4. Identities and Factorization 52

Commonly Used Formulas

1. Estimation, Approximation and Errors 2

2. Percentages 22

3. Polynomials and Formulas 38

4. Identities and Factorization 52

5. Equations and Inequalities 66

6. Rate and Ratio 82

7. Laws of Integral Indices and Surds 94

8. Basic Geometry 106

9. Symmetry and Transformation 132

10. Trigonometry 146

11. Mensuration 160

12. Coordinate Geometry 180

13. Probability and Statistical Diagrams 196

14. Measures of Central Tendency 230

Revision Test 252

Contents

Sample

Let’s Try

Smart Review

A. Trigonometric Ratios

1. In the figure, ABC is a right-angled triangle, where a is the opposite side（對邊）of i, b is the adjacent side（鄰邊）of i, c is the hypotenuse（斜邊）of i. The trigonometric ratios of sine（正 弦）, cosine（餘弦）and tangent（正切）are defined as follows.

(a) sin i = Opposite side of i Hypotenuse of i = ac (b) cos i = Adjacent side of i

Hypotenuse of i = bc (c) tan i = Opposite side of i

Adjacent side of i = ab For example:

A

B i C

z

4 5

3

(i) In the figure, find the values of sin i, cos i and tan i.

sin i =

54 , cos i =

53 , tan i = 34 .

(ii) In the figure, find the values of sin z, cos z and tan z.

sin z = 5

3, cos z =

4 , tan z = 5 43 . 2. Trigonometric ratios of special angles

Referring to the two right-angled triangles on the right.

i

Trigonometric ratio

30˚ 45˚ 60˚

sin i 2

1

2 1 2

c 2 m

2 3

cos i

2 3

2 1 2

c 2 m 2

1

1. In the figure, find i, correct to 1 decimal place.

A

C

7 4

B i

2. In the figure, find the value of cos i.

A

15 C 8

B

i

3. In the figure, find the length of BC, correct to 3 significant figures.

A

B C 37˚

5 cm

10

Opposite side a Hypotenuse c

Adjacent side b B

A i C

30˚

2 1

3

2 1

Trigonometry

Trigonometric ratios

Sample

10

+ =?

Worked Examples

Conventional Questions Section A(1)

1. In the figure, find the bearing of B from A. (Give the answer correct to 3 significant figures if necessary.) (3 marks)

A

B 170 m

N

120 m E

Solution

Let BAC = i.

A

B 170 m

N

120 m C E i

cos i = 170120

[1M]

i ≈ 45.1˚ (cor. to 3 sig. fig.) [1A]

a 90˚ - 45.1˚ = 44.9˚

` The bearing of B from A is N44.9^˚E. [1A]

． Candidates should only round off the numerical answer to the required significant figures in the final step.

． The question requires to find the bearing of ‘B from A’, thus mark a cross at A.

Reference: HKCEE 04 I Q5

Since no specific bearing is mentioned in the question, candidates can give 044.9˚ or N44.9˚E as the answer.

Candidates should let the required angle be i before doing the calculation.

Otherwise, 1 mark will be deducted for undefined symbol.

Instant Drill

In the figure, find the bearing of A from B. (Give the answer correct to 3 significant figures if necessary.) (3 marks)

A B

N

E 200 m

160 m

Solving Strategy

! Mind the Trap

! Mind the Trap

Sample

Instant Drill Instant Drill

HKDSE Exam Series — Integrated Exam Revision Exercises for Mathematics (Junior Secondary Topics) (Upgraded Edition)

Multiple-choice Questions Section A

6. cos

tan A

A 90c -

^ h=

A. sinA B. cosA C. sin A 1 D. cos A

1

Solution

cos

tan A

A 90c -

^ h

= tan cosA A 1

= sin cos

cos A A

A

# 1

= sin A 1

The answer is C.

tan sin

A A 90 90 c

c =

- -

^

^

h h

A. sinA B. cosA C. sin A 1 D. cos A

1

Reference: HKCEE 08 II Q23

In the figure, Katy measured the angle of elevation of the top of a building from the top of another building as 25˚. She also measured the angle of depression of the bottom of the same building as 36˚. The height of the measured building is 190 m. Find the distance between the two buildings. (Give the answer correct to 3 significant figures.)

(4 marks)

A l t e r n a t ive M e t h o d ( f o r multiple-choice questions only):

Set any value for A and put it into each expression and then compare the results. For example, let A = 20˚, then the value of the given expression is 2.923….

Option A: 0.342…. 8 Option B: 0.939…. 8 Option C: 2.923…. 4 Option D: 1.064…. 8 190 m 25˚

36˚

MC Shortcut

Sample

10

Mock Questions

(In the following questions, unless otherwise specified, give the answer correct to 3 significant figures if necessary.)

