Mathematics

PENERBIT
ILMU
BAKTI SDN. BHD.

Contents

PENERBITChapter 1 Number System
ILMU
BAKTI SDN. BHD.1.1 Real Numbers 1

1.2 Complex Numbers 5

Summative Practice 1 12

Answers 13

Chapter 2 Equations, Inequalities and Absolute Values

2.1 Equations 14

2.2 Inequalities 27

2.3 Absolute Values 30

Summative Practice 2 35

Answers 36

Chapter 3 Sequences and Series

3.1 Sequences and Series 37

3.2 Binomial Expansion 45

Summative Practice 3 51

Answers 52

Chapter 4 Matrices and Systems of Linear Equations

4.1 Matrices 53

4.2 Determinants of Matrices 60

4.3 Inverse of a Matrix 69

4.4 System of Linear Equations up to Three Variables 77

Summative Practice 4 90

Answers 92

Chapter 5 Functions and Graphs

5.1 Functions 94

5.2 Composite Functions 108

5.3 Inverse Functions 111

5.4 Exponential and Logarithmic Functions 116

5.5 Trigonometric Functions 121

Summative Practice 5 129

Answers 130

Chapter 6 Polynomials

6.1 Polynomials 133

6.2 Remainder Theorem, Factor Theorem and Zeroes of Polynomials 138

6.3 Partial Fractions 142
PENERBIT
ILMUSummative Practice 6 151
BAKTI SDN. BHD.
Answers 152

Chapter 7 Trigonometric Functions

7.1 Trigonometric Ratios and Identities 153

7.2 Compound Angles 157

7.3 Solution of Trigonometric Equations 161

Summative Practice 7 174

Answers 175

Chapter 8 Limits and Continuity

8.1 Limits 178
8.2 Asymptotes 188
8.3 Continuity 196
Summative Practice 8 199
Answers 201

Chapter 9 Differentiation

9.1 Derivative of a Function 202

9.2 Rules of Differentiation 206

9.3 Differentiation of Exponential, Logarithmic and Trigonometric Functions 210

9.4 Implicit Differentiation 214

9.5 Parametric Differentiation 220

Summative Practice 9 223

Answers 224

Chapter 10 Applications of Differentiation

10.1 Extremum Problems 227

10.2 Rate of Change 244

Summative Practice 10 248

Answers 249

Semester Examination 250
Answers 252

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PENERBIT1CHAPTER
ILMU
BAKTI SDN. BHD.Number System

1.1 Real Numbers

 Learning Outcomes

● nDuemfinbeernsa, truartaiol nnaulmnbuemrsbe(Nrs),(Qw)haonlednirurmatbioenrasl(nWu)m, binetersge(Q–rs).(Z), prime

● Represent rational and irrational numbers in decimal form.

● Represent the relationship of number sets in a rQea∪l nQ–um=bRe.r system
diagrammatically showing N ⊂ W ⊂ Z ⊂Q and

● Represent open, closed and half-open intervals and their representations

on the number line.

● Find union, ∪, and intersection, ∩, of two or more intervals with the aid of a

number line.

Classification of Numbers

 1 Numbers that are used for counting objects are known as natural numbers, N.
N = {1, 2, 3, …}

 2 Natural numbers together with zero are known as whole numbers, W.
W = {0, 1, 2, 3, …}

 3 The positive and negative numbers together with zero are known as integers, Z.
Z = {…, −3, −2, −1, 0, 1, 2, 3, …}

(a) Positive integers, Z+ = {1, 2, 3, …}

(b) Negative integers, Z− = {−1, −2, −3, …}

 4 Natural numbers that are only divisible by the number itself and 1 are known as

prime numbers.

Prime numbers = {2, 3, 5, 7, …}

 5 Numbers that can be written in the form of p are known as rational numbers, Q.
q
  Q = x : x = p , p, q ∈ Z, q ≠ 0
q 1

 Number System

Alternative (b) (3 + 3i)(a + bi) = 3 + 18i 1 + 2 : 2a = 7
Method 3a + 3bi + 3ai + 3bi2 = 3 + 18i
3a − 3b + (3a + 3b)i = 3 + 18i a = 7
3 + 18i 2
a + bi = 3 + 3i Equating the real and the
imaginary parts, 7
=33++138ii 3 – 3i Substitute a = 2 into 1,
3 – 3i 3a − 3b = 3
× a − b = 1......... 1 BAKTI SDN. BHD.  7 
2
=9 + 945–i –5 4i2 3a + 3b = 18 − b = 1
9i2 a + b = 6......... 2
+ 45i b = 5
a + bi = 63 18 2

