Chapter 1 Number Systems 1A p

1

Chapter 1 Number Systems

1A p.2

1B p.14

1C p.22

Chapter 2 Equations of Straight Lines

2A p.42

2B p.52

2C p.60

2D p.66

Chapter 3 Quadratic Equations in One Unknown

3A p.76

3B p.83

3C p.94

3D p.106

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2

F5A: Chapter 1A

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Consolidation Exercise


(Full Solution)

Maths Corner Exercise 1A Level 1


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5A Lesson Worksheet 1.0 (Refer to Book 5A P.1.3)

Objective: To review solving simultaneous linear equations in two unknowns by the graphical method and

the algebraic method, and solving quadratic equations in one unknown by the graphical method

and the algebraic method.

Solving Simultaneous Linear Equations in Two Unknowns by the Graphical Method

In each of the following, use the given figure to solve the simultaneous equations. [Nos. 1–2]

1.

=+

=+

2

32

yx

yx 2.

=−

−=+

445

623

yx

yx �Review Ex: 1

From the figure, the point of intersection of

the graphs is ( , ).

∴ The solution is x = ( ), y = ( ).

Solving Simultaneous Linear Equations in Two Unknowns by the Algebraic Method

Solve the following simultaneous equations algebraically. [Nos. 3–4]

3.

+=

=−

(2)...............

(1)...............

83

12

yx

yx 4.

=−

=+

(2)..............

(1)...............

543

74

yx

yx �Review Ex: 2–4

Substitute ( ) into ( ).

−1 0 −2 1

−1

−2

−3

5x − 4y = 4 3x + 2y = −6

x

y

1

0 1 2 3

2

3

y

x

2x + y = 3

x + y = 2

4

Solving Quadratic Equations in One Unknown by the Graphical Method

In each of the following, use the given figure to solve the quadratic equation. [Nos. 5–6]

5. x2 + 4x + 3 = 0 6. –x2 + x + 5 = 0 �Review Ex: 11

From the figure, the x-intercepts are ( )

and ( ).

∴ The solutions are .

Solving Quadratic Equations in One Unknown by the Algebraic Method

Solve the following equations by the factor method. [Nos. 7–8]

7. x2 + 2x – 8 = 0 8. x2 – 8x + 15 = 0 �Review Ex: 5–7

Solve the following equations by the quadratic formula. [Nos. 9–10]

9. 2x2 – 3x + 1 = 0 10. 9x2 + 6x + 1 = 0 �Review Ex: 8–10

��Level Up Question��

11. The figure shows the graph of x + y = 1. Solve

+=

=+

3

1

xy

yx by adding

a suitable straight line to the figure.

0

1

2

–3 –2 –1

–1

–4

y

x

y = x2 + 4x + 3

a = b =

c =

x =a

acbb

2

42−±−

a = b = c =

y

0

1

–2 –1

x + y = 1

x 1

2

3

2

4

−2 −1 0

1 2 3

6

4

2

x

y

y = −x2 + x + 5

5


Objective: To use the graphical method to solve simultaneous equations in two unknowns (one linear and

one quadratic in the form y = ax2 + bx + c).

Solving Simultaneous Equations (One Linear and One Quadratic) by the Graphical Method

Step 1: Draw the graphs of the linear equation and the quadratic equation in the same rectangular

coordinate plane.

Step 2: Read the points of intersection of the graphs.

Step 3: Write down the solutions of the simultaneous equations.

Instant Example 1 Instant Practice 1

The figure shows the graphs of y = 0.5x2 − 2x and

y = x – 4. Use the figure to solve the

simultaneous equations

−=

−=

4

25.0 2

xy

xxy.

From the figure, the points

of intersection are

(2.0 , −2.0) and (4.0 , 0.0). � ∴ The solutions of the simultaneous equations are

x = 2.0, y = −2.0 and x = 4.0, y = 0.0.

The figure shows the graphs of y = –x2 + 2 and

2x + 3y + 2 = 0. Use the figure to solve the

simultaneous equations

=++

+−=

0232

22

yx

xy.

From the figure, the points of

intersection are ( , )

and ( , ). ∴ The solutions of the simultaneous equations

are x = ( ), y = ( ) and x = ( ),

y = ( ).

1. The figure shows the graphs of y = –x2 – 3x and y = −x − 2.

Use the figure to solve the simultaneous equations

−−=

−−=

2

32

xy

xxy.

From the figure, the points of intersection are

.

Correct the values of y to the nearest 0.2.

−1

0 1 2 3 4

−2

y

x

y = 0.5x2 – 2x

y = x − 4

According to the scale of the figure, correct the values of x and y to the nearest 0.1.

−2 −1 0 1 2

2

−2

y

x

y = −x2 + 2 2x + 3y + 2 = 0

2

1

−1

0

−2

−3

y

x

y = −x − 2

y = −x2 − 3x

−4 −3 −2 −1 1

6

The figure shows the graph of y = 3 – x2. Solve each of the following simultaneous equations by adding a

suitable straight line to the figure. [Nos. 2–3] �Ex 1A: 5–7

2.

=

−=

3

3 2

y

xy 3.

=

−=

xy

xy

2

3 2

Add the graph of Add the graph of

to the given figure. to the given figure.

x –2 1 2 x –2 0 2

y y

The figure shows the graph of y = x2 + 4x + 4. Solve each of the following simultaneous equations by

adding a suitable straight line to the figure. [Nos. 4–5] �Ex 1A: 8–10

4.

=+

++=

24

442

yx

xxy 5.

−=−

++=

1

442

yx

xxy


6. The figure shows the graph of y = 2x2 – 3x – 1. Do the simultaneous

equations

=−−

−−=

0323

132 2

yx

xxy have two real solutions? Explain your answer.

If the two graphs have no points of intersection, the corresponding simultaneous equations have no real solutions.

Use three points to ensure that a correct straight line is drawn.

x

–2 –1 1 0

1

–1

2

2

3

y = 3 – x2

y

3

2

1

−4 −3 −2 −1 0

x

y y = x2 + 4x + 4

−1

1

−1 0 1 2

2

−1

−2

x

y

y = 2x2 − 3x − 1

7

New Century Mathematics (Second Edition) 5A

1 More about Equations

Consolidation Exercise 1A

Level 1

In each of the following, use the given graphs to solve the simultaneous equations. [Nos. 1–4]

1. (a)

=

++=

3

124 2

y

xxy (b)

−=

+−−=

xy

xxy

1

132

2. (a)

−=

+−=

2

752

xy

xxy (b)

+=

+−−=

9

532

xy

xxy

4

3

2

1

−3 −2 −1 0

1

y = −x2 − 3x + 1

y = 1 − x

x

y

4

3

2

1

0 1 2 3 4

y = x2 − 5x + 7

y = x − 2 x

y

4

3

2

1

−1 0

1 2 3

y = 4x2 + 2x + 1

y = 3

x

y

y = −x2 − 3x + 5

y = x + 9

y

x 0

−2 −1 −3 −4

2

4

8

6

8

3. (a)

=

+−−=

5

322

y

xxy (b)

−=

−−−=

xy

xxy

4

132

4. (a)

+=

+−=

6109

222

xy

xxy (b)

−=

+−=

xy

xxy

4257

862

5. Using the figure on the right, solve the following

simultaneous equations.

(a)

−−=

−−=

23

232 2

xy

xxy

(b)

−=

−−=

xy

xxy

2

232 2

(c)

−=

−−=

94

232 2

xy

xxy

1

2

1

−1

−2

−3

−4

−5

−4 −3 −2 −1 0

y = −4x

y = −x2 − 3x − 1

x

y

y = −x2 − 2x + 3

y = 5 5

4

3

2

1

−3 −2 −1 0

1 x

y

10

8

6

4

2

0 1 2 3 4 5 6

y = x2 − 6x + 8

x

y

7y = 25 − 4x

8

6

4

2

−1 0

1 2 3

9y = 10x + 6

y = x2 − 2x + 2

x

y

y = −3x − 2

y = 2x2 − 3x − 2 6

4

2

−2

−4

−2 −10

1 2 3

y = 2 − x

y = 4x − 9

x

y

9

10

5

−5

−10

−15

−4 −3 −2 −1 0

1

y = 7x + 20

y = −x − 8

y = −4x2 − 9x + 4

y = 5 − 10x

x

y

1

−1

−2

−3

−2 −1 0

1 2

y = x2 − 3

x

y

−4

4

3

2

1

−1

−2

0 1 2 3 4 5

y = −x2 + 7x − 8

x

y

6. Using the figure on the right, solve the following

simultaneous equations.

(a)

−−=

+−−=

8

494 2

xy

xxy

(b)

−=

+−−=

xy

xxy

105

494 2

(c)

+=

+−−=

207

494 2

xy

xxy

The figure shows the graph of y = x2 − 3. Solve each of the

following simultaneous equations by adding a suitable

straight line to the figure. [Nos. 7–9]

7.

−=

−=

2

32

y

xy

8.

−−=

−=

42

32

xy

xy

9.

−=

−=

73

32

xy

xy

The figure shows the graph of y = −x2 + 7x − 8. Solve each

of the following simultaneous equations by adding a

suitable straight line to the figure. [Nos. 10–12]

10.

−=

−+−=

3

872

xy

xxy

11.

−=

−+−=

44

872

xy

xxy

12.

−=

−+−=

xy

xxy

212

872

10

Level 2

13. The figure shows the graph of y = 3x2 + 5x − 1. Solve

each of the following simultaneous equations by adding


(a)

−=+

−+=

2

153 2

yx

xxy

(b)

=−

−+=

435

153 2

xy

xxy

14. The figure shows the graph of y = −5x2 − 7x + 6. Solve

each of the following simultaneous equations by adding


(a)

−=

+−−=

xy

xxy

3172

675 2

(b)

−=+

+−−=

1923

675 2

yx

xxy

15. Solve the simultaneous equations

=−

+−=

72

122

1 2

yx

xxy for −1 ≤ x ≤ 5 graphically.

[Unit length for both axes: 10 divisions (1 cm) ]


=+

++−=

62

142 2

yx

xxy for −1 ≤ x ≤ 3 graphically.

[Unit length for x-axis: 10 divisions (1 cm)

Unit length for y-axis: 5 divisions (0.5 cm)]


=−

+−=

1326

362 2

yx

xxy for 0 ≤ x ≤ 3 graphically.

[Unit length for both axes: 10 divisions (1 cm)]

1

−1

−2

−3

−3 −2 −1 0

1

y = 3x2 + 5x − 1

x

y

10

5

−5

−10

−3 −2 −1 0 1

x

y

y = −5x2 − 7x + 6

11


=+

++−=

218

584 2

yx

xxy graphically.

[Unit length for x-axis: 10 divisions (1 cm)

Unit length for y-axis: 2 divisions (0.2 cm)]


−=+

++=

33

542

yx

xxy graphically.

[Unit length for both axes: 5 divisions (0.5 cm)]


=+

+−−=

124

52 2

yx

xxy graphically.

[Unit length for both axes: 10 divisions (1 cm)]

21. The figure shows the graph of y = −x2 + 4x + 4 for −1 ≤ x ≤ 5.

(a) Use the figure to solve the simultaneous equations

−=

++−=

232

442

xy

xxy.

(b) Judy claims that there is only a set of real values of x and y satisfying both y = −x2 + 4x + 4 and

2y = 3x − 2. Do you agree? Explain your answer.

8

7

6

5

4

3

2

1

−10

1 2 3 4 5

y = −x2 + 4x + 4

y

x

12

22. The figure shows the graph of y = 2x2 − 3x − 5 for −1.5 ≤ x ≤ 3.

(a) Use the figure to solve the simultaneous equations

−=

−−=

1962

532 2

xy

xxy.

(b) Is it possible that more than one set of real values of x and y satisfy both y = 2x2 − 3x − 5 and

2y = 6x − 19? Explain your answer.

*23. The figure shows the graph of y = −2(x − 3)2 + 6.

(a) Write down the coordinates of the vertex of the graph.

(b) Consider the simultaneous equations (*)......6)3(2 2

+=

+−−=

bmxy

xy, where m and b are real numbers.

Determine whether each of the following statements is correct. Explain your answers with the

help of the figure.

(i) If m > 0 and b = 6, then (*) must have no real solutions.

(ii) If m = 0 and b < 6, then (*) must have two real solutions.

(iii) If m < 0 and b > 6, then (*) must have two real solutions.

4

2

−2

−4

−10

3

x

y

y = 2x2 − 3x − 5

1 2

−6

4

2

−2

−4

−10

3

x

y

y = 2x2 − 3x − 5

1 2

−6

y = −2(x − 3)2 + 6

x

y

O

13

Answers


1. (a) x = −1.0, y = 3.0; x = 0.5, y = 3.0

(b) x = −2.0, y = 3.0; x = 0.0, y = 1.0

2. (a) x = 3.0, y = 1.0

(b) x = −2.0, y = 7.0

3. (a) no real solutions

(b) no real solutions

4. (a) x = 0.5, y = 1.2; x = 2.6, y = 3.6

(b) x = 1.0, y = 3.0; x = 4.4, y = 1.0

5. (a) x = 0.0, y = −2.0

(b) x = −1.0, y = 3.0; x = 2.0, y = 0.0

(c) no real solutions

6. (a) x = −3.0, y = −5.0; x = 1.0, y = −9.0

(b) no real solutions

(c) x = −2.0, y = 6.0

7. x = −1.0, y = −2.0; x = 1.0, y = −2.0

8. x = −1.0, y = −2.0

9. no real solutions

10. x = 1.0, y = −2.0; x = 5.0, y = 2.0


12. x = 4.0, y = 4.0; x = 5.0, y = 2.0

13. (a) x = −1.8, y = −0.2; x = −0.2, y = −1.8

(b) x = −1.8, y = −0.3; x = 0.3, y = 1.0

14. (a) no real solutions

(b) x = −2.4, y = −6.0; x = 1.3, y = −11.5

15. x = 4.0, y = 1.0


17. x = 1.7, y = −1.4; x = 2.8, y = 1.9

18. x = 2.0, y = 5.0


20. x = −1.2, y = 3.3; x = 0.8, y = 2.8

21. (a) x = 3.8, y = 4.7

(b) no

22. (a) x = 1.5, y = −5.0

(b) no

23. (a) (3 , 6)

(b) (i) yes (ii) yes

(iii) no

14

F5A: Chapter 1B

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Book Example 5


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Book Example 8


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Book Example 9


(Video Teaching)



(Full Solution)

Maths Corner Exercise 1B Level 1



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15

1B Level 2 ○ Problems encountered ○ Skipped

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Maths Corner Exercise 1B Multiple Choice


___________ ( )



_________

16


Objective: To use the algebraic method to solve simultaneous equations in two unknowns (one linear and

one quadratic).

Review: Relations between the Discriminant and the Nature of Roots

1. For a quadratic equation ax2 + bx + c = 0 (a ≠ 0), discriminant (∆) = b2 – 4ac.

Discriminant ∆ > 0 ∆ = 0 ∆ < 0

Number of real roots

Solving Simultaneous Equations (One Linear and One Quadratic) in Two Unknowns by the

Algebraic Method

Step 1: Make one of the unknowns (e.g. y) the subject of the linear equation.

Step 2: Substitute the subject (y) in Step 1 into the quadratic equation in two unknowns to eliminate the

unknown (y).

Step 3: Solve the quadratic equation in one unknown obtained.

Step 4: Substitute the root(s) obtained in Step 3 into the equation in Step 1 to find the value(s) of the

other unknown (y).

Step 5: Write down all the solutions of the simultaneous equations.


Solve the following simultaneous equations

algebraically.

−=

−+=

(2).....................

(1)............

xy

xxy

5

732

Substitute (2) into (1). � Step 2

5 – x = x2 + 3x – 7

x2 + 4x – 12 = 0

(x + 6)(x – 2) = 0

x = –6 or 2 � Step 3

Substitute x = –6 into (2). � Step 4

y = 5 – (–6) = 11

Substitute x = 2 into (2).

y = 5 – 2 = 3 ∴ The solutions of the simultaneous

equations are x = –6, y = 11 and

x = 2, y = 3. � Step 5

Solve the following simultaneous equations

algebraically.

−−=

−=

(2)........

(1)................

10

2

2xxy

xy

Substitute (1) into (2).

x – 2 = ( )

( ) = 0

( )( ) = 0

x = ( ) or ( )


Step 1: y is the subject of

the equation (2).

( ) ( )

x ( )

x ( )

17

2.

−=

+−=

(2)..................

(1)............

113

452

xy

xxy 3.

++=

−=+

(2).............

(1)..................

202

710

2xxy

xy �Ex 1B: 1–12

Find the number of real solutions of each of the following simultaneous equations. [Nos. 4–5]

4.

+=

−+=

(2)........................

(1)...............

xy

xxy

5

262

5.

=+

+−=

(2)...................

(1)............

xy

xxy

26

542

�Ex 1B: 16–18

Substitute ( ) into ( ).

( ) = ( )

( ) = 0

∆ =


6. Gary claims that k is negative when the following simultaneous equations have one real solution.

Do you agree? Explain your answer.

−=

++=

(2)....................

(1)...............

xky

xxy

3

52

Make y the

subject of (1).

For ax2 + bx + c = 0

(a ≠ 0), ∆ = b2 – 4ac.

18



Consolidation Exercise 1B

Level 1


1.

−=

+−=

xy

xy

1

12

2.

=

−=

xy

xxy 22

3.

+=

+−−=

62

142

xy

xxy

4.

−=

+−=

xy

xxy

37

952

5.

−=

−+−=

44

442

xy

xxy 6.

++=

−−=

25

42

2xxy

xy

7.

=−+

+−−=

022

622

yx

xxy 8.

=−

=−+

64

02

yx

xxy 9.

−=+

−=−

62

222

yx

xyx

10.

=+−

=−−

24

043

2xyx

xy 11.

=++

+−=−

094

732

yx

yxx 12.

−=−

=+

yxx

yx

22

54

2

13.

=−

−=+

42

43 2

xy

yxx 14.

=−

+=−

32

14 2

yx

xxy 15.

=−−

=+++

042

0232

xy

yxx

16.

=−+

+=+

043

12

yx

xyx 17.

=+−

+=+

014

16 2

yx

xyx 18.

−=−

−=+

246

45

xyx

yx

19. y = 4 − x2 = 3x 20. x = x2 − 2 = y + 1 21. x2 = 4(x + 3) = 8 − y

22. 3x + 2 = x − x2 = y + 7 23. x2 + 4x = y − 3x = 6 − x 24. y − x2 + 4x = 4x + y = 6

Find the number of real solutions of each of the following simultaneous equations. [Nos. 25–28]

25.

−=

−=

74

52 2

xy

xy 26.

+=

−+=

87

143 2

xy

xxy

27.

−=

−+=

xy

xxy

26

432 2

28. y = 2x2 + 4x − 6 = 5x − 2

Level 2


29.

=+

++=

122

32 2

yx

xxy 30.

=+

+=+

xy

xyx

453

1343 2

19

31.

−=+

=−

9432

24 2

yx

xyx 32.

=+

+=+

73

442 2

yx

xyx

33. 2x − 4y2 = 6x + 5y = −1 34. 3x2 + 2x + 8 = −1 − 4x + 4y = 5y − 3x

35. 2x2 − x(3y − 4x) = 4x − 3y + 11 = 14 36. x2 + 9x + 6y = 6x + 4y = −18

37. x(x − 5) + (3x − 2y)x = 2y − 3x − 11 = −10 38. (5y + 2x)x + 3x2 − 2 = 10 − 10y − 4x = 6

39. 2x2 − y2 − 3 = 8x + 6y = 2x 40. 7x2 + y2 = 9(x + y + 1) = 36x − 45

41.

=+−

=−−

033

6232 22

xy

yxyx 42.

=−

=−+

345

3843 22

xy

xxyy

43.

=−

=−

kxy

kxy

2

114 222

(where k is a constant) 44.

=−

=++

myx

myxyx

2

37 222

(where m is a constant)

45. If p and q satisfy the equations p(2p + 1) = 4q − 10 and 3q − 4p = 7, find the values of p and q

algebraically.

46. If a and b satisfy the equations a2 + 3a(a + b − 2) = 1 and 2a + 4b + 5 = 0, find the values of a and b

algebraically.

47. If the simultaneous equations

=+−

−−=

033

223 2

kyx

xxy have only one real solution, find the value of k.


=+−

+=

042

32 2

ykx

kxy have only one real solution, find the values of k.


−=

=−

xky

xyx

24

1222

have no real solutions, find the range of values of k.


=+

−+=

62

224 2

yx

kxxy have two real solutions, find the range of values of k.

51. Show that the simultaneous equations

=+

=+

63

222

ykx

yx have real solution(s) when k ≤ −3 or k ≥ 3.

52. Consider the simultaneous equations

−=

=−−+

xy

xkykx

61

0322 22

, where k ≠ 0.

20

D

A E

B C

x cm

y cm

x cm y cm C A B

P

Q R

C: y = x2 − 3x − 4

y = 3x + k

O

y

x

P

L: 2x + y = 8

C: y = −x2 + 2x + p

x

y

O

A

(a) If the simultaneous equations have real solution(s), find the range of values of k.

