Semester 1 Mathematics STPM (Teacher's Edition)

1

ii

UCAPAN ALU-ALUAN TUAN PENGARAH
JABATAN PENDIDIKAN NEGERI PULAU PINANG

Bismillahirrahmanirrahim.
Assalamualaikum Warahmatullahi Wabarakatuh dan Salam Sejahtera.
Saya merakamkan jutaan terima kasih kepada Unit Sains dan Matematik, Sektor Pembelajaran, Jabatan
Pendidikan Negeri Pulau Pinang atas penghasilan Matematik (T) dan Matematik (M) pada tahun 2021
ini. Penerbitan modul ini sangat tepat pada masanya selaras dengan usaha Jabatan Pendidikan Negeri
Pulau Pinang dalam meningkatkan pencapaian akademik para pelajar khususnya dalam peperiksaan
STPM.
Modul ini ditulis hasil gabungan kepakaran Rakan Pembimbing Guru (SISC+), Jurulatih Utama (JU)
dan guru-guru yang berwibawa serta berpengalaman dalam mata pelajaran berkenaan. Semoga dengan
usaha murni ini akan memperoleh kejayaan yang lebih baik pada masa hadapan umpama pepatah
menyatakan, apa yang kita semai hari ini, itulah yang akan kita tuai pada masa hadapan.
Penghasilan modul ini juga diharap dapat menjadi salah satu medium tambahan kepada pembelajaran
pelajar kita. Saya berharap dengan terhasilnya modul ini, akan menjadi titik permulaan untuk
penghasilan modul-modul yang lain bagi membantu para guru dan pelajar menguasai mata pelajaran
Matematik (T) dan Matematik (M) ke tahap yang lebih baik. Sesungguhnya, kejayaan pelajar berkait
rapat dengan komitmen, kerajinan serta kesungguhan para guru dalam melaksanakan tugas mendidik
dengan penuh dedikasi. Saya berharap modul ini akan digunakan seoptimum yang mungkin sebagai
bahan bantu dalam pengajaran dan pembelajaran (PdP) mata pelajaran Matematik (T) dan Matematik
(M).
Akhir kata, saya mengucapkan setinggi-tinggi penghargaan dan terima kasih kepada Unit Sains dan
Matematik, Sektor Pembelajaran Jabatan Pendidikan Negeri Pulau Pinang yang begitu proaktif
menerbitkan modul ini. Begitu juga ucapan terima kasih saya rakamkan kepada semua Rakan
Pembimbing Guru (SISC+), Jurulatih Utama (JU) serta guru-guru mata pelajaran Matematik (T) dan
Matematik (M) yang telah memberikan sumbangan idea, tenaga dan masa bagi merealisasikan modul
yang sangat berharga ini. Semoga Allah SWT sentiasa memberkati usaha murni yang kita lakukan.

“PULAU PINANG PENERAJU TRANSFORMASI PENDIDIKAN NEGARA”
Sekian, terima kasih.

(ABDUL RASHID BIN ABDUL SAMAD)
Pengarah Jabatan Pendidikan Negeri Pulau Pinang

iii

SENARAI PEGAWAI DAN PENGGUBAL MODUL MATEMATIK T DAN MATEMATIK M
SEMESTER 1

BIL NAMA JABATAN /PPD/SEKOLAH

1 ROZITA BINTI KATAN JABATAN PENDIDIKAN NEGERI PULAU
PINANG

2 ASMAH BINTI OMAR SISC+ PPD SEBERANG PRAI TENGAH

3 ANBU CHELIAN A/L SOUNDARAJAN SISC+ PPD TIMUR LAUT

4 ONG KHYE CHING SISC+ PPD SEBERANG PRAI TENGAH

5 Dr SUZLIPAH SANUSI SISC+ PPD BARAT DAYA

6 TEH MEOW LEE SMK TINGGI BUKIT MERTAJAM
(KETUA KUMPULAN 1)

7 KHOR AI NI SMJK HENG EE, PULAU PINANG.
(KETUA KUMPULAN 2)

8 MEOR MOHD FAIZAL BIN ABU BAKAR KOLEJ TINGKATAN ENAM HAJI

(KETUA KUMPULAN 3) ZAINUL ABIDIN

9 MUMINAH BT ABDUL RAHMAN KOLEJ TINGKATAN ENAM HAJI
(KETUA KUMPULAN 4) ZAINUL ABIDIN

10 TAN SYNN CHEAK SMJK CHUNG LING PULAU PINANG

11 LOOI LI FOONG SMK TUN SYED SHEH BARAKBAH

12 NOR HABSAH BINTI MAT SALLEH SMK SIMPANG EMPAT
13 INTAN SHARIZA BINTI SHAFIE
14 KEK PENG PENG KOLEJ TINGKATAN ENAM DESA
MURNI

SMK ST XAVIER

iv

15 LOW YEE YEAN SMJK CHUNG HWA CONFUCIAN
16 OOI GHIM LOOI
17 TING PING PING SMJK PEREMPUAN CHINA PULAU
PINANG

SMK (L) METHODIST

18 ONG SWAD MEY SMJK JIT SIN

19 NURUL AINI BINTI ARNIS SUTAN SATI SMK PENANTI

20 NORHAYATI BINTI SALLEH SMK TELOK AIR TAWAR
21 ROSMAWATI BINTI KADIR
22 PARAMESWARY TIYAGARAJAN SMK TUNKU ABDUL RAHMAN,
NIBONG TEBAL

SMJK CHUNG LING BUTTERWORTH

23 ABDUL SALAM BIN ABDUL RAHIM PENANG FREE SCHOOL
24 SIM KAH SENG
25 TAN GAIK CHOO SMJK PEREMPUAN CHINA PULAU
PINANG

SMK TINGGI BUKIT MERTAJAM

26 SAKTHIYAVAANI A/P BALAKRISHNAN SMK SEBERANG JAYA

27 SIM KAH SENG SMJK PEREMPUAN CHINA PULAU
PINANG

28 TAN SYNN CHEAK SMJK CHUNG LING PULAU PINANG

v

ISI KANDUNGAN

TOPIC 1 : FUNCTIONS page 1
1.1 FUNCTIONS

1.2 POLYNOMIAL AND RATIONAL FUNCTIONS page 41

1.3 EXPONENTIAL AND LOGARITHMIC FUNCTIONS page 76

1.4 TRIGONOMETRIC FUNCTIONS page 100

TOPIC 2 : SEQUENCES AND SERIES page 137
2.1 SEQUENCES

2.2 SERIES page 146

2.3 BINOMIAL EXPANSIONS page 190

TOPIC 3 : MATRICES

3.1 MATRICES page 205

3.2 SYSTEMS OF LINEAR EQUATIONS page 253

1

1.1(a) Learning Outcome:
FUNCTIONS a) State the domain and range of a function, and find composite functions.

