IT Add Maths F5 Answer

61

TOPICAL TEST 1UJIAN TOPIKAL 1

PAPER 1 KERTAS 1 1 T3 – T2 = T2 – T1

(2h + 1) – (2h – 2) = (2h – 2) – h 2h + 1 – 2h + 2 = 2h – 2 – h 3 = h – 2 h = 5

2 Tn = a + (n – 1)d T3 = a + 2d = 1 …… T5 = a + 4d = 9 …… Solve the equation and Selesaikan persamaan dan 2d = 8 Common difference, d = 4 Beza sepunya, The first term, a = 1 – 2(4) Sebutan pertama, = –7

3 Tn + 1 – Tn =Tn – Tn – 1

T3 – T2 = T2 – T1

3x 2 + 7 – (6x) = 6x – (3 + x) 3x 2 + 7 – 6x = 5x – 3 3x 2 + 7 – 6x – 5x + 3 = 0 3x 2 – 11x + 10 = 0 (3x – 5)(x – 2) = 0 3x – 5 = 0, or / atau x – 2 = 0 x = 5

3 x = 2

4 Sn = n2 [2a + (n – 1)d]

a = –20, d = –16 – (–20) = 4 Sn > 0 n

2 [2(–20) + (n – 1)4] > 0 n

2 [–40 + 4n – 4] > 0 n(–20 + 2n – 2) > 0 n(2n – 22) > 0 n > 0, 2n – 22 > 0 2n > 22 n > 11 n = 12

5 Tn = a + (n – 1)d a = 1 026 d = 1 011 – 1 026 = – 15 n = 50 T50 = 1 026 + 49(–15) = 1 026 – 735 = 291

6 3k + 1 – 2k + 1 = 2k – 1 – 4 k + 2 = 2k – 5 k = 7

7 d = Tn + 1 – Tn

=T2 – T1

= 8 – 4 = 4

8 (a) Tn + 1 – Tn = Tn – Tn–1

T3 – T2 = T2 – T1

9 – x = x – 3 9 – x – x + 3 = 0 12 – 2x = 0 12 = 2x x = 6

(b) Tn + 1

Tn = Tn

Tn – 1

T3

T2 = T2

T1

9x = x

3 x 2 = 27 x = √27 = 3√3

9 T9 = a + 8(2) = 3 + 3p a = 3p – 13 ……

Sn = n2 [2a + (n – 1)d]

S4 = 2(2a + 6) = 2p – 10 2a + 6 = p – 5 a = p – 11

2 ……

Substitute into Gantikan dalam

p – 112 = 3p – 13

p – 11 = 6p – 26 15 = 5p p = 3

10 7 – 4 + k = 2k – 4 – 7 3 + k = 2k – 11 k = 14 Common difference = T2 – T1

Beza sepunya = 7 – (4 – 14) = 7 + 10

= 17

11 a = 52 d = 48 – 52 = – 4 Tn < 0 52 + (n – 1)(–4) < 0 52 – 4n + 4 < 0 56 – 4n < 0 –4n < –56 n > 56

4 n > 14 n = 15

12 Sum of arithmetic progression Hasil tambah janjang aritmetik

Sn = n2 [2a + (n – 1)d]

S3 = 32 [2a + 2d)

S3 = 3(a + d)

a = – 38 , d = – 3

16 + 38 = 3

16

S3 = 3(– 38 + 3

16 ) = – 916

or / atau

S3 = – 38 + (– 3

16 ) + 0 = – 916

13 Tn = a + (n – 1)d a = 5, d = 8 – 5, Tn = 131 = 3 131 = 5 + (n – 1)3 3(n – 1) = 126 n – 1 = 42 n = 43

ANSWERSJAWAPAN

14 Sn = a(1 – r n)1 – r

S5 = a(1 – r 5)1 – r

S5 = 7 2227 = 211

27

r = 23

21127 =

a[(1 – ( 23 )5]

