chapter 1 functions fungsi - pelangibooks.com standard math/online... · graf (a) {(4, 1), (4, 2),...

© Penerbitan Pelangi Sdn. Bhd.1

FunctionsFungsi1

CHAPTER

1. Ordered pairsPasangan bertertib

Arrow diagramGambar rajah anak panah

GraphGraf

(a) {(4, 1), (4, 2), (4, 4), (8, 1), (8, 2), (8, 4), (8, 8)} 4

8

1

2

4

8

Set X Set Y

Set X84

Set

Y

2

1

4

8

(b) {(3, 3), (6, 2), (6, 3), (9, 3)}3

6

9

2

3

Set X Set Y

3

3

2

6 9

Set

Y

Set X

(c) {(1, 2), (1, 3), (2, 3)}1

2

3

1

2

3

Set X Set Y1

Set

Y

Set X32

1

2

3

2. DomainDomain

CodomainKodomain

ObjectsObjek

ImagesImej

RangeJulat

(a) {r, s, t, u} {5, 7, 9} r, s, t, u 5, 7, 9 {5, 7, 9}

(b) {0, 3, 4, 7, 8} {2, 5, 6, 9, 10} 0, 3, 4, 7, 8 2, 5, 6, 9, 10 {2, 5, 6, 9, 10}

(c) {1, 3, 5} {1, 3, 4, 6} 1, 3, 5 1, 3, 4 {1, 3, 4}

3. (a) Many-to-one relation(b) One-to-one relation(c) One-to-many relation

Additional Mathematics Form 4 Chapter 1 Functions

© Penerbitan Pelangi Sdn. Bhd. 2

4. Yes/ NoYa / Bukan

ReasonSebab

(a) No The relation is not a function because the object 4 has two images.

(b) No The relation is not a function because the object 4 has no image.

(c) Yes The relation is a function because every object has only one image.

5. RelationHubungan

Function notationTatatanda fungsi

(a) square of f : x → x2

orf(x) = x2

(b) divided by 2 f : x → 12 x

orf(x) = 1

2 x

(c) multiplied by –2

f : x → –2x orf(x) = –2x

6. (a) (b) (c)

ObjectsObjek –2, –1, 1 –3, 0, 6 –1, 1, 8

ImagesImej 1 –1, 1, 5 –1, 1, 2

DomainDomain

{–2, –1, 1} {–3, 0, 6} {–1, 1, 8}

RangeJulat {1} {–1, 1, 5} {–1, 1, 2}

7. (a) (i) f(0°) = 2 tan 0° = 0

(ii) f(45°) = 2 tan 45° = 2(1) = 2

(b) (i) f(–1) = –1–1 – 3

= 14

(ii) f(1) = 11 – 3

= – 12

(c) (i) f(–2) = –2[2 – 3(–2)] = –16

(ii) f( 13 ) = ( 1

3 )[2 – 3( 13 )]

= 13

8. (a) (i) f(x) = 2

x + 5

2 = 2

x + 5 = 4 x + 5 = 16 x = 11

(ii) f(s) = –3

s + 5

2 = –3

s + 5 = –6 s + 5 = 36 s = 31

(b) (i) f(x) = –2

|x – 5|x

= –2 |x – 5| = –2x x – 5 = –2x or x – 5 = 2x 3x = 5 x = –5

x = 53

(ii) f(t) = 4

|t – 5|t

= 4

|t – 5| = 4t t – 5 = –4t or t – 5 = 4t 5t = 5 3t = –5 t = 1 t = – 5

3



(c) |f(u)| = 12

2u + 5u – 1 = 1

2

2u + 5u – 1 = 1

2

2(2u + 5) = u – 1 4u + 10 = u – 1 3u = –11

u = – 113 or

2u + 5u – 1 = – 1

2

2(2u + 5) = –(u – 1) 4u + 10 = –u – 1 5u = –9

u = – 95

9. (a) (i) gf(x) = g(3x – 2) = 4 – 5(3x – 2)2

= –45x2 + 60x – 16

(ii) fg(x) = f(4 – 5x2) = 3(4 – 5x2) – 2 = –15x2 + 10

(iii) f 2(x) = f(3x – 2) = 3(3x – 2) – 2 = 9x – 8

(b) (i) fgh(x) = fg(2x2 + 1)

= f 1 12x2 + 1 2

= 21 12x2 + 1 2 + 3

= 22x2 + 1 + 3

(ii) hgf(x) = hg(2x + 3)

= h1 12x + 3 2

= 21 12x + 3 2

2 + 1

= 2(2x + 3)2 + 1, x ≠ – 3

2

10. (a) (i) f 2(2) = f 1 22(2) + 1 2

= f 1 25 2

=

25

21 25 2 + 1

= 25 × 5

9

= 29

(ii) gk1 14 2 = g1 1

4 + 42 = g1 9

2 2 = 3 – 9

2

= – 32

(iii) g(|q2 – 5|) = –1 3 – |q2 – 5| = –1 |q2 – 5| = 4

q2 – 5 = 4 or q2 – 5 = –4 q = ±3 q = ±1

Since q . 0, q = 1, 3.

(b) (i) fk1 14 2 = f 1 1

4 + 42 = f1 9

2 2

=

92

21 92 2 + 1

= 920

(ii) k(3 – 2r) = 5 3 – 2r + 4 = 5 3 – 2r = 1 3 – 2r = 1

Since 3 – 2r . 0, 3 – 2r = 1 2r = 22 r = 1



(iii) k(3 – m) = 2 3 – m + 4 = 2 3 – m = –2 3 – m = 4 m = –1

11. (a) k(x) + 2 = xx – 1

k(x) = xx – 1 – 2

= x – 2(x – 1)x – 1

= x – 2x + 2x – 1

k(x) = 1 2 – xx – 1 2

2 , x ≠ 1

(b) 3[k(x)]2 + 2 = x2 – 2 3[k(x)]2 = x2 – 4

k(x) = x2 – 43

(c) 12k(x) + 1 = 3

x – 3

2k(x) + 1 = x – 33

2k(x) = x – 33 – 1

= x – 63

k(x) = x – 66

= x6 – 1

12. (a) n1 2x 2 = 1 + 4x

Let y = 2x

, x = 2y

n(y) = 1 + 41 2y 2

= 1 + 8y

n(x) = 1 + 8x

, x ≠ 0

(b) n(2x + 1) = x – 2

Let y = 2x + 1, x = y – 12

n(y) = y – 12 – 2

= y – 52

n(x) = x – 52

(c) n(2 + 3x) = 2x + 2

Let y = 2 + 3x, x = y – 23

n(y) = 2y – 2

3 + 2

= 21 3y + 4 2

= 6y + 4

n(x) = 6x + 4 , x ≠ –4

13. (a) (i) g(u) = –1 |2u + 3| – 4 = –1 |2u + 3| = 3 2u + 3 = –3 2u = –6 u = –3 or 2u + 3 = 3 2u = 0 u = 0

(ii) g(v) = 2 |2v + 3| – 4 = 2 |2v + 3| = 6 2v + 3 = 6 v = 3

2

or 2v + 3 = –6 v = – 9

2

(b) (i) h(s) = 1

2s + 11 – 2s

= 1

2s + 1 = 1 – 2s 4s = 0 s = 0



(ii) h(t) = –3

2t + 11 – 2t

= –3

2t + 1 = –3 + 6t 4t = 4 t = 1

14. (a) (i) Let y = q–1(x) q(y) = x

3y + 7y – 5 = x

3y + 7 = xy – 5x xy – 3y = 7 + 5x y(x – 3) = 7 + 5x

y = 7 + 5xx – 3 , x ≠ 3

q–1(x) = 7 + 5xx – 3 , x ≠ 3

(ii) q–1(–2) = 7 + 5(–2) –2 – 3

= 35

(b) (i) (a) Let y = r –1(x) r(y) = x

2y

= x

y = 2x

r –1(x) = 2x

, x ≠ 0

(b) rr –1(x) = r 1 2x 2

= 22x

= x

(c) r –1r(x) = r –11 2x 2

= 22x

= x

(ii) rr –1(x) = r –1r(x) = x

SPM Practice 1

Paper 1

1. (a) {(–3, 2), (–2, 5), (1, 2)}(b) {–3, –2, 1}

2. h(2) = 10 2(2) + m = 10 m = 6

3. (a) {4, 6, 10}(b) Many-to-one relation

4. fg(x) = f(4x – 5) = 2(4x – 5) + 3 = 8x – 7

5. f(a) = 8 a – 5a = 8 –4a = 8 a = –2

6. (a) 3(b) {–2, –1, 1, 2}

7. (a) hg(2) = 2(2) – 1 = 3

(b) hg(x) = 2x – 1 3g(x) + 4 = 2x – 1 3g(x) = 2x – 5

g(x) = 2x – 53

8. (a) f(4) = 4 – 3 = 1

(b) 3f –1(r) = 1

f –1(r) = 13

f 1 13 2 = r

= 13 – 3

= –2 23



9. (a) Let y = h–1(x)

y = 23 – x

3y – xy = 2 xy = 3y – 2

x = 3 – 2y

h(y) = x h(y) = 3 – 2

y

h(x) = 3 – 2x , x ≠ 0

(b) 3 – 2x

= –2

5 = 2x

x = 25

10. (a) 3(–1) = –1

(b) {(–1, –1), (0, 0), (1, 1)}

11. fg(2) = f [3(2) – 1] = f(5) = 2(5)+ 5 = 15

12. (a) f(q)(b) p

13. fg(1) = 3 f[a + b(1)] = 3 2(a + b) = 3

a + b = 32

a = 32 – b

14. (a) Let y = f –1(x) f(y) = x 2 + 5y = x

y = x – 25

f –1(x) = x – 25

(b) gf –1(7) = g( 7 – 25 )

= g(1) = 4 + 3(1) = 7

15. gf(14) = g(14 – 6) = g(8)

= 83(8) – 4

= 820

= 25

16. (a) gf(x) = g(3x – 2) = 6(3x – 2) = 18x – 12

(b) gf(x) = 12 f(x)

18x – 12 = 12 (3x – 2)

2(18x – 12) = 3x – 2 36x – 24 = 3x – 2 33x = 22

x = 23

17. (a) –2(b) 6

18. (a) When f(x) = 0, x = q –|2q + 1| = 0 –(2q + 1) = 0 –2q = 1

q = – 12

(b) f(3) = –|2(3) + 1| = –7 The range of f(x) is 0 < f(x) < –7.



19. g2(x) = nx + 2 …… 1 g2(x) = g(3x – m) = 3(3x – m) – m = 9x – 4m …… 2

1 = 2, 9x – 4m = nx + 2 9x = nx and –4m = 2 n = 9 m = – 1

2

20. f(x) = mx + c f 2(x) = f(mx + c) = m(mx + c) + c = m2x + mc + c Compare with f 2(x) = 16x – 9, m2 = 16 m = ± 4 Since m . 0, m = 4 When m = 4, mc + c = –9 4c + c = –9 5c = –9 c = – 9

5

Paper 2

1. (a) (i) g–1(x) = y g(y) = x x = 2y + 3

y = x – 32

g–1(x) = x – 32

(i) h(2x + 3) = 10x + 7 Let y = 2x + 3

x = y – 32

h(y) = 101 y – 32 2 + 7

= 5y – 15 + 7 = 5y – 8 h(x) = 5x – 8

(b) (5x – 8) = 6x – 1 2(5x – 8) + 3 = 6x – 1 10x – 13 = 6x – 1 4x = 12 x = 3

2. (a) h[g(x)] = x2 – 4x + 1 2 – g(x) = x2 – 4x + 1 g(x) = 1 + 4x – x2

(b) gh(2) = g(2 – 2) = g(0) = 1 + 4(0) – 02 = 1

3. (a) Let f –1(x) = y f(y) = x

y2 – 3 = x

y2 = x + 3

y = 2x + 6 f –1(x) = 2x + 6 Let g–1(x) = z g(z) = x 4z + 1 = x

z = x – 14

g–1(x) = x – 14

f –1g–1(x) = f –1( x – 12 )

= 2( x – 12 ) + 6

= x – 12 + 6

= x – 1 + 122

= x + 112

(b) fg(x) = f(4x + 1)

= 4x + 12 – 3

= 4x + 1 – 62

= 4x – 52

Let (fg)–1(x) = y (fg)(y) = x

4y – 52 = x

4y – 5 = 2x

y = 2x + 54

(fg)–1(x) = 2x + 54



(c) f –1g–1(2x) = (fg)–1(x)

2x + 112 = 2x + 5

2 2(2x + 11) = 2x + 5 4x + 22 = 2x + 5 2x = –17

x = – 172

Additional Mathematics Form 4 Chapter 2 Quadratic Equations


1. General formBentuk am

Values of a, b and cNilai-nilai bagi a, b dan c

(a) (4 – 3x)(4x + 1) = 0 16x + 4 – 12x2 – 3x = 0 12x2 – 13x – 4 = 0

a = 12, b = –13, c = –4

(b) (2x – 3)(2x – 3) = 6 4x2 – 12x + 9 = 6 4x2 – 12x + 3 = 0

a = 4, b = –12, c = 3

(c) 2 – x—4

= 3x2

8 – x = 12x2 12x2 + x – 8 = 0

a = 12, b = 1, c = –8

2. (a) Substitute x = 3 into x2 – 2x – 15. (3)2 – 2(3) – 15 = 9 – 6 – 15 = –12≠0 Therefore, 3 is not a root of the quadratic

equation.

(b) Substitute x = –1 into 2x2 + 5x – 12. 2(–1)2 + 5(–1) – 12 = 2 – 5 – 12 = –15≠0 Therefore, –1 is not a root of the quadratic

equation.

(c) 6x2 + x – 1 = 0 Substitute x = – 1—

2 into 6x2 + x – 1.

61– 1—2 2

2 + 1– 1—

2 2 – 1 = 3—2 – 1—

2 – 1

= 0

Therefore, – 1—2

is a root of the quadratic equation.

Quadratic EquationsPersamaan Kuadratik2

CHAPTER

3. (a) When x = 3—2

,

2x – 3 = 21 3—2 2 – 3 = 0

So, x = 3—2

is a root of the equation.

When x = – 3—5

,

3 + 5x = 3 + 51– 3—5 2 = 0

So, x = – 3—5 is a root of the equation.

(c) When x = –4, x + 4 = –4 + 4 = 0

So, x = –4 is a root of the equation.

When x = 3, x + 4 = 3 + 4 4 – 3x = 4 – 3(3) =7≠0 =–5≠0

So, x = 3 is a root of the equation.



4. (a) 4x2 – 7x – 2 = 0 Try x = –1, 0, 1

x 4x2 – 7x – 2–1 90 –21 –5

Try x = – 1––4

,

4x2 – 7x – 2 = 41– 1—4 2

2 – 71– 1—

4 2 – 2

= 1—4

+ 7—4

– 2

= 0

Therefore, – 1—4

is a root of the quadratic

equation 4x2 – 7x – 2 = 0.

(b) 3x2 + 14x – 5 = 0

Try x = 0, 1––2

, 1

x 3x2 + 14x – 50 –51––2

2 3––4

1 12

Try x = 1––3

,

3x2 + 14x – 5 = 31 1—3 2

2 + 141 1—

3 2 – 5

= 1—3

+ 14—3

– 5

= 0

Therefore, 1—3

is a root of the quadratic

equation 3x2 + 14x – 5 = 0.

5. (a) x2 + 3x – 10 = 0 (x + 5)(x – 2) = 0 x + 5 = 0 or x – 2 = 0 x = –5 x = 2

(b) 4x2 – 19x + 12 = 0 (x – 4)(4x – 3) = 0 x – 4 = 0 or 4x – 3 = 0 x = 4 x = 3

4

(c) x2 – 3x – 18 = 0 (x + 3)(x – 6) = 0 x + 3 = 0 or x – 6 = 0 x = –3 x = 6

(d) 12x2 – 17x + 6 = 0 (4x – 3)(3x – 2) = 0 4x – 3 = 0 or 3x – 2 = 0 x = 3

4 x = 23

(e) 42 + 5x – 2x2 = 0 2x2 – 5x – 42 = 0 (x – 6)(2x + 7) = 0 x – 6 = 0 or 2x + 7 = 0 x = 6 x = – 7

2

6. (a) x2 + 8x – 3 = 0

x2 + 8x + 1 82 2

2 = 3 + 1 8

2 22

x2 + 8x + (4)2 = 3 + (4)2 (x + 4)2 = 3 + 16 x + 4 = ± 19 x = –4 ± 19

x = –4 + 19 or x = – 4 – 19 = 0.3589 = –8.359

\ x = 0.3589 or –8.359

(b) x2 – 5x + 3 = 0

x2 – 5x + 1–52 2

2 = –3 + 1–5

2 22

x2 – 5x + 1–52 2

2 = –3 + 25

4

1x – 52 2

2 = 13

4

x – 52 = ±

132

x = 5 ± 132

x = 5 + 132

or x = 5 – 132

= 4.303 = 0.6972

\ x = 4.303 or 0.6972



(c) x2 – 8x – 7 = 0

x2 – 8x + 1–82 2

2 = 7 + 1–8

2 22

x2 – 8x + (–4)2 = 7 + (–4)2

(x – 4)2 = 7 + 16 = 23 x – 4 = ± 23 x = 4 ± 23

x = 4 + 23 or x = 4 – 23 = 8.796 = –0.7958

\ x = 8.796 or –0.7958

(d) 2x2 = 3 – 10x 2x2 + 10x = 3

x2 + 5x = 32

x2 + 5x + 1 52 2

2 = 3

2 + 1 52 2

2

1x + 52 2

2 = 31

4

x + 52 = ±

312

x = –5 – 312

x = –5 + 312

or x = –5 – 312

= 0.2838 = –5.284

\ x = 0.2838 or –5.284

(e) 3x2 = 5x + 8

3x2 – 5x = 8

x2 – 53 x = 8

3

x2 – 53 x + 1–5

6 22 = 8

3 + 1–56 2

2

1x – 56 2

2 = 8

3 + 2536

= 12136

x – 56 = ± 11

6

x = 5 ± 116

x = 5 + 116 or x = 5 – 11

6 = 2.667 = –1

\ x = 2.667 or –1

(f) 2x(x + 3 ) = 5 2x2 + 6x = 5 x2 + 3x = 5

2

x2 + 3x + 1 32 2

2 = 5

2 + 1 32 2

2

1x + 32 2

2 = 19

4

x + 32 = ±

192

x = –3 ± 192

x = –3 + 192

or x = –3 – 192

= 0.6794 = –3.679

\ x = 0.6794 or –3.679

(g) 4x(x – 5) = 3

x(x – 5) = 34

x2 – 5x = 34

x2 – 5x + 1–52 2

2 = 3

4 + 1–52 2

2

1x – 52 2

2 = 28

4 = 7 x – 5

2 = ± 7

x = 52 ± 7

x = 52 + 7 or x = 5

2 – 7

= 5.146 = –0.1458

\ x = 5.146 or –0.1458



7. Values of a, b and cNilai-nilai bagi a, b dan c

The roots using the formulaPunca-punca dengan menggunakan rumus

x = –b ± b2 – 4(a)(c)2(a)

(a) a = 2; b = 4; c = –3 x = – 4 ± 42 – 4(2)(–3)2(2)

= – 4 ± 404

= – 4 + 404

or – 4 – 404

= 0.5811 or –2.581

(b) a = 3; b = –7; c = 3 x = –(–7) ± (–7)2 – 4(3)(3)2(3)

= 7 ± 136

= 7 + 136

or 7 – 136

= 1.768 or 0.5657

(c) a = 4; b = –5; c = –2 x = –(–5) ± (–5)2 – 4(4)(–2)2(4)

= 5 ± 578

= 5 + 578

or 5 – 578

= 1.569 or –0.3187

(d) 6x = 2x2 + 3 2x2 – 6x + 3 = 0

a = 2; b = –6; c = 3

x = –(–6) ± (–6)2 – 4(2)(3)2(2)

= 6 ± 124

= 6 + 124

or 6 – 124

= 2.366 or 0.6340

(e) 2x(x + 4) = 5 2x2 + 8x = 5 2x2 + 8x – 5 = 0

a = 2; b = 8; c = –5

x = –8 ± (8)2 – 4(2)(–5)2(2)

= –8 ± 1044

= –8 + 1044

or –8 – 1044

= 0.5495 or –4.550



7. Values of a, b and cNilai-nilai bagi a, b dan c

The roots using the formulaPunca-punca dengan menggunakan rumus

x = –b ± b2 – 4(a)(c)2(a)

(f) (4 – 3p)(2 + 5p) = 3p 8 + 20p – 6p – 15p2 = 3p 15p2 – 11p – 8 = 0

a = 15; b = –11; c = –8

p = –(–11) ± (–11)2 – 4(15)(–8)2(15)

= 11 ± 60130

= 11 + 60130

or 11 – 60130

= 1.184 or –0.4505

(g) 4n – 3n

= 2

4n2 – 3 = 2n 4n2 – 2n – 3 = 0

a = 4; b = –2; c = –3

n = –(–2) ± (–2)2 – 4(4)(–3)2(4)

= 2 ± 528

= 2 + 528

or 2 – 528

= 1.151 or –0.6514

8. (a) x = 3 or x = 5 x – 3 = 0 or x – 5 = 0 (x – 3)(x – 5) = 0 x2 – 8x + 15 = 0

(b) x = –1 or x = 6 x + 1 = 0 or x – 6 = 0 (x + 1)(x – 6) = 0 x2 + 5x – 6 = 0

(c) x = 3 or x = –4 x – 3 = 0 or x + 4 = 0 (x – 3)(x + 4) = 0 x2 + x – 12 = 0

(d) x = –5 or x = –2 x + 5 = 0 or x + 2 = 0 (x + 5)(x + 2) = 0 x2 + 7x + 10 = 0

(e) x = 1—2

or x = 3

x – 1—2

= 0

2x – 1 = 0 or x – 3 = 0 (2x – 1)(x – 3) = 0 2x2 – 7x + 3 = 0

(f) x = 3—5

or x = – 1—4

x – 3—5

= 0 x + 1—4

= 0

5x – 3 = 0 or 4x + 1 = 0 (5x – 3)(4x + 1) = 0 20x2 – 7x – 3 = 0



9. Quadratic equationPersamaan kuadratik

New rootsPunca-punca baharu

New quadratic equationPersamaan kuadratik baharu

(a) x2 – 2x + 6 = 0 a + b = 2 ab = 6

[2a , 2b]

2a + 2b = 2(a + b) = 2(2) = 4

2a(2b) = 4ab = 4(6) = 24

x2 – (2a + 2b)x + 2a(2b) = 0 x2 – 4x + 24 = 0

(b) x2 + 5x + 4 = 0

a + b = –5 ab = 4

[3a , 3b]

3a + 3b = 3(a + b) = 3(–5) = –15

3a(3b) = 9ab = 9(4) = 36

x2 – (3a +3b)x +3a(3b) = 0 x2 – (–15)x + 36 = 0 x2 + 15x + 36 = 0

(c) 2x2 – 4x – 3 = 0

x2 – 2x – 32 = 0

a + b = 2

ab = – 32

[3a , 3b]

3a + 3b = 3(a + b) = 3(2) = 6

3a(3b) = 9ab

= 91– 32 2

= – 272

x2 – (3a +3b)x +3a(3b) = 0

x2 – 6x + 1– 272 2 = 0

2x2 – 12x – 27 = 0

(d) 3x2 – 2x + 6 = 0

x2 – 23 x + 2 = 0

a + b = 23

ab = 2

1 a2 , b

2 2

a2 + b

2 = 1 a + b2 2

= 13

1 a2 21 b

2 2 = ab4

= 24

= 12

x2 – 1 a + b2 2x + 1 ab

4 2 = 0

x2 – 13 x + 1

2 = 0

6x2 – 2x + 3 = 0



Quadratic equationPersamaan kuadratik

New rootsPunca-punca baharu

New quadratic equationPersamaan kuadratik baharu

(e) 4x2 + 3x – 2 = 0

x2 + 34 x – 1

2 = 0

a + b = – 34

ab = – 12

1 2a

, 2b 2

2a

+ 2b

= 2(a + b)ab

= 21– 34 21– 2

1 2 = 3

1 2a 21 2

b 2 = 4ab

= 4(–2) = –8

x2 – 2(a + b)ab

x + 4ab

= 0

x2 – 3x – 8 = 0

10.

(a) 3x2 + px – q = 0 ; 3 1

2 , –44

x2 + p3 x – q

3 = 0

Let a = 12 and b = –4.

a + b = – p3

12 + (–4) = – p

3 – 7

2 = – p3

p = 212

ab = – q3

1 12 2(–4) = – q

3 q = 6

(b) 4x2 – (p + 2)x = q ; 34, – 23 4

x2 – (p + 2)4 x – q

4 = 0

Let a = 4 and b = – 23 .

a + b = p + 24

4 – 23 = p + 2

4

103 = p + 2

4

p + 2 = 403

p = 403 – 2

= 343

ab = – q4

(4)1– 23 2 = – q

4

q = 323



(c) 2x2 + (1 – p)x – 5 = 0 ; [–q, 2]

x2 + (1 – p)2 x – 5

2 = 0

Let a = –q and b = 2.

a + b = – (1 – p)2

–q + 2 = p – 12

–2q + 4 = p – 1 p = 5 – 2q …… 1

ab = – 52

(–q)(2) = – 52

q = 54

Substitute q = 54 into 1.

p = 5 – 21 54 2

= 52

11. (a) 9x2 + kx + 2 = 0

x2 + k9 x + 2

9 = 0

Let a and 2a be the roots.

a + 2a = – k9 , a(2a) = 2

9

3a = – k9 a2 = 1

9

k = –27a a = ± 13

When a = 13 , k = (–27)1 1

3 2 = –9

When a = – 13 , k = (–27)1– 1

3 2 = 9

(b) 4x2 + (3k – 2)x + 3 = 0

x2 + (3k – 2)4 x + 3

4 = 0


a + 3a = – (3k – 2)4 , a(3a) = 3

4

4a = 2 – 3k4 a2 = 1

4

16a = 2 – 3k a = ± 12 k = 2 – 16a

3

When a = 12 , k =

2 – (16)1 12 2

3 = –2

When a = – 12 , k =

2 – (16)1– 12 2

3 = 10

3

(c) 5x(x – 4) = 4(x + k) 5x2 – 20x = 4x + 4k 5x2 – 24x – 4k = 0

x2 – 245 x – 4

5 k = 0


a + 5a = 245 , a(5a) = – 4

5 k 25a2 = – 4k

6a = 245

a = 45

k = – 254 a2

= –1254 21 4

5 22

= – 4



12. Value of b2 – 4acNilai bagi b2 – 4ac

Type of rootsKeadaan punca

(a) (–12)2 – 4(9)(4) = 144 – 144 = 0

Since b2 – 4ac = 0, the quadratic equation has two equal roots.

