Form 5 Additional Maths Note

zefry@sas.edu.my 1
FORM FIVE
ADDITIONAL MATHEMATIC NOTE
CHAPTER 1: PROGRESSION
Arithmetic Progression
Tn = a + (n – 1) d
Sn =
2
n
[2a + (n – 1)d]
=
2
n
[a + Tn]
S1 = T1 = a
T2 = S2 – S1
Example : The 15th term of an A.P. is 86 and
the sum of the first 15 terms is 555. Find
(a) the first term and the common
difference
(b) the sum of the first 20 terms
(c) the sum from the 12th to the 20th term.
(a) T15 = 86
Sn =
2
n
[a + Tn]
555 =
15
2
[a – 86]
74 = a  86
a = 12
a + 14d = 86
12 + 14d = 86
14d = 98
d = 7
(b) S20 =
20
2
[2(12) + 19(7)]
= 10[24 – 133]
= 1090
(c) Sum from 12th to 20th term = S20 – S11
= 1090 
11
2
[24 + 10(7)]
= 1090 –(253)
= 837
Geometric Progression
Tn= arn-1
Sn =
(1 )
1
n
a r
r


=
( 1)
1
n
a r
r


For 1 < r < 1, sum to infinity
1
a
S
r
 

Example: Given the 4th term and the 6th term of
a G.P. is 24 and 10
2
3
respectively. Find the 8th
term given that all the terms are positive.
ar3
= 24 ------------------(1)
ar5
= 10
2
3
=
32
3
--------(2)
(2) (1)
5
3
32 1
3 24
ar
ar
 
r2
=
4
9
r =
2
3

Since all the terms are positive, r =
2
3
a ×
3
2
3
 
 
 
= 24
a = 24 ×
27
8
= 81
T8 = 81
7
2
3
 
 
 
= 4
20
27
Example: Find the least number of terms of the
G.P 18, 6, 2,
2
3
, ... such that the last term is less
than 0.0003. Find the last term.
a = 18, r =
6 1
18 3

Tn < 0.0003
18 ×
1
1
3
n
 
 
 
< 0.0003
1
1
3
n
 
 
 
<
0.0003
18
(n – 1) log
1
3
< log
0.0003
18
n – 1 >
0.0003
log
18
1
log
3
[Remember to change the sign as log
1
3
is
negative]
n – 1 > 10.01

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n > 11.01
 n = 12.
T12 = 18×
11
1
3
 
 
 
= 0.0001016
Example: Express each recurring decimals
below as a single fraction in its lowest term.
(a) 0.7777....
(b) 0.151515...
(a) 0.7777...= 0.7 + 0.07 + 0.007 + ..
a = 0.7
r =
0.07
0.7
= 0.1
0.7 0.7 7
1 1 0.1 0.9 9
a
S
r
    
 
(b) 0.151515... = 0.15 + 0.0015 + 0.000015....
a = 0.15, r =
0.0015
0.15
= 0.01
0.15 0.15 15 5
1 0.01 0.99 99 33
S    

CHAPTER 2: LINEAR LAW
Characteristic of The Line of Best Fit
1. Passes through as many points as possible.
2. All the other points are as near to the line of
best fit as possible.
3. The points which are above and below the
line of best fit are equal in number.
To Convert from Non Linear to Linear Form
To convert to the form Y = mX + c
Example:
Non
Linear
Linear m c
y = abx
log y = log
b(x) + log a
log b log a
y = ax2
+ bx
y
ax b
x
 
a b
y = ax +
b
x
xy = ax2
+ b a b
Example: The diagram shows the line of best fit
by plotting
y
x
against x.
Find the relation between y and x.
Solution:
m =
5 1 1
6 2 2