=27 + 52 i ∴a= 7 , b = 5
2 2

Chapter 1 ∴a= 7 , b = 5
2 2

Example 5

Given z = 2 + i , find z‾ in the form a + bi.
3 − i

Solution

z = 2 + i × 3 + i
3 − i 3 + i

= 6 + 2i + 3i + i2
9 − i2

= 5 + 5i
10

= 1 + 1 i
2 2
ILMU
∴ z‾ = 1 − 1 i
2 2

Example 6

Given z = 4 + 2i, express each of the following in the form a + bi.
PENERBIT
(a) (z + 1)(z − 1) (b) z
z–1

Alternative Solution
Method
(a) (z + 1)(z − 1) (b) z z 1
(z + 1)(z − 1) = [(4 + 2i) + 1][(4 + 2i) − 1] –
= z2 − 1 = (5 + 2i)(3 + 2i)
= (4 + 2i)2 − 1 = 15 + 10i + 6i + 4i2 = (4 4 + 2i 1
= 16 + 16i + 4i2 − 1 = 11 + 16i + 2i) −
= 15 + 16i − 4
= 11 + 16i = 4 + 2i × 3 − 2i
3 + 2i 3 − 2i
8
= 12 − 8i + 6i − 4i2
9 − 4i2

= 16 − 2i
13

= 16 − 123i
13

Summative Practice 1

 1 (a) Express z = − 5 − 5 i in polar form. BAKTI SDN. BHD.

(b) Write zz‾ − 2i in the form a + bi, where z = 1 − 3i and z‾ is the conjugate of z.
1+i

Chapter 1  2 It is given that z = 3i .
5 +i

(a) Express z in the form a + bi, where a and b ∈ R.
(b) Determine the modulus and argument of z.

  3 Given (α + βi) = (5 + −9 − i3), determine the values of α and β if α ≠ β.
2i
EXAM

CLONE

 4 The complex number, z = a + bi satisfies the equation z2 = 6 + 8i.
(a) Find all the possible values of z.
(b) Hence, express z in polar form.

  5 The complex number z is given by z = 2 − i.

EXAM (a) Write z − 1 in the form a + bi, where a and b are real numbers.
CLONE z

 (b) Determine z − 1 . Hence, find the values of the real numbers p and q if
z
z z 1 2
p − qi = − 1 − ILMU
z z .

  6 Given z1 = 1 + 2i and z2 = 3 − 3i, express z2 + i3 in Cartesian form.
z‾1 −z2

  7 The complex numbers z1 and z2 are given by z1 = −i and z2 = 3 + i 2 .
EXAM
CLONE (a) Express z12 and z‾2 in the form a + bi, where a, b ∈ R.

PENERBIT (b) From 7(a), find w = z12 + z‾2 . Hence, find |w| and arg w.
z1

 8 It is given that z1 = 1 − 3i and z2 = 2 + i. Without using a calculator, find z3 = z‾1 . Hence,
evaluate |z3|. z2

 9 Simplify ([3, 8) ∩ (−5, 7)) ∪ {2, 6 ,11}.

10 Find the values of a and b that satisfy the equation 1 + ai = 1 − 3i .
b−i

12

Answers

PENERBIT Quick Check 1.1  6 (a) (0, 4) Summative Practice 1
ILMU (b) [−2, 3)
BAKTI SDN. BHD. (c) [−5, 0)    1 (a) z = 10
 1 (a) True (d) [0, ∞) cos − 3 π +
4
(b) False  7 (a) [−5, 9) i sin − 34 π
(b) [−2, 3)
(c) True (c) [−8, 9)
(d) [−2, 0]
(d) False (b) 4 − 6i

(e) True  2 (a) 1 + 5  i
2 2
(f) False

 2 (a) {4, sin 90°} (b) |z| = 3
2
(b) {0, 4, sin 90°} Quick Check 1.2

(c) {−7, 0, 4, sin 90°}  1 (a) 3i arg z = 1.1503

(d) {−7, −0.36, 0, 1 , sin 90°, 4} (b) 21i  3 α = −8, β = 10
(e) {− 5 , e, π} 2
(c) −42  4 (a) z = 2 2 + 2 i,