➢ DEFINITION
A function is a special type of relation where each and every element in the domain has only
one image in the range.

➢ NOTES
The types of relations that are connected as a function are:

a) One-to-one relation b) Many-to-one relation

a. .d a.
b. .e b. .d
c. .f
c. .e

Domain Range Domain Range

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 ∶ →

If ∈ and ∈ such that y is assigned to x under f, we say f maps x to y and write
f ∶ → 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.

( For more information, scan this QR code.)

Composite Functions
If f is a function which maps set X to set Y and g is a function which maps set Y to set Z,
Then ∘ is a composite function which maps set X directly to set Z, as illustrated in figure.

∘

For more

information,

xf y g z scan this
=f(x) =g(y) QR code.

=gf(x)

XY Z

For ∘ to be defined, the range of f must be a subset of the domain of g, ⊆ .

2

EXAMPLE 1
The domain of a function h, where ℎ: → 2 , ∈ ℝ

(a) Find the images of -1 and 6.

(b) Find the element in the domain with images 16.

(c) Find the range of the function h.

Solution:

(a) For x = −1

ℎ(−1) = 2−1

= 1
2

For x = 6,
ℎ(6) = 26

= 64

(b) ℎ( ) = 16
2 = 16
2 = 24

= 4

(c) The range is ℝ+.

EXAMPLE 2

State the set of value of x for which each of the following function is defined.

(a) : → √ − 1
1
(b) ℎ: → −1

(c) : → lg( − 2)

Solution:
(a) For f defined, − 1 ≥ 0

≥ 1
Hence, the domain of the function f is { : ≥ 1}

(b) For g defined, − 2 > 0
> 2

Hence, the domain of the function g is { : > 2}

(c) For g defined, − 1 ≠ 0
≠ 1

Hence, the domain of the function g is { : ≠ 1}

EXAMPLE 3
Given the function ( ) = √ and ( ) = 2 + 9
(a) Determine the domain and the range of f and g.
(b) Explain why ∘ exist.
(c) Find ∘ and state its domain and range.

3

Solution:
(a) For f defined, ≥ 0

Then, the domain of f is { : ≥ 0}
The domain of g is { : ∈ ℝ}.

(b) The range of g is { : ≥ 9}
The domain of f is { : ≥ 0}
Since ⊆ , ∘ exist.

(c) ∘ (x) = [ ( )]
= ( 2 + 9)
= √ 2 + 9

The domain of ∘ is the same as the domain of g, that is { : ∈ ℝ}.
The range of ∘ is { : ≥ 3}.

EXAMPLE 4
If ( ) = cos and ( ) = 1 + , ∈ ℝ, find
(a) ∘ ( )
(b) ∘ ( )
(c) ∘ ( )

Solution:
(a) ∘ ( ) = [g( )]

= (1 + )
= cos (1 + )

(b) ∘ ( ) = [ ( )]
= ( cos )
= cos (cos )

(c) ∘ ( ) = [ ( )]
= (cos )
= 1 + cos

EXAMPLE 5
If ( ) = 2 − 3 and ∘ ( ) = 2 + 1 , find g(x)

Solution:
If ∘ ( ) = 2 + 1

[ ( )] = 2 + 1
2 ( ) − 3 = 2 + 1

2 ( ) = 2 + 4
( ) = + 2

4

EXAMPLE 6

Given that the function : → 2 + 1, find the function g if the composite function
∘ ∶ → 1 , ≠ 1.

−1

Solution:

If ∘ ( ) =1

−1
[ ( )] = 1
−1
(2 + 1) = 1
−1

Let = 2 + 1

Then = 1 ( − 1)
2
1
Hence, ( ) = 21( −1)−1

=2
−3

Then, ( ) = 2 , ≠3

−3

WORKSHEET 1.1

Q1 Given that : → 2 + + 1 and ∈ X, where = {−2, −1, 0, 1, 2}.
(a) Sketch the diagram to represent g.
(b) State the domain of g.
(c) State the range Y of g.
(d) Is g a one-to-one function?

Solution:
1(a)

-2 .
.1

-1 .
.3

0.
1 .7

2

X Y
(b) Domain of g is {-2, -X1, 0, 1, 2}

(c) Range of g is {1, 3, 7}

5
(d) No. g is many-to-one function.

Q2 Function f is defined by : → 3 − 5 , ∈ ℝ. Find
(a) f (2)
(b) the value of x whose image is 13
(c) ( 2)
(d) f (2 − 1)
(e) the value of x if ( 2) = (2 − 1)

Solution:
2(a) (2) = 3 − 5(2)
= −7

(b) ( ) = 3 − 5
3 − 5 = 13
−5 = 10
= −2

(c) ( 2) = 3 − 5 2

(d) (2 − 1) = 3 − 5(2 − 1)
= 8 − 10

(e) ( 2) = (2 − 1)
3 − 5 2 = 8 − 10

5 2 − 10 + 5 = 0
2 − 2 + 1 = 0
( − 1)2 = 0
= 1

Q3 Determine the domain of f such that f is a function:

(a) ∶ → ln ( + 2) +1

1 √1−
2+2 −8
(b) ∶ → + √

Solution:

3(a) For f defined:
+ 2 > 0 and 1 − x > 0
> −2 and < 1
Then, domain of f is { : − 2 < < 1}

(b) For f defined:
2 + 2 − 8 > 0 and ≥ 0

≠ −4 or 2 and ≥ 0
Then, domain of f is { : 0 ≤ < 2 or > 2}

6

Q4 Given that : → + , ∈ ℝ, g (1) = −3 and g (−1) = 1, find

(a) the values of a and b,
(b) the values of n if ( 2 + 1) = 5 − 6

Solution: (1) = −3
4(a) + = −3

(−1) =1
+ (−1) = 1

= 1 +

(1 + ) + = −3
= −2

− 2 = −3
= −1

Then, = −1 and = −2

(b) ( 2 + 1) = 5 − 6
−1 − 2( 2 + 1) = 5 − 6
2 2 + 5 − 3 = 0

( + 3)(2 − 1) = 0
= −3 or = 1

2

Q5 The functions f and g are defined by ( ) = 2 − 4 and ( ) = √ − 2 respectively.
(a) Determine the domain and the range of
(i) f
(ii) g
(b) Explain why ∘ exist.
(c) Find ∘ and determine its domain.

