −1 c l 0°, 165.6°, 360°
⎛2⎞ i ±90°, 19.5°, 160.5° j 0°, ±180°, ±41.4° 2 2 2 2 3 −1 m 0°, 78.5°, 281.5°, 360°
b A translation of ⎜ 0 ⎟ followed by a stretch
⎝ ⎠ k ±60° l ±45°, ±135° n 0°, 120°
3 −1 e − 3 +1 2 2
( )
1 d f o 90°, 180°, 270°
scale factor parallel to the y-axis 2 a ± p , ± 3p b ± p , ± 5p 3 +1 3 −1 3 −1
2 4 4 6 6 p No solutions in range
c A stretch (scale factor 2) parallel to the y-axis p
c ± ,± 2p p , 2p
d 0, ±p, ± ± 6 a 63 b −
63 c −
65 q 19.1°, 70.9°, 199.1°, 250.9°
3 3 3 3 65 16 16 r 112.8°, 247.2°
followed by a reflection in the x-axis, and then a
45°, 225°, 71.6°, 251.6° 7 a 13 b 84 c 13 s 35.3°, 144.7°, 215.3°, 324.7°
⎛0⎞ 3 a − − −
translation of ⎜ ⎟ 85 85 84 t 23.6°, 156.4°, 90°, 270°
⎝1⎠ b 45°, 225°, 116.6°, 296.6°
⎛ −2 ⎞ c 90°, 221.8°, 318.2° 8 a 12 + 5 5 b 2+ 5 u 90°, 323.1°
18 a 2 x + 3 c T1: A translation of ⎜ ⎟ d 60°, 300° 30
11 tan 3 A = 3 tan A − tan A
3
x+2 ⎝ 0⎠
e 0°, 360°, 138.6°, 221.4° 5 + tan b 1 − 3 tan A
2
T2: A reflection in the x-axis 9 a 1 b 1 c
f 26.6°, 206.6° 13 2 1 − 5 tan b Exercise 2.6
⎛0⎞ g 0°, 360° 11 a 17.1°, 197.1° b 113.8°, 293.8°
T3 : A translation of ⎜ ⎟ 1 a 5, 36.9° b 13, 67.4° c 5, 63.4°
⎝2⎠ h 45°, 225° c 160.9°, 340.9° d 77.9°, 147.1°, 257.9°, 327.1° d 2 5, 26.6° e 5, -53.1° f 17, -61.9°
i 30°, 150° e 106.1°, 286.1° f 38.2°, 141.8° 2 a 4.9°, 129.9° b 17.6°, 229.8°
Before you start Answers
4 a x2 = 4y2 + 16 b 4x2 = 9y2 + 36 g 52.5°, 142.5°, 232.5°, 322.5° c 82.1°, 334.1° d 102.3°, 195.7°
Chapter 2 c x2y2 = 144 - 9x2 d (x - 1)2 + (y - 1)2 = 1 h 114.3°, 335.7°
1 a 3 b 3 +1 c 1 d 0 2 2 13 a i +1, 50° ii -1, 230° 5 a p b 7p , − p
e x2(y2 - 8y + 17) = 9 f x2 + b 2 = 1 3 12 12
a y b i +1, 340° ii -1, 160°
2 a 15 b 8 c p , -0.927 d p , 0.643
17 15 5 a A stretch parallel to the x-axis of scale factor 2 1 33 2 2
3 a 17.5°, 162.5° b 116.6°, 296.6° 16 a 3 b 65 c p
c 75.5°, 284.5° b An enlargement, centre (0, 0) and scale factor 1 2 6 a 63.4°, -116.6°, 0°, 180°, 360°
2 b -11.3°, -131.3°, 108.7°, -36.5°, 83.5°, -156.5°
− 21 ⎛ −90° ⎞ Exercise 2.5
4 a c A translation of ⎜ ⎟ and a stretch parallel to c -38.8° d 180°
5 ⎝ 0 ⎠ 1 a sin 46° b cos 84° c tan 140° 1
d cos 100° e sin 6q f cos 8q 7 a 5, 50.8° b 5, 50.8° c , 50.8°
the y-axis of scale factor 2 5
Exercise 2.1 1
d A rotation of 180° about the origin g cos2 20° h 2cos2 q i sin 2q 8 a 10, 36.9° b -10, 71.6° c 10, 161.6°
1 a -1.06 b -0.839 c -2.92
( )( )
2
d -3.24 e -1.73 f 1.56 6 a (±180°, 0), ±60°, 1 , ±300°, 3
318 2 2 319

Answers Answers

9 a 13, 67.4° b Max 13, 157.4°; Min -13, 337.4° 6 a i p ii p iii p 5 a Q(x) = 2x + 5, R(x) = 2x - 4 b l = -11, m = 10 15 a y
c 33, 157.4° d 10, 53.1° 6 3
6 a 5sin (q + 36.9°) b 113°, 353° f(x) = |2x + 1|
e -2, 233.1° f 1, 157.4° b i 1 ii 1 iii 3 4
2 2 7 a R = 17, a = 76°
10 a 13 sin (q − 33.7° )
7 b 0°, 228.2°, 131.8° b 119° (2.08 rad), 33° (0.576 rad) 2
b y
8 8 a R = 13 b q = 157° g(x) = |x – 1|
√13 y
9 a f(x) 9 b Does not exist c -40
y = sec x 3
10 a (x - 2)2 - 3 –1 O 1 5 x
O x 2
–326.3 –146.3 33.7 213.7 360 y = sec (x – 90º) b x = 0, -2
–2 b i f(x) -3 ii f −1 (x ) = 2 + x + 3 , x -3
1
– √13
y = cos x 16 a i 3, − 7 ii 3 iii 3
c y 3
O x
c A stretch parallel to the y-axis of scale factor 13 –180º –90º
−1
90º 180º
b i x < −7, x > 3 ii x Î R iii x > 0
3
⎛ 33.7° ⎞ 4
and a translation of ⎜ ⎟ −2 17 a y
⎝ 0 ⎠
11 2 5, t = 1.29
12 r1 = r2 = r3 = r4 = 5 2 y = |f(x)|
⎛ 90° ⎞ 3
a1 = 36.9° a2 = -36.9 a3 = 233.1° a4 = 126.9° A translation of ⎜ ⎟
⎝ 0 ⎠ y = f(x)
Review 2 9 a 84 13 c 36 –3 O 1 2 3 4 5 x
b −
1 5 85 85 77
1 a i ii 5 iii
2 2 −2
10 a 1 2 ( 3 − 1) b 17 − 7 3 O 3 x
b i 13
ii − 12 iii − 13 4 26 2
5 5 12
2 a 14.5°, 165.5° b 174° c 24.2°, 114.2° 11 a 19.1°, 199.1° b 90°, 270°, 120°, 240° 11 a Stretch parallel to the y-axis by scale factor 2,
d 3.2°, 50.2°, 123.2°, 170.2° c 30°, 150°, 270° d 60°, 300° reflection in the x-axis then translation by 1 unit y
b
e 22.5°, 67.5°, 112.5°, 157.5° f 120° e 60°, 300° f 0°, 180°, 30°, 150°, 210°, 330°, 360° parallel to the y-axis
C3

C3
4 a 45°, 63.4°, -116.6°, -135° g 0°, 360°, 124.8°, 235.2° h 90°, 199.5°, 340.5°
i 0, 90°, 360° b y = x2 + 4x + 8 y = f(|x|)
b 30°, 150°, -19.5°, -160.5° 3
c ±90°, 19.5°, 160.5° 13 a 0° b 0°, ±180°, 76.0°, -104° 12 a -2.25 f(x) 4 b k=1
d ±75.5°, ±120° c ±18.4°, ±161.6°
14 a y = 1 - 2x2 b 2x2(y + 1) = 1 c x(1 - y2) = 2y 13 a f −1 (x ) = x , x ∈
e ±90°, 56.3°, -123.7° 5
f 45°, -135°, 63.4°, -116.6° 15 a 3 4 b gf (x ) = 3 x 2 − 2, gf −1 (x )
−1
−2 –3 O 3 x
7 25
⎛−p ⎞
5 a i A translation of ⎜ 2 ⎟ and a stretch 16 a p b 2p c 2 d 2
⎜ ⎟ 3 3 13 5 14 a i y
⎝ 0⎠
(scale factor 2) parallel to the y-axis c y
17 3 + 8 2
( )
ii A stretch scale factor 1 parallel to the x-axis,
2 18 60°
15
y = f(x – 3)
followed by a reflection in the x-axis, and a
19 b 24° c 99.7°, 170.3°, 279.7°, 350.3°
⎛0⎞
a translation of ⎜ ⎟
⎝ 3⎠ 20 b 0°, 180°, 360°, 26.6°, 206.6°
⎛p ⎞ 22 a 13, 22.6° b 17, 61.9°
iii A translation of ⎜ 4 ⎟ , followed by a stretch x O 6 x
⎜ ⎟ c 5, 26.6° d 2, 45°
O b
⎝0⎠
(scale factor 2) parallel to the y-axis, and a 23 a 5, 53.1° b 5 c 103.3°, 330.5°
⎛0⎞ 24 a i ± 10, 71.6°, 251.6° ii ± 5, 296.6°, 116.6° ii y
translation of ⎜ ⎟ 18 a y
⎝1⎠ b i 20.8°, 122.4° ii 0°, 233.1°
b 4b
y
25 a 4 10, 18.4° b 38.0°, 285.2° y = |3x – 2|
4 c i −4 10 ii 161.57°
2
3
1
2 Revision 1 y=
x
1 1 x+4 O x
5
(1 + x )(2 + x )2 O x
2a
–180º –90º O 90º 180º x x ( x + 2) x2 + 4 x + 2 −2
−1 2 a b
x +1 ( x + 1)( x + 2) b a = 3, b = 9, 0 f(x) 9 b There is only one intersection.
(-90°, 0), (90°, 0) 3 Q(x) = x2 + 3x - 2, R(x) = -3 c gf(x) = 4 - x c x=1
320 (0, 1), (0, 2) 4 Q(x) = x2 - x + 4, R(x) = x + 1 321

Answers Answers

19 a y Exercise 3.1 ⎛2⎞ b y
⎛0⎞ c Reflection in y-axis; translation of ⎜ ⎟ ;
y = |f(x)| 1 a Reflection in y-axis; translation of ⎜ ⎟ ⎝0⎠ y = f –1(x)
P ⎝1 ⎠ ⎛ 0⎞
2 b Reflection in y-axis; reflection in x-axis; translation of ⎜ ⎟
⎝ −3 ⎠ In 4 y = f(x)
⎛0⎞
translation of ⎜ ⎟ Exercise 3.2
⎝1 ⎠
O x 1 a i 4.48 ii 0.223 iii 0.223
3 c Stretch (scale factor 3) parallel to y-axis; iv 0.405 v -0.405 vi 2.47 –4 –3 O In 4 x
⎛0⎞ b i 5 ii x iii 5
b y translation of ⎜ ⎟
⎝2⎠ iv x v 5 + ln 2 vi 25
y = f(|x|) –3
d Stretch (scale factor 3) parallel to y-axis; reflection ⎛0⎞
2 a Translation of ⎜ ⎟ ; x > 0, x Î R
2 ⎛0⎞ 2 ⎝ ⎠ –4
in x-axis; translation of ⎜ ⎟ ⎛0⎞
⎝2⎠ b Reflection in x-axis; translation of ⎜ ⎟ ;
( 2)
x > 0, x Î R ⎝3⎠ f -1(x) = e x - 4
x e Stretch scale factor 1 parallel to x-axis; stretch
–3 O 3 Domain x Î R
(scale factor 3) parallel to y-axis ⎛ −2 ⎞ Range y > -4
c y c Translation of ⎜ ⎟ ; x > -2, y Î R
⎛ −1 ⎞ ⎝ 0⎠
y = f(x + 3) f Translation of ⎜ 0 ⎟ c y
⎝ ⎠ ⎛2⎞
2 d Translation of ⎜ ⎟ ; x > 2, y Î R
⎛2⎞ ⎝0⎠
g Translation of ⎜ ⎟
⎝0⎠ 4 y = f(x)
e Stretch (scale factor 3) parallel to y-axis;
⎛2⎞ 3
O x h Reflection in y-axis; translation of ⎜ ⎟ ⎛0⎞
–3 ⎝0⎠ translation of ⎜ ⎟ ; x > 0, y Î R
⎝1 ⎠
2 A = 100
20 a y
y = f(x + 1) t 0 5 10 15 20 ( )
f Stretch scale factor 1 parallel to x-axis;
2
O x
3 4
P 100 128 165 212 272 ⎛0⎞
C3

