KSSM Form 4 Additional Mathematics Notes (Chapter 1-5)

FORM 4
LAIZHIJUN
LIM JUNHAO
OOIHIANGEE
OOIMINGYANG
SIAW JIAQI

i
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Form 4 Additional Mathematics / Lim Jun Hao, Lai Zhi Jun, Ooi Hian Gee, Ooi Ming
Yang, Siaw Jia Qi
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ii
Table of Contents
Chapter 1 Functions
1.1 Functions................................................................................................................................ 1
1.2 Composite Functions.............................................................................................................. 7
1.3 Inverse Functions ............................................................................................................... 12
Summative Exercises ................................................................................................................ 18
Chapter 2 Quadratic Functions
2.1 Quadratic Equations and Inequalities ............................................................................... 19
2.2 Types of Roots of Quadratic Equations ............................................................................. 24
2.3 Quadratic Functions ........................................................................................................... 26
Summative Exercises ............................................................................................................... 38
Chapter 3 Systems of Equations
3.1 Systems of Linear Equations in Three Variables .............................................................. 39
3.2 Simultaneous Equations Involving One Linear Equation & One Non-Linear Equation . 45
Chapter 4 Indices, Surds and Logarithms
4.1 Laws of Indices.................................................................................................................... 50
4.2 Laws of Surds ..................................................................................................................... 52
4.3 Laws of Logarithms ........................................................................................................... 55
4.4 Applications of Indices, Surds and Logarithms ................................................................ 62
Chapter 5 Progressions
5.1 Arithmetic Progressions ..................................................................................................... 64
5.2 Geometric Progressions ..................................................................................................... 71
Solutions
References

CHAPTER 1 FUNCTIONS
1
Chapter 1 Functions
1.1 Functions
1.1.1 What is a function?
Let’s take an example.
If the radius of a circle is 2 cm, then the circumference, C of the circle is C = 2πr = 4π cm.
From this example, we know that the radius of a circle will affect the circumference of the circle;
therefore, the circumference of a circle is a function of radius, such that C = 2πr.
A relation can be represented by an arrow diagram as shown below.
1. f is a function from set X to set Y .
2. The element in set X is known as object, whereas the element in set Y is known as image.
Note that the term “preimage” is in common usage in math instead of the term “object”.
3. Example, the element 3 in set X and the element 1 in set Y correspond to each other whereby 3 is the
object of 1 and 1 is the image of 3.
4. Any element x in set X that is mapped to one element y in set Y by a function is written in function
notation as 𝒇 ∶ 𝒙 → 𝒚 or 𝒇(𝒙) = 𝒚 , where x is the object and y is the image.
5. A function maps an object to an image only. Therefore, the relation is called a function if the relation
is either one-to-one relation or many- to-one relation.
3
5
8
1
3
6
Set X Set Y
Look at the graph of y = x − 2 on the right. The relation
between the value of 3 on the x-axis and the value of 1 on the
y-axis can be written as 3 → 1.
We can say that 3 is mapped to 1, and likewise for 5
→ 3 and 8 → 6. Every point (x, y) on the line corresponds to
the mapping of x → y where the values of x on the x-axis are
mapped to the values of y on the y-axis.
A relation from x-axis to y-axis is called a function if
each element in the x-axis is related to exactly one element in
the y-axis. Hence, this is a function since each element from
X is related to only one element in Y.
(3,1)
(5,3)
(8,6)

CHAPTER 1 FUNCTIONS
2
(b) a 5
b 4
c 10
EXAMPLE 1
Which of the following is a function? Explain.
(a)
12 8
20 6
8 12
1
Solution:
This is a function since each object is mapped to
only one image although the element 1 has no
object.
This relation is known as one-to-one function.
Solution:
This is a function since each object has only one
image although the element 10 has no object.
This relation is known as many-to-one relation.
Solution:
This is not a function since one object has two
images. (Does not satisfy the condition of a
function)
This relation is known as one-to-many relation.
Solution:
This is not a function since each object has more
than one image. (Does not satisfy the condition of
a function)
This relation is known as many-to-many relation.
(c) 3
9 -3
1 1
-1
(d)
2 1
4 2
4

CHAPTER 1 FUNCTIONS
3
To determine whether a graph represents a function or not, we use vertical line test.
1.1.2 Introducing asymptotes
The asymptote of a curve is a line such that the distance between the line and the curve approaches zero
as one or both of the x or y coordinates tend to infinity. However, the asymptote of a curve will not touch
the curve.
The function is undefined at x = 3, therefore the graph will not touch the line x = 3.
Is there a way to calculate the vertical asymptote?
Take note that the function is undefined when denominator is 0.
In this example, the denominator of the function is x−3. We will have the function undefined when
x−3 = 0, therefore x = 3.
The vertical line intersects more than
1 point of the graph.
Hence the graph does not represent a
function.
The vertical line intersects only 1
point of the graph.
Hence the graph represents a function.
Graph of 𝑦 =
1
𝑥−3
Vertical asymptote x=3
From the graph, we know that:
1. When the x-value approaches 3 from the left
( 3−
), the corresponding y-value approaches
negative infinity (−ꚙ).
2. When the x-value approaches 3 from the right
( 3+
), the corresponding y-value approaches
infinity (ꚙ).
Function is undefined when denominator = 0,
because division by zero makes an operation
undefined.
𝑥 − 3 = 0
𝑥 = 3
y
x

CHAPTER 1 FUNCTIONS
4
1.1.3 Introducing the absolute-value function
1.1.4 How to determine the domain and the range of a discrete function?
Let’s take back Example 1(a).
a)
12 8
20 6
8 12
1
1.1.5 How to determine the domain and the range of a continuous function?
In this example, the domain is −5 ≤ 𝑥 ≤ 6 whereas the range is −1 ≤ 𝑓(𝑥) ≤ 1.
Graph of y=|x|
The absolute value |x| is expressed as:
For instance, |1| = 1; |−1| = 1.
|x| =
x when x ≥ 0
−x when x < 0
Set X Set Y
Domain Codomain
1. The domain of a function is the set of all possible input
values.
2. The codomain of a function is the set into which all of the
output function is forced to fall.
3. The range of a function is the set of all possible output
values.
In this example,
Domain = {12, 20, 8}
Codomain = {8, 6, 12, 1}
Range = {8, 6, 12}
Range
Domain
𝑓( 𝑥) = 𝑠𝑖𝑛 𝑥
y
x
x
y

CHAPTER 1 FUNCTIONS
5
Sketch the graph of f : x → |2x–4| in the domain −3 ≤ 𝑥 ≤ 3.
Solution:
Step 1: You need to have 4 points in order to construct a graph of an
absolute-value function.
The four points are:
(a) vertex,
(b) y-intercept,
(c) left endpoint of the graph,
(d) right endpoint of the graph.
Step 2: You need to plot the points onto a graph paper and indicate the x-axis and y-axis.
Step 3: Connect the points in the shape of V with vertex as the turning point.
x |2𝑥 − 4|
-3 10
0 4
2 0
3 2
The vertex can be obtained by equating the
algebraic expression inside the modulus
sign with 0 and solving it.
In this example,
2x – 4 = 0
2x = 4
x = 2
when x = 2,
f(2) = |2(2) – 4| = 0
Hence the coordinate of the vertex is (2,0).
EXAMPLE 2
𝑓( 𝑥) = |2𝑥 − 4|
(-3,10)
(0,4)
(2,0)
(3,2)
This is the vertex.

CHAPTER 1 FUNCTIONS
6
EXAMPLE 3
A function f is defined by f : x → 2x + 2, find
(a) the image of 2,
(b) the object that have the image 4.
EXAMPLE 4
A function f is defined by f : x → for all values of x, except x = b, where a is a constant.
(a) State the value of b.
(b) Find the value of a given that the value 5 is mapped to itself under f.
1.1.6 How to determine the image of a function when its domain is given and vice
versa?
Solution:
(a) f : x → 2x + 2
f (x) = 2x + 2
When x = 2,
f (2) = 2(2) + 2
= 6
Hence, the image of 2 is 6.
(b) When f (x) = 4
2x + 2 = 4
2x = 2
x = 1
Hence the object that has the image 4 is 1.
Solution:
(a) Function is undefined when:
denominator = 0
x – 2 = 0
x = 2
It is given that function is undefined
when x = b, therefore b = 2.
(b) 5 is mapped to itself, therefore
𝑓(5) = 5
2(5) + 𝑎
5 − 2
= 5
10 + 𝑎 = 5(3)
𝑎 = 5
Note: For a function, for example f : x → 2x + 2,
if x = 1, then f (1) = 2(1) + 2 = 4;
if x = 2, then f (2) = 2(2) + 2 = 6;
if x = u, then f (u) = 2(u) + 2 = 2u + 2.
2𝑥 + 𝑎
𝑥 − 2

CHAPTER 1 FUNCTIONS
7
1.2 Composite Functions
1.2.1 What are composite functions?
For example, let x be a tree. In order to produce furniture, the tree (x) undergoes cutting process (function
f) to obtain wood (f (x)). The wood (f (x)), is then manufactured (Function g) to produce furniture (gf (x)).
fg(x) is read as “f composed with g of x” . fg(x) = f[g(x)].
gf(x) is read as “g composed with f of x” . gf(x) = g[f(x)].
Note: 𝒇 𝟐( 𝒙) = 𝒇𝒇( 𝒙).
1.2.2 How to determine composite functions?
Two functions f and g are defined by f (x) = x + 1 and g (x) = 𝑥2
. Find the following composite
functions:
(a) 𝑓2 ( 𝑏) 𝑔2 ( 𝑐) 𝑓𝑔 ( 𝑑) 𝑔𝑓
Solution:
(a) 𝑓2( 𝑥) = 𝑓𝑓( 𝑥)
= 𝑓[𝑓(𝑥)]
= 𝑓(𝑥 + 1)
= ( 𝑥 + 1) + 1
= 𝑥 + 2
(b) 𝑔2( 𝑥) = 𝑔𝑔( 𝑥)
= 𝑔[𝑔(𝑥)]
= 𝑔( 𝑥2)
= (𝑥2
)2
= 𝑥4
(c) 𝑓𝑔( 𝑥) = 𝑓[𝑔( 𝑥)]
= 𝑓(𝑥2
)
= 𝑥2
+ 1
(d) 𝑔𝑓( 𝑥) = 𝑔[𝑓( 𝑥)]
= 𝑔(𝑥 + 1)
= (𝑥 + 1)2
= 𝑥2
+ 2𝑥 + 1
x f (x) gf (x)
f g
EXAMPLE 5
f (x) = x + 1
f (x + 1) = (x + 1) + 1

CHAPTER 1 FUNCTIONS
8
1.2.3 How to determine the objects or the images of composite functions?
Two functions f and g are defined by 𝑓( 𝑥) = 𝑥 + 2 and 𝑔( 𝑥) = 𝑥2
, find 𝑔𝑓(𝑥). Then, find the value
of 𝑔𝑓(2).
Solution:
𝑔𝑓( 𝑥) = 𝑔[𝑓( 𝑥)]
= 𝑔(𝑥 + 2)
= (𝑥 + 2)2
= 𝑥2
+ 4𝑥 + 4
𝑔𝑓(2) = 22
+ 4(2) + 4
= 16
Two functions f and g are defined by f (x) = x − 2 and g (x) = 𝑥2
+ 2. Find the values of x when
gf (x) = 6.
Solution:
𝑔𝑓( 𝑥) = 𝑔[𝑓( 𝑥)]
= 𝑔(𝑥 − 2)
= (𝑥 − 2)2
+ 2
= (𝑥2
− 4𝑥 + 4) + 2
= 𝑥2
− 4𝑥 + 6
𝑔𝑓( 𝑥) = 6
𝑥2
− 4𝑥 + 6 = 6
𝑥2
− 4𝑥 = 0
( 𝑥)( 𝑥 − 4) = 0
𝑥 = 0, 𝑥 = 4
EXAMPLE 7
EXAMPLE 6
𝑓( 𝑥) = 𝑥 + 2
Tip: The word “values”
indicates that there will
be at least 2 values of x

CHAPTER 1 FUNCTIONS
9
1.2.4 How to determine a function when the composite function and one of the
functions are given?
(a) A function f is defined as 𝑓 ∶ 𝑥 → 𝑥 + 2. Find the function g such that 𝑓𝑔 ∶ 𝑥 → 3𝑥 + 4.
(b) A function f is defined as 𝑓 ∶ 𝑥 → 𝑥 + 2. Find the function g such that 𝑔𝑓 ∶ 𝑥 → 𝑥 + 5.
Solution:
(a) (Case where the function determined is
situated “inside” the composite function)
𝑓𝑔 ∶ 𝑥 → 3𝑥 + 4
𝑓𝑔 ( 𝑥) = 3𝑥 + 4
𝑓𝑔 ( 𝑥) = 𝑔( 𝑥) + 2
3𝑥 + 4 = 𝑔( 𝑥) + 2
𝑔( 𝑥) = 3𝑥 + 2
(b) (Case where the function determined is
situated “outside” the composite function)
𝑔𝑓 ( 𝑥) = 𝑥 + 5
𝑔( 𝑥 + 2) = 𝑥 + 5
𝑔( 𝑢) = ( 𝑢 − 2) + 5
𝑔( 𝑢) = 𝑢 + 3
𝑔( 𝑥) = 𝑥 + 3
EXAMPLE 8
How to change
𝒈(𝒙 + 𝟐) to 𝒈(𝒙)?
𝐿𝑒𝑡 𝑥 + 2 = 𝑢
𝑥 = 𝑢 − 2
Note: For a function, for example f : x → 2x + 2,
if x = 1, then f (1) = 2(1) + 2 = 4;
if x = 2, then f (2) = 2(2) + 2 = 6;
if x = u, then f (u) = 2(u) + 2 = 2u + 2.

CHAPTER 1 FUNCTIONS
10
1.2.5 How to determine the domain of a composite function?
If ƒ and g are functions, the composite function (“ƒ composed with g”) is defined by ƒ(g(x)).
The domain of this composite function consists of the numbers x in the domain of g for which g(x) lies
in the domain of ƒ.
Similarly, if ƒ and g are functions, the composite function (“g composed with f”) is defined by g(f(x)).
The domain of this composite function consists of the numbers x in the domain of f for which f(x) lies in
the domain of g.
Two functions f and g are defined by 𝑓(𝑥) = √ 𝑥 and 𝑔( 𝑥) = 𝑥 − 2. Find 𝑔𝑓(𝑥) and its domain.
Solution:
𝑓𝑔( 𝑥) = 𝑓(𝑥 − 2)
= √𝑥 − 2
Notice that 𝑔(𝑥) = 𝑥 − 2 is defined for all real x but 𝑔(𝑥) belongs to the domain of ƒ only if x ≥ 0.
The domain of the composite function 𝑔𝑓( 𝑥) = √𝑥 − 2 is 𝑥 ≥ 2 .
1.2.6 Further examples and applications on composite functions
A function f is defined as 𝑓: 𝑥 →
1
𝑥
, 𝑥 ≠ 0. Find 𝑓2
, 𝑓4
, 𝑓5
and 𝑓20
.
Solution:
𝑓( 𝑥) =
1
𝑥
𝑓2( 𝑥) = 𝑓𝑓( 𝑥)
= 𝑥
𝑓4( 𝑥) = (𝑓2
)2( 𝑥)
= 𝑓2
[𝑓2( 𝑥)]
= 𝑓2( 𝑥)
= 𝑥
𝑓5( 𝑥) = 𝑓[𝑓4( 𝑥)]
= 𝑓( 𝑥)
𝑓20 ( 𝑥) = 𝑓4
𝑓4
𝑓4
𝑓4
𝑓4( 𝑥)
= 𝑥
EXAMPLE 9
𝑓2( 𝑥) = 𝑥
= 𝑓 (
1
𝑥
)
=
1
1
𝑥 =
1
𝑥
EXAMPLE 10
𝑥 − 2 ≥ 0

CHAPTER 1 FUNCTIONS
11
In a grocery store, the discounted price, q of a product A is denoted as 𝑞( 𝑥) =
1
2
𝑥, where x is the
original price of the product. Upon checking out, the product will be charged service tax, p and is
denoted by 𝑝( 𝑞) = 1.2𝑞. How much does product A costs if the original price of the product is
RM20?
Solution:
The question is to find 𝑝𝑞(20).
𝑝𝑞( 𝑥) = 𝑝 (
1
2
𝑥)
= 1.2 (
1
2
𝑥)
= 0.6𝑥
𝑝𝑞(20) = 0.6(20)
= 12
Therefore product A costs RM12.
EXAMPLE 11
pq

CHAPTER 1 FUNCTIONS
12
1.3 Inverse Functions
1.3.1 What are inverse functions?
Let’s take an example of ice and water.
Water undergoes freezing process to become ice. Freezing process is denoted by the function f.
Ice undergoes melting process to become water again. Melting process is the inverse of freezing process
and is denoted by the inverse function 𝑓−1
.
1.3.2 What are the characteristics of an inverse function?
1) A function f that maps set X to set Y has an inverse function, 𝑓−1
if f is a one-to-one function.
(Inverse function, 𝑓−1
is valid when each element in set Y is mapped onto one and only one element in
set X and the function f is valid)
2) The inverse of an inverse function gives back the original function.
Example 𝒇𝒇−𝟏( 𝒙) = 𝒙 and 𝒇−𝟏
𝒇( 𝒙) = 𝒙
3) If two functions f and g are inverses of each other:
a) The domain of f = the range of g.
b) The domain of g = the range of f.
c) The graph of function f is the reflection of graph of function g across the line y = x.
4) If (a,b) lies on the graph f , where a and b are real numbers, then (b,a) lies on the graph 𝒇−𝟏
since
graph f is the reflection of graph 𝑓−1
at line y = x.
Note:
1.3.3 How to determine whether the function is one-to-one?
A function ƒ(x) is one-to-one on a domain D if 𝑓(𝑥1) ≠ 𝑓(𝑥2) whenever 𝑥1 ≠ 𝑥2 in D.
For example, 𝑓( 𝑥) = 𝑥3
is one-to-one on any values of x in the domain D because 𝑥1
3
≠ 𝑥2
3
whenever 𝑥1 ≠ 𝑥2
in D.
Water Ice
f
𝑓−1
𝒇−𝟏( 𝒙) ≠
𝟏
𝒇( 𝒙)
𝑓(𝑥)
x
𝑓( 𝑥) = 𝑥3
𝑥2𝑥1
𝑓(𝑥1)
𝑓(𝑥2)

CHAPTER 1 FUNCTIONS
13
1.3.4 How to determine whether the graph of function has an inverse function?
We use horizontal line test to test whether the graph of function has an inverse function.
Given that 𝑓 (3) = 5 and 𝑓 (1) = 3. Determine 𝑓−1(5) and 𝑓−1(3).
Solution:
𝑓 (3) = 5 therefore 𝑓−1(5) = 3
𝑓 (1) = 3 therefore 𝑓−1(3) = 1
Given that 𝑓: 𝑥 →
1
𝑥
, 𝑥 ≠ 0. Find 𝑓−1(3).
Solution:
𝐿𝑒𝑡 𝑎 = 𝑓−1
(3)
𝑓( 𝑎) = 3
𝑦 = 𝑥2
𝑦 = 𝑥2
is a function.
However, it does not have an inverse
function since horizontal line cuts the
graph of function at least 2 points.
𝑦 = 𝑥 + 2
𝑦 = 𝑥 + 2 is a function.
It has an inverse function since horizontal
line cuts the graph of function at only 1
point.
x
x
y y
EXAMPLE 13
𝑓( 𝑎) =
1
𝑎
𝑓( 𝑥) = 𝑦
𝑓−1( 𝑦) = 𝑥
EXAMPLE 12