(f) {−7, − 5, −0.36, 0, e, 4, 1 , (d) −3 5 z = −2 2 − 2 i
2
π, sin 90°}  2 (a) 7 + 11i (b) z = 10 [cos (0.4636) +
i sin (0.4636)] and
(b) 4 − i
z = 10 [cos (−2.6779) +
(g) { } (c) 26 + 7i i sin (−2.6779)]
 3 (a) (−∞, −2]
 3 (a) 1 8 6
(b) (8, ∞) 5 5
(c) (−5, −1) (b) i  5 (a) − i
(d) (0, 2]
(e) [−4, 5) (c) −1 (b) p = 5256 , q = 192
 4 (a) x  3 25
(b) 5  x  8 (d) 3
(c) x  9
(d) −4  x  3  4 (a) 6 + 3  i  6 49 − 23 i
(e) 6  x  10 5 5 30 30
 5 (a)
(b) 21 + 20  i  7 (a) z12 = −1
29 29 zˉ2 = 3 − i 2

(c) 14 + 7  i (b) w = 2 + 2i
5 5
|w| = 6
3 4 Answers
(d) 25 + 25  i arg w = 0.9553

6  5 (a) a = −1, b = 2  8 z3 = 1 + i
(b) |z3| = 2
(b) a = 17, b = −7
 9 [3, 7) ∪ {2, 11}
(c) a= 3 , b = 8
8 5 5 10 a = 2, b = −1 and
(c) a = −1, b = 2
 6 zˉ = 2 − 3  i
13 13
0 6
(d) 6 12
 7 5 − 5  i

−2 2

13

PENERBIT
ILMU
BAKTI SDN. BHD.
2CHAPTER

Equations, Inequalities and
Absolute Values

2.1 Equations

 Learning Outcomes

● Express the rules of indices.
● Explain the meaning of a surd and its conjugate.
● Perform algebraic operations on surds.
● Express the laws of logarithms.
● Change the base of logarithms.
● Find the equations involving surds, indices and logarithms.

Rules of Indices

 1 A number that has been raised to a power is known as an index number. The
power, also known as the index, indicates how many times the number must be
multiplied by itself.

an = a × a × a × … × a
        n factors

 2 An index number is written as

Base an Index

where a is a real number and n is a positive integer.
 3 an is read as ‘a to the power of n’. For example, the number 56 is read as

‘5 to the power of 6’.

14

Equations, Inequalities and Absolute Values 

6  Common Error
–
(b) x 1 x Students conduct cross

6 − x  0 Rearrange the inequality so that the right multiply in inequalities.
– hand side is zero
x 1 Find the common denominator x 6  x
−
x –6 1 − x(x − 1)  0 Factorise to simplify 1
x –1 6  x(x − 1)
PENERBIT 6  x2 − x
ILMU6−x2 + x x2 − x − 6  0
BAKTI SDN. BHD. x–1 0
(x − 3)(x + 2)  0
−x2 + x + 6 x = 3 or x = −2
x–1  0

(−x + 3)(x + 2)  0
x–1
Chapter 2
x = 3, x = −2 or x = 1

Table of signs:

(−∞, −2) (−2, 1) (1, 3) (3, ∞)

(x + 2) −+++

(x − 1) −−++

(−x + 3) +++−

(−x + 3)(x + 2) + − + −
x–1

Since (−x + 3)(x + 2)  0, where the inequality sign is , the “+” sign is
x–1
chosen for final answer.

∴ The solution set is {x: x  −2 ∪ 1  x  3}.

Quick Check 2.2

 1 Solve each of the following inequalities.

(a) 3x + 6  18 (c) 3x  x +2
4 4
(b) 5 − 9x  2x + 6 (d) x + 7  2x + 5  7x + 3

 2 Solve each of the following inequalities.

(a) (x − 5)(x + 3)  0 (c) −x2 + 3x + 4  0

(b) 3(x2 − 1)  6x

 3 Solve each of the following inequalities.

(a) x − 7  0 (c) x + 2  8
x + 3 x

(b) x2 − x − 6  0 (d) x 1 2  x 2 3
x + −

 4 Solve each of the following inequalities.

(a) −2  3 − 5m  1 (c) 3  5 − 4x
4 3 1−x
(b) −6x2 − 8x − 2  0

 5 It is given that ℎ(x) = mx2 − 2x + m. Find the range of values of m so that ℎ(x) is
always positive.

29

Equations, Inequalities and Absolute Values 

6  Common Error
–
(b) x 1 x Students conduct cross

6 − x  0 Rearrange the inequality so that the right multiply in inequalities.
– hand side is zero
x 1 Find the common denominator x 6 1  x
−
x –6 1 − x(x − 1)  0 Factorise to simplify 6  x(x − 1)
x –1
PENERBIT 6  x2 − x
ILMU6 − x2 + x x2 − x − 6  0
BAKTI SDN. BHD.x–1  0
(x − 3)(x + 2)  0
−x2 + x + 6 x = 3 or x = −2
x–1  0

(−x + 3)(x + 2)  0
x–1
Chapter 2
x = 3, x = −2 or x = 1

Table of signs:

(−∞, −2) (−2, 1) (1, 3) (3, ∞)

(x + 2) −+++

(x − 1) −−++

(−x + 3) +++−

(−x + 3)(x + 2) + − + −
x–1

Since (−x + 3)(x + 2)  0, where the inequality sign is , the “+” sign is
x–1
chosen for final answer.

