Semester 1 Mathematics STPM (Teacher's Edition)

46

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

2 − = 2 -------(1)

(−2) = 2(−8) − 3 (4) − 2 + = −54
14 − = 38 ------ (2)

(2) – (1)
12 = 36
= 3
= 4

( ) = 2 3 − 9 2 + 3 + 4
Let 2 3 − 9 2 + 3 + 4 ≡ ( − 1)(2 2 + − 4)

Comparing coefficients of 2 : − 9 = − 2
= −7

∴ ( ) = 2 3 − 9 2 + 3 + 4
= ( − 1)(2 2 − 7 − 4)
= ( − 1)(2 + 1)( − 4)

Given 4 3 + 3 2 − 9 + 2
= 2(2 3 − 9 2 + 3 + 4) + 21 2 − 15 − 6
= 2( − 1)(2 + 1)( − 4) + 3(7 2 − 5 − 2)
= ( − 1)(4 − 1)( + 2)

4 Given 4 − 10 2 + 1 = ( 2 + + 1)( 2 + + 1)
Comparing coefficients of 3:
+ = 0
= − ------(1)

Comparing coefficient of 2:
−10 = + 2

= −12 ------- (2)

(1) → (2) 2 = 12
= ±2√3

When = 2√3, = −2√3
When = −2√3, = 2√3

∴ 4 − 10 2 + 1 = ( 2 − 2√3 + 1)( 2 + 2√3 + 1)

When 4 − 10 2 + 1 = 0
( 2 − 2√3 + 1)( 2 + 2√3 + 1) = 0

For 2 − 2√3 + 1 = 0
= √3 ± √2

47

For 2 + 2√3 + 1 = 0
= −√3 ± √2

Hence, the solutions are √3 ± √2, −√3 ± √2

5 4 + 2 3 + 5 2 − 16 − 20 = ( 2 + + )2 − 4( + )2

Comparing coefficients of 2:
2 − 3 = 5

= 4

Comparing coefficient of :
2 − 8 = −16
When = 4, = 3

For ( ) = 0
( 2 + + 4)2 − 4( + 3)2 = 0

( 2 + + 4)2 = 4( + 3)2
2 + + 4 = 2( + 3)
2 − − 2 = 0

( + 1)( − 2) = 0
= −1 or = 2

Hence the real roots are = −1 or = 2 .

1.2(b) Learning Outcome:
INEQUALITIES a) Solve polynomial and rational inequalities
b) Solve inequalities involving modulus signs in simple cases

➢ DEFINITION
An inequality is similar to an equation except that the statement is that two expressions have a
relationship other than equality, such as <, <, >, or >.

➢ NOTES
To solve an inequality means to find all values of the variable that make the inequality true.

Linear Inequalities
A linear inequality is one in which each term is constant or a multiple of the variable.

Nonlinear Inequalities
The sign of a product or quotient
If a product or quotient has an even number of negative factors, then its value is positive.
If a product or quotient has an odd number of negative factors, then its value is negative.

48

➢ CONCEPT

1. A < B  A + C < B + C

2. A < B  A – C < B – C

3. If C > 0, then A < B  CA < CB

4. If C < 0, then A > B  CA > CB

5. If A > 0 and B > 0, then A < B  1  1
A B

6. If A < B and C < D, then A + C < B + D

To Solve a Nonlinear Inequality:
1. Write the inequality so that all nonzero terms are on side of the inequality sign.

If there are fractions, write the expression with a single fraction.
2. Factor the nonzero side of the inequality.
3. Determine the values for which each factor is zero and divide the number line into

the intervals.
4. Use test values to make a diagram using the number line, showing the sign of each

factor on each interval.
5. Determine the solution from the table of signs. Check the endpoints of the intervals

with the inequality.

Absolute Value Inequalities
Properties of Absolute Value Inequalities
1. x  c  ̶ c < x < c ; x  c  ̶ c < x < c

2. x  c  x < ̶ c or x > c ; x  c  x < ̶ c or x > c

3. x − a  c  ̶ c < x – a < c → a – c < x < a + c

4. x − a  c  x – a < ̶ c or x – a > c → x < a – c or x > a + c

Linear inequalities involving modulus can be solved by
(a) applying the properties of modulus
(b) squaring both sides of the inequality
(c) using the graphical method

EXAMPLE 1
Find the set of values for x for which
(a) 3x + 2  6

(b) 4 − 3x  5

Solution:
(a) 3x + 2  6

− 6 < 3x + 2 < 6

− 8 < 3x < 4

− 8 <x< 4
3 3

Therefore the solution set is { x │ − 8 <x < 4 }
3 3

49

EXAMPLE 2

Find the set of values for x where
(a) 2x +1  1

1+ x 2

(b) x + 3  3x − 5

x

(c) ( x + 3 ) ( 2 – x ) ( x – 1 ) > 0

Solution:

(a) 2x +1  1 -- + 3x + 1 > 0
1+ x 2
+ 1+ x > 0
2x +1 − 1  0 -+ 1 x
1+ x 2 −3

2(2x +1) − (1+ x)  0 +
2(1+ x)

3x +1  0 +-
2(1+ x)

Therefore the solution set is { x │ −1 <x < − 1 }
3

(b) x + 3  3x − 5

x
x + 3 −  3x − 5   0

x
x(x + 3) − (3x − 5)  0

x
x2 + 5  0
x

x2 + 5  0 x

Therefore the solution set is { x │ x > 0 }

(c) ( x + 3 ) ( 2 – x ) ( x – 1 ) > 0
(x+3)(x–2)(x–1)<0

-+ ++ x +3 > 0
-- ++ x ̶ 1>0
-- -+ x ̶ 2>0
x
- ̶3 + 1- 2 +

Therefore the solution set is { x │x < ̶ 3 or 1 < x < 2 }

50

EXAMPLE 3
Find the set of real values of x where 3x3 + 4 < x2 + 12x

Solution: -+ ++ x+2>0
3x3 + 4 < x2 + 12x -- ++ 3x ̶ 1 > 0
-- -+ x ̶ 2>0
3x3 – x2 – 12x + 4 < 0 x
- ̶ 2+ 1
Let P(x) = 3x3 – x2 – 12x + 4 2+
When x = 2 3-

P(2) = 3(2)3 – (2)2 – 12(2) + 4 = 0
Therefore, (x – 2) is a factor of P(x).

When x = –2
P(–2) = 3(–2)3– (–2)2 – 12(–2) + 4 = 0

Therefore, (x + 2) is a factor of P(x).

