Pra U STPM 2022 Penggal 1 - Mathematics (T)

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PRE-U PELANGI BESTSELLER
STPM Text PRE-U STPM Text Mathematics (T) CC039332a PRE-U

Term 1 specially designed for
students who are sitting for the STPM


notes and practices are based on
Mathematics (T) examination. The comprehensive
the latest syllabus and exam format
set by Majlis Peperiksaan Malaysia. PRE-U STPM
This book will provide you with the Text
necessary skills and strategies to
excel in the subject.

TERM 1 Our Pre-U & STPM Titles: STPM Text Mathematics (T)

› Success with MUET
› MUET My Way
FEATURES › Pengajian Am Penggal 1, 2, & 3

› Bahasa Melayu Penggal 1, 2, & 3
■ Comprehensive Notes and Practices › Biology Term 1, 2, & 3 Lee Yoon Woh
› Physics Term 1, 2, & 3
■ Useful Features like Bilingual Tan Guan Hin 1
Keywords, Learning Outcomes and › Chemistry Term 1, 2, & 3 Tey Kim Soon
Worked Examples › Mathematics (T) Term 1, 2, & 3
› Sejarah Penggal 1, 2, & 3 TERM
■ Summary › Geografi Penggal 1, 2, & 3 Mathematics (T)
› Ekonomi Penggal 1, 2, & 3
■ STPM Practices › Pengajian Perniagaan Penggal 1, 2, & 3

■ STPM Model Paper Term 1


■ Complete Answers


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CVR_PreU_STPM_2021 Mathematics T (Penggal 1).indd 1-3
CVR_PreU_STPM_2021 Mathematics T (Penggal 1).indd 1-3 27/07/2021 8:17 AM

CONTENTS







Chapter
1 FUNCTIONS 1
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
1.1 Functions 2
1.2 Polynomial and Rational Functions 21
1.3 Exponential and Logarithmic Functions 43
1.4 Trigonometric Functions 56

Chapter
2 SEQUENCES AND SERIES 92
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
2.1 Sequences 93
2.2 Series 99
2.3 Binomial Expansions 119


Chapter
3 MATRICES 136
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

3.1 Matrices 137
3.2 Systems of Linear Equations 160


Chapter
4 COMPLEX NUMBERS 172
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
4.1 Complex Numbers 173


Chapter
5 ANALYTIC GEOMETRY 192
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
5.1 Analytic Geometry 193

Chapter
6 VECTORS 225
• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •
6.1 Vectors in Two and Three Dimensions 226
6.2 Vector Geometry 246

STPM Model Paper (954/1) 266

Answers 268




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Mathematics Term 1 STPM

Mathematical Notation







Symbols sin, cos, tan, trigonometric functions
= is equal to csc, sec, cot
≠ is not equal to sin , cos , tan ,
–1
–1
–1
 is identical to or is congruent to csc , sec , cot –1 inverse trigonometric functions
–1
–1
≈ is approximately equal to
, is less than
< is less than or equal to Matrices
. is greater than A a matrix A
> is greater than or equal to 0 null matrix
∞ infinity I identity matrix
\ therefore A T transpose of matrix A
A inverse of a non-singular square matrix A
–1
Set notation det A determinant of a square matrix A
 is an element of
 is not an element of Complex numbers
∅ empty set i square root of –1
{x | …} set of x such that … z a complex number z
N set of natural numbers, {0, 1, 2, 3, …} |z| modulus z
Z set of integers
+
Z set of positive integers arg z argument of z
Q set of rational numbers z* complex conjugate of z
R set of real numbers
[ a, b ] closed interval {x | x  R, a < x < b} Geometry
( a, b ) open interval {x | x  R, a , x , b} AB length of the line segment with end points
[ a, b ) interval {x | x  R, a < x , b} A and B
( a, b ] interval {x | x  R, a , x < b} ∠BAC angle between the line segments AB and AC
 union ∆ABC triangle whose vertices are A, B and C
 intersection
// is parallel to
 is perpendicular to
Functions
f a function f
f(x) value of a function f at x Vectors
f : A → B f is a function under which each a a vector a
element of set A has an image in set B |a| magnitude of a vector a
f : x ↦ y f is a function which maps the element i, j, k unit vectors in the directions of the Cartesian
x to the element y coordinates axes
–1
f inverse function of f →
f g composite function of f and g which AB vector represented in magnitude and
°
direction by the directed line segment
is defined by f g(x) = f[g(x)]
°
x
e exponential function of x → from point A to point B
→
log x logarithm to base a of x |AB| magnitude of AB
a
ln x natural logarithm of x, log x a · b scalar product of vectors a and b
e
a × b vector product of vectors a and b
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Mathematics Term 1 STPM Chapter 1 Functions

CHAPTER
1 FUNCTIONS 1












Subtopic Learning Outcome

1.1 Functions (a) State the domain and range of a function, and find composite functions.
(b) Determine whether a function is one-to-one, and find the inverse of a one-to-one
function.
(c) Sketch the graphs of simple functions, including piecewise-defined functions.
1.2 Polynomial and (a) Use the factor theorem and the remainder theorem.
rational functions (b) Solve polynomial and rational equations and inequalities.
(c) Solve equations and inequalities involving modulus signs in simple cases.
(d) Decompose a rational expression into partial fractions in cases where the
denominator has two distinct linear factors, or a linear factor and a prime
quadratic factor.
1.3 Exponential (a) Relate exponential and logirithmic functions, algebraically and graphically.
and logarithmic (b) Use the properties of exponents and logarithms.
functions (c) Solve equations and inequalities involving exponential or logarithmic
expressions.

1.4 Trigonometric (a) Relate the periodicity and symmetries of the sine, cosine and tangent functions
functions to their graphs, and identify the inverse sine, inverse cosine and inverse tangent
functions and their graphs.
(b) Use basic trigonometric identities and the formulae for sin (A ± B), cos (A ± B)
and tan (A ± B), including sin 2A, cos 2A and tan 2A.
(c) Express a sin θ + b cos θ in the forms of r sin (θ ± ∝) and r cos (θ ± ∝).
(d) Find the solutions, within specified intervals, of trigonometric equations and
inequalities.




Bilingual Keywords

composite – gubahan logarithmic function – fungsi logaritma
domain – domain partial fraction – pecahan separa
equation – persamaan polynomial function – fungsi polinomial
exponential function – fungsi eksponen range – julat
factor theorem – teorem faktor rational function – fungsi nisbah
function – fungsi remainder theorem – teorem baki
inequality – ketaksamaan trigonometric function – fungsi trigonometri
inverse – songsangan














01a STPM Math T T1.indd 1 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

1.1 Functions


1
Functions
Suppose we have set A = {1, 4, 9, 16, 25} and set B = {1, 2, 3, 4, 5}. Each element in set B is the square root
of a corresponding element in set A. Figure 1.1 shows the relation ‘is the square root of’ between the elements
in set A and set B.
A 'is the square root of' B
1 1
4 2
9 3
16 4
25 5
Figure 1.1
The relation between set A and set B can also be represented by the set of ordered pairs as follows:
{(1, 1), (4, 2), (9, 3), (16, 4), (25, 5)}
In the relation between set A and set B, each element in set A is connected to a unique element in set B. A
relation such as this is known as a one-to-one relation.

Now, suppose set P = {–2, –1, 1, 2, 3,} and set Q = {1, 4, 9, 16}. The relation between set P and set Q is as
shown in the following diagram:
P 'is the square of' Q
–2
1
–1
4
1
9
2
16
3
Figure 1.2
Notice that in the relation between set P and set Q, the elements –2 and 2 of set P are connected to the element
4 of set Q, and the elements –1 and 1 of set P are connected to the element 1 in set Q. A relation such as this
is known as a many-to-one relation.
A relation whereby each element in a set X is connected to one and only one element in a set Y is known as
a function. Hence, one-to-one relations and many-to-one relations are functions. If a relation is one-to-one,
then the function is one-to-one.
A function f from a set X to a set Y is defined as a rule that associates exactly one element of Y with each
element of X. We say, f maps X into Y and we write f : X → Y.
If x  X and y  Y such that y is assigned to x under f, we say f maps x to y and we write f : x ↦ y or y =
f(x), where y is the image of x.

In general, a function f is defined for certain values of x only. This set of values of x for which f is defined is
called the domain of f. The set of values of f(x) for a given domain is called the range of f.
In the previous example, let g be the function “is the square of’ from set P to set Q.
We write g : x ↦ x or g(x) = x .
2
2
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01a STPM Math T T1.indd 2 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

Set P is the domain and set Q is the codomain. The set of images {1, 4, 9} is the range of this function.

P 'is the square of' Q
1
–2
–1 1
1 4
2 9
3 16 Range
Domain Codomain
Figure 1.3


Example 1


X Y
–2
0
–1
1
0
2
1
5
2
f
(a) Show that the diagram above defines a function f from set X to set Y.
(b) Find f(x).
(c) State the domain of f and its range.
Solution: (a) For each element x in set X, there exists a unique image y in set Y. Hence, f
can be defined as a function from set X to set Y.
(b) f(0) = 1 = 0 + 1
2
f(–1) = f(1) = 2 = 1 + 1
2
f(–2) = f(2) = 5 = 2 + 1
2
Hence, f(x) = x + 1.
2
(c) Domain of f is {–2, –1, 0, 1, 2}
Range of f is {1, 2, 5}


Example 2

Determine whether each equation defines y as a function of x. If so, determine whether the function is
one-to-one.
(a) y = x + 2,
2
(b) y = x + 3x + 2,
(c) y = x – 2.
2
Solution: (a) y = x + 2
For each value of x  R, the value of y is unique.
Thus y is a function of x.
No two values of x  R have the same image.
Thus the function is one-to-one.



3





01a STPM Math T T1.indd 3 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions


2
(b) y = x + 3x + 2
For each value of x  R, the value of y is unique.
1 Thus y is a function of x.
When x = 0, y = 0 + 0 + 2 = 2.
When x = –3, y = 9 – 9 + 2 = 2.
There exist two values of x  R with the same image.
Thus the function is not one-to-one.
2
(c) y = x – 2
y = ± x – 2
For each value of x  R with x . 2, there exist two possible values of y.
For example, y = ±2 when x = 6.
Thus y is not a function of x.




Algebraic operations on functions

Algebraic operations can be applied to functions.
If f : x ↦ f(x) and g : x ↦ g(x), then
(i) f + g : x ↦ f(x) + g(x),
(ii) f – g : x ↦ f(x) – g(x),
(iii) f · g : x ↦ f(x) · g(x),
(iv) f : x ↦ f(x) , g(x) ≠ 0,
g g(x)
(v) af : x ↦ af(x), a  R.