(b) If the simultaneous equations have real solution(s), how many positive integral values of k are

there? Explain your answer.

53. Consider the simultaneous equations

=−

=+

xyk

kxy

25

45 2

.

(a) If the simultaneous equations have no real solutions, find the range of values of k.

(b) If the simultaneous equations have real solution(s), find the maximum integral value of k.

54. The figure shows a logo which is formed by a rectangle and an

equilateral triangle. AB = x cm, AE = y cm and x > y. If the area of the

rectangle is 24 cm2 and the perimeter of the logo is 30 cm, find the

values of x and y.

55. In the figure, AB, AC and BC are diameters of the semi-circles APB,

AQC and CRB respectively, where AC > BC. The area and the perimeter

of the shaded region are 12π cm2 and 16π cm respectively. Let AC = x

cm and BC = y cm. Find the lengths of AC and BC.

* 56. In the figure, the straight line y = 3x + k touches the curve C: y = x2 − 3x −

4 at the point P.

(a) Find the value of k.

(b) Find the coordinates of P.

(c) The straight line y = 3x + k + m cuts C at two distinct points Q

and R, where m > 0. Express the coordinates of the mid-point of

QR in terms of m.

* 57. In the figure, the straight line L: 2x + y = 8 touches the curve C: y = −x2 + 2x +

p at the point A.

(a) Find the coordinates of A.

(b) L1 is a straight line passing through A and with slope 2. L1 cuts C

again at another point B.

(i) Find the coordinates of B.

(ii) Suppose D is a point lying in quadrant IV. Are the areas of △AOD and △BOD the same? Explain your answer.

21

Answers


1. x = 0, y = 1; x = 1, y = 0

2. x = 0, y = 0; x = 3, y = 3

3. x = −5, y = −4; x = −1, y = 4


5. x = 0, y = −4

6. x = −6, y = 8; x = −1, y = −2

7. x = −2, y = 6; x = 2, y = −2

8. x = 2, y = 2; x = 3, y = 6

9. x = −2, y = −2

10. x = −3, y = −5; x = 2, y = 10


12. x = −3, y = 17; x = 1, y = 1

13. x = 0, y = 4; x = −5, y = −6


15. x = −3, y = −2; x = −2, y = 0

16. x = −3, y = 13; x = 1, y = 1

17. x = 0, y = 1; x = 10, y = 41

18. x = 0, y = −4; x = −11, y = 51

19. x = −4, y = −12; x = 1, y = 3

20. x = −1, y = −2; x = 2, y = 1

21. x = −2, y = 4; x = 6, y = −28


23. x = −6, y = −6; x = 1, y = 8

24. x = 0, y = 6

25. one real solution

26. two real solutions


28. two real solutions

29. x = −3, y = 18; x =2

3, y = 9


31. x = −4

3, y =

4

15

32. x = −3

1, y =

9

22; x =

2

5, y =

2

3

33. x =18

7, y = −

3

2; x = −

8

3, y =

4

1


35. x = −2

7, y = −

3

17; x = 2, y =

3

5

36. x = −3, y = 0; x = 3, y = −9


38. x = −2, y =5

6; x =

3

4, y = −

15

2

39. x = −1, y = 1; x = 3, y = −3

40. x =4

9, y =

4

3


42. x =2

1, y = 1; x =

3

4, y =

3

5

43. x = −3k, y = −5k; x = −5

k, y =

5

3k

44. x =3

m, y = −

3

5m

45. p =6

1, q =

9

23; p = 2, q = 5

46. a = −10

1, b = −

5

6; a = 4, b = −

4

13

47. −9

10

48. 2, 4

49. k < −2

50. k > −2

7

52. (a) k ≤ 3 and k ≠ 0

(b) 3

53. (a) k >5

1

(b) 0

54. x = 8, y = 3

55. AC = 12 cm, BC = 4 cm

56. (a) −13

(b) (3 , −4)

(c) (3 , m − 4)

57. (a) (2 , 4)

(b) (i) (−2 , −4)

(ii) yes

22

F5A: Chapter 1C

Date Task Progress

Lesson Worksheet


(Full Solution)

Book Example 10


(Video Teaching)

Book Example 11

○ Complete ○ Problems encountered ○ Skipped (Video Teachin

g)

Book Example 12


(Video Teaching)

Book Example 13


g)

Book Example 14


(Video Teaching)

Book Example 15


g)

Book Example 16


(Video Teaching)

23

Book Example 17


g)

Book Example 18


(Video Teaching)

Book Example 19


g)

Book Example 20


(Video Teaching)

Book Example 21


g)

Book Example 22


(Video Teaching)

Book Example 23


g)

Book Example 24


(Video Teaching)

24

Book Example 25


g)

Book Example 26


(Video Teaching)

Book Example 27


g)



(Full Solution)

Maths Corner Exercise 1C Level 1



___________ ( )




___________ ( )

Maths Corner Exercise 1C Multiple Choice


___________ ( )



_________

25

5A Lesson Worksheet 1.3A (Refer to Book 5A P.1.22) Objective: To solve fractional equations which can be transformed into quadratic equations.

Steps of Solving Fractional Equations

Step 1: Express each fraction with the same denominator, which is the L.C.M. of the denominators in the

fractions.

Step 2: Multiply both sides of the equation by the L.C.M. to eliminate all the denominators.


Solve

+

x

21 (3 + x) = –2.

+

x

21 (3 + x) = –2 � x ≠ 0

x

+

x

21 (3 + x) = –2x �Multiply both sides by x.

(x + 2)(3 + x) = –2x

x2 + 5x + 6 = –2x

x2 + 7x + 6 = 0

(x + 1)(x + 6) = 0

x = –1 or –6 �

Solve (x + 4)

−

x

31 = 5.

(x + 4)

= 5

( )(x + 4)

= 5( )

(x + 4)( ) = 5( )

x2 + ( ) − ( ) = 5( )

x2 − ( ) − ( ) = 0

( )( ) = 0

x = or

Solve the following fractional equations. [Nos. 1–4]

1. x – 4 =x

4− 2. x –

1

4

−x = 1 �Ex 1C: 1–3

( )(x − 4) = −4

3. 2

5

+x = 3 – 2x 4.

−1

4

x(x – 9) = 1

Remember to check whether the answers obtained satisfy the original equation.

Multiply both

sides by x − 1.

Multiply both sides by ( ).

( ) ( )

x ( )

x ( )

26


Solve

2

3

−x –

x

2 = 1.

2

3

−x –

x

2 = 1 � x ≠ 0 and x ≠ 2.

)2(

)2(23

−

−−

xx

xx = 1 �

3x – 2(x – 2) = x(x – 2) �

3x – 2x + 4 = x2 – 2x

x2 – 3x – 4 = 0

(x + 1)(x – 4) = 0

x = –1 or 4

Solve

1

2

+x –

x

6 = 1.

)(

2

−

)(

6

= 1

))((

)(6)(2

− = 1

2( ) − 6( ) = ( )( )

( ) = ( )2 + ( )

( ) = 0

( )( ) = 0

x = or


5. 2

8

+x –

x

2 = 1 6.

x−1

15 +

x

12 = –2 �Ex 1C: 4

))((

)(2)(8

− = 1

8( ) − 2( ) = ( )( )

7. 1

1

−x –

5

3

+x =

12

1− 8.

16

72

−x +

4

11

+x = 2 �Ex 1C: 19(b)


9. Solve

3

4

−x +

3

5

+x = −3.

Step 1: The denominator x(x – 2) is the L.C.M. of x and x – 2.

Step 2: Multiply both sides by x(x – 2).

The L.C.M. of x + 1 and x

is ( ).

The L.C.M. of x + 2 and x is ( ).

x2 − 16

= ( )( )

27

5A Lesson Worksheet 1.3B & C (Refer to Book 5A P.1.25)

Objective: To solve equations of higher degree and equations with radical signs which can be transformed

into quadratic equations.

Equations of Higher Degree

Some equations of degrees higher than 2 can be transformed into quadratic equations and then be solved.


Find the real roots of x4 – 7x2 – 18 = 0.

Let u = x2. Then u2 = (x2)2 = x4. � (am)n = amn

The original equation becomes

u2 – 7u – 18 = 0

(u – 9)(u + 2) = 0

u = 9 or –2 �

Since u = x2,

x2 = 9 or –2 (rejected) �

x = 3±

Find the real roots of x4 – 13x2 + 36 = 0.

Let u = ( ). Then u2 = ( )2 = ( ).


( ) = 0

( )( ) = 0

u = ( ) or ( )

Since u = ( ),

x2 = ( ) or x2 = ( )

x = or x =

Find the real roots of the following equations. [Nos. 1–4]

1. x4 + 2x2 – 24 = 0 2. x5 + x3 – 2x = 0 �Ex 1C: 5, 6

Let u = ( ). Then u2 = ( )2 = ( ).


3. (x3)2 + 7x3 – 8 = 0 4. 3x4 – 26x2 – 9 = 0 �Ex 1C: 21

Don’t stop here. We need to find the values of x.

For any real number x,

x2 ≥ 0.

• Don’t cancel ‘x’.

• If abc = 0, then a = 0 or b = 0 or

c = 0.

Use a new variable u to replace x3.

28

Equations with Radical Signs

We can solve equations with radical signs by using the method of change of variable or by squaring both

sides of equations.


Find the real root of x – 4 x – 5 = 0.

Let u = x . Then u2 = ( x )2 = x.


u2 – 4u – 5 = 0

(u – 5)(u + 1) = 0

u = 5 or –1

Since u = x ,

x = 5 or –1 (rejected) � Note that x ≥ 0.

x = 25

Find the real root of x + 5 x – 6 = 0.

Let u = ( ). Then u2 = ( )2 = ( ).


( ) = 0

( )( ) = 0

u = ( ) or ( )

Since u = ( ),

( ) = ( ) or ( ) ( )

x =

Find the real root(s) of the following equations. [Nos. 5–6]

5. x – 6 x + 9 = 0 6. 552−+ xx = 3 �Ex 1C: 7, 8


7. Find the real root of x + 1+x = 5.


Square both sides of the equation.

29

5A Lesson Worksheet 1.3D (Refer to Book 5A P.1.27)

Objective: To solve exponential equations which can be transformed into quadratic equations.

[In this worksheet, give the answers correct to 2 decimal places if necessary.]

Exponential Equations


Solve 22x – 9(2x) + 8 = 0.

Let u = 2x. Then u2 = (2x)2 = 22x. � (am)n = amn


u2 – 9u + 8 = 0

(u – 8)(u – 1) = 0

u = 8 or 1 �

Since u = 2x,

2x = 8 or 2x = 1 � 8 = 23, 1 = 20

x = 3 or x = 0

Solve 42x – 6(4x) + 8 = 0.

Let u = ( ). Then u2 = ( )2 = ( ).


( )2 − 6( ) + 8 = 0

( )( ) = 0

u = ( ) or ( )

Since u = ( ),

( )x = ( ) or ( )x = ( )

x = or x =

1. Solve (5x)2 – 5x – 20 = 0. 2. Solve 4(22x) – 9(2x) + 2 = 0. �Ex 1C: 9, 10

Let u = ( ). Then u2 = ( )2.


3. Solve 32x + 1 – 10(3x) + 3 = 0. 4. Solve 16x – 17(4x) + 16 = 0. �Ex 1C: 25, 26

For any real number x, 5x > 0.

22 = 4

2 = 21

4

amn = (am)n = (an)m

ma

1= a−m

16x = [( )2]x = [( )x]2


32x + 1 = ( )(32x)

30


Solve 3x – 4 − x3

12 = 0.

Let u = 3x. The original equation becomes

u – 4 −u

12 = 0

u2 – 4u – 12 = 0 � Multiply both sides by u (u ≠ 0).

(u – 6)(u + 2) = 0

u = 6 or –2

Since u = 3x,

3x = 6 or –2 (rejected) �

log 3x = log 6 � Take logarithms on both sides.

x log 3 = log 6 � log an = n log a

x =

3log

6log

= 1.63, cor. to 2 d.p.

Solve 2x + 2 − x2

15 = 0.

Let u = ( ). The original equation becomes

( ) + 2 −

) (

15 = 0

( )2 + 2( ) − ( ) = 0

( )( ) = 0

u = ( ) or ( )

Since u = ( ),

( )x = ( ) or ( ) (rejected)

log ( )x = log ( )

( ) log ( ) = log ( )

x =

)(log

)(log

= , cor. to 2 d.p.

Solve the following exponential equations. [Nos. 5–6]

5. 4x = 3 + x4

18 6. 5x +

x5

6 = 5 �Ex 1C: 27


7. Does the equation 2(32x) + 3x + 2 + 4 = 0 have at least 1 real root? Explain your answer.

For any real number x, 3x > 0.

31

5A Lesson Worksheet 1.3E (Refer to Book 5A P.1.30)

Objective: To solve logarithmic equations which can be transformed into quadratic equations.

Logarithmic Equations


Solve (log3 x)2 – 4 log3 x + 3 = 0.

Let u = log3 x. Then u2 = (log3 x)2 .


u2 – 4u + 3 = 0

(u – 1)(u – 3) = 0

u = 1 or 3 �

Since u = log3 x,

log3 x = 1 or log3 x = 3

x = 3 or x = 33 � If loga x = y, then x = a y.

x = 3 or x = 27

Solve (log2 x)2 + log2 x – 6 = 0.

Let u = ( ). Then u2 = ( )2.


( )2 + ( ) − 6 = 0

( )( ) = 0

u = ( ) or ( )

Since u = ( ),

log2 ( ) = ( ) or log2 ( ) = ( )

x = ( )( ) or x = ( )( )

x = or x =

1. Solve (log x)2 + 4 log x + 4 = 0. 2. Solve (log2 x)2 + 9 log2 x + 20 = 0. �Ex 1C: 11

Let u = ( ). Then u2 = ( )2.



Solve log2 (x + 3) + log2 (x + 1) = 3.

log2 (x + 3) + log2 (x + 1) = 3

log2 [(x + 3)(x + 1)] = log2 8 � ∴ (x + 3)(x + 1) = 8

x2 + 4x + 3 = 8

x2 + 4x – 5 = 0

(x – 1)(x + 5) = 0

x = 1 or –5 (rejected)

Solve log3 (x – 3) + log3 (x + 5) = 2.

log3 ( ) + log3 ( ) = 2

log3 [( )( )] = log3 ( ) ∴ ( )( ) = ( )

x2 + ( )x − ( ) = ( )

x2 + ( )x − ( ) = 0

( )( ) = 0

x = or ( ) ( )

a–m =m

a

1

log2 a + log2 b

= log2 ab

2 = log3 32

When x = –5, log2 (x + 3) = log2 (–2) and log2 (x + 1) = log2 (–4) are undefined.

3 = log2 23 = log2 8


32

Solve the following logarithmic equations. [Nos. 3–4]

3. log x + log (x – 4) = log 12 4. log (x + 4) + log (x − 11) = 2 �Ex 1C: 12

log [( )( )] = log ( )


Solve log2 (x2 – 2) – log2 (x + 3) = 1.

log2 (x2 – 2) – log2 (x + 3) = 1

log2

3

22

+

−

x

x = log2 2 � ∴

3

22

+

−

x

x = 2

x2 – 2 = 2x + 6

x2 – 2x – 8 = 0

(x – 4)(x + 2) = 0

x = 4 or –2

Solve log3 (x + 3) – log3 (x2 + 1) = 0.

log3 ( ) − log3 ( ) = 0

log3

)(

)(

= log3 1 ∴

)(

)(

= 1

( ) = x2 + ( )

x2 − ( ) − ( ) = 0

( )( ) = 0

x = or


5. 2 log5 x – log5 (6 − x) = log5 3 6. log (x – 2) – log (x2 + 4) = –1 �Ex 1C: 29

log5 ( ) − log5 (6 − x) = log5 3

log5

) (

) ( = log5 3


7. Fanny claims that all the real roots of log4 x – log4 (x2 + 2) + 1 = 0 are integers. Do you agree?

Explain your answer.

log2 a – log2 b

= log2b

a

0 = log3 1

Don’t write

x + (x − 4) = 12.

2 = log 102

1 = log2 2


−1 = log 10−1

33

5A Lesson Worksheet 1.3F (Refer to Book 5A P.1.31)

Objective: To solve trigonometric equations which can be transformed into quadratic equations.

Trigonometric Equations


Solve 2 cos2 θ – 1 = 0, where 0° ≤ θ ≤ 360°.

Let u = cos θ. Then u2 = cos2 θ.


2u2 – 1 = 0

u2 =

2

1

u =

2

1± �

When u =

2

1,

cos θ =

2

1

θ = 45° or 360° – 45°

= 45° or 315°

When u =

2

1− ,

cos θ =

2

1−

θ = 180° – 45° or 180° + 45°

= 135° or 225°

Solve 3 tan2 θ + tan θ = 0, where 0° ≤ θ ≤ 360°.

Let u = ( ). Then u2 = ( ).


3 ( )2 + ( ) = 0

( )( ) = 0

u = ( ) or

When u = ( ),

tan θ = ( )

θ = or or

When u =

,

tan θ =

θ = 180° − ( ) or 360° − ( )

= or

Solve the following trigonometric equations, where 0° ≤ θ ≤ 360°. [Nos. 1–2]

1. cos2 θ – cos θ = 0 2. 2 sin2 θ + sin θ – 1 = 0 �Ex 1C: 13–15

Let u = ( ). Then u2 = ( ).

∵ cos θ > 0 ∴ θ lies in

quadrant I or quadrant IV. ∵ cos θ < 0 ∴ θ lies in quadrant II

or quadrant III.

When 0° ≤ θ ≤ 360°, how many

roots does tan θ = 0 have?

∵ tan θ ( > / < ) 0 ∴ θ lies in quadrant

or quadrant .

When sin θ > 0, θ lies in quadrant or quadrant .

Don’t stop here.

We need to find θ.

34


Solve cos2 θ + 2 sin θ – 1 = 0, where 0° ≤ θ ≤ 360°.

cos2 θ + 2 sin θ – 1 = 0

1 − sin2 θ + 2 sin θ – 1 = 0 � cos2 θ = 1 – sin2 θ

sin2 θ − 2 sin θ = 0 �

sin θ (sin θ − 2) = 0

sin θ = 0 or sin θ = 2 (rejected)

θ = 0° or 180° or 360°

Solve 2 sin2 θ – 3 cos θ − 3 = 0, where 0° ≤ θ ≤ 360°.

2( ) − 3( ) − 3 = 0

2[1 − ( )] − 3( ) − 3 = 0 �

2 − 2( ) − 3( ) − 3 = 0

2( ) + 3( ) + ( ) = 0 �

( )( ) = 0

( ) + 1 = 0 or 2( ) + 1 = 0

cos θ = ( ) or cos θ =

θ = ( ) or θ = ( ) or

( ) ∴ θ = or or


(Give the answers correct to the nearest 0.1°.)

3. 3 sin2 θ + 7 cos θ – 5 = 0 4. 4 cos2 θ – 5 sin θ + 2 = 0 �Ex 1C: 30(a)–(c)

3[1 – ( )] + 7( ) – 5 = 0


5. For 0° ≤ θ ≤ 360°, how many roots does the equation 3 sin θ tan θ = –8 have? Explain your answer.

u2 − 2u = 0,

where u = sin θ.

For any angle θ,

−1 ≤ cos θ ≤ 1.

2u2 + 3u + 1 = 0,

where u = cos θ.

sin2 θ = 1 – ( )

For any angle θ, −1 ≤ sin θ ≤ 1.

