Semester 1 Mathematics STPM (Student's Edition)

46

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:

47
Q3 Determine, in each case, the set of values of x where

(a) x2 > 3x
(b) x +1  1

x3
Solution:

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

5+ x

Solution:

48

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:

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
ANSWER

49 1
The set of values of x where 1  1 is { x : -1 < x < 1}
Section x +1 x −1
1.2

Exercise
1.5

2
y

y = –x + 1 y=x–1

= √3 −

–1 0 1 2 x

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

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

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

x

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

6 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) POLYNOMIAL Learning Outcome:
AND RATIONAL a) decompose a rational expression into partial fractions in cases where
FUNCTIONS
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.

50

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:

3 + 4
(1) ( + 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)

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

51

When = 2, 5 = 30 − 5(2)
= 4

When = −3, 5 = 30 − 5(−3)
= 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

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

52

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)

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

53

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

( − 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

54

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:

55

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

(a) 2 2−5 +7
( −2)( −1)2

(b) +1
( +3)2

Solution:

56

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

(a) 6−
(1− )(4+ 2)

(b) 3 2
( −1)( 2+ +1)

Solution:

57

4. Express in partial fractions (Improper Fractions) :

(a) 3 2−2 −7
( −2)( +1)

(b) 3 3
( +1)( −2)

Solution:

58

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 2 + 4 2
Section ( + 1)( + 2) = + 1
1.2
2 31 1
Exercise 2 − 9 = 2( − 3) − 2( + 3)
1.6

3 5( + 1) 3 2
25 − 2 = 5 − − 5 +

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

5 9 11 3
( − 1)( + 2)2 = − 1 − + 2 − ( + 2)2

59

6 1 11 1
2( 2 − 1) = − 2 − 2( + 1) + 2( − 1)

7 2 + 1 11
2 − 1 = 1 + − 1 − + 1

8 2 + − 1 11
( + 1)( + 2) = 1 − + 1 − + 2

9 10 − 11 −2 2 − 3
( − 4)( 2 + 1) = − 4 + 2 + 1

10 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

60

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 a = −15

1.2 b = 5

Clone 1(a) Factor = (x - 2)
STPM 1(b) ( ) = ( + 1)( − 2)(2 − 1)(4 + 1)
1(c) { : < −1 or − 1 < < 1 or > 2}

42

2 The solution set is { x : –4 < x < 0 or x > 5 }

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

4 = 1, = 1, = −1, = −1

5(a) o =

−1 = 1 , ≠ 1
2 − 2

5(b) = 1, = 0, = −1

5(c) ℎ( ) = o ( )

 6 Solution for inequality 2 − x  2 + 1 : x : x  2 − 5
x

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.

61

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) = −4
Section = 2
1.2
( ) = (2 + 1)( − 1)( 2 + 2 + 2)
STPM
Pass Year 1(b) 1
∴ {− 2 < < 1}
1(c)

2(a) g is not one-to-one function.
2(b)

( )
= | − 3| − 1

= 2 2
1

12 4

(3, −1)

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

62

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.

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

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

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

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

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

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

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

upwards.

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

65
WORKSHEET 1.7

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

(a) = +2 (b) = − + 3

(c) = − − −1 (d) = 1− + 2

66

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

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

67

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

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

68

i.e. 9y − 1 = 0 or y − 3 = 0

9y = 1 or y = 3

y=1

9
3 = 1 or
9 3 = 3
1 3 = 31
3 = 32 or
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 = 0 = 3
Section 2 = 0 = 2
1.3 3 = 2 =1
4 = −1 = 2
Exercise 5 = 2 = 1
1.7 6 = −1 or = 1

22

69

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− = 20
23

(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

70

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 x=-3, y=3
Section
1.3 2

Exercise 2 x=0,y=2
1.8 3 x=5,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]

71

ANSWER 1.(a) Point of intersection is ( 0 , 4 )
Section
1.3

Clone
STPM

ANSWER 1 (a)
Section
1.3

STPM
Past Year

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

72

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)

GRAPH OF LOGARITHMIC FUNCTIONS

Functions Graph

Domain: > 0

1. Range: ∈ ℝ

= ln −intercept = 1
Asymptote: = 0
=

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

As if graph = ln has
been reflected about
−axis

73 = 2 + Domain: > 0
Range: ∈ ℝ
3. −intercept = −2
= 2 + ln
= 0.1353 (4 . )
Asymptote: = 0