Solution:
5(a) (i) Domain of f is { : ∈ ℝ}
Range of f is { : ≥ −4}

(ii) Domain of g is { : ≥ 2}
Range of g is { : ≥ 0}

(b) Because ⊆

7

(c) ( ) = (√ − 2)
= (√ − 2)2 − 4
= − 6

Domain of ∘ is { : ≥ 2}

Q6 If ( ) = √ , ≥ 0 and ( ) = 1 − 2, ∈ ℝ, find
(a) ∘ ( )
(b) ∘ ( )
(c) ∘ ( )

Solution:
6(a) ( ) = (1 − 2)
= √1 − 2, ≤ −1  and   ≥ 1

(b) ( ) = g(√ )
= 1 − (√ )2
= 1 −

(c) ( ) = g(1 − 2)
= 1 − (1 − 2)2
= 1 − (1 − 2 2 + 4)
= 2 2 − 4

EXERCISE 1.1

Q1 Given that ( ) = √ + 1, ≥ −1 ,
(a) Find the value of
(i) f (8)
(ii) f (0)

(b) Find the value of x such that
(i) ( ) = 5
(ii) ( ) = 10

Q2 Determine the domain of f such that f is a function:

(a) : → √16 − 2 − ln ( 2 − − 2)

(b) : → 4 − ln (1 − )
√ +4

Q3 The functions f and g are defined by : → and : → +4 respectively.
+2

(a) State the domain of f and g.

(b) Find the composite function ∘ and state its domain.

Q4 The functions g and h are defined by ( ) = 25 , 1 < < 9, ∈ ℝ and
3 −2

8

( ) = 2, 1 < < 3, ∈ ℝ .
(a)Determine the range of

(i) f
(ii) g

(b)Explain why ∘ exists. Hence, find ∘ g, stating its domain.

Q5 The functions f and g are defined by : → 2 − 1, ∈ ℝ and : x → 2 + 3, ∈ ℝ respectively.
Find
(a) ( ∘ )(3)
(b) ( ∘ ) (1)

2

ANSWER 1(a) (i) (8) = √8 + 1
Section =3
1.1
(ii) (0) = √0 + 1
Exercise =1
1.1
(b) (i) ( ) = 5
√ + 1 = 5
+ 1 = 25
x = 24

(ii) ( ) =10
√ + 1 =10
+ 1 =100

x = 99

2(a) For f defined:
16 − 2 ≥ 0 and 2 − − 2 > 0

(4 − )(4 + ) ≥ 0 and ( + 1)( − 2) > 0
−4 ≤ ≤ 4 and < −1 or > 2

Then, domain of f is { : − 4 < < −1 or 2 < ≤ 4}

(b) For f defined:
+ 4 > 0 and 1 − > 0
> −4 and < 1
> −4 and < 0
Then, domain of f is { : − 4 < < 0}

3(a) Domain of f is { : ∈ ℝ, ≠ −2}
Domain of g is { : ∈ ℝ, ≠ 0}

(b) ( ) = ( )

+2

9

= + 2+4

+2

= 5 + 8 , ≠ 0 ……. ①



The domain of ∘ f follows the domain of f, that is
{ : ∈ ℝ, ≠ −2}. ………②

Combining ①and ②, the domain of ∘ f is
{ : ∈ ℝ, ≠ −2, ≠ 0}

4(a) (i) Range of f is { : 1 ≤ ≤ 25}
(ii) Range of g is { : 1 ≤ ≤ 9}

(b) Because ⊆ .
( ) = ( 2)

= 25 , ≠ ±√2
3 2−2
3

Domain of ∘ g is { : 1 ≤ ≤ 3}

5(a) Method 1
(3) = (32 − 1)

= g(8)

= 2(8) + 3

= 19

Method 2

( ) = [ ( )]
= ( 2 − 1)
= 2( 2 − 1) + 3
= 2 2 + 1

(3) = 2(3)2 + 1

= 19

(b) Method 1

(1) = f (2 (1) + 3)

2 2

= (4)

= 42 − 1

= 15

Method 2

( ) = (2 + 3)

= (2 + 3)2 − 1

= 4 2 + 12 + 8

(1) = 4 (1)2 + 12 (1) + 8
22 2

= 15

10

1.1(b) ONE-TO- Learning Outcome:
ONE b) Determine whether a function is one-to-one, and find the inverse of a
FUNCTIONS one-to-one function

➢ DEFINITION
One-to-one function is a function that maps distinct elements of its domain to distinct elements of its
codomain.

➢ NOTES
1. Every element of the function’s codomain is the image of at most one element of its domain.

A function f : D → S is a one-to-one function if and only if a,bD, f (a) = f (b) implies that

a =b.

2. An inverse function (or anti-function) is a function that “reverses” another function. If the function
f applied to an input x gives a result of y , then applying its inverse function g to y gives the result

x , that is g( y) = x if and only if f (x) = y . The inverse function of f is also denoted as f −1 .

➢ CONCEPT

1. To determine whether or not a function is a one-to-one function, we can use the horizontal
line test.

2. If there is only one intersection point, the function is a one-to-one function.
3. If f is a one-to-one function:

• the domain of f −1 is the range of f
• the range of f −1is the domain of f
4. The graph of f and f −1 are symmetrical about the straight line y=x
5. Inverse functions ONLY exist for one-to-one functions.
6. A function which is NOT one-to-one can have an inverse function if we RESTRICT the
domain of the function so that it is one-to-one.
****USE COMPLETING THE SQUARE****

REMEMBER THIS !

: → 2, ∈ ℝ (not one-to-one)
: → 2, ∈ ℝ, ≥ 0 (one-to-one)

EXTRA KNOWLEDGE!

11

EXAMPLE 1
Determine whether each of the following functions is one-to-one.
(a) f (x) = x3

(b) g(x) = x

Solution:
(a) Method 1 (using definition)

Let, x1 = x2
Then, x13 = x23

f (x1) = f (x2 )
 f (x) = x3 , is one-to-one function

Method 2 (graphical method)

Since no horizontal lines intersect the graph of f (x) = x3 more than once, by the Horizontal
Line Test, the function f is one-to-one.
(b) Method 1 (using definition)
Let 1 = 2
Then, √ 1 = √ 2

g(x1) = g(x2 )
∴   ( ) = √ is one-to-one function

12
Method 2 (graphical method)

Since no horizontal lines intersect the graph of g(x) = x more than once, by the Horizontal

Line Test the function g is one-to-one.

EXAMPLE 2

A function is defined by : f : x → 3 − x ,
(a) find f −1

(b) determine the domain and the range of
(i) f
(ii) f −1

(c) sketch the graph of y = f (x) and y = f −1(x) on the same axes.

Solution: = −1( )
(a) Let ( ) =

√3 − =

3− y = x2

y = 3− x2

∴   −1: → 3 − 2

(b) (i) domain = x x  3
range = y y  0

(ii) domain = x x  0
range = y y  3

13
(c)

EXAMPLE 3

The function g is defined by: g:x→ 2.
x2 +1

(a) Sketch the graph of y = g(x)

(b) State the domain of g so that g −1 exists. Determine the corresponding range of g .