CHAPTER 1 FUNCTIONS
14
To determine whether two functions f and g are inverses of each other, we need to make sure both the
domain for 𝑓𝑔(𝑥) and 𝑔𝑓(𝑥) are the same.
Determine whether the function has an inverse of .
Solution:
Check the domain of the function 𝑓(𝑥).
Function is undefined when denominator = 0
𝑥 + 2 = 0
𝑥 = −2
Therefore the domain of the function 𝑓(𝑥) is 𝑥 < −2 and 𝑥 > −2.
Check the domain of the function 𝑔(𝑥).
Function is undefined when denominator = 0
𝑥 − 1 = 0
𝑥 = 1
Therefore the domain of the function 𝑔(𝑥) is 𝑥 < 1 and 𝑥 > 1.
Check the domain of the function 𝑓𝑔(𝑥).
Notice that is defined for all real x except 𝑥 = 1 but 𝑔(𝑥) belongs to the domain of
ƒ only if 𝑥 ≠ −2.
The domain of the composite function 𝑓𝑔( 𝑥) = 𝑥 is 𝑥 < −2, −2 < 𝑥 < 1, 𝑥 > 1.
Check the domain of the function 𝑔𝑓(𝑥).
Notice that is defined for all real x except 𝑥 = −2 but 𝑓(𝑥) belongs to the domain of 𝑔
only if 𝑥 ≠ 1.
The domain of the composite function 𝑔𝑓( 𝑥) = 𝑥 is 𝑥 < −2, −2 < 𝑥 < 1, 𝑥 > 1.
EXAMPLE 14
𝑓( 𝑥) =
𝑥 − 3
𝑥 + 2
𝑔( 𝑥) = −
3 + 2𝑥
𝑥 − 1
𝑓( 𝑥) =
𝑥 − 3
𝑥 + 2
𝑔( 𝑥) = −
3 + 2𝑥
𝑥 − 1

CHAPTER 1 FUNCTIONS
15
Since 𝒇𝒈( 𝒙) = 𝒈𝒇( 𝒙) = 𝒙 and both the domain for 𝑓𝑔(𝑥) and 𝑔𝑓(𝑥) are the same, therefore
is an inverse of .
A function f is defined as 𝑓: 𝑥 → 𝑥2
+ 1 for the domain 0 ≤ 𝑥 ≤ 3.
(a) Sketch the graphs of 𝑓 and 𝑓−1
on the same graph.
(b) State the domain and image of 𝑓 and 𝑓−1
.
Solution:
(a)
The graph of f is a part of the quadratic curve y = 𝑥2
+ 1.
The graph of 𝑓−1
is a reflection of the graph f by the line y=x
(b) Domain of f is 0 ≤ 𝑥 ≤ 3.
Range of f is 1 ≤ 𝑓( 𝑥) ≤ 10.
Domain of 𝑓−1
is 1 ≤ 𝑥 ≤ 10.
Range of 𝑓−1
is 0 ≤ 𝑓−1
(𝑥) ≤ 3
x 0 1 2 3
y 1 2 5 10
EXAMPLE 15
f
𝑓−1
y=x
(0,1)
(1,0)
(3,10)
(10,3)
𝑔𝑓( 𝑥) = 𝑔 (
𝑥 − 3
𝑥 + 2
)
=
−5𝑥
𝑥 + 2
×
𝑥 + 2
−5
= 𝑥
𝑥 − 3 − (𝑥 + 2)
𝑥 + 2
= −
3 + 2 (
𝑥 − 3
𝑥 + 2
)
𝑥 − 3
𝑥 + 2 − 1
=
−3( 𝑥 + 2) − 2(𝑥 − 3)
𝑥 + 2
÷
𝑔( 𝑥) = −
3 + 2𝑥
𝑥 − 1
𝑓( 𝑥) =
𝑥 − 3
𝑥 + 2

CHAPTER 1 FUNCTIONS
16
1.3.5 How to determine the inverse function?
There are a few steps to determine the inverse function:
Step 1: Change the function from 𝑦 = 𝑓(𝑥) to 𝑥 = 𝑓(𝑦).
Step 2: Write x as 𝑓−1
(𝑦).
Step 3: Substitute the variable y with x.
If 𝑓( 𝑥) = 3𝑥 + 1, find 𝑓−1( 𝑥) and hence find 𝑓−1(4).
Solution:
𝑓( 𝑥) = 3𝑥 + 1
𝐿𝑒𝑡 𝑦 = 3𝑥 + 1
3𝑥 = 𝑦 − 1
Always do silly mistakes?
You can check the validity of the inverse function by using the characteristic of inverse function.
You must make sure that 𝒇𝒇−𝟏( 𝒙) = 𝒙 and 𝒇−𝟏
𝒇( 𝒙) = 𝒙.
1.3.6 Further examples on inverse functions
Inverse functions do apply in our everyday life.
1) For example, you are running a marathon. You travelled 10 miles per hour constantly, and want to
know how far you have gone in x hours. This could be represented by the function 𝑓( 𝑥) = 10𝑥. Now if
I know I have travelled x miles, then how long I have been travelling for? This could be represented by
the inverse function of .
2) For example, 3 Malaysian Ringgit (MYR) is equivalent to 1 Singaporean Dollar (SGD). The currency
from x SGD to MYR could be represented by the function 𝑓( 𝑥) = 3𝑥. Now if I have x MYR and I
want to convert it to SGD, how much is it? This could be represented by the inverse function of
.
EXAMPLE 16
Step 1: Change the function from 𝑦 = 𝑓(𝑥) to 𝑥 = 𝑓(𝑦).
Step 2: Write x as 𝑓−1
(𝑦).
Step 3: Substitute the variable y with x.
𝑓−1( 𝑦) =
𝑦 − 1
3
𝑥 =
𝑦 − 1
3
𝑓−1( 𝑥) =
𝑥 − 1
3
𝑓−1(4) =
4 − 1
3
= 1
𝑓−1( 𝑥) =
𝑥
10
𝑓−1( 𝑥) =
𝑥
3

CHAPTER 1 FUNCTIONS
17
Further Exploration
Functions (Part I)
https://youtu.be/oCf4-G6wojY
Functions (Part 2)
https://youtu.be/g7RZ5aV7RG4
Composite Functions (Part 1)
https://youtu.be/TIdFFbM2MNA
https://youtu.be/SpK4wsd_W6g
https://youtu.be/uZTuBdNYFxQ
Inverse Functions
https://youtu.be/LsXrVOdAgOI

CHAPTER 1 FUNCTIONS
18
1) A function f is defined by 𝑓: 𝑥 → 𝑥2
+ 4. State
(a) the image of 4,
(b) the object that have the image 4.
2) A function f is defined by f : x → for all
values of x, except x = q and p is a constant.
(a) State the value of q.
(b) Find the value of p given that the value 2 is
mapped to itself under f.
3) Sketch the graph of 𝑓: 𝑥 → |𝑥 + 3| in the domain
−4 ≤ 𝑥 ≤ 1. Hence, state the range of the
function.
1) Given the functions 𝑓( 𝑥) = 2𝑥 + 1 and
𝑓𝑔( 𝑥) = 2𝑥2
+ 5. Find 𝑔𝑓( 𝑥). Hence, find the
value of x when 𝑔𝑓( 𝑥) = 2.
2) Given the function , 𝑥 ≠ 1. Find
(a) 𝑓2( 𝑥),
(b) 𝑓3( 𝑥),
(c) 𝑓6( 𝑥).
3) Roland got lost from Robin Hood, his dad in the
woods. Robin Hood is determining the area in
which to search. Roland can walk at an estimated
5 miles per hour. [𝑈𝑠𝑒 𝑡ℎ𝑒 𝑠𝑦𝑚𝑏𝑜𝑙 𝜋]
(a) Write the function 𝑟(𝑥) which would be the
distance Roland could walk in x hours.
(b) Write the function 𝑝(𝑟) which would give the
potential search area given Roland walked r
miles.
(c) Find the composite function of area for Robin
Hood to search for him.
1.1 Functions
1. 2 Composite Functions
Summative Exercises
4)
The diagram shows a container having the shape
of a cylinder. It had 1540𝑐𝑚3
of water and is
being filled with water at a constant rate. The
height of the water in the container increases at a
constant rate of 1𝑐𝑚 per second as water was
poured in. The radius and height of the container
is 7𝑐𝑚 and 50𝑐𝑚 respectively.
(a) Write the function for the height h of water
after t seconds.
(b) Write the function for the volume V of the
water in terms of h, height of water.
(c) Find the function 𝑉ℎ(𝑡).
(d) Hence, find the volume of water in 𝑐𝑚3
after
10 seconds.
1) Given the functions , 𝑥 ≠ 2 and
𝑔: 𝑥 → 3 − 𝑥. Find
(a) 𝑓−1
(b) 𝑔−1
(c) 𝑔−1
𝑓−1
(d) (𝑓𝑔)−1
(e) (𝑔𝑓)−1
Hence, find the function which is the same as
𝑔−1
𝑓−1
.
2) Given the function 𝑔: 𝑥 → 𝑥 − 4.
(a) Find the expressions for 𝑔2
and 𝑔−1
.
(b) Show that (𝑔−1
)2
= (𝑔2
)−1
.
(c) What is the value of x when
𝑔𝑔−1 ( 𝑥) = 𝑔−1(5𝑥 + 2).
2𝑝 + 3
𝑥 − 4
𝑓 ∶ 𝑥 ⟶
1 + 𝑥
1 − 𝑥
7cm
25cm
[𝑈𝑠𝑒 𝜋 =
22
7
]
𝑓: 𝑥 →
2
𝑥 − 2

CHAPTER 2 QUADRATIC FUNCTIONS
19
CHAPTER 2 Quadratic Functions
2.1.1 How to solve quadratic equations?
• By using Completing The Square method
EXAMPLE 1
By using the completing the square method to solve the following equation.
2𝑥#
+ 7𝑥 + 6 = 0.
2𝑥#
+ 7𝑥 + 6 = 0
𝑥#
+
7
2
𝑥 + 3 = 0
𝑥#
+
7
2
𝑥 = −3
𝑥#
+
7
2
𝑥 + +
,
7
2
-
2
.
#
= −3 + +
,
7
2
-
2
.
#
𝑥#
+
7
2
𝑥 + ,
7
4
-
#
= −3 +
49
16
,𝑥 +
7
4
-
#
=
1
16
𝑥 +
7
4
= ±3
1
16
𝑥 = −
3
2
𝑜𝑟 𝑥 = −2
Hence, the solutions of the equation 2𝑥#
+ 7𝑥 + 6 = 0 are −
7
#
and −2.

Move the constant term to the RHS
of the equation
Add the term
8
9:;<<=9=;>? :< @
#
A
#
to
the both RHS and LHS
of the equation
( 𝑥 + 𝑎)#
= 𝑥#
+ 2𝑎𝑥 + 𝑎#
Factorise the equation by 2 so that the
coefficient of 𝑥#
becomes 1
2.1 Quadratic Equations and Inequalities

20
• Formula method
is used to solve a quadratic equation 𝑎𝑥#
+ 𝑏𝑥 + 𝑐 = 0
2.1.2 How to form quadratic equations from given roots?
Assume the given roots of a quadratic equation are 𝛼 and 𝛽 then we calculate the sum of roots and the
product of roots.
Therefore, the quadratic equation with the given roots a and b can be written as:
𝑥 =
−𝑏 ± I 𝑏# − 4𝑎𝑐
2𝑎
EXAMPLE 2
By using the formula to solve 2𝑥#
+ 7𝑥 + 6 = 0.
Compare the given equation 2𝑥#
+ 7𝑥 + 6 = 0 with the general form
equation 𝑎𝑥#
+ 𝑏𝑥 + 𝑐 = 0. Hence, 𝑎 = 2, 𝑏 = 7 and 𝑐 = 6.
𝑥 =
−7 ± J(7)# − 4(2)(6)
2(2)
𝑥 =
−7 ± I1
4
𝑥 = −
3
2
𝑜𝑟 𝑥 = −2
Hence, the solutions of the equation 2𝑥#
+ 7𝑥 + 6 = 0 are −
7
#
and −2.
Sum of roots = 𝛼 + 𝛽 = −
K
L
Product of roots = 𝛼𝛽 =
9
L
𝑥#
− M𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠S𝑥 + M𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠S = 0

21
EXAMPLE 3
Form a quadratic equation where the roots are 5 and -7.
𝛼 = 5 𝑎𝑛𝑑 𝛽 = −7
Sum of roots = 5 + (−7) = −2
Product of roots = 5 × (−7) = −35
𝑥#
− M𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠S𝑥 + M𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠S = 0
𝑥#
− (−2) 𝑥 + (−35) = 0
𝑥#
+ 4𝑥 − 35 = 0
EXAMPLE 4
Assume a and b are the roots of the quadratic equation 2𝑥#
− 3𝑥 − 5 = 0,
form a quadratic equation with the following roots:
(a) 2𝛼, 2𝛽
2𝑥#
− 3𝑥 − 5 = 0 where 𝑎 = 2, 𝑏 = −3 𝑎𝑛𝑑 𝑐 = −5
𝛼 + 𝛽 = −
𝑏
𝑎
=
3
2
𝛼𝛽 =
𝑐
𝑎
= −
5
2
Sum of roots: Product of roots:
2𝛼 + 2𝛽 = 2M𝛼 + 𝛽S (2𝛼)M2𝛽S = 4𝛼𝛽
= 2 8
7
#
A = 4 8−
Z
#
A
= 3 = −10
Thus, the quadratic equation with the roots 2a and 2b is
𝑥#
− 3𝑥 − 10 = 0.

22
2.1.4 How to solve quadratic inequalities?
• By using graph sketching method
• Number line method
• Table method
EXAMPLE 5
Find the range of values of 𝑥 for the following quadratic inequalities
𝑥#
− 11𝑥 + 24 ≥ 0 by using graph sketching, number line and table method.
Graph sketching method:
𝑥#
− 11𝑥 + 24 ≥ 0
( 𝑥 − 8)( 𝑥 − 3) ≥ 0
When ( 𝑥 − 8)( 𝑥 − 3) = 0, 𝑥 = 8 𝑎𝑛𝑑 𝑥 = 3
Thus, the graph will intersect the 𝑥-axis at the point 𝑥 = 8 𝑎𝑛𝑑 𝑥 = 3. Since
( 𝑥 − 8)( 𝑥 − 3) ≥ 0, hence the values of 𝑥 is determined on the graph above
the 𝑥-axis. Therefore, the range of values of 𝑥 is 𝑥 ≤ 3 𝑜𝑟 𝑥 ≥ 8.
𝑦
𝑥
3 8
+ +
-
𝑦 = ( 𝑥 − 8)(𝑥 − 3)

23
Number line method:
Test point 2: Test point 5: Test point 9:
(2 − 8)(2 − 3) ≥ 0 (5 − 8)(5 − 3) ≤ 0 (9 − 8)(9 − 3) ≥ 0
Since ( 𝑥 − 8)( 𝑥 − 3) ≥ 0, then the range of values of 𝑥 is determined on the
positive part of the number line. Hence, the range of values of 𝑥 is
𝑥 ≤ 3 𝑜𝑟 𝑥 ≥ 8.
Table method:
Range of 𝑥
𝑥 ≤ 3 3 ≤ 𝑥 ≤ 8 𝑥 ≥ 8
( 𝑥 − 3) − + +
( 𝑥 − 8) − − +
( 𝑥 − 8)( 𝑥 − 3) + − +
Since ( 𝑥 − 8)( 𝑥 − 3) ≥ 0, then the range of values of 𝑥 is determined on
the positive part of the number line. Hence, the range of values of 𝑥 is
𝑥 ≤ 3 𝑜𝑟 𝑥 ≥ 8.
+ +-
3 8
𝑥 ≤ 3 𝑥 ≥ 83 ≤ 𝑥 ≤ 8

24
2.2.1 What are the types of roots of quadratic equations and the value of
discriminant?
What is a discriminant? A discriminant is the part under the square root in the quadratic formula
𝑏#
− 4𝑎𝑐. The discriminant will tell us whether there are two solutions, one solution or no solution in
the quadratic equation.
In general:
• 𝑏#
− 4𝑎𝑐 > 0, the equation has two different solutions or real roots.
• 𝑏#
− 4𝑎𝑐 < 0, the equation has no real roots.
• 𝑏#
− 4𝑎𝑐 = 0, the equation has one solution.
2.2.2 How to solve the problems involving types of roots of quadratic
equations?
The discriminant besides that telling us the types of roots of the quadratic equation, it can be
used to:
• Find the value of a variable in the quadratic equation
• Derive a relation
𝑥 =
−𝑏 ± I 𝑏# − 4𝑎𝑐
2𝑎
discriminant
EXAMPLE 6
Determine the type of roots for the quadratic equation 6𝑥#
+ 10𝑥 − 1 = 0.
𝑎 = 6, 𝑏 = 10 𝑎𝑛𝑑 𝑐 = −1
𝑏#
− 4𝑎𝑐 = (10)#
− 4(6)(−1)
= 124
𝑏#
− 4𝑎𝑐 > 0
Thus, the quadratic equation 6𝑥#
+ 10𝑥 − 1 = 0 has two different real roots.
2.2 Types of Roots of Quadratic Equations

25
EXAMPLE 7
Find the value of a variable in the quadratic equation.
The quadratic equation 4𝑥#
+ 2𝑥 + 9 = 𝑘𝑥, where k is a constant, it has two
real roots. Find the possible values for k.
4𝑥#
+ 2𝑥 + 9 = 𝑘𝑥
4𝑥#
+ (2 − 𝑘)𝑥 + 9 = 0
𝑎 = 4 , 𝑏 = 2 − 𝑘, 𝑐 = 9
Since it has two real roots, we use 𝑏#
− 4𝑎𝑐 > 0.
𝑏#
− 4𝑎𝑐 > 0
(2 − 𝑘)#
− 4(4)(9) > 0
4 − 8𝑘 + 𝑘#
− 144 > 0
𝑘#
− 8𝑘 − 140 > 0
( 𝑘 − 14)( 𝑘 + 10) > 0
𝑘 = 14 𝑜𝑟 𝑘 = −10
Has two real roots.
EXAMPLE 8
Derive a relation.
The quadratic equation 𝑥#
− 6𝑚𝑥 + 11𝑛 = 0 has one real root, express 𝑚 in
terms of 𝑛.
𝑥#
− 6𝑎𝑥 + 11𝑏 = 0
𝑎 = 1 , 𝑏 = −6𝑚, 𝑐 = 11𝑛
Since it has only one real root, we use 𝑏#
− 4𝑎𝑐 = 0
𝑏#
− 4𝑎𝑐 = 0
(−6𝑚)#
− 4(1)(11𝑛) = 0

26
2.3.1 What are the effect of changes on the shape and position of the quadratic
equation 𝑓( 𝑥) = 𝑎𝑥#
+ 𝑏𝑥 + 𝑐 when the values of 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 change?
36𝑚#
− 44𝑛 = 0
36𝑚#
= 44𝑛
𝑚#
=
bb
c
𝑛
𝑚 = ±J
bb
c
𝑛
EXAMPLE 9
The diagram shows the sketch of graph for 𝑓( 𝑥) = 2𝑥#
+ 4𝑥 + 5 where
𝑎 = 2, 𝑏 = 4 𝑎𝑛𝑑 𝑐 = 5. Resketch the shape and position of the graph when
the following values changes.
5
3
1
𝑓(𝑥)
𝑥
𝑓( 𝑥) = 2𝑥#
+ 4𝑥 + 5
2.3 Quadratic Functions

27
(a) The values of 𝑎 become 1 and 6
(b) The values of 𝑐 become 3 and 7
𝑓(𝑥)
𝑥
3
5
1
𝑓( 𝑥) = 2𝑥#
+ 4𝑥 + 5
𝑓( 𝑥) = 6𝑥#
+ 4𝑥 + 5
𝑓( 𝑥) = 𝑥#
+ 4𝑥 + 5
𝑓(𝑥)
𝑓( 𝑥) = 2𝑥#
+ 4𝑥 + 5
3
5
1
𝑓( 𝑥) = 2𝑥#
+ 4𝑥 + 7
𝑓( 𝑥) = 2𝑥#
+ 4𝑥 + 3