, passing through (6, 5)
5 =
1
(6)
2
+ c c = 2
The equation is
1
2
2
y
x
x
  , or
y =
1
2
x2
+ 2x.
Example: The table below shows values of two
variables x and y, obtained from an experiment.
It is known that y is related to x by the equation
y = abx
.
x 1 2 3 4 5 6
y 4.5 6.75 10.1 15.2 22.8 34.1
(a) Explain how a straight line can be obtained
from the equation above.
(b) Plot the graph of log y against x by using a
scale of 2 cm to 1 unit on the x-axis and 2
cm to 0.2 unit on the y-axis.
(c) From your graph, find the value of a and b.
Solution:
(a) y = abx
log y = log b(x) + log a
By plotting log y against x, a straight line is
obtained.
(b)
x 1 2 3 4 5 6
log
y
0.65 0.83 1.00 1.18 1.36 1.53

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(c) c = log a = 0.48
a = 3
log b =
1.00 0.65
3 1


= 0.175
b = 1.5
CHAPTER 3: INTEGRATION
1
1
n
n x
x dx c
n

 

The area between the graph and the x-axis
A = ydx
The area between the graph and the y-axis
A = xdy
The volume generated when a shaded region is
rotated through 360o
about the x-axis
V =
2
y dx
Volume generated when a shaded region is
about the y-axis
V = 2
x dy
Example: Find
(a) 2
3 2 3x x dx 
(b)
4
6
2x x
dx
x


(a)
2
3 2 3x x dx 
=
3 2
3 2
3
3 2
x x
x c  
= x3
+ x2
+ 3x + c
(b)
4
6
2x x
dx
x

 =
4
6 6
2x x
dx
x x

= 2 5
2x x dx 

=
1 4
2
1 4
x x
c
 
 
 
= 4
1 1
2x x
  + c
The Rule of Integration
1. ( ) ( )
b b
a a
kf x dx k f x dx 
2. ( ) ( ) ( ) ( )
g b b
a a a
f x g x dx f x dx g x dx    
3. ( ) ( ) ( )
b c c
a b a
f x dx f x dx f x dx   
4. ( ) ( )
b a
a b
f x dx f x dx 
Example:
Given
3
1
( ) 9f x dx , find the value of
(a)
3
1
4 ( )f x dx
(b)
3
1
[5 ( )]f x dx
(a)
3
1
4 ( )f x dx = 4 ×
3
1
( )f x dx
= 4 × 9 = 36
(b)
3
1
[5 ( )]f x dx =
3
1
5dx
3
1
( )f x dx
=  3
1
5x + 9
= [15 – 5] + 9 = 19
Area Below a Graph
1. The area below a graph and bounded by
the line x = a, x = b and the x-axis is

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A =
b
a
y dx
2. Area between the graph and the line y = c,
y = d and the y-axis is
A =
d
c
xdy
Example:
Given A is the point of intersection between the
curve y = 5x – x2
and the line y = 2x, find the
area of the shaded region in the diagram below.
y = 2x
y = 5x – x2
2x = 5x – x2
x2
 3x = 0
x(x − 3) = 0
x = 0 or 3.  A(3, 6)
5x – x2
= 0
x(5 – x) = 0 , x = 0 or x = 5
Area under a curve = Area of triangle +
5
2
3
5x x dx =
1
2
× 3 × 6 +
52 3
3
5
2 3
x x 
 
 
= 9 +
125 125 45
[ ] 9]
2 3 2
 
   
 
= 16
1
3
unit2
Volume of Revolution
revolved through 360o
about the x-axis is
V = 2
b
a
y dx
about the y-axis is
V = 2
d
c
x dy
CHAPTER 4: VECTORS
Addition of Two Vectors
(a) Triangle Law
AB BC AC 
(b) Parallelogram Law

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AB AD AC 
Parallel Vectors
AB is parallel to PQ if
AB
k
PQ
 , where k is a
constant.
If
AB
k
BC
 , since B is a common point, A, B
and C are collinear.
Vector on Cartesian Plane
OA xi yj 
2 2
OA x y  = magnitude of vector
OA
Unit vector in the direction of
2 2
xi yj
OA
x y