(b) 22 – 4(3)(4) = 4 – 48 = –44 , 0

Since b2 – 4ac , 0, the quadratic equation has no real roots.

(c) (–3)2 – 4(5)(6) = 9 – 120 = –111 , 0

Since b2 – 4ac , 0, the quadratic equation has no real roots.

(d) 2x(4x – 3) = 5 8x2 – 6x = 5 8x2 – 6x – 5 = 0(–6)2 – 4(8)(–5) = 36 + 160 = 196 . 0

Since b2 – 4ac . 0, the quadratic equation has two different roots.

13.

(a) 2x2 + px + 8 = 0

b2 – 4ac = 0 (p)2 – 4(2)(8) = 0 p2 – 64 = 0 p2 = 64 p = ±8 p = –8 or 8

(b) 3x2 – 2px + p = 0

b2 – 4ac = 0 (–2p)2 – 4(3)(p) = 0 4p2 – 12p = 0 4p(p – 3) = 0

4p = 0 or p – 3 = 0 p = 0 p = 3

(c) px2 – 4x + 3p – 4 = 0

b2 – 4ac = 0 (–4)2 – 4(p)(3p – 4) = 0 16 – 12p2 + 16p = 0 12p2 – 16p – 16 = 0 3p2 – 4p – 4 = 0 (3p + 2)(p – 2) = 0

3p + 2 = 0 or p – 2 = 0 p = – 2

3 p = 2

(d) 2(p + 1)x2 – 3(p + 1)x + 9 = 0 ; p≠–1

b2 – 4ac = 0 [–3(p + 1)]2 – 4[2(p + 1)](9) = 0 9(p + 1)2 – 72(p + 1) = 0 9(p + 1)[(p + 1) – 8] = 0 9(p + 1)(p – 7) = 0

p + 1 = 0 or p – 7 = 0 p = –1 p = 7 Since p≠–1,hencep = 7.

(e) x2 – 2x + 7 = p(1 – 2x)

x2 – 2x + 7 = p – 2px x2 + 2px – 2x + 7 – p = 0 x2 + 2(p – 1)x + (7 – p) = 0

b2 – 4ac = 0 [2(p – 1)]2 – 4(1)(7 – p) = 0 4(p2 – 2p + 1) – 28 + 4p = 0 4p2 – 8p + 4 – 28 + 4p = 0 4p2 – 4p – 24 = 0 p2 – p – 6 = 0 (p + 2)(p – 3) = 0

p + 2 = 0 or p – 3 = 0 p = –2 p = 3



14. (a) x2 + 6x + p – 3 = 0

b2 – 4ac . 0 (6)2 – 4(1)(p – 3) . 0 36 – 4p + 12 . 0 48 – 4p . 0 – 4p . – 48 4p , 48 p , 12

(b) 2x2 – 3x + p – 2 = 0

b2 – 4ac . 0 (–3)2 – 4(2)(p – 2) . 0 9 – 8p + 16 . 0 25 – 8p . 0 –8p . –25 8p , 25

p , 258

(c) px2 + 3x – 6 = 0

b2 – 4ac . 0 (3)2 – 4(p)(– 6) . 0 9 +24p . 0 24p . –9 p . – 9

24

p . – 38

(d) 5 – 2x = (p – 3)x2

(p – 3)x2 + 2x – 5 = 0

b2 – 4ac . 0 (2)2 – 4(p – 3)(–5) . 0 4 + 20p – 60 . 0 –56 . –20p 56p , 20p

p . 5620

p . 145

(e) 4x2 – 4px + p2 = 5x

4x2 – 4px – 5x + p2 = 0 4x2 – (4p + 5)x + p2 = 0

b2 – 4ac . 0 [–(4p + 5)]2 – 4(4)(p2) . 0 16p2 + 40p + 25 – 16p2 . 0 40p . –25

p . – 2540

p . – 58

15. (a) x2 + 3x – 4 + h = 0

b2 – 4ac , 0 (3)2 – 4(1)(–4 + h) , 0 9 + 16 – 4h , 0 – 4h , –25 4h . 25

h . 254

(b) x2 + 2hx + (2 – h)2 = 0

b2 – 4ac , 0 (2h)2 – 4(1)(2 – h)2 , 0 4h2 – 4(4 – 4h + h 2) , 0 –16 + 16h , 0 –16 , –16h 1 . h h , 1

(c) hx(x + 2) = 3 – h – 4x

hx2 + 2hx + 4x + h – 3 = 0 hx2 + (2h + 4)x + (h – 3) = 0

b2 – 4ac , 0 (2h + 4)2 – 4(h)(h – 3) , 0 4h2 + 16h + 16 – 4h2 + 12h , 0 28h + 16 , 0 28h , –16

h , – 1628

h , – 47



16. (a) 2px2 + qx + 2p = 0

b2 – 4ac = 0 (q)2 – 4(2p)(2p) = 0 q2 – 16p2 = 0 q2 = 16p2

p2 = q2

16

p = ± q4

p = – q4 , q

4

(b) (q – 2p)x2 – 4px + p = 0, p≠0

b2 – 4ac = 0 (–4p)2 – 4(q – 2p)(p) = 0 16p2 – 4pq + 8p2 = 0 24p2 – 4pq = 0 6p2 – pq = 0 p(6p – q) = 0

p = 0 or 6p – q = 0 p = q

6

Since p≠0,hencep = q6 .

(c) 2px2 + 4(p – q)x = q – p

2px2 + 4(p – q)x + p – q = 0

b2 – 4ac = 0 [4(p – q)]2 – 4(2p)(p – q) = 0 16(p2 – 2pq + q2) – (8p)(p – q) = 0 16p2 – 32pq + 16q2 – 8p2 + 8pq = 0 8p2 – 24pq + 16q2 = 0 p2 – 3pq + 2q2 = 0 (p – q)(p – 2q) = 0 p – q = 0 or p – 2q = 0 p = q p = 2q

SPM Practice 2

Paper 1

1. Since the equation has no roots, then b2 – 4ac , 0 (–4)2 – 4(2 – k)(6) , 0 16 – 48 + 24k , 0 24k , 32

k , 43

2. (a) Substitute x = 3 into the quadratic equation, 2(3)2 + h(3) – 27 = 0 18 + 3h – 27 = 0 3h = 9 h = 3

(b) 2x2 + hx – 27 = 0

x2 + h2 x – 27

2 = 0

Sum of roots = – h2

Hence, – h2 = –2

h = 4

3. (a) x(x – 3) = 6 x2 – 3x = 6 x2 – 3x – 6 = 0

(b) Sum of roots = 3

(c) b2 – 4ac = (–3)2 – 4(1)(–6) = 9 + 24 = 33 . 0 Hence, the equation has two distinct roots.

4. b2 – 4ac , 0 (–4)2 – 4(k)(–3) , 0 16 + 12k , 0 12k , –16

k , – 43



5. Sum of roots = 4 + h + 6 = h + 10Hence, –(k – 1) = h + 10 –k + 1 = h + 10 h + k = –9 …… 1 Product of roots = 4(h + 6)Hence, 12 = 4(h + 6) h = –3Substitute h = –3 into 1, –3 + k = –9 k = –6

6. x(x – 8) = 2p – qx 2 – 8x + q – 2p = 0For equal roots, b2 – 4ac = 0(–8)2 – 4(1)(q – 2p) = 0 64 – 4q + 8p = 0 8p = 4q – 64

p = 12 q – 8

7. px2 + (1 – 4p)x + 4p – 3 = 0 b2 – 4ac = 0 (1 – 4p)2 – 4(p)(4p – 3) = 0 1 – 8p + 16p2 – 16p2 + 12p = 0 1 + 4p = 0 4p = –1

p = – 14

8. (x + 2p)2 = 25 (–3 + 2p)2 = 25 9 – 12p + 4p2 = 25 4p2 – 12p – 16 = 0 p2 – 3p – 4 = 0 (p + 1)(p – 4) = 0p = –1 or p = 4

9. 2x2 + x = 2p(2x – p) 2x2 + x – 4px + 2p2 = 0 2x2 + (1 – 4p)x + 2p2 = 0 b2 – 4ac . 0 (1 – 4p)2 – 4(2)(2p2) . 0 1 – 8p + 16p2 – 16p2 . 0 1 . 8p

p , 18

Gantikan x = –3

10. x2 – 6x + 4 = x2 – 6x + 1– 62 2

2 – 1– 6

2 22 + 4

= (x – 3)2 – (–3)2 + 4 = (x – 3)2 – 9 + 4 = (x – 3)2 – 5

x2 – 6x + 4 = 0Hence, (x – 3)2 – 5 = 0 (x – 3)2 = 5 x – 3 = ± 5 x = 3 ± 5 = 5.236 or 0.7639

11. x = 14 , x = – 2

5 4x – 1 = 0 , 5x + 2 = 0

(4x – 1)(5x + 2) = 0 20x2 + 8x – 5x – 2 = 0 20x2 + 3x – 2 = 0

12. 3x2 + 4px – 2p = 0

x2 + 4p3 x – 2p

3 = 0

Let the roots be a and 3a.

a + 3a = – 4p3

4a = – 4p3

a = – p3 …… 1

a(3a) = – 2p3

3a2 = – 2p3

a2 = – 2p3 …… 2

Substitute 1 into 2.

1– p3 2

2 = – 2p

9

p2

9 = – 2p9

p2 = –2p p2 + 2p = 0 p(p + 2) = 0 p = –2



13. 5x2 – (2h + 3)x – k – 4 = 0

x2 – 2h + 35 x – (k + 4)

5 = 0

Sum of roots = 45 + (–3) = 2h + 3

5 – 11

5 = 2h + 35

2h + 3 = –11 h = –7

Product of roots = 45 (–3) = – (k + 4)

5

– 125 = – (k + 4)

5 k + 4 = 12 k = 8

Paper 2

1. (a) Total area = x2 + (x + 2)2

= x2 + x2 + 4x + 4 340 = 2x2 + 4x + 4 2x2 + 4x – 336 = 0 x2 + 2x – 168 = 0 (x – 12)(x + 14) = 0 x = 12

(b) Total perimeter = 4x + 4(x + 2) = 4(12) + 4(12 + 2) = 104 cm

2. (a) x2 + 3(4x + p) = 0 x2 + 12x + 3p = 0 Sum of roots = r + 3r = –12 4r = –12 r = –3 Product of roots = r(3r) = 3p p = r2

= (–3)2 = 9

(b) r + 7 = –3 + 7 = 4 r – 2 = –3 – 2 = –5 Sum of roots = 4 + (–5) = –1 Product of roots = 4(–5) = –20 Hence, the quadratic equation is

x2 + x – 20 = 0.

3. (a) x(x – 4) = 2k – 5 x2 – 4x = 2k – 5 x2 – 4x – 2k + 5 = 0

Since a ≠b, different roots, b2 – 4ac . 0 (–4)2 – 4(1)(–2k + 5) . 0 16 + 8k – 20 . 0 8k . 4

k . 12

(b) x2 – 4x – 2k + 5 = 0 a + b = – (– 4) = 4 ab = –2k + 5

3x2 + hx – 2 = 0

a3 + b

3 = – h3

a + b = –h 4 = –h h = – 4

1a3 21 b

3 2 = – 23

ab9 = – 2

3 ab = –6 –2k + 5 = –6 2k = 11

k = 112

4. Let the squares of the two pieces of wire be square A and square B.

Square A:Let the length of each side = x cmPerimeter = 4x cmArea = x2

Square B:Perimeter = (92 – 4x) cm

Length of each side = 92 – 4x4 = (23 – x) cm

Area = (23 – x)2

Total area = 265 x2 + (23 – x)2 = 265 x2 + 529 – 46x + x2 – 265 = 0 2x2 – 46x + 264 = 0 x2 – 23x + 132 = 0 (x – 11)(x – 12) = 0 x = 11 or x = 12

Gantikan ab = –2k + 5



When x = 11,The perimeter of square A = 4(11) = 44 cmThe perimeter of square B = 92 – 4(11) = 48 cm

When x = 12,The perimeter of square A = 4(12) = 48 cmThe perimeter of square B = 92 – 4(12) = 44 cmTherefore, the lengths of the two pieces of the cut wire are 44 cm and 48 cm.

5. Let the usual walking speed of Jeffrey be x m per minute.Time taken when walking at usual speed = 600

xTime taken when walking at increased speed = 600

x + 5Difference in time taken,

600x

– 600x + 5 = 20

30x

– 30x + 5 = 1

30(x + 5) – 30x = x(x + 5) 150 = x2 + 5x x2 + 5x – 150 = 0 (x – 10)(x + 15) = 0 x = 10 or x = –15 (Not acceptable)Hence, the usual walking speed of Jeffrey is 10 m per minute.

6. (a) 2x2 + hx + k = 0 When x = 5, 2(5)2 + h(5) + k = 0 50 + 5h + k = 0 5h + k = –50 …… 1 When x = –2, 2(–2)2 + h(–2) + k = 0 8 – 2h + k = 0 –2h + k = –8 …… 2 1 – 2: 7h = –42 h = –6 Substitute h = –6 into 1. 5(–6) + k = –50 –30 + k = –50 k = –20

(b) 2x2 – 6x – 20 = p 2x2 – 6x – 20 – p = 0 2x2 – 6x – (20 + p) = 0 b2 – 4ac , 0 (–6)2 – 4(2)[–(20 + p)] , 0 36 + 160 + 8p , 0 8p , –196 p , –24.5


Quadratic FunctionsFungsi Kuadratik3

CHAPTER

1. Quadratic function (Yes/ No)Fungsi kuadratik (Ya/ Bukan)

Values of a, b and c (If the function is a quadratic function)

Nilai-nilai a, b dan c (Jika fungsi ialah fungsi kuadratik)

(a) Yes a = 3, b = –2, c = 6

(b) No –

(c) Yes g(x) = –3x2 + 5xa = –3, b = 5, c = 0

(d) Yes h(x) = –x2 – 2x + 3a = –1, b = –2, c = 3

(e) Yes f(x) = –5x2 + 7a = –5, b = 0, c = 7

2. Coefficient of x2 (positive or negative)

Pekali x2 ( positif atau negatif)

Shape of graph (maximum or minimum point)

Bentuk Graf (titik maksimum atau titik minimum)

(a) a = –3 < 0 (negative)

A parabola with a maximum point.

(b) g(x) = –x2 – 3x + 10

a = –1 < 0 (negative)

A parabola with a maximum point.

(c) h(x) = 2x2 – 15x – 8 a = 2 . 0 (positive)

A parabola with a minimum point.

(d) a = 5 > 0 ( positive)

A parabola with a minimum point.

Additional Mathematics Form 4 Chapter 3 Quadratic Functions


3. (a) Two equal roots.(b) No real roots.(c) Two equal roots.(d) No real roots.(e) Two different roots.

4. Type of the rootsKeadaan punca

Position of the graphKedudukan graf

(a) b2 – 4ac = (4)2 – 4(–3)(2) = 40 . 0

Therefore, f(x) = 0 has two different roots.

a = –3 , 0, the shape of f(x) is a parabola, , with a maximum point and intersects the x-axis at two points.

x

(b) b2 – 4ac = (–1)2 – 4(1)( 14 )

= 0

Therefore, g(x) = 0 has two equal roots.

a = 1 . 0, the shape of g(x) is a parabola, , with a minimum point and touches the x-axis at one point.

x

(c) b2 – 4ac = (3)2 – 4(–4)(–1) = –7 , 0

Therefore, h(x) = 0 has no real roots.

a = –4 , 0, the shape of h(x) is a parabola, , with a maximum point and does not meet the x-axis.

x

(d) f(x) = –25x2 + 30x – 9

b2 – 4ac = (30)2 – 4(–25)(–9) = 0

Therefore, f(x) = 0 has two equal roots.

a = –25 , 0, the shape of f(x) is a parabola, , with a maximum point and touches the x-axis at one point.

x

(e) b2 – 4ac = (11)2 – 4(5)(7) = –19 , 0

Therefore, g(x) = 0 has no real roots.

a = 5 . 0, the shape of g(x) is a parabola, , with a minimum point that does not meet the x-axis.

x



5. (a) f(x) = x2 – 4x + 7

= x2 – 4x + 1–42 2

2 – 1–4

2 22 + 7

= (x – 2)2 – (–2)2 + 7 = (x – 2)2 + 3

a = 1 . 0

Therefore, the minimum value of f(x) is 3 when x = 2.

(b) f(x) = x2 – 2x + 6

= x2 – 2x + 1–22 2

2 – 1–2

2 22 + 6

= (x – 1)2 – (–1)2 + 6 = (x – 1)2 + 5

a = 1 . 0

Therefore, the minimum value of f(x) is 5 when x = 1.

(c) g(x) = x2 + 3x – 1

= x2 + 3x + 1 32 2

2 – 1 3

2 22 – 1

= 1x + 32 2

2 – 1 3

2 22 – 1

= 1x + 32 2

2 – 13

4

a = 1 . 0

Therefore, the minimum value of g(x)

is – 134 when x = – 32 .

(d) f(x) = –2x2 – 4x + 5

= –23x2 + 2x – 52 4

= –23x2 + 2x + 1 22 2

2 – 1 2

2 22 – 5

2 4 = –23(x + 1)2 – (1)2 – 5

2 4 = –23(x + 1)2 – 7

2 4 = –2(x + 1)2 + 7

2

a = –2 , 0

Therefore, the maximum value of

f(x) = 72 when x = –1.



6. (a) f(x) = x2 – 4x – 5

a = 1 . 0, the shape of f(x) is <.

b2 – 4ac = (–4)2 – 4(1)(–5) = 36 . 0

Therefore, the graph of f(x) is a parabola with a minimum point and intersects the x-axis at two points.

f(x) = x2 – 4x – 5

= x2 – 4x + 1–42 2

2 – 1–4

2 22 – 5

= (x – 2)2 – (–2)2 – 5 = (x – 2)2 – 9

Therefore, the minimum value of f(x) = –9 when x = 2.

When f(x) = 0, x2 – 4x – 5 = 0 (x + 1)(x – 5) = 0

x + 1 = 0 or x – 5 = 0 x = –1 x = 5

When x = 0 , f(0) = –5

f (x) = x2 –4x –5

x

f(x)

0–1

–5

(2,–9)

5

(b) f(x) = –x2 – 2x – 1 a = –1 , 0, the shape of f(x) is >.

b2 – 4ac = (–2)2 – 4(–1)(–1) = 0

Therefore, the graph of f(x) is a parabola with a maximum point and touches the x-axis at one point.

f(x) = –[x2 + 2x + 1]

= –[x2 + 2x + 1 22 2

2 – 1 2

2 22 + 1]

= –[(x + 1)2 – (1)2 + 1] = –[(x + 1)2 – 0] = –(x + 1)2

Therefore, the maximum value of f(x) = 0 when x = –1.

When x = 0, f(0) = –1

x

f(x)

0

–1

(–1,0)

f (x) = –x2 –2x –1



(c) f(x) = 4x2 – 6x + 3

a = 4 . 0, the shape of f(x) is <.

b2 – 4ac = (–6)2 – 4(4)(3) = –12 , 0

Therefore, the graph of f(x) is a parabola with a minimum point and does not touch the x-axis.

f(x) = 4x2 – 6x + 3

= 43x2 – 32 x + 3

4 4 = 43x2 – 3

2 x + 1–34 2

2 – 1–3

4 22 + 3

4 4 = 431x – 3

4 22 + 3

16 4 = 41x – 3

4 22 + 3

4

Therefore, the minimum value of f(x)

= 34 when x = 3

4 .

When x = 0, f(0) = 3

x

f(x)

0

3

( , )34

34

f (x) = 4x2– 6x + 3

7. (a) x2 + 2x , 15

x2 + 2x – 15 , 0 (x + 5)(x – 3) , 0

x3–5

Therefore, –5 , x , 3.

(b) 3x(x + 1) > 24 – 3x

3x2 + 3x > 24 – 3x 3x2 + 6x – 24 > 0 x2 + 2x – 8 > 0 (x – 2)(x + 4) > 0

x2–4

Therefore, x < –4 or x > 2.

(c) 7 – 2x > (x + 4)2

7 – 2x > x2 + 8x + 16 x2 + 10x + 9 < 0 (x + 1)(x + 9) < 0

x–9 –1

Therefore, –9 < x < –1.

(d) x(8 – x) , 12

8x – x2 – 12 , 0 x2 – 8x + 12 . 0 (x – 2)(x – 6) . 0

x62

Therefore, x , 2 or x . 6.



SPM Practice 3

Paper 1

1. (a) f(x) = 1x – 52 2

2 – 9

4 By comparing f(x) = (x + p)2 + q,

p = 52 , q = – 9

4 The coordinates of the minimum point

= 1 52 , – 9

4 2

(b) x = 52

(c)

x41

The range is x , 1 or x . 4.

2. f(x) = –1x + 52 22 + 2p

By comparing with f(x) = –(x + a)2 + b,The maximum point = (–a, b)

(a) k = –a = – 52

(b) 2p = 16 p = 8

(c) f(0) = q

–10 + 52 2

2 + 16 = q

q = 9 34

3. f(x) = –4x2 + 9x + 12 –4x2 + 9x + 12 < 3 4x2 – 9x – 9 > 0 (4x + 3)(x – 3) > 0

x3– 34

Therefore, x < – 34 or x > 3.

4. (a) f(x) = (x – 4)2 + 2k – 1 Minimum value of f(x) = 2k – 1 2k – 1 = 9 k = 5

(b) x = 4

5. x(5x – 1) , x2 – 7x + 10 5x2 – x , x2 – 7x + 10 4x2 + 6x – 10 , 0 2x2 + 3x – 5 , 0 (2x + 5)(x – 1) , 0

x1– 52

Therefore, – 52 , x , 1.

6. (a) Since f(x) = hx2 – 12x + 3k has a minimum point, h . 0 Given – 2 , h , 2, h = {–1, 0, 1}, Therefore, h = 1.

(b) f(x) = x2 – 12x + 3k Since f(x) has same root, b 2 – 4ac = 0 (–12)2 – 4(1)(3k) = 0 144 – 12k = 0 12k = 144 k = 12

7. 3x2 + 13x < 10 3x2 + 13x – 10 < 0 (3x – 2)(x + 5) < 0

–5 2—3

Therefore, –5 < x < 23 .



8. (a) f(x) = –2x2 + 4x + 6 = –2[x2 – 2x – 3]

= –23x2 – 2x + 1–22 2

2 – 1–2

2 22 – 34

= –2[(x – 1)2 – 1 – 3] = –2[(x – 1)2 – 4] = –2(x – 1)2 + 8 p = 2 and q = 8

(b) When f(x) = 0, –2x2 + 4x + 6 = 0 2x2 – 4x – 6 = 0 (x + 1)(2x – 6) = 0 x = –1 or x = 3

When x = 0, f(0) = 6

x

y

0 3

6

–1

(1,8)

9. (a) From the graph, the maximum point is (5, 0). By comparing f(x) = –(x + p)2 + q The maximum point = (–p, q) p = –5 and q = 0

f(x) = –(x – 5)2 + 0 When x = 0 f(0) = –(0 – 5)2

= –25 Therefore, (0, –25) is the intersection point

at y-axis, k = –25.

(b) x = 5

10. (a) f(x) = 18x – 3x2

= –3x2 + 18x = –3[x2 – 6x]

= –33x2 – 6x + 1–62 2

2 – 1–6

2 22

4 = –3[(x – 3)2 – 9] = –3(x – 3)2 + 27 The coordinates of the maximum point

= (3, 27).

(b) When f(x) = 0 –3x2 + 18x = 0 –3x(x – 6) = 0 x = 0 or x = 6

x60

The range is 0 , x , 6.

11. (a) f(x) = –(x + h)2 + k The maximum value of f(x) = k

when x = –h. The axis of symmetry is x = –h f(x) = –(x – 2)2 – 4

(b)

x

–8

0 (2,–4)

f(x)



12. (a) p(x) = –3x2 + 48x + 12 = –3[x2 – 16x – 4]

= –3[x2 – 16x + 1–162 2

2 – 1–16

2 22 – 4

= –3[(x – 8)2 – (–8)2 – 4] = –3[(x – 8)2 – 68] = –3(x – 8)2 + 204

(b) (i) x = 8, 8 units must be sold per month tomaximiseprofit.

(ii) Maximumprofit = 204 × 1 000 = RM204 000

13. (a) s(t) = –5t2 + 60t = –5[t2 – 12t]

= –53t2 – 12t + 1–122 2

2 – 1–12

2 22

4 = –5[(t – 6)2 – 36] = –5(t – 6)2 + 180 The time for the projectile to reach the

maximum height is 6 s. The time for the projectile to hit the ground, 6 × 2 = 12 s

(b) Maximum altitude = 180 m


Simultaneous EquationsPersamaan Serentak4

CHAPTER

1. (a) x + y = 1 …… 1 x2 – xy – y2 = 1 …… 2

Rewrite 1 in the equivalent form, y = 1 – x …… 3

Substitute 3 into 2, x2 – x(1 – x) – (1 – x)2 = 1 x2 – x + x2 – (1 – 2x + x2) = 1 x2 – x + x2 – 1 + 2x – x2 – 1 = 0 x2 + x – 2 = 0 (x – 1)(x + 2) = 0 x = 1 or –2

Substitute x = 1 into 3, y = 1 – 1 = 0

Substitute x = –2 into 3, y = 1 – (–2) = 3

\ The solutions are x = 1, y = 0; x = –2, y = 3.