3x3 – x2 – 12x + 4 = (x + 2)(x – 2)(3x – 1) by inspection
3x3 – x2 – 12x + 4 < 0

(x + 2)(x – 2)(3x – 1) < 0

Therefore the solution set is { x │ x < − 2 or 1 < x < 2 }

3

EXAMPLE 4

Find the set of values of x for which x +1  4 x −1

Solution:

(x + 1)2 < 16(x – 1)2 - - + 5x ̶ 3 > 0
x2 + 2x + 1 < 16(x2 – 2x +1) + + 3x ̶ 5 > 0
x
x2 + 2x + 1 < 16x2 – 32x + 16 - 3- 5
5 3+
15x2 – 34x + 15 > 0

(5x – 3)(3x – 5) > 0

Therefore the solution set is { x │ x < 3 or x > 5 } +
5 3

EXAMPLE 5 y y = 1
Find the set of values of x for which 2x −1  1
y = –2x + 1
x 1

Solution: y = 2x – 1
x
Sketch y = 2x −1 and y = 1 0 11

x 2

y = 2x – 1 ….(1)

y=1 ….(2)
x

Substituting (1) into (2)

2x – 1 = 1

x

51

x(2x – 1) = 1
2x2 – x – 1 = 0
(2x + 1)(x – 1) = 0
x = − 1 or x = 1

2
The set of values of x for which 2x −1  1 is { x │ x < 0 or x > 1 }

x

EXAMPLE 6

Sketch, on the same axes, the graphs of y = 2x +1 and y = 1 – x2. Hence, solve the inequality

2x +1 > 1 – x2.

Solution: y = 2x +1 y 1 x
Point of intersection for x  − 1 1
-1 0
2
– (2x + 1) = 1 – x2
x2 – 2x – 2 = 0

x = 2 4+8
2

=1− 3

For 2x +1 > 1 – x2,

The solution set is { x │x > 0 or x < 1− 3 }

WORKSHEET 1.5

Q1 Find, in each case, the set of values of x for
(a) x ( x – 2 ) > x + 4

(b) x – 2 > x + 4

x

Solution: - - + x–4>0
1(a) x ( x – 2 ) > x + 4 - + + x+1>0
x2 – 3x + 4 > 0 x
(x–4)(x+1)>0 –1 - 4

The solution set is { x : x < –1 or x > 4 } + +

52

1(b) x – 2 > x + 4 -+ ++ x+1>0
-- ++ x>0
x -- -+ x–4>0
x −2−(x + 4)  0
- –1 + 0- 4 + x
x
x2 − 3x − 4  0

x
(x − 4)(x +1)  0

x

The solution set is { x : –1 < x < 0 or x > 4 }

Q2 Sketch, on the same diagram, the graph of y = 1 and the graph of 6y = 1 + ︱x︱.
x

Find the set of values of x such that 1 + ︱x︱> 6
x.

Solution:
2 6y = 1 + x ……(1)

y = 1 ……….(2) y
x

From (1) and (2) B● ●A
x2 + x – 6 = 0
0 x
(x+3)(x–2)=0

x = -3 or x = 2

At point A , x > 0
Therefore x = 2

6y = 1 + x ……(3)

y =−1 ……...(4)
x

From (3) and (4)
x2 – x – 6 = 0

(x+2)(x–3)=0
x = –2 or x = 3

At point B , x < 0
Therefore x = –2

The set of value of x such that 1 + ︱x︱> 6 is { x : x < –2 or x > 2 }.
x

53

Q3 Determine, in each case, the set of values of x where
(a) x2 > 3x

(b) x + 1  1

x3

Solution: x2 > 3x - - + x-3>0
3(a) x(x–3)>0 - + + x>0
x
The set of value of x such x2 > 3x 3+
is { x : x < 0 or x > 3 } + 0-

3(b) x + 1  1 -- + x>0
-+ + 2x + 3 > 0
x3 x
x +1− 1 0 3 0+
+ −2 -
x3
2x + 3 0

3x

The set of values of x where x +1  1 is { x : x < − 3 or x > 0 }
x3 2

Q4 Find the solution set for the inequality x  2 .

5+ x

Solution: x 2 -- + 3x + 10 > 0
4 5+ x -+ + x + 10 > 0
x
 x 2  4 + −10 - 10
5+ x −3 +

x2 < 4(5 + x)2
3x2 + 40x + 100 > 0

(3x + 10) (x + 10) > 0

The set of values of x where x  2 is { x : x < -10 or x > − 10 }
5+ x 3

54

Q5 Find the set of values of x which satisfies the inequality 4x −11 x − 3 .

x+6 2

Hence solve the inequality 4 x −15  x −4
x +5 2.

Solution:

5 4x −11  x − 3 -- + x−1>0
+ x+1>0
x+6 2 x
1+
x − 3 − 4x −11  0 -+
2 x+6

x2 −5x + 4  0 + −1 -
2(x + 6)

(x −1)(x − 4)  0
(x + 6)

The solution set is { x : x < −6 or −1 < x < 4 }…….(1)

4 x −15  x − 4 -+ + + x+6>0
x +5 2 -- + + x−1>0
-- - +
From (1) , by deduction : x–4>0
- -6 + 1- 4+ x
x −1 −6

x  − 5 (undefined)

1 x −1 4

4  x  25

The solution is { x : 4 < x < 25 }

EXERCISE 1.5

Q1 Find the set of values of x that satisfy the inequality 1  1 .

x +1 x −1

Q2 On the same diagram, sketch the graph of y = | x – 1 | , x  and the graph of
y = 3 − x , x ≤ 3. Hence, or otherwise, solve the inequality | x – 1 | > 3 − x .

Q3 Find the solution set for the inequality x − 6  3 .

x+2

Q4 Find the solution set for the inequality 4x − 5  x − 2 .

x

Q5 Find the solution set for the inequality 1  1 .

2−x x−3

Q6 Find the set of values of x such that -16 < x3 – 4x2 + 4x – 16 < 0

55 1 11

ANSWER x +1 x −1
Section 1 − 1 0
1.2 x +1 x −1

Exercise −2 0
1.5 (x +1)(x −1)

2 0
(x +1)(x −1)

The set of values of x where 1  1 is { x : -1 < x < 1}

x +1 x −1

2
y

y = –x + 1 y=x–1

= √3 −

| x – 1 | > 3−x –1 0 12 + x
+
(x – 1)2 > 3 – x - - x–2>0
x2 – 2x + 1 > 3 – x - + x+1>0
x2 – x – 2 > 0
(x + 1) (x – 2) > 0 + −1 - 2+ x

Since x < 3, the solution set is { x : x < –1 or 2 < x < 3 }.

3 x−6 3
x+2
+
-
x − 6 −30 - + + x+6>0
x+2 - x+2>0
x
x − 6 − 3(x + 2) 0 + −6 - −2 +
x+2
− 2x −12 0
x+2

x+6  0
x+2

The set of values of x is { x : -6 < x < -2 }.