2
For example, let f(x) = x + 1 and g(x) = x , where the domain is the set of real numbers, R.
2
Then, f + g : x ↦ (x + 1) + x , x  R
2
f – g : x ↦ (x + 1) – x , x  R
2
f · g : x ↦ (x + 1)x , x  R
f : x ↦ x + 1 , x ≠ 0
g x 2
2f : x ↦ 2(x + 1), x  R






Composite functions
Suppose that the functions f and g are defined for the set of real numbers R, with
f : x ↦ y and g : y ↦ z
This means that we can create a new function which maps x directly to z. This function is called a composite
function and is written as g f, where g f(x) = g[f(x)].
º
°
Consider the diagram below, which shows the mapping of f, g and g f.
°


4





01a STPM Math T T1.indd 4 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions


y

f 1
g


x z
g f
°
Figure 1.4
For example, if
f : x ↦ x – 2 and g : x ↦ 3x,
i.e. f(x) = x – 2 and g(x) = 3x,
then g[f(x)] = 3f(x)
= 3(x – 2)
Hence, g f : x ↦ 3(x – 2)
°
For example, if x = 5,
f(5) = 5 – 2 = 3
g(3) = 3(3) = 9
i.e. g[f(5)] = g(3) = 9


3

f
g

5 9
g f
°
Figure 1.5

Example 3

2
If f(x) = x and g(x) = x + 1, find
(a) f g(x) (b) g f(x)
°
°
(c) f f(x) (d) g g(x)
°
°
Solution: (a) f g(x) = f[g(x)] (b) g f(x) = g[f(x)]
°
°
2
= f(x + 1) = g(x )
= (x + 1) 2 = x + 1
2
(c) f f(x) = f[f(x)] (d) g g(x) = g[g(x)]
°
°
= f(x ) = g(x + 1)
2
= (x ) = (x + 1) + 1
2 2
= x 4 = x + 2
Note: 1. f g ≠ g f, from (a) and (b).
°
°
2. f[g(x)] ≠ f(x) · g(x) since f[g(x)] = (x + 1) whereas f(x) · g(x) = x (x + 1).
2
2
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01a STPM Math T T1.indd 5 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

Example 4


1 Given that the function f : x ↦ 2x + 1, find the function g if
(a) the composite function f g is f g : x ↦ 6x + 11,
°
°
1
(b) the composite function g f is g f : x ↦ x – 1 , x ≠ 1.
°
°
Solution: (a) Given that f(x) = 2x + 1.
If f g(x) = 6x + 11.
°
f[g(x)] = 6x + 11.
∴ 2g(x) + 1 = 6x + 11
2g(x) = 6x + 10
g(x) = 3x + 5
Hence, g : x ↦ 3x + 5

1
(b) If g f(x) = x – 1 .
°
g[f(x)] = 1 .
x – 1
∴ g(2x + 1) = 1
x – 1
Let u = 2x + 1
then x = 1 (u – 1)
2
Hence, g(u) = 1
1 (u – 1) – 1
2
= 2
u – 3

i.e. g : u ↦ 2 , u ≠ 3 or g : x ↦ 2 , x ≠ 3.
u – 3 x – 3




Exercise 1.1


1. If f(x) = x – 9, x  R, find
2
(a) f(0), (b) f(4), (c) f(3), (d) f(–3).
2. Given that f(x) = x + 1 , x  –1,
(a) find the value of
(i) f(8), (ii) f(0), (iii) f(–1),
(b) find the value of x such that
(i) f(x) = 5, (ii) f(x) = 10.
3. By drawing a suitable sketch graph, find the range of each of the following functions.
(a) f(x) = 2x, x  [–1, 3],
(b) f(x) = 2 – x, x  [–1, 3],
(c) f(x) = x – x – 2, x  [–2, 3].
2


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01a STPM Math T T1.indd 6 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

4. Let X = {1, 2, 3, 4,}. Show whether each of the following relations, which define the set of ordered pairs,
represent a function from X to X.
(a) f : {(2, 3), (1, 4), (2, 1), (3, 2), (4, 4)}
(b) g : {(3, 1), (4, 2), (1, 1)} 1
(c) h : {(2, 1), (3, 4), (1, 4), (2, 3), (4, 4)}
x
5. The domain of a function h, where h : x ↦ 2 , is R.
(a) Find the images of 0, 2, –1 and 6.
(b) Find the element in the domain with image 16.
(c) Find the range of the function h.

6. If f(x) = 2x – 3 and f g(x) = 2x + 1, find g(x).
°
2
7. If f(x) = x – 1 and g f(x)= 3 + 2x – x , find g(x).
°
8. If f(x) = cos x and g(x) = 1 + x, x  R, find
(a) f g(x) (b) f f(x) (c) g f(x)
°
°
°

2
9. If f(x) =  x , x  0 and g(x) = 1 – x , x  R, find
(a) f g(x) (b) g f(x) (c) g g(x)
°
°
°
10. If f : R → R and g : R → R are defined by f(x) = x + 3x + 1 and g(x) = 2x – 3, find
2
(a) f g (b) g f (c) g g (d) f f
°
°
°
°
2
11. If f : R → R and g : R → R are defined by f(x) = 2x – 3 and g(x) = x + 5, find
(a) g f(2) (b) f g(3) (c) f g(a – 1)
°
°
°
(d) g f(x) (e) f g(x + 1) (f) g g(x)
°
°
°
Inverse functions
Let f and g be two functions defined respectively by
1
f : x ↦ 2x + 1 and g : x ↦ (x – 1)
2
1
i.e. f(x) = 2x + 1 g(x) = — (x – 1)
123
2
1423
Object of f Image of f Object of g Image of g
Then the composite function g f is defined as
°
g f(x) = g[f(x)]
°
= g(2x + 1)
= 1 [(2x + 1) – 1]
2
= x
i.e. we get back the object for the function f, i.e. x.
Similarly, f g(x) = f[g(x)]
°
1
= f[ (x – 1)]
2
1
= 2[ (x – 1)] + 1
2
= x
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01a STPM Math T T1.indd 7 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

Hence, g is the inverse function of f and is written as f . The inverse of a function f can be obtained by
–1
inverting the direction of the arrow in the diagram, such as the one shown below.
1 f

a b

f –1
Domain of f Range of f
–1
–1
(Range of f ) (Domain of f )
Figure 1.6
For the function f : x ↦ 2x + 1, the inverse of f is f : x ↦ 1 (x – 1).
–1
2
1
x f 2x +1 f –1 – (x–1)
x 2
–1 –1 –1 –1
0 1 1 0
1 3 3 1

2 5 5 2
A B B A
Figure 1.7(a) Figure 1.7(b)
The diagram in Figure 1.7(a) shows the function f : x ↦ 2x + 1. The diagram in Figure 1.7(b) shows the inverse
of the function f, which maps each element in the set B to only one element in the set A. This relation is a
one-to-one relation. Hence, the inverse of the function f is also a function, and is called the inverse function
of f.
Now consider the function f below,
2
i.e. f : x ↦ x , x  R
–1
–1
2
The inverse of the function f is f : x ↦ x or f : x ↦ ± x .
x f x 2 x f –1 + – x
–2 4 4 –2
–1 1 –1
1 1 1
2 2
P Q Q P
Figure 1.8(a) Figure 1.8(b)
2
The diagram in Figure 1.8(a) shows the function f : x ↦ x . In the inverse of function f, as shown in Figure
1.8(b), the elements 4 and 1 in the set Q are each mapped to two elements in the set P. Each of the elements
4 and 1 does not have a unique image. Hence, the inverse of the function f is not a function.
Now consider the function f again, where
2
f : x ↦ x , x  R +
The inverse of this function is
–1
f : x ↦  x (positive square root only)

+
2
Hence, the inverse of the function f is also a function, i.e. if f : x ↦ x , x  R 0 , then the inverse function of
+
–1
f is f : x ↦  x, x  R 0 .
8




01a STPM Math T T1.indd 8 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

From the above examples, we make the following conclusions:
(a) The inverse of a function is not necessarily also a function.
(b) The inverse of a given function is also a function if the given function is a one-to-one function.
(c) The inverse of a many-to-one function can be a function if we restrict the domain. 1


Example 5

Find the inverse of each of the following functions, indicating its domain.
(a) f : x ↦ 3x + 5, x  R (b) f : x ↦ 2 , x ≠ 3
x – 3
Solution: (a) f(x) = 3x + 5
–1
Let f (x) = a
∴ f(a) = x
3a + 5 = x
a = x – 5
3
x – 5
–1
Hence f (x) = , x  R
3
1
–1
or f : x ↦ (x – 5), x  R
3
(b) f(x) = 2 , x ≠ 3.
x – 3
–1
Let f (x) = b
∴ f(b) = x
2 = x
b – 3
b = 2 + 3x
x
Hence, f (x) = 2 + 3x , x ≠ 0
–1
x
2 + 3x
–1
or f : x ↦ , x ≠ 0.
x


Exercise 1.2

1.
X Y
1
a
2 b
3
c
4
d
5

(a) Show that the elements in sets X and Y define a function f from set X to set Y.
(b) Find f(1) and f(3).
(c) State the domain of f.
(d) State the range of f.
(e) Is f a one-to-one function?


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01a STPM Math T T1.indd 9 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

2
2. Given that g : x ↦ x + x + 1 and x  X, where X = {–2, –1, 0, 1, 2}.
(a) Sketch a diagram to represent g.
(b) State the domain of g.
1 (c) State the range Y of g.
(d) Is g a one-to-one function?
3. Given that h : x ↦ 2x – 3, where x  {1, 3, 5, 7, 9} and that the range is {–5, –3, –1, 3, 5, 7, 11, 15}.
(a) Sketch a diagram to represent h.
(b) Find h(1) and h(9).
(c) Is h a one-to-one function?

2
4. The function f is defined by f : x ↦ x – x, x  R. State whether f is one-to-one, giving a reason for the
statement. If the domain of f is restricted to the subset of R for which x  k, find the least value of k for
which f is one-to-one.
5. The function f is defined by
f : x ↦ x , x  R
2
x + 1
1
If a  R and a ≠ 0, find the image of — under f.

a
Deduce that f is not one-to-one.
Show that if a, b  R with a . b  1, then f(b) . f(a).
Deduce that, if the domain of f is restricted to the subset of R given by {x : x  1}, then f is one-to-one.
State the range of f in this case.