As if graph = ln has
been shifted up 2 units

4. Domain: > −3
= 2 + ln( + 3) Range: ∈ ℝ
−intercept = −3 + −2 =
= −3 −2.865 (4 . )
= 2 + ( + 3) −intercept = 2 + ln 3
Asymptote: = −3

As if graph = ln has been
translated 3 units to the left
and 2 units up

5. = ( + 3) Domain : > −3 OR (−3, ∞)
Range : ∈ ℝ OR (−∞, ∞)
( ) = ln( + 3)
Domain −1 : ∈ ℝ OR
Graph of = ( ) (−∞, ∞)
and inverse, Range −1 : > −3 OR (−3, ∞)
= −1( )
= − 3 graph = ( ) and =
To find the inverse: −1( ) is a reflection of each

Let = ln( + 3) across on the line =
+ 3 =
= − 3
( ) =
−1( ) =
−1( ) = − 3
∴ −1( ) = − 3

74

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
1.
Domain :
( ) = 2 ln Range :
−intercept =
−intercept =
Asymptote:

Domain −1:
Range −1:

∴ −1( ) =

2.
( ) = 2 ln( + 3)

75
3.

( ) = ln( − 1)

4.
( ) = 3 + (1 − )

76

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
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) ∴ −1( ) = 1 −
Section Domain −1 = { : < ln 2}
1.3 Range −1 = { : − 1 < < 1}

Exercise 1(b)
1.9
2(a)

= ln( − 1) = √ − 2

2(b)
2(c) ∴ ∘ does not exist.

77
3(a)

= 2 ln

3(b)

∴ −1( ) = 2

Domain −1 = ℝ

3(c)

∘ −1( ) = 4, ∈ ℝ

Range ∘ −1 = { : > 0}

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

78 Quick Notes

EXAMPLE 3 Change the inequality
Solve the inequality (2 +5)(42 −3) > 3 +6 sign when divide or
multiply by negative
Solution: value.
(2 +5)(42 −3) > 3 +6 In this case, the inequality
lg(2 +5 ∙ 42 −3) > lg 3 +6 sign remains the same
lg(2 +5) + lg(42 −3) > lg 3 +6 since the value of (lg 2 +
( + 5) lg 2 + (2 − 3) lg 4 > ( + 6) lg 3 2 lg 4 − lg 3) caries
lg 2 + 5 lg 2 + 2 lg 4 − 3 lg 4 > lg 3 + 6 lg 3 positive sign.
(lg 2 + 2 lg 4 − lg 3) > 6 lg 3 − 5 lg 2 + 3 lg 4
6 lg 3 − 5 lg 2 + 3 lg 4
> lg 2 + 2 lg 4 − lg 3
> 3.077

EXERCISE 1.10

1. For each of the following, express in terms of and ,

a) = 2(3 + )

b) ln( − 2) − ln( + 2) =

2. Solve the equation (3 − log3 ) log3 3 = 1

3. Given that log = 3 log 2 − log ( − 2 ) , express in terms of .
2

4. Find the value of such that (9 − log2 ) log2 8 = 2

5. Solve the equation ln + ln( + 2) = 1

6. Solve the simultaneous equations,

log3 = 5 and 2 = 2
log9

7. Solve the simultaneous equation

log4 = 1 and (log2 )(log2 ) = −2
2

8. Given that log (3 − 4 ) + log 3 = 2 + log (1 − 2 ) , where 0 < < 1,
log2
2

Find the value of .

79

ANSWER 1(a) 6
Section = ln [ − 2 ]
1.3
2 + 2
Exercise 1(b) = 1 −
1.10

2 = 3

3 , ∴ = 4

4 = 32

5 ∴ = −1 + √1 −

6 = 27 , = 9

7 = 1 , =4
2

8 2
= 3

1.4(a) Learning Outcome:
TRIGONOMETRIC a) Relate the periodicity and symmetries of sine, cosine and tangent
FUNCTIONS functions to their graphs and identify the inverse sine, inverse cosine
and inverse tangent function and their graphs
b) Use basics trigonometric identities and the formulae for sin (A ± B),
cos (A ± B) and tan (A ± B) including sin 2A, cos 2A and tan 2A

1.4 TRIGONOMETRIC FUNCTIONS

1. Graphs of basic trigonometric functions

Functions Graph Examples
Period : 2
= sin Amplitude : 1
Domain : (−∞, ∞)
Range : [-1, 1]