C3
translation of ⎜ ⎟ ; x > 0, y Î R
1 ⎝ ⎠
13.9 weeks ⎛0⎞
–2 O 2 x g Reflection in x-axis; translation of ⎜ ⎟ ; y = f –1(x)
⎝1 ⎠
3 M0 = 6 x > 0, y Î R
f -1(x) = -ln (x - 3)
(0, –a) t 0 5 10 15 20 Domain x > 3
⎛1 ⎞
b y M 6 3.64 2.21 1.34 0.81
h Reflection in y-axis; translation of ⎜ ⎟ ; stretch Range y Î R
⎝0⎠
y = f(|x|) (scale factor 2) parallel to y-axis; x < 1, y Î R
Half-life = 6.93 sec d y
4 a 0.1733 b 113 c 7 weeks
d There is a limit to the number of organisms ⎛ −3 ⎞
–3 O 3 x possible in a unit volume of water. 3 a Translation of ⎜ ⎟ ; stretch (scale factor 3) parallel
–b ⎝ 0⎠ 2
5 b 332.2 hours 2In 2
⎛0⎞
(–1, –a) (1, –a) 6 a N to y-axis; translation of ⎜ ⎟ 1
⎝1 ⎠ 2In 2
1 200 b (-2.4, 0), (0, 3.2) O x
c i a = 2, b = 1 ii − 6 1 2
4 a y
22 a 0°, 60° b 45°, 71.6°
y = f –1(x) y = f(x)
c 70.9° d 90°, 120°
150
23 b 59.0°, 63.4°, 239.0°, 243.4° y = f(x) y= f –1(x)
24 b 22.5°, 67.5° -1
f (x) = 2ln (2 - x)
25 a i p ii 3 b − 1 c −
36 O 2 4 t
4 2 2 85 11
Domain x < 2
26 a i 13cos (q + 22.6°) ii 49.5° b 53.1° b 150 c 200 e
–2
Range y Î R
d A stretch (scale factor 50) parallel to y-axis; O 1 x
–2 1
Before you start Answers ⎛ 0⎞ e
a reflection in x-axis; a translation of ⎜ ⎟ 5 a f −1 (x ) = 3 − 1 e x g-1(x) = 2ln x
2
⎝ 200 ⎠ x −1
Chapter 3 f −1 (x ) = e Domain x Î R Domain x > 0
( )
2
1 a 1 b 1 c 2 7 a Stretch scale factor 1 parallel to x-axis;
Domain x Î R Range y < 3 Range y Î R
9 3 2
2 x = 9, y = 10 − 1, z = 4 ⎛ −1.5 ⎞ ⎛0⎞ Range y > 0
translation of ⎜ ⎟ ; translation of ⎜ ⎟
3 a x − 5 , x Î R, y Î R b 2x + 3, x Î R, y Î R ⎝ 0 ⎠ ⎝4⎠

( )
3
c x2 + 1, x 0, y 1 b Stretch scale factor 1 parallel to x-axis;
2 2 2
x2 ⎛ 0.5 ⎞ ⎛ 0⎞
322 4 a 3x2 + 6 b −3 translation of ⎜ ⎟ ; translation of ⎜ ⎟ 323
4 ⎝0 ⎠ ⎝ −4 ⎠

Answers Answers

b y y b y = ex + 2 Before you start Answers
d
y
y = f –1(x) Chapter 4
3 y = f –1(x) 1 Gradient of chords will approach 2.
2 a 4 b 2 c -3
3 a y = 4x - 10 b y = 2x - 5
3 4 (-1, 6) maximum
O 3 x 2 (3, -26) minimum
y = f(x)
1 Exercise 4.1
y = f(x) 1 a 2x + cos x b 6 + sin x
O x O 2 3 x c cos x + sec2 x d cos x + 1

(
f(x) passes through 2 1 , 0 and (0, ln 6)
2 ) 3 y
a
e -3sin x - 4sec2 x f 6 x + 2 + 1 sin x
2
f (x) passes through (ln 6, 0) and 0, 2 1
-1
2 ( ) Domain x Î R
Range y > 2
2 a 4
d -1
b p
e 3
c 6

c gf (x ) = 6 − 2 x y = ln x 3 x + y 2 =1+ p
c y = -ln (x - 2) 4

( )
gf (-5) = 4 p
y 4 y 2 − x = 1 − , (−0.214, 0) i.e p − 1, 0
4 4
y = f –1(x)
Exercise 3.3
1 a 2.20 b 2.00 c 2.14
O 1 e2 x 5 y p
= x + 1 − , (0.571, 0) i.e
2 (p
2
− 1, 0 )
2
d 0.882 e -2.14 f -2.00 6 a p b 5p
3 6
g -0.405 h 0.996 i 0.870 y 3
c y = f(x)
7 a p b 0.464
j 0.0767 k 3 l 0.232 2 2
b
2 a 403 b 6.39 c 1 8 a p b 3p
d 0.859 e -4.39 f -5.56 4
9 5, -5
3 a 0, ln 2 b ln 3, ln 4 c ln 2 O x 10 a Acos t
2 3
C3

C3
d ln 2.5 e -ln 2, ln 1.5 f ln 4 11 a i q0 ii q0
1
4 a (0.434, 20) b ( −
1
3
,2 ) c (-0.564, 1) b i q0 ii -q0

( )
Domain x > 2 12 a 5 metres b 0.54, 3.68 s c -5.23 m s-2, t = 6.37 s
1
− ln 2 , − Range y Î R
O 1 x
5 and (0, 1) –2 1 1
13 a i x 0.78 radians
2 e e 2

6 a 2.5 b 2.68 1 ii x 0.51 radians
d y = 2 ln (1 − x )
7 a y=5
5 40
b (0, 8), ln 3 , 3 ( ) y
y
b k=1
2
8 (ln 2, 3) Exercise 4.2
9 1 ln 4 = 0.0575 y= f –1(x)
1 a y
y = 3x
5 3
2 1 20 y = ex
a 15 million b 26 million d
3
10 1 ln 10 = 0.0105, t = 65.8 years O 1 x y = 2x
10 9
11 193 years x
O 1
12 36.4° C
4 a ln 2 b ±ln 2
Review 3 5 a 1 (ln 6 − 3) b 1 (e4 − 2) y = f(x)
1 a 1.2 b 0.85 2 3
6 a 7.39, 0.69, 1.79 b ln x, ln 2(x + 1) 1
c 11 d 0.95 Domain x < 1
7 a y = e x-3 –3 O 3 x
2 y Range y Î R
y b
a y = ex b i 1 ii e = 2.718…
y = f –1(x) 8 A=5 c i y=x+1 ii y = ex
5
4 y = f(x) t 0 5 10 15 20 2 a 6x + e x b 4e x + 6
3 P 5 6.42 8.24 10.59 13.59 c 5cos x + 2e x d 3 sec 2 x − 1 e x
2
2 e–3 e 1 - ex f -e x
t = 13.9 days
3 a i e2 - 1 ii 6.39
1 O e–3 x 9 a T0 = 100 b 332 °C b i 6 - 4e ii -4.87
p
O x Domain x Î R 10 a 180 b 20 c 200 c i e2 + 1 ii 5.81
c Range y > 0 11 a 1.46 b 1 c 45 ° d i −
1
ii -0.368 e 3
2 e
324 325

Answers Answers

4 a y = 5x + 2 b y = 2x + 8 tan x + 2 x sec 2 x 2(2 x + 3) 3
n 4cos 4x cos x - sin 4x sin x o cos x (1 - 3sin2 x)
3x2ln x + x − 1
3 −
i j s − 2 t −3(2 x + 1) 2
x (x + 3 x − 1)2 p e x(3cos 3x + sin 3x) q -e-2x (sin x + 2cos x)
5 y = −2 x + 1 2 x
2 −
3
e3 x (3 x − 1) esin x ( x cos x − 1)
− x ( x2 + 1)
4 3 2 4 3 2 1
k 5x - 4x - 3x + 4x l 5x - 8x + 12x + 2x - 2 u 2 v − r s
8 a minimum at point (0, 2) x(ln x )2 x2 x2
2 p
b maximum at point (0, 0) 2
e -x(2cos 2x - sin 2x)
b y = -27e-3 2e x − t
c maximum at point (ln 2, 4ln 2 - 3)
3 a y=0 w − x 2 (2 x − 1) 3
4 x=1 (e x + 1)2 3 2 a 2xcot (x2) b 2xsec (x2) cot (x2)
9 y 5 b 0.6 to 1 d.p. 2 a 2q cos (q 2) b 2sin q cos q c 6sin 3x cos 3x d -12cos2 4x sin 4x
6 (0, 0) minimum e 4tan 2x sec2 2x f 3tan2 (x + 4)sec2 (x + 4)
cosq
O x (-0.2, 0.007) maximum c d 3q 2 sec2 (q 3) g sin 2x e sin2 x
h 2x
2 sinq
–1 (-1, 0) minimum
e 3tan2 q sec2 q f −
sec 2 q 3 a 4x ln 4 b 5x ln 5 c c x ln c
2x 2x
7 minimum value of − 1 when x = 3 1 tan 3 q d 2 ln 4 e 5 ln 25 f x x(1 + ln x)
3e e cos x x
− b e x × ee 1(
1 + 5)
( ) ( )
3p − 1
3 a 4 −
8
p 1
maximum at 4 , 2 ; minimum at 4 , 2 sin 2 x 2
1
8( x + 2) 4e x 5 4
c − d −
Exercise 4.5 ( x 2 + 4 x − 1)2 x2 6 -1
maximum at (0, -1) ln 2
a sin x − 2x cos x b 2 x tan x −2x sec x
2 2 1
asymptote y = x 1 e 2( x − 1) f -cosec x 7 = 0.431
sin x tan x ln 5
10 a (0, 0), (ln 2, 0) b minimum when x = ln 3 (2) c 1 − 12 d
1 − ln x 4 a k cos kx b -k sin kx c k sec2 kx 8 a 1 3
6 x+4
b x c 1
x x2 1 x2 − 3 2 x
11 y d e k e kx
e x (sin x − cos x ) x 1 1
e f − x 5 y = 24x - 164 9 a b
sin 2 x ex 12 y 2 + 1 12 sin 3 y cos y

g
sin x − 2 x cos x
h −
x sin x + 2 cos x
x3
6 ( 4 , 16 )
3 1
c 1 cot2 (2 y )cos2 (2 y )
30
d 1
1 + ln y
2 x sin 2 x
−2 4x 7 1 ( 2 − 1) e ey
f 1
1 i j − 2 2 cos y − sin y 12 cos 4 y − 8 sin 2 y
(1 + x )2 ( x − 1)2 y = 3 x + 16
8 2 y ( y − 2)
g
C3

C3
O x 6 x( x 3 − x + 1) x(2 ln x − 1) b2 x − ax 5y − 8
k l 9 a b
(3 x 2 − 1)2 (ln x )2 a2 + b2 x 2 10 6y = x + 4
minimum at (-0.35, 0.83) ( ax 2 + b )3
2 a 7 b 1 1
11 8y = ± (x + 4)
−
Exercise 4.3 3 (3, 0) c 4 a2 x(a2 x 2 − b2 ) 3 13 maxima at (1.02, 2.47) and (4.16, 57.2)
3
minima at (2.59, -11.9) and (5.73, -275.3)
1 a 2 b 2 c 3 d 1 4 x + 2y = 0 10 x = 0
x x x x 2 y
11 b a = 4
( )
5
e −1 f 3 g − 1 h 2 x3 13 minima at (np, 0), maxima at ⎛ n + 1 p , 1⎞ ,
⎜ ⎟
x 2x 2x ⎝ 2 ⎠
a x 2 ( x cos x − x sin x − sin x cos x )
1 2 2
7 n = 0, ±1, ±2,¼
i 1 +1 j 2
+2
x x
b tan x + x sec2 x 14 minimum at (0, 0)
y
2 a 21 b 1 c e+1 ex ⎡(1 + x ) ln x − 1⎤
2 2 c ⎣ ⎦ d ex (1 - ln x + x ln x)
(ln x )2 x2
O p 2p x
d 2 e 2
3 x (1 + 2ln x - x ln x) f ( x cos x + sin x )ln x − sin x (not to scale)
3 a minimum at (1, 1) e
ex (ln x )2
b minimum at (1, 1) and (-1, 1) O x
Exercise 4.6
4 y = 1 x + 1 − ln 2, (ln 4 - 2, 0)
2 1 a 2x cos(x2 + 1) b -3x2 sin (x3 - 1) 14 a (1, 2), (-1, -2)
Exercise 4.7
5 x + 6y = 1, 1 37
6 c 2x sec (x ) 2 2
d 22( x + 1) 2 2
1 a (x - 1) [6xsin x + (x - 1)cos x] 2
y 1
x + 2x + 3 y=x+
1 b (x3 + 1) [6x2tan x + (x3 + 1) sec2 x] x
6 3x2
4(2e + 1) e f 3 c e x (3x + 2)3 (3x + 14) 4
x3 + 1 x
d ( log 10 x ) e x (2 x + 3) 2 2 1
7 = 1 g cot x
1
h x cos (ln x) d e (x cos 2 x − sin 2 x ) x=y+
dx x ln10 2 x +1 x3 y
1 cos x esin x O x
i j x 2 (1 + ln x ) − 1 –10 –5 5 10
Exercise 4.4 x ln x f g sin (x2) + 2x2cos (x2) 1 –2
k e x+4 x2 x x2 −1 x=y+
1 a 1 + ln x b 2xsin x + x2cos x l 2xe y –4
c e x(sec2 x + tan x) d (x2 - x - 2)e x m 10x(x2 + 1)4 n 14(2x + 6)6 h sin2 x + 2x sin x cos x i cos 3x - 3x sin 3x 1
y=x+
j 2x (tan 2x + x sec2 2x) k (2x + 1)e2x+1
( ) x x
e e x ln x + 1 f sin x + cos xln x o -6x2(x3 - 1)-3 p 2x3
x x x2 − 1 l 2xln(x2 - 1) + c x = y + 1 is the inverse of y = x + 1
e x ( x − 1) 3x2 x2 − 1 y x
g − 13 (2 cos x + x sin x ) h q r −
1
m cos x cos 2x - 2sin x sin 2x Their graphs are reflections in the line y = x
x x2 2 x3 + 1 ( x − 1)2