∴ The solution set is {x: x  −2 ∪ 1  x  3}.

Quick Check 2.2

  1 Solve each of the following inequalities.

(a) 3x + 6  18 (c) 3x  x +2
4 4
(b) 5 − 9x  2x + 6 (d) x + 7  2x + 5  7x + 3

  2 Solve each of the following inequalities.

(a) (x − 5)(x + 3)  0 (c) −x2 + 3x + 4  0

(b) 3(x2 − 1)  6x

  3 Solve each of the following inequalities.

(a) x − 7  0 (c) x + 2  8
x + 3 x

(b) x2 − x − 6  0 (d) x 1 2  x 2 3
x + −

  4 Solve each of the following inequalities.

(a) −2  3 − 5m  1 (c) 3  5 − 4x
4 3 1−x
(b) −6x2 − 8x − 2  0

  5 It is given that ℎ(x) = mx2 − 2x + m. Find the range of values of m so that ℎ(x) is
always positive.

29

PENERBIT3CHAPTER
ILMU
BAKTI SDN. BHD.Sequences and Series

3.1 Sequences and Series

 Learning Outcomes

● Write nth term of simple sequences and series. (n 1)d
Fusinedththeesnutmh teformrmouflaa,riSthnm=e2ntic[2saeq+u(enn−ce1)adn]dorseSrnie=s,2nT(na=+al
● + − and
).

● Find the nth term of geometric sequences and series, Tn = arn − 1 and use the
sum formula, Sn =
a(1 − r n) for r ≠ 1.
1−r
a
● Find the sum to infinity, S∞ = 1 − r , |r|  1.

 1 A sequence is a set of numbers occurring in a definite order. The numbers are
arranged according to a particular rule.

 2 A finite sequence has a fixed number of terms, meanwhile an infinite sequence
has an infinite number of terms.

For example, 2, 4, 6, 8 is a finite sequence and 2, 4, 6, 8 … is an infinite sequence.

 3 Each number in a sequence is known as a term and is represented by Tn. From
the above example, 2 is the first term (T1), 4 is the second term (T2) and so forth.

2, 4, 6, 8, …
↓↓↓↓
T1, T2, T3, T4, …

 4 A series is the sum of the terms of a sequence. The sum of a fixed number of
terms is known as a finite series, meanwhile the sum of an infinite number of
terms is known as an infinite series.

For example, 2 + 4 + 6 + 8 is a finite series and 2 + 4 + 6 + 8 + … is an infinite
series.

37

 Sequences and Series

Smart Tips! Determine the Expansion of (1  x)n

 (a bx n  1 When n is negative or a fraction (n is not a positive integer), the binomial
+ bx)n = an 1 + a expansion for (1 + x)n is an infinite series, given by

(1 + x)n = 1 + nx + n(n − 1)x2 + n(n − 1)(n − 2)x3 +…
2! 3!
Factorise to get (1 + x)n BAKTI SDN. BHD.
before applying the  2 The expansion does not terminate and the series is convergent for |x|  1.
expansion

Example 5

Expand the following up to the term in x3 and state the range of values of x for

Chapter 3  which each expansion is valid. x 1  (c) x 1
(a) (1 + x)−2 (b) 1 − 2 2 1 − 3 2

Solution

(a) (1 + x)−2 = 1 + ( − 2)x + (−2)(−3)  x2 + (−2)(−3)(−4)  x3 + …
2! 3!
= 1 − 2x + 3x2 − 4x3 + …

= 1 − 2x + 3x2 − 4x3

The expansion is valid for |x|  1 or −1 x  1

 (b) 1 − x 1
2 2

 1 − 2x   1  1 − 1 − 2x 2  1  1 − 1 1 − 2 − 2x 3 + …
2 2 2 2 2 2
= 1 + + +
2! 3!
x x2 x3
ILMU= 1 − 4 − 32 − 128 + …

= 1 − x − x2 − x3
4 32 128
 
The expansion is valid for x 1
2
x
−1  2  1

PENERBIT   1 −2  x  2
2
(c) 1 − x
3
    1   1 − 2
1 − 3x  1 1 − − 3x 2 1 1 − 1 − 3x 3
2 2 2 2 2 2
= 1 + + + + …
1! 2! 3!