56

4 4x − 5  x − 2

x

4x −5 + 2 − x  0 - + ++ x>0
x - - ++ x 1>
- - -+ 0x 5 > 0
4x − 5 + x(2 − x)  0
x - 0+ 1- 5 + x

6x −5− x2  0
x

x2 − 6x + 5  0
x

(x −1)(x − 5)  0
x

The set of values of x such that 4x − 5  x − 2 is { x : 0 < x < 1 or x > 5 }

x

5 11


2−x x−3 - + + +
x ̶ 2>0
1 −1 -
2 − x x−3 0 - + +
2x ̶ 5 > 0

(x − 3) − (2 − x)  0 - - - + x ̶ 3>0

(2 − x)(x − 3) - 2+ 5 3+ x
2x −5
0 2-

(2 − x)(x − 3)

2x − 5  0
(x − 2)(x − 3)

The set of values of x such that 2 1 x  x 1 3 is { x : 2 < x < 5 or x > 3 }
− − 2

6 ̶ 16 < x3 – 4x2 + 4x – 16 < 0 + x ̶ 4>0
x>0
-- x

For x3 – 4x2 + 4x – 16 < 0 - + +
Let f (x) = x3 – 4x2 + 4x – 16
0- 4+
f (4) = 6 – 64 + 16 – 16 = 0

(x – 4) is a factor of f(x)

+

x3 – 4x2 + 4x – 16 = (x – 4)(x2 + 4)

by inspection
(x – 4)(x2 + 4) < 0

Since x2 + 4 > 0 for all real values of x.
x–4<0

x<4

⸫ x3 – 4x2 + 4x – 16 < 0 for { x : x < 4 }

57

For x3 – 4x2 + 4x – 16 > ̶ 16
x3 – 4x2 + 4x > 0

x (x2 – 4x + 4) > 0
x (x – 2)2 > 0

But (x – 2)2 > 0 for x ∈ R , x ≠ 2
Then x(x – 2)2 > 0 for x > 0 , x ≠ 2
⸫ x3 – 4x2 + 4x – 16 > ̶ 16 for { x : x > 0 , x ≠ 2 }

The set of values of x such that ̶ 16 < x3 – 4x2 + 4x – 16 < 0 is
{ x : 0 < x < 2 or 2 < x < 4 }

1.2(c) RATIONAL Learning Outcome:
FUNCTIONS a) 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

Rational Function

Rational function, ( ) = ( ) , ( )
( ) ( )

If < , ( ) is a proper function.

If ≥ , ( ) is an improper fraction.

R(x) could be expressed as a sum of simpler rational functions, called partial fractions.

Partial Fractions
Proper fractions in their lowest form, having only single factor in its denominator

Rules:

Type 1 Denominator with linear factors (Proper fraction)

Eg:

(1) 3 + 4
( + 1)( − 2) = ( + 1) + ( − 2)

3 + 4
(2) ( + 1)( − 2)( + 3) = ( + 1) + ( − 2) + ( + 3)

58

Type 2 Denominator with repeated linear factors (Proper fraction)

Eg:

3 + 4
(1) ( + 1)2( − 2) = ( + 1) + ( + 1)2 + ( − 2)

3 + 4
(2) ( + 1)3( + 3) = ( + 1) + ( + 1)2 + ( + 1)3 + ( + 3)

Type 3 Denominator with prime quadratic factors (Proper fraction)

Eg:

3 + 4 +
(1) ( + 1)( 2 + 1) = ( + 1) + ( 2 + 1)

3 2 + 5 + 4 + +
(2) ( 2 + 1)( 2 + 3)( + 1) = ( 2 + 1) + ( 2 + 3) + ( + 1)

Type 4 Improper fractions (*use long division method)

Eg:

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

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

EXAMPLE 1

Express 30−5 as partial fractions.
(2− )(3+ )

Solution:

30−5 = +
(2− )(3+ ) (2− ) (3+ )

(3 + ) + (2 − ) = 30 − 5

When = 2, 5 = 30 − 5(2)
= 4

When = −3, 5 = 30 − 5(−3)
= 9

Therefore,

30 − 5 49
(2 − )(3 + ) = (2 − ) + (3 + )

59

EXAMPLE 2

Express 4 2+ +1 as partial fractions.
( 2−1)

Solution:

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

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

When = 0, − = 1
= −1

When = −1, 2 = 4
= 2

When = 1, 2 = 6
= 3

Therefore,

4 2 + + 1 1 2 3
( 2 − 1) = − + ( + 1) + ( − 1)

EXAMPLE 3

Express 5 +1 as partial fractions.
( +1)2(2 +1)

Solution:

5 + 1
( + 1)2(2 + 1) = + 1 + ( + 1)2 + (2 + 1)

( + 1)(2 + 1) + (2 + 1) + ( + 1)2 = 5 + 1

When = −1, − = −5 + 1
= 4

1 13
When = − 2, 4 = − 2

= −6

Comparing the coefficient of 2,

2 + = 0, (−6)
= − 2 = − 2 = 3

Therefore,

5 + 1 34 6
( + 1)2(2 + 1) = + 1 + ( + 1)2 − (2 + 1)

60

EXAMPLE 4

Express 2 +1 as partial fractions.
( −2)( 2+1)

Solution:

2 + 1 +
( − 2)( 2 + 1) = − 2 + 2 + 1

( 2 + 1) + ( + )( − 2) = 2 + 1

When = 2, 5 = 5
= 1

Comparing the coefficient of 2,

+ = 0, = −

= −1

Comparing coefficient of constant, (when = 0)
− 2 = 1, 2 = 1 −

= 0

Therefore,

2 + 1 1
( − 2)( 2 + 1) = − 2 − 2 + 1

EXAMPLE 5

Express 3+4 2−6 as partial fractions.
2+2 −8

Solution:
Using long division method,

+ 2

2 + 2 − 8 3 + 4 2 + 0 − 6

3 + 2 2 − 8

2 2 + 8 − 6

2 2 + 4 − 16

4 + 10

3 + 4 2 − 6 4 + 10
2 + 2 − 8 = ( + 2) + 2 + 2 − 8

4 + 10 4 + 10
2 + 2 − 8 = ( + 4)( − 2)


= + 4 + − 2

61

( − 2) + ( + 4) = 4 + 10

When = −4, −6 = −6
= 1

When = 2, 6 = 18
Therefore, = 3

3 + 4 2 − 6 13
2 + 2 − 8 = ( + 2) + + 4 + − 2

WORKSHEET 1.6

1. Express in partial fractions (Denominator – Distinct Linear Factors):

(a) −11
( +3)( −4)

(b) 3 2−21 +24

( −2)( +1)( −3)

Solution:

1(a) − 11
( + 3)( − 4) = + 3 + − 4

( − 4) + ( + 3) = − 11

When = −3, −7 = −14
When = 4, = 2

7 = −7
= −1

Therefore,

− 11 21
( + 3)( − 4) = + 3 − − 4

1(b) 3 2 − 21 + 24
( − 2)( + 1)( − 3) = − 2 + + 1 + − 3

( + 1)( − 3) + ( − 2)( − 3) + ( − 2)( + 1) = 3 2 − 21 + 24

When = 2, −3 = −6
When = −1, = 2
When = 3,
12 = 48
= 4

4 = −12
= −3

Therefore,

3 2 − 21 + 24 243
( − 2)( + 1)( − 3) = − 2 + + 1 − − 3

62

2. Express in partial fractions (Denominator – Repeated Linear Factors):

(a) 2 2−5 +7
( −2)( −1)2

(b) +1
( +3)2

Solution:

2(a) 2 2 − 5 + 7
( − 2)( − 1)2 = − 2 + − 1 + ( − 1)2

( − 1)2 + ( − 2)( − 1) + ( − 2) = 2 2 − 5 + 7

When = 2, = 5

When = 1, − = 4
= −4

When = 0, + 2 − 2 = 7
Therefore, 5 + 2 + 8 = 7
= −3

2 2 − 5 + 7 5 3 4
( − 2)( − 1)2 = − 2 − − 1 − ( − 1)2

2(b) + 1
( + 3)2 = + 3 + ( + 3)2

( + 3) + = + 1

When = −3, = −2

When = 0, 3 + = 1
3 − 2 = 1

= 1

Therefore,

+ 1 1 2
( + 3)2 = + 3 − ( + 3)2

3. Express in partial fractions (Denominator – Prime Quadratic Factor) :

(a) 6−
(1− )(4+ 2)

(b) 3 2
( −1)( 2+ +1)

63

Solution:

3(a) 6 − +
(1 − )(4 + 2) = 1 − + 4 + 2

(4 + 2) + ( + )(1 − ) = 6 −

When = 1, 5 = 5
= 5

When = 0, 4 + = 6
= 6 − 4
= 6 − 4(1)
= 2

Comparing coefficient of 2,

− = 0, =

= 1

Therefore,

6 − 1 + 2
(1 − )(4 + 2) = 1 − + 4 + 2

3(b) 3 2 +
( − 1)( 2 + + 1) = − 1 + 2 + + 1

( 2 + + 1) + ( + )( − 1) = 3 2

When = 1, 3 = 3
= 1

When = 0, − = 0
= 1

Comparing coefficient of 2,
+ = 3
= 3 −
= 2

Therefore,

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

4. Express in partial fractions (Improper Fractions) :

(a) 3 2−2 −7
( −2)( +1)

(b) 3 3

( +1)( −2)

64

Solution:
4(a) 3 2 − 2 − 7 3 2 − 2 − 7
( − 2)( + 1) = 2 − − 2

Using long division method,

3
2 − − 2 3 2 − 2 − 7

3 2 − 3 − 6

− 1

3 2 − 2 − 7 − 1
( − 2)( + 1) = 3 + ( − 2)( + 1)

− 1
( − 2)( + 1) = − 2 + + 1

( + 1) + ( − 2) = − 1

When = 2, 3 = 1
When = −1, 1
Therefore,
= 3
−3 = −2

2
= 3

3 2 − 2 − 7 12
( − 2)( + 1) = 3 + 3( − 2) + 3( + 1)

4(b) 3 3 3 3
( + 1)( − 2) = 2 − − 2

Using long division method,

2 − − 2 3 + 3
3 3

3 3 − 3 2 − 6
3 2 + 6
3 2 − 3 − 6

9 + 6
3 3 9 + 6
( + 1)( − 2) = (3 + 3) + ( + 1)( − 2)

9 + 6
( + 1)( − 2) = + 1 + − 2

( − 2) + ( + 1) = 9 + 6

When = −1, −3 = −3
When = 2, = 1
Therefore,
3 = 24
= 8

3 3 1 8
( + 1)( − 2) = (3 + 3) + + 1 + − 2

65

EXERCISE 1.6

Express the following in partial fractions.

1. 2x+4
( +1)( +2)

2. 3
2−9

3. 5( +1)
25− 2

4. 4 +2
( +2)( 2−1)

5. 9
( −1)( +2)2

6. 1
2( 2−1)

7. 2+1
2−1

8. 2+ −1

( +1)( +2)

9. 10−11
( −4)( 2+1)

10. 1
( −1)( 2− +1)

ASNWER 1 2x + 4
Section ( + 1)( + 2) = + 1 + + 2
1.2
( + 2) + ( + 1) = 2 + 4
Exercise
1.6 When = −1, = −2 + 4
= 2

When = −2, − = −4 + 5
= 0

Therefore,

2 + 4 2
( + 1)( + 2) = + 1

66

2 3
2 − 9 = − 3 + + 3
( + 3) + ( − 3) = 3

When = 3, 6 = 3
When = −3, 1
Therefore,
= 2

−6 = 3
1

= − 2

31 1
2 − 9 = 2( − 3) − 2( + 3)

3 5( + 1)
25 − 2 = 5 − + 5 +
(5 + ) + (5 − ) = 5( + 1)

When = 5, 10 = 30
= 3

When = −5, 10 = −20
= −2

Therefore,

5( + 1) 3 2
25 − 2 = 5 − − 5 +

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

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

When = −2, (−1)(−3) = 4(−2) + 2
= −2

When = −1, (1)(−2) = −4 + 2
= 1

When = 1, (3)(2) = 4 + 2
C=1

Therefore,

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

67

5 9
( − 1)( + 2)2 = − 1 + + 2 + ( + 2)2

( + 2)2 + ( − 1)( + 2) + ( − 1) = 9

When = 1, (1 + 2)2 = 9
= 1

When = −2, (−2 − 1) = 9
= −3

When = 0, 4 − 2 − = 9
4(1) − 2 + 3 = 9
= −1

Therefore,

9 11 3
( − 1)( + 2)2 = − 1 − + 2 − ( + 2)2

6 1
2( 2 − 1) = + 2 + + 1 + − 1

( + 1)( − 1) + ( + 1)( − 1) + 2( − 1) + 2( + 1)

=1

When = 0, (1)(−1) = 1
= −1

When = −1, (−2) = 1
= − 1

2

When = 1, (2) = 1

= 1

2

Comparing the coefficient of : − = 0

= 0

Therefore,

1 11 1
2( 2 − 1) = − 2 − 2( + 1) + 2( − 1)

7 2 + 1 2 − 1 + 2 2
2 − 1 = 2 − 1 = 1 + 2 − 1

2
2 − 1 = − 1 + + 1
( + 1) + ( − 1) = 2

When = 1, 2 = 2
= 1

68

When = −1, −2 = 2
= −1

Therefore,

2 + 1 11
2 − 1 = 1 + − 1 − + 1

8 2 + − 1
( + 1)( + 2) = + + 1 + + 2

( + 1)( + 2) + ( + 2) + ( + 1) = 2 + − 1

Comparing the coefficient of 2, = 1

When = −1, (−1 + 2) = 1 − 1 − 1
= −1

When = −2, (−2 + 1) = 4 − 2 − 1
= −1

Therefore,

2 + − 1 11
( + 1)( + 2) = 1 − + 1 − + 2

9 10 − 11 +
( − 4)( 2 + 1) = − 4 + 2 + 1

( 2 + 1) + ( + )( − 4) = 10 − 11

When = 4, (16 + 1) = 10 − 44
= −2

When = 0, − 4 = 10

−2 − 4 = 10

= −3

Comparing the coefficients of 2, + = 0
−2 + = 0

= 2

Therefore,

10 − 11 −2 2 − 3
( − 4)( 2 + 1) = − 4 + 2 + 1

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

( 2 − + 1) + ( + )( − 1) = 1

When = 1, (1 − 1 + 1) = 1
= 1

69

When = 0, (1) − = 1
1 − = 1
= 0

Comparing the coefficients of 2, + = 0
1 + = 0

= −1

Therefore,
1 1

( − 1)( 2 − + 1) = − 1 − 2 − + 1

CLONE STPM

Q1 Given ( ) = 8 4 − 10 3 + 2 + + 2 where a and b are constant. ( + 1) is a factor
of ( ), and remainder 140 when ( ) is divided by ( + 2). Find the value of a and b.