326 327

Answers Answers

Review 4 2(cos x + x sin x ) Review 5 c y
16 a 3x2e3x(1 + x) b
1 a x3sec2 x + 3x2tan x b 3tan2 x sec2 x cos2 x
1 a y g(x) = x2 – 2
1
d x sec x − 3 tan x
2
c 3x2 sec2(x3) c 2tan x sec2x d −
2 y sin y 2 y = f(x) 4
x4 y = g(x)
e ex(sin x + cos x) f cos x esinx f(x) = In(x)
ii − 6 x sin 2 x 2+ cos 2 x
3 3 3
3x+2
17 a i xe (2 + 3x) 3
g −
cos x h 1 cos x − sin x ln x 3x
esin x x 1 O x
b –2 –1 1 2
i 3cos 3x j 6sec2 6x 2
2 (16 − x 2 )
2
k l 3e3x - 1
2x + 3 19 (0, -729) min, (3, 0) and (-3, 0) points of inflexion 1
m cos x − sin x
x
n ex(x2 - x + 1) –2
e Before you start Answers
1 − 3x2 –2 –1 O 1 2 x
o Two roots
2 x ( x 2 + 1)2 Chapter 5
d y
2 a ae ax b ae ax + b 1 a x3 - 1 b 2x4 - x2 + 1 Two roots
2 a -1.2, 3.2 b (1, 0) 4
c f ¢(x)ef(x) d acos (ax)
3 -2.1, 0.25, 1.9 b f(x) = |x + 2|
e acos (ax + b) f f ¢(x)cos [f(x)] y 3
g asec2 ax h asec2 (ax + b) 4 1
2
4 g(x) = x2 + 1
f ¢(x)sec2 [f(x)] 1
i j Exercise 5.1 1
x
1 a 2 b 1 c 2 d 2 3
a f ′( x )
k l e 3 f 1 g 3 h 2 –3 –2 –1 O 1 2 x
ax + b f (x) y = g(x)
2 a 1 b 2 c 1 2
3 a e x(cos 3x - 3sin 3x) b e3x(3tan x + sec2 x) Two roots
d 1 e 2 f 0 y = f(x)
1
c e3x(3 cos 2x - 2sin 2x) d 2 sec 2 ⎛ x ⎞ 3 b There is only one point of intersection. 3 a y
⎜ ⎟ g(x) = e–x
⎝2⎠ c [1, 2] d 1.9
e 3sin x + x cos x 1 4 c 0.703 O 3 x
f + 1 –1 1 2
x sin x x 2( x + 1) 5 b 3 c 2.13 2
One root
C3

C3
6(3 x + 1)( x + 2) 7 3.7 1 f(x) = 3 – x2
g − h 1 i 5sec2 5x
(2 x − 1)4 8 a = 1, x = 1.28
–2 –1 O
( 2) ( 2 )⎦
9 c [-2, -1], [1, 2], [3, 4] d 3.93 2 a y 1 2 x
j 2 x ⎡sin 2 x + p + x cos 2 x + p ⎤
⎢ ⎥ 10 b [-1, 0], [0, 1], [4, 5] c -0.88
⎣
11 a [2, 3] b 2.19 b x = 1.68 c x = -0.83
k 3e 3tan x
sec2 x l - 2 12 b 2.67
2
4 a 1.36 b 1.31 c 2.35 d -2.82
cos x 1
m cot x - xcosec2 x n sec x o 1 − x tan x 13 a 2 b 1.32
f(x) = x3 +2
g(x) =
x
5 c 2.20 d divergent
sec x 14 b 3.29 6 b 1.395
4 a 1(max) b p (min), − 3p (max) 15 a 1 c 6.846 7 a y
4 4 –3 –2 –1 O 1 2 3 x
16 3, 2.478
c 3p (min), − p (max) f(x) = 2 – e–x
4 4 17 0, (2n + 1)p, n Î Z 2
−q +
p 18 1.9 radians
5 y= –2 1 g(x) = √x
2
x x 1 Exercise 5.2
7 a 2 ln 2 b a ln a c
x ln 10 c 4.56
1
4 ( 4 )
O x
8 a y + x = 6+p
–1 1 2 3 4
b 4y + x = 4 + p , 4 + p , 0 c 2.46
2 d Diverges Two roots
4
b 1.466
3
9 a y = sin p x (180 ) b y = p cos ( p x ) = p cos(x °)
180 180 180
0.35
4
2.659
5
b y
8
d 3.921
b 1.58, 1.68, 1.70
1 2 g(x) = x3 + 1 d x = -1 because division by zero is impossible
10 a f ′(x ) = a l = 5, m = 7
6 c 5.25
sin x cos x a -0.856, +0.473
7 b 5, 6, 2 c 1.304
1 Revision 2
b f ′(x ) = cos x − sin x = 2 cot 2 x
2 2
x3 - 6x + 3, 2.145
8 f(x) = sin(x)
sin x cos x a x3 - 5x + 7 = 0
9 b x3
- 5x - 1 = 0 1 x −6
x−4
30 x − 1 c x3 - 8x + 2 = 0 d 3x4 = 100 O x
c f ′( x ) = d (2x - 1)(3x + 2)2(30x - 1)
b 3x − 1
–1 1 2 3 4
(2 x − 1)(3 x + 2) e x3 - 2x - 1 = 0 f x2 + 2e-x - 6 = 0 2 a 3x − 7
–1 x −2 x
12 4 10 a 30, 20 b The second 2x − 9
4 3 a b x = 6, 4
14 c 2.114 74 –2 ( x − 5)( x − 3)
(1 − x )3
11 c -1.3734 d -53.9° 4 a = 2, b = 0, c = -1, d = 1, e = 0
1 1 One root
15 a b 12 c 2.303 d 1.2
2 x 2 x −3

328 329

Answers Answers

5 a 2x − 9
y
b f −1 (x ) = 4 + 1 c x Î R, x ¹ 2 9 a y
y = g(x)
b i 24 a i e3x(sin x + 7cos x) ii 5x3
+ 3 x 2 ln(5 x + 2)
x−4 2−x g(x) = 3 – ln(x) 5x + 2
4
6 b (0, 4) −60
c ; 1, -3
y ( x + 1)4
y = f(x)

4
(0, 2a) 2 25 a (3, 1 ) , ( 3, 1 )
6
−
6
− b 18
(0, a) f(x) = ln(x) 26 a (1, e) min b (-1, e-1) min
(a, a)
3
a
(– 4 , a) O x O 5 x c ( )2
1
2(
1, 1 max, −1, − min ) d (0.464, 0.177) max
(-2.678, -94.74) min
2

1
b ( a , 95a )
5 c fg (x) = 4|x| + a d x = ±a
2
27 a xex(x + 2)
c ex(x2 + 4x + 2)
b (0, 0), (-2, 4e-2)

11 b 9 3 ⎛0⎞ d (0, 0) minimum, (-2, 4e-2) maximum
16 Reflection in x-axis translation ⎜ 3 ⎟ 1 + x2
O x ⎝ ⎠ 28 a b 9y + 5x = 16, 15y - 27x = 44
1 ( x 2 − 1)2
12 a − b ii p , 5p , 3p ii y
c y>0 d -0.418 3 3 6 6 2 c (-0.49, 2.05)
g(x) = 1 + 2ln(x)
29 b y = x
7 a 13 c 135° 1
y 30 a i x = ay c y= (x − 10 + 10 ln10)
10 ln 10
14 c p , 5p , 9p , 13p d B(10(1 - ln 10), 0)
8 8 8 8 2
15 a R = 13, a = 1.176 b x = 2.267, 0.085 31 a y
a c i 13 ii 1.176 f(x) = ln(x)
6 f(x) = x3
16 a i y O 1 x
–2 5
e
4
O a x 4
3 –2
2

( a , 0) , (0, a)
2 g(x) = 6 – x2
g(x) = 1 + 2ex Stretch parallel to y-axis by scale factor 2, 2
2 ⎛0⎞
f(x) = ex translation ⎜ ⎟

C4
C3

b y ⎝1⎠
–5 O x
iii y –2 O 2 4 x
a Stretch parallel to y-axis by scale factor 2,
⎛0⎞
translation ⎜ ⎟ 2
1 ⎝ ⎠ c [1, 2] d 1.5
ii y g(x) = 0.5ln(x + 2) 32 a [1, 2] b 1.32
O x 33 c 2.754
–2 5
O a x 34 a x3 + x - 3 = 0 b x3 - 5x + 10 = 0
3 f(x) = ln(x)
4 c x3 - 3x - 2 = 0
g(x) = 2 + e–x
2 35 a a = 12, b = 2 c 2.73
( a , 0) , (0, a)
4
Stretch parallel to y-axis by scale factor 0.5,
translation ⎜
⎛ −2 ⎞
⎟
36 b 1.395
f(x) = ex 37 a 2y = x + 1 c 2.1530
⎝ 0⎠
c a = 6, 10 O 5 x 38 b -2 ln 2
17 a 7 days b i 105 cells ii 80 cells
b f −1 (x ) = 1 + e , x ∈ R
x
8 a 2 e x1 = 4.9192, x2 = 4.9111, x3 = 4.9103
−3 + ln 7 2 ⎛0⎞ 18 b £670 c 15 years d 7.2%
Reflection in y-axis, translation ⎜ ⎟ 19 b 14 years 1
⎝2⎠ 39 a 3e x −
c y 20 a 425 °C b 7.49 mins c 1.64 °C/min 2x
iii y 21 a 0.405 b 4.39 c 0, 0.693 c x1 = 0.0613, x2 = 0.1568, x3 = 0.1425, x4 = 0.1445
4 g(x) = 3ex–2 22 a ex(sin x + cos x) b x2(1 + 3ln x)
2 e x (sin x − cos x )
Before you start Answers
c e-x(3 + 2x - x2) d
sin 2 x Chapter 6
2
e x − 1 2 2 x 2ln x
− 2
2 6x2 1 a 24 b 28
f
O 3 x 1 3
x( x − 1) ( x 3 + 1)2 2 a x(2x - 3)(2x + 3) b (x - 1)(x + 1)(x2 + 1)
x
f(x) = e 1
e or 2
2 3
g h 3x2 ex 3 a A = 5, B = 9 b A = 4, B = 1
(0, 2 ) O x sin x cos x sin 2 x
x=3 3x + 7 (2 x + 1)(3 x − 1)
3 2x − 2 −9 x 2 4 a b
Stretch parallel to y-axis by scale factor 3, i 2
j 3 ( x + 1)2 x( x 2 − 1)
3( x 2 − 2 x + 5)3 2( x 2 − 1)2
d x = 7 , 11 ⎛2⎞ 5 a x2 - 4x + 3 b x + 8 + 15
3 3 translation ⎜ ⎟ −ex x−2
0 ⎝ ⎠ k x − 1)2
l cos3 x - 2cos x sin2 x
(e Exercise 6.1
6 x 2 sin(2 x 3 ) + cos(2 x 3 ) 4
23 a i (2 + 3x)xe3x + 2 ii − 1 a 3
+ 1 b − 3
3x2 x + 2 x +1 x −3 x +4
1
b c 1
− 1 d 3+ 1
330 2 16 − x 2 4( x − 3) 4( x + 5) x x −1 331