35



Consolidation Exercise 1C

Level 1


1. (a) x

5+ 1 = 4x (b) 3 −

x

4=

2

x

2. (a) x−1

4− 3x = 5 (b) x +

4

6

+x= 3

3. (a) )3(15

+

− x

x= 4 (b) )4(3

2x

x−

+ = 8

4. (a)

−−

xx

87)1( = 9x (b)

++ 1

2)2(

xx = 2x + 1

5. (a) 3

3

+x+

x

4= 1 (b)

x

2−

2

1

−x= 3

6. (a) 2 −x−4

3=

x

1 (b)

1

15

+x= 1 +

x

8


7. (a) x4 + 12x2 − 64 = 0 (b) x4 − 10x2 + 9 = 0

8. (a) x6 − 5x4 − 36x2 = 0 (b) (x3)2 − 7x3 − 8 = 0

9. (a) (x3)2 + 26x3 − 27 = 0 (b) (x4)2 − 14x4 − 32 = 0

10. (a) x − x5 + 4 = 0 (b) x + x4 − 12 = 0

11. (a) 622−+ xx = 3 (b) 272

−+ xx = 4

12. (a) 2 − 162−− xx = 0 (b) 1832

+− xx − 6 = 0


13. (a) (3x)2 + 3x − 12 = 0 (b) (2x)2 − 2x + 2 − 32 = 0

14. (a) 22x + 2x + 1 − 24 = 0 (b) 42x + 4x + 2 + 15 = 0

36

15. (a) 32x − 2(3x + 1) − 27 = 0 (b) 52x + 2(5x + 1) − 11 = 0


16. (a) (log x)2 + log x − 6 = 0 (b) (log x)2 − 7 log x + 10 = 0

(c) (log3 x)2 − 2 log3 x − 8 = 0 (d) (log4 x)2 + 5 log4 x + 4 = 0

17. (a) log x + log (x + 6) = log 7 (b) log (x − 2) + log (x + 5) = log 8

(c) log5 (1 − x) + log5 (x + 7) = log5 15 (d) log7 (x + 4) + log7 (6 − x) = log7 9

18. (a) log (x − 2) + log (x − 11) = 1 (b) log2 (x + 1) + log2 (x − 3) = 5

(c) log3 (x + 2) + log3 (x + 8) = 3 (d) log4 (x + 3) + log4 (x + 15) = 3


19. (a) 4 cos2 θ − 3 = 0 (b) tan2 θ − 3 = 0

20. (a) 2 sin2 θ + sin θ = 0 (b) tan2 θ = tan θ

21. (a) cos2 θ − cos θ − 2 = 0 (b) 2 sin2 θ + sin θ − 1 = 0

22. (a) 2 tan θ − tan2 θ = 1 (b) 2 cos2 θ + 3 cos θ = −1

23. (a) tan2 θ + )13( − tan θ = 3 (b) 22 sin2 θ − )22( + sin θ = −1

24. The numerator of a fraction in its simplest form is 7. When both the numerator and the denominator of

the fraction are decreased by 3, the fraction is decreased by

9

1. Find the original fraction.

25. If the sum of a number and its reciprocal is

12

25, find all the possible numbers.

26. The sum of the common logarithms of two consecutive positive even numbers is log 24. Find the two

numbers.

27. Jack takes x hours to lay tiles on a floor of area n m2, where x > 2 and n > 0. John takes 1.2 hours less

than Jack to lay tiles on a floor of area n m2. If they work together, it takes 0.8 hour to lay tiles on a

floor of area n m2. Find the value of x.

28. Originally, the area of a triangle is 135 cm2. Then, there is a change so that the base of the triangle is

increased by 2 cm and the height is decreased by 3 cm. Let x cm be the base of the original triangle.

(a) Express the heights of the original triangle and the new triangle in terms of x.

(b) If the area of the new triangle is 15 cm2 less than that of the original triangle, find the value of x.

Level 2

37


29. (a) 4

2

+x−

5

5

−x= 2 (b)

5+x

x=

3

3

−x− 1

30. (a) 4

72

−x= 2 +

2

5

+x (b)

21

4

x−+

1+x

x= 4

31. (a) 3

2

−

+

x

x+

x

x−2= 2 (b)

2

1

+

−

x

x+ 2 =

3

1

+

+

x

x


32. (a) 2x4 − 17x2 − 9 = 0 (b) 9x4 − 37x2 + 4 = 0

33. (a) x6 + 7x3 − 8 = 0 (b) 27 + 26x3 − x6 = 0

34. (a) 2x8 − 29x4 − 48 = 0 (b) 4 − x4 − 3x8 = 0

35. (a) 2x − x5 − 12 = 0 (b) 3x − x13 − 10 = 0

36. (a) 6+x − x = 4 (b) x−3 − 3 = x

37. (a) 15 −x = x + 5 (b) x + x−26 = 10


(Give the answers correct to 2 decimal places if necessary.)

38. (a) 62x − 42(6x) + 216 = 0 (b) 22x + 3 − 3(2x + 1) + 1 = 0

39. (a) 16x − 5(4x) − 24 = 0 (b) 9x − 8(3x + 1) − 81 = 0

40. (a) 16x + 1 − 17(4x) + 1 = 0 (b) 9x + 1 − 28(3x + 1) + 27 = 0

41. (a) 3x +x3

54= 29 (b) 6x +

x6

21− 10 = 0


(Give the answers correct to 2 decimal places if necessary.)

42. (a) log (x − 5) − logx

2= log (9 − x) (b) log (x + 1) − log (x + 2) = log

x

3

43. (a) log (x2 − 8) − log (2x − 5) = log (x − 2) (b) log (x2 − 11) − log (2x − 9) = log (x + 1)

44. (a) log4 (x − 2) − log4 (x2 + 8) + 2 = 0 (b) log0.2 (2x + 7) − log0.2 (x + 3) = −2 − log0.2 x


38

(Give the answers correct to the nearest 0.1° if necessary.)

45. (a) 3

2cos2 θ = sin θ (b) sin2 θ − cos θ = −1

(c) 5 cos2 θ + 4 sin θ = 4 (d) 2 cos2 θ + 3 sin θ cos θ = 2 sin2 θ

46. (a) θsin

1− sin θ + 3 = 0 (b)

θcos

1− 2 cos θ = tan θ

(c) 2 tan θ +θtan

3= 5 (d)

θtan

1+

θcos

2= 0

47. (a) 6 sin2 θ − 7 cos θ sin θ − 3 cos2 θ = 0 (b) 6 sin θ tan θ + tan θ = 6 cos θ

(c) cos2 θ − 4 sin θ cos θ = 1 (d) θsin

1−

θtan

1= 4 sin θ

48. A group of members organize a party and share the party cost of $900 equally. If 5 more people join

the group and the party cost remains unchanged, each of the original members will pay $15 less. Find

the original number of members in the group.

49. Mike drives a car from place A to place B at a constant speed. If the speed of the car is decreased by

10 km/h, he will arrive at B 20 minutes later. If the distance between A and B is 100 km, find the

original driving speed of Mike.

50. The time taken by worker A and worker B together to produce 1 100 products is 5 days less than the

time taken by worker B alone to produce 900 products. If worker A alone produces 50 products per

day, does worker B alone produce more products per day than worker A alone? Explain your answer.

51. Amy spends $90 on buying apples every month. If the price of each apple is increased by $1.5, the

number of apples that she can buy in a month will be 5 less than before.

(a) Find the original price of each apple.

(b) Can Amy buy at least 12 apples in a month after the price increase? Explain your answer.

52. The profit $P of a shop can be represented by the formula P = 44 800 + 96 800(80.02x) − 38 000(640.02x),

where x (x ≥ 1) is the number of months elapsed since the start of the business of the shop. If the shop

will not make a profit or a loss k months after the start of the business of the shop, find the value of k.

(Give the answer correct to the nearest integer.)

53. Alex buys an antique car for $150 000 and an antique vase for $200 000. The values of the car and

the vase increase by 21% and 10% each year respectively. After t years, the value of the car will be

$50 000 more than that of the vase. Find the value of t.

(Give the answer correct to 3 significant figures.)

39

*54. According to a time schedule, a ship will leave town A at 3 p.m. on Monday. It will travel 198 km

upstream to town B at a constant speed and then will travel downstream back to town A at a constant

speed. If the speed of the current is 4 km/h, it takes 10 hours for the whole journey.

(a) When will the ship arrive at town B?

(b) Suppose it rains throughout the journey. When it rains, the speed of the current is increased by

1.5 km/h and the constant speed of the ship is decreased by 4.5 km/h for reasons of safety. Sam

claims that the ship will be back to town A at 2 a.m. on Tuesday when it rains. Do you agree?


*55. The straight line (2m − 9)x + (m − 1)y − 30 = 0 cuts the positive x-axis and the positive y-axis at the

points A and B respectively, where m is a constant and m > 5. It is given that OA + OB = 16, where O

is the origin.

(a) Express OA and OB in terms of m. Hence, find the value of m.

(b) Find the orthocentre and the circumcentre of △OAB.

*56. (a) Solve 32x + 3x − 42 = 0.

(b) Hence, solve log9 (42 − 32x) −2

x= 0.

(Give the answers correct to 3 significant figures.)

*57. Solve log (10x + 2) = 2x + 1.

(Give the answer correct to 3 significant figures.)

*58. Solve log5 x − 3 logx 25 = 1.

*59. In a city, the government launches a project to reduce the concentration of hazardous substances in

River A. The concentration N units/m3 of hazardous substances in River A can be modelled by

N = a(0.81)t + 10bt + 2, where t is the number of months elapsed since the start of the project, and a, b

are positive constants. When t = 0, N = 15. When t = 2, N = 12.068 3.

(a) Find the values of a and b.

(b) Find the concentration of hazardous substances in River A when t = 3, correct to 3 significant

figures.

(c) At the same time, the government launches another project to reduce the concentration of

hazardous substances in River B. The concentration N1 units/m3 of hazardous substances in River

B can be modelled by N1 = 10.8(0.9)k(t – 1) + 3, where t is the number of months elapsed since the

start of the project and k is a constant. It is known that N = N1 at the start of the two projects.

(i) Find the value of k.

(ii) Will the concentration of hazardous substances in River B be doubled that in River A at a

certain month since the start of the project? Explain your answer.

40

Answers


1. (a) −1,

4

5 (b) 2, 4

2. (a) −1,

3

1 (b) −3, 2

3. (a) −5, 3 (b) −3

4, 2

4. (a) −8,

2

1 (b) −1, 4

5. (a) −2, 6 (b) 1,

3

4

6. (a) 1, 2 (b) 2, 4

7. (a) −2, 2 (b) −3, −1, 1, 3

8. (a) −3, 0, 3 (b) −1, 2

9. (a) −3, 1 (b) −2, 2

10. (a) 1, 16 (b) 4

11. (a) −5, 3 (b) −9, 2

12. (a) −4, 5 (b) −3, 6

13. (a) 1 (b) 3

14. (a) 2 (b) no real

solutions

15. (a) 2 (b) 0

16. (a) 0001

1, 100 (b) 100, 100 000

(c) 9

1, 81 (d)

256

1,

4

1

17. (a) 1 (b) 3

(c) −4, −2 (d) −3, 5

18. (a) 12 (b) 7

(c) 1 (d) 1

19. (a) 30°, 150°, 210°, 330°

(b) 60°, 120°, 240°, 300°

20. (a) 0°, 180°, 210°, 330°, 360°

(b) 0°, 45°, 180°, 225°, 360°

21. (a) 180°

(b) 30°, 150°, 270°

22. (a) 45°, 225°

(b) 120°, 180°, 240°

23. (a) 45°, 120°, 225°, 300°

(b) 30°, 45°, 135°, 150°

24. 9

7

25. 4

3,

3

4

26. 4, 6

27. 5

12

28. (a) original height =x

270 cm,

new height =

− 3

270

xcm

(b) 18

29. (a) −2

5, 2 (b) −3, 5

30. (a) −5,

2

5 (b) 0, −

3

1

31. (a) 2

1, 6 (b) −

2

7, −1

32. (a) −3, 3 (b) −2, −3

1,

3

1, 2

33. (a) −2, 1 (b) −1, 3

34. (a) −2, 2 (b) −1, 1

35. (a) 16 (b) 25

36. (a) −2 (b) −1

37. (a) 5, 10 (b) −14, −2

38. (a) 1, 2 (b) −2, −1

39. (a) 2

3 (b) 3

40. (a) −2, 0 (b) −1, 2

41. (a) 0.63, 3 (b) 0.61, 1.09

42. (a) 6 (b) 3.65

43. (a) 3, 6 (b) 6.70

44. (a) 12.90, 3.10 (b) 12.10

45. (a) 30°, 150°

(b) 0°, 360°

(c) 90°, 191.5°, 348.5°

(d) 63.4°, 153.4°, 243.4°, 333.4°

46. (a) 197.6°, 342.4°

(b) 210°, 330°

(c) 45°, 56.3°, 225°, 236.3°

(d) 204.5°, 335.5°

47. (a) 56.3°, 161.6°, 236.3°, 341.6°

(b) 41.8°, 138.2°, 228.6°, 311.4°

41

(c) 0°, 104.0°, 180°, 284.0°, 360°

(d) 138.6°, 221.4°

48. 15

49. 60 km/h

50. yes

51. (a) $4.5 (b) yes

52. 26

53. 4.59

54. (a) 8:30 p.m. on Monday

(b) no

55. (a) OA =92

30

−m, OB =

1

30

−m, m = 6

(b) orthocentre: (0 , 0), circumcentre: (5 , 3)

56. (a) 1.63

(b) 1.63

57. −0.301

58. 25

1, 125

59. (a) a = 3, b = 0.9

(b) 10.9 units/m3

(c) (i) 1

(ii) no

42

F5A: Chapter 2A

Date Task Progress

Lesson Worksheet


(Full Solution)

Book Example 1


(Video Teaching)

Book Example 2


(Video Teaching)

Book Example 3


(Video Teaching)

Book Example 4


(Video Teaching)

Book Example 5


(Video Teaching)



(Full Solution)




___________ ( )




___________ ( )



___________ ( )

E-Class Multiple Choice ○ Complete and Checked Mark:

43

Self-Test ○ Problems encountered ○ Skipped _________

44


Objective: To review the basic properties of inequalities and solve linear inequalities in one unknown.

Basic Properties of Inequalities

In each of the following, fill in the blank with an appropriate inequality sign ‘>’, ‘<’, ‘≥’ or ‘≤’ according to

the given condition. [Nos. 1–3]

1. Let 0 < a < b.

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

2. Let a ≤ b < 0.

(a) 4a 7a (b) 2a + 1 2b + 1 (c) b

1

a

1

3. Let 0 < b ≤ a.

(a) a

2

b

2 (b) −3a −3b (c)

b4

1

a4

1

Linear Inequalities in One Unknown

Solve each of the following inequalities and represent the solutions on a number line. [Nos. 4–9]

4. 4x − 3 > 29 5. 3x − 4 ≤ x + 14 �Review Ex: 1−10

4x > ( )

x > ( )

Graphical representation: Graphical representation:

6. −5 – 2x < 3x + 10 7. 3(1 – 2x) ≥ 2x + 7

0 0

Use the symbol

‘’ when a value

is not included in

the solutions.

The inequality sign is

reversed when both

sides are divided by a

negative number.

Use the symbol

‘�’ when a value

is included in the

solutions.

3(1 − 2x)

45

8. 6(x − 3) >

2

92 +x 9.

2

2 x− ≤

3

7−x

10. (a) Solve 3(1 + x) >

5

18 +x. 11. (a) Solve

4

32 x− ≥ x – 7.

(b) Write down the smallest integer which (b) How many positive integers satisfy the

satisfies the inequality in (a). inequality in (a)?


12. (a) Solve 3(z – 4) ≤ 2(14 – z).

(b) Hence, solve 3[(x – 5) – 4] ≤ 2[14 – (x – 5)].

Multiply both sides of the inequality by the L.C.M. of 2 and 3.

46

5A Lesson Worksheet 2.1A (Refer to Book 5A P.2.5)

Objective: To solve compound linear inequalities in one unknown connected by ‘and’.

Compound Inequalities Connected by ‘and’

To solve a compound linear inequality in x connected by ‘and’, we need to find the range of possible

values of x which satisfy all the inequalities.

Complete the following table. [Nos. 1–4] �Ex 2A: 1

Compound inequality Number line Solutions Graphical representation

1.

x > –3 and x < 1

2.

x ≤ 4 and x ≤ 9

3.

x ≥ −6 and x < −1

4.

x < –5 and x > 2 Not applicable


Solve x + 2 > –1 and x – 1 ≤ 2, and represent the

solutions graphically.

Solving x + 2 > –1:

x > –3 …… (1) �

Solving x – 1 ≤ 2:

x ≤ 3 …..… (2) � ∵ x must satisfy (1) and (2). ∴ The solutions are –3 < x ≤ 3. �

Graphical representation:

Solve x – 4 > –2 and 2x ≥ 6, and represent the


Solving x – 4 > –2:

x > ( ) …… (1) �

Solving 2x ≥ 6:

x ≥ ( ) ……… (2) � ∵ x must satisfy (1) (2). ∴ The solutions are . �


0 3 −3

0 3 −3

0 3

0 −3

0

0

0

0 4 9

0 −1 −6

–3 0 1

0 2 −5

47

Solve each of the following compound inequalities and represent the solutions graphically. [Nos. 5–8]

5. x + 5 ≥ 3 and 2x – 7 ≤ 1 6. 3x ≥ 12 – x and 2x + 1 < x + 4 �Ex 2A: 2–5

Solving x + 5 ≥ 3:

x ≥ ( ) ……. ( )

Solving 2x – 7 ≤ 1:

2x ≤ ( )

……… ( )


7. –3 < x – 7 < 2 8. −5 ≤ 2x + 5 < 8 − x �Ex 2A: 6–9

Rewrite the compound inequality as

< x – 7 and x – 7 < .


9. How many integers satisfy the compound inequality −5x < −10 ≤ 1 – 7(x – 1)? Explain your answer.


0

�

0

�

48

2 Inequalities in One Unknown


Level 1

1. Represent the solutions of each of the following compound inequalities graphically.

(a) x > −4 and x ≤ 3 (b) x ≤ 2 and x < 5

(c) −6 < x < 8 (d) 9 ≥ x ≥ 3

(e)

<

−≥

6

2

x

x (f)

≥

>

5

7

x

x


2. 3x − 7 > 5 and x + 1 ≤ 6 3. 2 + 3x ≥ −7 and 4 ≤ x + 8

4. 3x + 4 ≥ x and 6x < 0 5. 4x − 2 ≤ x + 7 and 3x < 5 − 2x

6. 3x + 5 < −3 − x and 8 + 7x > 4x − 1 7. 7 + 2x < 4x − 3 and 1 − 8x < 5 − 9x

8. −5 ≤ 3x − 2 < 1 9. 11 ≥ 1 − 2x ≥ 3

10. 2x + 7 > 3 > x − 1 11. 8 + 3x < x ≤ 8 − 3x

12. 2x − 3 > 5 − 6x ≥ 13 − 10x 13. 3x + 8 < 1 − 4x < 1

14. x − 2 ≤ 3 − 4x ≤ 5x − 6 15.

+<

>−

954

1332

x

x

16.

−≥

+≤

345

1046

x

x 17.

≥−

>−

1143

721

x

x

18.

+<+

+≤−

234

364

xx

xx 19.

+<−

−<−

xx

xx

2594

629

20.

−>−

−>−

1357

2352

xx

xx 21.

−≤−

−≥−

xx

xx

238

2134

22. A panda eats x kg of bamboo every day.

(a) Express the total weight of bamboo that the panda eats every week in terms of x.

(b) If the total weight of bamboo that the panda eats this week is greater than 140 kg but cannot

exceed 196 kg, find the range of possible values of x.

23. Consider the compound inequality −1 ≤ 2x − 5 ≤ 1.

(a) Solve the compound inequality.

(b) If sin θ = 2x − 5, where θ is any angle, find the range of possible values of x.

49

24. In △ABC, ∠A = 30° and ∠B = x°, where ∠B is an acute angle.

(a) Express ∠C in terms of x.

(b) If ∠C is larger than ∠B and ∠C is less than ∠A + ∠B, find the range

of possible values of x.

Level 2

25. Represent the solutions of the compound inequality described by each of the following sentences

graphically.

(a) x is greater than 0 but less than 10.

(b) x is at least −2 but cannot exceed 1.

(c) x is not greater than 5 and is at most 3.


26. 3x < 10 − (x + 6) and 3x + 7 > 2(1 − x) 27. 13

2<

− x and

5

16 +x> 2

28. 123

)1(2≥−

+ xx and 4

4

)52(3

6

15<

−−

− xx 29. 2(x − 3) > 3(2x − 4) > 4(3x − 5)

30. 3x ≤ 52

3−

− x≤ 1 31. 1 ≤ 3

5

)13(2+

+x< 7

32.

−>+−

−≤−

51)34(2

)42(3)23(2

x

xx 33.

+>+

−>+−

543

15

3

4

25

xx

xxx

34.

−≤

+>−

4

3

2

)1(29

xx

xx

35.

>−−

−−

<+

2

1)31(21

13

26

2

9

x

xx

36. 7 − 3x ≥ 13 > x − 7 and x > −3 37. x ≤ 0 and −2 <

+−

2

313

x≤ 4

38. (a) Solve

21

5

)23(4 xx>−

−.

(b) Find all integers which satisfy both inequalities

21

5

)23(4 xx>−

− and 7

3

45≤

− x.

39. (a) Solve 2 − 3x ≥3

54 x− and

62

1

3

xx>+ .

(b) How many negative integers satisfy the compound inequality in (a)?

40. (a) Solve

≤−−−

+<−

1)1(22

31

)13(242

7

xx

xx

.

(b) Write down the smallest integer which satisfies the compound inequality in (a).

30° A B

C

x°

50

41. Find the range of values of k if each of the quadratic equations x2 + x − k − 1 = 0 and x2 − 4x + 2k − 3 = 0

has real root(s).

42. If both of the graphs of y = 2x2 − 4x + 10 − 2k and y = x2 + 2x + k + 3 have no x-intercepts, how many

possible integral values of k are there? Explain your answer.

43. The lengths of the three sides of a triangle are 3x + 3, x + 2 and 5x − 2, where x is an even number. By

using the triangle inequality, find all the possible values of x.