1 – 23

21181 = a( 211

243 ) a = 3

15 (a) r = Tn

Tn – 1 = 32

64 = 12

Sn = a(1 – r n)1 – r

Sn = 126, a = 64

126 = 64[(1 – ( 1

2 )n]1 – 1

2

63 = 64[1 – ( 1

2 )n ] 1 – ( 1

2 )n = 63

64

( 12 )n

= 164

n = 6

(b) S� = a1 – r

a = 64

r = 12

S� = 64

1 – 12

= 128

16 Tn = ar n – 1

Tn = – 24332

a = – 49

r = –1

– 23

= 32

– 24332 = – 4

9 ( 32 )n – 1

( 32 )n – 1

= 2 187128

( 32 )n

= 6 561256 = ( 3

2 )8

n = 8

17 a = 14

r =

1214

= 2 Tn = 64 Tn = arn – 1

64 = 14 (2)n – 1

2 n – 1 = 256 2n = 512

n = ln 512ln 2

= 9

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18 Tn

Tn – 1 = Tn + 1

Tn

– 8

3 p

= 89 p2

– 83

– 83p = – 1

3 p2

8 = p3

p = 2

r = – 8

32

= – 43 Tn = arn – 1

T4 = 2(– 43 )3

= 2(– 6427 )

= – 12827

19 (a) Tn + 1

Tn = T2

T 1

= 328

= 4

(b) Sn = a(r n – 1)r – 1

8(4 n – 1)

4 – 1 = 10 920

4n – 1 = 4 095 4n = 4 096 = 46

n = 6

20 r = 3654 = 2

3 S10 = a

1 – r

= 54

1 – 23

= 54

13

= 162

21 a = 4 , r = – 84

= –2 T4 = ar3

= 4(–2)3

= –32

22 a = 5 , S� = a1 – r

6 = 51 – r

1 – r = 56

r = 16

23 a = 0.05

r = 0.0050.05 = 0.1

S� = a1 – r

= 0.051 – 0.1

= 118

1p = 118 , p = 18

24 0.48484848… = 0.48 + 0.0048 + 0.000048 + … a = 0.48

r =0.00480.48 = 0.01

S� = a1 – r

= 0.481 – 0.01 = 16

33

25 S� = a1 – r

a = 4

r = – 8

94

= – 29

S� = 41 – (– 2

9 ) = 36

11

PAPER 2 KERTAS 2 1 (a) T1 = S1

= 12 + 3(1) = 4 (b) S2 = 22 + 3(2) = 10 T2 = S2 – S1 = 10 – 4 = 6 d = T2 – T1

= 6 – 4 = 2 (c) T10 = S10 – S9

= [102 + 3(10)] – [92 + 3(9)] = 100 + 30 – 81 – 27 = 22 (d) Tn = Sn – Sn – 1

= [n 2 + 3n] – [n – 1)2 + 3(n – 1)] = n 2 + 3n – (n 2 – 2n + 1 + 3n – 3) = 2n + 2

2 (a) T1 = S1

= 4(1) – 2(1)2

= 2 (b) T2 = S2 – S1

S2 = 4(2) – 2(2)2

= 0 T2 = 0 – 2 = –2 d = T2 – T1 = –2 – 2 = –4 (c) T = a + (n – 1)d T10 = 2 + 9(–4) = 2 – 36 = –34 (d) Tn = a + (n – 1)d = 2 + (n – 1)(–4) = 2 – 4n + 4 = 6 – 4n

3 (a) Tn = a + (n – 1)d T12 = 1 500 + 11(200) = RM3 700 (b) Total annual salary in n years Jumlah gaji tahunan dalam n tahun

= (T1 + T2 + …… + Tn) × 12 months / bulan = Sn × 12 In first 5 years Dalam 5 tahun pertama = S5 × 12

= 52 [2(15 000) + 4(200)] × 12

= 52 [38 000] × 12

=RM114 000

4 (a) Sn = n2 [2a + (n – 1)d]

S5 = 52 (2a + 4d) = 105

2a + 4d = 42 ……

S10 = 305 + 105 = 410 410 = 10

2 [2a + 9d]

2a + 9d = 82 …… – 5d = 40 d = 8 Substitute d = 8 into Gantikan d = 8 dalam 2a + 4(8) = 42 a = 5

(b) S20 = 202 [2(5) + 19(8)]

= 10[10 + 152] = 1 620

5 (a) S5 = 52 (2a + 4d) = 45

2a + 4d = 18 …… S10 = 10

2 (2a + 9d) = 120 + 45 2a + 9d = 33 …… – 5d = 15 d = 3 Substitute d = 3 into Gantikan d = 3 dalam 2a + 4(3) = 18 a = 3 (b) Sum of T11 to T20 = S20 – S10

Hasil tambah T11 hingga T20 S20 = 20

2 [2(3) + 19(3)]