6. Find the inverse of each of the following functions, stating its domain.
(a) f : x ↦ x – 2, x  R
2
(b) f : x ↦ x + 1, x  0
2
(c) f : x ↦ (x – 1) , x  1
(d) g : x ↦ (x – 2)(x – 4), x  3
+
(e) g : x ↦ (x – 3)(x + 3), x  R
(f) h : x ↦ 2 , x ≠ 3
x – 3
(g) h : x ↦ x + 2 , x ≠ 2
x – 2
2
7. Given that f(x) = x + 1 and x  2, find f (x).
–1
2x + 1
2
Prove that (f f)(x) = x.
–1
°
8. The functions f and g are defined for x  R (excluding –1, 0 and 1) by
f : x ↦ 1 + x and g : x ↦ 1
1 – x x
Show that (f g) = f g.
–1
°
°
9. Sketch the graph of each of the following functions and its inverse.
(a) f(x) = x + 3, x  R
(b) f(x) = 2x – 1, x  R
(c) f(x) = (x – 1)(x + 1), x  R
+
2
(d) f(x) = x + 3x – 4, x . – 3
2


10





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Mathematics Term 1 STPM Chapter 1 Functions
Graphs of linear, quadratic and cubic-functions

The simplest algebraic function is the linear function, usually in the form f(x) = ax + b, where a and b are
rational numbers, either positive or negative. The graph of a linear function is a straight line, and a is known 1
as the gradient and b the intercept on the y-axis. The graph of y = ax + b for different values of a and b are
as follows:
Linear function
f(x) = ax + b
y y y y



x x x x
O O O O

a . 0, b . 0 a . 0, b , 0 a , 0, b . 0 a , 0, b , 0

Figure 1.9
The shape of the graphs for a quadratic and a cubic function are as shown below.
Quadratic function Cubic function
2
3
2
f(x) = ax + bx + c f(x) = ax + bx + cx + d
or or


a . 0 a , 0 a . 0 a , 0

Figure 1.10
Notice that the graph of a quadratic function has one stationary point, whereas a cubic function has two
stationary points.

Example 6

Sketch the graphs of the following functions.
2
(a) f(x) = x + 4x + 5, (b) f(x) = –2x + 3x + 1.
2
Solution: (a) f(x) = x + 4x + 5 can be expressed as
2
f(x) = (x + 2) – 4 + 5, by completing the square
2
= (x + 2) + 1
2
The graph cuts the y-axis when x = 0, i.e. y = 5.
As x → `, y → `
Similarly when x → –`, y → `.
When x = –2, y = 1 is the minimum value of f(x).
2
Hence the graph of y = x + 4x + 5 is as shown below.
y
2
y = x + 4x + 5
5



1
x
–2 0

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Mathematics Term 1 STPM Chapter 1 Functions

(b) f(x) = –2x + 3x + 1 can be expressed as
2
2
f(x) = –2(x – 3 x) + 1
1 2
9
1
2
= –2 x – 3 2 – 16  + 1, by completing the square
4
1
2
= –2 x – 3 2 + 9 + 1
8
4
2
1
= –2 x – 3 2 + 17 .
8
4
The graph cuts the y-axis when x = 0, i.e. y = 1.
As x → `, y → –`.
Similarly when x → –`, y → –`.
When x = 3 , y = 17 is the maximum value of f(x).
4 8
2
Hence the graph of y = –2x + 3x + 1 is as shown below.
y
17
–
8
2
y = –2x + 3x + 1
1


x
0 3
–
4


Example 7

Sketch the graphs of
2
3
2
(a) y = x – 2x – 5x + 6 (b) y = x – x – x + 1
3
2
3
Solution: (a) Let y = x – 2x – 5x + 6
= (x + 2)(x – 1)(x – 3)
2
3
The graph of y = x – 2x – 5x + 6 is in the form .
The graph crosses the x-axis when y = 0,
i.e. (x + 2)(x – 1)(x – 3) = 0.
Hence, x + 2 = 0, x – 1 = 0 or x – 3 = 0.
i.e. x = –2, 1 or 3.
The graph crosses the y-axis when x = 0, i.e. y = 6
As x → ∞ , y → ∞.
As x → –∞ , y →–∞.


The maximum point must lie in the interval [–2, 1].
The minimum point lies in the interval [1, 3].




12





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Mathematics Term 1 STPM Chapter 1 Functions

With the information obtained above, the graph of y = x – 2x – 5x + 6 may
3
2
be drawn as shown below.
1
y
3
2
y = x – 2x – 5x + 6
6




x
–2 0 1 3

3
(b) Let y = x – x – x + 1
2
2
= (x + 1)(x – 2x + 1)
= (x + 1)(x – 1) 2
3
2
The graph of y = x – x – x + 1 is in the form .
The graph crosses the x-axis when y = 0,
i.e. (x + 1)(x – 1) = 0
2
2
Hence, x + 1 = 0 or (x – 1) = 0
i.e. x = –1, 1, 1 (repeated)
The graph crosses the y-axis when x = 0, i.e. y = 1.
As x → ∞ , y → ∞.


As x → –∞ , y → –∞.
The value x = 1 (repeated) shows that the curve touches the x-axis at the point
x = 1.
The minimum point is at (1, 0).
The maximum point lies in the interval [–1, 1].

3
The graph of y = x – x – x + 1 is as shown below.
2
y
3
2
y = x – x – x + 1
1


x
–1 0 1





Graphs of power functions


n
(a) If n is a positive integer (n  Z ), the curve y = kx meets the x-axis at y = 0, i.e. x = 0.
n
+
If n is even, the curve touches the x-axis at the origin (0, 0).
The curve has a minimum point if k . 0 and a maximum point if k , 0.


13





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Mathematics Term 1 STPM Chapter 1 Functions

The graph of y = kx (n even) is as follows:
n
k 0 k 0
1
y y
maximum point
x
O




x
O
minimum point
Figure 1.11(a) Figure 1.11(b)

Notice that the graph is symmetrical about the y-axis.

(b) If n is odd, the curve has a point of inflexion at the origin (0, 0). The graph of y = kx (n odd) is as follows:
n

k 0 k 0
y y

point of point of
inflexion inflexion
x x
O O



Figure 1.12(a) Figure 1.12(b)


Example 8


Sketch each of the following curves.
(a) y = 5x 4 (b) y = 3(x – 2) 5

5
Solution: (a) The graph of y = 5x has a (b) The graph of y = 3x has a point of
4
minimum point at the inflexion at the origin. The shape of
5
origin and is symmetrical the curve y = 3(x – 2) is similar of
about the y-axis. to that y = 3x but has a point of
5
inflexion at (2, 0).
y y 5 5
y = 3x y = 3(x – 2)
y = 5x 4

x
0 (2, 0)


x
O



14





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Mathematics Term 1 STPM Chapter 1 Functions
Graphs of rational functions


A rational function is a function in the form f(x) , where f(x) and g(x) are algebraic functions with g(x) ≠ 0.
g(x) 1
1
An example of a rational function is xy = 1, i.e. y = 1 or x = .
x y
Notice that when x → ±∞, y → 0 and when y → ± ∞, x → 0. The lines y = 0 and x = 0 are known as the
asymptotes for the curve xy = 1.


In general, a curve which approaches a straight line when x → ± ∞ or when y → ±∞ will have the straight
line as its asymptote.

y



xy = 1

x
O
asymptotes






Figure 1.13


Example 9

Sketch the curve with equation y = 2x – 1 .
x + 1
State the asymptotes of the curve.

Solution: y = 2x – 1
x + 1
1
x(2 – — )
Rewriting the equation as y = x ,
1
x(1 + — )
x
1
2 – —
x
We get y = .
1
1 + —
x
When x → ± ∞, 1 → 0 and y → 2.
x
Hence, y = 2 is an asymptote to the curve y = 2x – 1 .
x + 1
When x = –1, y = 2x – 1 is undefined.
x + 1
Hence, x = –1 is also an asymptote to the curve.
When x = 0, y = –1.
When y = 0, x = 1 .
2



15





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Mathematics Term 1 STPM Chapter 1 Functions


The graph of y = 2x – 1 is as shown below.
x + 1
1 The asymptotes to the curve are y = 2 and x = –1.
y
asymptote (x = –1)
2x – 1
y = –
x + 1
asymptote (y = 2)

2
x
–1 0 1
–
–1 2




Graphs of functions involving modulus signs

x, x  0,
The absolute value or modulus of x is defined by |x| =
–x, x , 0.
The graphs of y = x and y = |x| are as follows:
y y = x y y = |x |




x x
–3 –2 –1 0 1 2 3 –3 –2 –1 0 1 2 3


Figure 1.14(a) Figure 1.14(b)
For x , 0, the graph of y = |x| is the reflection of the graph y = x about the x-axis, i.e. the graph of y = –x.


Example 10

Sketch the graph of y = x – 3x + 2, noting the points of intersection of the graph with the x-axis.
2
2
On a separate diagram, sketch the graph of y = |x – 3x + 2|.
Solution: y = x – 3x + 2
2
At the point of intersection with the x-axis, y = 0. y
x – 3x + 2 = 0
2
(x – 1)(x – 2) = 0 2 y = x – 3x + 2
2
x = 1 or 2.
At the y-axis, x = 0 ⇒ y = 2. 1
The graph of y = x – 3x + 2 is as shown on x
2
the right. 0 1 2



16





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Mathematics Term 1 STPM Chapter 1 Functions

The graph of y = |x – 3x + 2| can be drawn by reflecting the section of the graph
2
that is below the x-axis, as shown below.
1
y
2
2 y = x – 3x + 2
1

x
0 1 2






Example 11


Sketch the graph of y = |2x – 3|.
By drawing the graph of y = x on the same diagram, determine the values of x such that |2x – 3| . x.
Solution: |2x – 3| = 0 when x = 3 .
2
The graph of y = |2x – 3| is symmetrical about x = 3 .
2
When x = 0, y = 3.
When x = 3, y = 3.
The graph of y = |2x – 3| is as shown below.

y
y = x
3 y = 2x – 3 A




B

x
0 1 3 _ 2 3
2
Notice that the graph of y = x intersects the graph of y = |2x – 3| at two points,
A and B.
At A, 2x – 3 = x
x = 3

At B, –(2x – 3) = x
–2x + 3 = x
3x = 3
x = 1

Hence, from the graph, |2x – 3| . x if x , 1 or x . 3.





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Mathematics Term 1 STPM Chapter 1 Functions
Graphs of piecewise functions

A function can be defined differently for different intervals in a given domain. We say that the function is
1 defined piecewise on a domain, or that the function is a piecewise function.
Consider the function f defined by

x + 1 , –1 < x , 0.
f(x) = x 2 , 0 < x , 1,
2 , 1 < x < 2.

We notice that the graph of f has 3 sections or pieces, as shown below. We say that the function f is piecewise
for the domain [–1, 2].

y
2

Sketch Graphs
1
INFO

x
–1 0 1 2
Figure 1.15
Note: • indicates that this point is included;
° indicates that this point is excluded.