Conventional Questions Section A(1)

1. Simplify tan cos

90c i i

-

^ h. (2 marks)

2. Simplify sin cos cos

sin i i

i

+ i. (2 marks)

3. Simplify 2sin^90c-ihcos30c-cosi. (3 marks)

4. Simplify cos

cos 90c tan 90c i

i i

- -

^ h ^ h

. (3 marks)

5. Simplify cos 90 tan

c i i

-

^ h

. (3 marks)

6. Simplify +1 sin 1-cos_^90c - _h . (3 marks)

7. Simplify 1 sin 90

1

2 c i

- ^ - h. (3 marks)

8. Simplify sin sin

sin sin

1 90 1 90

30 30

c c

c c

i i

- - -

- +

^ h ^ h. (4 marks)

9. Prove that sin sin tan

90

90 1

c

c /

i i

i -

-

^

^

h

h . (3 marks)

10. Prove that sin²i+cos²^90c-ihtan²^90c-ih/1. (3 marks)

11. Prove that cos sin

sin cos sin

1 2 ² / i i

i i

i +

- - . (3 marks)

12. Without using a calculator, find the value of sin tan cos

30 60

45

2 2

2

c c

c

- . (4 marks)

13. Without using a calculator, find the value of tan

tan sin

0 3

2 45 45

2 2

c

c- c

. (4 marks)

Sample

HKDSE Exam Series — Integrated Exam Revision Exercises for Mathematics (Junior Secondary Topics) (Upgraded Edition)

Multiple-choice Questions Section A

28. In the figure, cosi+ tani= A. a

b a c + . B. a

c a

+ b. C. a

b a + b. D. b

c b

+ a.

29. In the figure, tan i = A. 4

3. B. 5

3. C. 54

. D. 4

5.

30. In the figure, sin x = A. 17

12. B. 13

12. C. 13

5 . D. 12

5 .

31. In the figure, sin x = A. 34

. B. 4

3. C. 5

3. D. 54

.

32. cos sin

cos cos

1 90

30

1 90

60 c

c

c c

i i

- - -

+ - =

^ h ^ h

A. cos 1

2i B. tan

cos i i C. cos

tan i i D. cos tan

1

i i

33. If A and B are both acute angles and A + B = 90˚, then

cos A sin B 1

2 2 =

+ A. 1.

B. 2sin A 1

2 . C. 2cos A

1

2 . D. 2cos B

1

2 .

34. cos tan

A A

90c- =

^ h

A. sin A B. cos A C. sin A 1 D. cos A

1

35. If 0˚ < x < 90˚, which of the following must be correct?

I. sin x + sin (90˚ – x) > 0 II. cos x ÷ cos (90˚ – x) > 1 III. tan x × tan (90˚ – x) = 1

A. I only B. II only C. I and III only D. II and III only

c a

b i

Reference: HKCEE 08 II Q23

Reference: HKDSE 13 II Q23 Reference: HKCEE 09 II Q24 Reference: HKCEE 08 II Q24

Reference: HKCEE 06 II Q23

Reference: HKCEE 04 II Q22

Reference: HKDSE 12 II Q19

x

5 5

6 B

i

4 3

5 C

A

x 5

5

17 A B

C D

Sample

HKDSE Exam Series — Integrated Exam Revision Exercises for Mathematics (Junior Secondary Topics) (Upgraded Edition)

Conventional Questions

1. Simplify x y x y

4 2

2 3 5 -

^ - h

and express the answer with positive indices. (3 marks)

2. Make c as the subject of the formula c d

1 2

3 -

+ = 5d. (3 marks)

3. Factorize

(a) 4x² - 12xy + 9y²,

(b) 4x² - 12xy + 9y² - 2x + 3y.

(3 marks)

4. The cost of a watch is $ 1200. If the watch is sold at a discount of 20% of its marked price, the profit

percentage is 30%. Find the marked price. (4 marks)

5. The ratio of the costs of a bottle of orange juice to a bottle of milk is 5 : 3. If the total cost of 4 bottles of orange juice and 6 bottles of milk is $76, find the cost of a bottle of milk. (4 marks)

6. In a polar coordinate system, the polar coordinates of points A, B and C are (8, 123°), (7, 213°) and (6, 303°) respectively.

(a) Let O be the pole. Are A, O and C collinear? Explain your answer.

(b) Find the area of 3ABC.

(4 marks)

7. (a) Solve x 5 2 - 21

≤ 4x + 9.

(b) Find the number of negative integers that satisfy the inequality in (a) and write down the smallest one.

