Modul 1 Functions EDISI GURU

51

EXERCISE 1.2(a)

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.

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 (a) Solve the equation |3 + 4| = |3 − 11|.
(b) Hence, using logarithms, solve the equation |3 × 2y + 4| = |3 × 2 − 11|, giving the answer
correct to 3 significant figures.

52

ANSWER Module 1

Section 1. 4 − 5 3 + 2 + 7 + 1 = ( − 1) ( ) − 8
1.2 4 − 5 3 + 2 + 7 + 1 = (2 + 1) ( ) + 11

Exercise 2
1.2(a)
When = 1, − 5 + + 7 + 1 = −8

+ = −11 − − − (1)

When = − 1, (− 1)4 − 5 (− 1)3 + (− 1)2 + 7 (− 1) + 1 = 11

22 2 2 2 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
=

∴ = −

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,


= − , , , −

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

53 ( ) = 2 3 − 9 2 + 3 + 4
Let 2 3 − 9 2 + 3 + 4 ≡ ( − 1)(2 2 + − 4)
Section Comparing coefficients of 2 : − 9 = − 2
1.2
= −7
Exercise
1.2(a) ∴ ( ) = 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

For 2 + 2√3 + 1 = 0
= −√3 ± √2

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

5.(a) (3 + 4)2 = (3 − 11)2 Or 3 + 4 = −(3 − 11)

9 2 + 24 + 16 = 9 2 − 66 + 121
90 = 105
= 7

6

54 5.(b) |3 × 2y + 4| = |3 × 2 − 11|

Section Let 2 = 7
1.2 6
7
Exercise log 2 = log 6
1.2(a)
7
log 2 = log 6

= 0.222

55
Learning Outcome:
a)Solve polynomial and rational inequalities
b)Solve inequalities involving modulus signs in simple cases

I1N.2E(bQ)➢UALDAInETiInFEeISqNuaIliTtyIiOs sNimilar 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.

➢ 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 1  1
5. If A > 0 and B > 0, then A < B  AD B
6. If A < B and C < D, then A + C < B +

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 eachfactor 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

56

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

EXAMPLE 2

Find the set of values for x where

(a) 2x +1  1 (b) x + 3  3x − 5 (c) ( x + 3 ) ( 2 – x ) ( x – 1 ) > 0
1+ x 2 x

Solution :

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

57

(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 }

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
3x3 – x2 – 12x + 4 < 0 -+ x-2 > 0
Let p(x) = 3x3 – x2 – 12x + 4 -- x

When x = 2 1 5x - 3 > 0
p(2) = 3(2)3 – (2)2 – 12(2) + 4 = 0 3x - 5 > 0
Therefore, (x – 2) is a factor of p(x) 3-
x
When x = -2
p(-2) = 3(-2)3- (-2)2 – 12(-2) + 4 = 0

Therefore, (x+2) is a factor of p(x) - -2 + 2+

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

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

58

EXAMPLE 5

Find the set of values of x for which 2x −1  1 y
x
y= 1
Solution : y=- 1 y=2x-1
2x+1 x
Sketch y = 2x − 1 and y = 1
x 0 11
2
y = 2x – 1 ….(1)

y = 1 …….(2)
x

Substituting (1) into (2)

y = 2x – 1 = 1
x

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

1

x2 – 2x – 2 = 0

x = 2 4+8 -1 0 1 x
2

=1− 3

For 2x + 1 > 1 – x2 ,

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

59

WORKSHEET 1.2(b)
Q1 Find, in each case, the set of values of x for

(a) ( x -2 ) > x + 4
(b) x – 2 > x + 4

x

1.(a) x ( x – 2 ) > x + 4 - - + x-4>0
x2 – 3x + 4 > 0 + + x+1>0
(x–4)(x+1)>0
4 x
The solution set is { x : x < -1 or x > 4 } -
+
+ -1 -

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

x2 − 3x − 4  0 -- -+
x
x
(x − 4)(x +1)  0
x - -1 + 0 4+

-

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

2. 6y = 1 + x ……(1)

y=1 ……….(2) y
x

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

x = -3 or x = 2 B● ●A

At point A , x > 0 0

Therefore x = 2 x
6y = 1 + x ……(3)

y = − 1 ……...(4)
x

60

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

Q3 Determine, in each case, the set of values of x where
(a) x2 > 3x
(b) x +1  1
x3