Example 12

Sketch the graph of the function f defined by
2
x , –3 < x , 2.
f(x) = 5 , x = 2,
8 – x , 2 , x < 4.

Solution: For –3 < x , 2, the graph is quadratic.
For x = 2, y = 5.
For 2 , x < 4, the graph is straight line.
Hence, the graph of the function f is as shown below.

y
9

6 y = 8 – x
y = x 2
5 y = 5
4


x
–3 0 2 4


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Mathematics Term 1 STPM Chapter 1 Functions

Example 13


For x  R, [x] is defined as the greatest integer not exceeding x, 1
5
for example, 3 4 = 2, [3] = 3 and [–1.2] = –2.
2
Sketch the graph of f defined by f(x) = [x] for –2 < x , 3.

Solution: From the definition of [x].
–2 for –2 < x , –1,
–1 for –1 < x , 0,
f(x) = 0 for 0 < x , 1,
1 for 1 < x , 2,
2 for 2 < x , 3.

The sketch graph of f is as follows:
y
y = x[ ]
2
1

x
–2 –1 0 1 2 3

–1
–2





Exercise 1.3


1. Sketch the graphs of
(a) y = 4x + 3 (b) y = –3x + 8 (c) 2y = 5x + 6
2. Sketch the graphs of
(a) y = x + 3 (b) y = 3x – 1 (c) y = –4x + 5x + 2
2
2
2
2
3. Sketch the graphs of
1
3
3
3
(a) y = x – 1 (b) y = –2x + 1 (c) y = – x + 5
2
4. Sketch the graph for each of the following curves, indicating the points of intersection of the curve with
the axes.
2
(a) y = x(x + 1)(x – 2) (b) y = (x + 1) (x – 2)
3
(c) y = (x – 2) 3 (d) y = x – 2x
2
3
(e) y = x – 3x + 2x
5. Sketch the graph of
2
2
2
(a) y = x(1 – x ) (b) y = x(1 – x )
19




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Mathematics Term 1 STPM Chapter 1 Functions

6. Sketch the graph for each of the following curves, showing the asymptotes.
(a) y = 2 (b) y = 1 – x (c) y = x 2
1 1 + x 1 + 2x 1 + x
7. Sketch the graph of
(a) y = x (b) y = x
2
1 + x 1 + x
8. Sketch the graph of
(a) y = |2x – 1|
(b) y = |(x – 1)(x – 2)|
(c) y = 3 + |x + 1|
(d) y = |2x + 5| – 4

9. Sketch the graphs of y = |x + 1| and y = |3x – 5| on the same diagram. Find the set of values of x such
that |x + 1| . |3x – 5|.

10. Sketch the graphs of y = |x| and y = |x – 7x + 6| on the same diagram. Determine the set of values of
2
2
x such that |x| , |x – 7x + 6|.
11. The function f is defined on the domain [–4, 4] by f(x) = x + 5, –4 < x < –2; f(x) = x + 2,
2
–2 , x < 2; and f(x) = –x + 7, 2 , x < 4. Sketch the graph of f.

12. The function f is defined on the domain [–2, 5] as follows:
2
–2x + 3, –2 < x < 0,
f(x) = x + 1, 0 , x < 2,
2
–2x + 8, 2 , x < 5.
Sketch the graph of f.


13. The function f is defined by
3
|2x – 1|, – < x , 3 ,
2 2
f(x) = 4, 3 < x < 3,
2
– x 2 + 6, 3 , x < 5.
3
Sketch the graph of f.


14. The function f is defined by
x + 1, –4 < x , –2,
x + 2, –2 < x , 0,
f(x) =
x + 3, 0 < x , 2,
x + 4, 2 < x < 4.
Sketch the graph of f.







20





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Mathematics Term 1 STPM Chapter 1 Functions

1.2 Polynomial and Rational Functions


1
Polynomial functions
Consider the algebraic expression
3
f(x) = 3x + 4x – x + 5
2
This is an expression in terms of x, where the highest power of x is 3. An expression such as this is called a
4
3
function of x of degree 3. Similarly, g(x) = 7x – 5x + 2x + x – 3 is called a function of x of degree 4 as the
2
highest power of x is 4.
In general, a function of x in the form
n
P(x) ≡ a x + a x n – 1 + … + a x n – r + … + a x + a
0 1 r n – 1 n
+
where a  R and a ≠ 0, n  Z is called a polynomial function of degree n.
r
0
Polynomials of degrees 1, 2, 3 and 4 are also known as linear, quadratic, cubic and quartic functions respectively.
The value of a polynomial, P(x), when x = a is written as P(a).
For example,
if P(x) = 3x + 4x – 1
2
then when x = 1, P(1) = 3 + 4 – 1 = 6
and when x = 2, P(2) = 12 + 8 – 1 = 19



Algebraic operations on polynomials
Consider two linear polynomials f(x) and g(x), which are defined respectively as
f(x) = 5x + 4
and g(x) = 3x – 1

Notice that
f(x) + g(x) = (5x + 4) + (3x – 1)
= 5x + 4 + 3x – 1
= 8x + 3

f(x) – g(x) = (5x + 4) – (3x – 1)
= 2x + 5
f(x) · g(x) = (5x + 4)(3x – 1)
= 15x + 12x – 5x – 4
2
= 15x + 7x – 4
2

From the above results, we see that we can add, subtract or multiply two polynomials P(x) and Q(x) to obtain
a new polynomial.

The addition, subtraction and multiplication of two polynomials obey the commutative, associative and distributive
laws respectively.





21





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Mathematics Term 1 STPM Chapter 1 Functions

Example 14


2
2
3
1 If P(x) = 2x + 4x – x + 3 and Q(x) = 3x + x – 1, find
(a) P(x) + Q(x),
(b) P(x) – Q(x),
(c) P(x) · Q(x).
2
2
3
Solution: (a) P(x) + Q(x) = 2x + 4x – x + 3 + 3x + x – 1
3
= 2x + 7x + 2
2
2
3
2
(b) P(x) – Q(x) = 2x + 4x – x + 3 – (3x + x – 1)
2
3
2
= 2x + 4x – x + 3 – 3x – x + 1
3
= 2x + x – 2x + 4
2
2
3
(c) P(x) · Q(x) = (2x + 4x – x + 3) · (3x + x – 1)
2
2
3
2
2
= 2x (3x + x – 1) + 4x (3x + x – 1)
2
– x(3x + x – 1) + 3(3x + x – 1)
2
4
3
2
5
3
4
= 6x + 2x – 2x + 12x + 4x – 4x
– 3x – x + x + 9x + 3x – 3
3
2
2
3
2
= 6x + 14x – x + 4x + 4x – 3 Polynomial of degree 5.
5
4
Note: If P(x) is a polynomial of degree m and Q(x) is a polynomial of degree n, then P(x) · Q(x) is a polynomial
of degree (m + n).
For the division of two polynomials, the long division method may be used, as shown in the following
example.
Example 15
Determine the quotient and remainder when 2x – 7x – 9x + 38 is divided by (x – 3).
2
3
2
Solution: 2x – x – 12
3
2
x – 3 2x – 7x – 9x + 38
3
2x – 6x
2
2
– x – 9x
– x + 3x
2
– 12x + 38
– 12x + 36
2
Using the long division method, the quotient is 2x – x – 12 and remainder 2.
2
From Example 15 above, we know that a polynomial P(x), of degree m, may be divided by another polynomial,
Q(x), of degree n, only if m  n, m, n ∈ Z . If P(x) is divisible by Q(x) exactly, i.e. without any remainder,
+
then the quotient is another polynomial of degree (m – n).
22
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Mathematics Term 1 STPM Chapter 1 Functions

2
4
5
For example, if P(x) = 2x – x – 5x – 2x – 3 and Q(x) = x + x + 1, then P(x) may be found using long
2
division as shown below. Q(x)
2x – 3x + x – 3 1
3
2
3
5
2
3
4
2
x + x + 1 2x – x + 0x – 5x – 2x – 3 Add a term 0x to avoid
confusion during working.
2x + 2x + 2x 3
5
4
–3x – 2x – 5x 2
4
3
–3x – 3x – 3x 2
4
3
x – 2x – 2x
2
3
x + x + x
3
2
2
– 3x – 3x – 3
2
– 3x – 3x – 3
0
Hence, P(x) = 2x – 3x + x – 3, i.e. a polynomial of degree 3.
3
2
Q(x)
If it is known that the division of two polynomials is exact, the quotient may also be obtained by using the
method as shown in Example 16 below.
Example 16
2
3
Find the quotient if x – 4x + 5x – 2 can be divided by (x – 2) exactly.
Solution: Let the quotient be the polynomial q(x).
3
2
x – 4x + 5x – 2
Hence, ––––––––––––––– = q(x)
x – 2
3
2
i.e. x – 4x + 5x – 2 ≡ q(x) · (x – 2) Multiply both sides by (x – 2)
1442443

Polynomial of degree 3 Polynomial of degree 2
Since q(x) is a polynomial of degree 2 (i.e. a quadratic function), it must be of
2
the form ax + bx + c.
2
2
3
Hence, x – 4x + 5x – 2 ≡ (ax + bx + c)(x – 2)
2
2
3
= ax + bx + cx – 2ax – 2bx – 2c
2
3
= ax + (b – 2a)x + (c – 2b)x – 2c
3
Equating coefficients of x : 1 = a
2

Equating coefficients of x : –4 = b – 2a

–4 = b – 2(1)
b = –2
Equating coefficients of x: 5 = c – 2b
5 = c – 2(–2)
c = 1
2
Hence, a = 1, b = –2 and c = 1, and the quotient is x – 2x + 1.
23
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Mathematics Term 1 STPM Chapter 1 Functions

Example 17


1 Find the constants A, B and C, such that
2
x – 5x + 12 ≡ A (x – 1)(x – 2) + B(x + 1)(x – 1) + C(x – 2)(x + 3).
Solution: Since the given equation is an identity, we can substitute suitable values of x into
the identity to determine the values of A, B and C.
When x = 1, 1 – 5(1) + 12 = A(0) + B(0) – 4C

8 = –4C
C = –2
2
When x = 2, 2 – 5(2) + 12 = A(0) + 3B + C(0)
6 = 3B
B = 2
When x = 0, 12 = 2A – B – 6C
12 = 2A – 2 – 6(–2)
12 = 2A – 2 + 12
A = 1
Hence, A = 1, B = 2 and C = –2.


2
Note: We can also find the constants A, B and C in Example 17 by equating the coefficients of x , x and the
constants on both sides of the identity.