(c) Find g −1 , hence determine the domain and range of g −1 .

(d) Sketch the graph of y = g(x) and y = g −1(x) based on the domains and ranges determined in

(b) and (c).

Solution:
(a)

(b) In order for g −1 to exist, g must be one-to-one. Hence, the domain of g must be restricted to only

x x  0 and range of g is y 0  y  2 .

(c) Let y = g −1(x)

( ) =

2 =
2+1

2 = xy2 + x

14
2 = 2−



= √2−



∴   −1( ) = √2−



(d)

WORKSHEET 1.2
Q1 Determine whether each of the following functions is one-to-one.

(a) f (x) = 2x − 5
(b) h (x) = 5 − x2
(c) p (x) = 1

x
Solution:

(a) Method 1 (using definition)
Let 1 = 2, then,
2 1 = 2 2
2 1 − 5 = 2 2 − 5
( 1) = ( 2)

Therefore, f (x) = 2x − 5 is one-to-one.

15
Method 2 (graphical method)

Since no horizontal lines intersect the graph of f (x) = 2x − 5 more than once, by the

horizontal line test, the function is one-to-one.
(b) Method 1 (using definition)

Let 1 = −2 and 2 = 2
ℎ(−2) = 5 − (−2)2 = 1
ℎ(2) = 5 − (2)2 = 1
Here, −2 ≠ 2 but ℎ(−2) = ℎ(2) = 1
Thus, the function ℎ( ) = 5 − 2 is not one to one.
Method 2 (graphical method)

As seen in the above graph, the horizontal line intersects the graph of h(x) = 5 - x2 more
than once. By the horizontal line test, the function is not one to one.

16

(c) Method 1 (using definition)
Let 1 = 2,

then 1 = 1

1 2

( 1) = ( 2)

Method 2 (graphical method)

Since no horizontal lines intersect the graph of p(x) = 1 more than once,
x

by the horizontal line test the function is one to one.

Q2 Given the function f, f : x → x2 . Show that the function is not one-to-one.

Solution:
Since ̶ 1,1 ϵ ℝ(domain) and f ( ̶ 1) = 1 = f (1) we can see that the function f is not one to one.

Q3 Given that :

f (x) = 3x +1 , x0
2x −1 , x 0

Show that the above function is one to one .

Solution:
Suppose a, b ϵ ℝ (domain), there are 4 cases to consider:
Case 1: a. b ≥ 0

( ) = ( )
∴ 3 + 1 = 3 + 1
∴ =

Case 2: a ≥ 0, b < 0

( ) = ( )

∴ 3 + 1 = 2 − 1

2
∴ = 3 ( − 1)

∴ < 0

17

This contradicts the choice a ≥ 0. Therefore, when a ≥ 0, b < 0, f (a) ≠ f (b).

Case 3: a < 0, b ≥ 0
Argue as in case 2, we can see that in this case, f (a) ≠ f (b).

Case 4: a < 0, b < 0

f (a) = f (b)

2a −1 = 2b −1

 a=b
We see that in any case, f (a) = f (b) always implies a = b, so f is one to one.

Q4 Determine if f (x) = sin x − sin x , 0  x  2 , is one to one and give reason to your answer.

Solution:
If 0  x   ,sin x is positive or zero, so sin x = sin x, and f (x) = 0 .
If   x  2 , sin x is negative or zero, so sin x = −sin x, and f (x) = 2sin x .

From the graph, the function is not one-to-one since f ( ) = f (2 ) = 0

Q5 Given that the following functions are not one- to- one. Determine the maximum domain of

each function so that it can have an inverse.
(a) ( ) = 3 + 4 2, ∈ ℝ
(b) ( ) = 2 − 4 + 1, ∈ ℝ

(c) ( ) = 3 + 2 − 2, ∈ ℝ

(d) ( ) = 2( + 1) 2, ∈ ℝ

(e) ( ) = 1 , ∈ ℝ
2+1

Solution:
(a) x ϵ ʀ, x ≥ 0 or x ≤ 0
(b) x ϵ ʀ, x ≤ 2
(c) x ϵ ʀ, x ≤ 1
(d) x ϵ ʀ, x ≥ 0
(e) x ϵ ʀ, x ≥ 0 or x ≤ 0

18

EXERCISE 1.2

Q1 Given the function f (x) = 4 − x .

(a) Find the domain and range of the function.
(b) Determine whether f −1 exists. If it exists, find its expression, domain and range.

Q2 Given the function f (x) = x +1 , x  2 is one to one. Find f −1(x) .
x−2

Q3 A function f is defined by f : x → 2 + 2x −1, x 1.

(a) Find f −1(x) .
(b) Determine its domain and range.

Q4 The function f is defined by f (x) = ex − e−x , x . Find the inverse of f.
ex + e−x

Q5 The function f is defined by f (x) = x2 − 2x + 2, 0  x  1. Sketch the graphs of
y = f (x) and y = f −1(x) on the same axes.

ANSWER 1(a) 4 − x  0  x  4
Section Df = (−, 4]
1.1
4 − x  0  f (x)  0
Exercise Rf = [0,)
1.2
1(b) Let x1, x2  Df , where x1 = x2 , then
4 − 1 = 4 − 2

√4 − 1 = √4 − 2
( 1) = ( 2)

Thus, the function is one to one. Hence, f −1 exists.

Let :
= √4 −
2 = 4 −
= 4 − 2

−1( ) = 4 − 2

∴ −1( ) = 4 − 2

Domain of f −1= Range of f = [0, ) Range of f −1 = Domain of

f = (−, 4]

19

2 Let = +1

−2

− 2 = + 1

= 2 +1

−1

−1 ( ) = 2 + 1
− 1

∴ −1( ) = 2 −1 , ≠ 1

−1

3 Let = −1( )

( ) =

2 + √ − 1 =

− 1 = ( − 2)2
− 1 = 2 − 4 + 4

= 2 − 4 + 5
∴ −1( ) = 2 − 4 + 5

The domain of f −1is the same as the range of f that is

{ | ≥ 2, ∈ ℝ}.
The range of f −1is the same as the domain of f that is

{ | ≥ 1, ∈ ℝ}.

4 Let −1( ) = ⇒ ( ) =
− −
+ − =
2 − 1
2 + 1 =
2 − 1 = 2 +

(1 − ) 2 = 1 +

2 = 1 +
1 −
1 1 +
= 2 (1 − )
1 1 +
∴ −1( ) = 2 (1 − ) , ∈ ℝ, −1

5

20 Learning Outcome:
a) Sketch the graphs of simple functions, including piecewise-defined
1.1(c) GRAPH OF
FUNCTION functions.