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

+ 0-

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

The set of values of x where x +1  1
x3

is { x : x < − 3 or x>0}
2

61

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

4. x  2 -- + 3x+10 > 0
5+ x -+ + x+10 > 0

 x 2  4 + −10 - 10 x
5+ x
x2 < 4(5 + x)2 −3 +
x2 < 100 + 40x + 4x2
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

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

5. 4x − 11  x − 3 = x − 3 − 4x − 11  0
x+6 2 2 x+6
+
= x2 + 3x −18 − 8x + 22  0 - - + x-1 > 0
2(x + 6) - + x+1 > 0

= x2 − 5x + 4  0 x

2(x + 6) + −1 - 1

= (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 >
x +5 2 - - ++ 0x-1 > 0
- -+
From (1) , by deduction : - x
x −1 −6 -6 +

x  − 5 ( undefined )

1 x −1 4 - 1 4+
2 x 5
-

4  x  25

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

62

EXERCISE 1.2(b)

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

Exercise
16.63

ANSWER Module 1

Section 1. . 1  1
1.2 x +1 x −1
1 − 1 0
Exercise x +1 x −1
1.2(b) −2 0
(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. . | x – 1 | > 3 − x y

(x – 1)2 > 3 – x y = -x + 1 y=x-1
x2 – 2x + 1 > 3 – x
x2 – x – 2 > 0
(x+1)(x–2)>0

= √3 −

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

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

+ −1 - 2

+

3. x − 6  3
x+2

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

− 2x − 12 0 + −6 - −2
x+2
+

x+6  0
x+2

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

4. 4x − 5  x − 2

x

4x −5 + 2 − x  0
x

4x − 5 + x(2 − x)  0
x

64

6x −5− x2  0 - + + + x>0
x
-- + +

x-1 > 0

x2 −6x + 5  0 -- - +
x
x-5 > 0

(x −1)(x − 5)  0 - 0+ 1 5+
x
-

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

x

5. 1  1

2−x x−3

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

2x −5 0 2-
(2 − x)(x − 3)

2x −5  0
(x − 2)(x − 3)

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

6. . -16 < x3 – 4x2 + 4x – 16 < 0

For x3 – 4x2 + 4x – 16 < 0 +
+
Let f(x) = x3 – 4x2 + 4x – 16 - - x-4 > 0
+ 4 x>0
f(4) = 6 – 64 + 16 – 16 = 0 -
(x – 4) is a factor of f(x) + x

x3 – 4x2 + 4x – 16 = (x – 4)(x2 + 4)

by inspection + 0-
(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 }

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 }

65

1.2(c) Polynomial Learning Outcome:
and Rational a) decompose a rational expression into partial fractions in cases where
Functions
denominator has two distinct linear factors, or a linear factor and a pri

factor

Rational Function b)
c)

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

< , ( ) .

≥ , ( ) .

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)

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

Type 2 Denominator with repeated linear factors (Proper fraction)

Eg:

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

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

Type 3 Denominator with prime quadratic factors (Proper fraction)

Eg:

(1) 3 + 4 +
( + 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:

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

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

66

EXAMPLE 1
Express 30−5 as partial fractions.

(2− )(3+ )

SOLUTION:
30−5 = +

(2− )(3+ ) (2− ) (3+ )

(3 + ) + (2 − ) = 30 − 5

ℎ = 2, 5 = 30 − 5(2) = 20, = 4

ℎ = −3, 5 = 30 − 5(−3) = 45, = 9

Therefore,

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

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

ℎ = 0, − = 1, = −1

4
ℎ = −1, 2 = 4, = 2 = 2

6
ℎ = 1, 2 = 6, = 2 = 3

Therefore,

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

67

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

ℎ = −1, − = −5 + 1 = −4, = 4

11 3
ℎ = − 2 , 4 = − 2 , = −6
ℎ 2,

(−6)
2 + = 0, = − 2 = − 2 = 3

Therefore,

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

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

ℎ = 2,5 = 5, = 1

ℎ 2,

+ = 0, = − = −1, = −1

, ( ℎ = 0)