Example :
Given OA = x and OB y . P is a point on AB
such that AP : PB = 1 : 2 and Q is the midpoint
of OB. The line OP intersects AQ at the point E.
Given OE = kOP and AE hAQ , where h
and k are constants,
(a) find OQ and OP in terms of x and/or
y .
(b) Express OE in terms of
(i) k, x and y ,
(ii) h, x and y .
(c) Hence, find the value of h and k.
(a) OQ =
1 1
2 2
OB y
OP OA AP  =
1
3
x AB
=
1
( )
3
x y x  =
1
(2 )
3
x y
(b) (i) OE = kOP = k ×
1
(2 )
3
x y
=
2
3 3
k k
x y
(ii) OE OA AE  = OA + h AQ
=
1
( )
2
x h x y  
= (1 – h) x +
2
h
y
(c) Compare the coefficient of x and y
1 – h =
2
3
k
---------------(1)
and
2 3
h k
 , h =
2
3
k
-----(2)
Substitute in (1)
1 
2
3
k
=
2
3
k
1 =
4
3
k
 k =
3
4
h =
2
3
k
=
2
3
×
3
4
=
1
2
.
Example :
Given
3
4
OP
 
 
 
and
1
5
OQ
 
 
 
.
(a) Find OP
(b) Find the unit vector in the direction of
OP .
(c) Given OP = mOA  n OQ and A is the
point (2, 7). Find the value of m and n.
(a) OP = 2 2
3 ( 4) 25 5   
(b) Unit vector in the direction of OP =
3 4
5
i j
(c) OP = mOA  n OQ
3 2 1
4 7 5
m n
     
      
     

zefry@sas.edu.my 6
2m – n = 3 -------(1)
7m – 5n = 4 -----(2)
(1) × 5
10m – 5n = 15 ----(3)
7m – 5n = 4 ----(4)
17m = 19
m = 
19
17
substitute in (1),
38
17
n = 3
n =
38
3
17
 = 
13
17
CHAPTER 5: TRIGONOMETRIC
FUNCTION
Angles In The Four Quadrants
The Three Trigonometric Functions
Secant  = sec  =
1
cos
Cosecant  = cosec  =
1
sin
Cotangent  = cot  =
1
tan
The Relation Between Trigonometric
Functions
Sin  = cos (90o
− )
Cos  = sin (90o
− )
Tan  = cot (90o
− )
Tan  =
sin
cos


Graphs of Trigonometric Functions
y = asin bx
Amplitude = a
Number of periods = b
The Addition Formulae
Sin (A  B) = sin A cos B  cos A sin B
Cos (A  B) = cos A cos B sin A sin B
Tan (A  B) =
tan tan
1 tan tan
A B
A B

The Double Angle Formulae
Sin 2A = 2 sin A cos A
Cos 2A = 2 cos2
A − 1
= 1 – 2 sin2
A
Tan 2A = 2
2tan
1 tan
A
A
CHAPTER 6: PERMUTATION AND
COMBINATION
1. Arrangement of n different objects without
repetition.
n
nP = n!
2. Arrangement of r objects from n objects
!
( )!
n
r
n
P
n r


Example : Given the word ‘TABLES’. Find
(a) the number of ways to arrange all the letters
in the word.
(b) The number of ways of arranging the 6
letters such that the first letter is a vowel.
(c) The number of ways of arranging 4 letters
out of the 6 letters such that the last letter is
‘S’.
(a) Number of arrangement = 6
6P = 6! = 720
(b) Number of ways of arranging 1 vowel out
of 2 = 2
1P
Number or ways of arranging the remaining
5 letters = 5
5P .
Total arrangement = 2
1P × 5
5P = 240.
OR:
Total number of ways = 2 × 5! = 240
(c) If the last letter is ‘S’, the number of ways
of arranging 3 letters out of the remaining 5
letters = 5
3P = 60.
OR:
Number of ways = 5 × 4 × 3 × 1 = 60
2 5 4 3 2 1
5 4 3 1

zefry@sas.edu.my 7
2. Combination of r objects from n objects
is
!
!( )!
n
r
n
C
r n r