44. The dimensions of a cuboid in the figure are 8 cm × x cm × 6 cm. If its

total surface area is at most 320 cm2 and its volume is greater than 192 cm3,

how many possible integral values of x are there? Explain your answer.

45. A test consists of 10 multiple-choice questions. 3 marks will be awarded for a correct answer. 2 marks

will be deducted for a wrong answer or a blank answer. If Cathy gets at most 18 marks but has at least

3 correct answers, find all the possible numbers of correct answers that she has.

* 46. The following table shows the scores of Michael in three tests and the corresponding weights. If the

weighted mean scores of Michael in the three tests is at least 65 but at most 85, can the score of Test 2

be greater than 95? Explain your answer.

Test 1 Test 2 Test 3

Score 80 x + 30 x + 20

Weight 1 2 3

* 47. Mr Chan drives from place A to place C via place B. The following table shows the distances travelled

and the average speeds of the two journeys:

A to B B to C

Distance travelled (km) d d + 20

Average speed (km/h) 30 25

If the total time required is more than 3 hours and the difference of the time required in the two

journeys is not more than 1.2 hours, find the range of values of d.

* 48. Consider the straight line L: 2x + 3y − 5k + 4 = 0, where k is a constant. The x-intercept of L does not

exceed 6 and the y-intercept of L is greater than 2.

(a) Find the range of values of k.

(b) If k is an integer, find the possible area(s) of the region bounded by L and the two axes.

6 cm

8 cm

x cm

51

Answers


2. 4 < x ≤ 5 3. x ≥ −3

4. −2 ≤ x < 0 5. x < 1

6. −3 < x < −2 7. no solutions

8. −1 ≤ x < 1 9. −5 ≤ x ≤ −1

10. −2 < x < 4 11. x < −4

12. x ≥ 2 13. no solutions

14. x = 1 15. x > 8

16. −1 ≤ x ≤ 2 17. x < −3

18. 1 < x ≤ 3 19. 5 < x < 7

20. no solutions 21. x = 3

22. (a) 7x kg (b) 20 < x ≤ 28

23. (a) 2 ≤ x ≤ 3 (b) 2 ≤ x ≤ 3

24. (a) ∠C = 150° − x°

(b) 60 < x < 75

26. −1 < x < 1 27. x >2

3

28. x ≥ 2 29. x <3

4

30. −9 ≤ x ≤ −1 31. −2 ≤ x < 3

32. 0 < x ≤9

5 33.

2

5

2

1<< x

34. x ≤ 1 35. no solutions

36. −3 < x ≤ −2 37. −3

11≤ x ≤ 0

38. (a) 3

2<x (b) −4, −3, −2, −1, 0

39. (a) 2

13 ≤<− x (b) 2

40. (a) 55

12≤<− x (b) −2

41. 2

7

4

5≤≤− k 42. 5

43. 2, 4, 6 44. 4

45. 3, 4, 5, 6, 7 46. no

47. 30 < d ≤ 60

48. (a) 2 < k ≤5

16 (b)

12

121

52

F5A: Chapter 2B

Date Task Progress

Lesson Worksheet


(Full Solution)

Book Example 6


(Video Teaching)

Book Example 7


(Video Teaching)

Book Example 8


(Video Teaching)

Book Example 9


(Video Teaching)

Book Example 10


(Video Teaching)



(Full Solution)




___________ ( )




___________ ( )



___________ ( )

E-Class Multiple Choice ○ Complete and Checked Mark:

53

Self-Test ○ Problems encountered ○ Skipped _________

54

5A Lesson Worksheet 2.1B (Refer to Book 5A P.2.13)

Objective: To solve compound linear inequalities in one unknown connected by ‘or’.

Compound Inequalities Connected by ‘or’

To solve a compound linear inequality in x connected by ‘or’, we need to find the range of possible

values of x which satisfy at least one of the inequalities.

Complete the following table. [Nos. 1–4] �Ex 2B: 1

Compound inequality Number line Solutions Graphical representation

1.

x > 2 or x ≤ −2

2.

x ≤ 4 or x ≤ 9

3.

x > 6 or x < 6

4.

x ≥ −4 or x < 0


Solve x + 2 < 6 or x – 1 ≤ 1, and represent the


Solving x + 2 < 6:

x < 4 …… (1) �

Solving x – 1 ≤ 1:

x ≤ 2 …… (2) � ∵ x must satisfy (1) or (2). ∴ The solutions are x < 4. �


Solve x – 7 ≥ −5 or 4x ≥ 20, and represent the


Solving x – 7 ≥ −5:

x ≥ ( ) …… (1) �

Solving 4x ≥ 20:

x ≥ ( ) ……… (2) � ∵ x must satisfy (1) (2). ∴ The solutions are . �


0 4

0 4 2

0 2

0 4

0

0

0

0 −2 2

4 0 9

6 0

0 −4

55


5. 3x − 2 ≥ x − 6 or x − 3 > −2 6. x + 1 < 4x − 8 or 5x + 9 < 3x + 7 �Ex 2B: 2–11

Solving 3x − 2 ≥ x − 6:

2x ≥ ( )

… ( )

Solving x − 3 > −2:

x > ( ) …. ( )


7. 3(x − 2) < x + 2 or 4x − 5 > 11 8. 3(x + 5) > −2x or 7 – 6x ≥ −5


9. (a) Solve x −

3

9+x < −5 or 4 − x ≥ 8.

(b) Find the greatest integer satisfying the compound inequality in (a).

0 �

0 �

56




Level 1

1. Represent the solutions of each of the following compound inequalities graphically.

(a) x > 2 or x ≤ 1 (b) x < 2 or 3 < x

(c) x ≥ 7 or x > 5 (d) x ≤ −1 or 0 ≥ x

(e) 13 > x or x > 13 (f) x ≥ 2 or x ≤ 7


2. 3x < 0 or 3 + 2x > 5 3. 6x + 14 ≤ 2 or 3x − 17 > 1

4. 1 − 3x ≥ 13 or 3 ≤ 4x − 13 5. 3x + 6 ≤ x or 2x − 1 > 9

6. 4x + 3 > 7 or 3x − 5 ≥ 10 − 2x 7. 9x + 5 < −3x − 10 or 8 − 7x ≤ −6

8. 2x − 1 ≤ 7 − 2x or −6 − 5x > 9 9. 2x − 7 < x + 2 or 3x − 4 ≤ 5x + 2

10. 5x − 4 < 17 − 2x or −x ≤ 3x − 16 11. −2 + 3x < 4x − 9 or 3x − 5 ≥ 7x + 7

12. 2x > 12 + x or 4 − 2x ≤ 5x − 3 13. 11 − x < 7x − 5 or 2x − 3 < 13 − 2x

14. 3x − 6 < 6 − x or 5 − x ≥ x − 1 15. 9 − x < −3x + 1 or 7 − 5x > −5 − 8x

16. P(6 − 3x , −4) and Q(−3 , 8 − 2x) are two points on a rectangular coordinate plane.

(a) If at least one of these two points lies in quadrant III, find the range of values of x.

(b) If Q lies above the line y = 10 or below the line y = −14, find the range of values of x.

17. The present age of Creamy is 12. She will start studying at university when she is 18 and will graduate

from there when she is 22. Suppose she is not a university student after x years (x ≥ 0).

(a) Express the age of Creamy after x years in terms of x.

(b) Find the range of possible values of x.

18. Jason has a parcel containing 3 toy cars. Each toy car weighs x kg and

the package box weighs 0.3 kg.

(a) Express the weight of the parcel in terms of x.

(b) A courier company offers a special discount on parcels satisfying

the requirement on the right. If Jason can enjoy the discount, find

the range of values of x.

Special

Discount!!

For parcel’s weight:

(i) under 1.5 kg

57

Level 2

19. Represent the solutions of the compound inequality described by each of the following sentences

graphically.

(a) x is greater than 0 or less than −5.

(b) x is at most 3 or cannot exceed 4.

(c) x is at least 1 or not less than −2.


20. 3(3 − x) − 1 > x or −2x < x − 9 21. 4(3 − x) ≤ 4 − 3x or 3(x − 5) < 7 − 4(x + 2)

22. 2

59 x−>

3

101 x− or 8 − 3x ≤ 5x + 6 23.

3

241

2

3 xx −<+

− or −19 − x ≥ 2x + 8

24. 2

1− x <

2

7− 2x or

4

7

32

1+<

− xx 25. 2(3 + x) > 3 − x or 5

3

5−

− x≥

2

x

26. 3 − 2x ≤3

2+x or

3

12

2

−−

xx≥ 1 27.

−−

2

2113

x≥ 1 or

3

21

2

1 xx −+

−< 1

28. 514

233 >

+

− x or 2(1 − 5x) <

2

21 x− 29.

−+

3

212

x≤ x or

25

)3(2 xx−

−≥ 3

30. (a) Solve

2

1+x< 4 − 3x or

6

57

8

910

xx≤

−

−.

(b) Find the maximum value of x which satisfies the compound inequality in (a).

31. (a) Solve

3

2

2

3 −>

− xx or 4(5 − x) ≤ 2(3 − x).

(b) Write down the smallest odd number which satisfies the compound inequality in (a).

(c) Zoey claims that the value found in (b) also satisfies the compound inequality

3

2

2

3 −>

− xx and 4(5 − x) ≤ 2(3 − x). Do you agree? Explain your answer.

32. (a) Solve −9x < 27 or

3

1

2793

1−≤−

xx.

(b) Hence, solve

−≥

−

−≤−<−

314

3

26

3

1

2793

1 279

xx

xxx or

.

33. (a) Solve 5 − 4x > 3(2 − x) ≥ −x.

(b) Hence, solve 5 − 4x > 3(2 − x) ≥ −x or 4 − x ≥ 2.

58

34. (a) Solve

−>−

−≥−

3

2

4

3

61

26)23(4

xx

xx

.

(b) Hence, solve

−>−

−≥−

3

2

4

3

61

26)23(4

xx

xx

or 0 < 4 − 4x.

35. In the figure, the perimeter of the parallelogram ABCD is 80 cm. Let

AD = x cm.

(a) Express the length of AB in terms of x.

(b) If AB does not exceed 10 cm or AD is shorter than one-third of

AB, find the range of values of x.

36. In a convex n-sided polygon, the sum of the interior angles is less than 1 080° or the number of interior

angles is at most 6. Find all the possible values of n.

37. A nutritionist recommends that people weighing below 40 kg or above 90 kg should follow a diet plan.

Jacky’s weight is x kg. Daisy’s weight is less than twice Jacky’s weight by 20 kg.

(a) Express Daisy’s weight in terms of x.

(b) Suppose Daisy is recommended to follow a diet plan.

(i) Find the range of values of x.

(ii) If Jacky’s weight is greater than 30 kg but less than 65 kg, will the nutritionist recommend

Jacky to follow a diet plan? Explain your answer.

* 38. Ricky and Candy share an amount of $200. Let $x be the amount that Ricky has.

(a) Express the amount that Candy has in terms of x.

(b) Each of them spends $30 on snacks. If at least one of them has more than $130 left, find the

range of values of x.

* 39. The figure shows the graph for a car travelling from

town X to town Y and then returning to town X during

the period 8:00 to 9:40 in a morning. When the car

travels for t min, the distance between the car and town

X is less than 24 km. Find the range of values of t.

* 40. Consider the quadratic equation in x: 2x2 + (5 − 3k)x + 7k − 4 = 0. If the sum of roots of the equation

is greater than 2 or the product of roots does not exceed 5, how many integral values of k cannot

satisfy the conditions? Explain your answer.

Dis

tan

ce f

rom

tow

n X

(k

m)

Time

8:00 8:20 9:00 9:40 0

32

48

X

Y

A B

C D

x cm

59

Answers


2. x < 0 or x > 1 3. x ≤ −2 or x > 6

4. x ≤ −4 or x ≥ 4 5. x ≤ −3 or x > 5

6. x > 1 7. x <4

5− or x ≥ 2

8. x ≤ 2 9. all real numbers

10. x < 3 or x ≥ 4 11. x > 7 or x ≤ −3

12. x ≥ 1 13. all real numbers

14. x ≤ 3 15. x < −4 or x > −4

16. (a) x > 2 (b) x < −1 or x > 11

17. (a) x + 12 (b) 0 ≤ x < 6 or x > 10

18. (a) (3x + 0.3) kg

(b) 0 < x < 0.4 or x > 1.9

20. x < 2 or x > 3 21. x ≥ 8 or x < 2

22. x > −5 23. x < −7

24. all real numbers 25. x > −1 or x ≤ −4

26. x ≥ 1 or x ≤ −4 27. x ≥6

1−

28. x <6

1 or x >

6

1 29. x ≥ 2 or x ≤ −2

30. (a) x ≤ 195 (b) 195

31. (a) x > 5 (b) 7

(c) yes

32. (a) x > −3 (b) x ≥ 3

33. (a) x < −1 (b) x ≤ 2

34. (a) 2

1− < x ≤ 1 (b) x ≤ 1

35. (a) (40 − x) cm

(b) 0 < x < 10 or 30 ≤ x < 40

36. 3, 4, 5, 6, 7

37. (a) (2x − 20) kg

(b) (i) 10 < x < 30 or x > 55

(ii) no

38. (a) $(200 − x)

(b) 30 ≤ x < 40 or 160 < x ≤ 170

39. 0 ≤ t < 15 or 80 < t ≤ 100

40. 1

60

F5A: Chapter 2C

Date Task Progress

Lesson Worksheet


(Full Solution)

Book Example 11


(Video Teaching)

Book Example 12


(Video Teaching)

Book Example 13


(Video Teaching)

Book Example 14


(Video Teaching)



(Full Solution)




___________ ( )




___________ ( )



___________ ( )



_________

61


Objective: To solve quadratic inequalities in one unknown by the graphical method.

Solving Quadratic Inequalities in One Unknown by the Graphical Method

The general steps of solving a quadratic inequality graphically are as follows:

Step 1: Write the corresponding quadratic function of the given inequality.

Step 2: Find the x-intercepts of the graph of the quadratic function.

Step 3: Sketch the graph of the quadratic function to show where the graph cuts the x-axis.

Step 4: Read the solutions of the quadratic inequality from the graph.


Solve (x + 1)(x − 2) ≥ 0 graphically.

The corresponding quadratic function is

y = (x + 1)(x − 2).

When y = 0, (x + 1)(x − 2) = 0

x = −1 or 2

Since the coefficient of x2 is 1 (> 0), the graph of

y = (x + 1)(x − 2) opens upward.

Sketch:

From the sketch, the required solutions are

x ≤ −1 or x ≥ 2.

Solve (−x + 2)(x − 5) < 0 graphically.


y = ( )( ).

When y = 0, ( )( ) = 0

x = ( ) or ( )

Since the coefficient of x2 is ___ ( > / < 0), the graph

of y = (−x + 2)(x − 5) opens ( upward / downward ).

Sketch:


.

1. Solve x2 + 6x + 5 ≤ 0 graphically.


.

When y = 0,

Since the coefficient of x2 is ___ ( > / < 0), the

graph of ______________________________.

Sketch:

From the sketch,

2. Solve −x2 − 2x + 8 > 0 graphically. �Ex 2C: 5–11

x

y

2 −1 0

y = (x + 1)(x − 2)

� Step 1

� Step 2

� Step 3

� Step 4

x

y

0

� Step 1

� Step 2

� Step 3

� Step 4

y = (x + 1)(x − 2)

= x2 − x − 2 y = (−x + 2)(x − 5)

=

62


Solve x2 − 2x + 3 ≥ 0 graphically.


y = x2 − 2x + 3.

For the equation x2 − 2x + 3 = 0,

discriminant = (−2)2 − 4(1)(3)

= −8 < 0 ∴ The graph of y = x2 − 2x + 3 has no x-

intercepts.

Since the coefficient of x2 is 1 (> 0), the graph of

y = x2 − 2x + 3 opens upward.

Sketch:


all real numbers.

Solve −x2 + 4x − 7 > 0 graphically.


y = ( ).

For the equation −x2 + 4x − 7 = 0,

discriminant = ( )2 − 4( )( )

= ( ) ( > / < ) 0 ∴ The graph of y = −x2 + 4x − 7

( has / has no ) x-intercepts.

Since the coefficient of x2 is ___ ( > / < 0), the graph

of y = −x2 + 4x − 7 opens ( upward / downward ).

Sketch:

From the sketch,

.

3. Solve x2 + 3x + 5 < 0 graphically.


.

For the equation x2 + 3x + 5 = 0,

discriminant =

Since the coefficient of x2 is ___ ( > / < 0), the

graph of ______________________________.

Sketch:

From the sketch,

4. Solve −x2 − 2x − 6 ≤ 0 graphically. �Ex 2C: 23, 24


5. Kenneth claims that the inequality x2 + 4x + 4 ≤ 0 has no solutions. Do you agree? Explain your answer.

x

y

O

y = x2 − 2x + 3

� Step 2

� Step 3

� Step 4

x O

y

� Step 1

� For ax2 + bx + c = 0,

discriminant ∆

= b2 − 4ac

� Step 1

� Step 2

� Step 3

� Step 4

63




[In this exercise, leave the radical sign ‘√’ in the answers if necessary.]

Level 1

1. Use the given graph to solve the following quadratic inequalities.

(a) x2 − 8x + 15 ≤ 0

(b) x2 − 8x + 15 > 0


(a) 3 − 2x − x2 < 0

(b) 3 − 2x − x2 ≥ 0

Solve each of the following quadratic inequalities graphically and represent the solutions on a number line.

[Nos. 3–20]

3. 3x(x + 3) > 0 4. x(x − 12) ≤ 0

5. (x + 5)(x − 7) < 0 6. (x − 4)(2x − 5) ≥ 0

7. x2 − 4x > 0 8. x2 + 9x ≤ 0

9. x2 − 12x + 35 ≥ 0 10. x2 − 13x − 30 < 0

11. x2 > 4 12. 0 ≤ 25 − x2

13. 13x + x2 + 36 < 0 14. 4x − 32 + x2 ≥ 0

15. 6x + 27 − x2 > 0 16. 24 − x2 + 2x ≤ 0

17. x2 − 5x + 2 ≤ 0 18. 5x + x2 − 4 ≥ 0

19. −x2 − 3x − 1 > 0 20. 4 + x − x2 < 0

1 2 3 5 0 x

y y = x2 − 8x + 15

−3 x

y

1

y = 3 − 2x − x2

0

64


(a) −x2 + 8x − 16 ≥ 0

(b) −x2 + 8x − 16 < 0


(a) −3x − x2 − 3 > 0

(b) −3x − x2 − 3 ≤ 0

Level 2

Solve the following quadratic inequalities graphically. [Nos. 23–36]

23. 4x2 − 11x + 7 > 0 24. 30 − 3x2 − 13x ≥ 0

25. 5x + 6x2 − 6 ≤ 0 26. 4x + 1 − 5x2 < 0

27. 3x2 − 18x + 27 > 0 28. −12x − 9 − 4x2 ≥ 0

29. 5x2 + x + 5 ≤ 0 30. 3x − x2 − 10 < 0

31. −8x < 4 − 5x2 32. 4x2 − 2x ≥ 2x + 3

33. (x + 2)(x + 3) ≤ 2 34. 3(x + 2)(x − 3) > 2(x − 8)

35. 3 − 2x < (2x + 1)(x − 1) 36. (1 − 3x)(5 − x) ≤ 2 − 9x

37. (a) Solve 8x2 + 22x − 21 ≥ 0 graphically.

(b) Using the results of (a), solve 8(2y − 1)2 + 22(2y − 1) − 21 ≥ 0.

(c) Find the largest negative integer satisfying the inequality in (b).

* 38. (a) Solve x2 − 12x − 45 < 0 graphically.

(b) Using the results of (a), solve (4z − 1)2 − 48z + 12 < 45.

(c) If k is a positive integer satisfying the inequality in (b), find the smallest value of k.

−3

x

y

0

y = −x2 − 3x − 3

− 3

y = −x2 + 8x − 16

x

y

4 0

65

Answers


1. (a) 3 ≤ x ≤ 5

(b) x < 3 or x > 5

2. (a) x < −3 or x > 1

(b) −3 ≤ x ≤ 1

3. x < −3 or x > 0 4. 0 ≤ x ≤ 12

5. −5 < x < 7 6. x ≤2

5 or x ≥ 4

7. x < 0 or x > 4 8. −9 ≤ x ≤ 0

9. x ≤ 5 or x ≥ 7 10. −2 < x < 15

11. x < −2 or x > 2 12. −5 ≤ x ≤ 5

13. −9 < x < −4 14. x ≤ −8 or x ≥ 4

15. −3 < x < 9 16. x ≤ −4 or x ≥ 6

17. 2

175

2

175 +≤≤

−x

18. x ≤2

415 −− or x ≥

2

415 +−

19. 2

53

2

53 +−<<

−−x

20. x <2

171− or x >

2

171+

21. (a) x = 4

(b) all real numbers except 4

22. (a) no solutions

(b) all real numbers

23. x < 1 or x >4

7 24.