With the value of a and b,
(a) Find the factor of ( ) in the form of ( − ) where is a positive integer.
(b) Factorise ( ) completely.
(c) Find the set values of such that ( ) > 0.

Q2 Sketch on the same coordinates axes, the graph of y = x − 2 and y = x + 4 , x > -4.

Hence solve the inequality x − 2  x + 4 .

Q3 Sketch the graphs of y = 1 and y = x − 2 on the same axes.

x

Hence, solve the inequality 1  x − 2 .

x

Q4 Find the constants A, B, C and D such that

3 2 + 5
(1 − 2)(1 + )2 = 1 − + 1 + + (1 + )2 + (1 + )3

Q5 Functions f and g are defined by

f : x → , for x ≠ 1 ; g : x → ax2 + bx + c, where a, b and c are constants.
2 −1 2

(a) Find f o f, and hence, determine the inverse function of f.

(b) Find the values of a, b and c if g o f (x)= 3 2+4 −1
(2 −1)2

(c) Given that p(x) = x2 – 2, express ℎ( ) = 2−2 in terms of f and p.
2 2−5

70

Q6 Sketch on the same coordinates axes, the graphs of y = 2− x and y= 2+ 1 .
x

Hence, solve the inequality 2 − x  2 + 1 .
x

ANSWER

Section 1 (−1) = 0

1.2 8 + 10 + − + 2 = 0

− = −20 ……(1)

Clone

STPM (−2) = 140

128 + 80 + 4 − 2 + 2 = 140

2 − = −35…….(2)

(2) − (1)

a = −15
b=5

1(a) x = 2, (2) = 8(2)4 − 10(2)3 − 15(2)2 + 5(2) + 2
=0

Factor = (x - 2)

1(b) ( ) = 8 4 − 10 3 − 15 2 + 5 + 2
= ( + 1)(8 3 − 18 2 + 3 + 2)
= ( + 1)( − 2)(8 2 − 2 − 1)

= ( + 1)( − 2)(2 − 1)(4 + 1)

1(c) ( ) > 0
( + 1)( − 2)(2 − 1)(4 + 1) > 0

{ : < −1 or − 1 < < 1 or > 2}

42

2
y
B
A

x

-4 0 2

Finding the intersection point of y = x − 2 and y = x + 4 ;

y= x−2 = {− ( − −2 , ≥ 2
2) , < 2

71

Point A :

– (x – 2) = x + 4

[ – ( x – 2 ) ]2 = ( x + 4)2
x2 – 4x + 4 = x + 4
x2 – 5x = 0
x (x – 5) = 0
x = 0 or x = 5

When x = 0 , y = 2 ⸫ A (0, 2)
When x = 5 , y = 3 ⸫ B (5, 3)

From the graph, the solution set is { x : –4 < x < 0 or x > 5 }

3y

y = 1 = | − 2|


01 x

When 1=x–2

x2 – 2x – 1 = 0

x = 2  (−2)2 − 4(−1)
2

x= 2 8
2

x =1 2

Since x > 0 , x =1+ 2

When 1 = −x + 2

x

x2 – 2x + 1 = 0
(x – 1)2 = 0

x=1

The solution set is { x : 0 < x < 1 or 1 < x < 2 }

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

(1 + )3 + (1 − )(1 + )2 + (1 − )(1 + ) + (1 − )

= 3 2 + 5

When = −1, 2 = 3 − 5
= −1

72

When = 1, 8 = 3 + 5
= 1

Comparing the coefficient of 3,
− = 0
=
= 1

When = −1, 2 = 3 − 5
= −1

Therefore, = 1, = 1, = −1, = −1

5(a) o = [ ( )]


= (2 − 1)

= 2 − 1
2 (2 − 1) − 1

= 2 − 1
2 − 2 + 1

2 − 1
=1
=

Since o = ,
Hence −1 =

1
= 2 − 1 , ≠ 2

Therefore, −1: → x , x ≠ 1
2x −1 2

5(b) o ( ) = [ ( )]

= ((22 − −11))2
= + 1) +
(2 −
2
= (2 − 1)2 + 2 − 1 +

2 + (2 − 1) + (2 − 1)2
= (2 − 1)2

Given, o ( ) = −3 2+4 −1
(2 −1)2

−3 2 + 4 − 1 2 + (2 − 1) + (2 − 1)2
(2 − 1)2 =
(2 − 1)2

2 + (2 − 1) + (2 − 1)2 = −3 2 + 4 − 1

73

11 12 1
When = 2 , 4 = −3 (2) + 4 (2) − 1
1
=4
= 1

When = 0, = −1

Comparing coefficient of 2,
+ 2 + 4 = −3

2 = −3 − − 4
2 = −3 − 1 + 4
= 0
Therefore, = 1, = 0, = −1

5(c) Given ( ) = 2 − 2
2 − 2

ℎ( ) = 2 2 − 5
[ ( )] = ( 2 − 2)

2 − 2
= 2( 2 − 2) − 1

2 − 2
= 2 2 − 5
Therefore, ℎ( ) = o ( )

6

y

f(x) = 2-x

1
h(x) = 2+

x

2

x

1
(- ,0)

2

2 − x = − 2 + 1 
 x

x2 − 4x −1= 0

x=2 5

 Hence, from the graph, solution for inequality 2 − x  2 + 1 : x : x  2 − 5
x

74
STPM PAST YEAR

STPM MM 2015 [SECTION A]
Q1 The polynomial ( ) = 4 + 3 + 2 − 10 − 4, where a and b are constants has a factor

(2x + 1).
When ( ) is divided by ( − 1), the remainder is –15.

(a) Determine the values of a and b.
(b) Factorise ( ) completely.
(c) Find the set of values of x which satisfies the inequality ( ) < 0.

STPM MM 2020 [SECTION B]

Q2 The function is given by ( ) = { 2−2 , <2



2 − 3 , ≥ 2,

(a) Show that is not one-to-one function.

(b) Sketch the graph of .

(c) By sketching a suitable graph on the same axes, determine the number of roots of the equation

( ) + 1 = | − 3|.

(d) (Not related)

ANSWER 1(a) Given that (2 + 1) is a factor of ( ),
Section
1.2 By the factor theorem,

STPM (− 1) = 0
Pass Year
2

(− 1)4 + (− 1)3 + (− 1)2 − 10 (− 1) − 4 = 0
22 2 2

1 + 1 + 7 = 0

16 4 8

+ 4 = −14 ------- (1)

Given that when ( ) is divided by ( − 1) the remainder is −15.