80 Period : 2
= cos Amplitude : 1
= tan Domain : (−∞, ∞)
= cot Range : [-1, 1]

= csc Period :
= sec
Domain : All real numbers except
+ ,
an integer
2

Range : [−∞, ∞]

Period :

Domain : All real numbers except
, an integer
Range : [−∞, ∞]

Amplitude : defined

Period : 2

Domain : All real numbers except
, an integer
Range : ≤ −1 ≥ 1

Amplitude : defined

Period :

Domain : All real numbers except

2 + , an integer

Range : ≤ −1 ≥ 1

Amplitude : defined

81

2. Graphing y = k + A sin (Bx + C) and y = k + A cos (Bx + C)

| | = Amplitude

Solve Bx + C = 0 and Bx + C = 2π for period and these shift

Bx + C = 0 Bx + C = 2π

= − = − + 2



Phase shift period

∴ period = 2



Phase shift (horizontal translation)

If − is positive : shift to the right



If − is negative : shift to the left



| | is vertical translation

If k positive : translate vertically up k units

If k negative : translate vertically down k units

3. Graphing = tan( + ) or = cot( + )

Amplitude is undefined

Solve + = 0 and + = 2 for period and these shift

+ = 0 + = 2

= − = − +



Phase shift period

4. Graphing = sec ( + ) or = csc ( + )

Amplitude is undefined

Solve + = 0 and + = 2

= − = − + 2



5. Inverse trigonometric functions

(a) The inverse sine function sin-1 or arc sin is defined as the inverse of the restricted sine function
y = sin x for − ≤ ≤ , −1 ≤ ≤ 1 .

22

Thus y = sin-1 x = arc sin x is equivalent to x = sin y for −1 ≤ ≤ 1, − ≤ ≤

• sin ( −1 ) = , −1 ≤ ≤ 1 22

sin−1(sin ) = , − ≤ ≤

22

82

(b) The inverse cos function cos-1 or arc cos is defined as the inverse of the restricted cosine
function = cos for 0 ≤ ≤ , −1 ≤ ≤ 1 .

Thus = cos−1 =

cos is equivalent to = cos for −1 ≤ ≤ 1, − ≤ ≤

22

• cos ( −1 ) = , −1 ≤ ≤ 1

−1(cos ) = , 0 ≤ ≤

6. Basic Trigonometric Identities

Reciprocal Identities sec = 1 cot = 1
csc = 1
cos tan
sin
tan(− ) = − tan
Quotient Identities cot = cos
tan = sin
sin
cos

Identities for Negatives
sin(− ) = − sin cos(− ) = cos

Pythagorean Identities 1 + cot2 = csc2
sin2 + cos2 = 1 tan2 + 1 = sec2

Sum Identities and difference Identities
sin (x ± y) = sin x cos y ± cos x sin y

cos (x ± y) = cos x cos y ∓ sin x sin y

tan ± tan
tan( ± ) = 1 ∓ tan tan

7. Cofunction Identities sin ( − ) = cos tan ( − ) = cot
cos ( − ) = sin
2 2
2

8. Double-angle Identities
sin 2 = 2 sin cos

cos 2x = cos2 x – sin2 x
= 1 – 2 sin2 x
= 2 cos2 x – 1

2 tan
tan 2 = 1 − tan2

83

9. Half-angle Identities
sin = ±√1−cos

2

cos = ±√1+sin

2

tan = ±√11 − cos
+ cos

= sin = 1−cos
1+cos sin

EXAMPLE 1 Shrink the graph of
Graph the function = 1 sin(2 − ) y = sin x horizontally by a
factor of 2, shrink it
24 vertically also by a factor
of ½ and shift it unit to
Solution:
= 1 sin(2 − ) 8

24 the right

1
= 2 sin [2 ( − 8)]

11
Amplitude = |2| = 2

2
Period = | 2 | =


Phase shift = 4 = 8

2

EXAMPLE 2

Graph the function = −1+ 3 sin(2 + ).

2 4

Solution:

Amplitude = 3 Period = |2 | =

4 2

2x + π = 0 2x + π = 2π The graph completes one full cycle as x

x = − x = − + = varies over the interval [− , ]
2 2 2 2 2

phase shift period

∴ phase shift = −
2

We graph = 3 sin(2 + ) , then vertically translate the graph down ½ unit.