Answers Answers

3 1 3 1 Exercise 7.1 d y
e x + 5 + 2x − 1 f 2( x − 4) + 2( x − 2) c 1− 1 + 1 d 2+ 1 − 1 4
x +1 2x − 1 x −1 x +1
1 y
7
2
5 1
a x −3 − x −2 b
2
+ 3 e 1− + 3 f 4
− 1
x +1 x + 4 2( x + 3) 2( x − 1) 5(2 x − 1) 5(3 x + 1) t = –3 6 t=3
8
4
− 3 2
− 1 g x −1+ 2 − 1 h x +1+ + 1
c
x x +1
d
x −1 x − 6 x x +1 3( x − 2) 3( x + 1) t = –2 t=2
–9 t = –1 O x –4 O 4 x
3 6 9
1 1
i 1− − 2 +
2
j −
6
+ 3 + 6 + 3 –3 t=1 9
e 2( x − 1) − x − 2 + 2( x − 3) x x x −1 x x 2 x + 1 ( x + 1)2 t=0

1 1 1 k 1+ 1 − 3 l 1 − 12 + 2 − 2
f 2( x − 1) − 2( x + 1) + 2 − x x x +1 x x −1 x +1
y
2 –4
t=2
3 3 2 Review 6 8
3 a x −2 − x +1 b + 1 e y
x −1 x + 4
1 a 5
− 4 b 2− 3 c 1
− 1
x − 2 x −1 x x+4 x −3 x +3 3
1 1 3 1
c 2( x − 5) − 2( x − 3) d 2 x − 2( x − 2)
2 a
1 1
( + 1
2 x −1 x + 3 )
b 1− 5 + 5
x 4( x − 1) 4( x − 5) t=0
t=1
2 5 3 1 1
e x − x +1 + x + 2 f − + + 1 c − 1 − 1 + 2
O 12 x
x 2x + 1 2x − 1 t = –1
2( x + 1) 2( x − 1) x + 2 –5 O 5 x
−2 3 1 1
4 a + 4 b 10( x − 3) + 5( x + 2) − 2( x − 1)
x−2 x−4 3 a 2
− 2 b 1+ 5 − 2
x −1 x + 3 3( x − 2) 3( x + 1) –3
6
c − + 12 1 2
d 1− x + 1+ x 2
–8
t = –2
2x − 1 3x − 2 c − 2 − 1 y
x 3( x + 1) 3( x − 2) f
3 4 2
e − + + 1
x 2x − 1 2x + 1 4 a + 1 b 2
− 2 3 a y
2x + 1 2x − 1 3x − 2 3x − 1
15
1 4 3
f 3(3 − 2 x ) + 3(3 + 2 x ) c − 3 − 2 d − 1 − 22 + 1
x − 2 x − 1 ( x − 1)2 x x x−3

e 1− 2
+ 6 1
+ 2
+ 2
C4

C4
2 4 5
g x − 3(2 + x ) + 3(1 − x ) f − –1 O 1 x
x 2 x − 3 (2 x − 3)2 x + 1 x − 1 ( x − 1)2

4 1 1 5 A = 1, B = 2, C = -2
h − 15(1 + 2 x ) + 5(2 − x ) + 3(2 + x ) 2 3 –4 O 5 x
6 a 3− 3 b 2+ x +2 − x −3
x+2
1 1 1 1
i − + + −
x − 1 x + 1 2( x − 2) 2( x + 2) c 1 + 2 − 22 − 3 d x+2+ 3 + 1
x x x +1 x −2 x+2
4 a y + 2x = 9 b xy = 12
7 a=k+1 –15 3
Exercise 6.2 8 a (x - 2)(x - 1)(x + 1)
2
c y = x – 2x - 1 d y = x2
2 2 3 1 1 4 3
1 a x −1 − x − 2 + b x +1 − x −3 + b 1
− 1 + 1 c -1, 1 or 2 y e y = (x + 1)2 + 2 f xy2 = 4
( x − 2)2 ( x − 3)2 3( x − 2) 2( x − 1) 6( x + 1) b
2 2
− 3
1 4 x2 y2
c x +1 + d x−4 + 2
+ 1
⎡ 2
b f ′(x ) = − ⎢ + 1 2⎤ 1 g y=x+2 h + =1
( x + 1)2 2 x − 1 ( x − 4 )2 9 a
( x − 1) ⎥ 9 16
⎣ ( x + 4)
2
x + 4 x −1 ⎦
1 O 3 8 x i y = 1 – 2x2 j y = 10 x 2 − 5
2 5 2
e x + 2 − x −2 f + 1 + 3
10 a 1
+ 2 9
x 2x2 4 x 4 ( x − 2) x + 2 x −1 y2
2
k x − =1 3x + 1
l y=
4 9 2
g
1
− 3 + 3 11 a − 1 − 2 b 9 –3
x 3 x − 1 (3 x − 1)2 x −2 x +3 8 5 6
1 1 3 1 c y′ = 1
+ 2
> 0 for all x ∈ R 6 3
h 2( x + 2) + ( x + 2)2 + ( x + 2)3 − 2 x ( x − 2)2 ( x + 3)2 c y
4
7 2, -4
2 1 1 12 a 1 − 1 c 1
2 a 1− x + 2 b 1 + 2( x − 1) − 2( x + 1) r r +1 8 ± 2, y = 1 x 2 − 1
4
13 a 1 − 1 c 1 9 a (10, 0) b (9, 0)
r +1 r + 2
c 1− 1
− 7 d 1+ 1 − 2
4( x − 1) 4( x + 3) x x +1 c (13, 0), (-7, 0) d (±3, 0)
O x 10 a (0, 23) b (0, 5), (0, 6)
1
e x +1+ 1 f x+ + 1 Before you start Answers
9
c (0, 0), (0, 1) d (0, ±1)
x −1 2( x − 1) 2( x + 1)
11 5
3 Chapter 7
g x −4+ 5
+ 1 h −1 + + 3 12 5
x + 2 x −1 x +3 x −3 1 a 1 (4 x − 1) b 17x2 + 4x + 12
2 13 3
1 1 1
i x+ +
4 16(2 x − 1) 16(2 x + 1) 2 a x = 2, y = -1 b x = 2 ± 14
5
a 2 x − 23 ; x3
3
1 7
A = 1, B = 2 , C = 2 , D = 5 3 +x− 1 +c
x 3 x
x3 2
332 4 a
8
− 4 2
− 2 − 2 b 2 + 2x; x + x2 + 3 + c c − ;x− 1 +c 333
5(2 x − 3) 5( x + 1) b 3( x − 2) 3( x + 1) ( x + 1)2 x3 x

Answers Answers

14 a 29 b
2
5
c 5y = 2x + 33 c ( p2 , 1) , ( 32p , 1)
− d (2, 8), (2, 2) 10 a a = 1 + 6
p
c 3 d -3 - 4x - 4x2, |x| < 1
15 y = 2x + 1 4 y + 2x = 24 (18, -12) 11 a -2sin3 q cos q b 2y + x = 4 e 1 + x + 1 x 2, |x| < 1
2
16 x = 2cos q, y = 2sin 2q
17 x = tan q, y = sin q 5 y = 4 x − 45 ( 3
− , − 48
4 ) c y= 8
x2 + 4
,x 0 f 3 + 3x − 8 x2 , x < 1
3 2
b x = 1, y = 3 t2 − 1
Q( )
18 a x = sin q, y = 3cot q
t 6 y + 2x = 3 1
− ,4 g 4− 5x2 , x <1
2 Before you start Answers 2
19 a a = 3, b = 2 b x = 3 + cos q, y = 2 + sin q 3
20 a a = 1, b = 2, c = 2, d = 3 7 a AP = 3cos q + 5 Chapter 8 5
5 3
8
b x = 1 + 2sec q, y = 2 + 3tan q BP = 5 - 3cos q a (1 + x ) b (1 + x )
−
1 6 1 + 1 − 1 2 + 1 3 , |x| > 2
2 2

21 a p , 5p b b = 1.9; (0, -1) and (0, 1.6) 8 a 1
,y = x ⎛ ⎞
b ⎜1, 1 ⎟ c y(2x – 1)2 = x 121 112 30
x 2x 2x
3 3 t2 t2 ⎝ 2 t ⎠ 2 a b c 7 a = ±4, 1 + 2x - 2x2 + 4x3, 1 - 2x - 2x2 - 4x3
1 1 10 3 6
22 a x = 3 , y = 2 b x = t + 2, y = t (t + 2) 3 a 8 + 12x + 6x2 + x3
t t
Exercise 7.4 8 4, 1 , 4
b 56; 256 + 1024x + 1792x2 + 1792x3 + 1120x4 2
c x= 1 3,y= t 3 d x = 1 − 1, y = 1 − t 1 1 + 448x5 + 112x6 + 16x7 + x8
1−t 1−t t 1 a 19
2
b 25
3
c 144 9 Either k = 3, n = 1 , − x 2, x < 1
5 25 3 3
4 a − x b − x2
23 c y d 28 e 82 f 8ln 2 2 4
or k = − 3 , n = − 2 , 5 x 2, x < 2
3 2 1 4 2 1 2 3 4 3
5 a + b + +
2 19 1 , 12 4 1 − 2x 1 + x 3(1 − 2 x ) 3(1 + x ) (1 + x )2 1 1 1
5 5 10 1 + − + + ... ,
2 x 8 x 2 16 x 3
3 a t = -1, (0, -6) t = 0, (-1, 0) Exercise 8.1 1 3 5
t = 1, (0, 2) t = 2, (3, 0) 1 a 1 - 2x + 4x2 - 8x3, |x| < 2
1 1 + 1 x 2 − 1 x 2 + 1 x 2 + ...
O x 2 8 16
b 32 c 9 b 1 + 1 x − 1 x 2 + 1 x 3 , |x| < 1
3
28 16 Exercise 8.2
4 t = 0, ± 2 17 1 1 2 1 3,
c 1+ x − x + x
1
|x| < 2 1 a 3 + 9x + 21x2 + 45x3, |x| <
1
15 2 2 2

Asymptote x + y = -1 5 a (4 1 , 1) , (2, 2), (4 1 , 4)
4 4
b 41
2 d 1 + 3x + 6x2 + 10x3, |x| < 1 b 4 - 8x + 28x2 - 80x3, |x| <
1
3
C4

C4
Exercise 7.2 6 (0, 5), (0, 17), 16 e 1 + x − x2 + 5 x3 , x < 1 c 5 − 15 x + 65 x 2 − 255 x 3, x <1
1 a (9, -6), (1, 2) b ( 9 , 3) , ( 25 , 5)
4 4
− 7 a (2, 2) b 5.67
1 + x + 3 x 2 + 5 x 3,
3 3 2
3 3
4 8
9 2 15 3
2