Exercise 1.4


1. If P(x) = 2x – 3x + 4x + 1, find the values of
2
3
1
(a) P(0) (b) P 1 2 (c) P(2)
2
2. If Q(x) = 2x – 5x + x – 3, find the values of
4
2
3
(a) Q(–1) (b) Q 1 2 (c) Q(2)
2
3. If F(x) = x + x – 1 and G(x) = 1 + 2x, find
2
(a) F(x) + 2G(x) (b) F(x) – G(x) (c) 3F(x) + x · G(x)
(d) (1 + x) · F(x) (e) G(x) · F(x)
4. By substituting suitable values of x into each of the identities below, find the values of the constants A, B
and C.
(a) 3x + 3 ≡ A(x – 1) + B(2 + x)
(b) 7x + 6 ≡ A(x – 2) + B(x + 3)
(c) 2x + 5 ≡ A(x + 1) + B(x – 2)
(d) 2x – 5x + 7 ≡ A(x + 1)(x – 2) + B(x + 1)(x – 1) + C(x – 2)(x – 1)
2
2
2
(e) x – 6x – 19 ≡ A(x + 5)(x – 1) + B(x – 1) + C(x + 5)(x + 1)
5. Find the product of
2
3
2
(a) 2x – x + 7 and x + 2 (b) 3x – 2x + 5x – 1 and 2x – 3
2
(c) x + 3x – 2 and 4x – x + 1 (d) 5x – 2x + 3 and x – 1
2
3
2
3
4
(e) 3x – 2x + 6x – 4 and 3x + 2
24
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Mathematics Term 1 STPM Chapter 1 Functions

6. Find, using long division, the quotient and remainder for
3
2
(a) (3x – 2x + 5) ÷ (x + 1) (b) (4x – 4x + 5x + 1) ÷ (2x – 3)
2
(c) (x – 3) ÷ (x + 3) (d) (3x – x – 6x + 11x – 1) ÷ (3x – 1)
2
3
4
5
3
4
(e) (x – 2x + 6x – 5) ÷ (x – x – 1) 1
2
3
2
3
2
7. The expression ax – 8x + bx + 6 is divisible by x – 2x – 3. Find the values of a and b.
8. Find the values of A and B such that
2
x + 4x + 4x + 1 ≡ (x + 1)(x + Ax + B)
2
3
9. Find J(x) if
3
x + 3x + x – 2 ≡ (x + 2) · J(x)
2
10. Find numbers a, b and c, such that
2x + 2x + 5x + 3x + 3 ≡ (x + x + 1)(ax + bx + c)
4
2
2
3
2
for all values of x.
The remainder theorem
3
When the polynomial 2x – 7x + 11x – 7 is divided by (x – 2), the quotient is 2x – 3x + 5 with remainder 3.
2
2
We can write
2
3
2x – 7x + 11x – 7 2 3 Remainder
––––––––––––––––– = 2x – 3x + 5 + –––––
x – 2 14243 x – 2
Quotient

2
2
or 2x – 7x + 11x – 7 = (2x – 3x + 5)(x – 2) + 3 Remainder (A constant)
3
14424443 14243123
The dividend Quotient Divisor
(Polynomial (Polynomial (Polynomial
degree 3) degree 2) degree 1)

In general, if a polynomial of degree n, P(x), is divided by (x – a), the quotient, Q(x), is a polynomial of degree
(n – 1), and the remainder, R, is a constant, i.e.
P(x)
R
x – a = Q(x) + x – a
or P(x) ≡ Q(x) · (x – a) + R
By substituting x = a, we see that P(a) = R.
When a polynomial P(x) is divided by (x – a), the remainder is P(a).
Example 18
2
Find the remainder when the polynomial P(x) = 2x + 7x – 5x – 4 is divided by (x + 3).
3
Solution: P(x) = 2x + 7x – 5x – 4
2
3
Since x + 3 is the divisor, choose a value of x such that x + 3 = 0, i.e. x = –3.
3
2
P(–3) = 2(–3) + 7(–3) – 5(–3) – 4
= –54 + 63 + 15 – 4
= 20
Hence, the remainder when the polynomial P(x) is divided by (x + 3) is 20.
25




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Mathematics Term 1 STPM Chapter 1 Functions

Example 19

4
3
2
1 Find the remainder when the polynomial 2x – 5x + x – 7x + 1 is divided by (2x + 1).
4
2
3
Solution: Let P(x) = 2x – 5x + x – 7x + 1
Since 2x + 1 is the divisor, we choose a value of x such that 2x + 1 = 0,
1
i.e. x = – .

2
1
1
1
1
1
P – 1 2 = 2 – 1 2 4 – 5 – 1 2 3 + – 1 2 2 – 7 – 1 2 + 1
2
2
2
2
2
= 1 + 5 + 1 + 7 + 1
8 8 4 2
= 11 or 5 1
2 2
Hence, the remainder when the polynomial 2x – 5x + x – 7x + 1 is divided by
4
2
3
1
(2x + 1) is 5 .
2
1
When a polynomial P(x) is divided by (ax + b), the remainder is P – b 2 .
a
Example 20
4
3
2
The polynomial ax – 5x + bx – 7x + 1 leaves a remainder of –8 when it is divided by (x – 1), and a
remainder of 11 when divided by (2x + 1). Determine the values of a and b.
2
2
3
4
Solution: Let P(x) = ax – 5x + bx – 7x + 1
Hence, P(1) = –8
i.e. a – 5 + b – 7 + 1 = –8
a + b = 3 ………… 
1
Also, P – 1 2 = 11
2
2
1
1
1
1
i.e. a – 1 2 4 – 5 – 1 2 3 + b – 1 2 2 2 – 7 – 1 2 + 1 = 11
2
2
2
2
a + 5 + b + 7 + 1 = 11
16 8 4 2 2
a + 10 + 4b + 56 + 16 = 88
a + 4b = 6 ………… 
 – : 3b = 3
b = 1
Substitute b = 1 into :
a + 1 = 3
a = 2
Example 21
When the polynomial P(x) is divided by (x – 1), its remainder is 5. When P(x) is divided by (x – 2), its
remainder is 7. Given that P(x) may be written in the form (x – 1)(x – 2) · Q(x) + Ax + B, where Q(x) is
a polynomial, A and B are constants, find the remainder when P(x) is divided by (x – 1)(x – 2).
26



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Mathematics Term 1 STPM Chapter 1 Functions


Solution: Given that P(x) ≡ (x – 1)(x – 2) · Q(x) + Ax + B
When P(x) is divided by (x – 1), the remainder is 5.
Hence, P(1) = 5 1
i.e. 0 + A + B = 5
A + B = 5 ………… 
When P(x) is divided by (x – 2), the remainder is 7.
Hence, P(2) = 7
i.e. 0 + 2A + B = 7
2A + B = 7 ………… 
 – : A = 2
Substituting A = 2 into :
2 + B = 5
B = 3
Hence, P(x) ≡ (x – 1)(x – 2) · Q(x) + 2x + 3.
P(x) 2x + 3
and –––––––––––– = Q(x) + ––––––––––––
(x – 1)(x – 2) (x – 1)(x – 2)
Hence, when P(x) is divided by (x – 1)(x – 2), the remainder is (2x + 3).


Note: Example 21 above shows that when a polynomial P(x) is divided by a quadratic expression (x – a)(x – b),
the remainder is a linear function in the form Ax + B, where A and B are constants.


The factor theorem
From the remainder theorem, we have shown that when a polynomial P(x) is divided by (x – a), its remainder,
R, is P(a). On the other hand, if (x – a) is a factor of P(x), then its remainder is zero, i.e. R = P(a) = 0.
For a polynomial P(x), (x – a) is a factor of P(x) if and only if P(a) = 0.


This theorem is very useful in the factorisation of polynomials of degrees higher than 2. By using this theorem,
the linear factors of the polynomial may be obtained.

Example 22


Factorise P(x) = x – 7x – 6.
3
Solution: [Hints to obtain the factors:
3
Notice that the constant term for P(x) = x – 7x – 6 is –6.
Hence, if (x – a) were to be a factor of P(x), then a must be a factor of –6,

i.e. a = ±1, ±2, ±3 or ±6.
Try substituting these values into P(x) such that P(a) = 0.]
If x = 1, P(1) = 1 – 7 – 6
= –12 ≠ 0
Hence, (x – 1) is not a factor of P(x).

3
If x = –1, P(–1) = (–1) – 7(–1) – 6
= –1 + 7 – 6
= 0
Hence, (x + 1) is a factor of P(x).



27





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Mathematics Term 1 STPM Chapter 1 Functions

If x = 2, P(2) = 8 – 14 – 6
= –12 ≠ 0
1 Hence, (x – 2) is not a factor of P(x).
If x = –2, P(–2) = (–2) – 7(–2) – 6
3
= –8 + 14 – 6
= 0
Hence, (x + 2) is a factor of P(x).
If x = 3, P(3) = 3 – 7(3) – 6
3
= 27 – 21 – 6
= 0
Hence, (x – 3) is a factor of P(x).
Since P(x) is of degree 3, it has only three linear factors.
3
P(x) = x – 7x – 6
= (x + 1)(x + 2)(x – 3)


Apart from the above method of using trial and error to obtain all the factors of a polynomial, we can also use
long division method, as shown in Example 23 below.