By the remainder theorem,
(1) = −15

+ 1 + − 10 − 4 = −15
+ = −2 ------(2)

(1) − (2), 3 = −12
= −4
∴ = 2

1(b) ( ) = 2 4 + 3 − 4 2 − 10 − 4
= (2 + 1)( − 1)( 2 + + )
= (2 + 1)( − 1)( 2 + 2 + 2)

75

1(c) (2 + 1)( − 1)( 2 + 2 + 2) < 0

(2 + 1): −o + +
( − 1): − − o+
( 2 + 2 + 2): +
+ +

1 x

1
−2

∴ {− 1 < < 1}

2

2(a) 2 < 2
2(b) ( ) = { 2 − ,

2 − 3 , ≥ 2,

For < 2
When = 0, = 1
asymptote− ∶ = 0
asymptote− ∶ = 2

For ≥ 2, straight line where passes (2 , 1) and gradient = 2

Let horizontal line, = , the line will cut more than one point when ≥
2.

Therefore, the g is not one-to-one function.

( )
= | − 3| − 1

= 2 2
1

12 4

(3, −1)

76 ( )
2(c)

=
= 2

2
1 (2, 1)

12

( ) + 1 = | − 3|
( ) = | − 3| − 1
Let = | − 3| − 1

From the graph, ( ) + 1 = | − 3| has 2 roots.

1.3 EXPONENTIAL Learning Outcome:
AND a) Relate exponential and logarithmic functions, algebraicly and
LOGARITHMIC graphically.
FUNCTIONS b) Use the properties of exponents and logarithms.
c) Solving equations and inequalities involving exponential and
logarithmic expressions.

EXPONENTIAL FUNCTIONS

Definition
If b is any number such that b > 0 and b ≠ 1 then an exponential function is a function in the form,

( ) =
where b is called the base and x is called the index, can be any real number.

Properties of ( ) = :

1. The graph of f (x) will always intersect at point (0 , 1) .

2. For every possible b we have bx > 0 . Note that this implies that bx ≠ 0 .

3. If 0 < b < 1, then the graph of f (x) will decrease as we move from left to right.

77

4. If b >1 then the graph of bx will increase as we move from left to right.
5. If = , then x = y
The value of b which is commonly used is e
( ) = , where e = 2.718281828…..= 2.718 (correct to 4 d.p.)

EXAMPLE 1
Sketch the graph of ( ) = and ( ) = ( )− .

Solution
If we don’t have any knowledge on what these graphs look like we’re going to have to choose
some values of x and do some function evaluations.

X ( ) = ( ) = ( )−
-2 (−2) = (−2) = 0.135
-1 (−1) = (−1) = 0.368 (−2) = ( )−(−2) = 7.389
0 (0) = (0) = 1 (−1) = ( )−(−1) = 2.718
1 (1) = (1) = 2.718
2 (2) = (2) = 7.389 (0) = ( )−(0) = 1
(1) = ( )−1 = 0.368
(2) = ( )−2 = 0.135

Function Graph Properties

( ) = Domain: ∈ ℝ
Range: y > 0
y-intercept = 1
Asymptote: y = 0

( ) = ( )− Domain: ∈ ℝ
Range: y > 0
y-intercept = 1
Asymptote: y = 0

78

EXAMPLE 2
Sketch the graph of each of the following functions.
(a) ( ) = −
(b) ( ) = − 2
(c) g(x) = − − 2
(d) ℎ( ) = − + 2
(e) ( ) = +1 + 2

Solution Graph Properties Translation
Function

(a) ( ) = − Domain: ∈ ℝ The graph
Range: y < 0 ( ) = is
y-intercept = − 1
Asymptote: y = 0 reflected about

x-axis

Domain: ∈ ℝ The graph
( ) = is
Range: y > − 2
translated 2 units
y-intercept = − 1
downwards.
Asymptote:
(b) ( ) = − 2 y = −2
(c) g(x) = − − 2
Domain: ∈ ℝ The graph
( ) = ( )− is
Range: y > − 2
translated 2 units
y-intercept = − 1
downwards.
Asymptote:
y=−2

79 Domain: ∈ ℝ The graph
(d) ℎ( ) = − + 2 Range: y < 2 ( ) = − is
y-intercept = 1
Asymptote: y = 2 translated 2 units

upwards.

(e) ( ) = +1 + 2 Domain: ∈ ℝ The graph
( ) = is
Range: y > 2
translated 1 unit to
y-intercept
=e+2 the left and 2 units
= 4.72
upwards.

Asymptote: y = 2

WORKSHEET 1.7

1. Sketch the graph of each of the following exponential functions.

(a) = +2 (b) = − + 3

80 (d) = 1− + 2
(c) = − − −1

Six rules of the Law of Indices

Rule 1:
Any number, except 0, whose index is 0 is always equal to 1, regardless of the value of the base.

Example:
Simplify 20:

Rule 2:

Example:
Simplify 2-2:

2−2 = 1
22

=1
4

Rule 3:
To multiply expressions with the same base, copy the base and add the indices.

Example:

Simplify : (note: 5 = 51)

51  53 = 51+3

= 54

= 5555

= 625

81

Rule 4:
To divide expressions with the same base, copy the base and subtract the indices.

Example:

Simplify :

( ) ( )5 y9  y5 = 5 y9−5

= 5y4

Rule 5:
To raise an expression to the nth index, copy the base and multiply the indices.

Example:
Simplify (y2)6:

( )y2 6 = y26

= y12

Rule 6:

Example:
Simplify 1252/3:

( )2 2

1253 = 3 125

= (5)2

=25

Equations Involving Exponential Expressions

EXAMPLE 1
Solve the equation 22 − 5 . 2 + 4 = 0

Solution:

(2²) − 5 . 2 + 4 = 0 ……………..①
Let y = 2 and substituting into ①,

y² - 5y + 4 = 0

Hence, ( y − 1 )( y − 4 ) = 0

i.e. y − 1 = 0 or y − 4 = 0

y = 1 or y=4
2 = 1 or 2 = 4
2 = 20 or 2 = 22

x = 0 or x=2

82

EXAMPLE 2
Solve the equation 9 +1 − 3 +3 − 3 + 3 = 0

Solution :
9 +1 − 3 +3 − 3 + 3 = 0

9 . 91 − 3 . 33 − 3 + 3 = 0
(32 ).9 − (3 ).27 − 3 + 3 = 0………………….①
Let y = 3 and substituting into ①,

9y² − 27y − y + 3 = 0

9y² − 28y + 3 = 0

(9y − 1) (y − 3) = 0

i.e. 9y − 1 = 0 or y − 3 = 0

9y = 1 or y = 3

y = 1
9
1
3 = 9 or 3 = 3
3 = 31
3 = 1 or
32 x=1
3 = 3−2 or

x = −2

EXERCISE 1.7

Solve the following exponential equations
1. 22 − 9. 2 + 8 = 0
2. 32 − 10. 3 + 9 = 0
3. 4 − 3. 2 +1 + 8 = 0
4. 22 +1 + 4 = 2 +3 + 2
5. 32 −3 − 4. 3 −2 + 1 = 0
6. 16 − 5. 22 −1 + 1 = 0

ANSWER 1 22 − 9(2 ) + 8 = 0
Section Let = 2 ,
1.3 2 − 9 + 8 = 0
( − 1)( − 8) = 0
Exercise = 1 = 8
1.3.1 ⸫2 = 1 2 = 8
2 = 20 2 = 23
= 0 = 3