3

56 ≤≤− x

25. 2

3− ≤ x ≤

3

2 26. x <

5

1− or x > 1

27. all real numbers except 3

28. x =2

3−

29. no solutions

30. all real numbers

31. 5

2− < x < 2 32. x ≤

2

1− or x ≥

2

3

33. −4 ≤ x ≤ −1 34. x <3

1− or x > 2

35. x <4

331−− or x >

4

331+−

36. 6

137

6

137 +≤≤

−x

37. (a) x ≤2

7− or x ≥

4

3

(b) y ≤4

5− or y ≥

8

7

(c) −2

38. (a) −3 < x < 15

(b) 2

1− < z < 4

(c) 1

66

F5A: Chapter 2D

Date Task Progress

Lesson Worksheet


(Full Solution)

Book Example 15


(Video Teaching)

Book Example 16


(Video Teaching)

Book Example 17


(Video Teaching)

Book Example 18


(Video Teaching)

Book Example 19


(Video Teaching)

Book Example 20


(Video Teaching)

Book Example 21


(Video Teaching)

Book Example 22


(Video Teaching)

67



(Full Solution)

Maths Corner Exercise 2D Level 1



___________ ( )




___________ ( )

Maths Corner Exercise 2D Multiple Choice


___________ ( )



_________

68

5A Lesson Worksheet 2.3A (Refer to Book 5A P.2.30)

Objective: To solve quadratic inequalities in one unknown by using compound inequalities.

Solving Quadratic Inequalities in One Unknown by Using Compound Inequalities

Let m and n be two real numbers.

(a) If mn > 0, then

>

>

0

0

n

m or

<

<

0

0

n

m.

(b) If mn < 0, then

<

>

0

0

n

m or

>

<

0

0

n

m.

If the inequality signs ‘>’ and ‘<’ are replaced by ‘≥’ and ‘≤’ respectively, the above results are still true.


Solve (x − 1)(x − 3) > 0 by using compound

inequalities.

(x − 1)(x − 3) > 0

>−

>−

03

01

x

x or

<−

<−

03

01

x

x

>

>

3

1

x

x or

<

<

3

1

x

x

x > 3 or x < 1 ∴ The solutions are x > 3 or x < 1.

Solve (x − 2)(x + 5) ≥ 0 by using compound

inequalities.

( )( ) ≥ 0

≥

≥

0

0

or

≤

≤

0

0

≥

≥

or

≤

≤

or ∴ The solutions are .

Solve the following quadratic inequalities by using compound inequalities. [Nos. 1–4]

1. (2x − 1)(x + 2) > 0 2. x2 + x − 12 ≥ 0 �Ex 2D: 1, 3

3. 2x2 − 5x + 3 > 0 4. x2 + 10x + 8 ≥ −13


1 0 3

:

1 0 3

:

>

>

3

1

x

x

<

<

3

1

x

x

Factorize x2 + x − 12 first.

Make the R.H.S. of the inequality equal to 0 first.

69

Solve (x + 2)(x − 1) < 0 by using compound

inequalities.

(x + 2)(x − 1) < 0

<−

>+

01

02

x

x or

>−

<+

01

02

x

x

<

−>

1

2

x

x or

>

−<

1

2

x

x

−2 < x < 1 or no solutions ∴ The solutions are −2 < x < 1.

Solve (x − 3)(x + 5) ≤ 0 by using compound

inequalities.

(x − 3)(x + 5) ≤ 0

≤

≥

0

0

or

≥

≤

0

0

≤

≥

or

≥

≤

or ∴ The solutions are .


5. (x + 3)(3x + 2) ≤ 0 6. x2 − x − 30 < 0 �Ex 2D: 2

7. 4x2 + 5x + 1 ≤ 0 8. x2 + 7x − 6 < 2


9. If the quadratic equation x2 + kx + 2k − 3 = 0 has no real roots, find the range of values of k by using

compound inequalities.

0 −2 1 :

<

−>

1

2

x

x

0 −2 1 :

>

−<

1

2

x

x

(no overlapping

part)

70


Objective: To solve quadratic inequalities in one unknown by using the tabulation method.

Solving Quadratic Inequalities in One Unknown by Using the Tabulation Method

Step 1: Factorize the quadratic polynomial in an inequality if necessary.

Step 2: Divide the number line into intervals by using the roots of the corresponding quadratic equation in

Step 1. Use a table to determine the signs of the factors of the quadratic polynomial in each interval.

Step 3: Read the solutions from the table according to the inequality sign given.


Solve (x + 2)(x − 5) > 0 by using the tabulation

method.

(x + 2)(x − 5) > 0

x < −2 x = −2 −2 < x < 5 x = 5 x > 5

x + 2 − 0 + + +

x − 5 − − − 0 +

(x + 2)(x − 5) + 0 − 0 + ∴ The solutions are x < −2 or x > 5. � Step 3

Solve (x − 7)(x + 3) ≤ 0 by using the tabulation

method.

(x − 7)(x + 3) ≤ 0

x < −3 x = −3 −3 < x < 7 x = 7 x > 7

x − 7 − −

x + 3 − 0

(x − 7)(x + 3) + 0

∴ The solutions are . � Step 3

Solve the following quadratic inequalities by using the tabulation method. [Nos. 1–4] �Ex 2D: 7−10

1. (x + 4)(x + 1) ≥ 0 2. x2 − 11x − 12 > 0

x < −4 x = −4

x + 4

3. 3x2 − 5x − 2 < 0 4. x2 − 7x + 6 ≤ −6


The roots of (x − 7)(x + 3) = 0 are and .

Factorize x2 − 11x − 12 first.

Make the R.H.S. of the inequality equal to 0 first.

� Step 2 � Step 2

The roots of (x + 2)(x − 5) = 0

are −2 and 5.

71

Solve x2 − 4x + 4 ≥ 0 by using the tabulation method.

x2 − 4x + 4 ≥ 0

(x − 2)2 ≥ 0

x < 2 x = 2 x > 2

x − 2 − 0 +

(x − 2)2 + 0 + ∴ The solutions are all real numbers. � Step 3

Solve x2 + 6x + 9 > 0 by using the tabulation method.

x2 + 6x + 9 > 0

> 0

x < −3 x = −3 x > −3

x + 3 − 0

(x + 3)2 + ∴ The solutions are

. � Step 3

Solve the following quadratic inequalities by using the tabulation method. [Nos. 5–8] �Ex 2D: 11, 12

5. x2 − 10x + 25 < 0 6. x2 + 12x + 36 ≤ 0

< 0

7. 4x2 + 12x + 9 < 0 8. 16x2 − 8x + 1 ≤ 0


9. (a) Complete the following table.

x < −4 x = −4 −4 < x < 6 x = 6 x > 6

x + 4 − 0 +

x − 6 0 +

(x + 4)(x − 6)

(b) Using the above table, Sara claims that there are 4 negative integers satisfying the inequality

x2 − 2x − 24 < 0. Do you agree? Explain your answer.

Step 1: The repeated root of = 0 is .

� Step 2 � Step 2

Step 1: The repeated root of

(x − 2)2 = 0 is 2. � �

( Use ‘+’ and ‘−’ to denote ‘positive value’ and ‘negative

value’ respectively.)

72



Consolidation Exercise 2D

[In this exercise, leave the radical sign ‘√’ in the answers if necessary.]

Level 1


1. (x − 6)(x + 7) < 0 2. (4x + 5)(x + 8) ≥ 0

3. x2 − 9x + 18 > 0 4. x2 + 4x − 32 ≤ 0

5. x2 > 100 6. 28 − 3x − x2 ≤ 0

7. x2 − 8x + 5 ≥ 0 8. 11 + 4x − x2 > 0

Solve the following quadratic inequalities by using the tabulation method. [Nos. 9–16]

9. (x + 5)(3x + 2) < 0 10. (3 + x)(11 − x) ≥ 0

11. x2 − 6x − 16 ≤ 0 12. x2 − 13x + 40 > 0

13. 25 < x2 14. 30 + 7x − x2 ≥ 0

15. x2 + 12x + 36 ≥ 0 16. x2 + 9 < 6x

Solve the following quadratic inequalities algebraically. [Nos. 17–22]

17. (x + 3)(x + 7) < 0 18. x2 − 12x + 20 ≥ 0

19. x2 − 4x − 45 > 0 20. −16 ≤ −x2

21. 8 + 7x − x2 < 0 22. 9x2 + 6x + 1 ≤ 0

23. If the quadratic equation x2 + (5 − k)x + 1 = 0 has two distinct real roots, find the range of values of k.

24. If the quadratic equation 4x2 − 4kx + k + 12 = 0 has no real roots, find the range of values of k.

25. If x = −2 satisfies the inequality x2 + 4kx + 5k2 > 4, find the range of values of k.

73

26. A stone is thrown upward from the top of a building onto the ground. After t seconds, the height of the

stone above the ground is (30 + 10t − 5t2) m. Find the range of values of t when the height of the stone

above the ground is less than 15 m.

27. In the figure, ABCD is a square and EFGH is a

parallelogram. If the area of ABCD is less than that

of EFGH, find the range of values of x.

Level 2

Solve the following quadratic inequalities algebraically. [Nos. 28–37]

28. 4x2 − 4x − 3 ≤ 0 29. 4x2 − 19x + 12 > 0

30. x2 + 6x + 11 ≤ 0 31. 1 + x − 6x2 ≥ 0

32. 7 > 4x − 2x2 33. 65 ≤ 16(x2 − 1)

34. (3x − 2)2 ≤ 9 35. 4(x − 2)(x + 1) ≤ 7

36. (2x + 3)(x + 7) > 9x 37. (3x + 1)(2x + 3) < 5x

38. (a) Solve x2 − 19x + 48 < 0 algebraically.

(b) Find the largest and the smallest integers satisfying the inequality in (a).

39. (a) Solve the following inequalities algebraically.

(i) 3(x − 1) ≥ 2x (ii) (2x + 1)(x − 5) < 13

(b) Hence, find the range of values of x satisfying both inequalities in (a).

40. (a) Solve 2(4x + 3)(x − 2) < 13 algebraically.

(b) How many positive integers satisfy the inequality in (a)?

41. If the quadratic graph of y = x2 − 4kx + (k + 14) does not have any x-intercepts, find the range of

values of k.

42. It is given that the quadratic equation in x: (k − 5)x2 + (k − 5)x − k = 0 has real root(s).

(a) Can the value of k be equal to 5? Explain your answer.

(b) Find the range of values of k.

43. If (2k + 1)x2 − 3kx + 4 > 0 holds for all real numbers x, where k ≠ −2

1, find the range of values of k.

44. Two real numbers differ by 4. Their product is greater than their sum by at least 44. Let x be the

x cm

x cm

(x + 8) cm

4 cm

A B

C D

E F

G H

74

smaller number.

(a) Find the range of values of x.

(b) If the two numbers are positive, find the minimum value of the larger number.

45. The sum of two distinct positive numbers is 8. Let x be the smaller number.

(a) Can x be greater than 4? Explain your answer.

(b) If the sum of the squares of the two numbers is not less than 50, find the range of values of x.

46. The length and the width of a rectangular field are 12 m and 4 m respectively. If its length is increased

by 2x m and its width is increased by x m, the new area of the field will be at least two times its

original area. Find the range of values of x.

47. In the figure, the length, the width and the height of a cuboid

are (x + 2) cm, (x + 1) cm and x cm respectively.

(a) Express the total surface area of the cuboid in terms of x.

(b) If the total surface area of the cuboid does not exceed

214 cm2, find the range of values of x.

* 48. In the figure, ABCDEFGH is a solid right prism. The base ABCD

of the prism is a trapezium, where ∠ABC = ∠BCD = 90° and

∠ADC = 45°. It is given that AB = (k − 2) cm, BC = k cm and

CH = 4 cm.

(a) Express the length of DC in terms of k.

(b) If the volume of the prism is at most 320 cm3, find the

range of values of k.

* 49. Andy rides a bicycle at the speed of (x + 4) m/s for 10x seconds. The total distance travelled is less

than 600 m.

(a) Find the range of values of x.

(b) If the total distance travelled is at least 210 m, find the range of values of x.

* 50. The x-intercept and the y-intercept of the straight line L are −1 and c respectively, where c is a non-zero

constant. L cuts the graph of y = x2 + 3x + 3 − c at two distinct points A and B.

(a) Express the equation of L in terms of c.

(b) Find the range of possible values of c.

(c) Let M(h , k) be the mid-point of AB. Find the range of possible values of h.

x cm

(x + 2) cm

(x + 1) cm

A B

C D

E

F G

H

k cm

4 cm 45°

75

Answers


1. −7 < x < 6 2. x ≥4

5− or x ≤ −8

3. x > 6 or x < 3 4. −8 ≤ x ≤ 4

5. x > 10 or x < −10 6. x ≥ 4 or x ≤ −7

7. x ≥ 4 + 11 or x ≤ 4 − 11

8. 2 − 15 < x < 2 + 15

9. −5 < x < −3

2 10. −3 ≤ x ≤ 11

11. −2 ≤ x ≤ 8 12. x < 5 or x > 8

13. x < −5 or x > 5 14. −3 ≤ x ≤ 10

15. all real numbers 16. no solutions

17. −7 < x < −3 18. x ≤ 2 or x ≥ 10

19. x < −5 or x > 9 20. −4 ≤ x ≤ 4

21. x < −1 or x > 8 22. x = −3

1

23. k < 3 or k > 7 24. −3 < k < 4

25. k < 0 or k >5

8 26. t > 3

27. 0 < x < 8 28. 2

1− ≤ x ≤

2

3

29. x <4

3 or x > 4 30. no solutions

31. 3

1− ≤ x ≤

2

1 32. all real numbers

33. x ≤4

9− or x ≥

4

9 34.

3

1− ≤ x ≤

3

5

35. 2

3− ≤ x ≤

2

5 36. all real numbers

37. no solutions

38. (a) 3 < x < 16

(b) largest integer: 15, smallest integer: 4

39. (a) (i) x ≥ 3 (ii) 2

3− < x < 6

(b) 3 ≤ x < 6

40. (a) 4

5− < x <

2

5 (b) 2

41. 4

7− < k < 2

42. (a) no (b) k ≤ 1 or k > 5

43. 9

4− < k < 4

44. (a) x ≤ −8 or x ≥ 6

(b) 10

45. (a) no (b) 0 < x ≤ 1

46. x ≥ 2

47. (a) (6x2 + 12x + 4) cm2

(b) 0 < x ≤ 5

48. (a) (2k − 2) cm (b) 2 < k ≤ 8

49. (a) 0 < x < 6 (b) 3 ≤ x < 6

50. (a) cx − y + c = 0

(b) c < −3 or c > 1

(c) h < −3 or h > −1

76

F5A: Chapter 3A

Date Task Progress

Lesson Worksheet


(Full Solution)



(Full Solution)




___________ ( )




___________ ( )



___________ ( )


○ Complete and Checked ○ Problems encountered ○ Skipped Mark: _________

77


Objective: To review the basic concepts of functions and transformations in a rectangular coordinate plane.

Functions

1. Consider the function y = 5x.

(a) Complete the table below.

x −1 0 1 2

y

(b) Draw the graph of the function y = 5x for

−1 ≤ x ≤ 2.

2. Consider the function y = −x + 1. �Review Ex: 1

(a) Complete the table below.

x −1 0 1 2

y

(b) Draw the graph of the function y = −x + 1

for −1 ≤ x ≤ 2.

Find the domain of each of the following functions. [Nos. 3–6]

3. f(x) = 5−x ∵ The value of ( ) under the radical

sign cannot be ( ). ∴ ( ) ≥ 0

x ≥ ( )

∴ The domain is all real numbers greater

than or equal to ( ) .

4. g(x) =

x

x

−1

6 �Review Ex: 2 ∵ The value of the denominator ( )

cannot be ( ). ∴ ( ) ≠ 0

x ≠ ( )

∴ The domain is

.

5. f(x) = 32 −x

6. g(x) =

4

92

+

−

x

x

x –1 0 1 2

10

5

–5

y

Choose a suitable scale for the x-axis and the y-axis.

78

Transformations in a Rectangular Coordinate Plane

The following tables show the new position after a point (x , y) is translated by n units or reflected in the x-

axis / the y-axis.

(a) Translation

Direction of

translation

Coordinates of new

position

To the right (x + n , y)

To the left (x − n , y)

Upward (x , y + n)

Downward (x , y − n)

(b) Reflection

Axis of

reflection

Coordinates of new

position

The x-axis (x , −y)

The y-axis (−x , y)

In each of the following, find the coordinates of Q. [Nos. 7–10] �Review Ex: 3, 4

7. A point P(3 , 7) is translated upward by 10 units

to point Q.

Coordinates of Q = ( , )

= ) , (

9. A point P(2 , 1) is reflected in the x-axis to

point Q.

Coordinates of Q = ) , (

8. A point P(6 , −5) is translated to the left by

7 units to point Q.

10. A point P(−3 , 4) is reflected in the y-axis to

point Q.

11. A point X(−2 , 1) is translated downward by

4 units to point Y, and then translated to the

right by 6 units to point Z. Find the coordinates

of Y and Z.

12. A point A is translated to the left by 5 units to

point B, and then reflected in the y-axis to point

C(3 , 6). Find the coordinates of A.

Let (x , y) be the coordinates of A.


13. A point A(−3 , 1) is translated to the right by 9 units to point B, and then translated downward by

4 units to point C. A point P(5 , 3) is reflected in the x-axis to point Q. Is CQ a horizontal line?


10 7

79


3 More about Graphs of Functions


Level 1

1. Write down the domain of each of the following functions.

(a) (b) (c)

(d)

(e) (f)

2. In each of the following functions, determine whether its graph is a straight line or a curve. State

whether the graph passes through the origin or not.

(a) y = 5 − 5x (b) y = (x − 1)2 − 1

(c) y =

2

3

1

+x +

9

1 (d) y = sin x

(e) y = 3x (f) y = log4 x

3. In each of the following functions, determine whether the maximum value or minimum value exists.

If so, state whether it is the maximum value or the minimum value.

(a) y = x2 − 4x + 5 (b) y = cos x

(c) y = 6−x (d) y = log0.1 x

4. Write down the axis of symmetry of each of the following quadratic graphs.

(a)

(b)

x

y

O

y = 4 − (x + 2)2

y = −2x − 5

x

y

O

y = sin x

0 180° 360° −180°

x

y

−360°

x

y = tan x

0 180° 360°

y

90° 270°

y = 1.5x

O x

y

y = log0.5 x O

x

y

3

2

1

−1

0 1 2 3 4 5

x

y

3

2

1

−1 −3 −2 −1

0 1

x

y

80

y = logp x

O x

y

Q

y = cx

O x

y

D

5. Write down the period of each of the following functions.

(If the function does not have such a feature, write down the word ‘none’.)

(a) (b)

6. Write down the maximum value / minimum value of each of the following functions and the axis of

symmetry of the corresponding graph.

(a) y = 2x2 + 8 (b) y = −2(x − 3)2 − 6

(c) y = 3(x − 5)2 − 7 (d) y = −4(x + 1)2 + 13

For each of the following functions,

(a) find the x-intercept(s) and y-intercept of its graph,

(b) hence, sketch the graph of the function. [Nos. 7–10]

7. y = x 8. y = 4 − x

9. y = −3x + 2 10. y = 2x − 6

For each of the following functions,

(a) find the direction of opening, the coordinates of the vertex and the axis of symmetry of its graph,

(b) hence, sketch the graph of the function. [Nos. 11–14]

11. y = 3x2 − 3 12. y = −x2 + 1

13. y = 4 − (x + 3)2 14. y = −4(x − 1)2 − 6

graph 15. The figure shows the graph of the function y = cx, where c is a constant. The

cuts the y-axis at D(0 , d).

(a) Find the coordinates of D.

(b) Is it possible that c + d > 2? Explain your answer.

16. The figure shows the graph of the function y = logp x, where p is a constant. The

graph cuts the x-axis at the point Q.

(a) Find the coordinates of Q.

(b) Q is translated to the right by 2 units to point R(r , s). Is it true that

pr > 3? Explain your answer.

Level 2

y = cos x

−1

1

0

y

x

180° 360° 720° 540°

y = 2x

O x

y

81

x 0

P(0 , 1)

y

y = 0.25x

1

Sketch the graph of each of the following functions. [Nos. 17–26]

17. y = 5 − 6x 18. y =5

43 x+

19. y = 2(x + 3)2 + 2 20. y = 9 − 4(x − 1)2

21. y = 3(2x − 3)2 − 12 22. y = sin x, where 0° ≤ x ≤ 540°

23. y = 8x 24. y = 0.52x

25. y = log6 x 26. y = log0.5 x

27. Consider the function y = 4x2 − 8x + 3.

(a) Find the x-intercepts and the y-intercept of the graph of the function.

(b) By the method of completing the square, express the function in the form y = a(x − h)2 + k, where

a, h and k are constants.

(c) Hence, sketch the graph of y = 4x2 − 8x + 3.

28. Consider the function y = 13 + 12x − x2.

(a) Find the x-intercepts and the y-intercept of the graph of the function.