a ±2 2 , ⎛ 1 , 2 ⎞ , ⎛ − 1 , −2 ⎞
3 f |x| < 1 d − x+ x − x , |x| < 1
(1, 3), (-1, -1), ( , )
1 5 8 ⎜ ⎟ ⎜ ⎟ b - ln 2 2 8 16 2 4 8 16
2 a (-1, 1), (3, 5) b ⎝ 2 ⎠ ⎝ 2 ⎠ 4
8 4 1
c (-1, 1), (3, 3) g 1 + 6x + 27x2 + 108x3, |x| < 3 e 1 − 5 x + 19 x 2 − 65 x 3, |x| < 2
Review 7 6 36 216
3 (2, 3) and (2, -3) 1 a y = x2 + 2x + 2 h 1 + x − 3 x 2 + 7 x 3, x <1
( 25 5 )
b y=2–x 2 16 104 2 640 3
4 (3, 0),
81 12
,− 28 16 2 f + x+ x + x, x <1
x2 y2 2
d y = x −1 3 9 27 81 2
c + =1 3 2 5 3 1
5 a (0, -3), (3, 0) b (0, -5), (1, 0), (-5, 0) 16 9 2 i 1+ x + x + x , x < 2
g 4 + 4x + 8x + 8x , |x| < 1 3
3 2 2 2
c (10, 0) d (0, -3), (7, 0), (-9, 0) 2 a x = sin q, y = sin 2q 3 2 1 3 2 a A = 2, B = 11
2 j 1 + x + x + x , |x| < 2
e (0, 2 ) , (2, 0)
3
− f (4pn, 0), n Î Z b x = 2sin q, y = 5tan q
4 2
3 2 1 3 1 16 32
2 3 4
b 3 + 7 x + 25 x + 9 x + 47 x ,
256 256 4096
x <4
1 k 1 + 3x + x − x , x <
6 (1, -2), (4, 4) 3 −1, 2 2 2 c It converges more slowly to the correct answer as
3
1
(2 )
7 Q t 2 , 2t , y2 = 8x 4 a ( −8 + 3 14 , 22 − 6 14 ) , ( −8 − 3 14 , 22 + 6 14 )
1
l 1− x − x −
3
1 2
36
1 3
162
x , |x| < 2
3 a
|x| increases.
5
− 4
125
b ln 64 ( )
8 a Q(2t 2, 0), M(2t 2, 3t) b 2y2 = 9x b (4, 0), (-12, 0) 2 a i 1 - x + x2 - x3 + x4 1 − x 1 + 2x
9 (-4, 2), (16, -18) ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ii 1 + x + x2 + x3 + x4 c 1 + 13x - 11x2 + 37x3
c ⎜ 3 , 5 ⎟, ⎜ 3 , − 5 ⎟, ⎜− 3 , 5 ⎟, ⎜− 3 , − 5 ⎟ 1 1
10 1 ⎝ 2 2 ⎠ ⎝ 2 2 ⎠ ⎝ 2 2 ⎠ ⎝ 2 2 ⎠ b i ii Exercise 8.3
4 1+ x 1− x
5 a 2y – 3x = 1, 3y + 2x = 8 1 0.737 42
Exercise 7.3 b 12y = 5x + 49, 5y + 12x = 119 3 a 2 + 1 x − 1 x2 , |x| < 4 2 729 + 1458x + 1215x2 + 540x3 + 1354 + 18x5 + x6
1 a 12 b 6 c 8 d -5 c x – y = 4, y + x = 2
4 64 3 a 2.236 07 b 1.732 05
3
1 1 1 x x 2 4 3.162 28
e f d 2 3y = 2 x + 1, 2 y + 2 3 x = 3 3 b 2 − 4 + 8 , |x| < 2 5
16 2 5 a 1 + x - x2 + x 3 b 10.009 990
28 3
2 a y = 2 x + 1, 2 y + x = 4 1 6 a 19.5 b
2 3 c 1 + 3 x + 27 x 2, x <2 6 1 - 6x + 24x2 - 80x3 ; 0.994 023 92
3 4 4 16 3
b 2y = 3x - 6, 3y + 2x = 4 c 0.574 (3 s.f.) d
2 7 a 1 − 1 − 12 − 1 3 , |x| > 1 b 9.949 874
d 1 ⎛1 + x + x ⎞ , |x| < 9
2
x ⎜ ⎟ 2x 8x 16 x
c y = 2 x − 2, y = − +1 7 a 4cos q b y = 2 x 2 + 3, −3 x 3 3⎝ 18 216 ⎠
2 3 c x = -100, 10.049 876
d 3y = 4x - 5, 4y + 3x = 0 8 b 2
s+r
d ( 121 22
9
,−
3 ) 4 a 1 + 1 x − 5 x 2 , |x| < 1
2 8 9 1 + 3 x + 3 x2 − 1 x3
4 32 128
2 ⎞ ⎛ 1
a ⎛− ,
1 2 ⎞
3 ⎜ ⎟, ⎜ ,− ⎟ 2 5 11 2 , |x| < 1
⎝ 3 3 3⎠ ⎝ 3 3 3⎠ 9 a y =2+x− x b t = 1 + 7 or 2.18 s (3 s.f.) b 1− x + x
20 5 2 8 10 a 1 + 3 x + 7 x 2 b 5 + 15 x + 65 x 2 + 255 x 3
2 2 4 8 16
b (1, 2), (1, -2) c 21.8 m (3 s.f.) 5 2, |x| < 2
c 2 + 3x + x
334 335
2

Answers Answers

11 2, 4, 3 cos x 6 a 5, 37 b 3.19 c 2.72 Revision 3
d − 1 e 3 tan 3x tan 2y f e −( x + y ) − 1
1 cos( x + y ) 2 7 a 126 000, 0.0314
= 1 - 3x + 6x2 - 10x3. . . 2 1 3 2 1
12 for |x| < 1 1 a − b − + +
(1 + x )3 ye x + e y b 3956e0.0314t, 5415 people/year x − 2 x +1 x x −1 x +1
h y (1 − e ln y )
x
g − y
xe + e x ex c 323 200, 6.7% 6 4 2 1 3
Review 8 c − d − −
4 −
3 8 a 150, 100 b 200, 0.03 cells/hour x−3 x−2 x x + 1 ( x + 1)2
1 3 a b -3 c
1 a 1 + 3x + 9x2 + 27x 3, |x| < 3 7 9 a 20, 0.0575 4 2 8
3 e − x − 2 + 2x − 1
d 0 e 2 f -2 b i 12 h ii 40 h x
b 1 + x − 1 x 2 + 1 x 3, x <1 g y=x+3 h x=3 c 1.15 g h-1, 0.863 g h-1 2 1 1
2 2 2 f + −
5(1 − 2 x ) 5( x − 3) ( x − 3)2
c 1 + 1 x + 5 x 2 + 5 x 3, |x| < 2 i 7y = 33 - 13x j y = 1− x 10 a 70, 0.154
2 2 A = 1, B = -3, C = 2, D = 1
381 36 b 10.8° C/min, 4.99° C/min
4 a 10y = 7x - 54, 7y = -10x - 8 3 3 1 2
1⎛ x2
+ x ⎞, c T = 10 + 70e-0.154t
3
1
d ⎜3 + x − ⎟ x <9 d 5.5 min 3 a 1− + b 1− −
b y = x, y = p − x 2( x + 1) 2( x − 1) 3( x − 1) 3( x + 2)
1
3⎝ 3 54 486 ⎠ 2 c x = 2, y = 2 11 Both models give the same estimate of 3 360 000,
2

2( )
e 1 1 + 1 x + 1 x 2 + 1 x 3 , |x| < 2 d y = -x, y = x
because P0at = P0ekt if k is defined by a = ek c 1+ 2 − 2 + 1 d x+2− 1 + 9
2 4 8 x x −1 ( x − 1)2 2x 2( x − 2)
5 y=x+1
2( )
1 Exercise 9.4 e 2 − 2 − 12 + 6
f 1+ x + 3 x
1 2 + 5 x 3 , |x| < 4 6 26 2 1 0.2 m2 per min x 2x − 1
x
8 128 1024 3
7 a 5y + 8x = 21 b 41 2 a 47.1 cm2 per sec b 3.14 cm per sec f 1− 4 + 1 2 − 1
g 1 − 1 x − 3 x2 − 1 x3 , |x| < 1 5 3( x + 1) ( x + 1) 3(2 x + 1)
2 8 16 8 (2, 3) 3 a 32.4 mm3 per min b 21.6 mm2 per min
9 (6, 14) 4 2 cm3 per hour 4 p = 1, q = 0, r = 4, s = 3, t = 9
h -(3 + 4x + 4x2 + 4x3), |x| < 1
(2 )
2 4
a 40.2 mm3/sec 2
b 15.1 mm /sec 5 a x = 21 b k= − ,−
2 A = 0.5, B = 0, C = −
1, |x| < 1 10 1 1 , 0 , 63.4° 5 3 3
16 6 15 cm3/sec 2
5 11 (0, 0), (4, 2), 90°, 31° 6 a 3 b y = x −1
3 − 7 a 0.00 111 cm/sec b 0.333 cm2/sec 3
8
4 1 − 1 − 1 2 − 1 3 , |x| > 2
12 y = ±3
8 0.0265 cm/min 7 a ( 1 , 0)
3
b (0, -1) c (3, 2), (0, -1)
x 2x 2x
13 1 , 1 9 126 mm2 per sec 8 a 2y + x = 11 b (-1, 6)
3
5 a a = 3, n = -2 b -108 c |x| <
1
3 14 a ( ) 4, 1 , (4, 1)
3
b ( )
1, − 1 , (1, − 2)
2
10
11
0.305 cm/sec
9 m/sec
9 a (7, 3), (12, 4)
10 a t = 0, 3, -3
b 17.7
b 64.8, 129.6
6 a 1 + 2 , 3 − 3 x + 9 x 2 − 15 x 3, x < 1
C4

C4
1 − x 1 + 2x 2 15 a (-5, -2) min; (-5, -3) max 12 -0.1 N m-2/sec 11 a P(0, 4.25), Q(2, 2) b 52 c 41
3 2
( )
−k dL
12 a t = p
1 1 1 3 13 64
b + − , 1 + 2x - x2 + 4x3, |x| < 1 b −3, min; (-3, -3) max p r 2 L2 dt d −8 3
1 − x 1 + x (1 + x )2 2 3 3
where k is Boyle’s constant. Assumes temperature 1
7 a A = 1, B = 2 b 3 - x + 11x2 16 (1, 1) and (1, -1) 13 a 1 - 3x + 9x2 - 27x3, |x| <
remains constant. 3
1 1 17 a maximum of 3, minimum of -1
c No as x = 2 is outside |x| < 3
b maximum of -1, minimum of 3 Review 9 b 1 + 5 x − 25 x 2 + 125 x 3, x <1
9 3.162 28 (5 d.p.) 3 3 2 8 16 5
18 a 2 b 4 ,max 3x 2x + 3y 2 y3
1 a - b - c
11 a p = -13.5, q = 67.5 b 1.424 687 5
19
dy
=
x2 + 8x − 4 4y 3x + 2 y 3x3 c 3 + 1 x − 1 x 2 + 1 x 3, x <1
3 54 486 2
12 a A = 3, C = 4, B = 0 b 4 + 8 x + 111 x 2 + 161 x 3 dx ( x + 4)2 y 2 e x − 2 xe y
2 1
4 2 d
3
tan 2x tan 3y e
3 xy 2
f
x 2 e y − 2 ye x d x + 1 x 2 + 3 x 3 , |x| < 4
8 128
Before you start Answers Exercise 9.2
1 x = ±1 3y2 + 4 xy − 3x2 17 e 1 − 3 x + 9 x 2 − 27 x 3, x < 4
Chapter 9 2 − 4 16 64 256 3
2 y = x + 2 2, y = x − 2 2, 3 y 2 − 6 xy − 2 x 2 7
1 a y = 6x - 5 b y = 2(x - 4) 1 1 3 2 5
y = − x + 2 2, y = − x − 2 2 , area = 16 3 a 3y - 2x = 2 b 7y - 5x = 8 c y + 6x = 3 f + x+ x + x 3, x <4
−
1 2 16 256 2048
2 a 3tan2 x sec2 x b 3 x 2 (x 3 + 1) 2 5 x - 2y + 2 = 0
2 (-0.618, ±0.300) 3 9
3
6 b (2, 3), (2, 5) c (2 ± 4 2 , − 1) 14 a 1 − x − x 2, x <1
( )
c ex(sin x + cos x) d x2(1 + 3ln x) 2 8
1 −1
5 ,
1 5 10 7 4 y + 3x = 8 3 b 1− x + x , x < 1
1 15 2
3 a b 1 (3 + ln 20) c 50(1 - ln1.5) 2 8 3
ln 1.5 2 8 a y = 3x - 7 b 3y = 2x - 7
Exercise 9.3 3 9 27
, x >3
Exercise 9.1 15 1 − − −
1 a ln 3 ´ 3x b 2ln 3 ´ 32x - 1 9 y = bx 2 −b 2 x 8 x 2 16 x 3
c 3 − 2x
2 a
1 a x b x2 c 20ln 3 ´ 35x + 2 d -2ln 3 ´ 3-2x 16 a = ±8, 2 + 2x - x2, 2 - 2x - x2
2y ⎛ ⎞ ⎛ ⎞
10 ⎜ 1 , 3 3 ⎟ , ⎜ 1 , −3 3 ⎟
3y y
e 2ln 10 ´ 102x + 5 f -6ln 5 ´ 51 - 2x
⎝2 2 ⎠ ⎝2 2 ⎠
1+ y
( )
1 1
d 1 e −
y
f − 1 1
x +2 17 a + , 2 + x + 5 x 2 + 7 x 3, x <1
3 y ( y + 2) x 1+ x 2 ln 2 2 2 + 6 × 2 x , 16 ln 2 11 a 54y + x = 27 b t = -6 1 − 2x 1 + x 2
2
2x + y x(1 + y 2 ) x (3 − y 2 ) 12 a 4y = x + 15 b 1
+
1
,
3 11
+ x+
73 2 431 3
x + x, x <1
g - h − i 3 a y = 6.59x - 0.59 2 + x 8 − 3x 2 4 8 16 3
x + 2y y ( x 2 − 2)
13 a (0, 2), (0, -2) b x=0
y (1 + x 2 ) y = -0.15x + 6.15
14 a i 4xln 4 ii (24x + 22x +2)ln 2 18 a A = − 3 , B = 1 b -1 - x + 4x3
y2 3y3 b y = 4.16x + 6 2 2
j -1 k l − b i y = 5 + 2xln 2 ii y = 4 + 5xln 2
x2 4 x3 y = -0.24x + 6
15 a 13, 50 b 12.3 per sec 5y − 2x ey y − sin y
m tan x tan y n cot2 y 4 a 40, 83.2 b 5.28, 1.46 19 a b c
c 5.8 per sec d t = 1.5 2 y − 5x 3 y 2 − xe y x(cos y − 1)
y2 y c 13.4, 20.1 d 0.55, 0.22
1 b − c − x 16 a £1975 b -£801 per year
2 a 5 a 3.21, -0.71 b 5.80, -2.55 y 2 xy − sec 2 ( x + y )
2xy x(1 − y ln x ) c rate of depreciation d cot x cot y e f
336 c 1.98, -0.020 d 54.6, -109 x − x 2 + sec 2 ( x + y ) 337