GRAPH OF FUNCTION

➢ DEFINITION
The graph of a function f is the set of all points in the plane of the form (x, f(x)). We could also define
the graph of f to be the graph of the equation y = f(x).

➢ CONCEPT > 0 < 0
A. Basic shape of the graph

Type of function
Linear function
( ) = + , ≠ 0

Domain = ℝ Domain = ℝ
Range = ℝ Range = ℝ

Quadratic function
( ) = 2 + + , ≠ 0

Domain = ℝ Domain = ℝ
Range = [ , ∞) Range = (−∞ , ]

Cubic function Domain = ℝ
( ) = 3 + 2 + + , Range = ℝ

≠ 0

Domain = ℝ
Range = ℝ

Reciprocal function

( ) = 1 , ≠ 0
−

Domain = (−∞, ) ∪ ( , ∞) Domain = (−∞, ) ∪ ( , ∞)



Range = (−∞, 0) ∪ (0, ∞) Range = (−∞, 0) ∪ (0, ∞)

21

( ) = , ≠ 0
2

Domain = (−∞, 0) ∪ (0, ∞) Domain = (−∞, 0) ∪ (0, ∞)
Range = (0, ∞) Range = (−∞, 0)

Square root function
( ) = ±√ −

Domain = [ , ∞) Domain = [ , ∞)
Range = [0, ∞) Range = (−∞, 0]

( ) = ±√ −

Domain = (−∞, ] Domain = (−∞, ]
Range = [0, ∞) Range = (−∞, 0]

B. Graphs of modulus functions

A modulus function is a function which gives the absolute value of a number or variable. The outcome
of this function is always positive, no matter what input has been given to the function.

The graphs of = + and = | + | are as below:

For x < − , the graph = | + | is the reflection of the graph = + about x-axis.



22 Quick Notes 1.1(c)

C. Graphs of piecewise functions https://youtu.be/-
A piecewise-defined function is a function defined by P5VUP_NQVM
multiple sub-functions, where each sub-function
applies to a different interval in the domain.

➢ NOTES
Step:
To sketch the graph of the function, we need to perform the following:
1. Determine type of function and corresponding shape.

2. Find y-intercept (point where (0)).

3. Find x-intercepts (points where ( ) = 0).

4. Find what asymptotes does function have, if any.

5. Find stationary points (minimum, maximum and inflection point).

6. Use test to classify stationary points.

7. Add "control" points (some arbitrary points), if needed.

8. Draw "important" and "control" points and connect them by lines
taking into account found behaviour of the function.

9. If function is even, odd or periodic then perform corresponding
reflection.

10. If function is obtained by transforming simpler function, perform
corresponding shift, compressing /stretching.

LLET
DANCE
TOGETHE

R!

23

EXAMPLE 1

Sketch on separate diagram, the graphs of

(a) = 1
−2
(b) = | 1 |
−2
1
(c) 2 = −2

Solution:

(a) = 1
−2

a > 0 shape
The asymptotes are x = 2, y = 0

(b) = | 1 | = 2
= 2
−2 = 2

The asymptotes are x = 2, y = 0

Reflection y negative become y positive

(c) 2 = 1
−2

The asymptotes are x = 2, y = 0

Reflection x < 2 become x > 2

24

EXAMPLE 2 y-
Sketch the graph of = 2 + − 2, noting the points of intersection of the graph with x-axis and

axis.
On a separate diagram, sketch the graph of = |2 + − 2|

Solution:
= 2 + − 2

• a = −1 < 0 shape
• When y = 0 2 + − 2 = 0

(1 + )(2 − ) = 0

= −1 2

• When x = 0, y = 2
Graph of = 2 + − 2 shown in diagram (a)

Diagram (a)

The graph of = |2 + − 2|
can be drawn by reflecting the section
of graph that is below the x-axis
as shown in diagram (b).

Diagram (b)

EXAMPLE 3 Quick Notes

Sketch the graphs of Note:
● indicates this point
( ) = {−2 2++32, < 1 is included
, ≥ 1
o indicates this point
Solution: is excluded

• For the region x < 1,
✓ we have a straight line with slope 2 and y-intercept 3.
✓ As x approaches 1, the value of the function approaches 5
(but does not reach it because of the “ < ” sign)

25

• For the region ≥ 1
✓ When x =1,
the function has value (1) = −(1)2 + 2 = 1
✓ As we go further to the right,
the function takes value based on ( ) = − 2 + 2
✓ It is a parabola.

WORKSHEET 1.3 . ( ) = √ − 2

Sketch the graph of the following: Solution:
. ( ) = 3 2 − 2 − 1 • shape
• − 2 ≥ 0
Solution: ≥ 2
• a = 3 > 0 shape • When y = 0, = 2
• When y = 0 3 2 − 2 − 1 = 0 • When ≥ 0, ≥ 2
(3 + 1)( − 1) = 0
= − 1 or = 2

3

• When x = 0, = −1

26

3(a) ( ) = | + 1| 3(b) With the aid of graph of ( ) in 3(a),
sketch the graph of 2 ( ).
Solution:
Solution:
( ) = | + 1| = {− ( ++11,), > −1 y value of graph 3(a) multiple by 2
≤ −1

3(c) With the aid of graph of ( ) in 3(a), 3(d) With the aid of graph of ( ) in 3(a),
sketch the graph of ( + 2). sketch the graph of ( ) − 2.

Solution: Solution:
The graph of 3(a) transform 2 step to left The graph of 3(a) transform 2 step downward

4(a) = 4(b) 2 =

−2 −2

Solution: Solution:

shape The asymptotes are = 2, = ±1
The asymptotes are x = 2, y = 1
lim = +∞ Since 2 ≥ 0 , hence −2 ≥ 0

→2+ ≤ 0 , ≥ 2

lim = −∞ The curve is symmetrical about the x-axis.

→2−

= 1 = 1
= −1

= 2 = 2

27

2 , < 0
5. ( ) = { , 0 ≤ < 2

−2 , ≥ 2

Solution:
• For the region < 0
✓ When → 0 , → 0
✓ As we go further to the left,
the function takes value based on ( ) = 2
✓ It is a parabola.
• For the region 0 ≤ < 2,
✓ we have a straight line with slope 1.
✓ As → 2 , → 2
• For the region ≥ 2
✓ It is a horizontal line where ( ) = −2

EXERCISE 1.3

1. Sketch the graph of (b) = −3 + 4
(a) = 2 − 3

2. Sketch the graph of

(a) = 2 − 4 (c) 2 = 2 − 4

(b) = 1
2−4

3. Given ( ) = − 1 3 + 3 + 1. Sketch the following graphs separately
2 2

(a) = ( ) (c) = −3 ( )

(b) = ( + 1) (d) = (2 )

4. Sketch in separate diagrams, the graphs of

(a) = +1 (c) 2 = +1

−2 −2
= | +1|
(b)
−2

5. Sketch the graph of (c) = −| + 2| + 3
(a) = | |
(b) = 2| − 1| − 4

28

6. The function f is defined by ( ) = {−32 +−28, , < −2
Sketch the graph of f. > −2

7. The function f is defined by ( ) = {5 −− 1 2,, ≤ 2
Sketch the graph of f . > 2

8. The function f is defined on the domain [-3 , 3] as follows:

( ) = {−11 , −3 ≤ < 0
, 0 ≤ <3

Sketch the graph of f.