(b) x + y = 3 …… 1 x2 + y2 = 17 …… 2

Rewrite 1 in the equivalent form, x = 3 – y …… 3

Substitute 3 into 2, (3 – y)2 + y2 = 17 9 – 6y + y2 + y2 = 17 2y2 – 6y – 8 = 0 y2 – 3y – 4 = 0 (y – 4)(y + 1) = 0 y = 4 or –1

Substitute y = 4 into 3, x = 3 – 4 = –1

Substitute y = –1 into 3, x = 3 – (–1) = 4

\ The solutions are x = –1, y = 4; x = 4, y = –1.

Additional Mathematics Form 4 Chapter 4 Simultaneous Equations


(c) 2x – y = 2 …… 1 x2 + 2y2 = 12 …… 2

Rewirte 1 in the equivalent form, y = 2x – 2 …… 3

Substitute 3 into 2, x2 + 2(2x – 2)2 = 12 x2 + 2(4x2 – 8x + 4) = 12 x2 + 8x2 – 16x + 8 – 12 = 0 9x2 – 16x – 4 = 0 (x – 2)(9x + 2) = 0 x = 2 or – 2

9 Substitute x = 2 into 3, y = 2(2) – 2 = 2

Substitute x = – 29 into 3,

y = 21– 29 2 – 2

= – 229

\ The solutions are x = 2, y = 2;

x = – 29 , y = – 22

9 .

(d) 2x + 3y – 1 = 0 …… 1 x2 + 6xy + 5 = 0 …… 2

Rewrite 1 in the equivalent form,

y = 1 – 2x3 …… 3

Substitute 3 into 2,

x2 + 6x 1 1 – 2x3 2 + 5 = 0

x2 + 2x(1 – 2x) + 5 = 0 x2 + 2x – 4x2 + 5 = 0 –3x2 + 2x + 5 = 0 3x2 – 2x – 5 = 0 (3x – 5)(x + 1) = 0

x = 53 or –1

Substitute x = 53 into 3,

y = 1 – 21 5

3 2 3 = – 7

9

Substitute x = –1 into 3,

y = 1 – 2(–1)3 = 1

\ The solutions are x = 53 , y = – 7

9 ;

x = –1, y = 1.



2. (a) x + y – 3 = 0 ……… 1 x2 + 2y2 – 8 = 0 …… 2


Substitute 3 into 2, x2 + 2(3 – x)2 – 8 = 0 x2 + 2(9 – 6x + x2) – 8 = 0 x2 + 18 – 12x + 2x2 – 8 = 0 3x2 – 12x + 10 = 0

x = –(–12) ± (–12)2 – 4(3)(10)2(3)

= 12 ± 246

= 2.816 or 1.184

Substitute x = 2.816 into 3,y = 3 – 2.816 = 0.184

Substitute x = 1.184 into 3,y = 3 – 1.184 = 1.816

Therefore, the solutions are x = 2.816, y = 0.184; x = 1.184, y = 1.816.

(b) 2x + y = 1 ……………… 1 2x2 + y2 – xy – 6 = 0 …… 2

Rewrite 1 in the equivalent form, y = 1 – 2x …… 3

Substitute 3 into 2, 2x2 + (1 – 2x)2 – x(1 – 2x) – 6 = 0 2x2 + 1 – 4x + 4x2 – x + 2x2 – 6 = 0 8x2 – 5x – 5 = 0

x = –(–5) ± (–5)2 – 4(8)(–5)2(8)

= 5 ± 18516

= 1.163 or –0.538Substitute x = 1.163 into 3,y = 1 – 2(1.163) = –1.326

Substitute x = –0.538 into 3,y = 1 – 2(–0.538) = 2.076

Therefore, the solutions are x = 1.163, y = –1.326; x = –0.538, y = 2.076.



(c) x – 2y = 5 …………… 1 x2 + xy2 + 2y = 0 …… 2

Rewrite 1 in the equivalent form, x = 5 + 2y …… 3

Substitute 3 into 2, (5 + 2y)2 + (5 + 2y)y + 2y = 0 25 + 20y + 4y2 + 5y + 2y2 + 2y = 0 6y2 + 27y + 25 = 0

y = –27 ± (27)2 – 4(6)(25)2(6)

= –27 ± 12912

= –1.304 or –3.196

Substitute y = –1.304 into 3,x = 5 + 2(–1.304) = 2.392

Substitute y = –3.197 into 3,x = 5 + 2(–3.197) = –1.392

Therefore, the solutions are x = 2.392, y = –1.304; x = –1.392, y = –3.197.

3. (a)

y m

x m

Let the length and breadth of the rectangle be x cm and y cm respectively.

2x + 2y = 42 …… 1 x2 + y2 = 152 …… 2


y = 42 – 2x2

= 21 – x …… 3

Substitute 3 into 2, x2 + (21 – x)2 = 225 x2 + 441 – 42x + x2 = 225 2x2 – 42x + 216 = 0 x2 – 21x + 108 = 0 (x – 9)(x – 12) = 0

x – 9 = 0 or x – 12 = 0 x = 9 x = 12



Hence, the length is 12 cm and the breadth is 9 cm.



(b) (x + 3) + (y – 1) + 13 = 30 x + y = 15 y = 15 – x … 1 (x + 3)2 + (y – 1)2 = 132 x2 + 6x + 9 + y – 2y + 1 = 169 x2 + y2 + 6x – 2y – 159 = 0 … 2

Subsitute 1 into 2. x2 + (15 – x)2 + 6x – 2(15 – x) – 159 = 0

x2 + (225 – 30x + x2) + 6x – 30 + 2x – 159 = 0 2x2 – 22x + 36 = 0 x2 – 11x + 18 = 0 (x – 2)(x – 9) = 0 x = 2 or 9 Substitute x = 2 into 1. y = 15 – 2 = 13 Substitute x = 9 into 1. y = 15 – 9 = 6 Therefore, x = 9, y = 6; x = 2, y = 13

(c) Let the two numbers be x and y. xy = –48 …… 1

1x

+ 1y

= 16 …… 2


y + x = xy6 …… 3


y + x = – 486

y + x = –8 y = –x – 8 …… 4

Substitute 4 into 1, x(–x – 8) = – 48 –x2 – 8x = – 48 x2 + 8x – 48 = 0 (x + 12)(x – 4) = 0 x = 4 or x = –12

Substitute x = 4 into 4, y = – 4 – 8 = –12

Substitute x = –12 into 4, y = –(–12) – 8 = 4

The two numbers are 4 and –12.

SPM Practice 4

Paper 2

1. 2x + y – 1 = 0 ………… 1 2x2 – y2 – 3y + 7 = 0 …… 2

Rewrite 1 in the equivalent form,y = 1 – 2x …… 3

Substitute 3 into 2, 2x2 – (1 – 2x)2 – 3(1 – 2x) + 7 = 0 2x2 – (1 – 4x + 4x2) – 3 + 6x + 7 = 0 2x2 – 1 + 4x – 4x2 – 3 + 6x + 7 = 0 –2x2 + 10x + 3 = 0

x = –10 ± (10)2 – 4(–2)(3)2(–2)

= –10 ± 124–4

= –0.284 or 5.284

Substitute x = –0.284 into 3,y = 1 – 2(–0.284) = 1.568

Substitute x = 5.284 into 3,y = 1 – 2(5.284) = –9.568

Therefore, the solutions are x = –0.284, y = 1.568; x = 5.284, y = –9.568



2. 3x + y = 4 …………… 1 5x2 + xy + y = 10 ……… 2


Substitute 3 into 2, 5x2 + x(4 – 3x) + (4 – 3x) = 10 5x2 + 4x – 3x2 + 4 – 3x – 10 = 0 2x2 + x – 6 = 0 (x + 2)(2x – 3) = 0 x = –2 or 3

2

Substitute x = –2 into 3,y = 4 – 3(–2) = 10Substitute x = 3

2 into 3,

y = 4 – 31 32 2

= – 12

Therefore, the solutions are x = –2, y = 10;x = 3

2 , y = – 12 .

3. x + 2y = 1 …………… 1 x2 – 2xy – 3y2 = 0 …… 2

Rewrite 1 in the equivalent form,x = 1 – 2y …… 3

Substitute 3 into 2, (1 – 2y)2 – 2y(1 – 2y) – 3y2 = 0 1 – 4y + 4y2 – 2y + 4y2 – 3y2 = 0 5y2 – 6y + 1 = 0 (y – 1)(5y – 1) = 0 y = 1 or 1

5

Substitute y = 1 into 3,x = 1 – 2(1) = –1Substitute y = 1

5 into 3,

x = 1 – 21 15 2

= 35

Therefore, the solutions are x = –1, y = 1;x = 3

5 , y = 15 .

4. y – 4x – 2 = 0 ……………… 1 4x2 + 3y2 – 5xy – 16 = 0 …… 2

Rewrite 1 in the equivalent form,y = 2 + 4x …… 3

Substitute 3 into 2, 4x2 + 3(2 + 4x)2 – 5x(2 + 4x) – 16 = 0 4x2 + 3(4 + 16x + 16x2) – 10x – 20x2 – 16 = 0 4x2 + 12 + 48x + 48x2 – 10x – 20x – 16 = 0 32x2 + 38x – 4 = 0 16x2 + 19x – 2 = 0

x = –19 ± (19)2 – 4(16)(–2)2(16)

= –19 ± 48932

= 0.097 or –1.285

Substitute x = 0.097 into 3,y = 2 + 4(0.097) = 2.388Substitute x = –1.285 into 3,y = 2 + 4(–1.285) = –3.140

Therefore, the solutions are x = 0.097, y = 2.388; x = –1.285, y = –3.140.



5. 3x + 2y – 1 = 0 …… 1 x2 + y2 – 2xy – 6 = 0 …… 2

Rewirte 1 in the equivalent form,

y = 1 – 3x2 …… 3


x2 + 1 1 – 3x2 22 – 2x1 1 – 3x

2 2 – 6 = 0

x2 + 1 1 – 6x + 9x2

4 2 – x + 3x2 – 6 = 0

4x2 + 1 – 6x + 9x2 – 4x + 12x2 – 24 = 0 25x2 – 10x – 23 = 0

x = –(–10) ± (–10)2 – 4(25)(–23)2(25)

= 10 ± 2400250(25)

= 1.180 or –0.780

Substitute x = 1.180 into 3,

y = 1 – 3(1.180)2

= –1.270

Substitute x = –0.780 into 3,

y = 1 – 3(–0.780)2

= 1.670

Therefore, the solutions are x = 1.180, y = –1.270; x = –0.780, y = 1.670.

6. 3x + y = 2 …………… 1 2x – 3y – 4xy = 0 …… 2


Substitute 3 into 2, 2x – 3(2 – 3x) – 4x(2 – 3x) = 0 2x – 6 + 9x – 8x + 12x2 = 0 12x + 3x – 6 = 0 4x2 + x – 2 = 0

x = –1 ± (1)2 – 4(4)(–2)2(4)

= –1 ± 338

= 0.593 or –0.843

Substitute x = 0.593 into 3,y = 2 – 3(0.593) = 0.221Substitute x = –0.843 into 3,y = 2 – 3(–0.843) = 4.529

Therefore, the solutions are x = 0.593, y = 0.221; x = –0.843, y = 4.529.



7.x cm

y cm

Hypotenuse

x + y = 46 …… 1

12 xy = 240 …… 2

Rewrite 1 in the equivalent form,y = 46 – x …… 3


12 x(46 – x) = 240

46x – x2 = 480 x2 – 46x + 480 = 0 (x – 30)(x – 16) = 0 x = 16 or 30

Substitute x = 16 into 3,y = 46 – 16 = 30Substitute x = 30 into 3,x = 46 – 30 = 16

Hypotenuse = x2 + y2 = 162 + 302 = 34 cm

8. Let the two numbers be x and y and y . x.

x2 + y2 = 1254 …… 1

y = 2x …… 2


x2 + (2x)2 = 1254

4(x2 + 4x2) = 125 20x2 = 125

x2 = 254

x = ±2.5

Since x . 0, substitute x = 2.5 into 3,

y = 21 52 2

= 5Therefore, the two numbers are 2.5 and 5.

9.

x m

y m

Perimeter = 92 2x + 2y = 92 …… 1 xy = 480 …… 2


x = 480y

…… 3


2( 480y

) + 2y = 92

480y

+ y = 46

480 + y2 = 46y y2 – 46y + 480 = 0 (y – 30)(y – 16) = 0y – 30 = 0 or y – 16 = 0 y = 30 y = 16

Substitute y = 30 into 3,

x = 48030

= 16

Substitute y = 16 into 3,

x = 48016

= 30

Therefore, the length = 30 m and the breadth 16 m.



10.

8 cm

y cm

x cm

2x + 2y + 32 = 80 x + y = 24 ……… 1 2(8x) + 2(8y) + xy = 512 16x + 16y + xy = 512 ……… 2 From equation 1, x = 24 – y …… 3

Substitute equation 3 into equation 2, 16(24 – y) + 16y + y(24 – y) = 512 384 – 16y + 16y + 24y – y2 = 512 y2 – 24y + 128 = 0 (y – 16)(y – 8) = 0 y – 16 = 0 or y – 8 = 0 y = 16 y = 8

Substitute y = 16 into 3,x = 24 – 16 = 8

Substitute y = 8 into 3,x = 24 – 8 = 16

Therefore, the length = 16 cm and the breadth = 8 cm.

11. x – y = 12 …… 1 x2 + y2 = 1 130 …… 2

Rewrite 1 in the equivalent form, y = x – 12 …… 3

Substitute 3 into 2, x2 + (x – 12)2 = 1 130 x2 + x2 – 24x + 144 = 1 130 2x2 – 24x – 986 = 0 x2 – 12x – 493 = 0 (x – 29)(x + 17) = 0 x – 29 = 0 or x + 17 = 0 x = 29 x = –17 (rejected)Since x . 0, \ x = 29


Therefore, the two numbers are 29 and 17.



12. 2(2x + 20) + 2(y + 10) = 88 2x + 20 + y + 10 = 44 2x + y + 30 = 44 2x + y = 44 …… 1

(2x + 20)(y + 10) = 420 2xy + 20x + 20y + 200 = 420 20x + 20y + 2xy = 220 10x + 10y + xy = 110 …… 2

Rewrite 1 in the equivalent form, y = 14 – 2x …… 3 Substitute 3 into 2; 10x + 10(14 – 2x) + x(14 – 2x) = 110 10x + 140 – 20x + 14x – 2x2 = 110 2x2 – 4x – 30 = 0 x2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = –3 or x = 5 Since x . 0, therefore x = 5.

Substitute x = 5 into 3, y = 14 – 2(5) = 4

Therefore, the length = 2x + 20 = 2(5) + 20 = 30 cm and the breadth = y + 10 = 4 + 10 = 14 cm

13. y = 20 – 3x …… 1 y = 2x2 – 4x – 7 …… 2

Substitute 1 into 2, 20 – 3x = 2x2 – 4x – 72x2 – x – 27 = 0

x = –(–1) ± (–1)2 – 4(2)(–27)2(2)

= 1 ± 2174

= 1 – 2174

or 1 + 2174

= –3.4327 or 3.9327 ≈ –3.43 or 3.93

Since x = h and h . 0, \ x = 3.93

Substitute x = 3.9327 into 1, y = 20 –3(3.9327) = 8.2019 ≈ 8.20

Therefore, h = x = 3.93 and k = y = 8.20.



14. 2x + y = 36 …… 1

y cm

x cm

122 + 1 y2 2

2 = x2

144 + y2

4 = x2

576 + y2 = 4x2 4x2 – y2 = 576 ……2


Substitute 3 into 2, 4x2 – (36 – 2x)2 = 5764x2 – (1 296 – 144x + 4x2) = 576 –1 296 + 144x = 576 144x = 1 872 x = 13

Substiture x = 13 into 3, y = 36 – 2(13) = 10

Therefore, the lengths of all the sides of the triangle are 13 cm, 13 cm and 10 m.

15. xy = 450 …… 1

(x + 5)(y – 3) = 450 xy – 3x + 5y – 15 = 450 xy – 3x + 5y = 465 …… 2


y = 450x

…… 2


x1450x 2 – 3x + 51450

x 2 = 465

450 – 3x + 2 250x = 465

450x – 3x2 + 2 250 = 465x 3x2 + 15x – 2 250 = 0 x2 + 5x – 750 = 0 (x – 25)(x + 30) = 0 x = 25 or x = –30Since x . 0, therefore x = 25.


y = 45025

= 18

Therefore, the length = 25 m and the breadth = 18 m.



16. x + y = 2 …… 1 x2 – y2 = 3 …… 2


Substitute 3 into 2, x2 – (2 – x)2 = 3 x2 – (4 – 4x + x2) = 3 – 4 + 4x = 3 4x = 7

x = 74


y = 2 – 74

= 14

Therefore, the numbers are 74 and 1

4 .

17. Let x = radius of circle I y = radius of circle II

2πx + 2πy = 20π 2π(x + y) =20π x + y = 10 …… 1

πx2 +πy2 = 52π π(x2 + y2) =52π x2 + y2 = 52 …… 2


Substitute 3 into 2, x2 + (10 – x)2 = 52 x2 + 100 – 20x + x2 = 52 2x2 – 20x + 48 = 0 x2 – 10x + 24 = 0 (x – 4)(x – 6) = 0 x = 4 or x = 6

Substitute x = 4 into 3,y = 10 – 4 = 6


Therefore, the radii of the two circles are 4 cm and 6 cm.



18.

x cm

y cm

4 cm

Let x = length of the cuboid y = width of the cuboid

4xy = 120 xy = 30 …… 1

4x + 4y + 4(4) = 60 x + y + 4 = 15 x + y = 11 …… 2


Substitute 3 into 1, x(11 – x) = 30 11x – x2 = 30 x2 – 11x + 30 = 0 (x – 5)(x – 6) = 0 x = 5 or x = 6

Substitute x = 5 into 3,y = 11 – 5 = 6


Therefore, length = 6 cm and width = 5 cm.

19. 2x = 2 + y …… 1 (x + y)2 = 448 + (x – y)2 …… 2

Rewrite 1 in the equivalent form,y = 2x – 2 …… 3

Substitute 3 into 2, (x + 2x – 2)2 = 448 + (x – 2x + 2)2 (3x – 2)2 = 448 + (2 – x)2 9x2 – 12x + 4 = 448 + 4 – 4x + x2 8x2 – 8x – 448 = 0 x2 – x – 56 = 0 (x – 8)(x + 7) = 0 x = 8 or x = –7Since it is a positive number, therefore, x = 8.

Substitute x = 8 into 3,y = 2(8) – 2 = 14

Therefore, the two numbers are 8 and 14.

Additional Mathematics Form 4 Chapter 5 Indices and Logarithms


1. (a) 24 = 2 × 2 × 2 × 2 = 16

(b) 80 = 1

(c) 3–3 = 133

= 127

(d) 645—6 = (26)

5—6

= 25

= 32

(e) 8– 2—

3 = (23)– 2—

3

= 2–2

= 14

3 3 3(f) —— = ——— = — 4

3—2 (22)

3—2 23

= 38

2. (a) 32 × 33 = 32 + 3

= 35

= 243

8(b) —– 43

8 = ————— 4 × 4 × 4 1 = — 8

(c) (125)–1 × (25)2

1 = —– × (52)2

125 1 = — × 54

53

= 54 – 3

= 5

2 3(d) 1—23 ÷ 1—2–3 3 2 2 2 = 1—23 ÷ 1—23

3 3 2 = 1—23 – 3

3 2 = 1—20

3 = 1

(e) (6–3 × 62)4 = (6–1)4

= 6–4

1 = —– 64

1 = ——–– 1 296

(f) (33)2 × 20 = 36 × 1 = 729

3. (a) r4s5 ÷ 2r3s–2

1 = —r4 – 3s5 – (–2) 2 1 = —rs7 2

(b) m3n4 ÷ 3m2n–1

1 = —m3 – 2n4 – (–1) 3 1 = —mn5 3

Indices and LogarithmsIndeks dan Logaritma5

CHAPTER



(c) 52a – 1 ÷ (25)3a – 1

= 52a – 1 ÷ (52)3a – 1

= 52a – 1 ÷ 56a – 2

= 52a – 1 – 6a + 2

= 51 – 4a

(d) 32x – 5 ÷ 92 – 3x

= 32x – 5 ÷ 32(2 – 3x)

= 32x – 5 – 4 + 6x

= 38x – 9

2p + 3 × 82p – 3(e) —————– 4p + 5

2p + 3 × 23(2p – 3) = ——————– 22(p + 5)

= 2p + 3 + 6p – 9 – 2p – 10

= 25p – 16

(f) 82n + 1 + 26n

= (23)2n + 1 + 26n

= 26n + 3 + 26n

= 26n(23 + 1) = 9(26n)

4. (a) log2 m = 4

(b) logr m = s

(c) 5 = log3 243

(d) 5 = logy z

(e) 5 = a4

(f) q = pr

5. (a) log3 x = 2 x = 32

= 9

(b) log100 x = 1 x = 1001

= 100

(c) log2 x = –1 x = 2–1

= 12

(d) log9 x = 1—2 1 — x = 9 2 = 3

2(e) log1 000 x = — 3 2 — x = (1 000) 3

2 — = (103) 3

= 102

= 100

6. (a) log10 1 = 0

(b) log10 10 = 1

(c) log10 34 = 1.908

(d) log10 0.153 = –0.8153

7. (a) x = 100 000

(b) x = 24.38

(c) x = 2.858

(d) x = 0.07079

8. (a) log3 60 = log3(2 × 2 × 3 × 5) = log3 2 + log3 2 + log3 3 + log3 5 = 2(0.6309) + 1 + 1.465 = 3.7268

12(b) log3 2.4 = log3 — 5 = log3 12 – log3 5 = log3 (3 × 2 × 2) – log3 5 = log3 3 + 2 log3 2 – log3 5 = 1 + 2(0.6309) – 1.465 = 0.7968

8(c) log3 —— = log3 8 – log3 5 5 1 = log3 2

3 – — log3 5 2 1 = 3(0.6309) – — (1.465) 2 = 1.1602



9. (a) log2 5 – log2 20 5 = log2 –— 20 1 = log2 — 4 = log2 2

–2

= –2

(b) log4 144 – 2 log4 3 144 = log4 1–—2 9 = log4 16 = log4 4

2

= 2 log4 4 = 2

(c) log5 100 + log5 2 – log5 8 100 × 2 = log5 1———–2 8 = log5

25 = log5

52

= 2

(d) log6 18 – log6 2 – 2 log6 3 = log6 18 – log6 2 – log6 3

2

= log6 1 182 × 3 × 3 2

= log6 1 = 0

log8 132

+ 3 log8 4 (e) ———————— log8 64

log8 132

+ log8 43

= ———————– log8 2

6

log8 1 132

× 432 = ——————— 6 log8 2 log8 2 = ———— = 1

6 6 log8 2

10. (a) 3 log2 x – 2 log2 3x x3 = log2 1————2 3x × 3x x = log2 — 9

1 (b) — logp 81m – 3 logp b 2

= logp 81m – logp b3

= logp 9 m – logp b3

9 m = logp1——2 b3

a b(c) logx — + 2 logx —– – logx a3b b 2a

a b2 — × —– b 4a2 = logx(————–) a3b 1 = logx —– 4a4

11. log10 2.1(a) log5 2.1 = ———– log10 5 = 0.4610

log10 3.147(b) log2 3.147 = ————— log10 2

= 1.654

5 log101—2 5 7(c) log2 — = ———— 7 log10 2 = –0.4854

log10 1.285(d) logw3 1.285 = ————– log10 3 = 0.4565

log10 4.247(e) log2 4.247 = ————— log10 2 = 2.086



12. log5 70(a) log3 70 = ——— log5 3 log5 (2 × 5 × 7) = ——————— log5 3 log5 2 + log5 5 + log5 7 = ————–————––– log5 3 0.4307 + 1 + 1.209 = ————–———— 0.6826 = 3.867

log5 63(b) logw3 63 = ———– log5 w3 log5 (3 × 3 × 7) = —————–—– 1 — log5 3 2

2 log5 3 + log5 7 = ——————— 1 — log5 3 2 2(0.6826) + 1.209 = ————–———– 1 — (0.6826) 2 = 7.542

log5 12(c) log25 12 = ———– log5 25 log5 (2 × 2 × 3) = ——————— log5 5

2

2 log5 2 + log5 3 = ———————– 2 2(0.4307) + 0.6826 = ———————–— 2 = 0.772

13. (a) logm w243 log3 w243 = ————– log3 m 5 — log3 3 2 = ———— log3 m

52

log3 3 = ———— n 5 = — 2n

(b) log9 3m2

log3 3m2

= ———— log3 9 log3 3 + log3 m

2

= ——––———— log3 3

2

1 + 2 log3 m = ——––——— 2 log3 3 1 + 2n = –——–– 2 = 1

2 + n

(c) log3 3m2

= log3 3 + log3 m2

= 1 + 2 log3 m = 1 + 2n

(d) log3 3m = log3 3 + log3

m

= 1 + log3 m1—2

= 1 + 12

log3 m

= 1 + 12

n

(e) logm m9

– log3 m3

= logm m – logm 9 – (log3 m – log3 3) log3 9 = 1 – ———– – n + 1 log3 m

= 2 – 2n – n

14. (a) 3x – 2 = 81 3x – 2 = 34

x – 2 = 4 x = 6

(b) 25x = 41 + 2x

25x = 22(1 + 2x)

5x = 2 + 4x x = 2

(c) 24x = 128

4x = 164

4x = 4–3

x = –3



(d) 3x

27 = 92x

3x

33 = 32(2x)

3x – 3 = 34x

x – 3 = 4x 3x = –3 x = –1

3 —(e) x 2 = 8

(x1—2 )3 = 23

x1—2 = 2

x = 22

= 4

15. (a) 3x = 0.4 log10 3

x = log10 0.4 x log10 3 = log10 0.4 log10 0.4 x = ———– log10 3 = –0.8340

(b) 31 – 2x = 8 (1 – 2x) log10 3 = log10 8 log10 3 – log10 8 = 2x log10 3 log10 3 – log10 8 x = ——–————— 2 log10 3 = –0.4464

(c) 3.12x = 14.26 2x log10 3.1 = log10 14.26 log10 14.26 x = ————— 2 log10 3.1 = 1.174

(d) 32 – x = 2x

(2 – x) log10 3 = x log10 2 2 log10 3 – x log10 3 = x log10 2 x(log10 2 + log10 3) = 2 log10 3

x = 2 log10 3

log10 2 + log10 3 = 1.226

(e) 3x + 1(4x – 2) = 48 (x + 1) log10 3 + (x – 2) log10 4 = log10 48 x(log10 3 + log10 4) = log10 48 – log10 3 + 2 log10 4

x = log10 48 – log10 3 + 2 log10 4

log10 3 + log10 4 = 2.232

16. (a) log4 (3n – 2) = log4 4 3n – 2 = 4 3n = 6 n = 2

(b) log6 (u – 1)2 = log6 (3 × 12)

(u – 1)2 = 36 u – 1 = ± 36 u – 1 = 6 u = 7 or u – 1 = – 6 (Not accepted because log of a negative numberisundefined.)