Example 23


Show that (x + 2) is a factor of f(x) = 6x + 13x – 4.
3
2
Hence, factorise f(x) completely and find the values of x such that f(x) = 0.
3
2
Solution: f(x) = 6x + 13x – 4
2
3
f(–2) = 6(–2) + 13(–2) – 4

= –48 + 52 – 4
= 0
Hence, by factor theorem, (x + 2) is a factor of f(x).
Using long division,
2
6x + x – 2
3
2
x + 2 6x + 13x + 0x – 4
3
2
6x + 12x
x + 0x
2
2
x + 2x
– 2x – 4
– 2x – 4
0
2
f(x) = (x + 2)(6x + x – 2)
= (x + 2)(3x + 2)(2x – 1)
When x = –2, f(–2) = 0
2
1
When x = – , f – 2 2 = 0
3 3
1
When x = 1 , f 1 2 = 0
2 2
2
The values of x such that f(x) = 0 are –2, – and 1 .
3
2
2
Note that –2, – and 1 are called the zeros of f.
3 2
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Mathematics Term 1 STPM Chapter 1 Functions

Example 24


3
2
Given that (x – 2) is a factor of f(x), where f(x) ≡ ax – 10x + bx – 2, a, b  R. 1
When f(x) is divided by (x – 3), its remainder is 16. Find the values of a and b.
2
3
Solution: Given that (x – 2) is a factor of f(x) ≡ ax – 10x + bx – 2.
By the factor theorem,
f(2) = 0
a(2) – 10(2) + b(2) – 2 = 0
3
2
8a – 40 + 2b – 2 = 0
8a + 2b = 42
4a + b = 21 ………… 
Given that when f(x) is divided by (x – 3), the remainder is 16.
By the remainder theorem,
f(3) = 16
a(3) – 10(3) + b(3) – 2 = 16
3
2
27a – 90 + 3b – 2 = 16
27a + 3b = 108 ………… 
9a + b = 36
 – : 5a = 15
a = 3

Substituting a = 3 into ,
12 + b = 21
b = 9
2
Hence, f(x) = 3x – 10x + 9x – 2.
3



Exercise 1.5

1. Use the remainder theorem to find the remainder when each of the following polynomials is divided by
the linear polynomial given.
3
2
3
4
(a) x + 4x – 3x + 2; x – 2 (b) x – 3x + 2x – 1; x + 2
3
(c) x – x + 6; x + 1 (d) x + 2x + 1; 2x – 1
3
5
(e) x – 2x + x + 1; 2x – 3 (f) x + 15x – 1; 3x + 1
3
2
3
2. Determine whether each of the following linear polynomials is a factor of the polynomial given.
(a) x + 2; x + 4x + 4x 2 (b) x – 1; x + 3x – 6x + 3
5
2
3
4
3
3
2
(c) x + 1; x – 2x + 6x + 9 (d) x – 2; x – 4x + 3x + 2
2
3
(e) 2x + 1; 2x – 3x + 2x + 2 (f) 2x – 1; x + 4x – 3x – 1
3
2
2
3. Factorise each of the following polynomials.
3
2
2
(a) 2x – 3x + 1 (b) 3x – 2x – 7x – 2
3
3
2
(c) x – x – 72 (d) x + x + x
4
5
2
(e) 4x – 13x + 6 (f) 4x – 4x – 9x + x + 2
3
4
3
3
2
4. If (x – 2) is a factor of ax + 3x – 2x + a, find the value of a.
5. Show that (x – a) is a factor of x + (1 – a)x + (3 – a)x – 3a.
3
2
29
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Mathematics Term 1 STPM Chapter 1 Functions

2
6. When the polynomial x + ax + b is divided by (x – 1) and (x + 2), its remainder is 4 and 5 respectively.
Find the values of a and b.
1 7. When x + px + qx + 1 is divided by (x – 2), its remainder is 9; when it is divided by (x + 3), its remainder
2
3
is 19. Find the values of p and q.
3
2
8. The expression 2x + ax + b can be divided exactly by (x + 1), and its remainder is 16 when divided by
(x – 3). Find the values of a and b.
2
2
9. The expression ax – 8x + bx + 6 can be divided exactly by x – 2x – 3. Find the values of a and b.
3
3
3
2
2
10. When the polynomials x + 4x – 2x + 1 and x + 3x – x + 7 are divided by (x – p), the remainders are
equal. Find the possible values of p.
2
3
2
3
11. The expressions x – 4x + x + 6 and x – 3x + 2x + k have a common factor. Find the possible values
of k.
2
3
12. If (x – a) is a factor of the expression ax – 3x – 5ax – 9, find the possible values of a. Factorise the
expression for each of the values of a.
2
13. Given that f(x) ≡ x + kx – 2x + 1 has a remainder k when it is divided by (x – k), find the possible
3
values of k.
2
3
14. Find the value of k if (x + 1) is a factor of 2x + 7x + kx – 3. Using this value of k, solve the equation
3
2
2x + 7x + kx – 3 = 0.
2
15. Find the values of a and b if f(x) ≡ ax + bx + 12 can be divided exactly by both (x + 1) and (x – 2)
3
respectively. With these values of a and b, solve the equation f(x) = 0.
2
3
16. The expression x + ax + bx – 8 is divisible by (x + 1). When it is divided by (x – 2), its remainder is
42. Find the values of a and b and the value of the expression when x = 1.
Polynomial and rational inequalities
The relations such as x . 0, 2x – 1 < 0 and 3x + 2x . 1 are known as inequalities.
2
The basic rules governing inequalities involving real numbers are as follows:
For any a, b  R with a . b,
(a) a + c . b + c, c  R,
(b) ac . bc, c  R, c . 0,
(c) ac , bc, c  R, c , 0.

Inequalities involving algebraic polynomials also obey the basic rules of inequalities for real numbers. The set
of numbers which satisfy an inequality is called the solution set. For example, if 2x – 3 < 4x + 9, then the

solution set is {x : x  –6}.
An inequality may be solved according to the type of function given, using the analytical method or graphical
method, or by drawing the real number line.





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Mathematics Term 1 STPM Chapter 1 Functions

Example 25


If x . 0, find the range of values of x which satisfy each of the following inequalities. 1
(a) 5x – 2 , 3 x + 5 (b) 3x  2 + 1
2 8 x
3
Solution: (a) 5x – 2 , x + 5
2 8
5
5x – 3 x , + 2
2 8
7 x , 21
2 8
x , 3
4
Since x . 0, the range of values of x for 5x – 2 , 3 x + 5 is 0 , x , 3 .
2 8 4
(b) 3x  2 + 1
x
3x – 2 – 1  0
x
1
Since x . 0, 3x – 2 – 1 2 · x  0
x
i.e. 3x – 2x – 1  0
2
(3x + 1)(x – 1)  0
Since x . 0, (3x + 1) is always positive.
Hence, x – 1  0
or x  1
The range of values of x for the inequality
3x  2 + 1 is x  1.
x

Example 26


Show that the following inequalities are true for all x  R.
(a) 2x + 8x + 9 . 0 (b) –3x + 2x – 5 , 0
2
2
2
Solution: (a) Let h(x) = 2x + 8x + 9
1
2
= 2 x + 4x + 9 2
2
1
2
2
2
= 2 x + 4x + 2 – 2 + 9 2
2
3
2
= 2 (x + 2) + 1 4
2
= 2(x + 2) + 1
2
2
Now, for all x  R, (x + 2)  0
Hence, 2x + 8x + 9 . 0 for all x  R.
2
2
(b) Let k(x) = –3x + 2x – 5
1
2
= –3 x – 2 x + 5 3 2
3
1
3
1
2
= –3 x – 2 x + – 1 2 2 – – 1 2 2 + 5 4
3
3
3
3
31
= –3 x – 1 2 2 + 14 4
9
3
1
= –3 x – 1 2 2 – 14
3
3
31
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Mathematics Term 1 STPM Chapter 1 Functions


1
For all x  R, –3 x – 1 2 2 < 0.
3
2
1 Hence, –3x + 2x – 5 , 0 for all x  R.

Alternative Method:
2
(a) h(x) = 2x + 8x + 9
with a = 2, b = 8, c = 9. y
2
2
b – 4ac = 8 – 4(2)(9)
= 64 – 72
= –8 y = h(x)
2
Hence, b – 4 ac , 0 and the graph of y = h(x) does not intersect the x-axis.
x
Since a . 0, the graph of y = h(x) is always above the x-axis. 0
2
Hence, 2x + 8x + 9 . 0 for all x  R.
2
(b) k(x) = –3x + 2x – 5
with a = –3, b = 2 , c = –5.
2
2
b – 4ac = 2 – 4(–3)(–5) y
= 4 – 60 x
0
= –56
y = k(x)
2
Hence, b – 4ac , 0 and the graph of y = k(x) does not intersect the x-axis.
Since a , 0, the graph of y = k(x) is always below the x-axis.
2
Hence, –3x + 2x – 5 , 0 for all x  R.
Example 27


2
Find the set of values of x which satisfy the inequality 2x + x . 3.
2
Solution: Given 2x + x . 3
2
or 2x + x – 3 . 0 y
(2x + 3)(x – 1) . 0
2
f(x) = 2x + x – 3
Consider the graph of the function
f(x) = (2x + 3)(x – 1) x
3 _ 0 1
–
At the points of intersection with the x-axis, 2
(2x + 3)(x – 1) = 0
Hence 2x + 3 = 0 or x – 1 = 0
3

x = – or x = 1
2
3
From the graph of f(x), f(x) . 0 if x , – or x . 1.
2
2
Hence, the range of values of x which satisfy the inequality 2x + x . 3
3
is x , – or x . 1.
2
3
The solution set of the inequality is {x : x , – or x . 1}.
2

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Mathematics Term 1 STPM Chapter 1 Functions

Example 28

2
3
If f(x) = 2x – 9x + 3x + 14, factorise f(x). Sketch the graph of y = f(x). 1
Hence, find the set of values of x such that f(x)  0.
3
Solution: f(x) = 2x – 9x + 3x + 14
2
f(–1) = –2 – 9 – 3 + 14 y
= 0
Hence, (x + 1) is a factor of f(x).
Using long division method, y = f(x)
f(x) = (x + 1)(2x – 11x + 14)
2
= (x + 1)(x – 2)(2x – 7)
At the x-axis, f(x) = 0
i.e. (x + 1)(x – 2)(2x – 7) = 0 x
x = –1, 2 or — –1 0 2 7 _
7
2
2
So the graph of f(x) intersects the x-axis at x = –1, 2 and 7 .
The graph of f(x) is as shown on the right. 2
From the sketch graph of y = f(x),
we see that f(x)  0 if –1 < x < 2 or x  7 .
2
Hence, the set of values of x such that f(x)  0 is
{x : –1 < x < 2 or x  7 }.
2

Example 29

Find the set of values of x such that x . 6 + 1.
x
6
Solution: Given that x . x + 1
6
x – x – 1 . 0
2
x – x – 6
x . 0
(x – 3)(x + 2)
x . 0
Let f(x) = (x – 3)(x + 2)
x
Notice that f(x) changes sign when x passes through x = 3, 0 and –2.
We draw up a table for the sign of f(x) by considering the sign of each factor in
each interval of x, as shown below.
x , –2 –2 , x , 0 0 , x , 3 x . 3
(x + 2) – + + +
x – – + + The sign of f(x)
is obtained by
(x – 3) – – – + multiplying the
f(x) – + – + signs of the 3
factors in each
From the above table, we see that f(x) . 0 or x . 6 x + 1, interval.
when –2 , x , 0 or x . 3.
Hence, the set of values of x is {x : –2 , x , 0 or x . 3}.



33





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Mathematics Term 1 STPM Chapter 1 Functions

Example 30

1
x
1 If x + 8 < x – 1 , find the set of values of x which satisfy the inequality.