= 1 +  1 − 3x  +  1 − 12   x2  +  1 − 21 − 32  −  x3  + …
2 2 9 2 27
3!
2!
   
= 1 − x − 1   x2 − 1   x3 +…
6 8 9 16 27

= 1 − 1 x − 1 x2 − 1 x3 + …
6 72 432
  x
The expansion is valid for 3 1

−1  x  1
3
−3  x  3

48

4CHAPTER BAKTI SDN. BHD.

Matrices and Systems of
Linear Equations

4.1 Matrices

 Learning Outcomes

● Identify the different types of matrices.
● Perform operations on matrices.
● Find the transpose of a matrix.
ILMU
Types of Matrices

PENERBIT 1 A matrix is a rectangular array of numbers that are arranged in m rows and
n columns enclosed by a pair of brackets as shown in Figure 4.1.

a11 a12 … a1j … a1n
a21 a22 … a2j … a2n

  
ai1 ai2 … aij … ain
A=  m rows

   Smart Tips!
   2 Row 1 → 4 5 1
am1 am2 … amj … amn
rows Row 2 → 6 9 7
n columns
Figure 4.1  A matrix Column Column Column
123
 2 The size or order of a matrix is determined by the number of rows followed by 3 columns
 the number of columns. For example, the matrix
Order: 2 × 3 matrix
columns. Therefore, it is 2 × 3 matrix.
4 5 1 has 2 rows and 3 53
6 9 7

Matrices and Systems of Linear Equations 

Operations on Matrices

 1 If A and B are two matrices of the same order, the sum of the two matrices is
obtained by adding up the corresponding elements of A and B.

   a11  a12  a13
b11  b12  b13
If A = a21  a22  a23 and B = b21  b22  b23 , then
a31  a32  a33 b31  b32  b33BAKTI SDN. BHD.

a11 + b11  a12 + b12  a13 + b13
a21 + b21  a22 + b22  a23 + b23
a31 + b31  a32 + b32  a33 + b33
  A + B =

 2 The difference of two matrices, A and B, is obtained by subtracting the Chapter 4
corresponding elements of B from A.

   a11  a12  a13
b11  b12  b13
If A = a21  a22  a23 and B = b21  b22  b23 , then
a31  a32  a33 b31  b32  b33

 a11 − b11  a12 − b12  a13 − b13

A − B = a21 − b21  a22 − b22  a23 − b23
a31 − b31  a32 − b32  a33 − b33

 3 The addition and subtraction of matrices of different orders is not defined.

 4 When a matrix A is multiplied by a scalar k, each element in the matrix is
multiplied by the scalar.

 a11  a12  a13

If A = a21  a22  a23 , then
a31  a32  a33
ILMU
ka11  ka12  ka13
   a11  a12  a13
kA = k a21  a22  a23 = ka21  ka22  ka23
a31  a32  a33 ka31  ka32  ka33

 5 If A is an m × n matrix and B is an n × p matrix, then the multiplication of A
and B produce an m × p matrix.
PENERBIT
a11  a12  a13 b11  b12  b13
   a31  a32  a33
If A = a21  a22  a23 and B = b21  b22  b23 , then
b31  b32  b33
Recap!
a11  a12  a13
The product of the
  AB = a21  a22  a23 matrices are defined if
b11  b12  b13 the number of columns
b21  b22  b23 of matrix A is equal to the
a31  a32  a33 b31  b32  b33 number of rows of matrix
B.
 =a11b11 + a12b21 + a13b31  a11b12 + a12b22 + a13b32  a11b13 + a12b23 + a13b33
a21b11 + a22b21 + a23b31  a21b12 + a22b22 + a23b32  AB
a31b11 + a32b21 + a33b31  a31b12 + a32b22 + a33b32  a21b13 + a22b23 + a23b33
a31b13 + a32b23 + a33b33 m × n  n × p

 6 The following properties are applied for all the matrices A, B and C, if the sum equal
and the product of the matrices are defined. 