(c) log10 32 + log10

(m – 1) = –1 log10

9(m – 1) = log10 10–1

1 9(m – 1) = —– 10 1 m = —– + 1 90 1 = 1—– 90

(d) log5 22 + log5

(4x – 1) = log5 5 + log5

(x + 8) log5

[4(4x – 1)] = log5 [5(x + 8)] 16x – 4 = 5x + 40 11x = 44 x = 4

(e) logp 32 = 3 – log2

24

logp 32 = 3 – 4

logp 32 = –1

9 = p–1

1 p = — 9



SPM Practice 5

Paper 1

1. (2x 4y)5

8x3y2 = 32x20 y5

8x3 y2

= 4x17y3

2. (a) p2 = 16 p = ±4 Since p . 0, p = 4

(b) log16 1 1p 2 = –log16 p

= –log16 4

= –log16 (42)

12

= –log16 1612

= – 12

3. (a) logx q = logx x –2

= –2 logx loga a = 1

= –2

(b) 3 logq x = 3logx q

loga p = 1logp a

= – 32

4. log2 3 + log2 (x – 2) = 1

log2 3(x – 2) = log2 2 loga p + loga q= loga pq

3(x – 2) = 2

x = 23 + 2

= 2 23

5. 8(23x + 4) = 1

23x + 4 = 18

= 2–3 3x + 4 = –3 3x = –7 x = – 7

3

6. 1 + log3 (x – 6) = log3 x log3 x – log3 (x – 6) = 1

log3 x

x – 6 = log3 3

xx – 6 = 3

x = 3x – 18 2x = 18 x = 9

7. log3 x2

y = log3 x

2 – log3 y = 2 log3 x – log3 y = 2t – s

8. log7 4x3 = log7 4 + log7 x3

= logx 4logx 7

+ 3 logx xlogx 7

= logx 22

logx 7 + 3(1)

logx 7

= 2 logx 2logx 7

+ 3logx 7

= 2p + 3q

9. 23x + y = 5 + 8x 23x · 2y = 5 + 23x mn = 5 + m mn – m = 5 m(n – 1) = 5

m = 5n – 1



10. log9 50 = log3 50log3 9

= log3 (2 × 5 × 5)log3 3

2

= log3 2 + log3 5 + log3 52 log3 3

= p + 2q2

= 12 p + q

11. 2x – 2 × 4x = 128 2x – 2 × 22x = 27 2x – 2 + 2x = 2 7 3x – 2 = 7 x = 3

12. log27 a – log3 b = 0

log3 alog3 27

= log3 b

log3 alog3 3

3 = log3 b

log3 a = 3 log3 b log3 a = log3 b

3

a = b3

13. 32x(3) = 18 + 32x 32x(3) – 32x = 18 32x(3 – 1) = 18

32x = 182

32x = 32

x = 1

14. log9 27ab

= log3

27ab

log3 9

= log3 3 3 + log3 a – log3 blog3 3

2

= 3 + m – n2

15. log3 q2

p = log3 q

2 – log3 p = 2 log3 q – log3 p = 2s – r

Paper 2

1. (a) 81p + q

9p = 34(p + q)

32p

= 34p + 4q

32p

= (3p)4 × (3q)4

(3p)2

= a4b4

a2

= a2b4

(b) log9 9a3

b = log9 9a3 – log9 b

= log9 (32)(3p)3 – log9 3

q = log9 3

2 + 3p – log9 3q

= (2 + 3p) log10 3log10 9

– (q) log10 3log10 9

= (2 + 3p)1 12 2 – q1 1

2 2 = 1 + 3

2 p – q2



2. (a) log3 (x + 2) – 2 log9 x2 + 3 log3 x

= log3 (x + 2) – 2 log3 x2

log3 9 + 3 log3 x

= log3 (x + 2) – 2 (2 log3 x)log3 3

2 + 3 log3 x

= log3 (x + 2) – 2 (2 log3 x)2 log3 3

+ 3 log3 x

= log3 (x + 2) – 2 log3 x + 3 log3 x = log3 (x + 2) + log3 x = log3 x(x + 2)

(b) log3 (x + 2) – 2 log3 x2 + 3 log3 x = 1

log3 x(x + 2) = log3 3 x2 + 2x = 3 x2 + 2x – 3 = 0 (x + 3)(x – 1) = 0 x = 1 (x = –3 is not accepted because log of a

negativenumber is undefined.)

3. (a) (i) log5 4.2 = log5 3 × 7

5 = log5 (3 × 7) – log5 5 = log5 3 + log5 7 – log5 5 = 0.892

(ii) log3 75 = log5 75log5 3

= log3 (3 × 5 × 5)log5 3

= log5 3 + 2 log5 5log5 3

= 3.928

(b) 70 0001 78 2

t = 20 000

1 78 2

t = 2

7

t log10 78 = log10

27

t = log10

27

log10 78

= 9.38

\ t = 10 years.

4. (a) log3 r – log9 r = 1 12

log3 r – log3 rlog3 9

= 32

log3 r – 12 log3 r = 3

2

12 log3 r = 3

2 log3 r = 3 r = 33 = 27

(b) 22x × 44y = 1 22x × 28y = 20 2x + 8y = 0 x + 4y = 0 …… 1 125x × 25y = 0.04 53x × 52y = 5–2 3x + 2y = –2 …… 2 2 × 2, 6x + 4y = –4 …… 3 3 – 1, 5x = –4 x = – 4

5

From 1, 4y = –1– 45 2

y = 15



5. 2x = 3 – 21 12x 2

22x = 3 · 2x – 2 22x – 3 · 2x + 2 = 0

Let y = 2x, y2 – 3y + 2 = 0 (y – 1)(y – 2) = 0

y = 1 or y = 2 2x = 1 2x = 2 log10 2

x = log10 1 \ x = 1 x log10 2 = log10 1 x = 0

Therefore, x = 0, 1

6. T = 80(0.89)x When T = 37, 80(0.89)x = 37

0.89x = 3780

log10 0.89x = log10 13780 2

x log10 0.89 = log10 13780 2

x = log10 137

80 2log10 0.89

= 6.617

\ x = 6.617 minutes.

Additional Mathematics Form 4 Chapter 6 Coordinate Geometry


1. (a) Distance of PQ = (6 – 1)2 + [2 – (–7)]2

= 25 + 81 = 10.30 units

(b) Distance of AB = [5 – (–7)]2 + (8 – 3)2

= 144 + 25 = 13 units

(c) Distance of EF = = 25 + 36 = 7.810 units

2. (a) (k – 1)2 + (–2 – 10)2 = 13 k2 – 2k + 1 + 144 = 169 k2 – 2k – 24 = 0 (k – 6)(k + 4) = 0 k – 6 = 0 or k + 4 = 0 k = 6 k = –4

(b) [6 – (–9)]2 + (–2 – 2k)2 = 17 225 + 4 + 8k + 4k2 = 289 4k2 + 8k – 60 = 0 k2 + 2k – 15 = 0 (k + 5)(k – 3) = 0 k + 5 = 0 or k – 3 = 0 k = –5 k = 3

(c) (–3 – k)2 + (2k + 1)2 = 50 9 + 6k + k2 + 4k2 + 4k + 1 = 50 5k2 + 10k – 40 = 0 k2 + 2k – 8 = 0 (k + 4)(k – 2) = 0 k + 4 = 0 or k – 2 = 0 k = – 4 k = 2

(–3 – 2)2 + [1 – (–5)]2

3. 2 + 8 3 + (–1)(a) Midpoint of CD = 1–—–– , ––——–2 2 2 = (5, 1)

–2 + 3 4 + (–12)(b) Midpoint of PQ = 1—–—– , ––—––—–2 2 2 1 = 1––, –42 2

3 + (–7) –2 + 5(c) Midpoint of EF = 1–––––––– , –––—–2 2 2 3 = 1–2, ––2 2

4. h + (–1) k + 8(a) 1––——– , –—––2 = (1, 4) 2 2

h – 1 k + 8 –—–– = 1 –—— = 4 2 2 h = 3 k = 0

h + (–9) 2 + k 5(b) 1–––—–— , ––—–2 = 1– ––, 42 2 2 2

h – 9 5 2 + k –—–– = – — –—— = 4 2 2 2 h = 4 k = 6

5 + k 2h + 4 (c) 1––—– , –——–2 = (7, –1) 2 2

5 + k 2h + 4 –—–– = 7 –——– = –1 2 2 k = 9 h = –3

Coordinate GeometryGeometri Koordinat6

CHAPTER



5. 1(8) + 2(2) 1(9) + 2(3)(a) M = 1–––———– , –––———–2 1 + 2 1 + 2 12 15 = 1––– , –––2 3 3 = (4, 5)

2(4) + 3(–6) 2(10) + 3(5)(b) M = 1–––———–– , ––––———–2 2 + 3 2 + 3 10 35 = 1– ––– , –––2 5 5 = (–2, 7)

1(2) + 3(6) 1(–4) + 3(2)(c) M = 1–––———–– , ––––———–2 1 + 3 1 + 3 20 2 = 1–––, ––2 4 4 1 = 15, —2 2

6. 1(4) + 3(p) 1(q) + 3(3)(a) 1–––––——–, –––––——–2 = (4, 1) 1 + 3 1 + 3

4 + 3p q + 9 ––—–– = 4 ––—–– = 1 4 4 3p = 12 q = –5 p = 4

3(q) + 2(7) 3(p) + 2(2q)(b) 1–––———–, ––—–—––––2 = (1, –2) 3 + 2 3 + 2

3q + 14 3p + 4q –––––— = 1 ––—––– = –2 5 5 q = –3 3p + 4(–3) = –10 2 p = — 3

5(–10) + 2(q) 5(3q) + 2(p)(c) 1––––———––, –——–—–––2 = (–6, 2p) 5 + 2 5 + 2

–50 + 2q 15q + 2p ––——–– = –6 –––——– = 2p 7 7 q = 4 12p = 15(4) p = 5

7. (a) Area of ΔABC = Area of ACDF – Area of ABEF – Area of BCDE 1 1 1 = —(5 + 10)(7) – —(5 + 3)(2) – —(3 + 10)(5) 2 2 2 = 12 unit2

(b) Area of ∆ABC = Area of ABPM – Area of ACNM

– Area of CBPN

= 12 (5 + 10)(13) – 1

2 (5 + 3)(9)

– 12 (3 + 10)(4)

= 97.5 – 36 – 26 = 35.5 unit2

8. (a) Area of ΔPQR 1 2 6 8 2 = — 2 5 3 9 5 1 = —(6 + 54 + 40 – 30 – 24 – 18) 2 = 14 unit2’

(b) Area of ΔPQR 1 3 5 6 3= — 2 2 –2 0 2

1 = — (–6 + 0 + 12 – 10 + 12 – 0) 2 = 4 unit2

(c) Area of ΔPQR 1 –8 1 5 –8 = — 2 6 –3 2 6 1 = ––(24 + 2 + 30 – 6 + 15 + 16) 2 = 40.5 unit2

9. (a) Area of PQRS 1 7 –3 2 8 7 = –– 2 4 4 –1 3 4 1 = ––(28 + 3 + 6 + 32 + 12 – 8 + 8 – 21) 2 = 30 unit2

(b) Area of PQRS 1 4 –2 –1 8 4= — 2 7 5 –3 2 7

1 = –– (20 + 6 – 2 + 56 + 14 + 5 + 24 – 8) 2 = 57.5 unit2

(c) Area of PQRS 1 0 2 5 10 0 = –– 2 6 –3 –7 1 6 1 = ––(0 – 14 + 5 + 60 – 12 + 15 + 70 – 0) 2 = 62 unit2



10. (a) Area of ∆ABC

1 1 4 –2 1 = –– 2 –3 3 –9 –3 1 = ––(3 – 36 + 6 + 12 + 6 + 9) 2 = 0 Therefore, A, B and C are collinear.

(b) Area of ∆ABC 3 –12 6 3 = 1

2 1 14 – 4 1

= 12

[42 + 48 + 6 – (–12) – 84 – (–12)]

= 18 Therefore, points A, B and C are non

collinear.

(c) Area of ∆ABC

8 –4 6 8 = 12 1 –8 –0.5 1

= 12

[–64 + 2 + 6 – (– 4) – (– 48) – (– 4)]

= 0 Therefore, points A, B and C are

collinear.

11. (a) x-intercept = – 4 y-intercept = 3

(b) x-intercept = 2 y-intercept = 5

12. (a) Gradient of AB 6 – (– 6) = ———— 2 – (–1) = 4

(b) Gradient of PQ 5 – 3 = ——— –1 – 2 2 = – — 3

13. (a) Gradient = – 64 = – 32

(b) Gradient = – 1 12–3 2 = 4

14. (a) The equation of the straight line is y – 5 = 2(x – 4) y – 5 = 2x – 8 y = 2x – 3

(b) The equation of the straight line is y – (–3) = –4(x – 6) y + 3 = –4x + 24 y = –4x + 21

(c) The equation of the straight line is

y – 2 = 13 [x – (–6)]

y – 2 = 13 x + 2

y = 13 x + 4

15. (a) The equation of the straight line is y – (–2) 4 – (–2) ––––––– = —––––– x – 1 5 – 1 y + 2 3 ——– = — x – 1 2

y + 2 = 32 (x – 1)

y = 32 x – 3

2 – 2

y = 32 x – 7

2

(b) The equation of the straight line is y – (–1) –7 – (–1) ––––—–– = –––––—— x – 2 4 – 2 y + 1 = –3(x – 2) y + 1 = –3x + 6 y = –3x + 5

(c) The equation of the straight line is y – (–5) 1 – (–5) ––––––– = ––––––– x – 3 –6 – 3

y + 5 = – 23 (x – 3)

y = – 23 x – 3



16. (a) The equation of the straight line is x y –– + –– = 1 2 3

(b) The equation of the straight line is x y – –– + –– = 1 5 2

(c) The equation of the straight line is x y – –– – –– = 1 3 4

17. 1 1(a) y = ––x – –– 2 4 Thus, gradient = 1

2 and y-intercept = – 14 .

4(b) y = ––x – 3 5 Thus, gradient = 4

5 and y-intercept = –3.

3(c) y = – ––x + 4 2 Thus, gradient = – 32 and y-intercept = 4.

18. (a) 2x6

+ 3y6

= 66

x3

+ y2

= 1

Thus, x-intercept = 3 and y-intercept = 2.

2x 5y 10(b) ––– – ––– = ––– 10 10 10 x y –– + ––— = 1 5 –2 Thus, x-intercept = 5 and y-intercept = –2.

(c) – 3x + 5y = 15 3x 5y 15 – ––— + ––– = ––– 15 15 15 x y —–– + –– = 1 –5 3

Thus, x-intercept = –5 and y-intercept = 3.

19. (a) 3x – 5y + 2 = 0

(b) x – 8y – 8 = 0

(c) 3x + 4y – 12 = 0

20. (a) 4x – 3y = 17 …… 2x + y = 1 ……

× 2, 4x + 2y = 2 …… – , –5y = 15 y = –3

Substitute y = –3 into equation . 2x + (–3) = 1 2x = 4 x = 2

Therefore, the point of intersection of the two straight lines is (2, –3).

(b) 3x + y = –4 …… x + 2y = 7 ……

× 2, 6x + 2y = –8 …… – , 5x = –15 x = –3

Substitute x = –3 into equation . –3 + 2y = 7 y = 5

Therefore, the point of intersection of the two straight lines is (–3, 5).

(c) 4x – 3y = 21 ……………… x + 12y = 1 ………………

× 4, 4x + 48y = 4 …… – , –51y = 17

1 y = – — 3 1 Substitute y = – –– into equation . 3

x + 121– 13 2 = 1

x = 5

Therefore, the point of intersection 1 of the two straight lines is 15, – —2. 3



21. (a) 4x + 2y – 9 = 0 …… 2x + y + 6 = 0 ……

From equation a, 2y = –4x + 9 9 y = –2x + — 2 m1 = –2From equation b, y = –2x – 6 m2 = –2

m1 = m2, the two straight lines are parallel.

(b) 2x – 3y + 6 = 0 …… 4x – 6y – 3 = 0 ……

From equation , 3y = 2x + 6 2 y = ––x + 2 3 2 m1 = –– 3From equation , 6y = 4x – 3 2 1 y = ––x – –– 3 2 2 m2 = –– 3m1 = m2, the two straight lines are parallel.

(c) x + y – 5 = 0 …… x – y + 10 = 0 ……

From equation , y = –x + 5 m1 = –1

From equation , y = x + 10 m2 = 1

m1 ≠ m2, the two straight lines are not parallel.

22. (a) 8x + 2y – 3 = 0 …… px – y + 7 = 0 ……

From , 2y = –8x + 3 3 y = –4x + –– 2 m1 = – 4 Since the two straight lines are parallel, m1 = m2 –4 = p p = –4

From , y = px + 7 m2 = p

(b) 4x + 2y + 1 = 0 ……… px + 3y – 9 = 0 ………

From , 2y = –4x – 1 1 y = –2x – — 2 m1 = –2

Since the two straight lines are parallel, m1 = m2 p –2 = – –– 3 p = 6

(c) x – py – 2 = 0 …… 4x – 5y + 10 = 0 ……

From , py = x – 2 1 2 y = ––x – — p p 1 m1 = –– p

Since the two straight lines are parallel, m1 = m2

1 4 –– = –– p 5 5 p = –– 4

23. (a) 4y = 2x + 5 1 5 y = ––x + — 2 4 1 m1 = –– 2 1 m1 = m2 = –– 2

The equation of the straight line is 1 y – 2 = ––(x – 6) 2 1 y = ––x – 3 + 2 2

y = 1—2x – 1

From , 3y = –px + 9 p y = – ––x + 3 3 p m2 = – –– 3

From , 5y = 4x + 10 4 y = ––x + 2 5 4 m2 = –– 5



(b) 2y = 6x + 3 3 y = 3x + — 2 m1 = 3 m1 = m2 = 3

The equation of the straight line is y – 7 = 3[x – (–2)] y = 3x + 6 + 7 y = 3x + 13

(c) 3y = –2x + 8 2 8 y = – ––x + –– 3 3 2 m1 = – –– 3 2 m1 = m2 = – –– 3

The equation of the straight line is 2 y – (–1) = – ––(x – 6) 3

y = – 2—3x + 3

24. (a) 2x + y – 7 = 0 ………… x – 2y – 8 = 0 …………

From equation , y = –2x + 7 m1 = –2From equation , 2y = x – 8 1 y = ––x – 4 2 1 m2 = — 2 1m1m2 = (–2)1—2 2 = –1Therefore, the two straight lines are perpendicular.

(b) 3x + 4y – 8 = 0 …… 4x – 3y + 12 = 0 ……

From equation , 4y = –3x + 8 3 y = – ––x + 2 4 3 m1 = – –– 4From equation , 3y = 4x + 12 4 y = ––x + 4 3 4 m2 = –– 3 3 4m1m2 = 1– ––21––2 4 3 = –1Therefore, the two straight lines are perpendicular.

(c) x – 4y + 7 = 0 …… 4x – y – 10 = 0 ……

From equation , 4y = x + 7 1 7 y = ––x + –– 4 4 1 m1 = –– 4 From equation , y = 4x – 10 m2 = 4 1 m1m2 = 1––2(4) 4 = 1 Therefore, the two straight lines are not

perpendicular.

25. (a) 2x – 4y + 12 = 0 …… hx + 3y – 4 = 0 ……

From equation , 4y = 2x + 12 1 y = ––x + 3 2 1 m1 = –– 2 From equation , 3y = –hx + 4 h 4 y = – ––x + –– 3 3 h m2 = – –– 3 Since the two straight lines are

perpendicular, m1m2 = –1 1 h 1—21– ––2 = –1 2 3 h = 6



(b) 4x – y + 12 = 0 …… x – hy + 5 = 0 ……

From equation , y = 4x + 12 m1 = 4From equation , hy = x + 5 1 5 y = —x + — h h 1 m2 = — h

Since the two straight lines are perpendicular, m1m2 = –1 1 (4) 1—2 = –1 h h = –4

(c) (h – 1)x – y + 8 = 0 …… x + 2y – 5 = 0 ……

From equation , y = (h – 1)x + 8 m1 = h – 1 From equation , 2y = –x + 5 1 5 y = – ––x + — 2 2 1 m2 = – –– 2

Since the two straight lines are perpendicular,

m1m2 = –1 1 1h – 121– —2 = –1 2 h – 1 = 2 h = 3

26. (a) 3y = 2x + 12 2 y = ––x + 4 3 2 m1 = –– 3 m1m2 = –1 2 1––2m2 = –1 3 3 m2 = – –– 2 Therefore, the equation of the straight

line is 3 y – 1 = – ––(x – 4) 2

= – 3—2x + 6

y = – 3—2x + 7

(b) 4y = –x + 8 1 y = – ––x + 2 4 1 m1 = – –– 4 m1m2 = –1 1 1– ––2m2 = –1 4 m2 = 4 Therefore, the equation of the straight

line is y – (–2) = 4(x – 4) y + 2 = 4x – 16 y = 4x – 18

(c) 2y = 6x + 9 9 y = 3x + — 2 m1 = 3 m1m2 = –1 3m2 = –1 1 m2 = – –– 3



Therefore, the equation of the straight line is

1 y – 4 = – ––[x – ( –6)] 3

= – 1—3x – 2

y = – 1—3x + 2

27. (a) (i) 3x + 2y – 16 = 0 3 y = – ––x + 8 2 3 mAB = – –– 2 2 mBC = –– 3 The equation of BC is 2 y – 0 = ––(x – 1) 3 2 2 y = ––x – –– 3 3 3(ii) y = – —x + 8 ..........a 2 2 2 y = —x – — ..........b 3 3 a = b: 3 2 2 – —x + 8 = —x – — 2 3 3 x = 4 Substitute x = 4 into a. 3 y = – —(4) + 8 = 2 2 The coordinates of B = (4, 2)

(iii)

C(1, 0)

2 1

D(h, k) B(4, 2)

h + 2(4) k + 2(2) ———— = 1 , ———— = 0 3 3 h = –5 k = –4 The coordinates of D = (–5, –4)

(b) (i) 2x + y – 2 = 0 y = –2x + 2 mAB = –2 x – py – 6 = 0 py = x – 6 y = 1

p x – 6p

mAC = 1p

(–2)1 1p 2 = –1

p = 2

(ii) 2x + y – 2 = 0 …… x – 2y – 6 = 0 …… × 2, 4x + 2y – 4 = 0 …… + , 5x – 10 = 0 5x = 10 x = 2 Substitute x = 2 into equation . 2(2) + y – 2 = 0 y = –2 The coordinates of A = (2, –2)

(iii) From , y = 12 x – 3

\ C(0, –3)

mBC = 4 – (–3)–1 – 0

= –7 The equation of BC is y = –7x – 3.