Solution: Given that x < 1
x + 8 x – 1
x – 1 < 0
x + 8 x – 1
x(x – 1) – (x + 8) < 0
(x + 8)(x – 1)
x – 2x – 8 < 0
2
(x + 8)(x – 1)
(x + 2)(x – 4) < 0
(x + 8)(x – 1)

By considering each of the four factors as positive, we get
x + 2 . 0, i.e. x . –2
x – 4 . 0, i.e. x . 4
x + 8 . 0, i.e. x . –8
x – 1 . 0, i.e. x . 1
These ranges can be represented on the real number line as follows.

x + 2 > 0
x – 4 > 0
x
– 8 –2 0 1 4
x – 1 > 0
x + 8 > 0
The numerator (x + 2)(x – 4)  0 in {x : x < –2 or x  4}
The denominator (x + 8)(x – 1) . 0 in {x : x , –8 or x . 1}
Hence, (x + 2)(x – 4) < 0 if the numerator and denominator are of opposite signs,
(x + 8)(x – 1)
i.e. when –8 , x < –2 or 1 , x < 4.
Hence, the set of values of x such that x < 1
x + 8 x – 1
is {x : –8 , x < –2 or 1 , x < 4}.


Inequalities involving modulus signs

Inequalities involving absolute value or modulus signs may be solved using the analytical or graphical method.

Example 31

Find the values of x such that |2x + 1| , 3.

Solution: Given that |2x + 1| , 3
This means –3 , 2x + 1 , 3
–4 , 2x , 2

–2 , x , 1
Hence, the inequality is valid if –2 , x , 1, i.e. x  (–2, 1).



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Mathematics Term 1 STPM Chapter 1 Functions

Note: Some inequalities involving moduli may be solved by squaring both sides of the inequality. However, it
is only valid if both sides of the inequality are positive or zero for all x  R.
1
Example 32

Find the values of x such that 2|x – 1| < |x + 3|.


Solution: Given that 2|x – 1| < |x + 3|
2
4(x – 1) < (x + 3) 2 Both LHS and RHS are  0.
Squaring both sides of inequality.
4x – 8x + 4 < x + 6x + 9
2
2
3x – 14x – 5 < 0
2
f(x )
(3x + 1)(x – 5) < 0
2
f(x) = 3x – 14x – 5
From the sketch graph of
f(x) = 3x – 14x – 5,
2
1
we see that f(x) < 0 if – < x < 5, 1 _ 0 5 x
3 – 3
1
3
4
i.e. in the closed interval – , 5 .
3
Example 33
Using graphical method, find the range of values of x for which the inequality |2x – 1| < |x| + 3 is valid.
2
Solution: We first sketch the graphs of y = |2x – 1| and y = |x| + 3 as follows:
2
y
y = –2x + 1
3
3
y = –x + — y = x + — A
2
2
B
3
—
1 2 y = 2x – 1
x
1 O 1 5
– — — —
2 2 2
The graphs of y = |2x – 1| and y = |x| + 3 intersect at A and B.
2
For A, we solve y = 2x – 1 and y = x + 3 , i.e. 2x – 1 = x + 3 ⇒ x = 5 .
2 2 2
1
For B, we solve y = –2x + 1 and y = –x + 3 , i.e. –2x + 1 = –x + 3 ⇒ x = – .
2 2 2
From the graphs, we notice that |2x – 1| < |x| + 3 is valid in the range
2
1
– < x < 5 .
2 2



35





01a STPM Math T T1.indd 35 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

Exercise 1.6


1 1. With the aid of a sketch graph, solve each of the following inequalities.
(a) (x + 1)(x – 2) < 0 (b) (x – 3)(x + 5) . 0
(c) (2x – 3)(x + 4) , 0 (d) (2x + 1)(4x – 1)  0
1
(e) ( x + 5)(x – 3) < 0 (f) (x – 2)(5x + 2) . 0
2
2. Solve each of the following inequalities.
2
(a) x  9 (b) x + 2x + 1 . 0
2
(c) x(x + 1) < –2(2x + 3) (d) 5x < 3x + 2
2
2
2
2
2
(e) (x – 2) . 9x (f) 3x – 2x  x + 3x + 3
3. Find the range of values of x such that each of the following inequalities is valid.
(a) (x + 2)(x – 1)(x + 3) , 0 (b) (x – 2) (x + 1) < 0
2
(c) x + 3x – 4  0 (d) 2x + 3x – 3x , 2
3
2
3
2
3
2
2
2
(e) x(5x + 8) < 1 (47x – 48) (f) 2x  7x + 17x – 10
2
4. Find the range of values of x which satisfy each of the following inequalities.
(a) 4 . 2 – x (b) 4 – 5x . 3
x + 3 1 – 2x
(c) 14  2x – 1 (d) 13 – 4x , 35
x – 2 x – 1 x – 3
(e) 9 < 7x + 5 (f) x + 1 . 3
4 – x x + 3 2x – 1 x – 2
5. Find the set of values of x which satisfy each of the following inequalities.
(a) |x – 2| , 1 (b) |x – 3|  5 (c) |3x + 4| . 5
(d) |2x – 5| < 11 (e) |x|  |x – 1| (f) 2|x – 2| , |x – 3|
(g) 3|x + 2| < |x – 6| (h) 5|2x – 3| . 4|x – 5| (i) |2x + 1| , 3x + 2
2
(j)  x  , 2 (k)  x – 4  < 3 (l)  x + 1  , 1
x + 4 x x – 1
Partial fractions
2
Let f(x) = 2x + 1 and g(x) = x + 3x + 2. When f(x) is divided by g(x), the resulting function
2x + 1 2x + 1
–––––––––– ≡ ––––––––––––
2
x + 3x + 2 (x + 1)(x + 2)
is known as a rational function.
Notice that in the above case, f(x) is linear and g(x) is quadratic, i.e. the degree of f(x) is less than
the degree of g(x). If f(x) is a polynomial function of degree m, and g(x) is a polynomial of degree n,
f(x)
where m , n, then h(x) = — is considered a proper rational function.
g(x)
x + 1
2

2
x + 2

For example, –––––––––––– , –––––––––– and –––––– are all proper rational functions.
(x – 1)(x + 3) 2x + x + 1 3x + 1
3
f(x)
However, if m  n, then h(x) = — is considered an improper rational function.
g(x)
2
2
For example, ––––––––––– , ––––––––––– and ––––––––––––––– are all improper rational functions.
x + 2x + 3 x + 3x + 1
2x + x – 3x + 1
3
2
2
x + 1 x + 4x – 5 (x + 1)(x + 2)
36
01a STPM Math T T1.indd 36 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions
Denominator with linear factors

Consider the addition of two proper rational functions as follows.
f(x) = 2 + 3 1
x + 1 x + 2
The function f(x) can be expressed as a single rational function with a common denominator, i.e.
f(x) = 2 + 3
x + 1 x + 2
= 2(x + 2) + 3(x + 1)
(x + 1)(x + 2)

= 5x + 7
(x + 1)(x + 2)
Notice that the addition of two proper rational functions results in a single proper rational function.

From the operation of addition as illustrated above, we see that the inverse process can also be carried out,
5x + 7
i.e. expressing the proper rational function (x + 1)(x + 2) as the sum of two rational functions with denominators
(x + 1) and (x + 2) respectively.

5x + 7 A B
Let ≡ + , where A and B are constants.
(x + 1)(x + 2) x + 1 x + 2
Hence 5x + 7 ≡ A(x + 2) + B(x + 1)
(x + 1)(x + 2) (x + 1)(x + 2)
5x + 7 ≡ A(x + 2) + B(x + 1)
= (A + B)x + (2A + B)
Equating coefficients of x: 5 = A + B ………… 
Equating constants: 7 = 2A + B ………… 
 – : 2 = A
Substituting A = 2 into ,
5 = 2 + B
B = 3

This process is known as expressing the function 5x + 7 in partial fractions with terms 2 and
3 . (x + 1)(x + 2) x + 1
x + 2
Notice that these two partial fractions are proper rational functions with the numerator as a constant and the
denominator as a linear function.

Example 34


Express 2x + 3 in partial fractions.
(x – 1)(x + 2)
Solution: Let 2x + 3 ≡ A + B
(x – 1)(x + 2) x – 1 x + 2
A(x + 2) + B(x – 1)
≡
(x – 1)(x + 2)
2x + 3 ≡ A(x + 2) + B(x – 1) This is an identity and
is valid for all x ∈ R



37





01a STPM Math T T1.indd 37 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

When x = 1, 2 + 3 = A(3) + B(0)
5 = 3A
1 A = 5
3
When x = –2, –4 + 3 = A(0) + B(–3)
–1 = –3B
B = 1
3
2x + 3 5 1
Hence, ≡ +
(x – 1)(x + 2) 3(x – 1) 3(x + 2)


Denominator with prime quadratic factors
We have learnt that for a partial fraction with a linear denominator, its numerator is a constant and the partial
fraction is a proper rational function. In the same way, if the denominator of a partial fraction is a quadratic
function, its numerator must be a constant or a linear function such that the partial fraction is a proper rational
function.
Hence, if the denominator of a proper rational function has a prime quadratic factor in the form
ax + bx + c, where a, b and c are constants, the given partial fraction must be of the form Ax + B , where
2
2
ax + bx + c
A and B are suitable constants.
Example 35


Express 3x + 4 in partial fractions.
2
(x + 2)(x – x + 1)
Solution: Let 3x + 4 ≡ A + Bx + C
2
(x + 2)(x – x + 1) x + 2 x – x + 1
2
2
A(x – x + 1) + (Bx + C)(x + 2)
≡
2
(x + 2)(x – x + 1)
2
3x + 4 ≡ A (x – x + 1) + (Bx + C)(x + 2)

When x = –2, –6 + 4 = A(4 + 2 + 1) + 0
–2 = 7A
2
A = –
7
2
Equating coefficients of x : 0 = A + B
2
B =
7
Equating the constants: 4 = A + 2C
2
= – + 2C
7
30
2C =
7
15
C =
7
Hence, 3x + 4 ≡ – 2 + 2x + 15
2
(x + 2)(x – x + 1) 7(x + 2) 7(x – x + 1)
2
Notice that the two partial fractions are proper rational functions with the numerator
of one of them as a linear function, i.e. 2x + 15.
38





01a STPM Math T T1.indd 38 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions
Denominator with repeated linear factors

If one of the terms in the denominator of a proper rational function is a repeated linear factor such as
2
(x + 1) , this factor can be considered to be a quadratic factor. So, the partial fraction concerned is of the 1
form Ax + B .
(x + 1) 2
2
Suppose we want to express 3x + 6x – 1 in partial fractions.
(x – 1)(x + 1) 2
2
Then, 3x + 6x – 1 ≡ A + Bx + C
(x – 1)(x + 1) 2 x – 1 (x + 1) 2
2
≡ A(x + 1) + (Bx + C)(x – 1)
(x – 1)(x + 1) 2
2
3x + 6x – 1 ≡ A(x + 1) + (Bx + C)(x – 1)
2
When x = 1, 3 + 6 – 1 = 4A
8 = 4A
A = 2
2
Equating coefficients of x : 3 = A + B
3 = 2 + B
B = 1
Equating constants: –1 = A – C
–1 = 2 – C
C = 3
2
Therefore, 3x + 6x – 1 ≡ 2 + x + 3
(x – 1)(x + 1) 2 x – 1 (x + 1) 2
Notice that the partial fractions obtained is correct, except that it is not in its simplest form, because
x + 3 ≡ (x + 1) + 2
(x + 1) 2 (x + 1) 2
≡ 1 + 2
x + 1 (x + 1) 2