(b) By the method of completing the square, express the function in the form y = a(x − h)2 + k, where

a, h and k are constants.

(c) Hence, sketch the graph of y = 13 + 12x − x2.

0.25x and 29. In the figure, the graph of y = 0.25x cuts the y-axis at P(0 , 1). The graphs of y =

another function y = f(x) are the images of each other when reflected in the y- axis. The

graphs of y = f(x) and another function y = g(x) are the images of each other when

reflected in the line y = x.

(a) Write down the functions f(x) and g(x).

(b) Sketch the graphs of y = f(x) and y = g(x) in the same figure.

*(c) Q is a point on the graph of y = g(x), where Q has the same y-coordinate as P. R is a point

on the straight line

3x − 2y + 6 = 0. If PQ ⊥ QR, find the coordinates of R.

82

Answers


1. (a) all real numbers


(c) all angles

(d) all angles except ± 90°, ± 270°, …

(e) all real numbers

(f) all positive real numbers

2. (a) straight line, no (b) curve, yes

(c) curve, no (d) curve, yes

(e) curve, no (f) curve, no

3. (a) yes, minimum value

(b) yes, both maximum and minimum

values exist

(c) no (d) no

4. (a) x = 3 (b) x = −1.5

5. (a) 360° (b) none

6. (a) minimum value: 8,

axis of symmetry: x = 0

(b) maximum value: −6,


(c) minimum value: −7,


(d) maximum value: 13,

axis of symmetry: x = −1

7. (a) x-intercept: 0, y-intercept: 0

8. (a) x-intercept: 4, y-intercept: 4

9. (a) x-intercept:

3

2, y-intercept: 2

10. (a) x-intercept: 3, y-intercept: −6

11. (a) direction of opening: upward,

coordinates of vertex: (0 , −3),


12. (a) direction of opening: downward,

coordinates of vertex: (0 , 1),



coordinates of vertex: (−3 , 4),

axis of symmetry: x = −3


coordinates of vertex: (1 , −6),


15. (a) (0 , 1) (b) no

16. (a) (1 , 0) (b) yes

27. (a) x-intercepts:

2

1,

2

3; y-intercept: 3

(b) y = 4(x − 1)2 − 1

28. (a) x-intercepts: −1, 13; y-intercept: 13

(b) y = −(x − 6)2 + 49

29. (a) f(x) = 4x, g(x) = log4 x

(c) (4 , 9)

83

F5A: Chapter 3B

Date Task Progress

Lesson Worksheet


(Full Solution)

Book Example 1


(Video Teaching)

Book Example 2


(Video Teaching)

Book Example 3


(Video Teaching)

Book Example 4


(Video Teaching)



(Full Solution)




___________ ( )




___________ ( )



___________ ( )



84


Objective: To solve equations f(x) = k by using the graph of y = f(x).

Solving Equations by Using Graphs of Functions

(a) The roots of an equation f(x) = 0 are the x-intercepts of the graph of y = f(x).

(b) The roots of an equation f(x) = k are the x-coordinates of the points of intersection of the graphs of

y = f(x) and y = k.

(c) If the graphs of y = f(x) and y = k do not have any point of intersection, then the equation f(x) = k

has no real roots or no real solutions.

In each of the following, use the given graph(s) to find the roots of the corresponding equation. [Nos. 1–2]

1. −x2 + 6x − 5 = 0

From the graph, the x-intercepts are ( ) and

( ). ∴ The required roots are ( ) and ( ).

2. cos x = 0.5, where 0° ≤ x ≤ 360° �Ex 3B: 1, 2

The two graphs intersect at ( , ) and

( , ). ∴ The required roots are .


By adding a suitable straight line on the given graph,

solve x2 − 5x + 4 = −1.6.

Draw the straight line y = −1.6 on the graph.

The two graphs intersect at (1.7 , −1.6) and

(3.3 , −1.6). ∴ The required roots are 1.7 and 3.3.

By adding a suitable straight line on the given graph,

solve tan x = 2, where 0° ≤ x ≤ 360°.

Draw the straight line on the graph.

The two graphs intersect at ( , ) and

( , ). ∴ The required roots are ( ) and

( ).

x

1

y = −x2 + 6x − 5

y

2 3 4 5 6

2

4

0

−2

0

−1

1

y

x 240° 360°

y = 0.5

120°

y = cos x

x

–1

0 1 2 3 4

–2

y = x2 − 5x + 4

y = −1.6

y

4

–4

x

y = tan x

90° 180° 0

y

270° 360°

2

–2

85

Solve the following equations by adding suitable straight lines on the given graphs. [Nos. 3–6]

3. x3 + 2x2 − 2x − 3 = 0.8

Draw the straight line on the graph.

The two graphs intersect at

.

4. 2x = 1.5 �Ex 3B: 4−8

5. 5 log2 x = 1

6. f(x) + 0.7 = 0, where 0 ≤ x ≤ 5 �Ex 3B: 13


7. The figure shows the graph of y = f(x). On the graph,

add a straight line

(a) y = k1 such that f(x) = k1 has two real roots,

(b) y = k2 such that f(x) = k2 has one real root,

(c) y = k3 such that f(x) = k3 has no real roots.

x

y

1

−1

−2

−3

−2 −1 0 1 −3

y = x3 + 2x2 − 2x − 3

2

y

x 0

1

2

1 −1

y = 2x

0.5 1 1.5 x

y

−1

y = log2 x 1

0 2

y

x

y = f(x)

O

Rewrite the given equation in the form log2 x = k, where k is a constant.

Rewrite the given equation in the form f(x) = k, where k is a constant.

0 1 2 3 4 5

−1

y

x

y = f(x)

−2

1

86




Level 1

1. In each of the following, use the given graph to find the roots of the corresponding equation.

(a) −2x2 + 5x + 7 = 0 (b) −x3 + 9x = 0

(c) 3x3 + 4x2 − 7x = 0 (d) 2 sin x = 0, where −90° ≤ x ≤ 450°

2. In each of the following, use the given graphs to find the roots of the corresponding equation.

(a) x2 + 8x + 11 = 4 (b) −3x2 + 7x + 8 = 15

x

y

10

5

−5 −1

0 1 2 3 4

y = −2x2 + 5x + 7

−10

10

5

−10

−3 −2 −1 0

1 2 3

y = −x3 + 9x

x

y

−5

10

5

−5

−10

−2 −1 0

1 2

x

y

y = 3x3 + 4x2 − 7x

−3

2

1

−1

−2

0 90° 180° 270° 360°

y = 2 sin x

x

y

−90° 450°

y = x2 + 8x + 11

y = 4

x

y

4

2

−2

−4

−8 −6 −4 −2 0

2

15

10

5

−5 −1

0 1 2 3 4

y = 15

y = −3x2 + 7x + 8

x

y

87

y = −2x3 − 5x2 + 3x

5

−5

−10

−15

−20

−3 −2 −1 0

1 2

y

x

y = −19

15

10

5

−5

−10

−2 0

2 4 6 8

y = −x3 + 2x2 + 8x − 8

y = 13

y = −12.5

x

y

y = −x2 + 2x + 7 8

6

4

2

−2 −1

0 1 2 3 4

x

y

(c) 2x3 − 7x + 2 = 4 (d) −4 sin2 x = −2.5, where 0° ≤ x ≤ 360°

3. Use the given graphs to find the roots of the following equations.

(a) −2x3 − 5x2 + 3x = 0

(b) −2x3 − 5x2 + 3x = −19

4. Use the given graphs to find the roots of the following equations.

(a) −x3 + 2x2 + 8x − 8 = 13

(b) −x3 + 2x2 + 8x − 8 = −12.5

5. The figure shows the graph of y = −x2 + 2x + 7. Solve the following

equations by adding suitable straight lines on the given graph.

(a) −x2 + 2x + 7 = 7

(b) −x2 + 2x + 7 = 4

y = 4

6

4

2

−2

−2 −1 0

1 2

x

y

y = 2x3 − 7x + 2

−1

−2

−3

−4

0 90° 180° 270° 360°

y = −4 sin2 x

y = −2.5

x

y

88

1

0.5

−0.5

−180° −90° 0

90° 180°

y = sin 2x

−1

x

y

y = cos2 x 1

0.5

0 90° 180° 270° 360°

x

y

2

−2

−4

0 2 4 6

y = log0.7 x

x

y

y = x3 − 2x

x

y

0

1

2

−1

1 2 −1

10

5

2 0

1

y = 3x

x

y

6. The figure shows the graph of y = x3 − 2x. Solve the following

equations by adding suitable straight lines on the given graph.

(a) x3 − 2x = 1.5

(b) x3 − 2x = −1

7. The figure shows the graph of y = sin 2x, where −180° ≤ x ≤ 180°.

Solve the following equations by adding suitable straight lines on the

given graph.

(a) sin 2x = 0.8

(b) sin 2x = −0.6

8. The figure shows the graph of y = cos2 x, where 0° ≤ x ≤ 360°. Use

the graph to solve the following equations.

(a) cos2 x = 0.5

(b) cos2 x = 1.2

9. Use the given graph to solve the following equations.

(a) 3x = 7

(b) 3x = 2.5

10. Use the given graph to solve the following equations.

(a) log0.7 x = −2

(b) log0.7 x = 1.4

89

2

−2

0 90° 180° 270° 360°

x

y

y = 2 cos x

y = 2

y = −1.4

y = 0.25x

y = 0.75

y = 0.4

x

y

0 0.5 1 1.5

0.5

1

2

−2

−20

2 4

y = f(x)

x

y

11. The graph of y = 2x2 − 5x + 1 is drawn in the rectangular coordinate plane. For each of the following

equations, find a straight line which should be added on the graph of y = 2x2 − 5x + 1 in order to solve

the equation.

(a) 2x2 − 5x = 0

(b) 2x2 − 5x + 5 = 0

(c) 2x2 − 5x − 2 = 0

12. The graph of y = −x2 + 4x + 2 is drawn in the rectangular coordinate plane. For each of the following

equations, find a straight line which should be added on the graph of y = −x2 + 4x + 2 in order to solve

the equation.

(a) −x2 + 4x = 0

(b) −x2 + 4x − 2 = 0

(c) −x2 + 4x + 6 = 3

Level 2

13. The figure shows the graphs of y = 2 cos x (where 0° ≤ x ≤ 360°),

y = 2 and y = −1.4. Use the graphs to solve the following equations.

(a) 2 cos x − 2 = 0

(b) 10 cos x + 8 = 1

14. The figure shows the graphs of y = 0.25x, y = 0.4 and y = 0.75. Use

the graphs to solve the following equations.

(a) 0.25x − 0.4 = 0

(b) 0.25x − 1 − 3 = 0

15. The figure shows the graph of y = f(x), where −4 ≤ x ≤ 5. Add a

suitable straight line on the graph to solve the following equations.

(a) f(x) + 1.8 = 0

(b) −f(x) = −1.6

(c) 10f(x) − 5 = 1

90

−10

5

−5

−1 0

1 2

x

y y = 3x3 − 2x2 − 4

−3

−6

−9

0

2 4 −2 x

y y = x3 − 4x2

y = −x3 − 3x2 + 5 4

2

−2

−2 0

2 4

x

y

y = f(x)

x

y

0

5

4 2 6 −4 −2

−5

−10

16. The figure shows the graph of y = x3 − 4x2. Use the graph to solve the

following equations.

(a) x3 − 4x2 + 10.5 = 0

(b) x3 − 4x2 + 3 = 0

17. The figure shows the graph of y = 3x3 − 2x2 − 4. Use the graph to

solve the following equations.

(a) 3x3 + 5x2 = 7x2 + 9

(b) 3x2(x − 1) = −x2 − 4

18. The figure shows the graph of y = −x3 − 3x2 + 5. Use the graph to

solve the following equations.

(a) x3 + 3x2 − 4 = 2

(b) 2x3 − 3 = 5 − 6x2

19. The figure shows the graph of y = f(x), where −4 ≤ x ≤ 6.

(a) Add a suitable straight line on the graph to solve the

equation f(x) = −3.

(b) If the equation f(x) = k has two distinct real roots, where

k is a positive constant and −4 ≤ x ≤ 6, find the value of k.

91

20. The figure shows the graph of y = sin x − cos x, where 0° ≤ x ≤ 360°.

(a) Add a suitable straight line on the graph to solve the equation sin x − cos x = 0.8.

(b) If the equation sin x − cos x = k has three distinct real roots, where k is a constant and 0° ≤ x ≤ 360°,

find the value of k and solve the equation sin x − cos x = k.

21. The manager of a factory investigates the relationship between the profit ($y) and the number (x) of

machines operated per day. He finds that the relationship can be represented by

y = −5

3x3 + 34x2 − 275x − 1 800. By adding suitable straight lines on the graph of

y = −5

3x3 + 34x2 − 275x − 1 800 (where x ≥ 0) given below, answer the following questions.

(a) Find the number of machines operated per day if the factory

(i) makes a profit of $2 200,

(ii) loses $1 800.

(b) Can the factory make a profit of $5 000? Explain your answer.

x

5 000

4 000

3 000

2 000

1 000

−1 000

−2 000

0 10 20 30 40 50

y = −5

3x3 + 34x2 − 275x − 1 800

y

x

y

1

−1

0 60° 120° 180° 240° 300° 360°

y = sin x − cos x

92

5 cm

x cm

E

D G

F

B

C

10 cm A

22. In the figure, the rectangular cardboard ABCD with length (16 − 4x) cm and width x cm is folded up

and stuck on a square cardboard to form a box without a lid. Let V(x) cm3 be the capacity of the box.

(a) (i) Express V(x) in terms of x.

(ii) Find the range of possible values of x.

(b) The figure below shows the graph of y = V(x) for x ≥ 0.

Using the graph, find the values of x such that the capacity of the box is

(i) 4 cm3,

(ii) 7 cm3.

*23. In the figure, ABCD and DEFG are rectangles. AB = 10 cm,

BC = 5 cm, DG = 2DE and AE = x cm, where 0 < x < 5.

(a) Show that the area of rectangle DEFG is 2(x2 − 10x + 25) cm2.

(b) Show that the area of △CFG is (5x − x2) cm2.

(c) The figure below shows the graph of y = 3x2 − 25x + 50.

Using the graph, find the value of x such that

(i) the areas of rectangle DEFG and △CFG are the same,

(ii) the area of rectangle DEFG is greater than that of △CFG

by 9.5 cm2.

(16 − 4x) cm

x cm x cm A

D

B

C

A, B

C, D

y = V(x)

x

y

0

2

1 2 4

4

3 5

8

6

10

15

10

5

0 1 2 3 4 5 6

y = 3x2 − 25x + 50

x

y

93

Answers


1. (a) −1.0, 3.5 (b) −3, 0, 3

(c) −2.3, 0.0, 1.0 (d) 0°, 180°, 360°

2. (a) −7, −1

(b) no roots

(c) −1.7, −0.3, 2.0

(d) 54°, 126°, 234°, 306°

3. (a) −3.0, 0.0, 0.5 (b) 1.7

4. (a) −3.0 (b) −1.4, −0.8, 4.2

5. (a) 0.0, 2.0 (b) −1.0, 3.0

6. (a) 1.7 (b) −1.6, 0.6, 1.0

7. (a) −153°, −117°, 27°, 63°

(b) −72°, −18°, 108°, 162°

8. (a) 45°, 135°, 225°, 315°

(b) no roots

9. (a) 1.8 (b) 0.8

10. (a) 2.0 (b) 0.6

11. (a) y = 1 (b) y = −4

(c) y = 3

12. (a) y = 2 (b) y = 4

(c) y = −1

13. (a) 0°, 360° (b) 135°, 225°

14. (a) 0.65 (b) 0.20

15. (a) 4.2 (b) −3.0

(c) −2.0

16. (a) −1.4 (b) −0.8, 1.0, 3.8

17. (a) 1.7 (b) −0.9

18. (a) 1.2 (b) −2.0, 1.0

19. (a) 0.0, 5.2 (b) 3.5

20. (a) 78°, 192°

(b) k = −1; 0°, 270°, 360°

21. (a) (i) 22, 42 (ii) 0, 10, 47

(b) no

22. (a) (i) V(x) = x3 − 8x2 + 16x

(ii) 0 < x < 4

(b) (i) 0.3, 2.8 (ii) 0.6, 2.2

23. (c) (i) 3.3 (ii) 2.2

94

F5A: Chapter 3C

Date Task Progress

Lesson Worksheet


(Full Solution)

Book Example 5


(Video Teaching)

Book Example 6


(Video Teaching)

Book Example 7


(Video Teaching)

Book Example 8


(Video Teaching)



(Full Solution)




___________ ( )




___________ ( )



___________ ( )



95


Objective: To solve inequalities f(x) > k, f(x) < k, f(x) ≥ k and f(x) ≤ k by using the graph of y = f(x).

Solving Inequalities by Using Graphs of Functions

(a) The solutions of f(x) > k are the range of the corresponding values of x such that the graph of y = f(x)

is above the straight line y = k.

(b) The solutions of f(x) < k are the range of the corresponding values of x such that the graph of y = f(x)

is below the straight line y = k.

If the inequality sign ‘>’ or ‘<’ is replaced by ‘≥’ or ‘≤’ respectively, then the solutions also include the x-

coordinate(s) of the point(s) of intersection.

In each of the following, solve the inequality by using the given graphs. [Nos. 1–2] �Ex 3C: 3−6

1. 4x2 − 12x + 3 > −2 2. x3 + 2x2 − 5x − 1 ≤ 5

From the graphs, the required solutions are

.


The figure shows the graph of y = x2 − 4x. Use the graph

to solve x2 − 4x ≥ −2.

Draw the straight line y = −2 on the graph of

y = x2 − 4x.

From the graphs of y = x2 − 4x and y = −2, the

required solutions are x ≤ 0.6 or x ≥ 3.4.

The figure shows the graph of y = sin x, where

0° ≤ x ≤ 360°. Use the graph to solve sin x > 0.8.

Draw the straight line y = ( ) on the graph of

y = sin x.

From the graphs of y = sin x and ,

the required solutions are .

y = 4x2 − 12x + 3

y

x 1 2 3 0

–4

4

y = −2

y = x3 + 2x2 − 5x − 1 10

5

0 1 2 −3 −2 −1 x

y

y = 5

4

2

x

−2

−4

y

0 1 2 3 4

y = x2 − 4x

y = −2

1

−1

y

0 90° 180° 270° 360° x

y = sin x

96

Solve the following inequalities by adding suitable straight lines on the given graphs. [Nos. 3–6] 3. 5x − 2x2 > −1

Draw the straight line on the

graph of y = 5x − 2x2.

4. 5x > −0.6 �Ex 3C: 7−11

5. –1 ≤ tan x ≤ 2, where −90° < x < 90°

6. −2 cos x ≤ 1.4, where 0° ≤ x ≤ 360°


7. In the figure, the straight line y = 1 and the graph of y = log2 x intersect

at (2 , 1). John claims that the solutions for log2 x ≤ 1 are x ≤ 2. Do you

agree? Explain your answer.

2

1

x −1

y

0 1

y = 5x

2

−2

y

0 90° 180° 270° 360° x

y = −2 cos x

x

y

(2 , 1)

y = log2 x

y = 1

O

How many straight lines should be drawn?

2

x

−2

0 1 2 3

y = 5x − 2x2

y

2

−2

−45° 0 45°

y y = tan x

x 90° −90°

97

y = x − 5

x

y

O ① ③

②

y = x3 − 2x2 + 1

x

y ① ②

③ ④ O




Level 1

1. f(x) = x − 5 is a linear function. The figure shows the graph of y = f(x). Which

range(s) (①, ② or ③) of values of x give(s) the solutions of

(a) x − 5 ≤ 0?

(b) x − 5 ≥ 0?

2. Consider the function f(x) = x3 − 2x2 + 1. The figure shows the graph of

y = f(x). Which range(s) (①, ②, ③ or ④) of values of x give(s) the solutions of

(a) x3 − 2x2 + 1 > 0?

(b) x3 − 2x2 + 1 < 0?

3. Use the given graphs to solve the inequalities below.

(a) −x2 − 3x + 3 < 5 (b) x2 − 3x − 7 > −10

(c) x2 − 2x ≤ 3 (d) x3 − 9x + 15 ≥ 15

y = −x2 − 3x + 3

y = 5

x

y

0 −4 −2

4

2

6 y = x2 − 3x − 7

y = −10

5

−5

−10

−2 0

2 4

x

y

4

2

0 2 4

y = 3

y = x2 − 2x

x

y

20

10

−2 0

2 4

y = x3 − 9x + 15

y = 15

x

y

98

y = f(x)

0

−10

2 4 6

−5

x

y

5

y = −10

x

y

y = 8

y = f(x)

0 −5

5

5

10

−10

−5


(a) 3 cos 2x > −2.4, where 0° ≤ x ≤ 360° (b) 4 sin 2x ≥ 3.2, where 0° ≤ x ≤ 360°

(c)

x

4

1≤ 16 (d) 9x < 27


(a) f(x) < −10

(b) f(x) ≥ −10


(a) f(x) ≤ 8

(b) f(x) > 8

y = −2.4

y

2

−2

0 90° 180° 270° 360°

y = 3 cos 2x

x

y = 4 sin 2x

y = 3.2

x

y

0

−2

2

4

−4

90° 180° 270° 360°

15

10

5

−2 −10

1

y =

x

4

1

x

y

y = 16

30

20

10

0 0.5 1 1.5

y = 9x

y = 27

y

x

99

y = f(x)

y = −3

x

y

(3 , −3) (−1.4 , −3)

O

y = −x2 + 2x + 15

x

y

10

0 4 2

15

5

−2

x

y y = 2x3 − 4x2 + 4

−1

2

0

−2

2

4

1


(a) f(x) < 10 (b) f(x) ≥ 8

8. The figure shows the graphs of y = f(x) and y = −3. The two graphs

intersect at the points (3 , −3) and (−1.4 , −3). Write down the

solutions of each of the following inequalities.