Answers Answers

20 a y = 1 x - 1 3
x − 2 cos x − 1 sin 2 x + c
1 3 3
b y+x=2 k 5 ln sin 2 x + c l sin (2x + 3) + c 2 a 2 (1 + e x )2 + c b 2 (1 − cos x )2 + c e tan x − 2 ln sec x + c f
2 2 5 2 3 3 2 4
22 a −
2
tanq c 6cosec 2a d p,6 m − 1 tan (4x + 1) + c
1 1 3
g 1 tan5 x + c 1
h sin x − sin 3x +c
3 4 4
n 4 sec 4x + c c 2 (x + 2) x − 1 + c d (1 + x 4 )2 + c 5 3
23 21.4 3 6
3
24 a 60 664 b 485 per year c 87 years o
1
ln |sec 4x + tan 4x| + c e
1
(ln x)3 + c f 2 x − 4 ln ( x + 2 ) + c i 1
cos3 x − cos x + c j x − 1 sin 2 x + 1 sin 4 x + c
4 3 3 8 4 32
25 a 0.0637 m/min b 0.8 m2/min
26 a 0.002 55 cm/s, b 0.48 cm3/s 1
p - 4 ln |cosec 4x + cot 4x| + c g 2 x + 1 − ln (1 + x + 1 ) + c k sin x − sin3 x + 3 sin5 x − 1 sin7 x + c
5 7
Before you start Answers 1 −2 x 1 sec x + 1 tan2 x + c
q −2e + c r sin 3 x − 3 cos 1 x + c h − 1− x2 +c i -1
sin (x) + c l − ln
2
3 3
Chapter 10
k 1 ln ⎛ e x − 1 ⎞ + c
x
1 1 2x 1 −2 x 1 sec3 x + c
1 a 2x ln x + x b ex(3x2 + x3) s - 2 cosec 2x + c t e − 2x − e + c j
2 ⎜ e + 1⎟ 4 a − 1 cos 2 x + c b −
1
cos 4 x + c
2 2 3 ⎝ ⎠ 4 8
2x2 +1 1
c ex(tan x + sec2 x) d 2 a 0 b c 2 ⎛ ⎞ 1 1
x2 + 1 2 l −
4 − x2 + c m 2 x + 1 + 4 ln ⎜ x + 1 − 1 ⎟ + c c −
2
cos x + c d −
12
cos 6 x + c
1 − ln x 4x ⎝ x + 1 + 1⎠
e f cot x
( ) e e −1
4
x2 d 2 2 2 −1 f 1 ln 7
1 1
3 2e 3 3 a 2(e - 1) b
9
c 2 −1 e −x + tan 3 x + c f −x − cot 3 x + c
3 3
3 a 1
+
1
b −3 + 2 + 1 20
2( x − 5) 2( x + 5) x x −1 x + 2 3 a - 1 + tan x + c b ln x - cot x + c 31
3 7
x d 23.2 e 2 f
162 g − x − 1 cot 2 x + c h 2 tan ⎛ 1 x ⎞ + c
⎜ ⎟
c 1− 1 + 1 2 d 1− + 2 ⎝2 ⎠
x x − 1 ( x − 1) 2 x 2( x − 2) c 1 ( e2 x −1 − e1−2 x ) + c d 1 sec 3x + c g e −1 h 352
1
(Arbitrary constants are omitted from answers.) 2 3 2e 15 5 − cos 8 x − 1 cos 4 x + c
1 16 8
Exercise 10.1 e − 1 cosec 3x 4 a (2, 0) b 30
1
3 6 a − cos 5 x − 1 cos 3 x + c b 1 sin x + 1 sin 9 x + c
1 a 1.45 b 8.72 c 1.17 10 5 a (0, 0), (4, 0), 128 b (0, 0), (2, 0), 4 10 6 2 18
4 15 3
d 0.137 e 1.57 f 2.12 p
c 1 sin x − 1 sin 5 x + c
2 a 0.9943
d convex graph
b 1 c 0.57%
Exercise 10.4
(2 )
c (0, 0), p , 0 , (p , 0); 5
2
2 10
3 1 2
6 a, d, e, g, h on sight 7 a b c
3 a 0.9185 b 0.9116
( x − 1) + c 5 5
1 3 6 5
1 a ln |x3 - 1| + c b ( x 2 + 1)4 ( x 2 + 1)2
6 a 1 ( x 3 + 1)5 + c − +c
C4

C4
4 55.1, p » 3.06 3p ⎛ ⎞
b ln ⎜ 2 + 1 ⎟
b 4 2
5 8 a c 0
( x + 3x − 1) + c
1 2 5
5 I = 2.30, 18.4 c ln |x2 + 3x - 1| + c d 3 2 ⎝ 2 − 1⎠
5 3
2
6 a 50.5 b 51.2 c 1.4% c 2 (3 x − 2)(x + 1) + c 2 d (1 + tan x ) + c 2
9
1
1 1 2 15 3 3
7 Area = pab where a, b are the lengths of the e ln |x2 - 4x + 1| + c f (x - 4x + 1)4 + c
2 8 1
2
semi-axes. For a circle, a = b = r
1 sin (x2 + 1) + c
e (x 2 + 1)2 + c f
15
x + 1(3x2 - 4x + 8) + c 11 a 1 b 8
g ln |sin x + 1| + c h 2 3
Exercise 10.2 2
12 p + 1
3
2
g 1 e x +1 + c h (e x − 1)2 + c
2
c e −1
2
1 a 2 b 5 2 3 2 3 2 6
2 e i
3
( x − 1)2 + c
1 2
j x2 − 1 + c
13 a odd function b even function
d ln 2 e − 2 f 3 + ln 4 Exercise 10.6
1 x2
27 p k 1 ln |x2 - 1| + c l e +c 1 1 Exercise 10.7
g + ln 4 h + 3 2 2 1 a sin 3x + c b − cos 4x + c
2 3 3 4 x
3 1 a 2 ln +c b 2ln |x - 3| - ln |x + 1| + c
m − 1 e− x + c n 1 ( x 2 + 2 x + 3) + c x+2
( ) ( )
2
2
2 a 1 tan q + c b 1 x2 - 2ln x + c 2 3 c 2 sin 1 x + c d − 2 cos 3 x + c
2 2 2 3 2 c ln |(2x + 1)(x + 1)| + c d ln (x − 1)(x + 3)3 + c
c 5sin x + 3cos x + c o 1 ln |x2 + 2x + 3| + c p esin x + c 1 1
2 2 −
e ln 2 x − 1 + c
a e2 - 1 e sin (2x + 1) + c f − cos (3x - 2) + c
3 b ln 3 2 3 f 4 ln x 2 1 + c
1 x3
q e +c r ln |ln x| + c 2x + 1 x
⎛p 1 ⎞ 3 1 1
4 a ⎜ , ⎟ b 2− 2 c 2 −1 g 4 tan (4x) + c h 2 tan (2x - 3) + c
g 1 + ln x − 1 + c
⎝4 2⎠ x 1
3
1
ln 2 h ln − +c
2 x x x −1 x −1
5 e2 - e - ln 2 » 3.98 1
−
1 1
cot 4 x + c
i ln x − 2 + 3 + c
4 x + ln |x - 1| + c 2 a tan 3x + c b cos 3x + c c −
6 e4 3 3 4 i
1 n x +1 x +1
x + 1 x n −1 + + x + ln | x − 1| + c
() (7 )
10 1
7 n n −1 d tan x + c e − ln cosec x + cot x + c a ln 8 b ln 2 + 3 c ln 3
p 2 2
2
5
f sec x + c g ln |sec x + tan x| + c
5
Exercise 10.3 Exercise 10.5 h x + tan x + 2ln |sec x + tan x| + c 4 ln 2 − 3 ln 3
1 1 1 2 3 2 2
1 a sin 5x + c b − cos 4x + c 1 a (1 + x 3 )5 + c (1 + x ) + c
b 3 2
1
5 4 (sin 3 x − cos 3 x ) + c
a x + 3 ln x − 3 + c b x + ln x − 1 + c
15 9 i
3 5
c
1
tan 3x + c d 2sin 1 x + c 1 1 2 x+3 x +1
3 2 c sin5 x + c d tan2 x + tan x + c j -cot x - 4ln |cosec x + cot x| + 4x + c
5 2
e
1
− cot 4x + c f 1 4 x −3
e +c 1 k -2 cot x - 2 cosec x - x + c l −
1
cot 2 x + c c x + 2 ln x + 1 + c
4 4 e − 3 +c f ln (x + 5) + 5 + c 4 x
9 ( 3 x − 1) x+5
g
1
(3 x + 2)5 + c h 1 ln |sec 3x| + c 1 3 a 1 x + 1 sin 2 x + c b 1 x − 1 sin 6 x + c d 1 x 2 + 2 x + 37 ln | x − 3| + 3 ln | x + 2| + c
15 3 g 2ln (x - 1) + (x - 1) - +c 2 4 2 12 2 5 5
x −1
i 1
ln |3x - 1| + c j −
1
+c h −1
+
2
+c c 1 (x + sin x ) + c d 1 x − 1 sin (6 x + 2 ) + c 6 ln x 2 − 1
3 3 ( 3 x − 1) 4(e x + 2)4 5(e x + 2)5 2 2 12
338 339

Answers Answers

Exercise 10.8 Exercise 10.10 1
b 1 (x − 4)5 + 3 (x − 4)4
10 a 1 b 1 − 134 6 a −
( x + 3)
1 a sin x - xcos x + c b xe x - e x + c 2 4 279 32 5 4
e 4e 1 a 5 p b 4p c 5
p 3 3
1
c 1 x 2 ln x − 1 x 4 + c d 1 e2 x (2 x − 1) + c 11 −
1
cos x sin2 x − 2 cos x c 2 (e x + 3)2 d 2 (x + 1)2 − 2(x + 1)2
2 4
e xtan x - ln |sec x| + c f
4
-e-x(1 + x) + c 1
3
3
3
3
2 a 38 p
15
b p ln 3
2 ()
c 8p
15
3
7 0.183
3

− cos x sin 3 x − cos x sin x + x
p2
f p e2(3e2 - 1)
4 8 8 1
d e p 8
g 1 sin 2 x − 1 x cos 2 x + c 5p 4 4 3
4 2
16 p p 9 a ln |sin x| b tan x c ln |sec x + tan x|
g (7 − 12 ln1.5) h (18 ln 3 − 8 ln 2 − 5)
h − 1 2 ln x − 1 2 + c i 1 2 x +1
e (2 x − 1) + c
( ) ( )
6 4
2x 4x 4 d 0.5 x + 1 sin(6 x ) e 0.5 x − 1 sin 6 x
3 m= r
( ln x − 2 ) + c
1 3 2 3 Exercise 10.9 6 6
j x (3 ln x − 1) + c k x 2 h
f 3x - 4ln |cos x| + tan x
9 3 3 1 1 x4 − x3 + c 2 1 x4 − 2x3 + 9 x2 + c 4
64
p
4 4 2 g 1 (3 x + 4 sin x + 0.5 sin 2 x )
( 4) ( 4)
15
l x sin x − p + cos x − p + c 1 1 3 1
8

m sin ( x + p ) − x cos ( x + p ) + c
3
7
(x - 6)7 + c 4
8
(x - 2)8 + c 5 p (4 − p )
8 (2)
h 2 sin x − 2 sin3 x (2) i − cos x
2 1
+ cos3 x − cos5 x
6 a 243p
3 3 5
3 b (-4, 16), (3, 9), 868 14 p
5 2 ( x 3 + 1) + c
1
6 6 2 6 ln |x3 + 1| + c 5 15 j 1
(3 x − sin 4 x + 1 sin 8 x ) k x - 2tan x
9 3 8 8
80
n 1 x 2 + x tan x − ln sec x + c 7 a (-2, -4), (2, 4) b p
( )⎠
1 sin (2x3) + c 8 3 x + 3 sin 2 x + c
0.5 ⎛ x + 2 sin x ⎞
2 7 3
3 2 4 5p l ⎜ ⎟
8 ⎝ 2
o x(ln |x|)2 - 2x ln |x| + 2x + c 9 x − 1 sin 6 x + c 10 4tan x + c 14
1
6
9 p 10 a sin 2 x + 1 sin10 x b 0.3
5 4 20
2 a 12 ( nx sin(nx ) + cos(nx )) + c 11 sin 4 x + c 12 1 sin 2 x − 1 x cos 2 x + c 2
4 4 2 ( x − 3)
( )
3
n 10 a (0, 4), (1, 4) b 1.3p 11 a ln b ln (x + 2)2 2 x − 1
13 1 3
x (3 ln (2 x ) − 1) + c 14 3 ln (x2 + 7) + c 2 ( x − 1)
b 12 enx (nx − 1) + c
11 p (3(ln 3)2 − 6 ln 3 + 4)
9 2
n
15
1
(5x - 1)(x + 1)5 + c 16 ( x + 2) + c
1 2 6 2 c 7
ln x + 1 − 7
30 12 16 x − 3 4( x − 3)
x n +1
c ((n + 1)ln x − 1) + c 1 2x 12 1946 p 12 a 0.182 b 2.886
(n + 1)2 17 1 2 x2 + c
e 18 e (2x - 1) + c 3
4
4
13 a 381 p b 197 p 13 a A = 3, B = 2 b 4 ln 2 − 5 ln 3
C4