(a) A + B = B + A AB ⇒ m × p matrix

(b) (A + B) + C = A + (B + C)

(c) A + (−A) = (−A) + A = 0

55

Matrices and Systems of Linear Equations 

Operations on Matrices

 1 If A and B are two matrices of the same order, the sum of the two matrices is
obtained by adding up the corresponding elements of A and B.

   a11  a12  a13
b11  b12  b13
If A = a21  a22  a23 and B = b21  b22  b23 , then
a31  a32  a33 b31  b32  b33BAKTI SDN. BHD.

a11 + b11  a12 + b12  a13 + b13
a21 + b21  a22 + b22  a23 + b23
a31 + b31  a32 + b32  a33 + b33
  A + B =

 2 The difference of two matrices, A and B, is obtained by subtracting the Chapter 4
corresponding elements of B from A.

   a11  a12  a13
b11  b12  b13
If A = a21  a22  a23 and B = b21  b22  b23 , then
a31  a32  a33 b31  b32  b33

 a11 − b11  a12 − b12  a13 − b13

A − B = a21 − b21  a22 − b22  a23 − b23
a31 − b31  a32 − b32  a33 − b33

 3 The addition and subtraction of matrices of different orders is not defined.

 4 When a matrix A is multiplied by a scalar k, each element in the matrix is
multiplied by the scalar.

 a11  a12  a13

If A = a21  a22  a23 , then
a31  a32  a33
ILMU
ka11  ka12  ka13
   a11  a12  a13
kA = k a21  a22  a23 = ka21  ka22  ka23
a31  a32  a33 ka31  ka32  ka33

 5 If A is an m × n matrix and B is an n × p matrix, then the multiplication of A
and B produce an m × p matrix.
PENERBIT
a11  a12  a13 b11  b12  b13
   a31  a32  a33
If A = a21  a22  a23 and B = b21  b22  b23 , then
b31  b32  b33
Recap!
a11  a12  a13
The product of the
  AB = a21  a22  a23 matrices are defined if
b11  b12  b13 the number of columns
b21  b22  b23 of matrix A is equal to the
a31  a32  a33 b31  b32  b33 number of rows of matrix
B.
 =a11b11 + a12b21 + a13b31  a11b12 + a12b22 + a13b32  a11b13 + a12b23 + a13b33
a21b11 + a22b21 + a23b31  a21b12 + a22b22 + a23b32  AB
a31b11 + a32b21 + a33b31  a31b12 + a32b22 + a33b32  a21b13 + a22b23 + a23b33
a31b13 + a32b23 + a33b33 m × n  n × p

 6 The following properties are applied for all the matrices A, B and C, if the sum equal
and the product of the matrices are defined. 

(a) A + B = B + A AB ⇒ m × p matrix

(b) (A + B) + C = A + (B + C)

(c) A + (−A) = (−A) + A = 0

55

PENERBIT5CHAPTER
ILMU
BAKTI SDN. BHD.Functions and Graphs

5.1 Functions

 Learning Outcomes

● Define a function.
● Identify a function from a graph by using the vertical line test.
● Identify a one-to-one function by using an algebraic approach or the

horizontal line test.
● Sketch the graph of a function.
● State the domain and range of a function.

Define a Function

 1 A function is a relationship between two sets, X and Y, in which each element
of set X is mapped to one and only one element of set Y. In Figure 5.1 and

Figure 5.2, each element of X is mapped to only one element of Y.

X Y X Y

ap p a
q b
bq r c
s d
cs
Figure 5.2  Many-to-one function

Figure 5.1  One-to-one function

 2 If f is a function from set X to set Y and is defined by f (x) = y, set X is the
domain and each of its elements is known as an object. Meanwhile, set Y is
the codomain and each of its elements is known as an image. The elements in
set Y that are mapped from X, are the actual obtained values of function f and
known as the range (refer Figure 5.3).

94

PolynomialsPENERBIT6CHAPTER
ILMU
BAKTI SDN. BHD.
6.1 Polynomials

 Learning Outcomes

● Perform addition, subtraction and multiplication of polynomials.
● Perform division of polynomials.

 1 Polynomials are algebraic expressions that have variables and coefficients in
them. A polynomial of degree n can be written in the form
P(x) = anxn + an − 1 xn − 1 + an − 2 xn − 2 + … + a1x + a0

where n is a positive integer and an ≠ 0.
 2 The real numbers in the expression, an, an − 1, …, a0 are called the coefficients of

the polynomial.

Addition, Subtraction and Multiplication of Polynomials

 1 When adding polynomials, always add the like terms, that is, terms of the
same power.

 2 The addition of polynomials always produces a polynomial of the same degree.
 3 The only difference between subtracting and adding polynomials is the type of

operation. Therefore, to subtract polynomials, subtract the like terms.
 4 When multiplying two polynomials, multiply each term of one polynomial with

each term of the other polynomial. Then, combine terms of the same degree.
Let
P(x) = anxn + an − 1 xn − 1 + … + a1x + a0 and Q(x) = bnxn + bn − 1 xn − 1 + … + b1x + b0

133

PENERBIT7CHAPTER
ILMU
BAKTI SDN. BHD.Trigonometric Functions

7.1 Trigonometric Ratios and Identities

 Learning Outcomes

● State trigonometric ratios of sin θ, cos θ, tan θ, cosec θ, sec θ and cot θ.