(c) –(–9) 1(i) mCD = ––––– = –– 18 2 1 mAB = –– 2 The equation of AB is

y = 12 x + 6

(ii) (mBC) (mAB) = –1 mBC = –2

The equation of BC is y = –2x – 9



(iii) y = 12 x + 6 ……

y = –2x – 9 ……

= , 12 x + 6 = –2x – 9

52 x = –15

x = –6 Substitute x = –6 into equation .

y = 12 (–6) + 6

y = 3

The coordinates of B are (–6, 3)

(iv) A(0, 6), B(–6, 3), C(0, –9), D(18, 0). Area of ABCD 1 0 –6 0 18 0 = –– 2 6 3 –9 0 6 1 = ––(54 + 108 + 36 + 162) 2 = 180 unit2

28. (a) AP = 5 (x – 3)2 + [y – (–4)]2 = 5 (x – 3)2 + (y + 4)2 = 52

x2 – 6x + 9 + y2 + 8y + 16 = 25 x2 + y2 – 6x + 8y = 0

(b) AP = 4 [x – (–5)]2 + (y – 6)2 = 42

x2 + 10x + 25 + y2 – 12y + 36 = 16 x2 + y2 + 10x – 12y + 45 = 0

(c) AP = 8 (x – 2)2 + [y – (–7)]2 = 82

x2 – 4x + 4 + y2 + 14y + 49 = 64 x2 + y2 – 4x + 14y – 11 = 0

29. (a) 2AP = PB 4[(x – 3)2 + (y – 4)2] = (x – 1)2 + [y – (–5)]2

4(x2 – 6x + 9 + y2 – 8y + 16) = x2 – 2x + 1 + y2 + 10y + 25 4x2 + 4y2 – 24x – 32y + 100 = x2 + y2 – 2x + 10y + 26 3x2 + 3y2 – 22x – 42y + 74 = 0

(b) 2AP = 3PB 4[(x – 4)2 + (y – 3)2] = 9[(x – 2)2 + (y – 5)2] 4(x2 – 8x + 16 + y2 – 6y + 9) = 9(x2 – 4x + 4 + y2 – 10y + 25) 4x2 + 4y2 – 32x – 24y + 100 = 9x2 + 9y2 – 36x – 90y + 261 5x2 + 5y2 – 4x – 66y + 161 = 0



30. (a) (i) 3AP = 2PB 9[(x – 2)2 + (y – 4)2] = 4[(x – 5)2 + (y – 1)2] 9[x2 – 4x + 4 + y2 – 8y + 16] = 4(x2 – 10x + 25 + y2 – 2y + 1) 9x2 + 9y2 – 36x – 72y + 180 = 4x2 + 4y2 – 40x – 8y + 104 5x2 + 5y2 + 4x – 64y + 76 = 0

(ii) Assuming the locus of point P intersects the x-axis, hence y = 0. 5x2 + 4x + 76 = 0 b2 – 4ac = 42 – 4(5)(76) = –1 504 , 0 Therefore, the locus of point P does not intersect the x-axis.

(b) (i) 2AP = PB 4{[x – (–4)]2 + (y – 1)2} = (x – 4)2 + (y – 6)2

4(x2 + 8x + 16 + y2 – 2y + 1) = x2 – 8x + 16 + y2 – 12y + 36 4x2 + 4y2 + 32x – 8y + 68 = x2 + y2 – 8x – 12y + 52 3x2 + 3y2 + 40x + 4y + 16 = 0

(ii) 3x2 + 3y2 + 40x + 4y + 16 = 0 ………………a y = x ………………b

Substitute equation b into equation a. 3x2 + 3x2 + 40x + 4x + 16 = 0 6x2 + 44x + 16 = 0 3x2 + 22x + 8 = 0 –22 ± 222 – 4(3)(8) x = —————————— 2(3) = –6.950 or –0.3837 From , y = –6.950 or –0.3837 The points of intersection are (–0.3837, –0.3837) and (–6.950, –6.950).

(c) (i) PA = AB (x – 3)2 + (y – 2)2 = (0 – 3)2 + (–2 – 2)2 x2 – 6x + 9 + y2 – 4y + 4 = 25 x2 + y2 – 6x – 4y – 12 = 0 The equation of the locus of point P is x2 + y2 – 6x – 4y – 12 = 0.

(ii) x2 + y2 – 6x – 4y – 12 = 0 ………………a Substitute x = 0, y = p into equation . p2 – 4p – 12 = 0 (p – 6)(p + 2) = 0 p = 6 or –2



SPM Practice 6

Paper 1

1. (a) The equation of EF is x–6 + y

4 = 1

x-intercept = –6; y-intercept = 4

(b) Let the coordinates of Q = (h, k)

4 1Q(h,k) P(2,1)F(0,4)

4(2) + 1(h)4 + 1 = 0

8 + h = 0 h = –8

4(1) + 1(k)4 + 1 = 4

4 + k = 20 k = 16 The coordinates of Q is (–8, 16).

2. (a) The coordinates of N

= 1 –4 + 02 , 0 + (–6)

2 2 = (–2, –3)

(b) The gradient of PQ = 0 – (–6)–4 – 0 = – 3

2 The gradient of the line perpendicular to

PQ,

m1– 32 2 = –1

= 23

The equation of the straight line which is perpendicular to PQ and passing through N is

y – (–3) = 23

[x – (–2)]

y + 3 = 23

(x + 2)

3y + 9 = 2x + 4 3y – 2x + 5 = 0

3. 8

T(0,–4)

P(x,y)

The equation of the locus of P is (x – 0)2 + [y – (4)]2 = 8 x2 + (y + 4)2 = 64 x2 + y2 + 8y + 16 – 64 = 0 x2 + y2 + 8y – 48 = 0

4. (a) 2p = 10 p = 5

(b) 3q = –6 q = –2

5. (a) ax – (b – 1)y + 5 = 0 (b – 1)y = ax + 5

y = ax + 5b – 1

= ab – 1

x + 5b – 1

\ m = ab – 1

(b) m1 × m2 = –1

m2 = – 1m1

= – 1a

b – 1

= – b – 1a

= 1 – ba

6.

1

2

E(–9, 2)

A(x, y)

F(3, 8)

A(x, y) = 1 2(–9) + 1(3)2 + 1 , 2(2) + 1(8)

2 + 1 2 = (–5, 4)

Distance AE = (–5 + 9)2 + (4 – 2)2 = 20 = 4 × 5 = 25 units



7. The area of the triangle

= 12

–1 2 k –1 4 0 6 4

= 12

[12 + 4k – 8 + 6]

= 12

(4k + 10)

12

(4k + 10) = 13 4k + 10 = 26 4k = 16 k = 4or

12

(4k + 10) = –13 4k + 10 = –26 4k = –36 k = –9

8. (a) 1 2N(p,q)

Q(0,9)P(–6,0)

N = (p, q)

= 1 2(–6) + 1(0)2 + 1

, 2(0) + 1(9)2 + 1 2

= (–4, 3) The coordinates of N = (–4, 3).

(b) mPQ = 9 – 00 – (–6)

= 96

= 32

m1 32 2 = –1

m = – 23

y – 3 = – 23

(x + 4)

y = – 23

x – 83

+ 3

3y = –2x + 1

The equation of the perpendicular line is 3y = –2x + 1.

9. y

C

A0 B x

Q(5,3)

P(x,y)

(x – 5)2 + (y – 3)2 = 5 x2 – 10x + 25 + y2 – 6y + 9 = 25 x2 + y2 – 10x – 6y + 9 = 0

When y = 0, x2 – 10x + 9 = 0 (x – 1)(x – 9) = 0 x = 1 or 9

When x = 0 y2 – 6y + 9 = 0 (y – 3)(y – 3) = 0 y = 3Therefore, A(1, 0), B(9, 0), C(0, 3).

The area of triangle ABC

= 12

(9 – 1)(3)

= 12 unit2

Paper 2

1. (a) (i) 7y + 3x + 23 = 0

y = – 37

x – 237

mBC = – 37

mCD · 1– 37 2 = –1

mCD = 73

y = mCD x + 5

y = 73

x + 5

The equation of CD is y = 73

x + 5.



(ii) y = – 37

x – 237

……

y = 73

x + 5 …………

= ,

– 37

x – 237

= 73

x + 5

Multiply both sides by 21. –9x – 69 = 49x + 105 –58x = 174 x = –3 Substitute x = –3 into .

y = 73

(–3) + 5

= –2 The coordinates of C = (–3, –2).

(b) Substitute x = 4 into ,

y = 37

(4) – 237

= –5 The coordinates of B is (4, –5). B(4, –5), P(x, y) and PB = 6. (x – 4)2 + [y – (–5)]2 = 6 x2 – 8x + 16 + y2 + 10y + 25 = 36 x2 + y2 – 8x + 10y + 5 = 0 The equation of the locus of P is x2 + y2 – 8x + 10y = 5 = 0

2. (a) 2y – x + 10 = 0

y = 12

x – 5

mRS = mPQ = 12

y = mRS x + c

12 = 12

(4) + c

c = 10

The equation of RS is y = 12

x + 10.

(b) 1 12 2mQR = –1

mQR = –2

y = mQR x + c –6 = –2(–2) + c c = –10 The equation of QR is y = –2x – 10.

(c) y = 12

x + 10 ……

y = –2x – 10 ……

= :

12

x + 10 = –2x – 10

52

x = –20 x = –8 Substitute x = –8 into . y = –2(–8) – 10 = 6 The coordinates of R is (–8, 6).

(d) Area of rectangle PQRS = 2 × Area of ∆QRS

= 2 × 12

–2 4 –8 –2 –6 12 6 –6

= |(–24 + 24 + 48) – (–24 – 96 – 12)| = |48 – (–132)| = 180 unit2

3. (a) Gradient of BC

= 6 –52 – (–2)

= 14

Gradient of AB = –4 The equation of AB is y – 6 = – 4(x – 2) y = – 4x + 14

(b) y = – 4x + 14 …… 4x + 7y = 2 ……………

For equation , 4x + y = 14 …………… – , 6y = –12 y = –2

Substitute y = –2 into equation . 4x + 7(–2) = 2 4x = 16 x = 4 Coordinates of A are (4, –2).



(c) Let E(h, k).

B(2, 6)

A(4, –2)

E(h, k)1

3

(h, k) = 1 1(4) + 3(2)1 + 3 , 1(–2) + 3(6)

1 + 3 2 = 1 5

2 , 42 Coordinates of E = 1 5

2 , 42.(d) Let D = (x, y)

12 4 2 –2 x 4

–2 6 5 y –2 = 29.5

(24 + 10 – 2y – 2x) – (–4 – 12 + 5x + 4y) = 59 7x + 6y = –9 .................a 4x + 7y = 2 ...................b a × 4 28x + 24y = –36 ...........c b × 7 28x + 49y = 14 .............d d – c, 25y = 50 y = 2

Substitute y = 2 into equation b. 4x + 7(2) = 2 x = –3 Coordinates of D = (–3, 2).

4. (a) Area of ∆POQ

= 12 0 –6 1 0

0 3 –4 0 = 1

2 [(0 + 24 + 0) – (0 + 3 + 0)]

= 10.5 unit2

(b) Let R = (x, y)

1

2Q(1, – 4)

P(–6, 3)

R(x, y)

(x, y) = 1 1(1) + 2(–6)1 + 2 , 1(–4) + 2(3)

1 + 2 2

Coordinates of R = 1– 113 , 2

3 2

(c) XP = 2XQ (x + 6)2 + (y – 3)2 = 2(x – 1)2 + (y + 4)2

(x + 6)2 + (y – 3)2 = 4[(x – 1)2 + (y + 4)2] x 2 + y 2 + 12x – 6y + 45 = 4x 2 + 4y 2 – 8x + 32y + 68

3x 2 + 3y 2 – 20x + 38y + 23 = 0

Hence, the equation of the locus of points x is 3x 2 + 3y 2 – 20x + 38y + 23 = 0.

5. (a) (i) Area of OABC

= 12 0 10 8 –12 0

0 3 21 4 0

= 12 [(0 + 210 + 32 + 0)

– (0 + 24 – 252 + 0)] = 235 m2

(ii) Let R(x, y)

1

4

R(x, y) Q(–4, 8)

P(8, 16)

Q(–4, 8) = 14x + 1(8)4 + 1 , 4y + 1(16)

4 + 1 2 By comparing the x-coordinate,

4x + 85 = – 4

4x + 8 = –20 4x = –28 x = –7

By comparing the y-coordinate,

4y + 165 = 8

4y + 16 = 40 4y = 24 y = 6 \ R(–7, 6)

(b) Let W(x, y) WQ = 1.5 (x + 4)2 + (y – 8)2 = 3

2

x2 + 8x + 16 + y2 – 16y + 64 = 94

x2 + y2 + 8x – 16y + 80 = 94

4x2 + 4y2 + 32x – 64y + 320 = 9 4x2 + 4y2 + 32x – 64y + 311 = 0

The equation of the spray ring: 4x2 + 4y2 + 32x – 64y + 311 = 0



6. (a) (i) 3y – 2x – 15 = 0 3y = 2x + 15

y = 23 x + 5

y-intercept = 5 Hence, coordinates of G = (0, 5).

(ii) MEF = 23

Hence, the gradient of the perpendicular line = – 3

2 .

Equation of the straight line which is perpendicular to EF and passes point E is

y – 1 = – 32 [x – (–6)]

y = – 32 x – 8

(b) Let F(h, k).

E(–6, 1)

F(h, k)

G(0, 5)2

3

2h + 3(–6)2 + 3 = 0

2h – 18 = 0 h = 9

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

2k + 3 = 25 k = 11 Hence, coordinates of F = (9, 11).

7. (a) (i) PQ2 = (k – 6)2 + (2k)2

= k2 – 12k + 36 + 4k2 = 5k2 – 12k + 36 QR2 = (k + 2)2 + (2k)2 = k2 + 4k + 4 + 4k2 = 5k2 + 4k + 4

PQ2 + QR2 = PR2

5k2 – 12k + 36 + 5k2 + 4k + 4 = 82

10k2 – 8k – 24 = 0 5k2 – 4k – 12 = 0 (k – 2)(5k + 6) = 0 k – 2 = 0 or 5k + 6 = 0 k = 2 k = – 6

5 [Not accepted]

(ii) 3y = 3x – 63

y = 13

x – 23

When x = 0, y = –23 y-intercept of PS = –23 The coordinates of S is (0, –23 ).

(iii) mPS = 13

mRS · 13

= –1

mRS = –3

y = mRS x + c –23 = –3 (0) + c c = –23

The equation of RS is y = –3 x – 23.

(b)

(2,4)Q

(x,y)T

(2,0)A

y

PR

S

–2 60x

TA = 4 (x – 2)2 + (y – 0)2 = 42

x2 – 4x + 4 + y2 = 16 x2 + y2 – 4x – 12 = 0

The equation of the locus is x2 + y2 – 4x – 12 = 0 with constant distance 4 units from A(2, 0).

Additional Mathematics Form 4 Chapter 7 Statistics


1. (a) –10, –4, –2, 0, 0, 1, 1

–10 + (–4) + (–2) + 0 + 0 + 1 + 1Mean = ––––––––––––––––—————– 7 = –2

Median = 0 Mode = 0 and 1

(b) 10, 11, 11, 12, 12, 12, 13, 15 10 + 11 + 11 + 12 + 12 + 12 + 13 + 15 Mean = ––––––––––––––––—– 8 = 12

Median = 12 + 122

= 12

Mode = 12

(c) 2.8, 3.3, 3.5, 4.7, 4.7, 5.3 2.8 + 3.3 + 3.5 + 4.7 + 4.7 + 5.3 Mean = –––––––––––––––––––——–––— 6 = 4.05 3.5 + 4.7 Median = ———— 2 = 4.1 Mode = 4.7

2. (a) (i) Modal class = (55 – 59) km(ii)

0

2

4

6

8

10

12

39.5

44.5

49.5

54.5

Distance (km)

Freq

uenc

y

59.5

57.5

64.5

69.5

From the histogram, mode = 57.5 km

(b) (i) Modal class = (70 – 79) marks(ii)

0

10

20

30

40

50

49.5

59.5

69.5

74.5

79.5

Marks

Freq

uenc

y

89.5

99.5

From the histogram, mode = 74.5 marks

StatisticsStatistik7

CHAPTER



3. (a) Height (cm) Class mark Frequency Cumulative

frequency150 – 154 152 10 10155 – 159 157 35 45160 – 164 162 100 145165 – 169 167 210 355170 – 174 172 121 476175 – 179 177 24 500

152 × 10 + 157 × 35 + 162 × 100 + 167 × 210 + 172 × 121 + 177 × 24(i) Mean = ———————————————– 500 = 166.69 cm

(ii) Median class = (165 – 169) cm 500 ––– – 145 2 Therefore, median = 164.5 + ––––––—–2 × 5 210 = 167 cm

(b) Mass (g) Class mark Frequency Cumulative

frequency60 – 69 64.5 28 2870 – 79 74.5 22 5080 – 89 84.5 24 7490 – 99 94.5 21 95

100 – 109 104.5 25 120

28(64.5) + 22(74.5) + 24(84.5) + 21(94.5) + 25(104.5)(i) Mean = —————————————— 120 10 070 = ———– 120 = 83.92 g

(ii) Median class = (80 – 89) g 120 ––– – 50 2 Therefore, median = 79.5 + ––––––—–2 × 10 24 = 83.67 g



4. (a) Length(cm)

Upper boundary Frequency Cumulative

frequency16.7 – 17.1 17.15 0 017.2 – 17.6 17.65 2 217.7 – 18.1 18.15 8 1018.2 – 18.6 18.65 14 2418.7 – 19.1 19.15 18 4219.2 – 19.6 19.65 6 4819.7 – 20.1 20.15 2 50

From the ogive, median = 18.7 cm

(b) Time(minutes)


frequency20 – 29 29.5 0 030 – 39 39.5 3 340 – 49 49.5 5 850 – 59 59.5 9 1760 – 69 69.5 10 2770 – 79 79.5 4 3180 – 89 89.5 1 32

From the ogive, median = 58.5 minutes

0

Cum

ulat

ive

frequ

ency

Length (cm)

50

17.15 18.1517.65 18.65 19.65 20.1519.15

40

30

20

10

029.5 39.5 49.5 59.5 69.5 79.5 89.5

35

Time (minutes)

Cum

ulat

ive

frequ

ency 30

25

20

15

10

5

5. (a) (i) New mean = 18 + 4 = 22 New mode = 19 + 4 = 23 New median = 20 + 4 = 24

(ii) New mean = (18 × 2) – 3 = 33 New mode = (19 × 2) – 3 = 35 New median = (20 × 2) – 3 = 37

(b) New mean = 60 54 × 8 + a – 66 ––––––––———– = 60 8 a + 366 = 480 a = 114

(c) Median = 12 y + z ––––– = 12 2 y + z = 24 ……a

Mean = 11 17 + x + y + z + 8 + 2 –––––––––——––––––– = 11 6 x + y + z = 39 ……b

b – a, x = 15 Since mode = 15,

y = x = 15 From a, z = 9



6. 74 + 67 + 49 + 66 + 89 + 51 + 73(a) Mean = ––––––––––––————–––––––––– = 67 7 49, 51, 66, 67, 73, 74, 89 Median = 67 The mean and the median are suitable measures of central tendency as both the values are the

central values of the set of numbers.

450 + 400 + 380 + 360 + 420 + 470 + 390 + 460 + 480 + 490 (b) (i) Mean = ——————————————————————————— = 430 10 360, 380, 390, 400, 420, 450, 460, 470, 480, 490 420 + 450 Median = ————— = 435 2 Mean and median are both suitable.

(ii) The median is the most suitable measure of central tendency as there is an extreme value, RM150, which exists in the data.

7. (a) 2.4, 3.6, 5.1, 6.2, 7.8, 8.9 Q1 Q3

(i) Range = 8.9 – 2.4 = 6.5(ii) Interquartile range = 7.8 – 3.6 = 4.2

(b) Score 10 12 14 16 18Frequency 2 10 6 2 8Cumulative frequency 2 12 18 20 28

Q1 Q3(i) Range = 18 – 10 = 8

(ii) Q1 = 14

× 282 th value

= 7th value = 12

Q3 = 34

× 282 th value

= 21th value = 18

Interquartile range = Q3 – Q1 = 18 – 12 = 6



8. 110 + 119 60 + 69(a) (i) Range = ————— – ———— 2 2 = 50 hours

(ii) Lifespan (hours) 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109 110 – 119Frequency 9 13 20 14 18 6Cumulative frequency 9 22 42 56 74 80

↑Q1

↑Q3

1 — (80) – 9 4Q1 = 69.5 + ————— × 10 13 = 77.96 hours

3 — (80) – 56 4Q3 = 99.5 + —————– × 10 18 = 101.72 hours

Interquartile range = 101.72 – 77.96 = 23.76 hours

7.95 + 7.93 7.80 + 7.82(b) (i) Range = ————— – ————— 2 2 = 0.15 g

(ii) Mass (g) 7.80 – 7.82 7.83 – 7.85 7.86 – 7.88 7.89 – 8.01 8.02 – 8.04 8.05 – 8.07Frequency 3 6 69 92 6 4Cumulative frequency 3 9 78 170 176 180

↑Q1

↑Q3

1 — (180) – 9 4Q1 = 7.855 + —————– × 0.03 69 = 7.871 g

3 — (180) – 78 4Q3 = 7.885 + —————— × 0.03 92 = 7.904 g

Interquartile range = 7.904 – 7.871 = 0.033 g

9. (a) Amount (RM)


frequency61 – 70 70.5 0 071 – 80 80.5 6 681 – 90 90.5 12 1891 – 100 100.5 21 39101 – 110 110.5 25 64111 – 120 120.5 12 76121 – 130 130.5 4 80

Q1 = RM91.5, Q3 = RM108.5 Therefore, interquartile range = 108.5 – 91.5 = RM17

070.5 80.5 90.5

91.5 108.5

100.5 110.5 120.5 130.5

10

20

30

Cum

ulat

ive

frequ

ency

40

50

60

70

80

Amount (RM)



(b) Time(minutes)

Upper boundary

FrequencyCumulative frequency

25 – 34 34.5 0 0

35 – 44 44.5 20 20

45 – 54 54.5 240 260

55 – 64 64.5 212 472

65 – 74 74.5 96 568

75 – 84 84.5 24 592

85 – 94 94.5 8 600

0

Time (minutes)

Cum

ulat

ive

freq

uenc

y

600

500

400

300

200

100

34.5 44.5 54.5 64.5 74.5 84.5 94.5

Q1 = 50.5 minutes, Q

3 = 63.5 minutes

Therefore, interquartile range = 63.5 – 50.5 = 13 minutes



10. (a) 8, 6, 7, 5, 8, 8, 12

– 8 + 6 + 7 + 5 + 8 + 8 + 12 (i) x = ––––––––––––––––———— 7 = 7.714 446(ii) σ 2 = –––– – (7.714)2

7 = 4.208

(iii) σ = 4.208 = 2.051

(b) 28, 35, 21, 42, 65 _ 28 + 35 + 21 + 42 + 65(i) x = ––————–––––––––––– 5 = 38.2 8439(ii) σ 2 = –––— – (38.2)2

5 = 228.56

(iii) σ = 228.56

= 15.12

(c) 3.6, 4.2, 3.4, 4.8, 3.5 _ 3.6 + 4.2 + 3.4 + 4.8 + 3.5(i) x = –––––––———––––––––––– 5 = 3.9 77.45(ii) σ 2 = ––––– – (3.9)2

5 = 0.28

(iii) σ = 0.28 = 0.5292

_ Σfx (d) (i) x = —— Σf 66 = ––– 46

= 1.435 mobile phones Σfx2(ii) σ 2 = —— – x–2 Σf 126 = –––– – (1.435)2 46 = 0.6799

(iii) σ = 0.6799 = 0.8246 mobile phone

11. (a) Number of durians Class mark, x Frequency, f fx fx2

80 – 84 82 4 328 26 89685 – 89 87 1 87 7 56990 – 94 92 0 0 095 – 99 97 5 485 47 045

100 – 104 102 12 1 224 124 848105 – 109 107 8 856 91 592

∑f = 30 ∑fx = 2 980 ∑fx2 = 297 950

297 950 2 980(i) σ 2 = ––––—– – –––—22

30 30 = 64.56

(ii) σ = 64.56

= 8.035 durians



(b) Marks Class mark, x Frequency, f fx fx2

0 – 9 4.5 5 22.5 101.2510 – 19 14.5 10 145 2 102.520 – 29 24.5 11 269.5 6 602.7530 – 39 34.5 12 414 14 28340 – 49 44.5 4 178 7 92150 – 59 54.5 2 109 5 940.560 – 69 64.5 1 64.5 4 160.25

∑f = 45 ∑fx = 1 202.5 ∑fx2 = 41 111.25

41 111.25 1 202.5(i) σ 2 = ————– – ––——22

45 45 = 199.5

(ii) σ = 199.5

= 14.12 marks

14. (a) (i) Machine P: 2 500 Mean = ——– 10 = 250 ml Standard deviation

= 625 05610

– 2502

= 2.366 ml Machine Q: 2 500 Mean = ——– 10 = 250 ml Standard deviation

= 625 24010

– 2502

= 4.899 ml

(ii) The standard deviation of machine P is smaller than machine Q. Therefore, machine P is said to be more reliable than machine Q.

12. (a) (i) New interquartile range = 7 × 3 = 21(ii) New variance = 22 × 32

= 36(iii) New standard deviation = 2 × 3 = 6

10(b) (i) New range = — 2 = 5

5(ii) New variance = — 22

= 1.25

!w5(iii) New standard deviation = —— 2 = 1.118

13. (a) (i) The range will increase drastically from 7 to 99 as number 100 is an extreme value.

(ii) The variance wil l increase drastically as the range has become much greater.

(b) (i) The range will increase from 7 to 9.

(ii) The interquartile range will remain unchanged.

(iii) The variance will decrease as the dispersion of the data is reduced.



(b) –xA = 168

= 2

sA = 408

– 22

= 1

–xB = 168

= 2

sB = 468

– 22

= 1.323

Set A is to be preferred as the standard deviation of set A is smaller than set B.