83

2 32 − 10(3 ) + 9 = 0
Let = 3 ,
2 − 10 + 9 = 0
( − 1)( − 9) = 0
= 1 = 9
⸫3 = 1 3 = 9
3 = 30 3 = 32
= 0 = 2

3 4 − 3(2 +1) + 8 = 0
22 − 3(2 × 21) + 8 = 0
Let = 2 ,
2 − 3( × 2) + 8 = 0
2 − 6 + 8 = 0
( − 4)( − 2) = 0
= 4 = 2
⸫2 = 4 2 = 2
2 = 22 2 = 21
= 2 =1

4 22 +1 + 4 = 2 +3 + 2
(22 × 21) + 4 = (2 × 23) + 2
Let = 2 ,
( 2 × 2) + 4 = ( × 8) +
2 2 + 4 = 8 +
2y2 − 9y + 4 = 0

(2 − 1)( − 4) = 0

2 = 1 = 4
1

= 2 or = 4
⸫2 = 1 2 = 4

2

2 = 2−1 2 = 22

= −1 = 2

5 32 −3 − 4(3 −2) + 1 = 0

32 3
( 33 ) − 4(32) + 1 = 0
Let = 3 ,

( 2 72) − 4( ) + 1 = 0

9
y2 − 12y + 27 = 0

( − 9)( − 3) = 0

= 9 = 3

⸫3 = 32 3 = 31

= 2 = 1

84

6 16 − 5(22 −1) + 1 = 0

(24x) − 22 + 1 = 0
5( 21 )
Let = 22 ,

2 − 5(2) + 1 = 0

2y2 − 5y + 2 = 0

(2 − 1)( − 2) = 0

2 = 1 = 2

1 or y = 21
y=2
⸫22 = 2−1 22 = 21

2 = −1 2 = 1

−1 1
= 2 or = 2

EXAMPLE 3

Solve the following simultaneous equations:
3 −1 . 243 +2 = 81
83−
23 = 1

Solution:
3 −1 . 243 +2 = 81
3 −1 . (35) +2 = 34
3 −1 . 35 +10 = 34

(x −1) + (5y + 10) = 4

x + 5y = − 5
x = − 5 − 5y …………….①

83−
23 = 1
(23)3−
23 = 20

(2)9−3 = 20
23
29−3 −3 = 20

9 − 3y − 3x = 0

−3y − 3x = −9

y + x = 3 ……………②

Substituting ① into ②,

y − 5 −5y = 3

− 4y = 8

y=−2

Substituting y = −2 into ①,
x = − 5 − 5(−2)
x=5

85

EXERCISE 1.8

Solve the following simultaneous equations:

1. 3 × 9 = 1

22 × 4 = 1
8
2. 2 + 3 = 10

2 +1 + 3 +1 = 29

3. 2 − 5 = 7

2 −1 + 5 = 41

ANSWER 1 3 × 9 = 1
Section 3 × 32 = 30
1.3
x + 2y = 0
Exercise x = - 2y ………….①
1.8

22 × 4 = 1

8

22 × 22 = 2−3

2x + 2y = - 3…………….②

Sub. ① into ②,

2(- 2y) + 2y = - 3

- 4y + 2y = - 3

- 2y = - 3

y = 3
2

Sub.y = 3 into ①,
2

x = - 2( 3 )

2

x=-3

2 2 + 3 = 10
2 = 10 − 3 ………….①
2 +1 + 3 +1 = 29
(2 )(21) + (3 )(31) = 29 …………②

Sub. ① into ②,
(10−3 )(2) + (3 )(3) = 29
Let p = 3

(10 - p)(2) + (p)(3) = 29

20 - 2p + 3p = 29

p=9
⸫3 = 9
3 = 32

y=2

86

Sub. y = 2 into ①,
2 = 10 − 32
2 = 10 − 9
2 = 1
2 = 20

x =0

3 2 − 5 = 7
5 = 2 − 7………….①

2 −1 + 5 = 41.…………②

Sub. ① into ②, 2 −1 + 2 − 7 = 41
2
2 + 2 = 48

(23)2 = 48
2
2 = 48 × 3

2 = 32

2 = 25

x=5

Sub. x = 5 into②,
5 = 25 − 7
5 = 25

y=2

CLONE STPM

1. The function, f and g, are defined by ( ) = e− + 3 and ( ) = 2e + 2 respectively.

Find the intersection point and sketch, on the same coordinate axes, the graphs of f and g, for ∈ ℝ,

indicating the x- and y- intercept. [5 marks]

STPM PAST YEAR

STPM MT 2019(U)/2020 (Section A)
1. The function, f and g, are defined by ( ) = 6 − − 2 and ( ) = − 1 respectively.

(a) Sketch, on the same coordinate axes, the graphs of f and g, for ∈ ℝ, indicating the x-intercept

and y- intercept. [5 marks]

(b) Find the set of values of x which satisfies 6 − − 2 ≤ − 1. [5 marks]

87

ANSWER 1.(a) To find the point of intersection between f(x) and g(x),
Section
1.3 Let, f(x) = g(x)
e−x + 3 = 2ex + 2
Clone
STPM x ex, 1 + 3ex = 2e2x + 2ex

2e2x − ex − 1 = 0

Let y = ex, 2y2 − y − 1 = 0

(2y + 1)(y − 1) = 0

y = −1 or y = 2
2

⸫ ex = −1 or ex = 1
2

For all real value of x, ex = 1

x=0
⸫y = 2(1) + 2

y=4

Point of intersection is ( 0 , 4 )

ANSWER 1 (a)
Section
1.3

STPM
Past Year

88

(b) To find the point of intersection between f(x) and g(x),

Let, f(x) = g(x)
6 − − 2 = − 1

6 − − = 2 − 1

x , 6 − 2 =
Let y = , 2 + − 6 = 0

y² + y - 6 = 0

( y + 3 )( y - 2 ) = 0

y = - 3 or y = 2
⸫ = −3 or = 2
For all real value of x, = 2

x = ln 2

For 6 − − 2 ≤ − 1
f(x) ≤ g(x)
x ≥ ln 2

The set values of x is {x : x ≥ ln 2 }

LOGARITHMIC FUNCTIONS

RULES OF LOGARITHM

➢ log ==lolgo g + log Quick Notes
➢ log − log
Exponential Logarithmic
➢ log = log form form

Change of bases of logarithms The base can be any = log =
value 1 = 0 log 1 = 0
= 1 log = 1
log = log If the base is then, Alert with the location of
log the colours
log = log
1 log
log = log

NOTES
1. log =

2. ln =

3. log10 = log = lg (Common logarithms)
4. log = ln (Natural logarithms)

89

GRAPH OF LOGARITHMIC FUNCTIONS

Functions Graph

1. Domain: > 0
= ln Range: ∈ ℝ
−intercept = 1
= Asymptote: = 0

2. = − Domain: > 0
= − ln = 2 + Range: ∈ ℝ
−intercept = 1
3. Asymptote: = 0
= 2 + ln
As if graph = ln has
been reflected about
−axis