9. The function f is defined by
sin , < −2


( ) = { 2 − 2 , −2 ≤ < 2
2 − 8 + 10 , ≥ 2

Sketch the graph of f.

ANSWER

Section 1(a) = 2 − 3

1.1 • Linear graph

Exercise • Gradient = 2
1.3 • y-intercept = −3

(b) = −3 + 4
• Linear graph
• Gradient = 2
• y-intercept = −3

29

2(a) = 2 − 4

• a = 1 > 0 shape
• When y = 0 2 − 4 = 0

( − 4) = 0
= 0 or 4

• When x = 0, y = 0
• = 2 − 4

= ( − 2)2 − 4

The graph has minimum point
at (2 , ̶ 4)

2(b) = 1 = 1
2−4 ( −4)

• Since ( − 4) ≠ 0,
The asymptotes are
x = 0, x = 4 and y = 0

2(c) 2 = 2 − 4 = 4
Since 2 ≥ 0 = 4
Therefore 2 − 4 ≥ 0
( − 4) ≥ 0
≤ 0, ≥ 4

The curve is symmetrical about

the x-axis since the multiples of y

are even.

3(a) ( ) = − 1 3 + 3 + 1

22

• a = − 1 < 0 shape
2

• When f (x) = 0,

− 1 3 + 3 + 1 = 0

22

= −1 or 2

• When x = 0, y = 1

• ′( ) = − 3 2 + 3

22

When ′( ) = 0 ,

− 3 2 + 3 =0
2 2
= −1 or 1

30

When = −1 , ( ) = 0
When = 1 , ( ) = 2

Minimum point at (−1, 0)
Maximum point at (1, 2)

3(b) = ( + 1)
Graph 3(a) move 1 step to left

3(c) = −3 ( )
Stretch Graph 3(a) with
factor scale 3, then
reflect on x-axis

3(d) = (2 )
Compress Graph 3(a) with
scale factor 2.
In direction parallel with x-axis.

4(a) = +1

−2

a > 0 shape

The asymptotes are x = 2, y = 1

When y = 0, = −1

When x = 0, = −



31

= 1

= 2

4(b) + 1
= | − 2|
| +1|
= ≥ 0 for all values of x
−2

= 1

= 2

4(c) 2 = +1

−2

Since 2 ≥ 0 , hence +1 ≥ 0

−2

≤ −1 , ≥ 2

= 1
= −1

= 2

5(a) = | |

• Sketch line y = x

• For x < − ,


the graph is the reflection

about x-axis

32

5(b) = 2| − 1| − 4
• Sketch line = 2( − 1)
• for x < 1 ,
the graph is the reflection
about x-axis
• the graph move down 4 steps

5(c) = −| + 2| + 3
• Sketch line = −( + 2)
• for x < −1 ,
the graph is the reflection
about x-axis
• the graph move up 3 steps

6 ( ) = {−32 +−28, , < −2
> −2

• For the region x < −2, use ( ) = −2 − 8

✓ we have a straight line with slope −2.

✓ As x approaches −2, the value of the function approaches −4

(but does not reach it because of the “ < ” sign)

• For the region > −2, use ( ) = 3 + 2
✓ we have a straight line with slope 3.

✓ As x approaches −2, the value of the function approaches −4
(but does not reach it because of the “ > ” sign)

33

7 ( ) = {5 −− 1 2,, ≤2
>2

• For the region ≤ 2

✓ When x =2, the function has value (2) = 5 − (2)2 = 1

✓ As we go further to the left, the function takes value based on

( ) = 5 − 2

✓ It is a parabola with maximum point (0, 5).

• For the region x > 2,
✓ we have a straight line with slope 1.
✓ As x approaches 2, the value of the function approaches 1
(but does not reach it because of the “ > ” sign)

8 ( ) = {−11 , −3 ≤ < 0
, 0 ≤ <3

• For the region −3 ≤ < 0
✓ It is a horizontal line y = −1 including x = −3 and excluding

x=0

• For the region 0 ≤ < 3
✓ It is a horizontal line y = 1 including x = 0 and excluding

x=3

34

9 sin , < −2
2 −
( ) = { , −2 ≤ < 2
2

2 − 8 + 10 , ≥ 2

• For the region x < −2,
✓ we have a sin x curve with maximum point(−1.5 , 1), … and

minimum points (− , −1) (−2 , −1),…
✓ As x approaches -2, the value of the function approaches -

0.9093 (but does not reach it because of the “ < ” sign)

• For the region −2 ≤ < 2,

✓ we have a straight line with slope − 1 and y-intercept 2.
2
(−2)
✓ when x = −2, the function has value (−2) = 2 − 2 = 3

✓ As x approaches 2, the value of the function approaches 1

(but does not reach it because of the “ < ” sign)

• For the region ≥ 2
✓ When x = 2, the function has value (2) = 22 − 8(2) +

10 = −2
✓ As we go further to the right, the function takes value based

on ( ) = 2 − 8 + 10
✓ It is a parabola.

CLONE STPM

Q1 The function f and g are defined by ( ) = 2 ln , > 0 and ( ) = √ − 1, >1
(a) Sketch the graphs of f and g.
(b) Find −1 and −1, stating the domains and ranges of −1 and −1.
(c) Determine whether ∘ exist.

Q2 Functions f and g are defined by f(x) = ex+2 and (g ∘ f )(x) = x , for all x ≥ 0.
(a) Find the function g, and state its domain.
(b) Determine the value of (f ∘ g)(e3).

Q3 (a) The function is defined as: ℎ( ) = | + 3| − | − 3|, ∈ .
(i) Sketch the graph of h.
(ii) State the range of h.

35

(b) The functions f and g are defined as:

( ) = +| | , ∈
2
− 1, < 0,
( ) = { 2, > 0.

Sketch the graphs of the function f and g.

From the graphs, determine the continuity of the composite function f ∘ g and hence discuss

whether the composite function f ∘ g is defined.