4

84

Shrink the graph of y = sin x

horizontally by a factor of 2,

shrink it vertically also by a

factor of ½ and shift it 8 unit to

the right

EXAMPLE 3
Simplify the following

(a) sin (b) 2 − 1
cos tan(− ) 1 − 2

Solution: (b) = 2 − 1 2 −1
2
(a) sin = sin 1 − 2 2
1− 2
cos tan(− ) cos (− tan )

= − sin 2 − 2
cos tan
= 2
− sin 2 − 2
cos (csoins ) 2

= 1 ⋅ 2
2
= 1

= − sin = −1 = 2
sin 2

WORKSHEET 1.9 = 2

Sketch the graph of = sin for 1. = sin

0 ≤ ≤ 3600. Hence sketch the graph of
(a) = 1 + sin
(b) = sin ( + 30°)
(c) = sin 2

(a) = 1 + sin (b) = sin ( + 30°)

85
(c) = sin 2

EXERCISE 1.11

1. State the amplitude A and period P of each function and graph the function over the indicated

interval.
(a) = 3 sin , − 2 ≤ ≤ 2
(b) = − 1 cos , − 2 ≤ ≤ 2

2

2. State the amplitude and period of the function and graph the function over the indicated

interval.

(a) = 2 + 2 sin , −4 ≤ ≤4

2
(b) = 4 − 2 cos , −4 ≤ ≤ 4
2

3. Find the period and phase shift than graph each function.

(a) = tan(2 + ), −3 < < 3

44

(b) = −2 tan ( − ) , −1 < <7

44

4. Evaluate cos (sin−1 √2 + cos−1 53).

2

5. Solve = sin−1 √2

2

6. Simplify each of the following trigonometric expressions.

(a) 2 sin2 + sin − 3
1 − cos2 − sin

(b) sin + cot
1 + cos

86

7. Simplify the following

(a) sin2 − 9 ∙ 10 cos + 5
2 cos + 1
3 sin + 9

(b) 1 − 2
sin2 − cos2 cos sin

(c) csc4 − 1
csc2

ANSWER 1(a) = 3 sin Amplitude = |3| = 3 Period of = 2
Section
1.4 1(b) The graph of − 1 cos 2 is the graph of − 1 cos 2 reflected across the x-
2 2
Exercise
1.11 axis. Amplitude = |1| = 1 Period of =

22

2 (a) A = 2, period = 2 = 4



4

2(b) 2
= 4, period = = 4

2

3(a)
∴ phase shift = − 2 , period = 2

3(b) ∴phase shift=1 ,period=4

4 −√2
10

5
= 2

6 (a) 2+ 3

sin

(b) cosec x

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

(c) 2 + 2

87

CLONE STPM

1. Sketch the graph of = 2 cos ( − ) for 0 ≤ ≤ 2 .

3

2. Sketch the graph of = √13 cos( − 56.3°) for 0 ≤ ≤ 2 .

STPM PAST YEAR
STPM MT 2015 (Section B Q7(c) )

1. Sketch = 13 cos( + 0.4) for 0 ≤ ≤ 2 .

STPM MT 2018 (Section B Q7(b) )
2. Sketch the graph of = 2 sin( − ) for 0 ≤ ≤ 2 .

6

ANSWER 1
Section
1.4

Clone
STPM


= 2 cos ( − 3)
2

= √13 cos( − 56.3°)

88 1

ANSWER
Section
1.4

STPM
Past
Year

= 13 cos( + 0.4)

2


= 2 sin( − 6)

1.4 (b) Learning Outcome:
TRIGONOMETRIC Find the solution, within specified intervals, of trigonometric equations
EQUATIONS

Trigonometric Equations

Simple trigonometric equations

The simple trigonometric equations like sin = 1 or cos = − √3 or tan(3 + ) = 1 can
2 2

be solved in few steps:

1. Determine the quadrants, the angle should be in, based on the given trigonometric equation.

2. Find the basic angle by using a scientific calculator
3. Determine the range of values of the required angles, for examples the range of values for

or 3 .