Domain: > 0
Range: ∈ ℝ
−intercept = −2

= 0.1353 (4 . )
Asymptote: = 0

As if graph = ln has
been shifted up 2 units

90 Domain: > −3
Range: ∈ ℝ
4. −intercept = −3 + −2 =
= 2 + ln( + 3) −2.865 (4 . )
−intercept = 2 + ln 3
= −3 Asymptote: = −3
= 2 + ( + 3)
As if graph = ln has been
translated 3 units to the left
and 2 units up

5. = ( + 3) Domain : > −3 OR (−3, ∞)
( ) = ln( + 3) Range : ∈ ℝ OR (−∞, ∞)

Graph of = ( ) Domain −1 : ∈ ℝ OR
and inverse, (−∞, ∞)
= −1( ) Range −1 : > −3 OR (−3, ∞)

To find the inverse: = − 3 graph = ( ) and =
−1( ) is a reflection of each
Let = ln( + 3)
+ 3 = across on the line =
= − 3
( ) =
−1( ) =
−1( ) = − 3
∴ −1( ) = − 3

91

WORKSHEET 1.8

Graph of logarithmic functions and its inverse

1. For the following logarithmic functions, find the inverse of . Hence sketch the graph
= ( )and = −1( ) on the same axes. State the particular in the third column.

Functions Graph Domain : > 0
1. Range : ∈ ℝ
= √ −intercept = 1
( ) = 2 ln = 2 −intercept = non
Asymptote: = 0
let ( ) =
−1( ) = Domain −1: ∈ ℝ
2 ln = Range −1: > 0



= 2 = √
−1( ) = √
∴ −1( ) = √

2. = 2 ( + 3) Domain : > −3
= √ − 3 Range : ∈ ℝ
( ) = 2 ln( + 3) −intercept = −2
−intercept = 2 ln 3
let ( ) = Asymptote: = −3

−1( ) = Domain −1: ∈ ℝ
Range −1: > −3
2 ln( + 3) =



+ 3 = 2 = √
= √ − 3
−1( ) = √ − 3

∴ −1( ) = √ − 3

92 = + 1 Domain : > 1
= ln( − 1) Range : ∈ ℝ
3. −intercept = 2
−intercept = non
( ) = ln( − 1) Asymptote: = 1

let ( ) = Domain −1: ∈ ℝ
−1( ) = Range −1: > 1
ln( − 1) =
= + 1 = 3 + (1 − ) Domain : < 1
−1( ) = + 1 = 1 − −3 Range : ∈ ℝ
−intercept = 1 − −3 =
∴ −1( ) = + 1 0.9502
−intercept = 3
4. Asymptote: = 1

f(x) = 3 + ln (1 - x) Domain −1: ∈ ℝ
let ( ) = Range −1: < 1

−1( ) =
3 + ln(1 − ) =
1 − = −3
= 1 − −3
−1( ) = 1 − −3

∴ −1( ) = 1 − −3

EXERCISE 1.9

1. Function ( ) and ( ), each defined for −1 < < 1, are given by
( ) = ln(1 − ) and ( ) = 2
a) Find −1( ) and state its domain and range
b) Show that ( ) + (− ) = ∘

2. The functions and are defined by ( ) = ln( − 1) , where > 1 and
( ) = √ − 2, where ≥ 2
a) Sketch on separate diagrams, the graph of functions and
b) (i) Explain why −1exist.
(ii) Hence, determine −1 and state its domain.
c) Does composite function ∘ exist? Why?

3. The functions and are defined by ( ) = 2ln , where > 0
and ( ) = √ , where ≥ 0

93

a) Sketch the graph of , and give reason why the inverse function −1 exists?
b) Find −1 and state its domain.
c) Find the composite function ∘ −1, and state its range

ANSWER 1(a) let ( ) = Quick Notes
Section −1( ) =
1.3 Domain = ∈ (−1, 1)
ln(1 − ) = Range = ∈ (−∞, ln 2)
Exercise 1 − = Domain −1 = ∈ (−∞, ln 2)
1.9 = 1 − Range −1 = ∈ (−1, 1)
−1( ) = 1 −
∴ −1( ) = 1 −

Domain −1 = { : < ln 2} Domain −1 = Range
Range −1 = { : − 1 < < 1} Range −1 = Domain

1(b) LHS:

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

= ln[1 − ( )]

= ∘

2(a)

= ln( − 1) = √ − 2

2(b) −1exist because it is one-to-one function

( ) = ln( − 1)
let ( ) =
−1( ) =

ln( − 1) =
= + 1
−1( ) = + 1
∴ −1( ) = + 1
Domain −1 = ℝ

94

2(c) Range = [0, ∞) Quick Notes
Domain = (1, ∞)
[0, ∞) ⊈ (1, ∞) ∘ exists if ⊆
⊈ Always list out all the domains
∴ ∘ does not exist. and ranges in the interval form,
such that it is easier to determine
3(a) the subset.

= 2 ln

−1 exists because is one-to-one function
(any horizontal line, = , only cuts the curve once)

3(b) ( ) = 2ln

let ( ) =
−1( ) =

2 ln =



= 2



−1( ) = 2



∴ −1( ) = 2

Domain −1 = ℝ

3(c) ∘ −1( ) = [ −1( )]



= ( 2)

= √
2



= 4, ∈ ℝ

Range ∘ −1 = { : > 0}

95

EQUATIONS and INEQUALITY involving LOGARTHMIC EXPRESSIONS

EXAMPLE 1
Express 3 log 10 − 2 log 5 − log 4 as a single logarithm.

Solution:

3 log 10 − 2 log 5 − log 4
= log 103 − log 52 − log 4

103
= log (52)(4)
= log 10

EXAMPLE 2 Quick Notes
Solve the equation (3 +2)(5 −1) = 15
There is no common base,
Solution: hence log in base 10 is
(3 +2)(5 −1) = 15 chosen
lg(3 +2)(5 −1) = lg 15
lg(3 +2) + log(5 −1) = lg 15
( + 2) lg 3 + ( − 1) lg 5 = lg 15
(lg 3 + lg 5) = lg 15 − 2 lg 3 + lg 5
lg 15 − 2 lg 3 + lg 5
= lg 3 + lg 5

= 0.7829

EXAMPLE 3 Quick Notes
Solve the inequality (2 +5)(42 −3) > 3 +6
Change the inequality
Solution: sign when divide or
(2 +5)(42 −3) > 3 +6 multiply by negative
lg(2 +5 ∙ 42 −3) > lg 3 +6 value.
lg(2 +5) + lg(42 −3) > lg 3 +6 In this case, the inequality
( + 5) lg 2 + (2 − 3) lg 4 > ( + 6) lg 3 sign remains the same
since the value of (lg 2 +
lg 2 + 5 lg 2 + 2 lg 4 − 3 lg 4 > lg 3 + 6 lg 3 2 lg 4 − lg 3) caries
(lg 2 + 2 lg 4 − lg 3) > 6 lg 3 − 5 lg 2 + 3 lg 4 positive sign.

6 lg 3 − 5 lg 2 + 3 lg 4
> lg 2 + 2 lg 4 − lg 3

> 3.077