28
2.3.2 What is the relationship between the graph of quadratic function and type
of roots?
𝑏#
− 4𝑎𝑐 Types of roots and position
of graph
𝑎 > 0 𝑎 < 0
𝑏#
− 4𝑎𝑐 > 0 • Two real roots
• Two intersection
points on 𝑥-axis
𝑏#
− 4𝑎𝑐 = 0 • One real root
• One intersection
point on 𝑥-axis
𝑏#
− 4𝑎𝑐 < 0 • No real root
• Does not intersect on
𝑥-axis
a ab b
𝛼 = b
𝛼 = b
EXAMPLE 10
Determine types of roots of quadratic function 𝑓( 𝑥) = 6𝑥#
+ 10𝑥 − 1 when
𝑓( 𝑥) = 0. Hence, sketch the graph generally.
𝑓( 𝑥) = 6𝑥#
+ 10𝑥 − 1
𝑎 = 6, 𝑏 = 10, 𝑐 = −1
𝑏#
− 4𝑎𝑐 = 10#
− 4(6)(−1)
= 124 (> 0)
Thus, the quadratic function 𝑓( 𝑥) = 6𝑥#
+ 10𝑥 − 1 has two real roots and it
has to intersect two points on 𝑥-axis. Since 𝑎 > 0, the graph of 𝑓(𝑥) is a
parabola which passes through the minimum point.
𝑥
𝑥
𝑥
𝑥
𝑥
𝑥
𝑥

29
2.3.3 What is the relation exists between the vertex form of a quadratic function
𝑓( 𝑥) = 𝑎(𝑥 − ℎ)#
+ 𝑘 with the other form of quadratic functions (general form
and intercept form)?
What is a vertex form? A vertex form is a quadratic function given by 𝑓( 𝑥) = 𝑎(𝑥 − ℎ)#
+ 𝑘, where
𝑎, ℎ 𝑎𝑛𝑑 𝑘 are constants. (ℎ , 𝑘) is the vertex and it is symmetrical about the 𝑥-axis if 𝑥 = ℎ.
There are two form of quadratic functions:
• General form: 𝑓( 𝑥) = 𝑎𝑥#
+ 𝑏𝑥 + 𝑐 where 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 are constants. A vertex is at point
,−
K
#L
, 𝑓 8−
K
#L
A- and it is said to be symmetric about the 𝑥-axis when 𝑥 = −
K
#L
.
• Intercept form: 𝑓( 𝑥) = 𝑎(𝑥 − 𝑝)(𝑥 − 𝑞) where 𝑎, 𝑝 𝑎𝑛𝑑 𝑞 are constants. 𝑝 𝑎𝑛𝑑 𝑞 are the
roots or we called it as 𝑥-intercepts for 𝑓(𝑥). A vertex is at the point ,
fgh
#
, 𝑓 8
fgh
#
A- and it is
said to be symmetric about the 𝑥-axis when 𝑥 =
fgh
#
.
EXAMPLE 11
Find the values of 𝑘 when the 𝑥-axis is tangent to the graph of a quadratic
function 𝑓( 𝑥) = ( 𝑘 + 1) 𝑥#
+ 4( 𝑘 − 2) 𝑥 + 2𝑘.
Tangent to the graph = One real root
Since it has only one real root, we use 𝑏#
− 4𝑎𝑐 = 0.
𝑎 = 𝑘 + 1, 𝑏 = 4𝑘 − 8, 𝑐 = 2𝑘
𝑏#
− 4𝑎𝑐 = 0
(4𝑘 − 8)#
− 4( 𝑘 + 1)(2𝑘) = 0
16𝑘#
− 64𝑘 + 64 − 8𝑘#
− 8𝑘 = 0
8𝑘#
− 72𝑘 + 64 = 0
𝑘#
− 9𝑘 + 8 = 0
( 𝑘 − 1)( 𝑘 − 8) = 0
𝑘 = 1 𝑜𝑟 𝑘 = 8

30
Vertex Form
𝑓( 𝑥) = 𝑎(𝑥 − ℎ)#
+ 𝑘
General Form
𝑓( 𝑥) = 𝑎𝑥#
+ 𝑏𝑥 + 𝑐
Intercept Form
𝑓( 𝑥) = 𝑎(𝑥 − 𝑝)(𝑥 − 𝑞)
Expansion
Factorisation
or FormulaExpansion
Completing
the Square
EXAMPLE 12
Express quadratic function 𝑓( 𝑥) = 2𝑥#
+ 9𝑥 + 10 in the intercept form,
𝑓( 𝑥) = 𝑎(𝑥 − 𝑝)(𝑥 − 𝑞) where 𝑎, 𝑝 𝑎𝑛𝑑 𝑞 are constants and 𝑝 > 𝑞 . Hence,
state the values of 𝑎, 𝑝 𝑎𝑛𝑑 𝑞.
𝑓( 𝑥) = 2𝑥#
+ 9𝑥 + 10
𝑓( 𝑥) = (2𝑥 + 5)(𝑥 + 2)
𝑓( 𝑥) = 2(𝑥 +
5
2
)(𝑥 + 2)
𝑎 = 2, 𝑝 = −
5
2
𝑎𝑛𝑑 𝑞 = −2
General form
Intercept form

31
EXAMPLE 13
Express quadratic function 𝑓( 𝑥) = 2𝑥#
+ 9𝑥 + 10 in the form of
𝑓( 𝑥) = 𝑎(𝑥 − ℎ)#
+ 𝑘 where 𝑎, ℎ 𝑎𝑛𝑑 𝑘 are constants. Hence determine the
value of 𝑎, ℎ 𝑎𝑛𝑑 𝑘.
𝑓( 𝑥) = 2𝑥#
+ 9𝑥 + 10
= 2 j𝑥#
+
9
2
𝑥 + 5k
= 2
⎣
⎢
⎢
⎡
𝑥#
+
9
2
𝑥 + +
,
9
2
-
2
.
#
− +
,
9
2
-
2
.
#
+ 5
⎦
⎥
⎥
⎤
= 2 rj𝑥 +
9
4
k
#
−
1
16
s
= 2 j𝑥 +
9
4
k
#
−
1
8
Thus 𝑎 = 2, ℎ = −
c
t
𝑎𝑛𝑑 𝑘 = −
b
u

32
2.3.4 What are the effect of changes of 𝑎, ℎ 𝑎𝑛𝑑 𝑘 on the shape and position of
graph for 𝑓( 𝑥) = 𝑎(𝑥 − ℎ)#
+ 𝑘?
EXAMPLE 14
The diagram shows the sketch of graph for 𝑓( 𝑥) = 2(𝑥 + 2)#
+ 3 where 𝑎 =
2, ℎ = −2 𝑎𝑛𝑑 𝑘 = 3. Resketch the shape and position of the graph when the
following values changes.

(a) The values of 𝑎 change to
b
#
and 6.
𝑓(𝑥)
𝑥
11
3
−2
𝑓( 𝑥) = 2(𝑥 + 2)#
+ 3
𝑓(𝑥)
𝑥
𝑓( 𝑥) = 2(𝑥 + 2)#
+ 3
𝑓( 𝑥) =
1
2
(𝑥 + 2)#
+ 3
𝑓( 𝑥) = 6(𝑥 + 2)#
+ 3

33
(b) The values of ℎ change to -6 and 4.
(c) The values of k change to 1 and 8.
𝑓(𝑥)
−6 4
𝑥
𝑓( 𝑥) = 2(𝑥 + 2)#
+ 3𝑓( 𝑥) = 2(𝑥 + 6)#
+ 3
𝑓( 𝑥) = 2(𝑥 − 4)#
+ 3
1
9
8
16
𝑓(𝑥)
𝑥
𝑓( 𝑥) = 2(𝑥 + 2)#
+ 3
𝑓( 𝑥) = 2(𝑥 + 2)#
+ 3
𝑓( 𝑥) = 2(𝑥 + 2)#
+ 3

34
2.3.5 How to sketch the graphs of quadratic functions?
Determine the shape
of graph by
identifying the
value of 𝑎.
Determine the
position of graph by
finding the value of
discriminant.
Determine the
vertex.
Determine the
intersection point on
the 𝑥-axis by
solving 𝑓( 𝑥) = 0.
Determine the 𝑦-
intercept by solving
𝑓(0).
Plot the points on
Cartesian plane and
draw a parabola
which is
symmetrical at the
vertical line passing
through the vertex.

35
EXAMPLE 15
Sketch the graph of quadratic function 𝑓( 𝑥) = −3𝑥#
+ 𝑥 + 1.
Since 𝑎 < 0, so 𝑓(𝑥) has a maximum point.
𝑏#
− 4𝑎𝑐 = (1)#
− 4(−3)(1)
= 13 (> 0)
Since 𝑏#
− 4𝑎𝑐 > 0, the 𝑓(𝑥) has two real roots which it intersects the
𝑥-axis at two points.
𝑓( 𝑥) = −3𝑥#
+ 𝑥 + 1
= −3 8𝑥#
−
@
7
−
b
7
A
= −3 r𝑥#
−
@
7
+ j
8v
w
x
A
#
k
#
− j
8v
w
x
A
#
k
#
−
b
7
s
= −3 y8𝑥 −
b
z
A
#
−
b7
7z
{
= −3 8𝑥 −
b
z
A
#
+
b7
b#
Hence, the maximum point is 8
b
z
,
b7
b#
A and the axis of symmetry is 𝑥 =
b
z
.
𝑓( 𝑥) = 0
−3𝑥#
+ 𝑥 + 1 = 0
8𝑥 −
bv√b7
z
A 8𝑥 −
bg√b7
z
A = 0
𝑥 =
bv√b7
z
≈ −0.434 𝑜𝑟 𝑥 =
bg√b7
z
≈ 0.768
Thus, the intersection points on the 𝑥-axis are 𝑥 = −0.434 and 𝑥 = 0.768.
𝑓(0) = −3(0)#
+ 0 + 1 = 1
The graph intersect at 𝑦-axis at (0,1).

36
2.3.6 How to solve problems of quadratic functions?
𝑓(𝑥)
𝑥
𝑥 =
1
6
𝑓( 𝑥) = −3𝑥#
+ 𝑥 + 1
1
0.768−0.434
EXAMPLE 16
Alvin dives into the pool at a distance of 6 meters from the surface of pool.
The height of the Alvin is given by ℎ( 𝑥) = 4 + 8𝑥 − 5𝑥#
where ℎ is his
height in meters and 𝑥 is the time in seconds.
ℎ( 𝑥) = 4 + 8𝑥 − 5𝑥#
where 𝑎 = −5, 𝑏 = 8 𝑎𝑛𝑑 𝑐 = 4
(a) Find the maximum height achieved by Alvin.
𝑥 = −
𝑏
2𝑎
= −
8
2(−5)
= 0.8
Then substitute 𝑥 = 0.8 into ℎ(𝑥)
ℎ(0.8) = 4 + 8(0.8) − 5(0.8)#
= 7.2
Thus, the maximum height reached by Alvin is 7.2 meters at 0.8
seconds.

37
(b) Calculate the time for Alvin to reach the surface of the pool.
ℎ( 𝑥) = 0
−5𝑥#
+ 8𝑥 + 4 = 0
5𝑥#
− 8𝑥 − 4 = 0
(5𝑥 + 2)( 𝑥 − 2) = 0
𝑥 = 2 𝑜𝑟 𝑥 = −
#
Z
(ignored)
Thus, the time for Alvin to reach the surface of pool is 2 seconds.

38
1. Solve the quadratic equation 3𝑥( 𝑥 − 5) = 2𝑥 − 1 without using calculator and give your
answer in three decimal places.
2. Form an equation with roots 5 and 9.
3. If 𝛼 𝑎𝑛𝑑 𝛽 are the roots of the quadratic equation 2(𝑥 − 5)#
= 4(𝑥 + 7), form an equation
with the roots of 2𝛼 𝑎𝑛𝑑 2𝛽.
4. Given that the roots of the quadratic equation 2𝑥#
+ ( 𝑚 + 1) 𝑥 + 𝑛 = 2 are -3 and
b
#
. Find
the value of 𝑚 𝑎𝑛𝑑 𝑛.
5. Find the range of values of 𝑥 for the quadratic inequalities 𝑥#
− 4𝑥 + 3 > 0.
1. Determine the types of roots for the following quadratic equation:
(a) 𝑥( 𝑥 − 2) = 5
(b) 𝑥( 𝑥 + 5) = 2𝑥 − 14
2. The quadratic equation 𝑥#
+ 2ℎ𝑥 + 4 = 𝑥 where ℎ is a constant, has one real root. Find all
the possible values for ℎ.
3. Given the equation 𝑥#
+ 5𝑛 = 4𝑚𝑥 has only one real root, express 𝑚 in terms of 𝑛.
4. The quadratic equation 𝑥#
+ 𝑝𝑥 + 𝑞 = 0 where 𝑝 𝑎𝑛𝑑 𝑞 are positive integers, has a
discriminant of 16 and 𝑝 − 𝑞 = −4. Find the possible value for 𝑝 𝑎𝑛𝑑 𝑞.
1. Given the quadratic function 𝑓( 𝑥) = −𝑥#
+ 𝑥 + 6 where 𝑎 = −1, 𝑏 = 1 𝑎𝑛𝑑 𝑐 = 6 .
Sketch the graph generally when the values changes.
(a) The value of 𝑎 changes to −
b
#
(b) The value of b changes to −1
(c) The value of 𝑐 changes to 1
2. Determine the types of roots for the quadratic function 𝑓( 𝑥) = 2𝑥#
+ 𝑥 − 5 when 𝑓( 𝑥) =
0. Thus, sketch the graph generally at the 𝑥-axis.
3. Given the quadratic function 𝑓( 𝑥) = 𝑚𝑥#
+ 4𝑥 − 6 where 𝑚 is a constant, has no real
roots, find the range of values for 𝑚.
4. Find the range of values of 𝑘 if the given quadratic function has two real roots,
𝑓( 𝑥) = 5𝑥#
− ( 𝑘𝑥 + 4) 𝑥 − 2.
Summative Exercises

CHAPTER 3 SYSTEMS OF EQUATIONS
39
3.1 Systems of Linear Equations in Three Variables
EXAMPLE 1
Describe whether the following equations are systems of linear equations in three variables or not.
(a) x + 2y + 3z = 29
3x + 4y + z3 = 61
2x – 3y – z = –12
Solution: No, a linear equation is an equation where the power of the variable is 1, however there is an
equation in which the highest power of the variable is 3.
(b) 3a + b – c = 4
7b + 2c – 4a = 41
2(a + 6b) – 4c = 36
Solution: Yes, a linear equation is an equation where the power of the variable is 1. Since, all three
equations have three variables with power 1, therefore it is a system of linear equations in three
variables.
3.1.1 What is a system of linear equations?
A system of linear equation is the existence of two or more linear equations which contain the same
set of variables.
3.1.2 The General Form of a linear equation
A linear equation in two variables can be written in the form Ax + By = C, where A, B and C are
constant.
The General Form of a linear equation in three variables is
Ax + By + Cz = D
Note that:  A, B, C and D are constants
 A, B and C are not equal to zero
 D can be zero or non-zero
3.1.3 Determining systems of linear equation in three variables
A linear equation is an equation where the power of the variable is 1. If the power of any variables in
the system of equation is more than 1, then it is NOT a system of linear equation.

40
3.1.4 The Graphing of System of linear equations in Three Variables
In systems of linear equation in three variables, the equations are represented as three-dimensional
planes in Cartesian space. Each plane might be orientated at any angle.
A system of linear equations in three variables has three axes: x-axis, y-axis and z-axis. The three
linear equations will form a plane on each axis, therefore a 3D plane is formed.
3.1.4.1 Why each linear equation in three variables will form a plane on each axis?
Let’s set the coefficients of all but one of the variables to zero by
using the general form of a linear equation in three variables: Ax + By + Cz = D
Step 1: Begin by setting A and B to zero and C to one.
0x + 0y + 1z = D
This eliminates all the variables except z which becomes z = D
Step 2: Since D is a constant, if we set D to 3. We obtain z = 3
Since both x and y are free to take any value, the graph of this equation consists of every point in 3D
space where z=3 while x-coordinate and y-coordinate are any real numbers.
The graph of this equation is therefore a horizontal plane three units above the origin.
Step 3: Repeat step 1 by setting A and C to zero and B to one.
0x + 1y + 0z = D
This eliminates all the variables except y which becomes y = D
Step 4: Repeat step 2. Since D is a constant, if we set D to 2. We obtain y = 3
Since both x and z are free to take any value, the graph of this equation consists of
every point in 3D space where y=2 while x-coordinate and z-coordinate are any real numbers.
The graph of this equation is therefore a vertical plane two units to the right of the origin.
The same argument can be applied by setting all variables with any value.
Depending upon the values of A, B, C and D, the plane may lie in any position and orientation. Each
plane can be positioned anywhere in space relative to other planes.
Since each plane graphically represents the set of solutions to one to three equation, the points where
all three planes simultaneously intersect correspond to the solutions which simultaneously satisfy all
three equations.
These points therefore correspond to the solutions of the system.
y=2
z=3

41
To illustrate, below shows a system of linear equation in three variables.
2x + y + z = 20
x + y + 2z = 22
3x + 2y + 2z = 35
These equations formed a system of linear equations.
By graphing, each equation can be displayed in a 3D plane.
Therefore, three planes are formed.
3.1.5 Three Cases when Solving a System of linear equations
There are three possible results when solving a system of linear equations in three variables.
(a) The 3 planes intersect at only one point. The system is consistent and has only one solution.
(b) The 3 planes intersect along a line or a plane. The system is consistent and infinitely many
solutions.
(c) The 3 planes do not intersect or intersect with no common points. The system is inconsistent
and no solution.
Hence, the system of linear equation in three variables can have a single unique solution, no solution
or infinite number of solutions depending on the way in which the three planes are oriented.
Note that:  When there is no solution the system is called "inconsistent".
 One or infinitely many solutions the system is called "consistent”
3.1.6 Methods to solve the system of linear equations
• Substitution Method
• Elimination Method

42
EXAMPLE 2
Solve the following system of linear equations using the substitution method.
2x + y + z = 20
x + y + 2z = 22
3x + 2y + 2z = 35
Solution:
2x + y + z = 20 … ○1
x + y + 2z = 22 … ○2
3x + 2y + 2z = 35 … ○3
From ○1 , z = 20 – 2x – y … ○4
Substitute ○4 into ○2
x + y + 2(20 – 2x – y) = 22
x + y + 40 – 4x – 2y = 22
–3x – y = –18
y = –3x + 18 … ○5
3x + 2(–3x + 18) + 2[20 – 2x – (-3x+18)] = 35
3x – 6x + 36 + 2(2 + x) = 35
x = 5
y = –3(5) + 18
= 3
z = 20 –2(5) – 3
= 7
Thus, x = 5, y = 3 and z = 7 are the solutions to this system of linear equation.
Substitute equation ○4
and ○5 into equation ○3
Substitute x = 5
into equation ○5
Substitute x = 5 and
y = 5 into equation ○4

43
EXAMPLE 3
Solve the following system of linear equations using the elimination method.
x –2y + z = –8
x + 3y – 2z = 41
3x + 2y + 2z = 214
Solution:
Choose any two equations.
x – 2y + z = –8 … ○1
x + 3y – 2z = 41 … ○2
Since the coefficient of x in ○1 and ○2 are equal
Eliminate the variable x by subtracting ○1 from ○2
○2 –○1 : 5y – 3z = 49 … ○4
Choose another two sets of equation
5y – 3z = 49 … ○4
–8y + z = –238 … ○6
Multiple equation ○6 with 3 so that the coefficient of x is equal to ○4
○6 ×3: –24y + 3z = –714 ... ○7
Eliminate the variable y by adding ○7 with ○4
○4 +○7 : –19y = –665
y = 35
5(35) – 3z = 49
–3z = –126
z = 42
x – 2(35) + 42 = –8
x = 20
Thus, x = 20, y = 35 and z = 42 are the solutions to this system of linear equations.