− 2 = 1, 2 = 1 − = 1 − 1 = 0, = 0

Therefore,

2 + 1 1
( − 2)( 2 + 1) = − 2 − 2 + 1

68

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

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


= + 4 + − 2
( − 2) + ( + 4) = 4 + 10

ℎ = −4, −6 = −6, = 1

ℎ = 2, 6 = 18, = 3

ℎ ,

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

69

WORKSHEET 1.2(c)

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

(a) −11 (b) 3 2−21 +24
( +3)( −4)
( −2)( +1)( −3)

1.(a) − 11
( + 3)( − 4) = + 3 + − 4

( − 4) + ( + 3) = − 11

ℎ = −3, −7 = −14, = 2

ℎ = 4,7 = −7, = −1

Therefore,

− 11 21
( + 3)( − 4) = + 3 − − 4

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

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

ℎ = 2, −3 = −6, = 2

ℎ = −1,12 = 48, = 4

ℎ = 3,4 = −12, = −3

Therefore,

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

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

(a) 2 2−5 +7 (b) +1
( −2)( −1)2 ( +3)2

2(a) .

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

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

ℎ = 2, = 5

ℎ = 1, − = 4, = −4

ℎ = 0, + 2 − 2 = 7, 5 + 2 + 8 = 7,2 = −6, = −3

Therefore,

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

70

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

( + 3) + = + 1

ℎ = −3, = −2

ℎ = 0, 3 + = 1, 3 − 2 = 1,3 = 3, = 1

Therefore,

+ 1 1 2
( + 3)2 = + 3 − ( + 3)2

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

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

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

(4 + 2) + ( + )(1 − ) = 6 −

ℎ = 1,5 = 5, = 5

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

− = 0, = , = 1

Therefore,

6 − 1 + 2
(1 − )(4 + 2) = 1 − + 4 + 2

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

( 2 + + 1) + ( + )( − 1) = 3 2

ℎ = 1,3 = 3, = 1

ℎ = 0, − = 0, = 1

2,

+ = 3, = 3 − , = 2

Therefore,

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

71

4. Express in partial fractions (Improper Fractions) :

(a) 3 2−2 −7 (b) 3 3

( −2)( +1) ( +1)( −2)

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

1
ℎ = 2, 3 = 1, = 3

2
ℎ = −1, −3 = −2, = 3

Therefore,

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

(b) 3 3 3 3
( + 1)( − 2) = 2 − − 2

Using long division method,

3 + 3

2 − − 2 3 3

3 3 − 3 2 − 6

3 2 + 6

3 2 − 3 − 6

9 + 6

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

9 + 6
( + 1)( − 2) = + 1 + − 2


= + 4 + − 2
( − 2) + ( + 1) = 9 + 6

ℎ = −1, −3 = −3, = 1

ℎ = 2, 3 = 24, = 8

Therefore,

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

72

EXERCISE 1.2(c)

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)

73

ANSWER Module 1

Section 1 2x + 4
1.2G ( + 1)( + 2) = + 1 + + 2

Exercise ( + 2) + ( + 1) = 2 + 4
1.2G
ℎ = −1, A = −2 + 4, A = 2

ℎ = −2, −B = −4 + 5, B = 0

2x + 4 2
ℎ , ( + 1)( + 2) = + 1

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

1
ℎ = 3, 6A = 3, A = 2

1
ℎ = −3, −6B = 3, B = − 2
31 1
ℎ , 2 − 9 = 2( − 3) − 2( + 3)