(a) f(x) < −3

(b) f(x) ≥ −3

9. The figure shows the graph of y = −x2 + 2x + 15. Use the given graph

to solve the inequalities below.

(a) −x2 + 2x + 15 > 16

(b) −x2 + 2x + 15 < 7

10. The figure shows the graph of y = 2x3 − 4x2 + 4. Use the given graph

to solve the inequalities below.

(a) 2x3 − 4x2 + 4 > 4

(b) 2x3 − 4x2 + 4 ≤ 2

x

y

y = 10

y = f(x)

−9 0

1 5 −5

11

y = f(x)

y = 8

−10 0 2 13 21

x

y

100

x

y

y = 0.5x

−1 0

2

4

6

1 −2

y = log3 x

x

y

0

1

1 2 3

−1

−2

y = 2 sin x − 1

0 90°

1

−2

−1

180° 270° 360°

x

y

−3

y = 3 cos x

−2

0

2

4

90° −90° 180° −180°

−4

x

y 11. The figure shows the graph of y = 3 cos x, where −180° ≤ x ≤ 180°.

Solve each of the following inequalities by adding a suitable straight

line on the graph of y = 3 cos x.

(a) 3 cos x ≤ 1

(b) 3 cos x > −1.4

12. The figure shows the graph of y = 2 sin x − 1, where 0° ≤ x ≤ 360°.

Solve each of the following inequalities by adding a suitable straight

line on the graph of y = 2 sin x − 1.

(a) 2 sin x − 1 ≥ 0.4

(b) 2 sin x − 1 < −1.9

13. The figure shows the graph of y = 0.5x. Use the given graph to solve

the inequalities below.

(a) 0.5x < 2

(b) 0.5x ≥ 6

14. The figure shows the graph of y = log3 x. Use the given graph to solve

the inequalities below.

(a) log3 x > −1.5

(b) log3 x ≤ −0.5

101

y = x3 + 8x2 + 5x − 17

x

y

0

10

2

20

30

−2 −4 −6 −10

O x

y

y = −2

y = f(x)

P Q R S

y = x3 − x2 − 4x − 7

x

y

0

10

−2

−10

2 4

Level 2

15. The figure shows the graphs of y = f(x) and y = −2. The coordinates of the

4 points of intersection P, Q, R and S are (−13 , −2), (−9 , −2), (−3 ,

−2) and (2 , −2) respectively. Write down the solutions of each of the

following inequalities.

(a) f(x) + 2 ≤ 0

(b) f(x) + 2 > 0

16. The figure shows the graph of y = −5x2 + 8x + 18. By adding a suitable straight line on the given graph,

solve −5x2 + 8x + 36 ≤ 0.

17. The figure shows the graph of y = x3 + 8x2 + 5x − 17. By

adding a suitable straight line on the given graph, solve

x3 + 8x2 + 5x > 33.

18. The figure shows the graph of y = x3 − x2 − 4x − 7. By adding

suitable straight lines on the given graph, solve the following

inequalities.

(a) x3 − x2 − 4x ≥ 3

(b) x3 − x2 < 4(x − 1)

y = −5x2 + 8x + 18

−20

0 1 2 3 4 −2 −1

−10

10

20

x

y

102

y = −x3 + 7x2 − 6x − 11

x

y

0 2

10

−5

−10

4

5

6 8

y = −5

y = 2x2 − 4x − 11

x

y

O

B A

y = −2x3 + 11x2 − 13x − 11

x

y

0

−5

1 2 3 4

5

−10

−15

x

y

y = x3 − x2 − 6x + 4

y = 4

O

R Q P

19. The figure shows the graph of y = −x3 + 7x2 − 6x − 11. By

adding suitable straight lines on the given graph, solve the


(a) x3 − 7x2 + 6x + 6 > 0

(b) 2x3 − 14x2 + 12x ≤ −29

20. The figure shows the graph of y = −2x3 + 11x2 − 13x − 11. By

adding suitable straight lines on the given graph, solve the


(a) 11x2 − 13x > 2x3 + 5

(b) −4x3 − 31 ≤ 2x(13 − 11x)

21. In the figure, the graphs of y = 2x2 − 4x − 11 and y = −5 intersect at two points A and B.

(a) Find the coordinates of A and B.

(b) Hence, solve the following inequalities.

(i) 2x2 − 4x − 11 > −5

(ii) x2 − 2x ≤ 3

22. In the figure, the graphs of y = x3 − x2 − 6x + 4 and y = 4 intersect at three points P, Q

and R.

(a) Find the coordinates of P, Q and R.

(b) Hence, solve 2(x3 − 6x + 2) + 1 ≥ 2x2 + 5.

103

y = f(x)

10

0 5 15 10 20

5

15

x

y

x

y

y = x3 − 15x2 − 250x − 12 000

30 20 10

10 000

40 0

20 000

30 000

−10 000

23. Let f(x) = 2x3 + 8x2 − 2x + k, where k is a constant. The graph of y = f(x) passes through (2 , 33).


(b) The figure below shows the graph of y = f(x).

Use the given graph to solve the following inequalities.

(i) x3 + 4x2 − x ≥ 4

(ii) 4x3 + 16x2 − 4x < 31

24. The total amount y million dollars in an investment plan after

x years is given by y = f(x). The figure shows the graph of

y = f(x). If the total amount in the investment plan is more

than

13 million dollars, use the given graph to find the range of

values of x.

25. Let x be the number of workers in a factory and $y be the

corresponding daily profit. The relationship between x and y

can be represented by y = x3 − 15x2 − 250x − 12 000. The

figure shows the graph of the function.

(a) If the daily profit of the factory is positive, use the given

graph to find the minimum number of workers in the

factory.

(b) Can the factory lose more than $18 000 per day?


10

5

−5

−10

−4 −3 −2 −1 0

1

x

y

y = f(x)

104

7 cm

6x cm

6 cm

18 cm

Fig. II

Fig. I

3x cm

+ 6

7

1x cm

3x cm

Fig. III

100

80

60

40

20

0 1 2 3

y = 4x3 − x

x

y

*26. In Fig. I, BCDF is a rectangular card with perimeter 30 cm. It is divided into a triangle and a

trapezium, where BF = 2x cm, EF = x cm and 0 < x < 5. Let A cm2 be the area of the trapezium BCDE.

(a) Express A in terms of x.

(b) Fig. II shows the graph of y = −x2 + 6x + 2. By adding suitable straight lines on the graph, find the

range(s) of values of x such that the area of the trapezium BCDE is

(i) greater than 40 cm2, (ii) less than or equal to 35 cm2.

*27. Fig. I shows a cup in the shape of an inverted right circular cone of base radius 6 cm and height 18 cm,

and a cylindrical can of base radius 7 cm. The depth of water in the cup is 6x cm (where 0 < x < 3) and

the depth of water in the can is

+ 6

7

1x cm.

(a) Express, in terms of x and π,

(i) the volume of water in the cup, (ii) the volume of water in the can.

(b) The water in both the cup and the can shown in Fig. I will be poured into two empty bowls which

are in the shape of a hemisphere of radius 3x cm (see Fig. II). It is known that the two bowls will

not be both fully filled.

(i) Show that 4x3 − x − 42 > 0.

(ii) Fig. III shows the graph of y = 4x3 − x for x ≥ 0. By adding a suitable straight line on the graph,

find the range of values of x such that the two bowls will not be both fully filled.

Fig. I x cm

2x cm A cm2

B C

D E

F

Fig. II

y = −x2 + 6x + 2

x

y

10

0 1 2 3 4 5 6

5

105

Answers


1. (a) ①, ② (b) ③

2. (a) ②, ④ (b) ①, ③

3. (a) x < −2 or x > −1


(c) −1 ≤ x ≤ 3

(d) −3 ≤ x ≤ 0 or x ≥ 3

4. (a) 0° ≤ x < 72° or 108° < x < 252° or

288° < x ≤ 360°

(b) 27° ≤ x ≤ 63° or 207° ≤ x ≤ 243°

(c) x ≥ −2.0

(d) x < 1.50

5. (a) x < −1 or 3 < x < 5

(b) −1 ≤ x ≤ 3 or x ≥ 5

6. (a) −11.0 ≤ x ≤ −3.5 or x ≥ 3.0

(b) x < −11.0 or −3.5 < x < 3.0

7. (a) x < −5 or x > 1

(b) x ≤ −10 or 2 ≤ x ≤ 13

8. (a) −1.4 < x < 3

(b) x ≤ −1.4 or x ≥ 3

9. (a) no solutions

(b) x < −2.0 or x > 4.0

10. (a) x > 2.0

(b) x ≤ −0.6 or 1.0 ≤ x ≤ 1.6

11. (a) −180° ≤ x ≤ −72° or 72° ≤ x ≤ 180°

(b) −117° < x < 117°

12. (a) 45° ≤ x ≤ 135° (b) 207° < x < 333°

13. (a) x > −1.0 (b) x ≤ −2.6

14. (a) x > 0.2 (b) 0 < x ≤ 0.6

15. (a) −13 ≤ x ≤ −9 or −3 ≤ x ≤ 2

(b) x < −13 or −9 < x < −3 or x > 2

16. x ≤ −2.0 or x ≥ 3.6

17. −6.4 < x < −3.2 or x > 1.6

18. (a) x ≥ 2.8

(b) x < −2.0 or 1.0 < x < 2.0

19. (a) −0.6 < x < 1.8 or x > 5.8

(b) x ≤ −1.0 or 2.6 ≤ x ≤ 5.4

20. (a) x < −0.3 or 2.5 < x < 3.3

(b) x ≥ −0.7

21. (a) A(−1 , −5), B(3 , −5)

(b) (i) x < −1 or x > 3

(ii) −1 ≤ x ≤ 3

22. (a) P(−2 , 4), Q(0 , 4), R(3 , 4)

(b) −2 ≤ x ≤ 0 or x ≥ 3

23. (a) −11

(b) (i) −4.0 ≤ x ≤ −1.0 or x ≥ 1.0

(ii) x < −3.7 or −1.6 < x < 1.3

24. x > 18

25. (a) 34 (b) no

26. (a) A = −5x2 + 30x

(b) (i) 2.0 < x < 4.0

(ii) 0 < x ≤ 1.6 or 4.4 ≤ x < 5

27. (a) (i) 8πx3 cm3

(ii) 7π(x + 42) cm3

(b) (ii) 2.2 < x < 3

106

F5A: Chapter 3D

Date Task Progress

Lesson Worksheet


(Full Solution)

Book Example 9


(Video Teaching)

Book Example 10


(Video Teaching)

Book Example 11


(Video Teaching)

Book Example 12


(Video Teaching)

Book Example 13


(Video Teaching)

Book Example 14


(Video Teaching)

Book Example 15


(Video Teaching)

107

Book Example 16


(Video Teaching)



(Full Solution)




___________ ( )




___________ ( )

Maths Corner Exercise 3D Multiple Choice


___________ ( )



108

5A Lesson Worksheet 3.4A(I) (Refer to Book 5A P.3.39)

Objective: To recognize the translation of functions (upward or downward).

Translating Upward or Downward: f(x) → f(x) ±±±± k

Transformation on f(x) Algebraic expression Example

Translate upward by k units

(k > 0) f(x) + k

Translate downward by k units

(k > 0) f(x) − k

In each of the following, the graph of the function y = f(x) undergoes a translation to become the graph of

the function y = g(x). State the direction and distance of the translation, and express g(x) in terms of f(x).

[Nos. 1–4] �Ex 3D: 4, 5

1. 2.

Direction: Direction:

Distance: Distance:

g(x) = f(x) + g(x) =

3. 4.


Distance: Distance:

g(x) = g(x) =

y = f(x)

y

x O

y = f(x) + k

y = f(x) − k

k units

k units

y = f(x)

y

x O

y = g(x) 3 units

(1 , −1)

(1 , 2)

y = g(x)

y

x O

y = f(x)

(2 , 0)

(2 , 2)

y = f(x)

y

x O

y = g(x) (–90° , −3)

(–90° , 3)

y = f(x)

y = g(x) 3

2

1

x −2 −1

y

0 1

Point (2 , 2) is translated ( upward / downward ) by units to point (2 , 0).

109


The graph of y = f(x) = x2 + 2 becomes the graph of y =

g(x) after each of the following transformations. Find the

function g(x).

(a) Translate upward by 4 units.

(b) Translate downward by 3 units.

(a) g(x) = f(x) + 4

= x2 + 2 + 4

= x2 + 6

(b) g(x) = f(x) – 3

= x2 + 2 – 3

= x2 – 1

The graph of y = f(x) = x2 + 3x + 1 becomes the graph of

y = h(x) after each of the following transformations. Find

the function h(x).

(a) Translate downward by 5 units.

(b) Translate upward by 2 units.

(a) h(x) = f(x) ( + / − )

=

=

(b) h(x) = f(x) ( + / − )

=

=

5. The graph of y = f(x) = x − 2 is translated

upward by 3 units to become the graph of

y = g(x). Find the function g(x).

g(x) =

6. The graph of y = f(x) = x3 − 2x2 + x + 3 is

translated downward by 8 units to become the

graph of y = g(x). Find the function g(x).

�Ex 3D: 7

7. The graph of y = f(x) = 2 cos x – 8 becomes the

graph of y = g(x) after each of the following

transformations. Find the function g(x).

(a) Translate upward by 7 units.

(b) Translate downward by 2 units.

8. The graph of y = f(x) = 2x + 5 becomes the



(a) Translate downward by 1 unit.

(b) Translate upward by 6 units.


9. In the figure, the coordinates of the vertex of the graph of the quadratic

function y = f(x) are (0 , 1). The graph of y = f(x) is translated upward

by 3 units to become the graph of y = h(x).

(a) Find the coordinates of the vertex of the graph of y = h(x).

(b) Sketch the graph of y = h(x) on the given figure.

y = f(x)

y = g(x) y

x O

(a)

y = f(x)

y = g(x) y

x O

y = f(x)

y = h(x)

y

x O

(a)

0

2

−2

4

y

x 2

6

y = f(x)

110

5A Lesson Worksheet 3.4A(II) (Refer to Book 5A P.3.41)

Objective: To recognize the translation of functions (to the left or to the right).

Translating to the Left or Right: f(x) → f(x ±±±± k)


Translate to the left by k units

(k > 0) f(x + k)

Translate to the right by k units

(k > 0) f(x − k)


the function y = g(x). State the direction and distance of the translation, and express g(x) in terms of f(x).

[Nos. 1–4] �Ex 3D: 6

1. 2.


Distance: Distance:

g(x) = f(x + ) g(x) =

3. 4.


Distance: Distance:

g(x) = g(x) =

y = f(x) y

x

y = f(x + k) y = f(x − k)

k units

k units

O

y = f(x)

y

x O

y = g(x)

2 units (0 , 1) (2 , 1)

y

x O

(−1 , 1) (4 , 1)

y = g(x)

y = f(x)

y = g(x) y

x O

y = f(x)

(0 , 2) (−3 , 2)

2

1

x

−1−1 0 1 2

y

y = f(x)

y = g(x)

3

Point (−1 , 1) is translated to the ( left / right ) by units to point (4 , 1).

111


The graph of y = f(x) = x + 2 becomes the graph of

y = g(x) after each of the following transformations. Find

the function g(x).

(a) Translate to the left by 3 units.

(b) Translate to the right by 6 units.

(a) g(x) = f(x + 3)

= x + 3 + 2

= x + 5

(b) g(x) = f(x − 6)

= x − 6 + 2

= x − 4

The graph of y = f(x) = x − 3 becomes the graph of

y = h(x) after each of the following transformations.

Find the function h(x).

(a) Translate to the right by 4 units.

(b) Translate to the left by 8 units.

(a) h(x) = f(x ( + / − ) )

=

=

(b) h(x) = f(x ( + / − ) )

=

=

5. The graph of y = f(x) = 2x − 1 is translated to

the right by 3 units to become the graph of

y = g(x). Find the function g(x).

g(x) =

6. The graph of y = f(x) = x2 + 5 is translated to

the left by 4 units to become the graph of

y = g(x). Find the function g(x). �Ex 3D: 8, 9

7. The graph of y = f(x) = 3x + 2 becomes the



(a) Translate to the right by 2 units.

(b) Translate to the left by 1 unit.

8. The graph of y = f(x) = 1 − sin x becomes the



(a) Translate to the left by 80°.

(b) Translate to the right by 40°.


9. The graph of y = f(x) = x2 − 2x − 1 is translated to the right by 3 units to become the graph of y = g(x).

Then the graph of y = g(x) is translated upward by 4 units to become the graph of y = h(x). Find h(x).

y = h(x)

y

x O

y = f(x)

(a)

(a + b)2 = a2 + 2ab + b2

y = f(x)

y

x O

y = g(x)

(a)

113


Objective: To recognize the reflection of functions in the x-axis and the y-axis.

Reflection in the x-axis and the y-axis: f(x) → −−−−f(x) and f(x) → f(−−−−x)


Reflect in the x-axis −f(x)

Reflect in the y-axis f(−x)

In each of the following, the graph of the function y = f(x) undergoes a reflection to become the graph of

the function y = g(x). State the axis of reflection and express g(x) in terms of f(x). [Nos. 1–4] �Ex 3D: 10–13

1. 2.

Axis of reflection: Axis of reflection:

g(x) = g(x) =

3. 4.

Axis of reflection: Axis of reflection:

g(x) = g(x) =

y = −f(x)

y

x O

y = f(x)

y = f(−x) y

x O

y = f(x)

y = f(x)

y

x O

y = g(x)

(2 , −2)

(2 , 2)

y = f(x)

y

x O

y = g(x)

(3 , 1) (−3 , 1)

y = f(x) y

x O

y = g(x)

(a , b) (−a , b)

y = f(x)

y = g(x) 2

1

−1

−2

y

0 1 2 3 x

114


The graph of y = f(x) = 2x − 1 becomes the graph of y =

g(x) after each of the following transformations. Find the

function g(x).

(a) Reflect in the x-axis.

(b) Reflect in the y-axis.

(a) g(x) = −f(x)

= −(2x − 1)

= −2x + 1

(b) g(x) = f(−x)

= 2(−x) − 1

= −2x – 1

The graph of y = f(x) = 2 − 3x becomes the graph of y =

h(x) after each of the following transformations. Find the

function h(x).

(a) Reflect in the y-axis.

(b) Reflect in the x-axis.

(a) h(x) =

=

=

(b) h(x) =

=

=

5. The graph of y = f(x) = x2 + 3x − 6 is reflected

in the x-axis to become the graph of y = h(x).


h(x) =

6. The graph of y = f(x) = −2x2 − 4x + 1 is reflected

in the y-axis to become the graph of y = h(x).


7. The graph of y = f(x) = tan x + 4 becomes the




(b) Reflect in the y-axis.

8. The graph of y = f(x) = 1 – 2x becomes the




(b) Reflect in the x-axis.


9. The graph of y = f(x) = x2 − 4x + 5 is translated to the left by 4 units to become the graph of y = g(x).

Peter claims that the graph of y = g(x) can also be obtained by reflecting the graph of y = f(x) in the y-

axis. Do you agree? Explain your answer.

y = f(x)

y

x O

y = g(x)

(a) y

x O

y = h(x) y = f(x)

(a)

115

5A Lesson Worksheet 3.4C(I) (Refer to Book 5A P.3.45)

Objective: To recognize the enlargement or reduction of functions along the y-axis.

Enlargement or Reduction along the y-axis: f(x) → kf(x)

In each of the following [Nos. 1–4], the graph of the function y = f(x) is enlarged to k times the original or

reduced to k of the original along the y-axis to become the graph of the function y = g(x).

(a) State whether the transformation is an enlargement or a reduction.

(b) Write down the value of k.

(c) Express g(x) in terms of f(x).

1. 2.

(a) (a)

(b) k = (b) k =

(c) g(x) = (c) g(x) =

3. 4.