Example: The PTA committee of a school
consists of 8 members. The members are elected
from 7 parents, 6 teachers and the principal of
the school. Find the number of different
committees that can be formed if
(a) the principal is one of the member of the
committee.
(b) the committee consists of the principal, 3
teachers and 4 parents.
(c) the committee consists of at least 2
teachers.
(a) number of committees = 1 13
1 7C C = 1716
(elect 1 principal, and 7 committee
members from 13 = 7 parents and 6
teachers)
(b) number of committees =
1 6 7
1 3 4C C C   700
(c) number of committees = total number of
committees – number of committees with
no teacher – number of committees with 1
teacher.
= 14 6 8 6 8
8 0 8 1 7C C C C C    = 2954.
CHAPTER 8: PROBABILITY
DISTRIBUTION
1. Binomial Distribution
The probability of getting r success in n
trials where p = probability of success and
q = probability of failure
P(X = r) = n r n r
rC p q 
2. Mean ,  = np
3. Standard deviation = npq
Example:
In a survey of a district, it is found that one in
every four families possesses computer.
(a) If 6 families are chosen at random, find the
probability that at least 4 families possess
computers.
(b) If there are 2800 families in the district,
calculate the mean and standard deviation
for the number of families which possess
computer.
(a) P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)
= 6 4 2 6 5
4 5(0.25) (0.75) (0.25) (0.75)C C
+ 6
(0.25)
= 0.03760
(b) Mean,  = np = 2800 × 0.03760
= 105.3
Standard deviation = npq =
2800 0.3760 0.6240  = 25.63
2. Normal Distribution
Z =
X 


where
Z = standard normal score
X = normal score
 = mean
 = standard deviation
P( Z < a) = 1 – P(Z >a)
P(Z < a) = P(Z > a)
P(Z > a) = 1 – P(Z > a)
P(a < Z < b) = P(Z > a) – P(Z > b)
P(a < Z < b) = 1 – P(Z > a) – P(Z > b)
Example: The volume of packet drink produced
by a factory is normal distributed with a mean of
500 ml and standard deviation of 8 ml.
Determine the probability that a packet drink
chosen at random has a volume of
(a) more than 510 ml
(b) between 490 ml and 510 ml.
(a) P(X > 510) = P(Z >
510 500
8

)
= P(Z > 1.25) = 0.1056
(b) P(490 < X < 510)
= P(
490 500
8

< Z <
510 500
8

)
= P(1.25 < Z < 1.25)
= 1 – P(Z > 1.25) – P(Z > 1.25)
= 1 – 2(0.1056)
= 0.7888
CHAPTER 9: MOTION ALONG A
STRAIGHT LINE
1. Displacement (S)
Positive displacement  particle at the
right of O

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Negative displacement  particle at the
left of O.
Return to O again  s = 0
Maximum/minimum displacement
ds
dt
= 0
Distance travelled in the nth second
= Sn – Sn-1
Example: Distance travelled in the third
second = S3 – S2
Example: A particle moves along a straight line
and its displacement, s meter, from a fixed point
O, t seconds after leaving O is given by s = 2t –
t2
. Find
(a) the displacement of the particle after 5
seconds,
(b) the time at which it returns to O again.
(c) the distance travelled in the fourth second.
(a) s = 2t – t2
t = 5,
s = 10 – 25 = 15 m
(b) Return to O again  s = 0
2t – t2
= 0
t(2 – t) = 0
t = 0 or t = 2 second
 the particle returns to O again when
t = 2 s.
(c) Distance travelled in the 4th second = S4 –
S3 = (8 – 16) – (6 – 9) =  8 + 3 =  5 m.
 the distance travelled in the 4th second is
5 m.
2. Velocity (v)
Velocity is the rate of change of
displacement with respect to time.
v =
ds
dt
Initial velocity  the value of v when
t = 0
Instantaneously at rest/change direction of
movement  v = 0
Moves towards the right  v > 0
Moves towards the left  v < 0
Maximum/minimum velocity