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

ℎ = 5, 10A = 30, A = 3

ℎ = −5, 10B = −20, B = −2

5( + 1) 3 2
ℎ , 25 − 2 = 5 − − 5 +

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

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

ℎ = −2, (−1)(−3)A = 4(−2) + 2

3A = −6

A = −2

ℎ = −1, (1)(−2) = −4 + 2

= 1

ℎ = 1, (3)(2) = 4 + 2

C= 1

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

74

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

( + 2)2 + ( − 1)( + 2) + ( − 1) = 9

ℎ = 1, (1 + 2)2 = 9, A = 1

ℎ = −2, (−2 − 1) = 9, = −3

ℎ = 0, 4 − 2 − = 9

4(1) − 2 + 3 = 9

2 = −2

= −1

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

6 1
2( 2 − 1) = + 2 + + 1 + − 1

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

ℎ = 0, (1)(−1) = 1

= −1

ℎ = −1, (−2) = 1

= − 1

2

ℎ = 1, (2) = 1

= 1

2

ℎ : − = 0

A=0

1 11 1
ℎ , 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

ℎ = 1, 2 = 2

= 1

ℎ = −1, −2 = 2

= −1

2 + 1 11
ℎ , 2 − 1 = 1 + − 1 − + 1

8 2 + − 1
( + 1)( + 2) = + + 1 + + 2

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

ℎ 2, = 1

ℎ = −1, (−1 + 2) = 1 − 1 − 1

= −1

75

ℎ = −2, (−2 + 1) = 4 − 2 − 1

= −1

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

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

( 2 + 1) + ( + )( − 4) = 10 − 11

ℎ = 4, (16 + 1) = 10 − 44

= −2

ℎ = 0, − 4 = 10

−2 − 4 = 10

= −3

ℎ 2, + = 0

−2 + = 0

= 2

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

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

( 2 − + 1) + ( + )( − 1) = 1

ℎ = 1, (1 − 1 + 1) = 1

= 1

ℎ = 0, (1) − = 1

1 − = 1

= 0

ℎ 2, + = 0

1 + = 0

= −1

1 1
ℎ , ( − 1)( 2 − + 1) = − 1 − 2 − + 1

76

CLONE STPM 1.2

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

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

77 1(b) Module 1

ANSWER ( ) = 8 4 − 10 3 − 15 2 + 5 + 2
Section = ( + 1)(8 3 − 18 2 + 3 + 2)
1.2 = ( + 1)( − 2)(8 2 − 2 − 1)
= ( + 1)( − 2)(2 − 1)(4 + 1)
Clone
STPM when ( ) > 0
( + 1)( − 2)(2 − 1)(4 + 1) > 0
1(c) { : < − − < < > }



2.

y

B
A

-4 0 2 x

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

y= x−2 = {− ( − −2 , ≥ 2
2) , < 2

Point A :
- ( x – 2) = y = x + 4

[ -( x – 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 }

78 3. When 1 = x – 2

Section
1.2G x2 – 2x – 1 = 0

Clone x = 2  (−2)2 − 4(−1) y
STPM 2 y= 1

x= 2 8 = | − 2|
2 x
= 1 2

Since x > 0 , x = 1+ 2

When 1 = −x + 2 01

x

x2 – 2x + 1 = 0
( x – 1 )2 = 0

x=1

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

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

(1 + )3 + (1 − )(1 + )2 + (1 − )(1 + ) + (1 − ) = 3 2 + 5
ℎ = −1, 2 = 3 − 5, = −1
4. ℎ = 1, 8 = 3 + 5, = 1
ℎ 3,

− = 0, = , = 1
ℎ = −1, 2 = 3 − 5, = −1

ℎ , = 1, = 1, = −1, = −1

( ) = 1 , ( ) = 2 + +
2 −

◦ = [ ( )]


= (2 − 1)

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

2 − 1
=1
=

◦ = ,

−1 =

1
= 2 − 1 , ≠ 2

ℎ , −1: → x , x ≠ 1
2x −1 2

79

◦ ( ) = [ ( )]

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

2 + (2 − 1) + (2 − 1)2
= (2 − 1)2

, ◦ ( ) = −3 2+4 −1
(2 −1)2

5(b) −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

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

ℎ = 0, = −1

2,

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

ℎ , = 1, = 0, = −1

( ) = 2 − 2

ℎ( ) = 2 − 2
2 2 − 5
[ ( )] = ( 2 − 2)

2 − 2
= 2( 2 − 2) − 1

2 − 2
= 2 2 − 4 − 1

2 − 2
= 2 2 − 5
5(c) ℎ , ℎ( ) = ◦ ( )

80
6

f(x) = 2-x y

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

81
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)

82

ANSWER Module 1

Section 1. (a) Given that (2 + 1) is a factor of ( ),
1.2
By the factor theorem,
Clone
STPM 1
(− 2) = 0
1)4 1)3 1)2
(− + (− + (− − 10 (− 1) − 4 = 0
2 2 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
= −
∴ =

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

(a) (2 + 1)( − 1)( 2 + 2 + 2) < 0

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

1 1 x
−2
∴ {− < < }



2.(a)
(b) ( ) = { − , <

− , ≥ ,

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

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

83

( )


=

= 2 2
1 (2, 1)

Type eqTuyapteioenqhueartieo.n he1re. 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)

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

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