= 10(6 + 57) = 630 S20 – S10 = 630 – (120 + 45) = 465

6 (a) Sn = n2 (a + l)

n2 (–12 + 30) = 63

n2 = 7

2 n = 7 (b) T4 = a + 3d T7 = a + 6d = 30 –12 + 6d = 30 d = 7 Middle term = T4

Sebutan tengah T4 = –12 + 3(7) = 9

7 (a) Sn = n2 (a + l)

n2 (–8 + 43) = 315

n2 = 9

n = 18 (b) T18 = a + 17d 43 = –8 + 17d 17d = 51 d = 3

8 (a) S� = a1 – r

16 = a1 – 1

2

a = 16 × 1

2 = 8 (b) T4 = ar3

= 8( 12 )3

= 1

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(c) S6 = a(1 – r 6)1 – r

= 8[1 – ( 1

2 )6]1 – 1

2

= 8[1 – 164] × 2

= 15 34

9 (a) ar 2 = 243 …… ar5 = 9 …… ÷

r3 = 9243

r = ( 9243)

13

= ( 127 )

13

= 13

:

a( 13 )2 = 243

a = 243 × 9 = 2 187

(b) S7 = a(1 – r 7)1 – r

= 2 187[1 – ( 1

3 )7]1 – 1

3

= 2 187( 2 186

2 187 )23

= 3 279

(c) S� = a1 – r = 2 187

1 – 13

a = 6 5612

10 S5 = a(r 5 – 1)

r – 1 = 3 ……

S8 – S3 = 24

a(r 8 – 1)r – 1 – a(r 3 – 1)

r – 1 = 24

a(r 8 – r3)r – 1 = 24

ar 3(r 5 – 1)

r – 1 = 24 ……

÷ :

r3 = 243 = 8

r = 2 From / Dari :

a(25 – 1)2 – 1 = 3

a(31) = 3 a = 3

31

T6 = ar5

= 331 (25)

= 9631

11 (a) T4 + T6 = 40 ar3 + ar 5 = 40 …… T7 + 79 = 1 080 ar 6 + ar 8 = 1 080 …… ÷ :

ar6 + ar8

ar3 + ar5 = 1 08040

r3(ar3 + ar5)ar3 + ar5 = 1 080

40

r3 = 27 r = 3

(b) Substitute into Gantikan dalam a(3)3 + a(3)5 = 40 27a + 243a = 40 270a = 40 a = 4

27

(c) S16 = 427

[316 – 1]

3 – 1

=

427

[4 306 720]

2

= 3 188 645.93

12 (a) 160 = a1 – 1

4

a = 120

(b) T4 = 120( 14 )3

= 158

(c) 120[1 – ( 1

4 )n]1 – 1

4

> 150

160[1 – ( 14 )n] > 150

1 – ( 14 )n

> 150160

( 14 )n

< 116

( 14 )n

< ( 14 )2

n lg( 14 ) < 2 lg ( 1

4 ) n > 2 n = 3

13 (a) x + 106x + 4 = 2x – 1

x + 10

(x + 10)2 = (2x – 1)(6x + 4) x2 + 20x + 100 = 12x2 + 2x – 4 11x2 – 18x – 104 = 0 (x – 4)(11x + 26) = 0 x = 4 or / atau x = 26

11

Positive term, x = 4 Sebutan bernilai positif, x = 4

(b) r = 2x – 1x + 10

= 2(4) – 1 4 + 10

= 714

= 12

(c) S� = a 1 – r

= 6(4) + 41 – 1

2

= 2812

= 56

14 (a) a = 0.08

r = 0.0080.08

= 0.1

S� = a1 – r

= 0.081 – 0.1

= 445

Given 1p

= S�

Diberi

p =

1454

p = 1180

15 r = 5.22.6

= 2 Tn = arn – 1

= 2.6(2)n – 1

2.6(2)n – 1 = 41.6 2n – 1 = 16 2n – 1 = 24

n – 1 = 4 n = 5 Sn = a(rn – 1)