(4 marks)

8. In the figure, E is a point on CD such that AE = ED. Given that AB // CD, +BAD = 38° and +AEB = 50°. Find x, y and z.

(4 marks)

50˚

38˚

F z

A B

x y

E D

C

Marks: /100

Date:

Revision Test

Sample

4

4 Identities and Factorization

Let’s Try

1. L.H.S. = 3(5x – 2) – 4x = 11x – 6 R.H.S. = 2(6x – 5) = 12x – 10 a L.H.S. ≠ R.H.S.

` 3(5x – 2) – 4x = 2(6x – 5) is not an identity.

2. L.H.S. = (x + 1)(x + 3) = x² + x + 3x + 3 = x² + 4x + 3 = R.H.S.

` (x + 1)(x + 3) = x² + 4x + 3 is an identity.

Let’s Try

1. Comparing the like terms and the constant terms on the two sides, we have

a = –4 and b = 7.

2. Comparing the like terms and the constant terms on the two sides, we have

3t = –9 and s + t = 6, i.e., t = –3 and s = 9.

Guidelines

As the calculation involved is easier, frst find the value of t. Then use the result to find s.

Let’s Try

1. (3 + h)²

= 3² + 2(3)h + h² = h² + 6h + 9

Guidelines

Solutions are usually expressed in descending order of the variable.

2. (16 – 7y)(16 + 7y)

= 16² – (7y)² = 256 – 49y²

3. (5m – 4)(25m² + 20m + 16)

= (5m – 4)[(5m)² + (5m)(4) + 4²]

= (5m)³ – 4³ = 125m³ – 64

Let’s Try

1. 2ab² – 4b = 2b(ab – 2)

2. 6x²y³z + 18xy²z² – 9y³z = 3y²z(2x²y + 6xz – 3y)

Guidelines

The common factor of 6, 18 and 9 is 3, while the common factor of x²y³z、xy²z² and y³z is y²z.

Let’s Try

1. 5p + 5q + 2mp + 2mq

= 5(p + q) + 2m(p + q) = (p + q)(5 + 2m)

2. h² – 3jk – 3hj + hk = h² + hk – 3jk – 3hj = h(h + k) – 3j(k + h) = (h + k)(h – 3j)

Let’s Try

1. a –5 a +1

–5a +a = (–5 + 1)a = –4a

` a² – 4a – 5 = (a – 5)(a + 1)

Sample

HKDSE Exam Series — Integrated Exam Revision Exercises for Mathematics (Junior Secondary Topics) (Upgraded Edition) Solution Guide

2. 2x –1 3x +1

–3x +2x = (–3 + 2)x = –x

` 6x² – x – 1 = (2x – 1)(3x + 1)

3. 2m +3n

2m +5n

+6mn +10mn = (6 + 10)mn = 16mn

` 4m² + 16mn + 15n² = (2m + 3n)(2m + 5n)

Let’s Try

1. a² + 8a + 16 = a² + 2(4)a + 4² = (a + 4)²

2. u² – 10uv + 25v² = u² – 2(5v)u + (5v)² = (u – 5v)²

3. 9m² – 49n²

= (3m)² – (7n)² = (3m + 7n)(3m – 7n)

4. x³ – 8 = x³ – 2³

= (x – 2)[x² + x(2) + 2²] = (x – 2)(x² + 2x + 4)

Common Mistakes

Some candidates may confuse the identity of sum of two cubes with that of difference of two cubes. In each of these two identities, there is only one negative sign. In the identity of difference of two cubes, the first +/– sign is negative; while in the identity of sum of two cubes, the second +/– sign is negative.

5. 27k³ + 1 = (3k)³ + 1³

= (3k + 1)[(3k)² – (3k)(1) + 1²] = (3k + 1)(9k² – 3k + 1)

Concept Builder

1. False

If an equation holds for ‘any values’ of the unknowns, then the equation is an identity.

2. False

L.H.S. = 7(2x – 1) + 8 = 14x + 1

R.H.S. = 8x + 6(x – 2) + 5 = 14x – 7

a L.H.S. ≠ R.H.S.

` 7(2x – 1) + 8 = 8x + 6(x – 2) + 5 is not an identity.

3. False

In any identity, besides the constant terms, the like terms on the two sides are equal.

4. False

(7x + 10)² = (7x)² + 2(7x)(10) + 10² = 49x² + 140x + 100 5. True

6. True

4 – 4x = 4(1 – x) = –4(x – 1)

7. False

x –4

3x +5

–12x +5x = (–12 + 5)x = –7x

This method can be used to factorize 3x² – 7x – 20 only.

Sample