4. Determine the values of angles in those quadrants.

89

EXAMPLE 1

Solve the following trigonometric equations, for 0° ≤ ≤ 360°

a) tan = 1.0 b) cos = − 1

2

Solution: Solution:
tan = 1.0 1

Basic angle: cos = − 2
Basic angle:
.
= cos−1 1 = 60°
= tan−11 = 45o 2

tan x has positive sign in the first But cos x has negative sign and thus the angles
and third quadrants . are in the second and third quadrants

∴ = 45°, -135o or 225o ∴ = 120°, −120° 240

EXAMPLE 2
Solve sin − √3 cos = 0 for 0° ≤ ≤ 360°

Solution:
sin − √3 cos = 0
sin = √3 cos
tan = √3
= tan−1√3 = 60°

Since tan > 0, thus x falls in the I and III quadrants. Hence, = 60° , 240°

EXAMPLE 3
Solve the equation 3 2 − 17 cos + 10 = 0 for 0° ≤ ≤ 360°.

Solution:

3cos2 − 17 cos + 10 = 0

Let = cos , 3 2 − 17 + 10 = 0

(3 − 2)( − 5) = 10

= 2 = 5

3

Substitute back:

cos = 2 cos = 5 (undefined)
3

2
cos = 3 ∶ = 48.19° , 311.81

90

EXERCISE 1.12
1. Find the solution for the following trigonometric equations in the given interval
a) tan = 0.2 , 0 ≤ ≤ 360°
b) sin 2 = 0.5 , 0° ≤ ≤ 360°
c) cos(2 + 30°) = 0.5 , − 180° ≤ ≤ 180°

2. Solve the equation exactly for 0≤ ≤ 2
a) 2 cos − 3 = −5
b) cosec = −2
c) 2sin2 − 1 = 0

3. Solve 2(cos + 1) = 1 for 0≤ ≤ 2

4. Solve 4 tan + 2 = 2 tan

5. Solve the equation 2sin2 − 5 sin + 3 = 0 for 0 ≤ ≤ 2

6. Solve 5cos2 − 2 cos 2 = 3 for 0≤ ≤ 2

7. 5 tan − cot 2 + 5 = 0 for 0° ≤ ≤ 360°

8. Solve 5sin2 − 4 sin − 1 = 0 for 0° ≤ ≤ 360°

ANSWER 1(a) = 11.31° 191.31°
Section 1(b) = 15°, 75°, 195°, 255°
1.4 1(c) = −45° , 15°

Exercise
1.12

2(a) =

2(b) 7 11
= 6 , 6

2(c) 3 5 7
= 4 , 4 , 4 , 4

3 2 4
= 3 , 3

91

4 = 135° , 315

5
= 2

6 ∴ = , 0 , 2

7 ∴ = 5.19°, 185.19° ∴ = 135°, 315°

8 ∴ = 90°, 192°, 348°

CLONE STPM

1. Solve the following trigonometric equations for −180° ≤ ≤ 180°

a) sin = 1

5

b) cos = − √3

5

2. Find the solutions for cos ( + ) = sin ( + ) in [− , ].

63

3. By using t = tan , solve the following equation in the given interval
2

4 sin − 3 cos = 3 ; [− , )

4. Given that 4sin + 5cos = 0, find the value of tan .
Hence, solve the equation (1 − tan )(4sin + 5cos ) = 0 in the interval
0o ≤ ≤ 360o, giving your values of to the nearest 0.1o.

5. Find the solutions of 9sin2 + 10 sin cos − 2cos2 = 1 in the interval [0° , 360°).

ANSWER 1(a) = 11.54°
Section = 168.46°
1.4
1(b) = 110.27°
Clone = -110.27°
STPM

92

2 = − , 0,
3 = 73.74°
4 = 450, 2250, 128. 70, 308. 70
5 ∴ = 14.04°, 123.69°, 194.04°, 303.69°

1.4(c) EQUATION Learning outcome:
Express a sin Ѳ ± b cos Ѳ in the form r sin (Ѳ ± α) and
INVOLVING THE r cos (Ѳ ± α)

FORMS
a sin Ѳ ± b cos Ѳ

Expressing sin + cos in the form sin( ± ) or cos ( ± )
c)

For r > 0 and 0 < < 2

sin + cos = sin( + )

sin − cos = sin( − )

cos + sin = cos( − )

− sin = cos ( + )

Where = √ 2 + 2 = tan−1 ( )



EXAMPLE 1
Solve 4 cos − 3 sin = 1 for 0 ≤ ≤ 360°

Solution:
Let 4 cos − 3 sin = cos ( + )

4 cos − 3 sin = cos cos − sin sin

Compare both sides Remember to take
4 = cos …….. (1) only the numerical
3 = sin ……….(2) values. Ignore the
sign
(1)2 + (2)2 42 + 32 = 2
25 = R2

R=5

93

(2)(2) ÷ (1) 3 = tan

4

= 36.87°

Then 5cos ( + 36.87°)=1
cos( + 36.87) = 1

5

Basic angle : 78.46o
+ 36.87° = 78.46° 281.54°

Thus, = 41.59° 244.67°.