Q4 The function is defined by ( ) → 2 − 3 , ∈ ℝ.
(a) Sketch the graph of g(x) . Hence, show that g(x) is not a one-to-one function.

(b) Determine the set of values of domain g, Dg , for which g(x) is a one-to-one function.

Q5 The function is defined by : → √3 + 1, ∈ ℝ, ≥ − 1. Find f −1 and state its domain and

3

range.

Q6 The function is defined as follows:
: → 4 + ( − 1)2, ∈ ℝ

(a) Sketch the graph of f .

(b) State the range of f .

(c) Determine if f −1 exist.

Q7 The functions f is defined by

( ) = 4 − 2 , − 3 ≤ < −1
|2 − 1|, > −1

(a) Find lim ( ).

→ −1

(b) Determine whether ( ) is continous at = −1. Give your reason.

(c) Sketch the graph of ( ).

Q8 A functions f is defined by : → √ + 1, ∈ [−1, ∞)
(a) Define −1 in the similar form and state its domain and range.
(b) Sketch the graph of f and −1 on the same axes.

ANSWER 1(a) (i) (ii) y=g(x)
Section y y
1.1
y=f(x)
Clone
STPM

01 x

01 x

36

1(b) (i) Let = −1( )
( ) =
2 ln =



−1( ) = 2
−1 = { : ∈ ℜ}
−1 = { : ∈ ℜ, > 1}

(ii) Let = −1( )
( ) =

√ − 1 =
−1( ) = 2 + 1

−1 = { : ∈ ℜ, > 0}
−1 = { : ∈ ℜ, > 1}

1(c) = { : ∈ ℛ} ; = { : > 1}

1(d) Since ⊈ , g ∘ f does not exist.

2(a) Let f (x) = u

ex+2 = u

x + 2 = ln u
x = ln u − 2

g[f (x)] = x
g(u) = ln u − 2
g(x) = ln x − 2
g :x → ln x − 2 , x  e2

2(b) fg(e3) = f ( ln e3 − 2)

= f ( 3−2)

= f (1)
= e1+2
= e3

3(a) (i) ℎ( ) = | + 3| − | − 3|, ∈

| + 3| = {− +−33, , ≥<−−33

| − 3| = {− −+33, , ≥<33

For < −3,
ℎ( ) = (− − 3) − (− + 3)

37

= −6

For −3 ≤ <3,
ℎ( ) = ( + 3) − (− + 3)

= 2

For ≥ 3,
ℎ( ) = ( + 3) − ( − 3)

=6

−6, < −3
Thus ℎ( ) = { 2 , − 3 ≤ < 3

6, ≥ 3

Sketch graph of function h is

(ii) Range of h is { |−6 ≤ ≤ 6}

3(b) ( ) = +| | , ∈
2
+(− )
For < 0, ( ) = 2 = 0

For > 0, ( ) = +( ) =

2

Therefore, ( ) = {0 ,, < 0
≥ 0

g( ) = { − 1, < 0
2, > 0

38
Sketch of the graph of function f:

Sketch of the graph of function g:

Since there is a break in the graph of g at x = 0, then f ∘ g is
discontinuous (@not continuous)
The range of g is the subset of f , then the composite function f ∘ g is
defined.

4 (a)

( ) = 2 − 3
= ( − 3)

Since g(0) = 0 and g(3) = 0, the function g is not one to one.

39

4(b) ( ) = 2 − 3
32 9

= ( − 2) − 4

 Domain :Dg : −  Dg  3 or Dg : 3  Dg  
2  2

5 Let

= −1( ), ( ) =

√3 + 1 =

3 + 1 = 2

2 − 1
= 3
2
∴ −1( ) = − 1
3

Since f (x) = 3x +1 0 , we can see that f −1(x)  [0,) .

Domain f −1(x) = [0, ) and range of f −1(x) = [− 1 , )

3

6(a)

6(b) From the graph, f(1) = 4 is the minimum then range of f is [4, ) .

6(c) Inverse of f does not exist since f is not a one-to-one function as can be
seen from the graph.

7(a) lim ( ) = lim (4 − 2)
→−1− →1−

=3

lim ( ) = lim |2 − 1|
→−1+ →1−

=3

∴ lim ( ) = 3

→−1−

40
7(b) lim ( ) exist but (−1) is not defined, thus ( ) is not continuous at

→−1−

x = −1

7(c)

8(a) (i) : → √ + 1, ∈ [−1, ∞)
Let, = −1( ), ↔ = ( )
= √ + 1
2 = + 1
= 2 − 1

−1( ) = 2 − 1
∴ −1( ) = 2 − 1
Domain of −1 is [0, ∞) or { : ≥ 0} and range is
[−1, ∞)or { : ≥ 1}

8(b)
y=x

f
1

−1

-1 1

-1

41

1.2(a) Learning Outcome:
POLYNOMIAL a) use the factor theorem and the remainder theorem
FUNCTIONS b) solve polynomial and rational equations.

POLYNOMIAL FUNCTION
1. A function of x in the form

( ) ≡ + 1 −1 + ⋯ + − + ⋯ + −1 +
where ∈ ℝ and ≠ 0, ∈ ℤ+ is called a polynomial function of degree n.

2. Polynomials of degrees 1, 2, 3 and 4 are also known as linear, quadratic, cubic and quartic
functions respectively.

3. The value of a polynomial, ( ), when = is written as ( ).
For example,
If ( ) = 3 2 + 4 − 1
Then when = 1, (1) = 3(1)2 + 4(1) − 1 = 6
And when = 2, (2) = 3(2)2 + 4(2) − 1 = 19

4. A number ∈ ℝ is the zeros of ( )
Example: If ( ) = ( − 2)( + 1)(2 − 1), then zeros of ( ) are 2, −1, 1 .

2

5. When polynomial ( ) is divided by ( − ), then ( ) = ( − ) ( ) + , where
( ) = , =

EXAMPLE 1
Determine the quotient and remainder when 4 − 2 3 + 6 − 5 is divided by ( 2 − − 1).

Solution:
Using long division method,

2 −

2 − − 1 4 − 2 3 + 6 − 5

(−) 4 − 3 − 2

− 3 + 2 + 6

(−) − 3 + 2 +

5 − 5

Hence 4 − 2 3 + 6 − 5 = ( 2 − − 1)( 2 − ) + 7 + 5
( ) = 2 − and = 5 + 5

THE REMAINDER THEOREM

When the polynomial ( ) with degree n is divided by (x - a), the quotient, ( ), is a polynomial of

degree (n - 1), and the remainder R is a constant,

( ) = ( ) +
− −

42

By substituting = , we see that ( ) =

( ) ≡ ( )( − ) +

When a polynomial ( ) is divided by ( − ), the remainder is ( )

EXAMPLE 2
Find the remainder when a polynomial ( ) = 3 5 − 2 3 + 7 is divided by (x ̶ 2).