C4
c 16p
d −
cos nx
n
( ln sec nx + 1) + c 19 -2x + tan 2x + c 20
1
2
sin 2x + c 7 2 4
21p b sin 3 x + 1 cos 3 x
x
(2 )
14 a x sin x + cos x
21 2 ln sin 1 x + c 22 xln |2x| - x + c 14 3 9
3 a p −1
8
b 4 ln 2 − 15 c ln 27 − 1
2 16 4 c 1 e3 x (3 x − 1) d − x cos nx + 12 sin nx
23 2x(ln |x| - 1) + c 24 x + ln |x + 1| + c 15 p ⎛1 − 3 ⎞ 9 n
2⎜ 2 ⎟
n
d log10 4 −
3 log10 e ⎝ e ⎠ 1 2x 1 2x
( ) e (2 x 2 − 2 x + 1) e (2 sin x − cos x )
e 3log10 4 - 2log10 e
sin 2 x + p + c
e f
( )
1 4
4 25 x + x3 + 1 x2 + 3x + c 26 1 4 5
16 2p ln 3 − 1
( )
p 4 2 2 2 x4
f ln 2 + p − 3p 2 g 1 h 1 e2 + 1 1 1 2x
2 6 g 16 (4 ln x − 1)
4 2 2 27 ln sec (2 x − p ) + c 28 e (sin 2x - cos 2x) + c
2 4 17 4 p ab2
4 p, 3p 3 15 a 0.1517 b 0.7183 c 1.446 d 3.196
1 1
cos 2 x − 1 cos 6 x + c 8p
( )
29 − cos 4x + c 30 16 a 0.5x sin 2x + 0.25cos 2x
5 1, 1 , 1 − 6 8 4 12 18
15
e e5 1
(2 x 2 + 2 x sin 2 x + cos 2 x )
ln x − 4 + c
( )
1 x−2 b
1 31
x +1
32
2
ln
x
+c 19 p 3 −1− p 8
6 a −e b 1 12 1 1
3
20 Torus (doughnut), 4p 2 17 a x 6 − x 4 b 1 x4 − 2 x3 − 1 x2 + 2x
34 2 ( e x + 1) 2 + c
2
1 33 x − 2 (x + 4) + c 6 4 4 3 2
7 a (5 x − 1)(1 + x )5 + c 3 3 1 1 2
30 1 1 Review 10 c (x − 5) 7
d (x − 2) 6
35 ln (e2x + 1) + c 36 − sin-3 x + c 1 a 7.40 b 1.47 c 4.04 d 0.477 7 6
ex 2 2 3
( 2 x + 15 )(x − 2)
b (1 + x ) + c 8 3
2 a 1.008 b 1 c 0.8% d concave curve e 2 f 0.5ln |x2 - 1|
c 2 (3 x + 2)( x − 1 + c
3
)2 37 − 1 cos 7 x − 1 cos 3 x + c 38 1 sin8 x + c 3 a 0.5ln 1.5, 1.5ln 2.5 b i 1.792 ii 1.684
5
14 6 8 x2
g 0.5 ln x − 1 h ln x − 1
15
(4)
1 1 i e2
4 a 3 sin 3 x b 5 tan 5 x c -4cos x x +1
8 a e x(x2 - 2x + 2) + c 39 1 ln x − 3 + c 40 −
2
cos3 x + c x
6 x+3 3 x
b -x2cos x + 2x sin x + 2cos x + c d 1 (6x + 1)5 e 0.5ln |sin 2x| f 0.25ln |4x + 3| j 2e 2 ( x − 2) k 0.25x2(2ln |3x | - 1)
p 30
b ln 3
( )
41 a 1 3
c 1 e2 x x 2 − x + 1 + c 8 2 g −
1
h 0.25e4x l x (3 ln 3x − 1)
4(4 x + 3) 9
c ln ( 2 + 1)
2 2
d 1 18 a ln |sin x| b ln |sec x + tan x|
1 −3 x 4 i 0.5e2x + 2e x + x j -2cot (0.5x + 1)
d − e (9 x 2 + 6 x + 2) + c
27 42 a − 1 2 + c b 1 tan3 x - tan x + x + c k 0.5ln |sec 2x + tan 2x| l 1
sec 2 x c 1 sin 4 x − 1 sin3 4 x
2 tan x 3 2 4 12
e 1 e2 x (cos 2 x + sin 2 x ) + c 1 x3
c e +c
2 1
d sin x − sin3 x + sin5 x + c
1
5 a (x 2 + 3)6 b 0.5ln |x2 + 3| c 0.5sin (x2 + 1) d 1
(6 x − 8 sin x + sin 2 x ) e 3 x + 1 sin 2 x
4 3 3 5 6 2 4
16
f 1 x 2 sin 3 x + 2 x cos 3 x − 2 sin 3 x + c
3 9 27 4 2 (
e 1 3 x − sin 2 x + 1 sin 4 x + c
8 ) d 1 sin7x
7
e 0.5ln |x2 - 2x - 1|
3
f 1
8
1
sin 4 x − sin10 x g
20
1 x
10
e (sin 3x - 3cos 3x)
h 2 (1 + x 3 )2
2
1 3x −0.5e − x
g
13
e (3 sin 2 x − 2 cos 2 x ) + c f 2 ln x + c g ln |3 + sec x| + c h 3x
+c
f ln |1 + tan x| g
9 h 1 (8 x 2 cos 4 x + 4 x cos 4 x − sin 4 x )
ln 3 32
b e −5
2
340 9 a 1 (ep − 2) c 2(ln 2)2 − 2 ln 2 + 3 43 a negative b positive c zero
341
5 2 e2 4

Answers Answers

19 a 0.3096 b 1.071 c 0.125 g 3 y = 1 x2 − x + c h y + ln y − 1 = c − 1 x Before you start Answers Exercise 12.3
53 p 2 + 2p 2 2 1 a 24 b 2 c -2
20 a 15 p b Ax Chapter 12
8 i y= j y 2(2y + 3) = x2 (2x + 3) + c 2 a 3 b 6 c 9
x +1 ⎛1 ⎞ ⎛ −1⎞
ii ⎛ ⎞
1
21 V = 10 8 p k sec y = Asin x l y (x + 1) = Ae x 1 a i ⎜ ⎟ iii ⎜ ⎟
⎜− ⎟ d 9 e -3 f -3
15 ⎝2⎠ ⎝ ⎠ 5 ⎝ 3⎠ 3 a 62.8° b 73.6° c 39.7° d 69.5°
y m y = tan x – x + c n x = ysin y + cos y + c b square with vertices (9, 1), (11, 3), (9, 3), (11, 1) e 82.6° f 71.6° g 85.5° h 90°
2+ y 1
20 o c − 4e − x = ln 2 a 41 b 74 , 24 4 a -1 b
2− y 2
18 1 1 1 3 a x = 2, y = 0, z = 3 b x = 1, y = -1, z = 2 5 a -3 b ±4
g(x) = 5x p y = 2 ln |sec 2x| + 4 ln|sec 2x + tan 2x| - 2 x + c 6 a parallel b neither c perpendicular
16 Exercise 12.1 4 2
6 (y – 1)2 = 8x + 9 1 a 13 b 13 c 21 d 14 7 a -3 b c 10, -
14 3 5
7 k=2 8 a i 80.8° ii 76.1° iii 16.7°
12 2 a 70 b 3 5 c 141 d 0
8 y 2 = 2(xsin x + cos x + 1) b 54.7° with each axis
3
10 9 x = tan q 3 a i+ 4j b
5
i − 12 j 9 29.5°
5 5 13 13
8
10 2y 3 – 3y 2 + 6y = 3x 2 + c 10 47.6°
f(x) = x2 + 4 11 a
dy
=x
dz
+z c 1
i+ 1 j+ 1 k d 2i − 1 j+ 2k 12 a 70.7°, 23.3°, 86.0° b 38 , 6 , 34
6 dx dx 3 3 3 3 3 3
c 7.12 square units
4 Exercise 11.2 ⎛5 ⎞ 13 b i + 4j + 7k
⎜2 3⎟ ⎛ −1 ⎞
2 1 a v = 1 t2 + 1 t + 6 b 13.5 m s–1 4 a ⎜ ⎟ b ⎜ ⎟
⎜ 3⎟ 14 a 2p × r b 2p × r
200 10 ⎜5 ⎟ ⎝ ⎠ c |p|2 - |r|2 d (p - r)2
2 a 403 b 23 days ⎝2 ⎠
–4 –3 –2 –1 O 1 2 3 4 5 x 17 i - 2j + k
3 6.2 years 5 a 16i + 12j b 10 i − 10 j − 5 k
3 3 3 18 180°
22 a 63 3 p b p(2e - 1) 4 26.3 days
4 Rt
6 a ±2 b ± 6
1 19 11
5 a i = Ae
−
L b L ln100 7 a 33.7° b 34.5° c i− 3 j 15
Before you start Answers R 10 10
Exercise 12.4
6 a 2.56 m b 50 hours 8 a 9i + j b 2i + 6j - 4k c -5i - j - 2k
Chapter 11 1 a r = (1 + 2l)i + (2 - 3l)j + (-3 + l)k
7 c 10 760 years 9 a l = 6, m = -12 b -4
+ 1)3 b r = (4 - 2l)i + (-1 + l)j + (3 + 2l)k
C4

C4
1
1 a y = x3 − 5x + 2 b y = (x −4 8 b Further 12 mins c 30.1°C 10 a l = 8, m = -6 b l = -4, m = -3
c r = (3 + 4l)i + (1 + 2l)k
3 3 9 b 2 cm
⎛2⎞ ⎛ −1 ⎞
b ⎛ ⎞ c ⎛ ⎞
3 5 d r = (1 + l)i + (-1 + l)j - lk
(2 x + 1)4 11 a ⎜ ⎟ d ⎜ ⎟
b 1 sin 2 x + 1 tan 3 x 10 a y = 3(e − 1) ⎜ ⎟ ⎜ ⎟
t
2 a b 0.906 grams c 1 gram ⎝4⎠ 1 5 ⎝ 3⎠ e r = (2 + l)i + j + 3lk
8 2 3 ⎝ ⎠ ⎝ ⎠
t 3e − 1 f r = 2li + k
n × ln 10
c 1
ln x − 1 d xsin x + cos x 1 1 Number of shakes to reduce to 10 pins = ln 2 12 a ⎛ ⎞
5
⎜ ⎟ b ⎛ ⎞
8
⎜ ⎟ c ⎛ ⎞
7
⎜ ⎟ d ⎛ ⎞
3
⎜ ⎟ 2 a r = (2 + l)i + (3 - 3l)j + (1 + 3l)k
4 x+3 −1 −3
where n = number of shakes for the half-life ⎝6⎠ ⎝4⎠ ⎝ ⎠ ⎝ ⎠ b r = (2 + 3l)i + (l - 1)j + k
e 1 ln | x 2 + 3| f x + ln |x - 1| 14 4i + j - 3k c r = (1 + 3l)i + 5lj - 2lk
2
d r = li + 2lj + 3lk
15 a m = a + b p = 3a + b q = a + 3 b
Review 11
c ln ⎛
2 ek k ⎞ e r = (1 + l)i + (-2 + l)j + (-3 + l)k
3 a
1 − 2ek
b
1 − ek
⎜
⎝ k2
⎟
− 1⎠ 1 a y = 1 x 3 + 2 ln x + c b y = ln |x 3 – 1| + c 2 4 4
3 f r = (2 + 3l)i + (-3 + 3l)j - k
c y = ln |sin x| + c d y= 1 b Trisection 2a + b a + 2 b
(c − x ) 3 3 3 a r = (2 + l)i + j - 2k
Exercise 11.1
e 1 ln 3 x − c f y=1– ke-x c Quintisection 4 a + b 3a + 2 b 2a + 3 b a + 4b b r = 2i + j + (-2 + l)k
A and c are arbitrary constants 3 5 5 5 5 4 a 2 b -5 c 0
1 a y = 3 x5 + c b y = 1 sin 2 x + c 2 a y = 2 x3 − x2 + x − 8 b y2 = e2x – 1 +1 Exercise 12.2 5 a 75.3°, intersect at (4, 7, 4)
5 2 3
1 a 6 b 20 b 24.5°, intersect at (4, 8, -6)
c y = 1 ln(x 2 + 1) + c d y = x – ln |x + 1| + c
3
3 a y = 3(x + e ) + c x
b y=k x2 +1 c 0°, parallel d 75.0°, skew e 30.7°, skew
2 2 2 3, 3
2 a 3y 3(4x + c) + 1 = 0 b e 2y = 2x + c c tan y = sin x + c d y − ln y + 1 = 1 x + c 6 a Possible equations are
3 1 1 1 2 r = (-1 + 3t)i + (1 + 2t)j + (3 + t)k for PQ
3 a b c 4 d 3 (i.e n = 1.5)
c y = Ae x – 3 d Ae x cos y = 1 e y =kx −2 f ysin y + cos y = x2 + c 3 5
1 x2 x −1 4 n - m, 2n - 2m and r = si + (5 - s)j + (2 + s)k for RS
3 a y2
– =c x2 b y = Ax c y= Ae 2 4 a 3y4 = 4x3 + 44 b y = 3x – 1 AB is twice the length of MN and is parallel to MN b (2, 3, 4) c 123.0°
d y = Ax – 1 e y2 = 2e t + c f y = Atet 1
−
1 c 2 y2 – ln |y| = -cos(x - 1) + 3 - ln 2 5 a 2a + b b 3a + b c 3a + 2 b 7 a r = (1 + t)i + (2 - 3t)j + (3 - 4t)k
g x = Ae t − 1 h Atcos x = 1 3 4 5 b 2 c 3
5 k=p 5a + 3 b na + mb
4 a y = 1 x3 + 1 x2 + x + 2 b 1 y2 − y = x − 1 1 d e 8 l = − 8 , m = 3 , 2 42
3 2 2 6 y= 8 m+n 7 7 7
1 − x sin x − cos x q − p MN 1
3
1 6 1 + k , PQ = 1 + k , MN || PQ Review 12
c 3 y = 2 x 2 + 10 d − = ln(1 + x ) + 1 − 2 7 a i v = 60t + 100 ii v = 700
y 1+ x 1 a 13 b 1 (3i + 12j + 4k)
3 RS 3 13
e sin y = sin x – 1 f e y =3− 1
b v = 40 7 (a + b), = , RS || OC
5
dA 4 OC 4 c (3i + 12j + 4k) d 76.7°
ex 8 a = kA, a = 2.5p 20.1t, 10p m2 b 29 days 13
dt 8 1:3
5 a y = ln A(x + 1) b e3y = 3x + c
2 – 4y = x2 + 4x + c d sin y = Ae x 9 a k = -0.2ln (0.8) b 6 minutes 9 k=2
c y 2 AB = 2i + 2j + 3k
e y 2 = x2 + 2x + c f sin y = Ax dx
= kx 10 PQ = SR = 1 b
10 a
dt
b 11.6 minutes 2 3 a 1 38 21 b K(4, 8, -3)
14 2
342 11 a 2 b 343
3 4 b 72.5°