44
EXAMPLE 4
Solve the following system of linear equations.
x – 2y – 4z = 3
4x – 8y – 16z = –12
x – 4z = 3
Solution:
x – 2y – 4z = 3 … ○1
4x – 8y – 16z = –12 … ○2
x – 4z = 3 … ○3
Since equation ○3 only has two variables, elimination method is used to eliminate the variable x in
equations ○1 and ○2 .
○1 × 4: 4x – 8y – 16 z = 12 … ○4
○4 –○2 : 0x + 0y + 0z = 24
0 = 24
Hence, we obtain 0 = 24. This is a contradiction as 0 ≠ 24. Therefore, the system of linear equation has
no solution.
EXAMPLE 5
Solve the following system of linear equations.
3x + 2y + z = 3
2x + y + z = 0
6x + 2y + 4z = –6
Solution:
3x + 2y + z = 3 … ○1
2x + y + z = 0 … ○2
6x + 2y + 4z = –6 … ○3
○1 ×2: 6x + 4y + 2z = 6 … ○4
○4 –○3 : 2y – 2z = 12 … ○5
○2 ×3: 6x + 3y + 3z = 0 … ○6
○6–○3 : y – z = 6 … ○7
○7 ×2: 2y – 2z = 12 … ○8
○8 –○5 : 0y + 0z = 0
0 = 0
Since 0=0, the system of linear equations has infinite number of solutions.
Multiple ○1 with 4 so that the
coefficient of x are equal in
equation ○1 and ○2
Multiple ○1 with 2

45
3.2 Simultaneous Equations Involving One Linear Equation
and One Non-Linear Equation
3.2.1 What is non-linear equation?
A non-linear equation is an equation that does not form a straight line, but a curve. A non-linear
equation has the variable with power more than or equal to two.
3.2.2 Difference between linear equations and non-linear equations
Comparison between linear equations and non-linear equations:
Linear Equations Non-Linear Equations
The graph forms a straight line It does not form a straight line, but forms a curve
It has only one degree
(The power of variable = 1)
It has more than or equal to two degree
(The power of variable ≥ 2)
All these equations form a straight line in XY
plane. These lines can be extended to any
direction but in a straight line form.
It forms a curve and if the value of the degree
increases, the curvature of the graph increases.
Examples
• 6y + 2x + 8 = 0
• 5y = 9x
• 4x + 8 = 9y
Examples:
• x2 + y2 = 1
• x2 + 12xy + y2 = 0
• x2 + x + 4 = 0
3.2.3 Methods to solve these Simultaneous Equations
The methods to solve the simultaneous equations involving one linear equation and one non-linear
equation are
• Solving by Substitution
• Solving by elimination
• Graphical representation method

46
EXAMPLE 6
Solve the following simultaneous equations using substitution method.
y – 2 = x
y = x2
Solution:
Let y – 2 = x … ○1
y = x2 … ○2
From ○1 : y = x + 2 … ○3
x + 2 = x2
x2 –x – 2 = 0
(x – 2)(x + 1) = 0
x = 2 or x = –1
y = 2 + 2 or y = –1 + 2
= 4 = 1
Thus, the solutions to these simultaneous equations are x= 2, y = 4 and x = –1, y = 1.
EXAMPLE 7
Solve the following simultaneous equations using elimination method.
x + 2y = 3
x² + 3xy = 10
Solution:
x + 2y = 3 … ○1
x² + 3xy = 10 … ○2
Multiple ○1 with –3x
○1 ×–3x: –3x2 – 6xy = –9x … ○3
○2 × 2: 2x2 + 6xy = 20 … ○4
○3 + ○4 : –x2 = –9x + 20
x2 – 9x + 20 = 0
(x – 5)(x – 4) = 0
x = 5 or x = 4
5 + 2y = 3 or 4 + 2y = 3
y = –1 y = –
1
2
When x = 5, y = –1
When x = 4, y = –
1
2
Thus, the solutions to these simultaneous equations are x = 5, y = –1 and x = 4, y = –
1
2
.
Multiple equation ○2 with 2
Substitute x= 5 and
x = 4 into equation ○1
Compare equation
○2 and ○3

47
EXAMPLE 8
Solve the following simultaneous equations using the graphical representation method.
x² + y² = 20
x + y = 6
Solution:
For equation x² + y² = 20
x –4 –2 0 2 4
y ±2 ±4 ± √20 ±4 ±2
For equation x + y = 6
x –4 –2 0 2 4
y 10 8 6 4 2
Based on the graph, there are two points of intersection which is (2,4) and (4,2).
Thus, the solutions to the simultaneous equations are x = 2, y = 4 and x = 4, y = 2.
Tip: To determine the
points to be plotted
on the graph, students
are encouraged to
construct a table.
Sketch a graph
based on the table
constructed.

48
Further Exploration
Visualizing Linear Equations in Three Variables
https://www.youtube.com/watch?v=Wm27Y6hxbRs
Types of Linear Systems in Three Variables
https://www.youtube.com/watch?v=WAzUwzV1F3g
Solving Systems of Equations in Three Variables
https://www.youtube.com/watch?v=fjfmNfIgQ2Q
System of Linear Equations in Three Variables (Substitution method)
https://www.youtube.com/watch?v=pYrynnnqoEI
Three Variable Systems with Infinite or Null Solution Sets
https://www.youtube.com/watch?v=pOSxxRosKoQ
Solving Simultaneous Equation (One linear and one non-linear)
https://www.youtube.com/watch?v=cP9Zzzff2VE

49
Summative Exercises
3.1 Systems of Linear Equations in
Three Variables
1) Based on the following system of linear
equations
x − 2y + z = 10
2x + 2y + 3z = 13
–2x + 5y + z = –7
Find the value of x + y + z.
2) Solve the following system of linear
equations using the elimination method.
x + 2y – z = 2
2x – 3y + z = –1
5x – y – 2z = –3
3) Solve the following system of linear
equations.
x + y + z = 6
2 y + 5z = −7
2x + 5y − z = 27
4) Adam has 90 balls of four colours: yellow,
pink, blue, and black.60 of them are not pink,
25 are black, and 67 are not yellow. What is
the number of yellow, pink and blue balls
respectively?
5) The standard equation of a circle is
x2+y2+Ax+By+C=0. Find the equation of
the circle that passes through the points (1,1),
(2, –4) and (5,5).
6) A total of RM50,000 is invested in three
funds paying 6%, 8%, and 10% simple
interest. The yearly interest is RM3,700.
Twice as much money is invested at 6% as
invested at 10%. How much was invested in
each of the funds.
3.2 Simultaneous Equations Involving
One Linear Equation and One Non-
Linear Equation
1) Solve the following simultaneous equations
2x + y = 1
x2 + y2 = 1
2) Solve the simultaneous equations
x2 – y = 14
2y – 4 = 12x
3) Solve the simultaneous equation
x² − y² = 7
2y = 2 + x
4) Solve the following simultaneous equations
using the graphical representation method
x + 2y – 4 = 0
x2 + y2 = 7 – xy
5) A straight line with equation 4x + y – 5 = 0
intersects the curve 27x2 + 21xy = –2y2 at
points A and B. Find the coordinates of the
points A and B.
6)
Fish pond
Banana
30 m
Ms Wong has a rectangular plot of land. She
grows bananas trees and rears fish in the
regions as shown in the diagram. Given that
the region planted with bananas is 460m2 and
the perimeter of the rectangular fish pond is
48m, find the value of x and of y.
x m
10m
y m

50
Chapter 4: Indices, Surds and Logarithms
𝑎 𝑥
4.1.1 Simplifying algebraic expressions involving indices
By using the law of indices from the above, algebraic expressions can be simplified. We can apply
techniques like grouping like terms together when we are simplifying them.
Base
Some law of indices to recall:
1. 𝑎 𝑛
= 𝑎 × 𝑎 × … × 𝑎 × 𝑎
2. 𝑎−𝑛
=
1
𝑎 𝑛
3. 𝑎0
= 1
4. 𝑎
𝑚
𝑛 = ξ 𝑎 𝑚𝑛
5. 𝑎 𝑚
× 𝑎 𝑛
= 𝑎 𝑚+𝑛
6. 𝑎 𝑚
÷ 𝑎 𝑛
= 𝑎 𝑚−𝑛
7. ሺ𝑎 𝑚ሻ 𝑛
= ሺ𝑎 𝑛ሻ 𝑚
= 𝑎 𝑚𝑛
8. ሺ𝑎𝑏ሻ 𝑛
= 𝑎 𝑛
𝑏 𝑛
9. ቀ
𝑎
𝑏
ቁ
𝑛
=
𝑎 𝑛
𝑏 𝑛
n times of a
EXAMPLE 1
Simplify the following:
(i) ሺ4𝑥−3
𝑦3ሻ−2
(ii)
7𝑚3 𝑛4
𝑚2 𝑛−2
Solution:
(i) ሺ4𝑥−3
𝑦3ሻ−2
= 4−2
𝑥−3×−2
𝑦3×−2
=
1
42 𝑥6
𝑦−6
=
1
16
𝑥6
𝑦−6
Index
4.1 Law of Indices

51
4.1.2 Solving problems involving indices
All equations involving indices can be solved following the rule below:
(i)
7𝑚3 𝑛4
𝑚2 𝑛−2 = 7 ቀ
𝑚3
𝑚2ቁ ቀ
𝑛4
𝑛−2ቁ
= 7ሺ𝑚3−2ሻ൫𝑛4−ሺ−2ሻ
൯
= 7𝑚𝑛6
EXAMPLE 2
Show that 7 𝑥+3
+ 7 𝑥
− 7 𝑥+2
is divisible by 5 for all positive integer 𝑥.
Solution:
7 𝑥+3
+ 7 𝑥
− 7 𝑥+2
= 7 𝑥
73
+ 7 𝑥
− 7 𝑥
72
= 7 𝑥ሺ73
+ 1 − 72ሻ
= 7 𝑥ሺ343 + 1 − 49ሻ
= 7 𝑥ሺ295ሻ
Since 295 is a multiple of 5, therefore 7 𝑥+3
+ 7 𝑥
− 7 𝑥+2
is divisible by 5 for all
positive integer 𝑥.
If 𝒂 𝒎
= 𝒂 𝒏
, then 𝒎 = 𝒏 or if 𝒂 𝒎
= 𝒃 𝒎
, then 𝒂 = 𝒃 when 𝒂 > 𝟎 and 𝒂 ≠ 𝟏.
EXAMPLE 3
Solve the following equations:
(i) 49ሺ7−4ሻ = 72+𝑚
(ii)
34
9
=
1
3 𝑚−3
Solution:
(i) 49ሺ7−4ሻ = 72+𝑚
72ሺ7−4ሻ = 72+𝑚
7−2
= 72+𝑚
𝑚 + 2 = −2
𝑚 = −4
(ii)
34
9
=
1
3 𝑚−3
34
32 =
1
3 𝑚−3
34−2
= 3−ሺ𝑚−3ሻ
−ሺ𝑚 − 3ሻ = 2
𝑚 = 1
Equate both side by comparing the indices
Equate both side by comparing the indices

52
4.2.1 Comparing rational numbers and irrational numbers, and relating surds
with rational numbers
Rational numbers are numbers that can be expressed in the form of
𝑎
𝑏
in the simplest form, where 𝑎
and 𝑏 are integers and 𝑏 ≠ 0.
Decimals that can be converted into fractions are rational numbers, such as 0.111 … =
1
9
Decimals that cannot be converted into fractions are irrational numbers.
Surds are numbers that are in the square root form ൫ξ 𝑎൯, where 𝑎 is any positive integer. Surds have
infinite decimal places and are non-recurring. The following are some examples:
ξ2 = 1.41421 … has infinite decimal places and non-recurring. Hence ξ2 is a surd.
ξ4 = 2 is an integer. Hence ξ4 is not a surd.
Besides, ξ 𝑎
𝑛
is the surd 𝑎 of order 𝑛, which means the 𝑎 raise to the
1
𝑛
th power.
In short, when a number cannot be simplified by eliminating the root, it is classified as a surd.
Recurring decimals are decimals that have a number, or a group of numbers repeated infinitely in the
decimal places. They are all rational numbers. For example:
0.262626 … can be written as fraction ቀ
26
99
ቁ, and can also be denoted by 0. 2̇6̇ or 0. 26.
EXAMPLE 4
Convert the following recurring decimals into fractions:
(i) 0.535353 … (ii) 0. 128
Solution:
(i) Let 𝑥 = 0.535353 …
100𝑥 = 53.535353 …
100𝑥 − 𝑥 = 53.535353 … − 0.535353 …
99𝑥 = 53
𝑥 =
53
99
(ii) Let 𝑥 = 0. 128
1000𝑥 = 128. 128
1000𝑥 − 𝑥 = 128. 128 − 0. 128
999𝑥 = 128
𝑥 =
128
999
Multiply the decimals by
10 raise to how many
numbers in the group of
the recurring decimals (in
this case 2 because 53 is
repeated).
Subtract the two equations
of 100𝑥 and 𝑥.
4.2 Law of Surds

53
4.2.2 Making and verifying conjectures on ξ 𝑎 × ξ𝑏 and ξ 𝑎 ÷ ξ𝑏
For 𝑎 > 0 and 𝑏 > 0,
ξ 𝑎 × ξ𝑏 = ξ𝑎𝑏
ξ 𝑎 ÷ ξ𝑏 = √
𝑎
𝑏
4.2.3 Simplifying expression involving surds
For some surds ξ 𝑎
𝑛
, 𝑎 may divided by 𝑛 numbers of factor 𝑝, then the surds can be simplified. For
example:
ξ72 can be written as ξ6 × 6 × 2, the number 72 has 2 factors of 6. It can be further simplified.
ξ72 = ξ6 × 6ξ2
= 6ξ2
Besides, expressions involving surds can also be simplify by basic algebraic manipulations and
operations such as addition, subtraction or grouping likes term(surds).
By using law of indices
EXAMPLE 5
Simplify the following surds:
(i)
ξ6
ξ24
(ii)
ξ7×ξ6
ξ3
Solution:
(i)
ξ6
ξ24
= √
6
24
= √
1
4
=
1
2
(ii)
ξ7×ξ6
ξ3
= √
7×6
3
= ξ14
EXAMPLE 6
Simplify the following expressions:
(i) ξ150 − ξ24 (ii) ൫5 + 3ξ3൯൫2 − 2ξ3൯
Solution:
(i) ξ150 − ξ24 = ξ25 × 6 − ξ4 × 6
= 5ξ6 − 2ξ6
= 3ξ6

54
4.2.4 Rationalising the denominators for the expressions involving surds
Fractions in the form of
1
𝑚ξ 𝑎±𝑛ξ 𝑏
can be rationalised and be simplified to eliminate the surds from the
denominator. To rationalise means to multiply conjugate surds to the numerator and denominator.
Conjugate surds are the surds with similar value but with the opposite operator, the following tables
show the conjugate surds of each surd.
Surds Conjugate Surds
1
𝑚ξ 𝑎 + 𝑛ξ𝑏
1
𝑚ξ 𝑎 − 𝑛ξ𝑏
1
𝑚ξ 𝑎
1
𝑚ξ 𝑎
1
𝑚ξ 𝑎 − 𝑛ξ𝑏
1
𝑚ξ 𝑎 + 𝑛ξ𝑏
(ii) ൫5 + 3ξ3൯൫2 − 2ξ3൯ = 10 − 10ξ3 + 6ξ3 − 6൫ξ3൯
2
= 10 − 4ξ3 − 18
= −4ξ3 − 8
EXAMPLE 7
Rationalise the following:
(i)
1
4ξ7
(ii)
1
3ξ5−2ξ3
Solution:
(i)
1
4ξ7
=
1
4ξ7
×
4ξ7
4ξ7
=
4ξ7
4ξ7×4ξ7
=
4ξ7
112
=
ξ7
28
(ii)
1
3ξ5−2ξ3
=
1
3ξ5−2ξ3
×
3ξ5+2ξ3
3ξ5+2ξ3
=
3ξ5+2ξ3
൫3ξ5−2ξ3൯൫3ξ5+2ξ3൯
=
3ξ5+2ξ3
൫3ξ5൯
2
−൫2ξ3൯
2
=
3ξ5+2ξ3
45−12
=
3ξ5+2ξ3
33
Multiply conjugate
surds to both numerator
and denominator.

55
4.2.5 Solving problems involving surds
While solving problems involving surds, apply the law of indices, law of surds and techniques to
simplify and rationalise surds.
4.3.1 Relating the equations in index form with logarithmic form and determine
the logarithmic value of a number
For index form equation 𝑎 𝑥
= 𝑁, a function logarithm is introduced to help solve this type of equation.
𝑎 𝑥
= 𝑁 ⟺ log 𝑎 𝑁 = 𝑥 where 𝑎 > 0 and 𝑎 ≠ 1
Since 𝑎0
= 1 and 𝑎1
= 𝑎, apply it into the logarithm function we get:
log 𝑎 1 = 0 and log 𝑎 𝑎 = 1
If we were to find the reverse of a result from the logarithmic function, we use antilogarithm or
antilog for short.
Also, if a base is not mentioned in the logarithmic function, it is understood as base 10.
log10 𝑁 = 𝑥 ⟺ antilog 𝑥 = 𝑁
In short,
𝑁 = 𝑎 𝑥
⟺ log 𝑎 𝑁 = 𝑥
Suppose index form equation as a function:
𝑓ሺ𝑥ሻ = 𝑎 𝑥
,
𝑓−1ሺ𝑥ሻ = log 𝑎 𝑥
EXAMPLE 8
Solve 𝑥 − 5ξ 𝑥 + 6 = 0 for all real 𝑥.
Solution:
𝑥 − 5ξ 𝑥 + 6 = 0
൫ξ 𝑥 − 2൯൫ξ 𝑥 − 3൯ = 0
ξ 𝑥 − 2 = 0 or ξ 𝑥 − 3 = 0
ξ 𝑥 = 2 or ξ 𝑥 = 3
𝑥 = 4 or 𝑥 = 9
Base
IndexNumber
4.3 Law of Logarithms

56
EXAMPLE 9
Express 35
= 243 in logarithmic form.
Solution:
35
= 243
log3 243 = 5
Express log7 343 = 3 in index form.
Solution:
log7 343 = 3
73
= 343
EXAMPLE 10
1. Find the value of the following:
(i) log3 27 (ii) log10 100
Solution:
(i) Let log3 27 = 𝑥
3 𝑥
= 27
3 𝑥
= 33
𝑥 = 3
Thus, log3 27 = 3
(ii) Let log10 100 = 𝑥
10 𝑥
= 100
10 𝑥
= 102
𝑥 = 2
Thus, log10 100 = 2
2. Solve the following:
(i) log2 𝑥 = 4 (ii) log7 𝑦 = 2
Solution:
(i) log2 𝑥 = 4
24
= 𝑥
𝑥 = 16
(ii) log7 𝑦 = 2
72
= 𝑦
𝑦 = 49