SPM Practice 7

Paper 1

1. (a) Mean = 6

[6 + 9 + (x2 + 1) + 3 + 2]5

= 6 21 + x2 = 30 x2 = 9 x = ±3 Since x . 0, x = 3

(b) 2, 3, 6, 9, 10 Median = 6

2. (a) 5 + 6 + 8 + 7 + 4 = 30 workers

(b)

[(5 × 5.5) + (6 × 7.5) + (8 × 9.5) + (7 × 11.5) + (4 × 13.5)]

30 = 283

30 = 9.433 km

3. (a) Variance = 50010

= 50

(b) s2 = ∑x2

n – (x)2

50 = 86010

– (x)2

(x)2 = 86 – 50 = 36 x = 6

4. (a) x = ∑xn

90 = ∑x8

∑x = 720 g

(b) s = ∑x2

n – x2 5 = ∑x2

8 – 902

25 = ∑x2

n – 8 100

8 125(8) =∑x2 ∑x2 = 65 000 g2

5. (a) Mean = 2807

= 40 ml

(b) Standard deviation

= 11 3127

– 402

= 16 = 4

6. (a) Let µ be the mean of set A and set B. For set A, nA = 6 and sA = 4. sA

2 = 42

∑xA

2

nA – µ2 = 42

∑xA

2

6 – µ2 = 16

∑xA

2

6 = 16 + µ2

∑xA2 = 96 + 6µ2 …… 1



For set B, nB = 4 and sB = 5.

sB2 = 52

∑xB

2

nB – µ2 = 52

∑xB

2

4 – µ2 = 25

∑xB

2

4 = 25 + µ2

∑xB2 = 100 + 4µ2 …… 2

The variance of the comboned set A and set B

= ∑xA2 +∑xB

2

nA + nB

– µ2

= 96 + 6µ2 + 100 + 4µ2

6 + 4 – µ2

= 196 + 10µ2

10 – µ2

= 19.6 + µ2 – µ2 = 19.6

7. (a) p = 10

(b) 5 + p = 10 + 8 + 9 5 + p = 27 p = 22 5 + p + 10 = 8 + 9 15 + p = 17 p = 2 \ 2 < p < 22

8. (a) ∑x = 8 × 6 = 48

(b) 48 – t7

= 5.6

48 – t = 39.2 t = 8.8

9. N = 5, x = 120, median = 120Let a, b, c, d, e as the mass from lightest to heaviest.

a + e2

= 120

a + e = 240 …… 1

Range = 20 e – a = 20 e = a + 20 …… 2

Substitute 2 into 1, a + a + 20 = 240 2a = 220 a = 110 …… 3

Substitute 3 into 2, e = 110 + 20 = 130\ 110, b, 120, d, 130

Since there are 3 parcels with equal mass, then b = d.

x = 120

110 + b + 120 + d + 1305

= 120

b + d + 360 = 600 b + b + 360 = 600 2b = 240 b = 120 = d

Therefore the mass of the parcels are 110 g, 120 g, 120 g, 120 g and 130 g.

Paper 2

1. (a) (i) Mean of girls = 5 Number of girls = 20 The total number of stationery owned,

Mean = ∑xn

5 = ∑x20

∑x = 100



(ii) s2 = ∑x2

n – (x)2

22 = ∑x2

20 – (5)2

29 = ∑x2

20 ∑x = 580

(b) (i) Mean of boys = 2, total number of stationery owned by boys = 20 Let the number of boys be n.

Mean = ∑xn

2 = 20n

n = 10

Total number stationery owned by the class,

20 + 100 = 120 Mean of the class,

Mean = ∑xn

= 12030

= 4

Sum of squares of the stationery owned by the class,

170 + 580 = 750

Standard deviation = ∑x2

n – x2 = 750

30 2 – (4)2

= 3

2. (a) s = ∑x2

n – x2 = 29 045 400

6 – 13 200

6 22

= 4 840 900 – 2 2002 = 900 = RM30

(b) New mean = 2 200 + 200 = RM2 400 New standard deviation = RM30

3. (a) (i)

3 + x + 5 + (3x + 4) + 11 + 13

6 = 8

4x + 366

= 8 x = 3

(ii) 3, 3, 5, 11, 13, 13

s2 =

32 + 32 + 52 + 112 + 132 + 132

6 – 82

= 19.67

(b) (i) Mean = 2(8) + 3 = 19

(ii) Standard deviation = 219.67 = 8.87

4. (a)

2

4

6

8

10

12

14

16

40.5

45.5

50.5

55.5

60.5

65.5

54.2

5

Mass (kg)

Fre

quen

cy

0

Modal mass = 54.25 kg(b)

Mass (kg)

Midpoint, x f fx fx2

41 – 45 43 2 86 3 69846 – 50 48 6 288 13 82451 – 55 53 15 795 42 13556 – 60 58 12 696 40 36861 – 65 63 5 315 19 845

∑f = 40 ∑fx= 2 180

∑fx2 = 119 870

s = 119 87040

– 2 18040 22

= 5.148 kg



5. Median class = 40 – 44

Median = 39.5 + 12

(31) – 12

10 25

= 39.5 + 1.75 = 41.25

6. (a)

Marks Number of students0 – 9 5

10 – 19 9 – 5 = 420 – 29 21 – 9 = 1230 – 39 29 – 21 = 840 – 49 32 – 29 = 3

(b) Q1 = 9.5 + 324

– 5

4 2(10) = 17

Q3 = 29.5 + 34

(32) – 21

8 2(10) = 33.25

The interquartile range = 33.25 – 17 = 16.25

7. (a) (i) Mean of set A,

(60 + 67 + 54 + 59 + 68 + 70)6

= 3786

= 63 Mean of set B,

(48 + 53 + 62 + 68 + 69 + 78)6

= 3786

= 63(ii) Standard deviation of set A,

(602 + 672 + 542 + 592 + 682 + 702)6

– (63)2

= 24 0106

– (63)2

= 5.715

Standard deviation set B,

(482 + 532 + 622 + 682 + 692 + 782)6

– (63)2

= 24 4266

– (63)2

= 10.10

(b) Set A is more consistent. The standard deviation of set A is lower than set B.

Additional Mathematics Form 4 Chapter 8 Circular Measure


Circular MeasureSukatan Membulat8

CHAPTER

1. (a) 0.6 rad. = 0.6 × 180°p

= 34.38°

(b) 1.8 rad. = 1.8 × 180°p

= 103.1°

(c) 4p5

rad. = 45

× 180°

= 144°

(d) 4.25 rad. = 4.25 × 180°p

= 243.5°

2. (a) 158° = 158 × p180

= 2.758 rad.

(b) 42.6° = 42.6 × p180

= 0.7435 rad.

(c) 122.8° = 122.8 × p180

= 2.143 rad.

(d) 50°159 = 50.25 × p180

= 0.8770 rad.

(e) 252°129 = 252.2 × p180

= 4.402 rad.

3. (a) s = 6 × 0.8 = 4.8 cm

(b) s = 6 × 2.5 = 15 cm

(c) s = 8 × (100 × p180

)

= 13.96 cm

(d) s = 8.5 × (2p – 1.8) = 38.11 cm

4. (a) 12r = 2

5 p

= 9.549 cm

(b) 26r = 220 × p

180

= 6.771 cm

(c) 40r = 2p – 4

5 p

= 10.61 cm

5. (a) q = 6.88

= 0.85 rad.

(b) q = 208.5

= 2.353 rad.

(c) q = 28.45.5

= 5.164 rad.



6. (a) In ΔOPM,P

R

OM

10 cm

0.6 rad.

PM sin 0.6 rad. = 10 PM = 10 × sin 0.6 rad. = 5.646 cmLength of arc PQR = 10 × 1.2 = 12 cmPerimeter of the shaded region= 12 + (2 × 5.646)= 23.29 cm

(b) In ΔOEN,

8 cm1.2 rad.

O

NGE

EN sin 1.2 = 8 EN = 8 × sin 1.2 = 7.456 cmLength of arc EFG = 8 × 2.4 = 19.2 cm Perimeter of the shaded region = 19.2 + (2 × 7.456)= 34.11 cm

(c) In ΔOPN,

NP R

O

55°6 cm

PN sin 55° = 6 PN = 6 sin 55° = 4.915 cm

Length of arc PQR = 6 × 1110 × p180 2

= 11.52 cm Perimeter of the shaded region= 11.52 + (2 × 4.915)= 21.35 cm

7. (a) 80° = 1.396 rad. Length of arc PR = 5 × 1.396 = 6.98 cm

Length of arc QS = 12 × 1.396 = 16.75 cm

PQ = RS = 12 – 5 = 7 cm Therefore, perimeter of PQSR = 6.98 + 16.75 + 2(7) = 37.73 cm

(b) (i) OR = 15p – 1.8

Radius = 11.18 cm

(ii) Length of arc PQ = 11.18 × 1.8 = 20.12 cm

8. (a) A = 12

(8)2(1.4)

= 44.8 cm2

(b) A = 12

(10)2(0.86)

= 43 cm2

(c) A = 12

(7)2(110 × p180

)

= 47.04 cm2

(d) A = 12

(7.5)2(2p – 2p5

)

= 141.4 cm2

(e) A = 12

(9)2[(360 – 105) × p180

]

= 180.2 cm2



9. (a) r = 2(40)1.2

= 8.165 cm

(b) r = 2(60)0.8

= 12.25 cm

(c) r = 2(65)5p4

= 5.754 cm

(d) r = 2(20.5)30 × p

180 = 8.849 cm

(e) r = 2(88)120 × p

180 = 9.167 cm

10. (a) q = 2(43.2)62

= 2.4 rad.

(b) q = 2(48.5)62

= 2.694 rad.

(c) q = 2(256)102

= 5.12 rad.

11. (a)

M CA

O

30°9 cm

In ∆OAM, AM sin 30° = —– 9 AM = 4.5 cm OM cos 30° = —– 9 OM = 7.794 cm

1 Area of ∆OAC = —(AC)(OM) 2 1 = —(2 × 4.5)(7.794) 2 = 35.07 cm2

1 Area of sector OABC = —(9)2(60 × p180

) 2 = 42.41 cm2

Area of segment ABC = 42.41 – 35.07 = 7.34 cm2

(b)

10 cm

0.6 rad. O

G

E

M

In ∆OEM, EM sin 0.6 = —– 10 EM = 5.646 cm OM cos 0.6 = —– 10 OM = 8.253 cm

1 Area of ∆OEG = —(EG)(OM) 2 1 = —(2 × 5.646)(8.253) 2 = 46.60 cm2

1 Area of sector OEFG = — (10)2(1.2) 2 = 60 cm2

Area of segment EFG = 60 – 46.60 = 13.40 cm2



(c)

O M

A

C

9 cm3π rad.– 8

In ∆OAM, AM sin 3p

8 = —– 9

AM = 8.315 cm OM cos 3p

8 = —– 9

OM = 3.444 cm 1 Area of ∆OAC = —(2 × 8.315)(3.444) 2 = 28.637 cm2

1 Area of sector OABC = —(9)2( 3p4

) 2 = 95.426 cm2

Area of segment ABC = 95.426 – 28.637 = 66.79 cm2

12. p(a) (i) q = 60 × —— 180 = 1.047 rad.

120 × p(ii) Length of arc AC = 6 × ––——— 180 = 12.57 cm Length of arc AB = 12 × 1.047 = 12.56 cm The perimeter of the shaded region = 12.56 + 12.57 + 6 = 31.13 cm

(iii) Area of sector DAB 1 = —(12)2(1.047) 2 = 75.38 cm2

Area of sector OAC 1 = —(6)2(2.094) 2 = 37.69 cm2

CN = !w62 – 32 = 5.196 cm

3 cm

6 cm

NDO

C

Area of ∆OCD 1 = —(6)(5.196) 2 = 15.59 cm2

Area of shaded region = 75.38 – 37.69 – 15.59 = 22.10 cm2

(b) (i) The length of arc ABC 2p = 3(2p – –––) 3 = 12.57 cm

Area of sector OABC(ii) ––––––––––––––––— Area of sector OED 1 2p ––(3)2(2p – –––) 2 3 = –––––––––––––––— 1 2p —(9)2(–––) 2 3

2 = –– 9 The ratio is 2 : 9.



SPM Practice 8

Paper 1

1. Area of segment DEB= Area of sector ABD – Area of ∆ABD

= 1 12

× 92 × π2 2 – 1 1

2 × AB × AD2

= 63.6255 – 1 12

× 9 × 92 = 23.1255 m2

Area of the shaded region= 92 – 2(23.1255)= 34.749 m2

2. (a) 126° × π180°

= 2.1994 rad.

(b) Perimeter of the sector OEF = Arc EF + OE + OF = 9(2.1994) + 2(9) = 37.795 cm

3. (a) OG = 9 cm

EG = 23

(9)

= 6 cm

(b) Area of sector OGH = 12

(92)q

= 812

q

Area of sector EFG = 12

(62)q

= 18q

Area of shaded region = 29.25 cm2

812

q – 18q = 29.25

452

q = 29.25 q = 1.3 rad.

4. (a) sin q = ADOA

= 1213

q = 1.176 rad.

(b) Length of arc AB = 13 × 1.176 = 15.288

OD = 132 – 122 = 5 BD = 13 – 5 = 8

Perimeter of shaded region = Length of arc AB + BD + AD = 15.288 + 8 + 12 = 35.288 cm

5.

O

P

Q

θ

(a) ∠OQP = ∠OPQ = 30° q = 180° – 30° – 30° = 120°

= 120°180°

× π

= 120°180°

× 3.142

= 2.095 rad



(b)

O

P

Q

M60°

10 cm

In ∆OPM,

sin 60° = PM10

PM = 8.660 cm

cos 60° = OM10

OM = 5 cm

Area of ∆OPM = 12

(2 × 8.660)(5)

= 43.3 cm2

Area of sector OPQ = 12

(10)2(2.095)

= 104.75 cm2

Area of shaded region = 104.75 – 43.3 = 61.45 cm2

6. R

S

PT

U Q

Since the sides of ∆RPS are equal length,∠RPS = π

3.

Area of segment RUS = Area of sector PRS – Area of PRS

= 1 12

× 82 × π3 2 – 1 1

2 × 82 × sin 60°2

= 33.5145 – 27.7128= 5.802 cm2

The area of grey region= Area of RUS + Area of RTS + Area of two circle= 2(5.802) + 2(π × 0.72)= 14.683 cm2

Paper 2

1. (a) cos ∠BOX = 36

∠BOX = 60° = 1.047 rad.

(b) sin ∠BOX = BXOB

BX = 6 (sin 60°) = 5.196 cm

BC = (BX)2 + (CX)2 = (5.196)2 + 92 = 108 = 10.392 cm

Length of arc BC = 6 (∠BOC)

= 61120° × π180° 2

= 12.568 cm

The perimeer of the shaded region = BC + Arc BC = 10.392 + 12.568 = 22.96 cm

(c) Area of sector OBC

= 12

× 62 × 1120° × π180° 2

= 37.704 cm2

Area of ∆OBC

= 12

(6)(6) sin 120°

= 15.5885 cm2

The area of the shaded region = Area of sector OBC – Area of ∆OBC = 37.704 – 15.5885 = 22.116 cm2



2. Q

P O

U

T R

S

8 cm 4 cm

(a) cos ∠QOT = OTOQ

= 48

∠QOT = 60° × π180°

= 1.047 rad.

(b) Perimeter of the shaded region = Arc PQ + Arc OQ + OP

= 38 × 1120° × π180° 24 + (8 × 1.047) + 8

= 16.7573 – 8.376 + 8 = 33.133 cm

(c) Area of sector OPQ

= 12

× 82 × 1120° × π180° 2

= 67.0293 cm

Area of sector ROQ

= 12

× 82 × 1.047

= 33.504

Area of ∆ROQ

= 12

× OR × QT

= 12

× 8 × 882 – 42

= 27.7128

Area of segment OUQ = Area of sector ROQ – Area of ∆ROQ = 33.504 – 27.7128 = 5.7912 cm2

The area of the shaded region = Area of sector OPQ – Area of segment OQU = 67.0293 – 5.7912 = 61.238 cm2

∠QRT = ∠QOT

3. (a) OQ = 9 + 4 = 13 cm OR = 132 – 122

= 5 cm

tan q = 125

q = 1.176 rad.

(b) RS = 5 – 4 = 1 cm

Length of arc PS = 4 × 1.176 = 4.704 cm

Perimeter of shaded region = PQ + QR + RS + arc PS = 9 + 12 + 1 + 4.704 = 26.70 cm

(c) Area of shaded region = Area of ∆ORQ – Area of sector OPS

= 1 12

× 12 × 52 – 1 12

× 42 × 1.1762 = 20.59 cm2

4. (a) tan q = RTOT

= 93

= 3

q = 1.249 rad.

(b) Length of arc PQ = 12 × 1.249 = 14.988

Length of arc PR = 9 × π2

= 14.139

OR = RT 2 + OT 2

= 92 + 32

= 9.4868 OR = 12 – 9.4868 = 2.5132

Perimeter of shaded region = 14.988 + 14.139 + 2.5132 = 31.64 cm

!

!

!



(c) Area of sector OPQ = 1

2(12)2(1.249)

= 89.928

Area of circle quadrant = 1

2(9)2 π

2 = 63.626

Area of ∆ORT

= 12

(OT )(RT )

= 12

(12 – 9)(9)

= 13.5

Area of shaded region = 89.928 – 63.626 – 13.5 = 12.802 cm2

5. (a) w = the circumference of the base of the cylinder

= 2πr

= 2 × 227

× 72

= 22 cm

The arc of the sector of circle P = w rpq = 22

rp1 π2 2 = 22

rpπ = 44

rp1227 2 = 44

rp = 14

l = rp + height of cylinder = 14 + 10 = 24 Therefore, l = 24 and w = 22

(b) Area of rectangular card = 22 × 24 = 528 cm2

Area of sector of circle P

= 12

(14)21 π2 2

= 49π = 49 × 22

7 = 154 cm2

Area of rectangle Q = 22 × 10 = 220 cm2

Area of the card which is not used = 528 – 154 – 220 = 154 cm2

6. E

G

F

D

B

N

MA

C

0.3 m1.3 m

0.5 m0.5 m

(a) cos ∠ABN = 0.31.3

∠ABN = 76°40' = 1.338 rad. ∠ABD = ∠ABN = 1.338 rad.

(b) The longer arc of the metal foil = Length of arc MF = 0.8 × (3.142 – 1.338) = 0.8 × 1.804 = 1.443 m

(c) AE = 0.8 + 0.3 = 1.1 EF = AN = 1.32 – 0.32 = 1.2649

Area of trapezium, ABFE

= 12

(AE + BF) × EF

= 12

(1.1 + 0.8) × 1.2649

= 1.2017 m2

Area of the shaded region = Area of ABFE – Area of sector AGM

– Area of sector BFM

= 1.2017 – 1 12

× 0.52 × 1.3382 – 1 1

2 × 0.82 × 1.8042

= 1.2017 – 0.1673 – 0.5773 = 0.457 m2 The area of the metal foil is 0.457 m2.

Additional Mathematics Form 4 Chapter 9 Differentiation


1. (a) limx → 2 6x

= 6(2) = 12

(b) limx → –1 (4 – 2x)

= 4 – 2(–1) = 6

(c) limx → 3 1

x + 5

= 13 + 5

= 18

2. (a) y = 5x2 – 2; P(1, 3)

Coordinates of Q Gradient of the chord PQ

(1.1, 4.05) 4.05 – 3———— = 10.5 1.1 – 1

(1.05, 3.5125) 3.5125 – 3————— = 10.25 1.05 – 1

(1.01, 3.1005) 3.1005 – 3————— = 10.05 1.01 – 1

(1.001, 3.010005) 3.010005 – 3—————— = 10.005 1.001 – 1

Hence, gradient of the tangent at point P = 10 Therefore, dy

dx = 10

DifferentiationPembezaan9

CHAPTER



3. (a) y = 4x2 …… y + y = 4(x + x)2

= 4[x2 + 2xx + (x)2] = 4x2 + 8xx + 4(x)2 ........

– , y = 8xx + 4(x)2

y ––– = 8x + 4x x lim y lim ––– = [8x + 4x] δx → 0 x δx → 0

dy ––– = 8x dx

5(b) y = — ……………… x 5 y + y = ——— …………… x + x 5 5 – , y = –––—– – — x + x x –5x = ––––—— x(x + x) y –5 ––– = –––––—– x x(x + x) lim y lim –5 —– = ———— x → 0 x

x → 0 x(x + x) dy 5 ––– = – –– dx x2

3(c) y = – ––x2 …… 4 3 y + y = – ––(x + x)2

4 3 3 3 = – ––x2 – ––xx – ––(x)2 … 4 2 4 3 3 – , y = – ––xx – ––(x)2

2 4 y 3 3 ––– = – ––x – ––(x) x 2 4 lim y lim 3 3 ––– = [– ––x – ––(x)] x → 0 x x → 0 2 4 dy 3 ––– = – ––x dx 2

4. (a) dydx

= 0

(b) dydx

= 56

(6x 6 – 1)

= 5x 5

(c) f(x) = 2x –2

f 9(x) = 2(–2x –2 – 1) = – 4

x 3

5. (a) f ′(x) = 6(4x3) = 24x 3

1 1 f ′(––) = 24(––)3

2 2 = 3

(b) f(x) = 5x–3

f ′(x) = 5(–3x–4) 15 15 f ′(–2) = – –––– = – ––– (–2)4 16

3(c) f(x) = —x–2

4 3 3 f ′(x) = ––(–2x–3) = – —– 4 2x3

3 f ′(1) = – –– 2

6. dy 1(a) ––– = 5(2x) – –– (3x2) dx 3 = 10x – x2

= x(10 – x)

(b) y = 6x–1 – x dy ––– = 6(–x–2) – 1 dx 6 = – –– – 1 x2

(c) f ′(x) = 5(–2x–3) – 2(–x–2) 10 2 = – –– + –– x3 x2

2 5 = ––(1 – ––) x2 x

7. (a) Let u = x2, v = 3x – 2 du dv ––– = 2x, ––– = 3 dx dx dy dv du ––– = u––– + v––– dx dx dx = x2(3) + (3x – 2)(2x) = x(3x + 6x – 4) = x(9x – 4)



(b) Let u = x2 – 1, v = x + 2 du dv ––– = 2x, ––– = 1 dx dx dy dv du ––– = u––– + v––– dx dx dx = (x2 – 1)(1) + (x + 2)(2x) = x2 – 1 + 2x2 + 4x = 3x2 + 4x – 1

x(c) Let u = x + 4, v = x2 – –– 2 du dv 1 —– = 1, ––– = 2x – — dx dx 2

1 x f ′(x) = (x + 4)(2x – ––) + (x2 – —) 2 2 1 x = 2x2 – ––x + 8x – 2 + x2 – –– 2 2 = 3x2 + 7x – 2

8. (a) Let u = 3x, v = 2x – 5 du dv ––– = 3, ––– = 2 dx dx

du dv v––– – u––– dx dx dy

dx = ––––––––––––

v2

(2x – 5)(3) – 3x(2) = –––––––––––––––— (2x – 5)2

6x – 15 – 6x = ––––––––––– (2x – 5)2

15 = – –––––––– (2x – 5)2

(b) Let u = 3x2, v = 2x + 5 du dv ––– = 6x, ––– = 2 dx dx dy (2x + 5)(6x) – (3x2)(2) ––– = –––––––––––––––––––– dx (2x + 5)2

12x2 + 30x – 6x2 = ––––––––––––––– (2x + 5)2

6x2 + 30x = –––––––— (2x + 5)2

6x(x + 5) = –––––––— (2x + 5)2

(c) Let u = 4x2, v = 5 – 2x du dv ––– = 8x, ––– = –2 dx dx

dy (5 – 2x)(8x) – 4x2(–2) ––– = ––––––––––––———– dx (5 – 2x)2

40x – 16x2 + 8x2 = ––––––————– (5 – 2x)2

8x(5 – x) = –––––––— (5 – 2x)2

9. (a) Let u = 3x – 5, hence y = u3

du dy ––– = 3, ––– = 3u2 dx du dy dy du Therefore, ––– = ––– × ––– dx du dx = 3u2 × 3 = 9u2

= 9(3x – 5)2

(b) Let u = (2x + 3)3, v = (4 – x)2

du dv ––– = 3(2x + 3)2(2), ––– = 2(4 – x)(–1) dx dx = 6(2x + 3)2 = –2(4 – x)

Therefore, dv du f ′(x) = u––– + v––– dx dx = (2x + 3)3[–2(4 – x)] + (4 – x)2[6(2x + 3)2] = (4 – x)(2x + 3)2[–2(2x + 3) + 6(4 – x)] = (4 – x)(2x + 3)2(–4x – 6 + 24 – 6x) = (4 – x)(2x + 3)2(18 – 10x) = 2(4 – x)(9 – 5x)(2x + 3)2

(c) Let u = 4x, v = (3x2 + 5)–2

du dv ––– = 4, ––– = –2(3x2 + 5)–3(6x) dx dx

Therefore, f ′(x) = 4x[–2(3x2 + 5)–3(6x)] + (3x2 + 5)–2(4) = 4(3x2 + 5)–3(–12x2 + 3x2 + 5) = 4(3x2 + 5)–3(5 – 9x2) 4(5 – 9x2) = –––––––— (3x2 + 5)3



10. dy(a) ––– = 4x + 6 dx dy When x = 1, ––– = 4(1) + 6 dx = 10 Therefore, the gradient of the tangent when

x = 1 is 10.

dy(b) ––– = (2x + 1)(2x) + (x2 – 3)(2) dx At point (–1, 2), dy ––– = [2(–1) + 1][2(–1)] + [(–1)2 – 3](2) dx = –2

Therefore, the gradient of tangent at point (–1, 2) is –2.

dy (x2)(5) – (5x – 2)(2x)(c) ––– = –––––––––––––––––– dx x4

–5x2 + 4x = ——–––– x4

dy –5(2)2 + 4(2) At point (2, 2), ––– = ——––––––– dx 24 3 = – — 4 Therefore, the gradient of tangent at point 3 (2, 2) is – —. 4

11. dy(a) ––– = 2x + 3 dx dy At point (1, –1), ––– = 2(1) + 3 dx = 5 Gradient of the tangent = 5 Therefore, the equation of the tangent at

point (1, –1) is y – (–1) = 5(x – 1) y + 1 = 5x – 5 y = 5x – 6 1 Gradient of the normal = – — 5 Therefore, the equation of the normal at

point (1, –1) is 1 y – (–1) = – —(x – 1) 5 1 1 = – —x + — 5 5 1 4 y = – —x – — 5 5

dy(b) ––– = 2(2x2 – 7)(4x) dx = 8x(2x2 – 7) dy When x = 2, ––– = 8(2)[2(2)2 – 7] dx = 16 y = [2(2)2 – 7]2

= 1 Therefore, the equation of the tangent at

point (2, 1) is y – 1 = 16(x – 2) y = 16x – 31 1 Gradient of the normal = – ––– 16 Therefore, the equation of the normal at

point (2, 1) is 1 y – 1 = – – –– (x – 2) 16 1 2 y = – –––x + ––– + 1 16 16 1 9 y = – –––x + — 16 8

dy (2x – 5)(1) – (x + 3)(2)(c) ––– = —————–––––––––– dx (2x – 5)2

–11 = —––––– (2x – 5)2

dy –11 When x = 3, ––– = ––––—––– = –11 dx [2(3) – 5]2

3 + 3 y = —––––– 2(3) – 5 = 6 Therefore, the equation of the tangent at

point (3, 6) is y – 6 = –11(x – 3) y = –11x + 39 1 Gradient of the normal = –– 11 Therefore, the equation of the normal at

point (3, 6) is 1 y – 6 = — (x – 3) 11 1 63 y = — x + ––– 11 11



12. dy(a) ––– = 6x + 6 dx dy At turning point, ––– = 0 dx 6x + 6 = 0 x = –1 When x = –1, y = 3(–1)2 + 6(–1) – 2 = –5 Therefore, the turning point is (–1, –5).