Answers

b i -34 ii
22 015 Index
5 a AB = 3i + 4j - k BC = -2i + j - 4k ; 94.9° 27
b k = -14 11 b i 1 − 3 x − 9 x 2 − 27 x 3 ii 0.866 210 937
2 8 16 acceleration 268 half angle formula 62-3 graphs 122-6
1
6 (-i + 2j + 2k)
3 c 0.018% addition standard results 42 intersection of two 122
x 2 − 2 xy algebraic fractions 2 cotangent (cot) 40 intersection with x-axis 123-4
7 a −4 b 65
d 2:1 12 a
y
b
9 2 2y − x x2 − y2 numerical fractions 2 domains and ranges 41 location of roots on x-axis
8 a i - 2j + 4k + l(2i + 3j - k) y 2 (2 y + x ) y ( y − x ln y )
of vectors 277-8 formula 43 124-6
b i - 2j + 4k + m(-i + 3j - k) c -
x2 ( y + 2 x)
d x( x − y ln x ) algebraic fractions graph of 41
c 49.9°, 2.54 addition and subtraction 2 standard results 42 half angle formulae 62-3
9 b 8.3° c 1.95 13 a cos x
sin y
b ( p − 2p
2
,
3 2 3) (
and p , 2p ) division 2-3, 4-6
multiplication 2-3
cover-up rule 150, 153, 154, 184
cubic approximation 132
half-life 82, 87, 203, 269
horizontal asymptote 21
10 b (5, 0, 1) c 80.4° arccos 50, 51 hypocycloid 199
14 b 64ln 2
11 a 78.9° b skew dr 250 arcsine (arcsin) 50, 51 definite integral 229
15 a 4pr2 b =
12 a t = 3, s = -1 c 11 d 33.6° dt p r 2 (2t + 1)2 arctan 50, 51 dependent variable 8 identity, trigonometric 46-7
1000t asymptote differential equations implicit function
c V = 2t + 1 d i 4.77 cm
horizontal 21 applications 268-9 coordinate geometry 195
Revision 4
3
−
3 1 1 1 vertical 21, 40 first-order 262-5 differentiation 194-5
1 a x x+2
b − + +
2( x + 1) ( x + 1)2 2( x − 3)
16 a 1, 1.202 69, 1.414 21 b 0.8859 d 0.51%
second-order 262 improper fraction 153, 154, 156
3
2 x
(e − 2) 1
(x + 2)8 (8 x − 11) binomial expansion 180-2 differentiation 23, 91-120, 193-212 indefinite integral 228
c x+ 2 +
2 17 a 2 b
x−2 x+2 3 72 binomial series 180-2 implicit functions 194-5 independent variable 8
1 1 5 approximations 186-8 of parametric equations 168-9 initial fraction 148
2 a A = 1, B = 0, C = 2 c x + 2 sin 2x d 5
sin x
partial fractions 184 parametric functions 198 ‘inside’ function 226, 228

C3/C4
11 50
b + x + 110 x 2 + 1936 x 3 e ln |x - 1| + 8ln |x + 2| + x + 2
12 bowl, hollow, volume of 250 disc, volume of 248 integral
9 27 27 243
brackets, expanding 4, 5 displacement vector 276 definite 229
C4

3 a A = -1, B = 1 , C = 5 b p = -2, q = 3, r = 3 f e x(x + 2) g 2 x
e sin 2 x + 1 e x cos 2 x diverging sequence 130-1 indefinite 228
2 2 5 5
cancelling of fractions 2 division limits 172
( )
cos t 1 1
4 a b y= (x + 3) h ln|x3 + 3| i 1
x − 1 sin 2 x Cartesian equation 160, 161-2, 290 algebraic fractions 2-3, 4-6 integration
2 tan t sec 2 t 4 2 3 2 2 chain rule 110-11, 114-15, 168, 172, numerical fractions 2 of parametric equations 172-3
x
c y2 = 1 + x 18 1 194, 198, 200, 206, 207, 228 numerical long 4 by parts 240-2, 246
6 ln 2 in reverse, standard forms 226-7 domain 8-11 by substitution 228-9, 246
2 cos 2t p 3p 5p 7p
5 a −
sin t
b t= , ,
4 4 4 4
, 19 a l = 7, m = 3 b - 1 cos 7 x + 1 cos 3 x circle, equation of 161 dot product 286-7 as summation 222
14 6
coefficients, equating, method of 148 double angle formulae 60-2 systematic approach 246
⎛ 1 , ± 1⎞ , ⎛ −1 , ± 1⎞ 5 2
c ⎜ ⎟ ⎜ ⎟
c 21 collinear points 283 using partial fractions 238, 246
⎝ 2 ⎠ ⎝ 2 ⎠ completing the square method 20, 23 epicycloid 199 using standard forms 224
20 c y = -x + 2 - ln 2 component equal vectors 277 using trigonometric identities
e y = −2 x 1 − x 2
21 b p = 45, q = 7 form 287 equivalent forms 66-7 232-5, 246
6 a 45p 2 b 240(p - 1) c 13.6% of vector 276 exponential function (exp (x) or ex) intersection, points of 166
1
7 a 11.75 b 57 3
22 c p composite function 12-13, 110 80-2 inverse function 16-19
8 e6 −1 compound angle formulae 54-7 differentiation 98-9 algebraic method 17
a2 3 23 b 20p d a = 0.345 constant of integration 222 equations involving ln x and ex 86 graphical method 17
8 a 6y = 3a - 3x b 16
376
p continuous 124 gradient of curve 80 Inverse Laplace Transforms 158
24 a 6.75 b overestimate c 15
9 a 1 + 3x + 9x2 + 27x3, |x| < 1 converging sequence 130-1 graph of 80, 81 inverse trigonometric functions
3 −a
25 p ⎛ (2b +b1)(2a + 1) ⎞
⎜ ⎟
cosecant (cosec) 40 inverse 84 50-2
⎝ ⎠ domains and ranges 41 exponential growth and decay 200-3 irrational number 81, 98
b 1 + 1 x − 1 x 2 + 5 x 3, |x| < 2
6 36 648 graph of 41 iteration 130
26 a i t = 1, 0.5 ii V = 21 p
4 standard results 42 factorial 180 iterative formula 130-2
c 1 − 1 x − 5 x 2 + 3 x 3 , |x| < 1
2 8 16 cosine (cos) family of curves 262 iterative method 130-2
1
27 a i y = ln , k is a constant ii y = k(x + 1)
d 1 + 5 x + 11 x 2 + 17 x 3, |x| < 4 k − 2x cos (A ± B) 55 flow diagram 12
8 128 1024 derivative of 93-5 functions, mapping as 8-11 Laplace Transforms 158
1 x 2 (2 x − 3)
e 3 + 3 x + 9 x 2 + 15 x 3, |x| < 1 iii y = ke 6 iv y = sin-1 (ketan x) domains and ranges 40 limits of integral 172
2 4 8
double angle formulae 60-1 general solution 262, 263, 264, 268 logarithm 84
2
1 − 1 nx + 1 n(n − 1)x 2 − 1 n(x − 1)(n − 2)x 3, |x| < 3 b i y = ⎛ x + 1⎞
⎜
2
⎟ ii y = 10 x + 1
2
graph of 40 geometry of the sphere 207 logarithmic function (ln x) 84
10 a 3 18 162 ⎝ 4 ⎠ 2
345
344

Index

differentiation 102 Pascal's triangle 180, 181, 182 sphere, volume of 249
equations involving ex and 86 percentage errors 219 standard forms
lowest common denominator 2 polygon law 278 chain rule in reverse 226-7
position vector 282-3 integration using 224, 232
magnitude of vector 276 distance between two points 282 standard integrals 224
many-to-many function 9 midpoint of a line 283 standard trigonometric results
many-to-one function 9, 10 principal value 50, 51, 74 42-4
inverse 16 product rule 104-5, 108, 114-15, 194 straight line
mapping 8-11 proper fraction 148 cartesian equation 290
maximum 11, 95 Pythagoras’ theorem 42, 43, 276 intersection 291-2
method of equating coefficients 148 vector equation 290-2
method of substitution 149 quadratic formula 26 stretching functions 28, 29, 30, 31
minimum 11, 95 quotient rule 106-8, 114-15 substitution, method of 149
modulus 22 subtraction
equations and inequalities, range 8-11 algebraic fractions 2
solving 26 rate of change 206-7 numerical fractions 2
function 22-4 reciprocal 16 of vectors 277-8
vector 276 multiplying by 3
multiplication trigonometric functions 40-4 tangent
algebraic fractions 2-3 reflections 28, 29, 31 derivative of 93-5
numerical fractions 2 repeated linear factor 152 domains and ranges 41
by reciprocal 3 resultant 278 double angle formula 60
natural (Naperian) logarithm 84 resultant vector 278 formula 43, 168-9
C3/C4

negative vector 277 root 122, 124-6 graph of 41
number line 22 roulette 199 half angle formula 62
numerical equivalent 153 positive values 47
scalar 276 standard results 42
one-to-many function 9 scalar multiple 277 tan (A ± B) 55
inverse 16 scalar product 286-7 transformation of graphs of
one-to-one function 9, 10 secant (sec) 40 functions 28-31
inverse 16 domains and ranges 40 translations 28, 29, 30, 31
order of transformations 28, 29 graph of 40 trapezium rule 218-19
ordinate 218 sec x 94 triangle law 278
standard results 42 trigonometric equations 46-7
parabola 161 self-inverse 35 differentiation of 92-5
parallelogram law 278 separating the variables 264, 265
parameter 160 simultaneous equations 166 unit vector 276-7, 279
parametric equations sine
curve sketching and 160-2 derivative of 93-5 vector
curves with 250-1 domains and ranges 41 applications in geometry 282-3
differentiation 168-9, 198 double angle formula 60-1 basic definitions and notations
integration 172-3 graph of 41 276-9
partial fractions 148-58 half angle formula 62-3 equation of straight line 290-2
binomial series 184 sin (A ± B) 54, 93 vertical asymptote 21, 40
integration using 238, 246 standard results 42 volume of revolution 248-51
separating 148-50 skew 291
particular solution 262, 263-4, 268 special triangles 55 zero vector 276

346

Maths_2

More Related Content

Similar to Maths_2 (13)

Maths_2