dv
dt
= 0
s = vdt
Distance travelled in the time interval
t = a until t = b
(a) If the particle does not stop in the time
interval
Distance =
b
a
vdt
(b) If the particle stops in the time t = c
seconds where c is in the interval a 
b
Distance =
c b
a c
vdt vdt 
passing through a fixed point O. Its velocity, v m
s-1
, t seconds after passing through O is given by
v = 2t + 3. Find the displacement at the time of 4
second.
s = vdt = 2 3t dt = t2
+ 3t + c
If the particle passes through O when t = 0, , s
= 0 when t = 0.  c = 0
 s = t2
+ 3t
When t = 4 s,
s = 16 + 12 = 28 m
Example:
A particle moves along a straight line passing
through a fixed point O. Its velocity, v m s-1
, t
seconds after passing through O is given by v =
t2
+ t – 6. Find
(a) the initial velocity of the particle,
(b) the time when the particle is momentarily at
rest.
(a) initial velocity  t = 0
v =  6 m s -1
(b) momentarily at rest  v = 0
t2
+ t – 6 = 0
(t + 3)(t – 2) = 0
t = 3 or t = 2 s
Since the time cannot be negative,  t = 2
s.
3. Acceleration (a)
Acceleration is the rate of change of
velocity with respect to time.
a =
dv
dt
=
2
2
d s
dt
Initial acceleration  a when t = 0
Deceleration  a < 0
Positive acceleration  velocity increasing
Negative acceleration  velocity
decreasing

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Zero acceleration  uniform velocity.
Maximum/minimum velocity  a = 0
v = adt
passing through a fixed point O. Its velocity, v m
s-1
, t seconds after leaving O is given by v = t2
–
6t – 7. Find
(a) the initial acceleration of the particle
(b) the time when the velocity of the particle is
maximum.
(a) a =
dv
dt
= 2t – 6
t = 0, a = 6 m s-2
(b) Maximum velocity, a = 0
2t – 6 = 0
t = 3 s.
Example : A particle moves along a straight line
passing through a fixed point O with a velocity
of 5 m s-1
. Its acceleration, a ms-2
a the time t
second after leaving O is given by a = 2t – 4.
Find the maximum velocity of the particle.
a = 2t – 4
v = 2 4t dt = t2
– 4t + c
t = 0, v = 5,  c = 5
v = t2
– 4t + 5
For maximum velocity, a = 0
2t – 4 = 0, t = 2 s.
vmax = 4 – 8 + 5 = 1 m s-1
CHAPTER 10: LINEAR PROGRAMMING
Steps
1. Form the linear inequalities.
2. Construct the region which satisfies the
constraints.
3. Form the optimum equation
ax + by = k
4. Find the point in the region which gives the
maximum or minimum value.
5. Substitute the value of x and y to obtain the
optimum value of k.
Example:
School uniform Time of preparation
(minutes)
Shirt 10
Shorts 20
The table above shows the time taken by a tailor
to prepare a shirt and a shorts of a school
uniform. In a week, the tailor sells x shirts and y
shorts. Given that in a week, the number of shirts
and shorts sold must be at least 10. The time for
preparation is at most 800 minutes. The ratio of
the number of shorts to the shirts must be at least
1 : 2.
(a) Write three inequalities other than x ≥ 0
and y ≥ 0 which satisfy the constraints
above.
(b) By using a scale of 2 cm to 10 units on the
x- and y-axes, draw the graph of all the
inequalities above. Hence, shade the region
R which satisfies the constraints above.
(c) The tailor makes a profit of RM5 and RM3
in selling a shirt and a shorts respectively.
Find the maximum profit made by the tailor
in a week.
(a) the inequalities are:
(i) x + y ≥ 10
(ii) 10x + 20y  800
x + 2y  80
(iii)
1
2
y
x

y ≥
1
2
x
(b) To draw x + y = 10, x = 0, y = 10 and y = 0,
x = 10
To draw x + 2y = 80
x = 0, 2y = 80, y = 40
y = 0, x = 80
To draw y =
1
2
x
x = 0, y = 0
x = 40, y = 20.
(c) Profit , k = 5x + 3y
Let k = 150
5x + 3y =150
x = 0, 3y = 150, y = 50
y = 0, 5x = 150, x = 30

zefry@sas.edu.my 10
From the graph, maximum profit is achieved
when x = 40 and y = 20.
 maximum profit = 5 × 40 + 3 × 20 = RM 260.
ALL THE BEST FOR YOUR SPM EXAM.

Form 5 Additional Maths Note

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Form 5 Additional Maths Note