Value of x –1.1 –1 –0.9 dyValue of — dx –0.6 , 0 0 0.6 . 0

dySketch of — dx

Therefore, (–1, – 5) is a minimum point.

(b) y = 3 + 5x – 2x2

dy ––– = 5 – 4x dx dy At turning point, ––– = 0 dx 5 – 4x = 0 1 x = 1— 4 1 1 1 When x = 1—, y = 3 + 5(1—) – 2(1—)2 4 4 4 1 = 6— 8 1 1 Therefore, the turning point is (1—, 6—). 4 8

Value of x 1 11 — 4

11 — 2

dyValue of — dx 1 . 0 0 –1 , 0

dySketch of — dx

Therefore, (1 14

, 6 18

) is a maximum point.

13. (a) (i) 5x + 5x + 6x + y + y = 120 16x + 2y = 120 2y = 120 – 16x y = 60 – 8x ...a

1 A = 6x(y) + —(6x)(4x) 2 = 6xy + 12x2 ...............b

Substitute a into b. A = 6x(60 – 8x) + 12x2

= 36x(10 – x)

dA(ii) ––– = 360 – 72x dx For maximum or minimum value

of A, dA ––– = 0 dx 360 – 72x = 0 x = 5

Value of x 4.9 5 5.1 dAValue of — dx 7.2 . 0 0 –7.2 , 0

dASketch of — dx

Therefore, A is maximum when x = 5.

(iii) Maximum area = 36(5)(10 – 5) = 900 cm2



(b) (i) Total surface area of the box = 216 cm2

4xh + 2x2 = 216 216 – 2x2 h = —–––––– …… 4x V = x2h ……

Substitute into . 216 – 2x2 V = x2 1––––—–—2 4x 1 = 54x – ––x3

2 dV 3(ii) ––– = 54 – ––x2

dx 2 For maximum or minimum value of

V, dV ––– = 0 dx 3 54 – –– x2 = 0 2 3 –– x2 = 54 2 x2 = 36 x = 6

Value of x 5.9 6 6.1 dVValue of — dx 1.785 . 0 0 –1.815 , 0

dVSketch of — dx

Therefore, V is maximum when x = 6.(iii) Maximum value of V 1 = 54(6) – ––(6)3 2 = 216 cm3

14. (a) Let V = Volume of water h = Height of water

V = πr2h = π(15)2h = 225πh dV ––– = 225π dh dV dV dh ––– = ––– × ––– dt dh dt dh 25 = 225π × ––– dt dh 25 ––– = ––––– dt 225π 1 = ––– cm s–1

9π

16(b) A = 3π( + r2) r = 3π(16r–1 + r2) dA ––– = 3π(–16r–2 + 2r) dr dA dA dr ––– = ––– × ––– dt dr dt 16 2 = 3π(2r – ) (—) r2 3 2 16 = 3π(—)[2(4) – ] 3 42

= 14π cm2 s–1

15. (a) Volume of the cube, V = x3

dV ––– = 3x2

dx δx = 5.9 – 6 = –0.1 cm δV dV –––≈––– δx dx dV δV ≈ –––× δx dx = 3x2 × (–0.1) = 3(6)2(–0.1) = –10.8 cm3

Therefore, the small change in the volume of the cube is –10.8 cm3.

(b) Total surface area of the cylinder, A = 2πrh + 2πr2

dA ––– = 2πh + 4πr dr δr = 3.05 – 3 = 0.05 cm δA dA –––≈ ––– δr dr

δA ≈ dAdr

× δr

= [2π(15) + 4π(3)](0.05) = 2.1πcm2

Therefore, the small change in the surface area of the cylinder is 2.1π cm2.



1(c) Volume of cone, V = ––πr2h 3 1 = ––πr2(9) 3 = 3πr2 dV ––– = 6πr dr

δr = 3.96 – 4 = –0.04 cm

δVδr

≈ dVdr

dV δV ≈ –––× δr dr = 6πr × δr = 6π(4)(–0.04) = –0.96πcm3

Therefore, the small change in the volume of the cone is –0.96π cm3.

16. (a) y = x4

dy ––– = 4x3 dx dy When x = 3, ––– = 4(3)3 dx = 108 δx = 3.01 – 3 = 0.01 δy dy ––– ≈ ––– δx dx dy δy ≈ ––– × δx dx = 108(0.01) = 1.08 When x = 3, y = 34

= 81 Therefore, 3.014 ≈ y + δy = 81 + 1.08 = 82.08

(b) y = 7x –3

dy ––– = –21x –4 dx

When x = 2, dy –21 ––– = —– dx 24

21 = – —– 16

δx = 1.95 – 2 = –0.05 dy δy ≈ ––– × δx dx 21 = – –— × –0.05 16 = 0.06563 7 When x = 2, y = — 23

7 = — 8 7 7 Therefore, ––—– ≈—+0.06563 1.953 8 = 0.9406

17. dy(a) ––– = 24x3 + 9x2 dx d 2y ––— = 72x2 + 18x dx2

dy(b) ––– = 3(3x – 4)2(3) dx = 9(3x – 4)2

d2y –––– = 18(3x – 4)(3) dx2

= 54(3x – 4)

(c) f(x) = 5x 3 – 2x

2

= 5x – 2x –2

f 9(x) = 5 + 4x –3

f (x) = 4(–3)x –4

= – 12x 4



18. dy(a) ––– = 6x + 6 dx dy At turning point, ––– = 0 dx 6x + 6 = 0 x = –1

When x = –1, y = 3(–1)2 + 6(–1) + 5 = 2 Therefore, the turning point is (–1, 2). d2y ––— = 6 . 0 dx2

Therefore, (–1, 2) is a minimum point.

(b) y = x(27 – x2) = 27x – x3

dy ––– = 27 – 3x2 dx dy At turning point, ––– = 0 dx 27 – 3x2 = 0 x2 = 9 x = ±3 When x = 3, when x = –3, y = 54 y = –54 d2y d2y ––— = –6x ––— = –6x dx2 dx2

= –18 , 0 = 18 . 0

Therefore, (3, 54) is a maximum point and (–3, –54) is a minimum point.

x – 4(c) y = —––– x2

= x–1 – 4x –2

dy ––– = –x –2 + 8x –3 dx dy At turning point, ––– = 0 dx 1 8 – –– + — = 0 x2 x3

–x + 8 ——— = 0 x3 x = 8 8 – 4 1 y = ——— = —– 82 16 d2y 2 24 —– = — – — dx2 x3 x4

2 24 1 = — – — = – —— 83 84 512 1 Therefore, 18, —2 is a maximum point. 16



SPM Practice 9

Paper 1

1. Let A represent the area of the land. A = 5x(5 – x) = 25x – 5x2

dAdx

= 25 – 10x

At turning point, dAdx

= 0

25 – 10x = 0 10x = 25 x = 2.5

d 2Adx2 = –10 , 0

Hence, A is maximum when x = 2.5.

Total length of fencing material= 2[5(2.5) + (5 – 2.5)]= 2(12.5 + 2.5)= 30 m

2. (a) x = 3t 2 + 8

dxdt

= 6t

(b) dydt

= dydx

× dxdt

24t 2 = dydt

× 6t

dydt

= 24t 3

6t = 4t 2

3. (a) y = 2x2 – 7x dydx

= 4x – 7

At point Q(1, –5), dydx

= 4(1) – 7 = –3Hence, the gradient of the tangent to the curve at point Q = –3.

(b) The gradient of the normal to the curve at point Q = 1

3.

The equation of the normal at point Q(1, –5) is

y – (–5) = 13

(x – 1)

y + 5 = 13

(x – 1)

y = 13

(x – 1) – 5

= 13

x – 13

– 5

y = 13

x – 163

4. (a) k′(x) = 3px 2 + 10x – 4

(b) k′(x) = 3px 2 + 10x – 4 k ″(x) = 6px + 10 k ″(1) = 6p(1) + 10 7 = 6p + 10 6p = –3 p = – 1

2

5. y = x 2(hx + 3) = hx 3 + 3x 2dydx

= 3hx 2 + 6x

At (–2, 0), dydx

= 3h(–2)2 + 6(–2)

6 = 12h – 12

h = 1812

= 32

6. y = 11 – 18x

= 11 – 18x –1

dydx

= 18x –2

= 18x2



when y = 5, 5 = 11 – 18x

18x

= 6

x = 3

dydx

= 1832

= 2

When y change from 5 to 5 + k,dydx

≈ dydx

for small changes in x and y.

dx≈ dy

1 dydx 2

= k2

7.

h cm

2πr cm

h cm

r cm

Let length=2πr cm width = h cm

Perimeter = 274πr + 2h = 27

h = 27–4πr2

= 13.5–2πr

Volumeof cylinder=πr2h =πr2(13.5–2πr) = 13.5πr2 – 2π2r3

At maximum, dVdr

= 0

27πr – 6π2r2 = 0 3πr(9 – 2πr) = 0

3πr = 0or 9 – 2πr = 0 r = 0 2πr = 9 r = 9

2π

When r = 92π

, h =13.5–2π1 92π2

= 4.5The length=2πr = 2π1 9

2π2 = 9Therefore, the length is 9 cm and the width is 4.5 cm.

8. (a) y = 3x(x – 4) = 3x 2 – 12x dy

dx = 6x – 12

(b) dydx

= 0

6x – 12 = 0 6x = 12 x = 2

(c) y = 3(2)(2 – 4) = –12

9. V = 43

πr 3

dVdr

= 43

π(3r 2)

= 4 πr 2

dVdt

= dVdr

× drdt

16 π = 4 πr 2 × 0.25

r 2 = 16π4π× 0.25

= 16 r = 4 cm

10. Volume of block of ice, V = x 3

dVdx

= 3x 2

dVdt

= dVdx

× dxdt

14.7 = 3x 2 × dxdt

When x = 14, dxdt

= 14.73(14)2

= 0.025 cm s–1



Paper 2

1. (a) (i) y = x2(6 – x) – 34

= 6x2 – x3 – 34

dydx

= 12x – 3x2

The gradient function of the curve is 12x – 3x2.

(ii) At turning points, dydx

= 0

12x – 3x2 = 0 3x(4 – x) = 0 3x = 0 or 4 – x = 0 x = 0 x = 4

When x = 0, y = 6(0)2 – (0)3 – 34

= – 34

When x = 4, y = 6(4)2 – (4)3 – 34

= 96 – 64 – 34

= 31 14

Hence, the turning points are at

10, – 34 2 and 14, 31 1

4 2.

(b) dydx

= 12x – 3x2

d 2ydx2 = 12 – 6x = 6(2 – x)

At 10, – 34 2, d 2y

dx2 = 6(2 – 0)

= 12 . 0

Hence, 10, – 34 2 is a minimum point.

At 14, 31 14 2, d 2y

dx2 = 6(2 – 4)

= –12 . 0

Hence, 14, 31 14 2 is a maximum point.

2. (a) y = x 3 – 9x 2 + 24x – 12

dydx

= 3x 2 – 18x + 24

At P(1, 4), x = 1,

dydx

= 3(1)2 – 18(1) + 24 = 9

(b) Gradient of normal = – 19

Equation of the normal is

y – 4 = – 19

(x – 1)

y = – 19

x + 379

(c) At turning point, dydx

= 0

3x 2 – 18x + 24 = 0 x 2 – 6x + 8 = 0 (x – 4)(x – 2) = 0 x = 4 or x = 2 x = 4 is coordinate for A(4, 4). When x = 2, y = (2)3 – 9(2)2 + 24(2) – 12 = 8

Hence, the coordinates of B are (2, 8).

d 2ydx2 = 6x – 18

At B(2, 8), d 2ydx2 = 6(2) – 18

= –6 , 0 Therefore, B is a maximum point.



3. Let A be the area of the whole land. A = xy ....................

Total length of fencing material is,4y + 2x = 240 ...................

y = 240 – 2x4

= 60 – 12

x .......... 3

Subsitute 3 into ,

A = x160 – x2 2

= 60x – x2

2dAdx

= 60 – x

At turning point, dAdx

= 0

60 – x = 0 x = 60d 2Adx2 = –1 , 0

Therefore, A is maximum when x = 60.

From 3: When x = 60, y = 60 – 12

(60)

= 60 – 30 = 30

From : When x = 60, y = 30, A = 60(30) = 1 800 m2

The maximum area is 1 800 m2.

4. dydx

= px 2 – qx

At turning point, x = 2, dydx

= 0

p(2)2 – q(2) = 0 4p – 2q = 0 2p – q = 0 ……

At x = –2, dydx

= –24

p(–2)2 – q(–2) = –24 4p + 2q = –24 2p + q = –12 …… + , 4p = –12 p = –3

Substitute p = –3 into . 2(–3) – q = 0 q = – 6

5. (a) dydx

= 16x – 2x2

At turning point, x = h, dydx

= 0

16(h) – 2h2

= 0

16h = 2h2

h3 = 216

= 18

h = 12

(b) dydx

= 16x – 2x2

d 2ydx2 = 16 – 2(–2x –3)

= 16 + 4x3

4

When x = 12

, d 2ydx2 = 16 + 4

1 12 2

3

= 48 . 0 Hence, the turning point is a minimum

point.

Additional Mathematics Form 4 Chapter 10 Solution of Triangles


Solution of TrianglesPenyelesaian Segi Tiga10

CHAPTER

1. 7.6 3.4 (a) ——— = ————– sin 71° sin /ACB 3.4 sin 71° sin /ACB = ————— 7.6 /ACB = 25.02°

5.4 3.2(b) ———–– = ————– sin 116° sin ∠BCA 3.2 sin 116° sin ∠BCA = ———–—— 5.4 ∠BCA = 32.18°

(c) ∠A = 180° – 76° – 37° = 67° AC 6.5 ——— = ———– sin 76° sin 67° 6.5 sin 76° AC = ————— sin 67° = 6.852 cm

2. (a)

8.4 cm 6.6 cm6.6 cm

40°

E

G2G1F

sin ∠EGF8.4

= sin 40°6.6

sin ∠EGF = 8.4 sin 40°6.6

∠EGF = 54.90° or 125.10°

(b)

7.2 cm

25.9°C2 C1

B

A4.4 cm

4.4

cm

sin ∠ACB sin 25.9° ————— = ———— 7.2 4.4 ∠ACB = 45.62° or 134.38°

When ∠ACB = 45.62°, ∠BAC = 108.48° 4.4 sin 108.48° BC = ——————– sin 25.9° = 9.554 cm

When ∠ACB = 134.38°, ∠BAC = 19.72° 4.4 sin 19.72° BC = —————— sin 25.9° = 3.399 cm

3. (a) (i) ∠DAC = 180° – 35° – 77° = 68° CD 8 ––—— = —–—— sin 68° sin 35° CD = 12.93 cm

(ii) ∠BAC = 77° – 29° = 48° BC 8 ——— = –——— sin 48° sin 29° BC = 12.26 cm



sin ∠EGF sin 48°(b) (i) ————– = ——— 6.4 5.0

∠EGF = 72.03° ∠EGH = 180° – 72.03° = 107.97°

EH 5(ii) ————— = ——— sin 107.97° sin 24° EH = 11.69 cm

(c) (i) ∠ABD = 180° – 123° – 25° = 32° In ΔABD, AD 5.2 ——— = ——— sin 32° sin 25° AD = 6.520 cm BC = AD = 6.520 cm BD 5.2(ii) –——— = ——— sin 123° sin 25° BD = 10.32 cm

(d) (i) ∠QPR = 63° – 42° = 21° PR 7 —–—– = –——— sin 42° sin 21° PR = 13.07 cm

(ii) ∠RPS = 180° – 63° – 81° = 36° RS 13.07 —–—– = –——— sin 36° sin 81° RS = 7.778 cm

4. 8.32 + 5.12 – 6.72(a) cos ∠LKM = ———————– 2(8.3)(5.1) = 0.5907 ∠LKM = 53.79°

(b) ∠PRQ is the smallest angle. 112 + 132 – 8.52 cos ∠PRQ = ——————— 2(11)(13) = 0.7614 ∠PRQ = 40.41°

(c) AC2 = 7.42 + 5.52 – 2(7.4)(5.5) cos 68° = 54.52 AC = 7.384 cm

(d) LM2 = 7.32 + 5.42 – 2(7.3)(5.4) cos 126° = 128.79 LM = 11.35 cm

5. (a) (i) In ΔADC, 7.42 + 2.82 – 5.92 cos ∠ADC = –——————— 2(7.4)(2.8) ∠ADC = 47.89°

(ii) In ΔABD, 7.4 sin 47.89° sin ∠ABD = —————— 6.6 ∠ABD = 56.28°

(b) (i) In∆ABD, BD2 = 72 + 92 – 2(7)(9) cos 72° = 91.06 BD = 9.543 cm

(ii) In ΔBCD, 9.543 sin 52° sin ∠BCD = ———–—–— 12.3 ∠BCD = 37.69°

6. (a) Areaof∆KLM 1= —(8)(10) sin 53° 2= 31.95 cm2

(b) sin ∠QPR sin 75° ———–— = ——— 10.2 12.3 ∠QPR = 53.23°

∠QRP = 180° – 75° – 53.23° = 51.77°

Areaof∆PQR 1= —(12.3)(10.2) sin 51.77° 2= 49.28 cm2



(c) 9.72 + 12.22 – 14.92 cos ∠Y = ———–————— 2(9.7)(12.2) ∠Y = 84.93°

Area of ΔXYZ 1= —(9.7)(12.2) sin 84.93° 2= 58.94 cm2

7. (a) (i) In ΔCFG, CF2 = 42 + 52

= 41

In ΔEFH, FH2 = 42 + 82

= 80

In ΔCGH, CH2 = 52 + 82

= 89

In ΔCFH, 41 + 80 – 89 cos ∠CFH = ——————— 2(!w41)(!w80 ) ∠CFH = 73.78°

(ii) Areaof∆CFH 1 = —(CF)(FH) sin ∠CFH 2 1 = —(41 )(80 ) sin 73.78° 2 = 27.50 cm2

(b) (i) In ΔTPQ, TQ2 = 72 + 82 = 113

In ΔPQR, PR2 = 82 + 122 = 208

In ΔPRT, RT2 = 72 + 208 = 257

In ΔQRT, 113 + 257 – 122 cos ∠QTR = ——————— 2(!w113)(!w257) ∠QTR = 48.46°

(ii) Areaof∆QTR 1 = —(!w113)(!w257) sin 48.46° 2 = 63.78 cm2



SPM Practice 10

Paper 2

1. (a) (i) In∆CDE,

CEsin 105°

= EDsin 32°

CE = 5.4sin 32°

× sin 105°

= 9.843 cm

(ii) In∆ABC, BC2 = 4.52 + 8.72

– 2(4.5)(8.7) cos 143° BC2 = 158.473 BC = 12.589 cm BE = BC – CE = 12.589 – 9.843 = 2.746 cm

(iii) Areaof∆CDE

= 12

× ED × EC × sin ∠CED

= 12

× 5.4 × 9.843 × sin 43°

= 18.125 cm2

(b) (i)

32º

105º5.4 cm

E C

D

Eʹ

(ii) In∆CDE, ∠DEC = 180° – 105° – 32° = 43° ∠EE’D = ∠DEC = 43° ∠CDE9 + ∠DCE9 = ∠DE9E ∠CDE9 = 43 – 32° = 11°

2. (a) (i) In∆SRQ,

cos ∠RSQ = 72 + 132 – 82

2(7)(13) = 0.846 ∠RSQ = 32.2°

(ii) In∆PQS, ∠PQS = ∠RSQ = 32.2°

PSsin ∠PQS

= PSsin ∠PQS

PSsin 32.2°

= 13sin 118°

PS = 13 sin 32.2°sin 118°

= 7.846 cm

(b) (i)

S

P Q

R

118º

8 cm13 cm

7 cm Rʹ

(ii) In∆SQR9,

QR9sin ∠R9SQ

= SQsin ∠SR9Q

8sin 32.2°

= 13sin ∠SR9Q

sin ∠SR9Q = 13 sin 32.2°8

= 0.866 ∠SR9Q = 59.99° ∠RQR9 = 180° – 2(59.99°) = 60.02°

Areaof∆QRR9

= 12

(8)(8) sin 60.02°

= 27.718 cm2



3. (a) (i) Areaof∆PQR = 25

12

(PR)(QR) sin 36° = 25

QR = 25 × 28 sin 36°

= 10.633 cm

(ii) In∆PQR, PQ 2 = PR 2 + QR 2 – 2(PR)(QR) cos 36° = 82 + 10.632 – 2(8)(10.63) cos 36° PQ = 6.277 cm

(iii) PQsin 36°

= 8sin ∠PQR

sin /PQR = 8 sin 36°6.277

/PQR = 48.51°

(b) (i)

P

Q RQ�

36°48.51°

8 cm6.277 cm 6.277 cm

/Q′PR + /PRQ′ = /PQ′Q /Q′PR = 48.51° – 36° = 12.51°

(ii) Areaof∆PQ′R

= 12

(8)(6.277) sin 12.51°

= 5.439 cm2

4. (a) (i) In∆ABD,

sin ∠ADB11

= sin 93.58°14

sin ∠ADB = 11 sin 93.58°14

∠ADB = 51.64°

(ii) In∆ADC, ∠ADB = ∠ACD = 51.64° ∠CAD = 180° – 51.64° – 51.64° = 76.72°

CDsin 76.72°

= 8sin 51.64°

CD = 8 sin 76.72°sin 51.64°

= 9.93 cm

(iii) In∆ABC, ∠BAC = 93.58° – 76.72° = 16.86°

Areaof∆ABC = 1

2 (11)(8) sin 16.86°

= 12.762 cm2

(b)

9.93 cm

9.93 cm

76.72°

B

E

DC

A

11 cm

8 cm

93.58°

In ∆ADE, ∠EAD = ∠CAD = 76.72°

In∆ABE, ∠BAE = 93.58° + 76.72° = 170.3°

BE 2 = 112 + 82 – 2(11)(8) cos 170.3° BE = 18.934 cm



5. (a) (i) In∆ACD, ∠ACD = 180° – 82° – 35° = 63°

ACsin ∠ADC

= 8sin 35°

ACsin 63°

= 8sin 35°

AC = 8sin 35°

× sin 63°

= 12.43 cm

(ii) In∆ABC, 12.432 = 52 + 102 – 2(5)(10) cos ∠ABC 12.432 = 125 – 100 cos ∠ABC

cos ∠ABC = 125 – 12.432

100 cos ∠ABC = –0.2950 ∠ABC = 107.16°

(iii) Area of quadrilateral ABCD = Area of ABC + Area of ACD

= 12

(5)(10) sin 107.16° +

12

(8)(12.43) sin 82°

= 73.12 cm2

(b) (i)

A

B’

C

B

12.43 cm

10 cm5 cm

(ii) ∠C9B9A9 = 180° – 107.16° = 72.84°

6. (a) In∆ACD,

ACsin 68°

= 12.5sin 82°

AC = 11.704 cm

(b) In∆ACD, ∠ACD = 180° – 68° – 82° = 30°

In∆ABC, ∠CAB = ∠ACD = 30° (DC // AB) BC 2 = AC 2 + AB 2 – 2(AC)(AB) cos ∠CAB = 11.72 + 6.32 – 2(11.7)(6.3) cos 30° = 48.91 BC = 6.994 cm

(c) In∆ABC,

sin ∠ABC11.7

= sin 30°6.994

sin ∠ABC = 11.7 sin 30°6.994

∠ABC = 56.77° Since ∠ABC is obtuse, ∠ABC = 180° – 56.77° = 123.23°

(d) Areaof∆ABC

= 12

(6.3)(11.7) sin 30°

= 18.428 cm2

Additional Mathematics Form 4 Chapter 11 Index Number


Index NumberNombor Indeks11

CHAPTER

1. 16.38(a) I = ——– × 100 12.60 = 130

11 151(b) I = ——— × 100 8 260 = 135

2. 420(a) I = —— × 100 500 = 84

1 250(a) (i) I1 = ——– × 100 1 000 = 125

1 600(ii) I2 = ——– × 100 1 250 = 128

3. Q1(a) I = —– × 100 Q0

350 75 = —— × 100 Q0

350 × 100 Q0 = ———––– 75 = RM467

(b) Let Q0, Q1 and Q2 be the prices of item K in year 1996, year 2001 and year 2006 respectively.