57
4.3.2 Proving the laws of logarithms
The basic laws of logarithm are as follows:
If 𝑎, 𝑥, 𝑦 are positive and 𝑎 ≠ 1, then
(i) log 𝑎 𝑥 + log 𝑎 𝑦 = log 𝑎 𝑥𝑦
(ii) log 𝑎 𝑥 − log 𝑎 𝑦 = log 𝑎
𝑥
𝑦
(iii)log 𝑎 𝑥 𝑛
= 𝑛 log 𝑎 𝑥 for any real number 𝑛.
The above formulae can be proven by applying the law of indices:
Let 𝑥 = 𝑎 𝑚
and 𝑦 = 𝑎 𝑛
that 𝑎,𝑥,𝑦 are positive and 𝑎 ≠ 1, we have 𝑚 = log 𝑎 𝑥 and 𝑛 = log 𝑎 𝑦.
(i) 𝑥𝑦 = 𝑎 𝑚
× 𝑎 𝑛
𝑥𝑦 = 𝑎 𝑚+𝑛
log 𝑎 𝑥𝑦 = 𝑚 + 𝑛
log 𝑎 𝑥𝑦 = log 𝑎 𝑥 + log 𝑎 𝑦
(ii)
𝑥
𝑦
=
𝑎 𝑚
𝑎 𝑛
𝑥
𝑦
= 𝑎 𝑚−𝑛
log 𝑎
𝑥
𝑦
= 𝑚 − 𝑛
log 𝑎
𝑥
𝑦
= log 𝑎 𝑥 − log 𝑎 𝑦
(iii) 𝑥 𝑝
= ሺ𝑎 𝑚ሻ 𝑝
𝑥 𝑝
= 𝑎 𝑚𝑝
log 𝑎 𝑥 𝑝
= 𝑚𝑝
log 𝑎 𝑥 𝑝
= 𝑝 log 𝑎 𝑥
EXAMPLE 11
Find the value of the following:
(i) antilog 3 (ii) antilog 0.2346
Solution:
(i) antilog 3 = 1000
(ii) antilog 0.2346 = 1.7163

58
EXAMPLE 12
Given log2 5 = 2.322 and log2 9 = 3.170, determine the value of:
(i) log2 45 (ii) log2
9
5
Solution:
(i) log2 45 = log2 5 × 9
= log2 5 + log2 9
= 2.322 + 3.170
= 5.492
(ii) log2
9
5
= log2 9 − log2 5
= 3.170 − 2.322
= 0.848
EXAMPLE 13
Without using calculator, find the value of:
(i) log5 10 + log5 15 − log5 6
(ii) 2 log3 6 − log3 4
(iii) 3 log2 ξ4
3
Solution:
(i) log5 10 + log5 15 − log5 6 = log5
10×15
6
= log5 25
= log5 52
= 2 log5 5
= 2
(ii) 2 log3 6 − log3 4 + log3 9 = log3 62
− log3 4 + log3 9
= log3
36×9
4
= log3 81
= log3 34
= 4 log3 3
= 4
(iii) 3 log2 ξ4
3
= log2൫ξ4
3
൯
3
= log2 22
= 2 log2 2
= 2

59
4.3.3 Simplifying algebraic expressions using the law of logarithms
By using the law of logarithms as stated in 4.3.2, we can simplify algebraic expressions:
EXAMPLE 14
1. Simplify the following expressions into single logarithms:
(i) log5 𝑥2
+ log5 𝑦 (ii) log 𝑎 𝑝 + log 𝑎 𝑞2
− 4 log 𝑎 𝑝
Solution:
(i) log5 𝑥2
+ log5 𝑦 = log5ሺ𝑥2
× 𝑦ሻ
= log5 𝑥2
𝑦
(ii) log 𝑎 𝑝 + log 𝑎 𝑞2
− 4 log 𝑎 𝑝 = log 𝑎 𝑝 + log 𝑎 𝑞2
− log 𝑎 𝑝4
= log 𝑎 ቀ
𝑝𝑞2
𝑝4 ቁ
= log 𝑎 ቀ
𝑞2
𝑝3ቁ
2. Given 𝑝 = log2 5, 𝑞 = log2 7 and 𝑟 = log2 3, write the following
expressions in terms of 𝑝, 𝑞 and/or 𝑟.
(i) log2 35
(ii) log2
75
7
(iii) log2
21
ξ5
Solution:
(i) log2 35 = log2 5 × 7
= log2 5 + log2 7
= p + q
(ii) log2
75
7
= log2 ቀ
25×3
7
ቁ
= log2 ቀ
52×3
7
ቁ
= log2 52
+ log2 3 − log2 7
= 2 log2 5 + log2 3 − log2 7
= 2𝑝 + 𝑞 − 𝑟
(ii) log2
21
ξ5
= log2 ቀ
7×3
ξ5
ቁ
= log2 7 + log2 3 − log2 5
ቀ
1
2
ቁ
= log2 7 + log2 3 −
1
2
log2 5
= 𝑞 + 𝑟 −
1
2
𝑝

60
4.3.4 Proving the relationship of log 𝑎 𝑏 =
log 𝑐 𝑏
log 𝑐 𝑎
and determining the logarithm of
a number
Suppose we have 𝑎, 𝑏 and 𝑐 are positive numbers, 𝑎 ≠ 1 and 𝑐 ≠ 1, also let log 𝑎 𝑏 = 𝑥, then 𝑎 𝑥
= 𝑏.
log 𝑐 𝑎 𝑥
= log 𝑐 𝑏
𝑥 log 𝑐 𝑎 = log 𝑐 𝑏
𝑥 =
log 𝑐 𝑏
log 𝑐 𝑎
Therefore, we get
log 𝑎 𝑏 =
log 𝑐 𝑏
log 𝑐 𝑎
The following summarises the above proof:
If 𝒂,𝒃 and 𝒄 are positive numbers, 𝒂 ≠ 𝟏 and 𝒄 ≠ 𝟏, then
𝐥𝐨𝐠 𝒂 𝒃 =
𝐥𝐨𝐠 𝒄 𝒃
𝐥𝐨𝐠 𝒄 𝒂
Additionally, if we have 𝑏 = 𝑐, then
log 𝑎 𝑏 =
log 𝑐 𝑏
log 𝑐 𝑎
=
log 𝑏 𝑏
log 𝑐 𝑎
=
1
log 𝑐 𝑎
Often, we change the base of logarithms to base 10 or base 𝑒 conventionally to evaluate the value of
the logarithms. 𝑒 is a mathematical constant with the value of non-recurring decimal 2.71828 …
Logarithms with base 𝑒 are called the natural logarithms and are written as log 𝑒 or ln .
EXAMPLE 15
Given 𝑚 = log3 𝑎, express the following expressions in term of m by
changing the bases of the logarithmic functions:
(i) log27 𝑎2
(ii) log 𝑎 81𝑎3
Solution:
(i) log27 𝑎2
=
log3 𝑎2
log3 27
=
2log3 𝑎
log3 3
3
= 2𝑚
3log3 3
=
2
3
𝑚
(ii) log 𝑎 81𝑎3
=
log3 81𝑎3
log3 𝑎
=
log3 34+log3 𝑎3
log3 𝑎
=
4 log3 3+3 log3 𝑎
𝑚
=
4+3𝑚
𝑚

61
4.3.5 Solving problems involving the law of logarithms
By applying the law of logarithms, we can solve problems involving indices that cannot be solved by
the way of equating 𝑎 𝑥
= 𝑏 𝑥
or 𝑎 𝑥
= 𝑎 𝑦
.
EXAMPLE 16
Solve the following equations:
(i) 3 𝑥+1
= 7 𝑥−2
(ii) 32𝑥
− 3 𝑥+1
+ 2 = 0
Solution:
(i) 3 𝑥+1
= 7 𝑥−2
log 3 𝑥+1
= log 7 𝑥−2
ሺ𝑥 + 1ሻ log 3 = ሺ𝑥 − 2ሻ log 7
𝑥 log 3 + log 3 = 𝑥 log 7 − 2 log 7
𝑥 log 3 − 𝑥 log 7 = −2 log 7 − log 3
𝑥ሺlog 3 − log 7ሻ = −2 log 7 − log 3
𝑥 =
−2 log 7−log 3
log 3−log7
𝑥 = 5.890
(ii) 32𝑥
− 3 𝑥+1
+ 2 = 0
ሺ3 𝑥ሻ2
− 3ሺ3 𝑥ሻ + 2 = 0
ሺ3 𝑥
− 2ሻሺ3 𝑥
− 1ሻ = 0
3 𝑥
= 2 or 3 𝑥
= 1
log 3 𝑥
= log 2 or log 3 𝑥
= log 1
𝑥 log 3 = log 2 or 𝑥 log 3 = 0
𝑥 =
log 2
log 3
or 𝑥 = 0
𝑥 = 0.631 or 𝑥 = 0

62
4.4.1 Solving problems involving indices, surds and logarithms
EXAMPLE 17
(i) A colony of bacteria weighs 𝑊ሺ𝑡ሻ = 100 + 30.02𝑡
gram after time 𝑡 (in
minutes). Find the time needed for the bacteria’s mass to reach 180g.
Solution:
100 + 30.02𝑡
= 180
30.02𝑡
= 80
log 30.02𝑡
= log 80
0.02𝑡 log 3 = log 80
0.02𝑡 =
log 80
log 3
𝑡 = 199.43 minutes
(ii) The total number of people infected by a highly infectious virus follow
the equation 𝑀ሺ𝑡ሻ = 1500 + 𝑒0.9𝑡
after time 𝑡 (in days). Determine the
minimum number of days needed for the virus to infect more than 9500
people.
Solution:
1500 + 𝑒0.9𝑡
> 9500
𝑒0.9𝑡
> 8000
ln 𝑒0.9𝑡
> ln 8000
0.9𝑡 ln 𝑒 > ln 8000
𝑡 >
ln 8000
0.9
𝑡 > 9.986
We round the answer the nearest integer larger than 9.986, the minimum
number of days needed for the virus to infect more than 9500 people is 10
days.
4.4 Application of Indices, Surds and Logarithms

63
4.1 Law of Indices
1. Simplify the expressions below:
(i)
(4𝑥
1
2)
3
2√ 𝑥
(ii)
3 𝑚+1+9
3 𝑚−1+1
2. Show that:
(i) 3 𝑝
+ 3 𝑝+1
+ 3 𝑝+3
is divisible by 31.
(ii) 7 𝑞
+ 7 𝑞+2
+ 72
+ 1 is divisible by 25.
3. Solve the following equations:
(i) 3 𝑥−3
= 9 𝑥
(ii) 25(5 𝑥+1) = 52𝑥−1
4.2 Law of Surds
1. Convert the following recurring decimals
into fractions:
(i) 0. 14̅̅̅̅
(ii) 0. 1̅
(iii) 0.256̅̅̅̅
2. Rationalise the following expressions:
(i)
3
2√2
(ii)
1
3√7−2√2
(iii)
2√3−3√5
2√3+3√5
1. Solve the following equations:
(i) log2(log3(2𝑥 + 1)) = 2
(ii) log64(log2(3𝑥 − 2)) = log25 √5
3
(iii) 3log3(2𝑥−3)
= 15
2. Determine the value(s) of 𝑥 in the following
equations:
(i) log9(2𝑥 + 12) = log3(𝑥 + 2)
(ii)
2
log7 2
= log2(4𝑥 − 1)
(iii) 𝑒ln3𝑥+4
= 25log5 √5
3. Given that
2 log3(𝑥 − 𝑦) = 1 + log3 𝑥 + log3 𝑦
Show that 𝑥2
+ 𝑦2
= 5𝑥𝑦.
4. (i) Simplify the expression:
log2(𝑥 + 2) + 5 log4 𝑥 − 3 log2 √ 𝑥
(ii) Hence, solve the equation:
log2(𝑥 + 2) + 5 log4 𝑥 − 3 log2 √ 𝑥 = 3
5. Find the value of:
log2 3 × log3 4 × log4 5 × log5 6
4.4 Application of Indices, Surds
and Logarithms
1. The diagram below shows 3 circles.
All circles touch each other and lie on a
common tangent of PQ. Circle A has a radius of
12cm, and circle B has a radius of 6cm. Find the
radius of the circle C.
2. Given a car’s value will deplete over years,
the car’s value after t years is RM80000 (
8
9
)
𝑡
.
Find:
(i) the value of the car after 5 years.
(ii) number of years for the car to drop below
RM35000 for the first time.
Summative Exercises

CHAPTER 5 PROGRESSIONS
64
5.1 Arithmetic Progression
5.1.1 What is an Arithmetic Progression?
An arithmetic progression is a sequence of numbers with a common difference between any two
consecutive terms.
The difference between any two consecutive terms in an arithmetic progression is called the common
difference, indicated by the symbol d.
Let’s say we have an arithmetic progression as follow:
120, 128, 136, 144, …
To calculate d, we can apply the formula
For this case,
We can see that = = , thus this is an arithmetic progression.
(Arithmetic progression always has the same difference between any two consecutive terms.)
There are three cases for the values of d.
1. d > 0
The terms in the arithmetic progression are increasing constantly,
Eg: 20, 40, 60, 80, 100, …
2. d < 0
The terms in the arithmetic progression are decreasing constantly,
Eg: 100, 80, 60, 40, 20, …
3. d = 0
The terms in the arithmetic progression are neither increasing nor decreasing.
Eg: 20, 20, 20, 20, 20, …
/
=
where represents the 优
term, n = 1, 2, 3, 4, …
=
=
= 128 120
=
=
= 136 128
=
=
= 144 136
= 8 = 8 = 8

65
EXAMPLE 1
Given the following sequences, determine whether they are arithmetic sequences.
(a) 80,83,86,89, … (b) 865, 859, 853, 846, …
Solution:
(a) = 83 80
= 3
= 86 83
= 3
= 89 86
= 3
Since there is a common difference =
= , it is an arithmetic sequence.
Now, let’s explore the properties of arithmetic progression a little bit more.
Suppose we have an arithmetic progression a, b, c, …, we can find d by either subtracting a from b or
subtracting b from c. Thus,
b a = c b
From the following equation, we get the following relationship for any three consecutive terms
(a, b, c) in an arithmetic progression.
5.1.2 How to find the value of the 优
term?
Assuming that a means the first term of the arithmetic progression and d means the common
difference, we can see the following pattern in the arithmetic progression.
12,19,26,33, …
The first term 12 = 12 + 0 x 7 = a + 0 x d
The second term 19 = 12 + 1 x 7 = a + 1 x d
The third term 26 = 12 + 2 x 7 = a + 2 x d
The fourth term 33 = 12 + 3 x 7 = a + 3 x d
Hence, we can conclude that
= a + (n 1)d
(b) = 859 865
= 6
= 853 859
= 6
= 846 853
= 7
Since ≠ , it is not an arithmetic
sequence.
2b = c + a

66
term,
a represents the first term,
n represents the number of terms,
d represents the common difference.
EXAMPLE 2
Given the arithmetic sequence 63,69,75, …,
(a) find the 优
term,
(b) find the smallest term which is greater than 100 for this sequence, and
(c) find the value of n if it is known that the 优
term has the value 165.
5.1.3 How to find the sum of the first n terms?
Let’s say we have a finite arithmetic progression given by
a , a + d , a + 2d , a + 3d , a + 4d , … , l d , l
Solution:
First term, a = 63
Common difference, d = 69 63
= 6
(a) = a + (n 1)d
The 优
term, t = 63 + (25 1)6
= 63 + 144
= 207
(b) > 100
a + (n 1)d > 100
63 + (n 1)6 > 100
57 + 6n > 100
n >
100 7
6
n > 7.167
The smallest integer n after 7.167 is 8, hence the smallest term which is greater than 100 in
this sequence is the 优
term, .
= a + (n 1)d
The 优
term, = 63 + (8 1)6
= 63 + 42
= 105
(c) = a + (n 1)d
165 = 63 + (n 1)6
165 = 57 + 6n
n =
16 7
6
= 18

67
where a represents the first term,
d represents the common difference,
l represents the last term.
Assuming the arithmetic progression has n terms, we can calculate the sum of the first n terms, ,
Sum of the first n terms, = a + a + d + a + 2d + a + 3d + a + 4d + … + l d + l ○1
We can also write in the following way,
Sum of the first n terms, = l + l d + l 2d + l 3d + l 4d + … + a + d + a ○2
○1 + ○2 : 2 = n(a + l)
= (a + l) ○3
From ○3 , we can deduce the sum of the first n terms, as follow:
Sum of the first n terms, = (a + )
= [a + a + (n 1)d]
= [2a + (n 1)d] ○4
From above, we get two formulae,
= (a + l)
= [2a + (n 1)d]
where represents sum of the first n terms in a finite arithmetic progression,
n represents number of terms,
l represents the last term,
d represents the common difference.
EXAMPLE 3
Given the arithmetic sequence 28, 41, 54, …,
(a) find the sum of the first 14 terms.
(b) find the sum of the terms from the fifth term to the tenth term, and
(c) find the smallest value of n such that the sum of the first n terms is more than 500.

68
n < 10.58, n > 7.27
Since will be always positive, n < 10.58 is rejected, hence the smallest value of n such
that the sum of the first n terms is more than 500 is 8.
5.1.4 What are the applications of arithmetic progression?
So far, we have gone thru what actually arithmetic progression is and also know how to find the 优
term as well as the sum of the first n terms. However, do you even know what the applications of
arithmetic progression in our real life are? Here are some examples of the applications.
In some of the cinemas and symposium halls or even stadiums, the arrangement of the seats forms an
arithmetic progression where the number of seats increases constantly from the first row to the last
row. For instance, there are 8 seats for the first row, 9 seats for the second row, 10 seats for the second
row and so on. The design makes the cinemas, halls and stadiums to have a stylish appearance.
Solution:
First term, a = 28
Common difference, d = 41 28
= 13
(a) = [2a + (n 1)d]
Sum of the first 14 terms, =
14
[2(28) + (14 1)13]
= 7(225)
= 1575
(b) Sum of terms from 优
term to 10 优
term =
=
10
[2(28) + (10 1)13]
4
[2(28) + (4 1)13]
= 5(173) 2(95)
= 865 190
= 675
(c) > 500
[2a + (n 1)d] > 500
[2(28) + (n 1)13] > 500
(43 + 13n) > 500
13 + 43n 1000 > 0

69
With the passage of time, our asset like cars and houses may depreciate by a fixed amount, k per year.
The value of the asset each year can be well represented by an arithmetic progression p, p k, p
2k, … where p represents the initial value of the asset.
Do you know Old Faithful? It is a natural geyser situated at the Yellow Stone National Park that
produces long eruptions which can be predicted using the knowledge of arithmetic progression. The
time between eruptions is believed to be based on the length of previous eruption.
If an eruption lasts for 1 minutes, then the next eruption will occur in approximately 46 minutes.
From the information above, we can clearly see that the time for the next eruption follows an
arithmetic progression 46, 58, 70, …. with a common difference of 12.
Try to predict how many minutes after the previous eruption will the next eruption occur if the
previous eruption lasts for 5 minutes? Yes, it is the fifth term of the progression and it is 94 minutes.