EXAMPLE 2
Find the minimum and maximum values of

a) 8 cos
b) 5 cos 2

Solution:
(a) −8 ≤ 8 cos ≤ 8
Minimum value = - 8
Maximum value = 8

(b) −5 ≤ 5 cos 2 ≤ 5
Minimum value = - 5
Maximum value = 5

EXAMPLE 3
Express sin 3 − cos 3 in the form sin (3 − ) with R > 0 and 0 ≤ ≤ 90°.
Solve the equation sin 3 − cos 3 = 1 for 0° ≤ ≤ 180°.

Solution:

sin 3 − cos 3 = R sin (3 − )

sin 3 − cos 3 = R sin 3 cos − sin cos 3

Compare both sides, 1 = cos ______(1)
1 = sin ______(2)

(1)2 + (2)2, 2 = 2( 2 + 2 )
2 = 2

= √2

94

(2) , tan = 1

(1)

= 45°

∴ sin 3 − cos 3 = √2 sin(3 − 45°)

Solve √2 sin (3 − 45°) = 1

sin(3 − 45°) = 1 Basic angle :45o

√2

So, 3 − 45° = 45°, 135°, 405°, 495°

3 = 90°, 180°, 450°, 540°

= 30°, 60°, 150°, 180

EXERCISE 1.13

1. By reducing the following trigonometric equations into cos( ∓ ) = or

sin( ± ) = , find the solutions in the given intervals.

(a) 4 sin − 3 cos = 2 [−360°, 360°]

(b) 24 sin + 7 cos = 5 [0°, 360°]

2. Each of the following expressions can be written in the form R cos (x − α) with −π < α < π.
In each case determine the values of R and α (in radians) correct to 3 decimal places

(a) 5 cos x + 12 sin x
(b) 6 cos x + 5 sin x

3. Solve the following equations for 0 < x < 2π
(a) cos x − 2 sin x = 1

(b) cos x + 2 sin x = 1

4. Express 7 sin 2 − 24 cos 2 = 15 in the form sin( − ) = , find the solution for the
interval 0° ≤ ≤ 360°.

5. Express the equation 4 sin 2 + 3 cos 2 in the form cos (2 − ).
(a) Hence solve the equation 4 sin 2 + 3 cos 2 = 1 for 0 ≤ ≤ 2 .
(b) Find the maximum and minimum values of 4 sin 2 + 3 cos 2 + 3.

95

ANSWER 1(a) 5 sin( − 36.87°) = 2
Section 1(b) = −166.71°, −299.55°, 60.45°, 193.29°
1.4
25 sin( + 16.26°) = 5
Exercise = 152.2°, 355.28°
1.13

2(a) ∴ = 13 = 1.176
2(b) ∴ = √61 = 0.6947

3(a) = 4.069

3(b) = 2.2142

4 25 sin (2 − 73.74°) = 15
= 55.31°, 108.44°, 235.28°, 288.41°

5(a) ∴ 4 sin 2 + 3 cos 2 = 5 cos (2 − 0.9273)
= 1.15, 2.92, 4.29, 6.06

5(b) The maximum value is 5 + 3 = 8
The minimum value is – 5 + 3 = - 2

CLONE STPM

1. By first expanding cos ( + 45°), then express cos ( + 45°) − √2 sin in the
form cos( +∝), where R  0 and 0° < ∝< 90°. Give the value of R correct to 4 significant
figures and the value of ∝ correct to 2 decimal places.
Hence, solve the equation cos ( + 45°) − √2 sin = 2, for 0° < < 360°.

2. Express 24cos − 7sin in the form r cos( +  ) where r  0 and 0     .

2

Hence,

(a) State the minimum and maximum values of
1

30 + 24 − 7
(b) Solve the equation 24 cos − 7 sin = 0 , 0 ≤ ≤ 2 .

3. Given that 5 cos 2 − 8 sin 2 ≡ cos (2 + ), where > 0 and 0 < < , state the
2

value of R and find the value of α in radians to three decimal places.