● Express

(i) tan θ = sin θ (iii) cos (90° − θ) = sin θ
cos θ

(ii) sin (90° − θ) = cos θ (iv) tan (90° − θ) = cot θ

● Use some special angles.

● Find the angle of a trigonometric equation.

● Use the basic trigonometric identities.

(i) sin2 θ + cos2 θ = 1

(ii) 1 + cot2 θ = cosec2 θ

(iii) 1 + tan2 θ = sec2 θ

 1 Consider the right-angled triangle in Figure 90° − θ
7.1, where the longest side is known as the bc
hypotenuse and the sides opposite to the
hypotenuse are known as the adjacent side θ
and opposite side to angle θ. a
Figure 7.1  Right-angled triangle
 2 The trigonometric ratios can be defined as
follows:

(a) sin θ = b (d) sec θ = 1 θ = c
c cos a

(b) cos θ = a (e) cosec θ = 1 = c
c sin θ b

(c) tan θ = b (f) cot θ = 1 = a
a tan θ b

153

Trigonometric Functions 

The trigonometric ratios of the special angles, namely 0°, 30°, 45°, 60° and 90°
can be obtained from the two triangles as shown in Table 7.1.

Table 7.1  Trigonometric ratios of special angles

Ratio sin θ cos θ tan θ cosec θ sec θ cot θ
Angle
PENERBIT
ILMU0° 0 1 0 Undefined 1 Undefined
BAKTI SDN. BHD.
30° 1 31 2 2 3
22 3 3

45° 111 2 21 Chapter 7
22

60° 31 3 22 1
22 3 3

90° 1 0 0 1 Undefined Undefined

 5 The basic trigonometric identities are given by

sin2 θ + cos2 θ = 1
1 + tan2 θ = sec2 θ
1 + cot2 θ = cosec2 θ

Example 2

Prove the following identities. (d) 1 − sin2 x = cot2 x
(a) sin4 x − cos4 x = 2 sin2 x − 1 1 − cos2 x

(b) sin x (cosec x − sin x) = cos2 x (e) sin2 x (sec x + cosec x) = 1 + tan x
(c) tan4 x + 2 tan2 x + 1 = sec4 x cos x tan x

Solution

(a) sin4 x − cos4 x = (sin2 x + cos2 x)(sin2 x − cos2 x) Smart Tips!
= sin2 x − cos2 x
= sin2 x − (1 − sin2 x) • a2 − b2 = (a + b)(a − b)
= sin2 x − 1 + sin2 x • sin2 x + cos2 x = 1
= 2 sin2 x − 1 [Proven] • cos2 x = 1 − sin2 x

 (b) sinx x − sin x) = x  1
(cosec sin sin x − sin x Smart Tips!

 
= sin x  1 − sin2 x • cosec x = 1
sin x sin x
= 1 − sin2 x
• 1 − sin2 x = cos2 x
= cos2 x [Proven]

(c) tan4 x + 2 tan2 x + 1 = (tan2 x + 1)2 Smart Tips!
= [(sec2 x − 1) + 1]2
= (sec2 x − 1 + 1)2 • a4 + 2a2 + 1 = (a2 + 1)2
= (sec2 x)2 • tan2 x = sec2 x − 1
= sec4 x [Proven]

155

PENERBIT 8CHAPTER
ILMU
BAKTI SDN. BHD.Limits and Continuity

8.1 Limits

 Learning Outcomes

● State the limit of a function f (x) as x approaches a given value a, xli→ma f (x) = L.
● FSitnadtelxlxit→i→mmhaaegf b((xxa))siwc hperonplxie→mratife (xs)o=f limits.
● 0 and lxi→ma g(x) = 0 by the following methods:

(i) factorisation (ii) multiplication by a conjugate

● Find one-sided limits in: (ii) xli→ma− f (x) = M
(i) xli→ma+ f (x) = L
f (x) f (x).
● Determine the existence of the limit of a function xli→ma– = xli→ma+
● Find infinite limits:
(i) lxi→ma f (x) = +∞ (ii) lxi→ma f (x) = −∞

● Find limits at infinity: (ii) xl→im−∞ f (x) = M
(i) xl→im+∞ f (x) = L
● Apply the following limits:

  (i) xl→im+∞1 = 0  (ii) xl→im−∞1 = 0 for n  0.
xn xn

 1 If a function f (x) approaches a particular value L when x approaches a certain
value a, we say that L is the limit for f (x) and can be written as lim f (x) = L.

x→a

 2 Different methods can be used to determine the limit of a function f (x). One of
the methods is the intuitive method. The steps are:

(a) draw a table of values of x and f (x),

(b) the values of f (x) are observed to see if they approach a specific value.