70
1
Have you ever used ladder before to fix the lamps or perhaps clean the fan? Actually, the lengths of
the rungs of a ladder decrease uniformly from the bottom to the top. Take a simple example, the
bottom rung of a ladder has a length of 40 cm and the lengths of the rungs decline constantly by 3 cm
until the top rung. There are five rungs for the ladder, thus we can express the length of the rungs
from the bottom to the top in an arithmetic progression 37 cm, 34 cm, 31 cm, 28 cm, 25 cm.
Apart from the examples shown, the application of arithmetic progression still can be observed in
many other aspects of our daily life. Have you figured them out?
1
https://en.m.wikipedia.org/wiki/Old_Faithful
Rung

71
5.2 Geometric Progression
5.2.1 What is a Geometric Progression?
A geometric progression is a sequence of numbers in which any term after the first term is obtained
by multiplying the previous term by a non-zero constant.
The non-zero constant is called the common ratio, indicated by the symbol .
Let’s say we have a geometric progression as follow:
60, 180, 540, 1620, …
To calculate , we can apply the formula
For this case,
We can see that = = , thus this is a geometric progression.
(Geometric progression always has same non-zero common ratio between any two consecutive terms.)
There are five cases for the values of .
1. > 1
The terms in the geometric progression are increasing exponentially if the first term is
positive and vice versa.
Eg 1: 20, 40, 80, 160, 320, … (Increase exponentially)
Eg 2: 20, 40, 80, 160, 320, … (Decrease exponentially)
2. < 1
The terms in the geometric progression increases in their absolute values and alternate in
signs.
Eg: 20, 40, 80, 160, 320, …
3. = 1
The terms in the geometric progression stay the same.
Eg: 20, 20, 20, 20, 20, …
=
term, n = 1, 2, 3, 4, …
=
=
=
1 0
60
= 3
=
=
=
40
1 0
= 3
=
=
=
16 0
40
= 3

72
4. = 1
It forms a sequence of numbers in same magnitude with alternating signs.
Eg: 20, 20, 20, 20, 20, …
5. 1 < < 1 where 0
The terms of the geometric sequence show exponential decay to 0.
Eg: 20, 4, 0.8, 0.16, 0.032, …
EXAMPLE 4
Given the following sequences, determine whether they are geometric sequences.
(a) 25, 100, 400, 1600, … (b) 3600, 600, 100, 20, …
Now, let’s explore the properties of geometric progression a little bit more.
Suppose we have a geometric progression a, b, c, we can find by either dividing a from b or
dividing b from c. Thus,
=
From the following equation, we get the following relationship for any three consecutive terms
(a, b, c) in a geometric progression.
Solution:
(a) =
100
= 4
=
400
100
= 4
=
1600
400
= 4
Since = = , it is a geometric sequence.
(b) =
600
3600
=
1
6
=
100
600
=
1
6
=
0
100
=
1
Since , it is not a geometric
sequence.
= ac

73
5.2.2 How to find the value of the 优
term?
Assuming that a means the first term of the geometric progression and r means the common ratio,
we can see the following pattern in the geometric progression.
10,20,40,80, …
The first term, 10 = 10 x 0
= a x
The second term, 20 = 10 x 1
= a x
The third term, 40 = 10 x = a x
The fourth term, 80 = 10 x 3
= a x
= a
r represents the common ratio, where r 1,
represents the 优
term.
EXAMPLE 5
Given the geometric sequence 0.5, 1, 2, 4, …,
(a) find the 10 优
term,
(b) find the smallest term which is greater than 100 for this sequence, and
(c) find the value of n if it is known that the 优
term of the sequence has the value 1024.
Solution:
First term, a = 0.5
Common ratio, r =
1
0
= 2
(a) = a
The 10 优
term, = 0.5 10 1
= 0.5(512)
= 256
(b) > 100
a > 100
(0.5) 1
> 100
1
> 200
ln( 1
) > ln 200
(n 1) ln 2 > ln 200
n 1 >
ln 00
ln

74
5.2.3 How to find the sum of the first n terms?
Let’s say we have a geometric progression given by
a , ar , a , a , … , a , a
r represents the common ratio, where r 1,
n represents the number of terms.
From the geometric progression, we can calculate the sum of the first n terms, ,
= a + ar + a + a + … + a + a ○1
We multiply with the common ratio, r,
r = ar + a + a + a + … + a + a ○2
○1 ○2 ,
(1 r) = a a
=

75
=
○2 ○1 ,
(r 1) = a a
=
=
−
−
=
=
−
−
We can use either of these formulae to calculate . Two of the formulae will give the same results.
Note that,
=
1
1
. =
−
−
EXAMPLE 6
Consider the geometric sequence 13, 39, 117, …, 85293. Find the sum of the terms.
Solution:
First term, a = 13
Common ratio, r =
3h
13
= 3
First, since we know that the last term of the progression is 85923, we apply the formula
= a to find the value of n.
85293 = 13 3 1
h3
13
= 3 1
6561 = 3 1
3 = 3 1
8 = n 1
n = 9
where represents sum of the first n terms in a finite geometric progression,
r represents the common ratio where r 1,
n represents the number of terms.

76
5.2.4 Can we calculate sum to infinity of the geometric progression?
As we have discussed earlier, there are five cases totally for the value of common ratio, r as follows:
1. > 1
2. < 1
3. = 1
4. = 1
5. 1 < < 1 where 0
We shall investigate whether or not we can calculate sum to infinity of the geometric progression for
every case one by one.
When > 1 or < 1, the terms in the geometric progression keep getting larger and larger in
magnitude, the sequence of terms diverges towards infinity, hence we are unable to find the sum to
infinity for this case.
3, 9, 27, 81, 243, …
When = 1, all the terms in the geometric progression will be the same, the series diverges, hence we
are unable to find the sum to infinity for this case.
5, 5, 5, 5, 5, …
When = 1, it forms a sequence of two values in same magnitude with alternate signs. The sum to
infinity of this kind of geometric progression oscillates between two values which are 0 and the first
term of the sequence, hence we are unable to find the sum to infinity for this case.
4, 4, 4, 4, 4, …
When 1 < < 1, the value of the term in the geometric progression approaches 0 (negligible) when
the number of terms approaches infinity. The series is said to converge to 0, hence we are able to find
the sum of infinity for this case.
100, 10, 1, 0.1, 0.01, 0.001, …
Now, we know that n is 9, so we can apply the formula = to find the sum of
terms, .
Sum of terms, =
13 3 h 1
3 1
=
13 1h6 4
4
= 63973
S = 0
S = 4

77
Let’s have a visual look for this case.
2
The areas of the coloured squares follow a geometric progression
1
4
,
1
16
,
1
64
, … with a common ratio of
1
4
.
The areas of the squares become smaller and approach zero.
You can also have visual look on infinite geometric progression via this video:
https://youtu.be/-y1Ob0K63hc
Now, we shall explore how to calculate the sum to infinity of geometric progression with the
restriction that 1 < < 1.
We have learnt that = or =
−
−
in our previous subtopic. From the formulae,
Thus, we deduce a new formula here.
=
where represents the “sum” of an infinite geometric progression,
r represents the common ratio where 1 < r < 1, r 0.
2
https://en.m.wikipedia.org/wiki/Geometric_series
lim =
=
lim
−
−
=
=
When n approaches ,
approaches 0

78
EXAMPLE 8
5.2.5 What are the applications of geometric progression?
Geometric progression is an important field of mathematics and have crucial applications in various
fields like physics, biology, finance, etc. Here are some of the examples of the applications in
different fields.
Mr. Lee, an entrepreneur, has invested RM5000 in a fixed deposit plan with an interest rate at 3.7%
yearly. His investment each subsequent year after that follows a geometric progression
RM5000(1.037), RM5000 1 037 , RM5000 1 037 3
, …. By using knowledge of geometric
progression, he can easily estimate the amount of savings he will have after n years, .
Express the recurring decimal 0.65454… in the simplest fraction form.
Solution:
0.65454… = 0.6 + (0.054 + 0.00054 + 0.0000054 + … )
= 0.6 +
=
3
+
0 0 4
1 0 01
=
3
+
0 0 4
0 hh
=
3
+
3
=
36
An infinite geometric progression 0.054, 0.00054, 0.0000054
with = 0.054 and = 0.01
=
EXAMPLE 7
Given a geometric progression 0.9, 0.81, 0.729, …, find the sum to infinity.
Solution:
First term, a = 0.9
Common ratio, r =
0 1
0 h
= 0.9
Sum to infinity, =
=
0 h
1 0 h
=
0 h
0 1
= 9

79
Normally, bacteria will replicate by binary fission, a process by which a bacterium split into two. As a
result, the amount of the bacteria follows a geometric progression as the population of bacteria
doubles every generation time. For most of the bacteria, the generation time will be within 20 to 60
minutes under optimum conditions. Thus, the population of bacteria after a given time period, , the
initial population of the bacteria, 0 and number of divisions are related by the following equation:
= 0 x
Doesn’t it seem alike to the formula = 1
?
Now, let’s have an example to see how geometric progression is applied for this case. Supposed a
bacterial cell can split into two in 25 minutes, given initial population of the bacteria is 100, what will
be the population of the bacteria after 5 hours, ? Since the number of divisions, n that will happen
in 5 hours will be 12 (300 minutes/25 minutes), we can get 409600 (100 x 1
) bacteria after 5 hours.
The number of the bacteria after every division is illustrated as follow:
200, 400, 800, 1600, 3200, 6400, …
1st
year 2nd
year 3rd
year

80
Geometric progression is applied in fractal geometry. In mathematics, a fractal is a self-similar
geometric figure in which they exhibit similar pattern at increasingly small scales and also at different
levels of views. Koch snowflake is one of the examples for the fractal space.
3
The Koch snowflake is formed in a sequence of stage. First, start with an equilateral triangle. On the
middle third of each side, draw an equilateral triangle of
1
3
of the length of the previously drawn
triangle, bending outwards and then erase the original middle third of that side. You now should have
a 12-sided polygon with 6 vertices. Continue the process for each of the 12 sides again. You now
should have a 36-sided polygon with 18 vertices. Then, continue the process until you get a beautiful
snowflake.
There are other examples of fractal geometries as well.
You may check how geometric progression is applied in fractal geometry via these videos:
https://youtu.be/rqP_bvOoJak
https://youtu.be/ocswsR6qFll
3
https://en.m.wikipedia.org/wiki/Koch_snowflake

81
Now, consider an infinite geometric progression 0 1 3 4
where the values of x are greater
than 1 and less than 1 but not equal to zero. This geometric progression has 1 as its first term and x
as its common ratio. Since the values of x are restricted between 1 and 1, we can calculate the finite
sum of this progression, . The calculations are as below:
=
0 1 3 4
+ … =
1
1
A new relation is formed from the progression. This relation is very useful to derive other useful
formulae to be applied in the field of Calculus as well as Physics.
For example, if we differentiate the following relation, we can get the following formula:
t1
1
=
1
1
for 1 1 where 0
Arithmetic Progression (Part 1)
https://youtu.be/hEU3byoqyVY
Arithmetic Progression (Part 2)
https://youtu.be/Q98NyR50mHw
Geometric Progression (Part 1)
https://youtu.be/3AYb5tqPle4
Geometric Progression (Part 2)
https://youtu.be/RLZaGJ9nK2U
Arithmetic and Geometric Progression Formulae
https://youtu.be/gua96ju_FBk
1 1 3 4
+… =
1
1

82
1) Given an arithmetic progression in which the first
term is 1038, the third term is 964 and the last
term is 594. Find the number of terms for the
arithmetic progression.
2) Harry has RM x in his piggy bank. He saves RM12
on the first day, RM15 on the second day, RM18
on the third day and follows this pattern for the
subsequent days. On the tenth day, he saves his
money as usual and finds that his piggy bank has
a total of RM605 inside. Find the value of x.
3) It is known that the middle two terms of an
arithmetic progression add up to 268 and the third
term is 101. Given the progression has 16 terms,
find the last term.
4) Rashid draws a circle with a radius of 3 cm at
first. Then, he increases the radius to 6 cm and
draw the new circle besides the previous circle.
He keeps drawing by following the arithmetic
sequence until he had drawn 7 circles in total.
After that, he draws a rectangle to inscribe the
circles. Find the area of the rectangle.
5) The sum of the first five terms and the first eight
terms for an arithmetic progression are
16070
3
and
6
3
. Find the tenth term.
6) Given an arithmetic sequence 78, 82, 86, …
(a) find the sum of the 10 terms after the 优
term,
(b) find the smallest term which is more than 200,
(c) if the last second term of the sequence is 158,
find the sum of the last five terms.
Summative Exercises
1) Given the geometric sequence 20, 40, 80, …
(a) find the 10 优
term,
(b) if the sequence only applies to the numbers
less than 10000, find the last term,
(c) find sum of the terms from t
term to
7 优
term.
2) Puan Kiah invests RM 5000 in her fixed deposit
saving account with an annual interest rate of 3%.
How many years minimum does it take to have
more than RM 6000 if she does not withdraw the
money throughout the duration?
3) A square paper with length of 22 cm is folded in
half. The action is repeated for another 5 times.
Calculate the area of the folded paper at the end.
4) Given the first term is 2000, the last term is
1180.98 and there are six terms for a geometric
sequence. Find the sum of the first four terms.
5) A ball is dropped from a height of 24 cm. It
rebounds to a height of 21 cm and continues to
rebound to 0.875 of its height continuously.
Calculate the total distance it travels before it
comes to a rest.
6) Given 7 cylinders arranged in a row in which their
radius and height follow geometric progression.
The radius and height are incrementing by 2 times
and 1.4 times respectively. The first cylinder has
a radius of 1 cm and height of 5 cm. Calculate the
volume of the last cylinder.
7) Express the recurring decimal 5.3888…. in its
simplest fraction form.
5.2 Geometric Progression5.1 Arithmetic Progression

CHAPTER 1 FUNCTIONS
83
SOLUTIONS
Chapter 1 Functions
1.1 Functions
1) (a) 𝑓(𝑥) = 𝑥2
+ 4
4 is the object. Therefore,
𝑓(4) = 42
+ 4
= 16 + 4
= 20
(b) 4 is the image. Therefore,
𝑓(𝑥) = 4
𝑥2
+ 4 = 4
𝑥2
= 0
𝑥 = 0
2) (a) Function is undefined when denominator = 0
𝑥 − 4 = 0
𝑥 = 4
𝑞 = 4
(b) 𝑓(2) = 2
2𝑝 + 3 = −4
2𝑝 = −7
Whenever see “x” in the function 𝑓(𝑥) = 𝑥2
+ 4, substitute 𝑥 = 4
Since 𝑓(𝑥) = 𝑥2
+ 4
Since 𝑥 = 𝑞
2𝑝 + 3
2 − 4
= 2
2 is mapped onto itself under 𝑓
2𝑝 + 3
−2
= 2
𝑝 = −
7
2

CHAPTER 1 FUNCTIONS
84
3) Step 1: You need to have 4 points in order to construct a graph.
The four points are:
(a) left endpoint of the graph,
(b) vertex,
(c) y-intercept,
(d) right endpoint of the graph.
(a) When 𝑥 = −4,
𝑓(−4) = |−4 + 3|
= | − 1|
= 1
(b) When 𝑓(𝑥) = 0
|𝑥 + 3| = 0
𝑥 + 3 = 0
𝑥 = −3
(c) When 𝑥 = 0,
𝑓(0) = |0 + 3|
= |3|
= 3
(d) When 𝑥 = 1,
𝑓(1) = |1 + 3|
= |4|
= 4
Step 2: You need to plot the points onto a graph paper and indicate the x-axis and y-axis.
Step 3: Connect the points in the shape of V with vertex as the turning point.
x f (x)
−4 1
−3 0
0 3
1 4
𝑓(𝑥) = |𝑥 + 3|
y-intercept →𝑥 = 0
left endpoint →𝑥 = −4
right endpoint →𝑥 = 1
vertex →𝑓(𝑥) = 0
(-4,1)
(-3,0)
(0,3)
(1,4)
𝑓(𝑥)
x
Range of the function → 0 ≤ f(x) ≤ 4
0 and 4 are included in the range of function

CHAPTER 1 FUNCTIONS
85
1.2 Composite Functions
1) In order to find 𝑔𝑓(𝑥), you need to have the function 𝑓(𝑥) and 𝑔(𝑥).
Now, the question has given the function 𝑓(𝑥).
Therefore you need to find the function 𝑔(𝑥).
𝑓𝑔(𝑥) = 2𝑥2
+ 5
𝑓(𝑥) = 2𝑥 + 1
𝑓[𝑔(𝑥)] =2𝑔(𝑥) + 1
2𝑥2
+ 5 = 2𝑔(𝑥) + 1
2𝑔(𝑥) = 2𝑥2
+ 4
𝑔(𝑥) = 𝑥2
+ 2
𝑔𝑓(𝑥) = 𝑔[𝑓(𝑥)]
= 𝑔(2𝑥 + 1)
= (2𝑥 + 1)2
+ 2
= 4𝑥2
+ 4𝑥 + 1 + 2
= 4𝑥2
+ 4𝑥 + 3
𝑔𝑓(𝑥) = 2
4𝑥2
+ 4𝑥 + 3 = 2
4𝑥2
+ 4𝑥 + 1 = 0
(2𝑥 + 1)(2𝑥 + 1) = 0
𝑥 = −
1
2

CHAPTER 1 FUNCTIONS
86
2)
(a) 𝑓2(𝑥) = 𝑓𝑓(𝑥)
(b) 𝑓3(𝑥) = 𝑓2
𝑓(𝑥)
ALTERNATIVE METHOD
𝑓3(𝑥) = 𝑓𝑓2(𝑥)
𝑓(𝑥) =
1 + 𝑥
1 − 𝑥
, 𝑥 ≠ 1
= 𝑓 (
1 + 𝑥
1 − 𝑥
)
=
1 +
1 + 𝑥
1 − 𝑥
1 −
1 + 𝑥
1 − 𝑥
=
(1 − 𝑥) + (1 + 𝑥)
1 − 𝑥
(1 − 𝑥) − (1 + 𝑥)
1 − 𝑥
=
2
1 − 𝑥
−2𝑥
1 − 𝑥
=
2
1 − 𝑥
×
1 − 𝑥
−2𝑥
= −
1
𝑥
, 𝑥 ≠ 0
= 𝑓2
(
1 + 𝑥
1 − 𝑥
)
= −
1
1 + 𝑥
1 − 𝑥
= −
1 − 𝑥
1 + 𝑥
= 𝑓 (−
1
𝑥
)
=
1 + (−
1
𝑥
)
1 − (−
1
𝑥)
=
𝑥 − 1
𝑥
𝑥 + 1
𝑥
=
𝑥 − 1
𝑥 + 1
, 𝑥 ≠ −1
=
𝑥 − 1
𝑥 + 1
, 𝑥 ≠ −1

CHAPTER 1 FUNCTIONS
87
2) (c) 𝑓6(𝑥) = 𝑓2
𝑓2
𝑓2(𝑥)
ALTERNATIVE METHOD
𝑓6(𝑥) = 𝑓3
𝑓3(𝑥)
3) (a) 𝑟(𝑥) = 5𝑥 (Roland could walk 5x distance in x hours)
(b) 𝑝(𝑟) = 𝜋𝑟2
(c) 𝑝𝑟(𝑥) = 𝑝(5𝑥)
= 𝜋(5𝑥)2
= 25𝜋𝑥2
= 𝑓2
𝑓2
(−
1
𝑥
)
= 𝑓2
(−
1
−
1
𝑥
)
= 𝑓2(𝑥)
= −
1
𝑥
, 𝑥 ≠ 0
−
1
−
1
𝑥
+ 𝑥
= 𝑓3
(
𝑥 − 1
𝑥 + 1
)
=
𝑥 − 1
𝑥 + 1 − 1
𝑥 − 1
𝑥 + 1 + 1
=
(𝑥 − 1) − (𝑥 + 1)
𝑥 + 1
(𝑥 − 1) + (𝑥 + 1)
𝑥 + 1
=
−2
𝑥 + 1
2𝑥
𝑥 + 1
=
−2
𝑥 + 1
×
𝑥 + 1
2𝑥
= −
1
𝑥
, 𝑥 ≠ 0
Spot where Roland is missing
r
rThe furthest that Roland could
walk is by walking through a
straight line. The area that we
could possibly find Roland is
the area of the circle whereby
the radius of circle is r miles.