(a) (a)

(b) k = (b) k =

(c) g(x) = (c) g(x) =


Enlarge to k times (k > 1) the

original along the y-axis kf(x)

Reduce to k (0 < k < 1) of the

original along the y-axis kf(x)

y = f(x)

x O

y = g(x)

(1 , 9)

(1 , 3)

y

y = f(x)

O

y = g(x)

(−1 , 2)

x

y

(−1 , 4)

O x

y = g(x)

y = f(x)

(a , b)

(a , 3b)

y y

x

2

0

y = f(x)

360° 180°

y = g(x) −2

y = kf(x) (k > 1)

x

y = f(x)

y

O

y = kf(x) (0 < k < 1)

(i) The x-intercepts of the two graphs are

( the same / different ).

(ii) The y-coordinate of point (1 , 9) is ____ times that of point (1 , 3).

116


The graph of y = f(x) = x2 − 2 becomes the graph of

y = g(x) after each of the following transformations.

Find the function g(x).

(a) Enlarge to 3 times the original along the y-axis.

(b) Reduce to

2

1 of the original along the y-axis.

(a) g(x) = 3f(x)

= 3(x2 − 2)

= 3x2 − 6

(b) g(x) =

2

1f(x)

=

2

1(x2 − 2)

= 12

2

−x

The graph of y = f(x) = 4x2 + 8x + 4 becomes the graph of

y = h(x) after each of the following transformations. Find

the function h(x).

(a) Reduce to

4

1 of the original along the y-axis.

(b) Enlarge to 1.5 times the original along the y-axis.

(a) h(x) =

)(

)(

f(x)

=

)(

)(

( )

=

(b) h(x) = ( )f(x)

= ( )( )

=

5. The graph of y = f(x) = 3 sin x + 2 is enlarged

to 2 times the original along the y-axis to

become the graph of y = g(x). Find the function

g(x).

g(x) =

6. The graph of y = f(x) = 6 − 5x is reduced to

5

1 of

the original along the y-axis to become the


7. The graph of y = f(x) = log x is enlarged to 6

times the original along the y-axis to become

the graph of y = g(x). Find the function g(x).

8. The graph of y = f(x) = 2x is reduced to

2

1 of

the original along the y-axis to become the



9. The figure shows the graphs of y = f(x) and y = g(x). Sandy claims that the

graph of y = g(x) can be obtained by enlarging the graph of y = f(x) to

2 times the original along the y-axis. Do you agree? Explain your answer.

y

x

(1 , 1)

(1 , 2) y = f(x)

y = g(x)

O

am ÷ an = am − n

O

y = g(x)

y

y = f(x)

x

(a)

y = f(x)

y = h(x)

O x

y (a)

117

5A Lesson Worksheet 3.4C(II) (Refer to Book 5A P.3.47)

Objective: To recognize the enlargement or reduction of functions along the x-axis.

Enlargement or Reduction along the x-axis: f(x) → f(kx)

In each of the following [Nos. 1–4], the graph of the function y = f(x) is enlarged to k times the original or

reduced to k of the original along the x-axis to become the graph of the function y = g(x).

(a) State whether the transformation is an enlargement or a reduction.

(b) Write down the value of k.

(c) Express g(x) in terms of f(x).

1. 2.

(a) (a)

(b) k = (b) k =

(c) g(x) = (c) g(x) =

3. 4.

(a) (a)

(b) k = (b) k =

(c) g(x) = (c) g(x) =


Reduce to

k

1 (k > 1) of the

original along the x-axis

f(kx)

Enlarge to

k

1 times (0 < k < 1)

the original along the x-axis

f(kx)

y = f(x)

y

x O

y = g(x)

(6 , 3) (2 , 3)

(300° , 2)

y = g(x)

y

x O

y = f(x)

(150° , 2)

y

x −1 −4

y = f(x)

y = g(x)

0

4

3

2

1

x −2 −1

y

0 1 2

y = g(x)

y = f(x)

(i) The y-intercepts of the two graphs are

( the same / different ).

(ii) The x-coordinate of point (6 , 3) is ____ times that of point (2 , 3).

y = f(kx) (0 < k < 1)

x

y = f(x)

y

O y = f(kx) (k > 1)

118


The graph of y = f(x) = 9x + 5 becomes the graph of

y = g(x) after each of the following transformations. Find

the function g(x).

(a) Enlarge to 3 times the original along the x-axis.

(b) Reduce to

5

1 of the original along the x-axis.

(a) g(x) =

3

xf

= 53

9 +

x

= 3x + 5

(b) g(x) = f(5x)

= 9(5x) + 5

= 45x + 5

The graph of y = f(x) = 4 − x2 becomes the graph of y =

h(x) after each of the following transformations. Find the

function h(x).

(a) Reduce to

2


(b) Enlarge to 6 times the original along the x-axis.

(a) h(x) = f( x)

= 4 − ( )2

=

(b) h(x) = f

= 4 −

2

=

5. The graph of y = f(x) = log4 x is reduced to

3

1

of the original along the x-axis to become the


g(x) =

6. The graph of y = f(x) = cos (2x + 10°) is enlarged

to 2 times the original along the x-axis to become


7. The graph of y = f(x) = 2x2 + x − 8 is reduced to

4

1 of the original along the x-axis to become


8. The graph of y = f(x) = 1 − 32x is enlarged to

8 times the original along the x-axis to become



9. The graph of y = f(x) = x2 is reduced to

2

1 of the original along the x-axis to become the graph of

y = g(x). Can the graph of y = g(x) be obtained after the graph of y = f(x) is enlarged along the y-axis?


y = f(x) y

O

y = g(x)

(a)

x

y

x

y = f(x)

y = h(x)

O

(a)

119




Level 1

1. In each of the following, the graph of the function y = f(x) undergoes a transformation to become the

graph of the function y = g(x). Express g(x) in terms of f(x).

(a) The graph of y = f(x) is translated to the left by 4 units to become the graph of y = g(x).

(b) The graph of y = f(x) is translated upward by 2 units to become the graph of y = g(x).

(c) The graph of y = f(x) is reflected in the x-axis to become the graph of y = g(x).

(d) The graph of y = f(x) is reflected in the y-axis to become the graph of y = g(x).

(e) The graph of y = f(x) is reduced to

5

1 of the original along the x-axis to become the graph of

y = g(x).

(f) The graph of y = f(x) is enlarged to 4 times the original along the y-axis to become the graph of

y = g(x).

2. The graph of the function y = f(x) is translated downward by 3 units to become the graph of y = g(x).

(a) Express g(x) in terms of f(x).

(b) Complete the following table.

f(x) 4 0

g(x) 3 −7

3. The graph of the function y = f(x) is reflected in the x-axis to become the graph of y = g(x).



f(x) 6 −1

g(x) −4 5

4. The graph of the function y = f(x) is reduced to

3

1 of the original along the y-axis to become the graph

of y = g(x).



f(x) 9 6

g(x) −4 −6

120

5. Let f(x) = sin x. The graph of the function y = f(x) becomes the graph of the function y = g(x) after

each of the transformations below. Find g(x).

(a) Translate to the left by 30°.

(b) Reduce to

4


6. Let f(x) = log 5x. The graph of the function y = f(x) becomes the graph of the function y = g(x) after

each of the transformations below. Find g(x).


(b) Translate downward by 1 unit.

7. Let f(x) = 7x. The graph of the function y = f(x) becomes the graph of the function y = g(x) after each

of the transformations below. Find g(x).


(b) Enlarge to 2 times the original along the x-axis.


the function y = g(x). State the direction and the distance of the translation, and express g(x) in terms of f(x).

[Nos. 8–11]

8.

9.

10.

11.

y = f(x) y = g(x)

(4 , −2)

(2 , −2)

x

y

O y = f(x)

y = g(x)

(−1 , 2)

(−1 , 5)

x

y

O

y = f(x) y = g(x)

(0 , 3) (4 , 3)

x

y

O

(−4 , −3)

(−4 , 6) y = g(x)

x

y y = f(x)

O

121


the function y = g(x). Find the function g(x). [Nos. 12–15]

12.

13.

14.

15.


the function y = g(x). State the axis of reflection and express g(x) in terms of f(x). [Nos. 16–19]

16.

17.

18.

19.

y = f(x) = −x2 + 4x + 9

(4 , 2)

(4 , 9)

y = g(x)

x

y

O

y = f(x) = −x + 1

(−2 , 3)

(7 , 3)

y = g(x)

x O

y

O

y = g(x)

x

y

(2 , −5)

(2 , 1)

y = f(x) = x3 − 5x − 3

O

y = f(x) = 2x − 12

(4 , −4) (−0.5 , −4)

y = g(x)

x

y

y = f(x)

x

y

(4 , 3)

y = g(x)

(−4 , 3)

O

(2 , 8)

(2 , −8) y = g(x)

y = f(x)

x

y

O

y = f(x) x

y

(4 , −7)

O y = g(x)

(−4 , −7)

x O

y = g(x)

(a , −b)

(a , b)

y = f(x)

y

122

−8

1 2 4

x

y

0

y = f(x) = −x3 y = h(x)

y = g(x)

y y = f(x)

y = g(x)

(4 , 3)

(8 , 3)

x O

y = f(x) = x3 − 4x2 − 5x − 1

(4 , −21)

y = g(x) O x

y

(4 , −7)


the function y = g(x). Find the function g(x). [Nos. 20–21]

20.

21.

22. In the figure, the graphs of y = g(x) and y = h(x) are obtained by enlarging and

reducing the graph of y = f(x) = −x3 along the x-axis respectively. Find g(x) and h(x).

23. The figure shows the graph of the function y = g(x) after enlarging or

reducing the graph of the function y = f(x).

(a) State whether it is an enlargement or a reduction along the x-axis or y-

axis.

(b) Express g(x) in terms of f(x).

24. In the figure, the graph of the function y = f(x) = x3 − 4x2 − 5x − 1 is

enlarged or reduced to become the graph of y = g(x).

(a) Describe the transformation.

(b) Find g(x).

O x

y

(6 , 6)

y = g(x)

(−6 , 6)

y = f(x) = −2x − 6 (−4 , −13)

(−4 , 13)

x

y

O

y = f(x) = x2 − 3

y = g(x)

123

y = g(x)

x

y

(−1 , 4)

(−1 , 1) O

y = f(x) = x3 + kx2 − 2x − 1 25. In the figure, the graph of the function y = f(x) = x3 + kx2 − 2x − 1 is enlarged or

reduced to become the graph of y = g(x), where k is a constant.


(b) Find g(x).

Level 2

26. According to the information given below, the graph of the function y = f(x) undergoes a

transformation to become the graph of y = g(x). Does the transformation involve a translation, a

reflection or a reduction?

(a) f(x) = 9x + 6, g(x) = 3x + 2 (b) f(x) = 2x2 + 3x + 1, g(x) = 2x2 − 3x + 1

(c) f(x) = 4x, g(x) = 16x (d) f(x) = log1+x

x, g(x) = log

1−x

x

(e) f(x) = cos x, g(x) = cos (50° + x)

In each of the following, the graph of the function y = f(x) is first transformed to become the graph of y =

g(x) and then to the graph of y = h(x) (indicated by the dotted lines and arrows in the figure). Describe the

two transformations, and express g(x) and h(x) in terms of f(x). [Nos. 27–32]

27.

28.

29.

30.

y = f(x)

y = g(x)

y = h(x)

x

y

② ① (0 , −3)

(−2 , −3)

(−2 , 6)

O y = g(x) y = h(x)

①

(−5 , 0)

(−5 , 15) (10 , 15) ②

O x

y

y = f(x)

(−4 , −2) ②

y = g(x)

y = h(x)

①

y = f(x)

(6 , −2)

(6 , −10)

x O

y

y = f(x)

y = g(x) y = h(x)

(2 , 2)

(2 , −2) (−2 , −2)

x

y ① ②

O

124

(1 , 8) (2 , 8)

O

y = f(x) = 2x

y =h(x) y = g(x)

(2 , 4)

x

y

31.

32.

33. In the figure, the graph of y = g(x) is obtained by enlarging or reducing the graph of y =

f(x) = 2x. Then, the graph of y = g(x) is enlarged or reduced to become the

graph of y = h(x). Describe the two transformations and write down the

function h(x).

In each of the following, write down the function represented by the final graph. [Nos. 34–37]

34. The graph of y = sin x is first translated downward by 4 units and then translated to the left by 70°.

35. The graph of y = 5x is first translated to the right by 6 units and then translated upward by 3 units.

36. The graph of y = x2 − 7 is first enlarged to 3 times the original along the y-axis and then translated to

the left by 8 units.

37. The graph of y = 1 − (x + 1)2 is first reduced to

2

1 of the original along the x-axis, then translated

upward by 5 units and finally reflected in the y-axis.

38. The graph of y = f(x) = x2 + 6x + 4 is translated to the right by 6 units to become the graph of y =

g(x). Jim claims that the graph of y = g(x) can also be obtained by reflecting the graph of y = f(x) in the

y-axis. Do you agree? Explain your answer.

39. The graph of y = f(x) = log3 x is first reduced to

2

1 of the original along the x-axis and then translated

upward by 1 unit to become the graph of y = g(x).

(a) Find g(x).

(b) Kyle claims that the graph of y = g(x) can be obtained by transforming the graph of y = f(x)

only once. Do you agree? Explain your answer.

(4 , 8)

O

(4 , −8) (18 , −8)

① ②

y = f(x)

y = g(x)

y = h(x)

x

y

y = h(x)

y = g(x)

①

② (a , 7 − b) (−a , 7 − b)

(−a , −b)

O x

y = f(x)

y

125

x

y

20

15

10

5

−5

−10

−6 −4 −2 0

2 4 6

y = x2

In each of the following, write down the transformations involved in the graph of y = f(x). [Nos. 40–42]

40. A function is transformed from y = f(x) to y = f(x − 2) and then to y = f(x − 2) + 1.

41. A function is transformed from y = f(x) to y = f(x + 4) and then to y = f(4 − x).

42. A function is transformed from y = f(x) to y = f(3x) and then to y = 5f(3x).

Each of the figures below shows the graph of a function y = f(x). Sketch on the same figure the graph of

y = h(x) after two transformations of the graph of y = f(x) and mark on the figure the new position and

coordinates of the given point. [Nos. 43–44]

43. Reduce to

3

1 of the original along the y-axis 44. Reflect in the x-axis first. Then translate

first. Then translate to the right by 3 units. upward by 2 units.

45. Consider the function f(x) = 2x2 + 12x − 9.

(a) Rewrite f(x) into the form of a(x − h)2 + k (where a, h and

k are constants) by using the method of completing the

square.

(b) The figure shows the graph of y = x2. By considering the

transformations of the graph of y = x2, sketch the graph of

y = f(x).

(1 , 18)

y = f(x)

25

20

15

10

5

−5 −2 −1

0 1 2 3 4

x

y

x

y

6

4

2

−2

−4

−6

−3 −2 −1 0

1 2 3

y = f(x)

(1 , −2)

126

46. The figure shows the graph of y = a cos (x + b°), where a is a constant and −90 < b < 90. It is given

that the graph of y = cos x undergoes two transformations to become the graph of y = a cos (x + b°).

(a) Describe the two transformations.

(b) Write down the values of a and b.

* 47. In the figure, the graph of y = f(x) = a(x − 2)(x + b) cuts the x-axis at two distinct points, where a, b are

integers and −8 < b < a < 8. The graph of y = f(x) is first reflected in the x-axis, then translated to the

left by 5 units and finally translated upward by 1 unit to become the graph of y = k(x).

(a) Express k(x) in terms of a, b and x.

(b) If the graph of y = k(x) passes through (−5 , −15), how many possible sets of values of a and

b are there? Explain your answer.

* 48. Let f(x) = r(x − a)(x − b)(x − c), where a, b, c and r are real numbers, r ≠ 0 and a < b < c. The figure

shows the graph of y = f(x).

(a) Write down the values of a, b and c.

(b) The graph of y = f(x) is translated horizontally to become the graph of y = g(x). Explain how

the transformation can be performed such that the greatest root of the equation g(x) = 0 is 0.

(c) The graph of y = f(x) is translated downward by k units to become the graph of y = h(x), where

k > 0. If h(1) = −30 and h(−2) = 0, find h(x).

x

y y = a cos (x + b°)

0 110°

−4

x

y

y = f(x) = a(x − 2)(x + b)

O

y = f(x)

0 x

y

1 3 −4

127

* 49. In the figure, the graph of y = f(x) = −x2 + 6x − 5 cuts the x-axis at the points A and B.

(a) Find the length of AB.

(b) The graph of y = f(x) is translated upward by k units to become the graph of y = g(x), where k > 0.

The graph of y = g(x) cuts the x-axis at the points P and Q.

(i) Express g(x) in terms of k and x.

(ii) Express the length of PQ in terms of k.

(iii) Suppose AB + PQ = 12.

(1) Find the value of k.

(2) The graph of y = g(x) is enlarged to s times the original along the y-axis to become

the graph of y = h(x), where s > 1. The graph of y = h(x) cuts the x-axis at the points P′

and Q′. Describe how the length of P′Q′ changes when s increases. Explain your answer.

A

x

y

y = g(x)

y = f(x)

O B Q P

128

Answers


1. (a) g(x) = f(x + 4)

(b) g(x) = f(x) + 2

(c) g(x) = −f(x)

(d) g(x) = f(−x)

(e) g(x) = f(5x)

(f) g(x) = 4f(x)

2. (a) g(x) = f(x) − 3

(b) f(x) 6 4 0 −4

g(x) 3 1 −3 −7

3. (a) g(x) = −f(x)

(b) f(x) 6 4 −1 −5

g(x) −6 −4 1 5

4. (a) g(x) =3

1f(x)

(b) f(x) 9 6 −12 −18

g(x) 3 2 −4 −6

5. (a) g(x) = sin (x + 30°)

(b) g(x) = sin 4x

6. (a) g(x) = log (−5x)

(b) g(x) = log 5x − 1

7. (a) g(x) = −7x

(b) g(x) = 27

x

8. translated to the left by 2 units; g(x) = f(x + 2)

9. translated upward by 3 units; g(x) = f(x) + 3

10. translated to the right by 4 units; g(x) = f(x −

4)

11. translated downward by 9 units; g(x) = f(x) −

9

12. g(x) = −x2 + 4x + 2

13. g(x) = −x + 10

14. g(x) = x3 − 5x + 3

15. g(x) = 2x − 3

16. y-axis; g(x) = f(−x)

17. x-axis; g(x) = −f(x)

18. y-axis; g(x) = f(−x)

19. x-axis; g(x) = −f(x)

20. g(x) = 2x − 6

21. g(x) = −x2 + 3

22. g(x) = 3

8

1x− , h(x) = −8x3

23. (a) an enlargement along the x-axis

(b) g(x) =

xf

2

1

24. (a) reduced to

3

1 of the original along

the y-axis

(b) g(x) = 3

3

1x −

2

3

4x − x

3

5−

3

1

25. (a) 1

(b) g(x) = 4x3 + 4x2 − 8x − 4

26. (a) a reduction along the y-axis

(b) a reflection along the y-axis

(c) a reduction along the x-axis

(d) a reflection along the y-axis

(e) a translation along the x-axis

27. first translated to the left by 2 units,

then translated upward by 9 units;

g(x) = f(x + 2), h(x) = f(x + 2) + 9

28. first translated upward by 15 units,

then translated to the right by 15 units;

g(x) = f(x) + 15, h(x) = f(x − 15) + 15

29. first translated to the right by 10 units,

then translated downward by 8 units;

g(x) = f(x − 10), h(x) = f(x − 10) − 8

30. first reflected in the x-axis,

then reflected in the y-axis;

g(x) = −f(x), h(x) = −f(−x)

31. first reflected in the x-axis,

then translated to the right by 14 units;

g(x) = −f(x), h(x) = −f(x − 14)

32. first translated upward by 7 units,

then reflected in the y-axis;

g(x) = f(x) + 7, h(x) = f(−x) + 7

33. first enlarged to 2 times the original along

the

y-axis, then reduced to

2

1 of the original

along

129

the x-axis; h(x) = 22x + 1

34. y = sin (x + 70°) − 4

35. y = 5x − 6 + 3

36. y = 3(x + 8)2 − 21

37. y = 6 − (1 − 2x)2

38. yes

39. (a) g(x) = log3 (2x) + 1

(b) yes

40. first translated to the right by 2 units,

then translated upward by 1 unit

41. first translated to the left by 4 units,

then reflected in the y-axis

42. first reduced to

3

1 of the original along the

x-axis, then enlarged to 5 times the original

along the y-axis

45. (a) f(x) = 2(x − 3)2 − 9

46. (a) first enlarged to 4 times the original

along the y-axis, then translated to the

left by 70°

(or first translated to the left by 70°,

then enlarged to 4 times the original

along the y-axis)

(b) a = 4, b = 70

47. (a) k(x) = −a(x + 3)(x + 5 + b) + 1

(b) 1

48. (a) a = −4, b = 1, c = 3

(b) translated to the left by 3 units

(c) h(x) = (x + 4)(x − 1)(x − 3) − 30

49. (a) 4

(b) (i) g(x) = −x2 + 6x − 5 + k

(ii) k416 +

(iii) (1) 12

(2) remain unchanged as 8 units

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