178

PENERBIT 9CHAPTER
ILMU
BAKTI SDN. BHD.Differentiation

9.1 Derivative of a Function

 Learning Outcomes

● Find the derivative of a function f (x) using the first principles
f (x ℎ) f (x) .
f ′(x) = lim + ℎ −

ℎ→0

● Discuss the differentiability of a function at x = a.
f (x) f (a)
f ′(a) = lim x − a
−
x→a−

f ′(a) = lim f (x) − f (a)
x − a
x→a+

 1 Let y = f (x) be a function of x. Then, the derivative of the function with respect

to x is denoted by the symbol dy and is defined as
dx

dy = f ′(x) = lim f (x + ℎ) − f (x)
dx ℎ
ℎ→0

 2 This idea is known as differentiation using first principles.
 3 f ′(x) is the gradient of the tangent to the curve y = f (x).
 4 If f (x) is differentiable at x = a, then f (x) is continuous at x = a.

f ′(a) = lim f (x) − f (a) = lim f (x) − f (a)
x→a+ x−a x→a− x−a

 5 A function may be continuous at one point but not differentiable. This occurs at
points where there are vertical tangents or corners.

202

PENERBIT10CHAPTER
ILMU
BAKTI SDN. BHD.Applications of Differentiation

10.1 Extremum Problems

 Learning Outcomes

● Find the critical points.
● Find the relative extremum using the first derivative test and second

derivative test.
● Find points of inflection.
● Sketch the graph of a polynomial function.
● Solve optimisation problems.

Critical Point

 1 Let f (x) be a function and (c, f (c)) be a point in the domain of the function. The
point (c, f (c)) is known as a critical point of f if

(a) f ′(c) = 0,

(b) f  ′(c) is undefined.

 2 The value of c is known as a critical number and the value of f (c) is known as

a critical value.

 3 When dy = 0, the gradient of the curve is 0 and the tangent to the curve is
dx

parallel to the x-axis. This point is called a stationary point.

 4 When dy is undefined, the tangent to the curve is either parallel to the y-axis or
dx
dy
dx does not exist.

227

Semester Examination

Duration: 2 hours

Instructions: This question paper consists of Section A and Section B. Section A consists of
3 questions and Section B consists of 7 questions. Answer all questions. The full marks for each
question or section are shown in brackets at the end of the question or section. All steps must be
shown clearly. Only non-programmable scientific calculators can be used. Numerical answers may
be given in the form of π, e, surd, fractions or up to three significant figures, where appropriate,
unless stated otherwise in the question.
BAKTI SDN. BHD.
Section A [25 marks]

 1 Compute

(a) lim x2 − 7x + 12 [3 marks]
x→3 x− 3 [3 marks]
[3 marks]
(b) lim |4x − 12|
x→3− x−3

x
(c) lim 1 + 2x2
Semester Examination x→∞

 2 Differentiate each of the following with respect to x.

(a) y = 7x + x [2 marks]

(b) y = (sin x + 2)3 [3 marks]

(c) x2 + 2xy2 = 3y + 4 [4 marks]

ILMU
 3 The equation of a curve is given as y = 4x3 + 15x2 − 18x + 5. Find the coordinates of the

stationary points and determine the nature of each stationary point. [7 marks]

Section B [75 marks]

  1 The complex zn1u2zm+1 bzˉe2rs. z1 and z2 are given by z1 = −3i and z2 = 3 + i 2 . Find z12z+1 zˉ2 . [5
Hence, find
PENERBIT marks]

 2 Solve each of the following.

(a) 4x − 3 (4) 1 x + 1 = 4 [5 marks]
2 2 2 [6 marks]

(b) 3  x2 − 2x  x + 4

 3 Given that f (x) = 2 .
2+
2

2 + 2
x

 (a) Simplify f (x) and evaluate f  1 . [4 marks]
3

(b) The domain of f (x) is the set of all the real numbers except three numbers. Determine all

the three numbers. [4 marks]

250

PENERBIT
ILMU
BAKTI SDN. BHD.