CHAPTER 1 FUNCTIONS
88
4) (a) In order to write the function for the height h of water after t seconds, we need to have the initial
height of water first.
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 = 𝜋𝑟2
ℎ
ℎ(𝑡) = 10 + 𝑡
(b) 𝑉(ℎ) = 𝜋𝑟2
ℎ
= 154ℎ 𝑐𝑚
(c) 𝑉ℎ(𝑡) = 𝑉(10 + 𝑡)
= 154(10 + 𝑡)
= 1540 + 154𝑡
(d) When 𝑡 = 10,
𝑉ℎ(10) = 1540 + 154(10)
= 3080𝑐𝑚3
1)
𝑔(𝑥) = 3 − 2𝑥
(a)
𝑥𝑦 − 2𝑦 = 2
𝑥𝑦 = 2𝑦 + 2
(b) 𝐿𝑒𝑡 𝑦 = 3 − 2𝑥
2𝑥 = 3 − 𝑦
1540 =
22
7
× 72
× ℎ𝑖
ℎ𝑖 =
1540 × 7
22 × 72
ℎ𝑖 = 10𝑐𝑚
=
22
7
× 72
× ℎ
𝑓(𝑥) =
2
𝑥 − 2
𝐿𝑒𝑡 𝑦 =
2
𝑥 − 2
𝑥 =
2𝑦 + 2
𝑦
𝑓−1(𝑦) =
2𝑦 + 2
𝑦
𝑓−1(𝑥) =
2𝑥 + 2
𝑥
, 𝑥 ≠ 0
𝑥 =
3 − 𝑦
2
𝑔−1(𝑦) =
3 − 𝑦
2
𝑔−1(𝑥) =
3 − 𝑥
2

CHAPTER 1 FUNCTIONS
89
1) (c)
(d) In order to find (𝑓𝑔)−1(𝑥), we need to find 𝑓𝑔(𝑥) first.
𝑓𝑔(𝑥) = 𝑓(3 − 2𝑥)
𝑦 − 2𝑥𝑦 = 2
−2𝑥𝑦 = 2 − 𝑦
𝑔−1
𝑓−1(𝑥) = 𝑔−1
(
2𝑥 + 2
𝑥
)
=
3 −
2𝑥 + 2
𝑥
2
=
1
2
×
3𝑥 − (2𝑥 + 2)
𝑥
=
𝑥 − 2
2𝑥
, 𝑥 ≠ 0
=
2
3 − 2𝑥 − 2
=
2
1 − 2𝑥
, 𝑥 ≠
1
2
𝐿𝑒𝑡 𝑦 =
2
1 − 2𝑥
𝑥 =
2 − 𝑦
−2𝑦
(𝑓𝑔)−1(𝑦) =
𝑦 − 2
2𝑦
(𝑓𝑔)−1(𝑥) =
𝑥 − 2
2𝑥
, 𝑥 ≠ 0
𝑥 =
𝑦 − 2
2𝑦

CHAPTER 1 FUNCTIONS
90
(e)
𝑥𝑦 − 2𝑦 = 3𝑥 − 10
𝑥𝑦 − 3𝑥 = 2𝑦 − 10
CONCLUSION
Therefore, 𝑔−1
𝑓−1(𝑥) = (𝑓𝑔)−1(𝑥)
2) 𝑔(𝑥) = 𝑥 − 4
(a) 𝑔2(𝑥) = 𝑔𝑔(𝑥)
= 𝑔(𝑥 − 4)
= (𝑥 − 4) − 4
= 𝑥 − 8
𝐿𝑒𝑡 𝑦 = 𝑥 − 4
𝑥 = 𝑦 + 4
𝑔−1(𝑦) = 𝑦 + 4
𝑔−1(𝑥) = 𝑥 + 4
𝑔𝑓(𝑥) = 𝑔 (
2
𝑥 − 2
)
= 3 − 2 (
2
𝑥 − 2
)
=
3(𝑥 − 2) − 4
𝑥 − 2
=
3𝑥 − 10
𝑥 − 2
𝑥(𝑦 − 3) = 2𝑦 − 10
(𝑓𝑔)−1(𝑦) =
2𝑦 − 10
𝑦 − 3
(𝑓𝑔)−1(𝑥) =
2𝑥 − 10
𝑥 − 3
, 𝑥 ≠ 3
𝑥 =
2𝑦 − 10
𝑦 − 3
𝐿𝑒𝑡 𝑦 =
3𝑥 − 10
𝑥 − 2
𝑔−1
𝑓−1(𝑥) = (𝑓𝑔)−1(𝑥) =
𝑥 − 2
2𝑥

CHAPTER 1 FUNCTIONS
91
2) (b) 𝐿𝑒𝑡 𝑦 = 𝑥 − 8
𝑥 = 𝑦 + 8
(𝑔2
)−1(𝑦) = 𝑦 + 8
(𝑔2
)−1(𝑥) = 𝑥 + 8
𝐿𝐻𝑆 = (𝑔−1
)2(𝑥)
= 𝑔−1
𝑔−1(𝑥)
= 𝑔−1(𝑥 + 4)
= (𝑥 + 4) + 4
= 𝑥 + 8
= (𝑔2
)−1(𝑥)
= 𝑅𝐻𝑆
Therefore (𝑔−1
)2(𝑥) = (𝑔2
)−1
(𝑥).
(c) 𝑔𝑔−1 (𝑥) = 𝑔−1(5𝑥 + 2)
𝑥 = (5𝑥 + 2) + 4
𝑥 − 5𝑥 = 2 + 4
−4𝑥 = 6
𝑥 = −
3
2

92
Chapter 2 Quadratic Functions
1. 3𝑥(𝑥 − 5) = 2𝑥 − 1
3𝑥*
− 15𝑥 = 2𝑥 − 1
3𝑥*
− 17𝑥 + 1 = 0
3 .𝑥*
−
17
3
𝑥 +
1
3
/ = 0
3 0𝑥*
−
17
3
𝑥 + 1
2−
17
3
3
2
4
*
− 1
2−
17
3
3
2
4
*
+
1
3
5 = 0
3 6.𝑥 −
17
6
/
*
−
277
36
8 = 0
3 .𝑥 −
17
6
/
*
−
277
12
= 0
3 .𝑥 −
17
6
/
*
=
277
12
.𝑥 −
17
6
/
*
=
277
36
𝑥 −
17
6
= ±:
277
36
𝑥 = ±:
277
36
+
17
6
𝑥 = 0.059 𝑜𝑟 𝑥 = 5.607
2. Given 𝛼 = 5 𝑎𝑛𝑑 𝛽 = 9.
𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠, 𝛼 + 𝛽 = 5 + 9 = 14
𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠, 𝛼𝛽 = 5(9) = 45
𝑥*
− ( 𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠) 𝑥 + ( 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠) = 0
𝑥*
− 14𝑥 + 45 = 0
3. 2( 𝑥 − 5)*
= 4( 𝑥 + 7)
2(𝑥*
− 10𝑥 + 25) = 4𝑥 + 28
2𝑥*
− 20𝑥 + 100 = 4𝑥 + 28
2𝑥*
− 24𝑥 + 72 = 0
𝑥*
− 12𝑥 + 36 = 0
𝑎 = 1, 𝑏 = −12, 𝑐 = 36
𝛼 + 𝛽 = −
(−12)
1
= 12 𝛼𝛽 =
36
1
= 36
Roots: 2𝛼, 2𝛽
𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠 = 2𝛼 + 2𝛽 = 2( 𝛼 + 𝛽) = 2(12) = 24
𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠 = 2𝛼(2𝛽) = 4( 𝛼𝛽) = 4(36) = 144

93
𝑥*
− 24𝑥 + 144 = 0
4. Roots: −3 𝑎𝑛𝑑
P
*
𝑠𝑢𝑚 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠 = −3 +
1
2
= −
5
2
𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑟𝑜𝑜𝑡𝑠 = −3 .
1
2
/ = −
3
2
𝑥*
− .−
5
2
/ 𝑥 + .−
3
2
/ = 0
𝑥*
+
5
2
𝑥 −
3
2
= 0
2𝑥*
+ 5𝑥 − 3 = 0
Compare 2𝑥*
+ 5𝑥 − 3 = 0 and 2𝑥*
+ ( 𝑚 + 1) 𝑥 + 𝑛 = 2
𝑚 + 1 = 5
𝑚 = 4
𝑛 − 2 = −3
𝑛 = −5
5. 𝑥*
− 4𝑥 + 3 > 0
( 𝑥 − 1)( 𝑥 − 3) > 0
When ( 𝑥 − 1)( 𝑥 − 3) = 0, 𝑥 = 1 𝑜𝑟 𝑥 = 3.
Since ( 𝑥 − 1)( 𝑥 − 3) > 0, thus the range of values of 𝑥 is determined on the graph above the
𝑥-axis. Hence, the range of values of 𝑥 is 𝑥 < 1 𝑎𝑛𝑑 𝑥 > 3.

𝑦
𝑥
1 3
+ +
−

94
1. 𝑥( 𝑥 − 2) = 5
𝑥*
− 2𝑥 − 5 = 0
𝑎 = 1, 𝑏 = −2, 𝑐 = −5
𝑏*
− 4𝑎𝑐 = (−2)*
− 4(1)(−5) = 24 (> 0)
Since the discriminant is > 0. Thus, 𝑥( 𝑥 − 2) = 5 has two real roots.
𝑥( 𝑥 + 5) = 2𝑥 − 14
𝑥*
+ 5𝑥 − 2𝑥 + 14 = 0
𝑥*
+ 3𝑥 + 14 = 0
𝑎 = 1, 𝑏 = 3, 𝑐 = 14
𝑏*
− 4𝑎𝑐 = (3)*
− 4(1)(14) = −47(< 0)
Since the discriminant is < 0. Thus, 𝑥( 𝑥 + 5) = 2𝑥 − 14 has no real root.
2. 𝑥*
+ 2ℎ𝑥 + 4 = 𝑥
𝑥*
+ (2ℎ − 1) 𝑥 + 4 = 0
𝑎 = 1, 𝑏 = 2ℎ − 1, 𝑐 = 4
𝑏*
− 4𝑎𝑐 = 0
(2ℎ − 1)*
− 4(1)(4) = 0
(2ℎ − 1)*
− 16 = 0
2ℎ − 1 = ±√16
2ℎ − 1 = 4 2ℎ − 1 = −4
2ℎ = 5 2ℎ = −3
ℎ =
5
2
ℎ = −
3
2
3. 𝑥*
+ 5𝑛 = 4𝑚𝑥
𝑥*
− 4𝑚𝑥 + 5𝑛 = 0
𝑎 = 1, 𝑏 = −4𝑚, 𝑐 = 5𝑛
𝑏*
− 4𝑎𝑐 = 0
(−4𝑚)*
− 4(1)(5𝑛) = 0
16𝑚*
− 20𝑛 = 0
16𝑚*
= 20𝑛
𝑚*
=
5
4
𝑛
𝑚 = ±
1
2
√5𝑛
4. 𝑥*
+ 𝑝𝑥 + 𝑞 = 0
𝑎 = 1, 𝑏 = 𝑝, 𝑐 = 𝑞
𝑏*
− 4𝑎𝑐 = 16
𝑝*
− 4(1)( 𝑞) = 16
𝑝 − 𝑞 = −4
𝑝 = 𝑞 − 4
( 𝑞 − 4)*
− 4𝑞 = 16
1
2

95
𝑞*
− 8𝑞 + 16 − 4𝑞 = 16
𝑞*
− 12𝑞 = 0
( 𝑞 − 0)( 𝑞 − 12) = 0
𝑞 = 12 𝑜𝑟 𝑞 = 0 (ignored)
𝑝 = 𝑞 − 4
𝑝 = 12 − 4
𝑝 = 8
1.
2. 𝑓( 𝑥) = 0
2𝑥*
+ 𝑥 − 5 = 0
𝑎 = 2, 𝑏 = 1, 𝑐 = −5
𝑏*
− 4𝑎𝑐 = (1)*
− 4(2)(−5) = 41(> 0)
Thus, the quadratic function 𝑓( 𝑥) = 2𝑥*
+ 𝑥 − 5 has two real roots. Since 𝑎 > 0, hence the
graph is a parabola which passes through minimum point and intersects two points on 𝑥-axis.
𝑓( 𝑥) = −𝑥*
+ 𝑥 + 6
( 𝑎) 𝑓( 𝑥) = −
1
2
𝑥*
+ 𝑥 + 6
( 𝑏) 𝑓( 𝑥) = −𝑥*
− 𝑥 + 6
( 𝑐) 𝑓( 𝑥) = −𝑥*
+ 𝑥 + 1
𝑓( 𝑥)
𝑥

96
3. 𝑓( 𝑥) = 𝑚𝑥*
+ 4𝑥 − 6
𝑎 = 𝑚, 𝑏 = 4, 𝑐 = −6
𝑏*
− 4𝑎𝑐 < 0
(4)*
− 4( 𝑚)(−6) < 0
16 + 24𝑚 < 0
24𝑚 < −16
𝑚 < −
2
3
4. 𝑓( 𝑥) = 5𝑥*
− ( 𝑘𝑥 + 4) 𝑥 − 2
𝑓( 𝑥) = 5𝑥*
− 𝑘𝑥*
− 4𝑥 − 2
𝑓( 𝑥) = (5 − 𝑘) 𝑥*
− 4𝑥 − 2
𝑎 = 5 − 𝑘, 𝑏 = −4, 𝑐 = −2
𝑏*
− 4𝑎𝑐 > 0
(−4)*
− 4(5 − 𝑘)(−2) > 0
16 + 40 − 8𝑘 > 0
56 > 8𝑘
𝑘 < 7

97
3.1 Systems of Linear Equations in Three Variables
1) x − 2y + z = 10 … ○1
2x + 2y + 3z = 13 … ○2
–2x + 5y + z = –7 … ○3
○1 ×2: 2x – 4y + 2z = 20 … ○4
○2 –○4 : 6y + z = –7
z = –7 – 6y … ○5
○2 +○3 : 7y + 4z = 6
4z = 6 – 7y … ○6
4(–7 – 6y) = 6 – 7y
–28 – 24y = 6 – 7y
17y = –34
y = –2
Substitute y = –2 into ○5
z = –7 – 6(–2)
z = 5
Substitute y = –2 and z = 5 into ○1
x – 2(–2) + 5 = 10
x = 1
So, x + y + z = 1 + (–2) + 5
= 4
2) x + 2y – z = 2 … ○1
2x – 3y + z = –1 … ○2
5x – y – 2z = –3 … ○3
○1 +○2 : 3x – y = 1 … ○4
○2 ×2: 4x – 6y + 2z = –2 … ○5
○3 +○5 : 9x – 7y = –5 … ○6
○4 ×7: 21x – 7y = 7 … ○7
○7 –○6 : 12x = 12
x = 1
Substitute x = 1 into ○4
3(1) – y = 1
y = 2
Substitute x = 1 and y = 2 into ○1
1 + 2(2) –z =2
z = 3

98
3) x + y + z = 6 … ○1
y + 5z = −7 … ○2
2x + 5y − z = 27 … ○3
From ○2 : y = –7 – 5z …○4
○1 ×2: 2x + 2y + 2z = 12 … ○5
○3 –○5 : 3y – 3z = 15
y = 5 + z … ○6
Compare ○5 and ○6
–7 – 5z = 5 + z
6z = –12
z = –2
Substitute z = –2 into ○4
y = –7 – 5(–2)
y = 3
Substitute y = 3 and z = –2 into ○1
x + 3 + (–2) = 6
x = 5
4) Let the variables
Y: The number of yellow balls
P: The number of pink balls
B: The number of blue balls
There are 90 balls altogether and 25 are black, so the total number of yellow, pink and blue balls
are 90 – 25 = 65. The equation written as:
Y + P + B = 65 … ○1
60 of the balls are not pink and 25 are black, so the number of not pink ball excluded black balls:
60 – 25 = 35
35 balls are the total number of yellow balls and blue balls, the equation written as:
Y + B = 35 … ○2
67 are not yellow ball and 25 are black, so the number of not yellow balls excluded black balls:
67 – 25 = 42
42 balls are the total number of pink balls and blue balls, the equation written as:
P + B = 42 … ○3
From ○1 and ○2 we can deduce that
○1 –○2 : P = 30
Substitute P = 30 into ○3
30 + B = 42
B = 12
Substitute B = 12 into ○2
Y + 12 = 35
Y = 23
Thus, Adam has 23 yellow balls, 30 pink balls and 12 blue balls respectively.

99
5) Using the standard equation given: x2 +y2 + Ax + By + C = 0
Substitute 3 coordinates point (1,1), (2, –4) and (5,5) into the standard equation given to form
three new equations.
For (1,1): 1 + 1 + A + B + C = 0
A + B + C = –2 … ○1
For (2, –4): 22 + (–4)2 + A(2) + B(–4) + C = 0
20 + 2A – 4B + C = 0
2A – 4B + C = –20 … ○2
For (5,5): 52 + 52 + 5A + 5B + C = 0
50 + 5A + 5B + C = 0
5A + 5B + C = –50 … ○3
○1 ×2: 2A + 2B + 2C = –4 … ○4
○4 –○2 : 6B + C = 16
C = 16 – 6B … ○5
○1 ×5: 5A + 5B + 5C = –10 … ○6
○6 –○3 : 4C = 40
C = 10
Substitute C = 10 into ○5
10 = 16 – 6B
B = 1
Substitute B = 1 and C = 10 into ○1
A + 1 + 10 = –2
A = –13
Hence, the equation of the circle that passes through the points (1,1), (2, –4) and (5,5) are
x2 + y2 – 13x + y + 10 = 0
6) Let the variables
x: The amount of money invested at 6%
y: The amount of money invested at 8%
z: The amount of money invested at 10%
The total amount of money invested at three different rate are RM50000
The equation can be written as:
x + y + z = 50000 … ○1
The yearly interest is RM3700. Simple interest in percentage form can be convert to decimal
form. For example, 6% simple interest is equivalent to 0.06 times the amount of money invested
at 6%.
0.06x + 0.08y + 0.10z = 3700 … ○2
Amount of money invested at 6% is twice as much as invested at 10%.
x = 2z
x – 2z = 0 … ○3
From ○3 : x = 2z … ○4
○1 ×0.08: 0.08x + 0.08y + 0.08z = 4000 … ○5
○5 –○2 : 0.02x – 0.02z = 300 … ○6
0.02(2z) – 0.02z = 300

100
z = 15000
Substitute z = 15000 into ○4
x = 2(15000)
= 30000
Substitute x = 30000 and z = 15000 into ○1
30000 + y + 15000 = 50000
y = 5000
Thus, RM30,000 is invested at 6%, RM5,000 is invested at 8% and RM15,000 is invested at
10%.