Hence, the partial fractions in its simplest form is
2
3x + 6x – 1 ≡ 2 + 1 + 2
(x – 1)(x + 1) 2 x – 1 x + 1 (x + 1) 2
2
This means that the function 3x + 6x – 1 can be expressed as partial fractions in the following way.
(x – 1)(x + 1) 2
2
Let 3x + 6x – 1 ≡ A + B + C
(x – 1)(x + 1) 2 x – 1 x + 1 (x + 1) 2
2
≡ A(x + 1) + B(x – 1)(x + 1) + C(x – 1)
(x – 1)(x + 1) 2
2
Hence, 3x + 6x – 1 ≡ A(x +1) + B(x – 1)(x + 1) + C(x – 1)
2
When x = 1, 3 + 6 – 1 = 4A
8 = 4A
A = 2
When x = –1, 3 – 6 – 1 = –2C

–4 = –2C
C = 2


39





01a STPM Math T T1.indd 39 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

Equating the coefficients of x : 3 = A + B
2
3 = 2 + B
B = 1
1
2
Therefore, 3x + 6x – 1 ≡ 2 + 1 + 2
(x – 1)(x + 1) 2 x – 1 x + 1 (x + 1) 2
In the same way, if the denominator of a proper rational function contains a repeated linear factor such as
3
3
(x + 1) , the denominator of the partial fractions concerned are (x + 1), (x + 1) and (x + 1) respectively.
2
Example 36


2
Express x + 1 in partial fractions.
(x – 1)(x + 1) 3
3
Solution: Since the denominator contains the repeated linear factor (x + 1) , the partial fractions
consist of terms with denominators (x +1), (x + 1) and (x + 1) .
3
2
2
Hence, x + 1 ≡ A + B + C + D
(x – 1)(x + 1) 3 x – 1 x + 1 (x + 1) 2 (x + 1) 3
Multiply both sides of the identity with (x – 1)(x + 1) ,
3
2
3
x + 1 ≡ A(x + 1) + B(x – 1)(x + 1) + C(x – 1)(x + 1) + D(x – 1)
2
When x = 1, 2 = 8A
A = 1
4
When x = –1, 2 = –2D
D = –1

Equating the coefficients of x , 0 = A + B
3
0 = 1 + B
4
B = – 1
4
When x = 0, 1 = A – B – C – D
= 1 + 1 – C + 1
4 4
C = 1
2

2
Hence, x + 1 ≡ 1 – 1 + 1 – 1 .
(x – 1)(x + 1) 3 4(x – 1) 4(x + 1) 2(x + 1) 2 (x + 1) 3


Degree of denominator less than or equal to numerator


The rational function f(x) , where the degrees of f(x) and g(x) are m and n respectively, with m  n, is
g(x)
considered to be an improper rational function. In this case, f(x) must first be divided by g(x) before expressing
in terms of partial fractions.



40





01a STPM Math T T1.indd 40 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

Example 37

2
Express x + 2x + 3 in partial fractions. 1
(x + 2)(x – 1)
2
x + 2x + 3 x + 2x + 3
2
Solution: ≡
2
(x + 2)(x – 1) x + x – 2
2
≡ (x + x – 2) + (x + 5)
x + x – 2
2
= 1 + x + 5
2
x + x – 2
= 1 + x + 5
(x + 2)(x – 1)
14243 x + 5
–––––––––– can be expressed
(x +2)(x – 1)
Let x + 5 ≡ A + B as partial fractions.
(x + 2)(x – 1) x + 2 x – 1
≡ A(x – 1) + B(x + 2)
(x + 2)(x – 1)
x + 5 ≡ A(x – 1) + B(x + 2)
When x = 1, 6 = 0 + 3B
B = 2
When x = –2, –2 + 5 = –3A + 0
3 = –3A
A = –1
x + 5 ≡ – 1 + 2
(x + 2)(x – 1) x + 2 x – 1
2
Hence, x + 2x + 3 ≡ 1 – 1 + 2
(x + 2)(x – 1) x + 2 x – 1

Example 38

3
Express x + 1 in partial fractions.
(x – 1)(x + 2)
3
3
Solution: We have x + 1 ≡ x + 1
2
(x – 1)(x + 2) x + x – 2
By using long division,
x – 1
2
x + x – 2 x + 0x + 0x + 1 Write x + 1 as x + 0x + 0x + 1.
2
3
3
2
3
x + x – 2x
2
3
– x 2 + 2x + 1
– x 2 – x + 2
3x – 1
3
Hence, x + 1 ≡ (x – 1) + 3x – 1
(x – 1)(x + 2) (x – 1)(x + 2)
≡ (x – 1) + A + B
x – 1 x + 2
x + 1 ≡ (x – 1) (x + 2) + A(x + 2) + B(x –1)
3
2
41



01a STPM Math T T1.indd 41 3/28/18 4:20 PM

Mathematics Term 1 STPM Chapter 1 Functions

When x = 1, 1 + 1 = 0 + 3A + B(0)
2 = 3A
1 A = 2
3
When x = –2, –8 + 1 = 0 + A(0) + B(–3)
–7 = –3B
B = 7
3
3
Hence, x + 1 = x – 1 + 2 + 7
(x – 1)(x + 2) 3(x – 1) 3(x + 2)





Exercise 1.7



Express each of the following expressions in partial fractions.

1. 2x + 4 2. x – 4 3. 3
(x + 1)(x + 2) (x + 2)(x – 1) x – 9
2

4. x 5. x + 1 6. x + 3
2
2
x – 1 x – 4 x + 5x + 4
2
2
7. 4x + 10 8. 4x + 2 9. x – 2x + 4
(2 + x)(3 + x)(4 + x) (x + 2)(x – 1) 2x(x – 3)(x + 1)
2
2
10. x – 4 11. x 12. 9
x(x + 1)(x + 3) (1 + x) 2 (x – 1)(x + 2) 2

2
2
13. 3x + x – 2 14. x – 1 15. 1
2
(2x – 1)(x – 2) 2 x (2x + 1) x (x – 1)
2
2
2
2
2
16. x + 1 17. x + x – 1 18. 2x + 3x + 2
2
2
x – 1 (x + 1)(x + 2) x + 3x + 2
3
3
19. x + x + 1 20. x + 3 21. 2x – 1
(x + 1) 2 (x + 1)(x – 1) (x + 2)(x + 1)
2
22. x + 1 23. 3x 24. 1
2
2
x(x + x + 1) (x – 1)(x + x + 1) (x – 1)(x – x + 1)
2
25. x
(x + 1)(x – 1) 3


42





01a STPM Math T T1.indd 42 3/28/18 4:20 PM

ANSWERS




Chapter 1 Functions Exercise 1.2
1. (b) c, d (c) {1, 2, 3, 4, 5}
Exercise 1.1 (d) {a, b, c, d} (e) No
1. (a) –9 (b) 7 (c) 0 (d) 0 2. (a) X Y
2. (a) (i) 3 (ii) 1 (iii) 0 –2 3
(b) (i) 24 (ii) 99 –1
3. (a) [–2, 6] or –2 < f(x) < 6 0 1 1
f(x ) 2 7
6 (b) {–2, –1, 0, 1, 2}
(c) {1, 3, 7}
(d) No
3. (a) Y
x
–1 0 3 –5
X –3
–2 1 –1
(b) [–1, 3] or –1 < f(x) < 3 3 3
5 5
f(x )
7 7
3 9 11
15
(b) –1, 15
(c) Yes
x
–1 0 3 1
4. No. f(0) = f(1) = 0,
–1
2
1
9
9
(c) 3 – , 44 or – < f(x) < 4 5. a 2 ; {g : 0 , y < }
4 4 1 + a 2
f(x ) 6. (a) f : x ⟼ x + 2, x ∈ R
–1
–1
(b) f : x ⟼ x – 1 , x ∈ R, x > 1
4
–1
(c) f : x ⟼ x + 1, x > 0
(d) g : x ⟼ 3 + x + 1 , x > –1
–1
(e) g : x ⟼ x + 9 , x . –9
–1
x
–2 0 3 2 + 3x
–1
(f) h : x ⟼ , x ≠ 0
x
– 9 _ (g) h : x ⟼ 2(x + 1) , x ≠ 1
–1
4 x – 1
1
1 – x
–1
4. (a) No (b) Yes (c) No 7. f (x) = 2x – 1 , , x < 1
2
1
5. (a) 1, 4, , 64 (b) 4 (c) R 9. (a) (b)
+
2
6. x + 2 f(x ) f f(x ) f –1
7. 4 – x 2 f –1 f
8. (a) cos (1 + x) (b) cos (cos x) x 0 x
(c) 1 + cos x 0
2
9. (a) 1 – x , –1 , x , 1 (b) 1 – x
(c) 2x – x 4
2
2
10. (a) 4x – 6x + 1 (c) f(x ) f (d) f(x ) f
2
(b) 2x + 6x – 1 f –1 f –1
(c) 4x – 9
(d) x + 6x + 14x + 15x + 5 0 x 0 x
4
3
2
11. (a) 6 (b) 25
2
2
(c) 2a – 4a + 9 (d) 4x – 12x + 14
4
2
(e) 2x + 4x + 9 (f) x + 10x + 30
2
268
Answers STPM Math T S1.indd 268 3/28/18 4:25 PM

Mathematics Term 1 STPM Answers

Exercise 1.3 4. (a) y (b) y
1. (a) y (b) y
–1
x
0 2
x
3 8 –1 0 2 –2


x x (c) y (d) y
0 0 8
3 –
– – 3
4
x
(c) y 0 2 x
0
– 2 2
8
-8
(e) y
x
0
6
– –
5
2. (a) y (b) y
x
0
1 2
x
1 0 1
3 – – –
– 3 3 5. (a) y
2 x
0 –1
(c) y x
-1 0 1
57
–
16
2
(b) y
x
0 5
–
8
x
-1 0 1
3. (a) y (b) y

x
0 1
6. (a) y (b) y
–1 1
x
0 1
–– x
3 0
x _ _ 1
3
-1 0
2 _ _ 1
2
(c) y
(c) y
x
5 –1 0
x
0 3
10


269





Answers STPM Math T S1.indd 269 3/28/18 4:25 PM

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