Q1 —– × 100 = 110 …… Q0

Q2 —– × 100 = 130 …… Q0

Q1 110 —– : —– = —— Q2 130 110 Q1 = —— × 78 130 = RM66



4. – ∑Iiwi(a) Composite index, I = ——– ∑wi

110 × 4 + 118 × 8 + 120 × 2 + 125 × 6 = ———————————————————— 4 + 8 + 2 + 6 = 119

– ∑Iiwi(b) Composite index, I = ——– ∑wi

125 × 3 + 120 × 4 + 110 × 2 + 105 × 1 = ———————————————————— 3 + 4 + 2 + 1 = 118

5. – ∑Iiwi(a) Composite index, I = ——– ∑wi

90 × 1 + x × 2 + 125 × 2 + 112 × 5 111 = ——————————————–—— 1 + 2 + 2 + 5 900 + 2x = ———— 10 x = 105

– ∑Iiwi(b) Composite index, I = ——– ∑wi

112 × 4 + 130 × 1 + 120 × 2 + x × 3 119 = ——————————————–—— 4 + 1 + 2 + 3 818 + 3x = ———— 10 1 190 = 818 + 3x 3x = 372 x = 124



6. 36(a) (i) x = —– × 100 30

= 120 y 124 = —– × 100 50 y = 62 120 125 = —— × 100 z z = 96

– 120 × 3 + 124 × 5 + 125 × 2(ii) I = ————————————— 3 + 5 + 2 = 123

(b) Price index for B, 34.20 h = ——– × 100 30 = 114

_ 136 × k + 114 × 2 + 108 × 3 I = –———————————— k + 2 + 3 136k + 552 120 = ————— k + 5 120k + 600 = 136k + 552 16k = 48 k = 3

(c) (i) 1.56x × 100 = 104

x = 1.50

y = 5.584.50

× 100

= 124

(ii) Ingredient Price

index, IWeightage,

w Iw

Flour 120 0.4 48Sugar 104 0.2 20.8Butter 124 0.3 37.2Pineapple 150 0.1 15

∑Iw = 121

I– =

∑Iw∑w

= 1211

= 121



SPM Practice 11

Paper 2

1. (a) Let Q0 = Price of component A in 2012, Q1 = Price of component A in 2014.

105 = Q1

RM12 × 100

Q1 = 105100

× RM12

= RM12.60

(b) 108 = 105(2) + 90(1) + x(4) + 112(3)10

1 080 = 636 + 4x 4x = 1 080 – 636

x = 4444

= 111 Hence, the percentage of price change for component C is 11%.

(c) (i) The composite index for the expenses in 2015 based on 2012

= 108 × 109100

= 117.72 = 118

(ii) Let P0 = Price of the toy in the year 2012, P1 = Price of the toy in the year 2015.

118 = P1

RM60 × 100

P1 = 118 × RM60100

= RM70.80



2. (a) (i) 125 = RM8.00Price of P in 2008

× 100

Price of P in 2008 = RM8 × 100125

= RM6.40

120 = Price of R in 2010RM4.50

× 100

Price of R in 2010 = 120100

× RM4.50

= RM5.40

(b) Composite index for 2010 based on 2008.

124 = 125(3) + 115(1) + 120(4) + 2n10

124 = 970 + 2n10

2n = 124(10) – 970

n = 2702

= 135

(c) Ingredients Price index for the year 2010 based on the year 2008

Pric index for the year 2012 based on the year 2008 Weightage

P 125 125 × 96100

= 120 3

Q 115 115 1

R 120 120 × 110100

= 132 4

S 135 135 2

Composite index for 2012 based on 2008,

120(3) + 115(1) + 132(4) + 135(2)10

= 127.3

(d) Let Q1 be the cost in year 2012, Q0 be the cost in year 2008.

Q1

RM26 × 100 = 127.3

Q1 = 127.3100

× RM26

= RM33.10



3. (a) Price index for transportation of 2010 based on 2008

= Transportation cost in 2010Transportation cost in 2008

× 100

140 = RM1 750Transportation cost in 2008

× 100

Transportation cost in 2008

= RM1 750 × 100140

× 100

= RM1 250

(b) Ratio of electricity : water : transportation = 72 : 180 : 108 = 2 : 5 : 3

Item Price index, I

Weightage, w Iw

Electricity 125 2 250Water 110 5 550

Transportation 140 3 420Total 10 1 220


I– = 1 22010

= 122

(c) Total expenditure in 2010Total expenditure in 2008

× 100

= 122

Total expenditure in 2010

= 122100

× RM25 000

= RM30 500

(d)

Item Price index, I

Weightage, w Iw

Electricity 125 × 110100

= 137.5

2 275

Water 110 5 550Transportation 140 × 125

100 = 175

3 525

Total 10 1 350


I– = 1 35010

= 135

4. (a) 3.00x

× 100 = 150

x = 3.00 × 100150

x = 2.00

y = 3.202.50

× 100

y = 128

z8.00

× 100 = 115

z = 115 × 8.00100

z = 9.20

(b) Composite index 17(150) + 12(125) + 25(128) + 46(115) = —––———————————––––––– 100 = 12 540

100 = 125.40

(c) Corresponding cost in the year 2008

= 125.40100

× RM25.00

= RM31.35

(d) Composite index for the year 2010 based on the year 2006

= 105100

× 125.4

= 131.67



5. (a) (i) h = 17.5014.00

× 100

= 125

(ii) 110 = P2012

14.00 × 100

P2012 = 110100

× 14.00

= RM15.40

(b) (i)

I w IwXYZ

105110112

k15

105k110560

∑ k + 6 670 + 105k

Composite Index, –I = 109

∑Iwi

∑wi

= 109

670 + 105kk + 6

= 109

670 + 105k = 109k + 654 4k = 16 k = 4

(ii) Corresponding price in the year 2010

= 100109

× RM48.50

= RM44.50

(c) For ingredient z.

I2014/2012 = I2014/2010

I2012/2010

× 100

= 120112

× 100

= 107.1

6. (a) x1.80

× 100 = 120

x = 1.80 × 120100

= 2.16

2.802.00

× 100 = y y = 140

(b) Angle of sector for component D = 360° – 160° – 96° – 24° – 32° = 48° A : B : C : D : E = 96 : 24 : 32 : 48 : 160 = 12 : 3 : 4 : 6 : 20

Component I w IwA 110 12 1 320B 120 3 360C 150 4 600D 116 6 696E 140 20 2 800

Total 45 5 776 –

I = 5 776

45 = 128.4

(c) Production cost of watch in year 2006

= RM40 × 128.4100

= RM51.36

Production cost of watch in year 2009

= RM51.36 × 110100

= RM56.50


Paper 1

1. (a) The relation is "is twice of".(b) Domain = {2x, 2y, 2z}

2. (a) g(x) = 4x + 1 g(–1) = 4(–1) + 1 = –3

(b) h(x) = px – q h(2) = 2p – q gh(2) = 7 g(2p – q) = 7 4(2p – q) + 1 = 7 8p – 4q + 1 = 7 8p – 4q = 6 4p – 2q = 3

p = 2q + 34

3. g : x → x2 + kg : 3 → 1332 + k = 13 k = 4

4. (a) 32

(b) (i) Let y = 1 + x2x – 3

y(2x – 3) = 1 + x 2xy – 3y = 1 + x 2xy – x = 1 + 3y x(2y – 1) = 1 + 3y

x = 1 + 3y2y – 1

h–1(y) = 1 + 3y2y – 1

h–1(x) = 1 + 3x2x – 1

, x ≠ 12

(ii) h–1(4) = 1 + 3(4)2(4) – 1

= 137

5. (a) fg(–1) = f(2–1)

= f 1 12 2

= 5 + 21 12 2

= 6

(b) Let y = 5 + 2x

x = y – 52

f –1(y) = y – 52

f –1(x) = x – 52

f –1g(x) = 32

f –1(2x) = 32

2x – 52

= 32

2x = 8 2x = 23 x = 3

6. b2 – 4ac , 0 (–3)2 – 4(2 – p)(2) , 0 9 – 16 + 8p , 0 8p , 7

p , 78

7. h = 50t – 5t2 = –5(t2 – 10t) = –5[t2 – 10t + (–5)2 – (–5)2] = –5[(t – 5)2 – 25] = –5(t – 5)2 + 125The maximum height = 125 m.

End-of-Year Assessment

Additional Mathematics Form 4 End-of-Year Assessment


8. (2x + 3)2 – 27 . x(2x + 3) 4x2 + 12x + 9 – 27 . 2x2 + 3x 2x2 + 9x – 18 . 0 (2x – 3)(x + 6) . 0

x–6 3

2

Therefore, x , –6 or x . 32

.

9. (a) The coordinates of the maximum point is (–1, 9).

(b) Equation of the axis of symmetry of the curve is x = –1.

(c) The range is x , –4 or x . 2.

10. 2x2 – (k + 1)x + 2 = 0Let the roots be a and 4a.

a + 4a = – ba

= – –(k + 1)2

5a = (k + 1)2

k = 10a – 1 …… 1

a × 4a = ca

= 22

4a2 = 1

a2 = 14

a = ± 12

Subsitute a = 12

into 1,

k = 101 12 2 – 1 = 4

Subsitute a = – 12

into 1,

k = 101– 12 2 – 1 = –6

Since k . 0, k = 4.

11. Let the number be x. x + x2 = 156 x2 + x – 156 = 0

x = –(1) ± (1)2 – 4(1)(–156)2(1)

= –1 ± 252(1)

= –1 – 252

or –1 + 252

= –13 or 12

12. (4p2q5)3

8p–2 q9 = 64p6q15

8p–2 q9

= 8p6 – (–2) q15 – 9 = 8p8q6

13. log3 116k2 2

= log3 16 – log3 k2

= logk 16logk

3 – logk k

2

logk 3

= 2 logk 4logk

3 – 2 logk k

logk 3

= 2nm

– 2m

= 2(n – 1)m

14. log10 {log10 [2 + log2 (x + 1)]} = 0 log10 {log10 [2 + log2 (x + 1)]} = log10 1 log10 [2 + log2 (x + 1)] = 1 2 + log2 (x + 1) = 10 log2 (x + 1) = 8 x + 1 = 28 x = 28 – 1 = 255

15. 52x – y = 25–x + 10 52x

5y = 152x + 10

ab

= 1a

+ 10

(× ab) on both sides of the equation. a2 = b + 10ab b(1 + 10a) = a2 b = a2

1 + 10a



16. (x – 3)2 + (y + 4)2 = 62 x2 – 6x + 9 + y2 + 8y + 16 = 36 x2 + y2 – 6x + 8y – 11 = 0

The equation of the locus of P is x2 + y2 – 6x + 8y – 11 = 0

17.

N(h, k)

y

xO

B( 10, 3)

A(2, 9)

1

2

Let both the ants meet at N(h, k).

h = 2(–10) + 1(2)2 + 1

= –20 + 23

= –6

k = 2(–3) + 1(9)2 + 1

= –6 + 93

= 1

Distance of ant A from its initial location, NA= [2 – (–6)]2 + (9 – 1)2 = 82 + 82 = 128= 82 units

18. 2y = –5x + 8

m = – 52

…… 1

y = (h – 4)x + 9 m = h – 4 …… 2

1 = 2, – 52

= h – 4

h = – 52

+ 4

= 32

19. (a)

5

5

0

x + y = 5

7x – 3y = 3y

x

x + y = 5 y = 5 – x …… 1 7x – 3y = 3 …… 2 Substitute 1 into 2, 7x – 3(5 – x) = 3 10x = 18 x = 1.8 Substitute x = 1.8 into 1, y = 5 – 1.8 = 3.2 Therefore, the coordinates of the

intersection point is (1.8, 3.2).

(b) The height of the isosceles triangle is 1.8 units.

20. (a) x , 15 The maximum value of x is 14.

(b) 6 + x , 15 + 14 + 8 6 + x , 37 x , 31 6 + x + 15 . 14 + 8 21 + x . 22 x . 1 Therefore, the range of x is 1 , x , 31.

21. (a) Total number of student = 4 + 7 + 8 + 10 + 6 = 35

(b) Mean score

=

(4 × 3.5) + (7 × 5.5) + (8 × 7.5) + (10 × 9.5) + (6 × 11.5)

35

= 276.535

= RM7.90



22. ∠TOP = p – 2(0.8727) = 1.3966 rad. = 80°

sin ∠TOP = PTOP

PT = 3.5 sin 80° = 3.4468 cm

cos ∠TOP = OTOP

OT = 3.5 cos 80° = 0.6078 cm

QT = OQ – OT = 3.5 – 0.6078 = 2.8922

Area of ∆PQT

= 12

× QT × PT

= 12

× 2.8922 × 3.4468

= 4.9844 cm

Area of the sector OPQ

= 12

× 3.52 × 1.3966

= 8.5542 cm

Area of the region with green colour= 8.5542 – 4.9844= 3.57 cm2

23. (a) x = 2t + 5

dxdt

= 2

(b) dydx

= dydt

× dtdx

= (12t – 4) × 12

= 3121 x – 52 2 – 44 × 1

2

= [6(x – 5) – 4] × 12

= (6x – 34) × 12

= 2(3x – 17) × 12

= 3x – 17

24. Surface area of a sphere, S = 4pr2, δr = 6.02 – 6 = 0.02 cmdSdr

= 8pr

δS = dSdr

× δr

= 8pr × 0.02 = 8p(6) × 0.02 = 3.016 cm2

The small change in the surface area is 3.016 cm2.

25. Let the length of the rectangle = y cm the width of the rectangle = x cm the radius of the cylinder = r cm

x cm x cm

r cmy cm

2x + 2y = 48 x + y = 24 x = 24 – y …… 1 2pr = y r = y

2p …… 2

Volume of cylinder, V = pr 2x …… 3 Substitute 1 and 2 into 3.V = p1 y

2p22(24 – y)

= y2

4p (24 – y)

= 6y2

p – y3

4p dVdy

= 12yp

– 3y2

4p

When dVdy

= 0,

12yp

– 3y2

4p = 0

48y – 3y2 = 0 3y(16 – y) = 0

y = 0 (not accepted),y = 16x = 24 – 16 = 8Therefore, the length is 16 cm and the width is 8 cm.



Paper 2

1. 3x – y – 2 = 0 …… 1

2x2 + 3y2 – xy – 6 = 0 …… 2

Rewrite 1 in the equivalent form, y = 3x – 2 …… 3

Substitute 3 into 2, 2x2 + 3(3x – 2)2 – x(3x – 2) – 6 = 0 2x2 + 3(9x2 – 12x + 4) – 3x2 + 2x – 6 = 0 2x2 + 27x2 – 36x + 12 – 3x2 + 2x – 6 = 0 26x2 – 34x + 6 = 0 13x2 – 17x + 3 = 0

x = –(–17) ± (–17)2 – 4(13)(3)2(13)

= 17 ± 13326

x = 17 + 13326

or x = 17 – 13326

= 1.097 = 0.210


y = 3(1.097) – 2 = 1.292


y = 3(0.210) – 2 = –1.369

Therefore, the solutions arex = 1.097, y = 1.292; x = 0.210, = –1.369

2. (a) g(x) = 4 – 3x Let y = 4 – 3x

Then x = 4 – y3

g–1(y) = 4 – y3

g–1(x) = 4 – x3

Hence, the function which maps from set Y to set X is g–1(x) = 4 – x

3 .

(b) h(4 – 3x) = 9 – 6x Let y = 4 – 3x

x = 4 – y3

h(y) = 9 – 61 4 – y3 2

= 9 – 2(4 – y) = 1 + 2y Hence, h(x) = 1 + 2x

(c) gh(x) = 2x + 3 g(1 + 2x) = 2x + 3 4 – 3(1 + 2x) = 2x + 3 4 – 3 – 6x = 2x + 3 8x = –2 x = – 1

4

3. (a) 2x2 – 4x + 5 = p 2x2 – 4x + 5 – p = 0 a = 2, b = – 4, c = 5 – p Using b2 – 4ac . 0, (– 4)2 – 4(2)(5 – p) . 0 16 – 40 + 8p . 0 8p – 24 . 0 p . 3

(b) a + b = – 1– 42 2

a + b = 2 …… 1

ab = 5 – p2

…… 2

4x(x – q) – 3 = 0 4x2 – 4qx – 3 = 0

a4

+ b4

= – 1– 4q4 2

a + b4 = q

a + b = 4q …… 3

a4

× b4

= –34

ab = –12 …… 4

Compare 1 and 3. 4q = 2

q = 12



Compare 1 and 2,

5 – p2

= –12

5 – p = –24 p = 29

4. (a) 9x + 2y

3x = 9x × 92y

3x

= 32x × (9y)2

3x

= 3x × (9y)2 = mn2

(b) 3x = m x logp 3 = logp m

9y = n y logp 9 = logp n

log3 13nm2 2

= log3 3 + log3 n – 2 log3 m = log3 3 + y log3 9 – 2(x log3 3)

= log10 3

12

log10 3 + y1 log10 9

12

log10 32

– 2x1 log10 312

log10 32

= 2 + y1 2 log10 312

log10 32 – 2x(2)

= 2 + y(4) – 4x = 2 + 4y – 4x

5. (a) Area of ∆PQR

= 12

7 k –2 7 6 2k + 2 3 6

= 12

(14k + 14 + 3k – 12) – (6k – 4k – 4 + 21)

= 12

17k + 2 – (2k + 17)

= 12

15k – 15

(b) (i) 12

15k – 15 = 30 15k – 15 = 60

15k – 15 = 60 or 15k – 15 = –60 15k = 75 k = –3 k = 5

(ii) 12

(15k – 15) = 0

15k – 15 = 0 15k = 15 k = 1

(c) Q = (5, 12) Midpoint of PR,

T = 1 7 – 22

, 6 + 32 2 = 1 5

2, 9

2 2

mQT = 12 – 9

2

5 – 52

= 3

y – 12 = 3(x – 5) y – 12 = 3x – 15 y = 3x – 3 The equation of the median is y = 3x – 3.

6. (a) Median class = (45 – 49)

Median = 44.5 + 312

(48) – 23

124(5)

= 44.92

(b) Mean =

(32 × 5) + (37 × 8) + (42 × 10) + (47 × 12) + (52 × 9) + (57 × 4)

5 + 8 + 10 + 12 + 9 + 4

= 2 13648

= 44.5



7. (a) (i) Area of the turf

= 12

u–8–8

120

1613

–45

–8–8u

= 12

[(0 + 156 + 80 + 32) – (–96 + 0 – 52 – 40)]

= 12

[268 – (–188)]

= 228 m2

(ii) Let the coordinates of E = (p, q)

C(16, 13)

E(p, q)

B( 4, 5)

2

3

p = 2(16) + 3(–4)2 + 3

= 205

= 4

q = 2(13) + 3(5)2 + 3

= 415

= 8 15

The coordinates of E are 14, 8 15 2.

(b) (i) Midpoint of AC = 1 –8 + 16

2, –8 + 13

2 2

= 14, 52 2

Let (x, y) is a point which moves such that its distance from the midpoint of AC is always 2 m.

The equation of the track of powder is

(x – 4)2 + 1y – 52 2

2 = 2

(x – 4)2 + 1y – 52 2

2 = 22

1x2 – 8x + 16 + y2 – 5y + 254 2 = 4

x 2 + y2 – 8x – 5y + 734

= 0

4x2 + 4y2 – 32x – 20y + 73 = 0

(ii) On the y-axis, x = 0. 4y2 – 20y + 73 = 25 b2 – 4ac = (–20)2 – 4(4)(73) , 0 The equation has no roots. Thus, the circle does not intersect the

y-axis.

8. (a) Volume of cylinder, pr2h = 750p

h = 750r2

Cost of materials, P = 2prh × 0.02 + pr2 × 0.03

= 2pr1 750r2 2 × 0.02 + pr2 × 0.03

= 30pr

+ 0.03pr2

(b) P = 30pr

+ 0.03pr2

dPdr

= – 30pr2 + 0.06pr

When dPdr

= 0,

– 30pr2 + 0.06pr = 0

30pr2 = 0.06pr

r3 = 30p0.06p

r = 3500 = 7.937 cm

When d2Pdr2 = 60p

r3 + 0.06p = 0.566 . 0

r = 7.937 when P is minimum,

h = 7507.9372 = 11.906 cm

The radius = 7.937 cm and the height = 11.906 cm.

(c) The minimum cost, P

= 30p7.937 + 0.03p × 7.9372

= RM17.81



9. (a) (i) Mean = 4 + 5 + 8 + 3 + 10 + 6 + 6 + 8 + 8 + 910

= 6710

= 6.7 hours

Variance = 42 + 52 + 82 + 32 + 102 + 62 + 62 + 82 + 82 + 92

10 – (x)2

= 49510

– 6.72

= 4.61

(ii) New standard deviation = 2s = 24.61 = 4.294

(b) Numberof hours

Class mark,x f fx fx2

2 – 4 3 5 15 45

5 – 7 6 12 72 432

8 – 10 9 15 135 1 215

11 – 13 12 8 96 1 152

∑ f = 40 ∑ f = 318 ∑ fx2 = 2 844

Mean = ∑ fx∑ f

= 31840

= 7.95 hours Standard deviation = 7.8975 = 2.810

Variance = ∑ fx2

∑ f – x 2

= 2 84440

– 7.952

= 7.8975



10.

A

B

1.4 cm5

cm

AB D

C

θ 82.9°

(a) AB = 52 + 1.42

= 5.192 cm

Circumference of the base of the cone

= 2 × 227

× 1.4 = 8.8 cm

Length of arc BC = 8.8 cm 5.192 × q = 8.8

q = 8.85.192 = 1.695 rad

= 1.695 × 180°p

= 97.1° ∠CAD = 180° – 97.1° = 82.9°

tan 82.9° = CDAD

AD = 5.192tan 82.9°

= 0.647 cm Length = BA + AD = 5.192 + 0.647 = 5.839 cm Therefore, the length and the width of

the card are 5.839 cm and 5.192 cm respectively.

(b) The area of the net of the cone

= 12

× 5.1922 × 1.695

= 22.85 cm2

(c) The area of the card unused = 5.839 × 5.192 – 22.85 = 7.466 cm2

11. (a) (i) f(x) = hx + kx

f ’(x) = h – kx2

At turning point, f ’(x) = 0, x = 23

f ’(x) = h – kx2

0 = h – k(23 )2

h = k12

When f(x) = 243 , x = 23

f(x) = hx + kx

243 = h(23 ) + k23

243 = k12

(23 ) + k36

24 = k6

+ k6

k = 72

h = k12

= 7212

= 6

(ii) When f ’(x) = 0

6 – 72x2 = 0

x = ±12 = ± 23

At the second turning point, x = –23 ,

f(x) = 6x + 72x

f(–23 ) = 6(–23 ) + 72(–23 )

= –123 – 363

= –123 – 3633

= –243 The second turning point is

(–23 , –243 ).



(b) f ’(x) = 6 – 72x2

f ’’(x) = 144x3

f ’’(23 ) = 144(23 )3

= 3.464 . 0

Hence, (23 , 243 ) is a minimum point.

f ’’(–23 ) = 144(–23 )3

= –3.464 , 0

Hence, (–23 , –243 ) is a maximum point.

12. (a) (i) PRsin 105°

= 4.8sin 36°

PR = 7.888 cm

(ii) RT2 = 6.52 + 9.22 – 2(6.5)(9.2) cos 126° RT = 14.042 PT = RT – PR = 14.042 – 7.888 = 6.154 cm

(iii) ∠RPQ = 180° – 105° – 36° = 39° Area of ∆PQR

= 12

(7.888)(4.8)(sin 39°)

= 11.914 cm2

(b) (i)

4.8 cm105º

P

36º

39º

Pʹ

Qʹ

Rʹ

(ii) ∠P′Q′R′ = 105° – (180° – 39° – 39°) = 3°

13. (a) (i) cos ∠ACD = 12.62 + 82 – 72

2(12.6)(8) = 0.8619 ∠ACD = 30.47° = 0.532 rad.

(ii) ∠BCD = 180° – 125° = 55° ∠BAC = ∠ACD = 30.47°

BCsin 30.47°

= 12.6sin 125°

BC = 7.80 cm

(b) (i)

CDE

A B

125º

12.6 cm7 cm

8 cm

(ii) cos ∠ADC = 72 + 82 – 12.62

2(7)(8)

= –0.4086

∠ADC = 114.12° ∠ADE = 180° – 114.12° = 65.88° ∠DAE = 180° – 2(65.88°) = 48.24°

Area of ∆ADE = 12

(7)(7) sin 48.24°

= 18.276 cm2



14. (a) (i) Let Q2015 = Cost of ingredient C in the year 2015 Q2014 = Cost of ingredient C in the year 2014

Price index = Q2015

Q2014

× 100

85 = Q2015

24 × 100

Q2015 = 20.40

Therefore, the cost of ingredient C in the year 2015 is RM20.40.

(ii) Composite index = ∑Iiwi

∑wi

112 = (110 × 8) + (k × 5) + (85 × 3) + (107.5 × 4)8 + 5 + 3 + 4

112 = 1 565 + 5k20

k = (112)(20) – 1 5655

= 135

(b) (i) Composite index = 112 × 105100

= 117.6

(ii) Price of ice-cream = RM28 × 117.6100

= RM32.93

15. (a) (i) Let Q2013 = The price of material E in the year 2013 Q2011 = The price of material E in the year 2011

Price index = Q2013

Q2011

× 100

120 = 48Q2011

× 100

Q2011 = 40 The price of ingredient E in the year 2011 is RM40.

(ii) Let Q2013 = The price of material G in the year 2013 Q2011 = The price of material G in the year 2011

Price index = Q2013

Q2011

× 100

115 = Q2013

25 × 100

Q2013 = 28.75

The price of ingredient G in the year 2013 is RM28.75.



(b) Composite index = ∑Iiwi

∑wi

124 = (120 × 2) + (n × 4) + (115 × 3) + (95 × 1)2 + 4 + 3 + 1

124 = 680 + 4n10

4n = 124(10) – 680 n = 140

(c) Composite index = ∑Iiwi

∑wi

= (138 × 2) + (147 × 4) + (115 × 3) + (104.5 × 1)2 + 4 + 3 + 1

= 1 313.510

= 131.35 (d) The cost of a ceiling fan in the year 2015

= RM88 × 131.35100

= RM115.59

chapter 1 functions fungsi - pelangibooks.com standard math/online... · graf (a) {(4, 1), (4, 2),...

Documents