101
3.2 Simultaneous Equations Involving One Linear Equation
and One Non-Linear Equation
1) 2x + y = 1 … ○1
x2 + y2 = 1 … ○2
From ○1 : y = 1 – 2x … ○3
x2 + (1 – 2x)2 = 1
x2 + (1 – 4x + 4x2) = 1
5x2 – 4x = 0
x(5x – 4) = 0
x = 0 or
4
5
Substitute x = 0 and x =
4
5
into ○3
y = 1 – 2(0) or y = 1 – 2(
4
5
)
= 1 = –
3
5
When x = 0, y = 1
When x =
4
5
, y = –
3
5
2) x2 – y = 14 … ○1
2y – 4 = 12x … ○2
From ○2 : 2y = 12x + 4
y = 6x + 2 … ○3
x2 – (6x + 2) = 14
x2 – 6x – 16 = 0
(x – 8)(x + 2) = 0
x = 8 or –2
Substitute x = 8 and x = –2 into ○3
y = 6(8) + 2 or y = 6(–2) + 2
= 50 = –10
When x = 8, y = 50
When x = –2, y = –10
3) x² − y² = 7 … ○1
2y = 2 + x … ○2
From ○2 : x = 2y – 2 … ○3
(2y – 2)2 – y2 = 7
4y2 – 8y + 4 – y2 = 7
3y2 – 8y – 3 = 0
(y – 3)(3y + 1)= 0
y = 3 or –
1
3

102
Substitute y = 3 and y = –
1
3
into ○3
x = 2(3) – 2 or x = 2(–
1
3
) – 2
= 4 = –
8
3
When x = 4, y = 3
When x = –
8
3
, y = –
1
3
4) For equation x + 2y – 4 = 0
x –3 –2 –1 0 1 2 3
y 3.5 3 2.5 2 1.5 1 0.5
For equation x2 + y2 = 7 – xy
x –3 –2 –1 0 1 2 3
y 1, 2 –1, 3 –2, 3 ± √7 –3, 2 –3, 1 –2, –1
Based on the graph, there are two points of intersection which is (–2,3) and (2,1).
Thus, the solutions to the simultaneous equations are x = –2, y = 3 and x = 2, y = 1.
5) 4x + y – 5 = 0 … ○1
27x2 + 21xy = –2y2 … ○2
From ○1 : y = –4x + 5 … ○3
27x2 + 21x(–4x + 5) = –2(–4x + 5)2
27x2 – 84x2 + 105x = –2(16x2 – 40x + 25)
–57x2 + 105x = –32x2 + 80x – 50
25x2 – 25x – 50 = 0
x2 – x – 2 = 0
(x – 2)(x + 1) = 0

103
x = 2 or x = –1
Substitute x = 2 and x = –1 into ○3
y = –4(2) + 5 or y = –4(–1) + 5
= –3 = 9
When x = 2, y= –3
When x = –1, y = 9
Therefore, the coordinates of point A is (2, –3) and point B is (–1,9)
[Alternative answer: Coordinate of point A is (–1,9) and point B is (2, –3)]
6) Area of the rectangular plot: 30 × x = 30x
Area of the fish pond: (x – 10) × (30 – y) = 30x – xy – 300 + 10y
The equation of area of the region grew with bananas can be written as:
30x – (30x + 10y – xy – 300) = 460
–10y + xy + 300 = 460
xy – 10y = 160 … ○1
The perimeter of the rectangular fish pond: 2(x – 10) + 2(30 – y) = 2x – 2y + 40
The equation of the perimeter of the fish pond can be written as :
2x – 2y + 40 = 48 … ○2
We obtain two equation
From ○2 : x – y + 20 = 24
x = y + 4 … ○3
y(y + 4) – 10y = 160
y2 + 4y – 10y – 160 = 0
y2 – 6y – 160 = 0
(y – 16)(y + 10) = 0
y = 16 or y = –10 (rejected since area cannot be negative)
Substitute y = 16 into ○3
x = 16 + 4
= 20
Thus, x = 20 and y = 16.

104
1. (i)
(4𝑥
1
2)
3
2√ 𝑥
=
43(𝑥
1
2)
3
2𝑥
1
2
=
64𝑥
3
2
2𝑥
1
2
= 32𝑥
3
2
−
1
2
= 32𝑥
(ii)
3 𝑚+1+9
3 𝑚−1+1
=
3 𝑚31+9
3 𝑚3−1+1
=
3(3 𝑚+3)
3−1(3 𝑚+3)
= 9
2. (i) 3 𝑝
+ 3 𝑝+1
+ 3 𝑝+3
= 3 𝑝
+ 3(3 𝑝) + 33(3 𝑝)
= 3 𝑝(1 + 3 + 33)
= 3 𝑝(31)
Since 3 𝑝
+ 3 𝑝+1
+ 3 𝑝+3
can be written as 3 𝑝(31), which means it has a factor of 31. Thus, the
expression is divisible by 31.
(ii) 7 𝑞
+ 7 𝑞+2
+ 72
+ 1 = 7 𝑞
+ 72(7 𝑞) + 49 + 1
= 7 𝑞(1 + 72) + 50
= 7 𝑞(50) + 50
= 50(7 𝑞
+ 1)
Since 7 𝑞
+ 7 𝑞+2
+ 72
+ 1 can be written as 50(7 𝑞
+ 1), which means it has a factor of 50. Thus, the
expression is divisible by 50.
3. (i) 3 𝑥−3
= 9 𝑥
3 𝑥−3
= (32) 𝑥
3 𝑥−3
= 32𝑥
By comparison,
𝑥 − 3 = 2𝑥
𝑥 = −3
(ii) 25(5 𝑥+1) = 52𝑥−1
52(5 𝑥+1) = 52𝑥−1
5 𝑥+1+2
= 52𝑥−1
5 𝑥+3
= 52𝑥−1
By comparison,
𝑥 + 3 = 2𝑥 − 1
𝑥 = 4
4.1 Law of Indices
Simplify the expression by
using the law of indices
Notice 3 𝑚
+ 3 is the common factor
for the numerator and denominator.
We can factor and cancel it out.
Factor out the common factor
3 𝑝
. We find 31 is one of the
factors of the expression.
Make the bases for both sides
equal by using the law of indices.

105
1. (i) Let 𝑥 = 0. 14̅̅̅̅
100𝑥 = 14. 14̅̅̅̅
100𝑥 − 𝑥 = 14. 14̅̅̅̅ − 0. 14̅̅̅̅
99𝑥 = 14
𝑥 =
14
99
∴ 0. 14̅̅̅̅ =
14
99
(ii) Let 𝑥 = 0. 1̅
10𝑥 = 1. 1̅
10𝑥 − 𝑥 = 1. 1̅ − 0. 1̅
9𝑥 = 1
𝑥 =
1
9
∴ 0. 1̅ =
1
9
(iii) Let 𝑥 = 0.056̅̅̅̅
10𝑥 = 0. 56̅̅̅̅
1000𝑥 = 56. 56̅̅̅̅
1000𝑥 − 10𝑥 = 56. 56̅̅̅̅ − 0. 56̅̅̅̅
990𝑥 = 56
𝑥 =
56
990
=
28
495
2. (i)
3
2√2
=
3
2√2
×
√2
√2
=
3√2
4
(ii)
1
3√7−2√2
=
1
3√7−2√2
×
3√7+2√2
3√7+2√2
=
3√7+2√2
(3√7)
2
−(2√2)
2
=
3√7+2√2
63−8
=
3√7+2√2
55
(iii)
2√3−3√5
2√3+3√5
=
2√3−3√5
2√3+3√5
×
2√3−3√5
2√3−3√5
=
(2√3−3√5)
2
(2√3)
2
−(3√5)
2
=
(2√3)
2
−2(2√3)(3√5)+(3√5)
2
12−45
=
12−12√15+45
−33
= −
57−12√15
33
=
4√15−19
11
4.2 Law of Surds
Form two equations and solve for 𝑥.
Take note that the recurring decimal does not
start directly after the decimal point. Therefore,
forming two new equations of 10𝑥 and 1000𝑥
is necessary to solve for the fraction.
Multiply the conjugate for the surds in the
denominator in order to rationalise the fractions.

106
1. (i) log2(log3(2𝑥 + 1)) = 2
22
= log3(2𝑥 + 1)
34
= 2𝑥 + 1
81 = 2𝑥 + 1
𝑥 = 40
(ii) log64(log2(3𝑥 − 2)) = log25 √5
3
log64(log2(3𝑥 − 2)) =
log5 √5
3
log5 25
log64(log2(3𝑥 − 2)) =
log5 5
1
3
log5 52
log64(log2(3𝑥 − 2)) =
1
3
log5 5
2 log5 5
log64(log2(3𝑥 − 2)) =
1
6
log2(3𝑥 − 2) = 64
1
6
log2(3𝑥 − 2) = 2
3𝑥 − 2 = 22
𝑥 = 2
(iii) 3log3(2𝑥−3)
= 15
log3 15 = log3(2𝑥 − 3)
By comparison,
2𝑥 − 3 = 15
𝑥 = 9
2. (i) log9(2𝑥 + 12) = log3(𝑥 + 2)
log3(2𝑥+12)
log3 9
= log3(𝑥 + 2)
log3(2𝑥+12)
log3 32 = log3(𝑥 + 2)
log3(2𝑥+12)
2 log3 3
= log3(𝑥 + 2)
log3(2𝑥 + 12) = 2 log3(𝑥 + 2)
log3(2𝑥 + 12) = log3(𝑥 + 2)2
By comparison,
2𝑥 + 12 = (𝑥 + 2)2
2𝑥 + 12 = 𝑥2
+ 4𝑥 + 4
𝑥2
+ 2𝑥 − 8 = 0
(𝑥 + 4)(𝑥 − 2) = 0
𝑥 = −4 𝑜𝑟 2
(ii)
2
log7 2
= log2(4𝑥 − 1)
2
(
1
log2 7
)
= log2(4𝑥 − 1)
2 log2 7 = log2(4𝑥 − 1)
log2 72
= log2(4𝑥 − 1)
By comparison,
4𝑥 − 1 = 49
𝑥 = 12.5
Apply the law of logarithm by converting
logarithmic expression to indices form.
Changing the base of logarithms to make
both sides have the same base.

107
3. 2 log3(𝑥 − 𝑦) = 1 + log3 𝑥 + log3 𝑦
log3(𝑥 − 𝑦)2
= 1 + log3 𝑥 + log3 𝑦
= log3 3 + log3 𝑥 + log3 𝑦
= log3 3𝑥𝑦
By comparison,
(𝑥 − 𝑦)2
= 3𝑥𝑦
𝑥2
− 2𝑥𝑦 + 𝑦2
= 3𝑥𝑦
𝑥2
+ 𝑦2
= 5𝑥𝑦 (shown)
4. (i) log2(𝑥 + 2) + 5 log4 𝑥 − 3 log2 √ 𝑥
= log2(𝑥 + 2) +
5 log2 𝑥
log2 4
− 3 log2 √ 𝑥
= log2(𝑥 + 2) +
5
2
log2 𝑥 − log2 𝑥
3
2
= log2(𝑥 + 2) + log2 𝑥
5
2 − log2 𝑥
3
2
= log2 [
(𝑥+2)𝑥
5
2
𝑥
3
2
]
= log2[(𝑥 + 2)𝑥]
(ii) log2(𝑥 + 2) + 5 log4 𝑥 − 3 log2 √ 𝑥 = 3
log2[(𝑥 + 2)𝑥] = 3
𝑥(𝑥 + 2) = 23
𝑥2
+ 2𝑥 − 8 = 0
(𝑥 + 4)(𝑥 − 2) = 0
𝑥 = −4 𝑜𝑟 2
5. log2 3 × log3 4 × log4 5 × log5 6 =
log 3
log 2
×
log 4
log 3
×
log 5
log 4
×
log 6
log 5
=
log 6
log 2
= 2.5850
Apply the law of logarithms
to simplify the expressions.
Notice the pattern of the expression
where terms can be cancelled out
after changing their bases.

108
1. Suppose AMON be a rectangle with triangle ABC inscribed in it as follows:
Let the radius of circle A, B, and C be 𝑟𝐴, 𝑟𝐵, and 𝑟𝐶 respectively, we have:
(𝑟𝐴 + 𝑟𝐶)2
= 𝐴𝑁2
+ 𝑁𝐶2
(𝑟𝐶 + 12)2
= 𝐴𝑁2
+ 𝑁𝐶2
(𝑟𝐵 + 𝑟𝐶)2
= 𝐵𝑂2
+ 𝐶𝑂2
(𝑟𝐶 + 6)2
= 𝐵𝑂2
+ 𝐶𝑂2
(𝑟𝐴 + 𝑟𝐵)2
= 𝐴𝑀2
+ 𝑟𝐵
2
(12 + 6)2
= 𝐴𝑀2
+ 62
𝐴𝑀2
= 288
We know that 𝐴𝑁 = 𝑟𝐴 − 𝑟𝐶 = 12 − 𝑟𝐶, 𝐵𝑂 = 𝑟𝐵 − 𝑟𝐶 = 6 − 𝑟𝐶, and 𝐴𝑀 = 𝑁𝐶 + 𝐶𝑂
From equation 1,
𝑁𝐶2
= (𝑟𝐶 + 12)2
− 𝐴𝑁2
𝑁𝐶2
= 𝑟𝐶
2
+ 24𝑟𝐶 + 144 − (144 − 24𝑟𝐶 + 𝑟𝐶
2)
𝑁𝐶2
= 48𝑟𝐶
From equation 2,
𝐶𝑂2
= (𝑟𝐶 + 6)2
− 𝐵𝑂2
𝐶𝑂2
= 𝑟𝐶
2
+ 12𝑟𝐶 + 36 − (36 − 12𝑟𝐶 + 𝑟𝐶
2)
𝐶𝑂2
= 24𝑟𝑐
Therefore from 𝐴𝑀 = 𝑁𝐶 + 𝐶𝑂, we have
√288 = √48𝑟𝐶 + √24𝑟𝐶
√24√12 = √24√2𝑟𝐶 + √24√𝑟𝐶
√12 = √ 𝑟𝐶(√2 + 1)
12 = 𝑟𝐶(3 + 2√2)
𝑟𝐶 =
12
3+2√2
= 2.059𝑐𝑚
4.4 Application of Indices, Surds and Logarithms

109
2. (i) 𝑡 = 5, Let the value of the car after 5 years be 𝑥, we have
𝑥 = 𝑅𝑀80000 (
8
9
)
5
= 𝑅𝑀44394.32
(ii) 𝑅𝑀80000 (
8
9
)
𝑡
< 𝑅𝑀35000
(
8
9
)
𝑡
<
35000
80000
log (
8
9
)
𝑡
< log
7
16
𝑡 log
8
9
< log
7
16
𝑡 >
log
7
16
log
8
9
𝑡 > 7.019
8 years will be needed for the car to drop below RM35000 for the first time.

1) a = 1038
= 964
l = 594
= a + (3 1)d
964 = 1038 + 2d
d =
‫ﮠ‬ਤ
= 37
= a + (n 1)d
594 = 1038 + (n 1)( 37)
594 = 1075 - 37n
n =
ਤ
= 13
Since the last term is the term, the arithmetic progression has 13 terms.
2) Money Harry saves every day follows the arithmetic sequence,
12, 15, 18, …
a = 12
d = 15 12
= 3
Total amount of money saved till the tenth day, = [2(12) + (10 1)3]
= 5(51)
= RM255
x = 605 255
= 350
3) The middle two terms are the term and the term.
+ = 268
a + (8 1)d + a + (9 1)d = 268
2a + 15d = 268 ○1
= 101
a + (3 1)d = 101
a + 2d = 101 ○2
From ○2 ,
a = 101 2d ○3
5.1 Arithmetic Progression

Substitute ○3 into ○1 ,
2(101 2d) + 15d = 268
202 + 11d = 268
d =
‫ﮠ‬
= 6
Substitute d = 6 into ○3 ,
a = 101 2(6)
= 89
The last term, ‫ﮠ‬ = a + (16 1)d
= 89 + (15)6
= 179
4) The radius of the circles follows the arithmetic sequence,
3, 6, 9, ….
a = 3
d = 6 3
= 3
Sum of the seven terms, = [2(3) + (7 1)3]
= (24)
= 84
Length of the rectangle = 84 x 2
= 168 cm
Radius of the seventh circle, = 3 + (7 1)3
= 21 cm
Height of the rectangle = 21 x 2
= 42 cm
Area of the rectangle = 168 cm x 42 cm
= 7056 cm
5) =
‫ﮠ‬
[2a + (5 1)d] =
‫ﮠ‬
2a + 4d =
‫ﮠ‬
x
2a + 4d =
‫ﮠ‬ਤ
○1
=
‫ﮠ‬
[2a + (8 1)d] =
‫ﮠ‬
2a + 7d =
‫ﮠ‬
x
2a + 7d =
‫ﮠ‬
○2

○2 ○1 ,
7d 4d =
‫ﮠ‬ ‫ﮠ‬ਤ
3d = 13
d =
Substitute d = into ○1 ,
2a + 4( ) =
‫ﮠ‬ਤ
2a =
‫ﮠ‬ਤ
( )
a =
‫ﮠ‬
= 1080
= 1080 + (10 1)( )
= 1080 + ( 39)
= 1041
6) Given the following arithmetic sequence,
78, 82, 86, …
a = 78
d = 82 78
= 4
(a) Sum of the 10 terms after the term =
= [2(78) + (18 1)4] [2(78) + (8 1)4]
= 9(224) 4(184)
= 2016 736
= 1280
(b) > 200
78 + (n 1)4 > 200
74 + 4n > 200
4n > 126
n >
‫ﮠ‬
ਤ
n > 31.5
⸫ The smallest term which is more than 200 for this sequence is the term.
= 78 + (32 1)4
= 78 + 124
= 202
(c) First, we need to know the number of terms for the sequence.

= 158
78 + (n 1)4 = 158
74 + 4n = 158
4n = 84
n =
ਤ
ਤ
= 21
Since 158 is the last second term, the sequence has a total of (21 + 1) terms.
Sum of the last five terms =
= [78 + (158 + 4)] [2(78) + (17 1)4]
= 11(240) (220)
= 2640 1870
= 770
1) Given the geometric sequence.
20, 40, 80, …
a = 20
r =
ਤ
= 2
(a) = 20
= 10240
(b) < 10000
20 < 10000
< 500
ln < ln 500
(n - 1) ln 2 < ln 500
n 1 <
ln
ln
n < 8.97 + 1
n < 9.97
⸫ The last term of the sequence will be the term.
= 20
= 5120
(c) The sum of the term from term to term = a
The last term

=
o h
20
= 2520
2) The saving amount follows a geometric sequence,
5000, 1.03 x 5000, 1.03 x 1.03 x 5000, …
a = 5000
r = 1.03
> 6000
5000 > 6000
> 1.2
ln > ln 1.2
(n 1) ln 1.03 > ln 1.2
n >
ln ln
ln
n > 7.168
Since starting from the eighth term only the saving amount is more than RM6000, the minimum
years required for the fixed deposit is 7 years.
3) Area of the squared paper = 22 x 22
= 484 cm
When the paper is folded in half again and again, its area decreases in a geometric sequence,
484, 0.5 x 484, 0.5 x 0.5 x 484, …
a = 484
r = 0.5
Since the paper is folded in half for six times, we have to find the value of
= 484o h
= 7.5625
⸫ The area of the folded paper is 7.5625 cm .
4) a = 2000
‫ﮠ‬ = 1180.98
n = 6
‫ﮠ‬ = 1180.98
2000o h‫ﮠ‬
= 1180.98
= 0.59049
ln = ln 0.59049
5 ln r = ln 0.59049

= 5.3 +
= 5.3 +
= 5.3 +
ਤ
ਤ
= +
ਤ
ਤ
=
ਤ
+
=
ਤ
=
ln r = 0.10536
r = ‫ﮠ‬
= 0.9
Sum of the first four terms =
o ਤh
= 6878
5) The height of the ball follows the geometric progression,
21, 0.875 x 21, 0.875 x 0.875 x 21, …
a = 21
r = 0.875
=
= 168
Total distance travelled = 24 + 2 x168
= 24 + 336
= 360 cm
6) The radius and height of the cylinders follows geometric progression, G1 and G2 respectively.
G1: 1, 2 x 1, 2 x 2 x 1, …
G2: 5, 1.4 x 5, 1.4 x 1.4 x 5, …
For G1,
= 1
= 64
For G2,
= 5 ਤ
= 37.65
Volume of the last cylinder = π h
= πo‫ﮠ‬ਤh (37.65)
= 154214.4π cm
7) 5.3888… = 5.3 + (0.08 + 0.008 + 0.0008 + …)

g REFERENCES
Elsaket, N. (2016, Mac 20). Real-life application of arithmetic and
geometric sequence. Retrieved from
https://prezi.com/jw4lnrb6waos/real-life-application-of-arithmetic-
and-geometric-sequence/
Geometric Sequences and Series. Retrieved from
https://courses.lumenlearning.com/boundless-
algebra/chapter/geometric-sequences-and-series/
Geometric Series and Three Applications. Retrieved from
http://www.math.niu.edu/~richard/Math302/ch1sec5.pdf
Wong, M. K., Zaini bin Musa, Azizah binti Kamar, Saripah binti Ahmad,
Nurbaiti binti Ahmad Zaki, & Burham@Borhan, Z. H. b. (2019).
KSSM Additional Mathematics Form 4 Textbook. Johor Bahru:
Ministry of Education Malaysia
Wong, P. W., & Wong, S. M. (2015). Success Additional Mathematics SPM.
Selangor: Oxford Fajar Sdn.Bhd.

KSSM Form 4 Additional Mathematics Notes (Chapter 1-5)

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