Solution:
(2) = 3(2)5 − 2(2)3 + 7

= 87

EXAMPLE 3
Find the value of a and b when a polynomial ( ) = 3 − 6 2 + is divided by

( − 1)( + 2) , its remainder is 8x ̶ 16.

Solution:
3 − 6 2 + ≡ ( − 1)( + 2) ( ) + (8 − 16)

When = 1, − 6 + = 8 − 16
+ = −2 -----(1)

When = −2, − 8 − 24 − 2 = 8(−2) − 16
8 + 2 = 8
4 + = 4 -----(2)

(2) − (1), 3 = 6
= 2
∴ = −4

THE FACTOR THEOREM
From the remainder theorem, we have shown that when a polynomial ( ) is divided by (x - a),
its remainder, R is P(a). On the other hand, if (x - a) is a factor of ( ), then its remainder is zero,
R = P(a) = 0.

For a polynomial P(x), (x - a) is a factor of ( ) if and only if ( ) = 0

For a polynomial ( ), ( − ) is a factor of ( ) if and only if ( ) = 0



EXAMPLE 4
Given that( − 1) and (3 + 1) are factors of polynomial ( ) = 3 3 − 2 + + 2 , find the value

of a and b.

43

Solution:
Given that ( − 1) is a factor of ( ),

By the factor theorem,
(1) = 0

3 − + + 2 = 0
− + = −5 ------- (1)

(− 1) = 0

3

13 12 1
3 (− 3) − (− 3) + (− 3) + 2 = 0

− 1 − − + 2 = 0

993

− − 3 = −17 ------- (2)

(1) − (2), 4 = 12
∴ = 3
∴ = 8

EXAMPLE 5
Given that ( − 3) is a factor of ( ) = 3 + 2 − − . When the polynomial is divided

by( − 1), the remainder is 24. Find the value of p and q.

Solution:
Given that ( − 3) is a factor of ( ),

By the factor theorem,
(3) = 0

27 + 9 − 3 − = 0
9 − = −24 ------- (1)

Given that when ( ) is divided by ( − 1) the remainder is 24.
By the remainder theorem,

(1) = 24
1 + − 1 − = 24

− = 24 ------ (2)

(1) − (2), 8 = −48
∴ = −6
∴ = −30

WORKSHEET 1.4

Answer the following question. Q2 Find the value of a if the remainder 18
Q1 When the polynomial ( ) is divided by
when a polynomial
( + 1), its remainder is 3, and the ( ) = 2 3 + 2 − 5 + 6 is divided by
remainder is 1 when it is divided by ( − 1). ( − 4).

Find the remainder of the polynomial when
It is divided by 2 − 1.

44

Solution: Solution:
( ) = ( 2 − 1) ( ) + + ( ) = 2 3 + 2 − 5 + 6
( ) = ( − 1)( + 1) ( ) + +
(4) = 18
when = −1, (−1) = 3 2(4)3 + (4)2 − 5(4) + 6 = 18
− + = 3 -----(1)
16 = −96

= −6

when = 1, (1) = 1
+ = 1 -----(2)

(1) + (2), 2 = 4
= 2
= −1

Hence, the remainder is − + 2

Q3 Given the function Q4 Given that (x + 2) is a factor of the polynomial
f(x) = (2x - l)(x - 2)(x + 3). ( ) = 2 3 − 2 + − 6. Determine the
value of a. Factorise ( ) completely.
Find the value of constant h such that (x + 2)
is a factor of f(x) + hx. Solution:
( ) = 2 3 − 2 + − 6
Solution:
Given ( ) = (2 − 1)( − 2)( + 3) (−2) = 0
Let ( ) = ( ) + ℎ 2(−2)3 − (−2)2 + (−2) − 6 = 0

= (2 − 1)( − 2)( + 3) + ℎ 2 = −26
= −13
Given ( + 2) is a factor of ( ),
∴ (−2) = 0 ( ) = 2 3 − 2 − 13 − 6
= ( + 2)(2 2 − 5 − 3)
(2(−2) − 1)(−2 − 2)(−2 + 3) + ℎ(−2) = 0 = ( + 2)(2 + 1)( − 3)
ℎ = 10

EXERCISE 1.4

Q1 The polynomial 4 − 5 3 + 2 + 7 + 1 leaves a remainder of −8 when it is divided by

( − 1), and a remainder of 11 when divided by (2 + 1). Determine the values of a and b.
2

Q2 Given ( ) = 6 4 + 5 3 + 2 + 4, where is a constant, find the value of given (2 + 1)
is a factor of ( ). Assuming this value of , factorise ( ) completely and solve the equation
( ) = 0.

Q3 The polynomial 2 3 − 3 2 + + has the factor ( − 1) and when divided by + 2,

it results in remainder −54. Find a and b. Then factorise this polynomial.
Hence, find all the factors of 4 3 + 3 2 − 9 + 2.

45

Q4 Find all the real values of a and b that satisfy the equation
4 − 10 2 + 1 = ( 2 + + 1)( 2 + + 1)

Hence, determine all the real roots of the equation
4 − 10 2 + 1 = 0

Q5 If ( ) = 4 + 2 3 + 5 2 − 16 − 20, show that ( ) can be expressed in the form
( 2 + + )2 − 4( + )2, where a and b are constants to be determined. Hence, find both the
real roots of the equation ( ) = 0.

ANSWER 1 4 − 5 3 + 2 + 7 + 1 = ( − 1) ( ) − 8
Section 4 − 5 3 + 2 + 7 + 1 = (2 + 1) ( ) + 11
1.2
2
Exercise
1.4 When = 1,
− 5 + + 7 + 1 = −8

+ = −11 − − − (1)

When = − 1,

2
1 4 1 3 1 2 1 11
(− 2) − 5 (− 2) + (− 2) + 7 (− 2) + 1 = 2
1 + 5 + 1 − 7 + 1 = 11
16 8 4 2 2
1 + 1 = 59
16 4 8

+ 4 = 118 − − − (2)

(2) − (1), 3 = 129
∴ = 43
∴ = −54

2 1
(− 2) = 0
6 (− 1)4 + 5 (− 1)3 + (− 1)2 + 4 = 0

222

6 − 10 + 4 + 64 = 0

= −15

( ) = 6 4 + 5 3 − 15 2 + 4
( ) = (2 + 1)(3 3 + 2 − 8 + 4)
( ) = (2 + 1)( − 1)(3 2 + 4 − 4)

( ) = (2 + 1)( − 1)(3 − 2)( + 2)

When ( ) = 0, = − 1 , 1, 2 , −2

23