r – 1

S5 = 2.6(25 – 1)2 – 1

= 80.6


PAPER 1 KERTAS 1 1

2

3

4 m = 2 – 64 – 2 = –4

2 = –2, c = 10 y = mx + c y = –2x + 10

5 m = –8 + 20 + 8

= –68

= – 34 , c = –8

y = mx + c y = – 3

4 x – 8

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6 m =

12 – 5

8 – 12

=

– 92

152

= – 92 × 2

15

= – 35

y = – 35 lg x + c ……

Substitute ( 12 , 5) into

Gantikan ( 12 , 5) dalam

5 = – 35 ( 1

2 ) + c

c = 5310

y = – 35 lg x + 53

10

7 m = 12 – 412 – 4

= 88 = 1

yx = x2 + c ……

Substitute (4, 4) into Gantikan (4, 4) dalam 4 = 4 + c c = 0 y

x = x2

y = x3

8 m = 9 –512 – 2

= 410 = 2

5

y = 25 x + c ……

Substitute (2, 5) into Gantikan (2, 5) dalam

5 = 25 (2) + c

c = 215

y = 25 x + 21

5

When y = 5 / Apabila y = 5

5 = 25 x + 21

5

25 x = 4

5 x = 2

9 m = 7 – 147 – 3

= – 74

y2 = – 74 x + c ……


14 = – 74 (3) + c

c = 774

y2 = – 74 x + 77

4 When y = 4.5 / Apabila y = 4.5

(4.5)2 = – 74 x + 77

4 7

4 x = –1

x = – 47

10 m = 7 – 44 – (–2)

= 36

= 12

y = 12 x + c ……

Substitute (–2, 4) into Gantikan (–2, 4) dalam

4 = 12 (–2) + c

c = 5 y = 1

2 x + 5

When x = 10 / Apabila x = 10

y = 12 (10) + 5 = 10

11

When x = 17, y = 21 / Apabila x = 17, y = 21

12

When y = 18, x = 9 / Apabila y = 18, x = 9

13 y = 4x2 – 7x y

x = 4x – 7

Y = yx

X = x m = 4 c = –7

14 log y = log(abx – 2) = log a + (x – 2)log b log y = (x – 2)log b + log a Y = log y

X = x – 2 m = log b c = log a

15 y = ax – b

1y = x – b

a

1y = 1

a x – ba

Y = 1y

X = x m = 1

a c = – b

a

16 y = ax + bx

xy = ax2 + b Y = mX + c m = 3 – 6

5 – 1 = – 34

a = – 34

Substitute (1, 6) into Y = mX + c Gantikan (1, 6) dalam Y = mX + c

6 = – 34 (1) + c

c = 274

b = 274

17 y√x = px + q√x

y = px

√x + q

y = p√x + q Y = mX + c

m = 14 – 712 – 4 = 78

p = 78


7 = 78 (4) + c

q = 72

18 y = pqx

lg y = lg(pqx) lg y = lg p + x lg q lg y = lg q x + lg p Y = mX + c

m = 10 – 610 – 4 = 4

6 = 23

lg q = 23

q = 4.6416 Substitute (4, 6) into Y = mX + c Gantikan (4, 6) dalam Y = mX + c

6 = 23 (4) + c

c = 103

lg p = c = 103

p = 2154.4347

19 y = ax2 + bx

xy = ax + b

Y = mX + c

m = 4 – 97 – 3

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= –54 = – 5

4

a = – 54


9 = – 54 (3) + c

c = 514

b = 514

20 y = 2 + 3yx

y – 3yx = 2

xy – 3y = 2x y(x – 3) = 2x

y = 2xx – 3

1y = x

2x – 32x

1y = – 3

2 ( 1x ) + 12

m = 3 – rs – 2 = – 3

2 3 – r = –3 s – 2 = 2 r = 6 s = 4

21 (a) Y = mX + c Given / Diberi y = ax4

log y = 4 log x + log a with / dengan Y = log y, X = log x, m = 4, c = log a (b) c = log a = 4 m = 4 = k – 4

6 – 0

k – 4 = 24 k = 28

22 y = –7x2 + 4 y

x2 = –7 + 4x2

yx2 = 4( 1

x2 ) – 7

Y = 4X + c Y = y

x2 and / dan X = 1x2

23 y2 = 4x(7 – x)

y 2

x = 4(7 – x)

y 2

x = –4x + 28

Substitute (a, 0) into Y = mX + c Gantikan (a, 0) dalam Y = mX + c 0 = –4a + 28 a = 7 Substitute (2, b) into Y = mX + c Gantikan (2, b) dalam Y = mX + c b = –4(2) + 28 = 20

24 y = k7x

log10 y = log10 k – log10 7x

log10 y = log10 k – x log10 7 log10 y = x(–log10 7) + log10 k Y = mX + c c = –4 = log10 k k = 10–4

25 qx + px = y

xy = qx2 + p, Y = xy, X = x2, m = q, c = p

q = m = 4 – 013 – 7 = 4

6 = 23


0 = 23 (7) + p

p = – 143

PAPER 2 KERTAS 2 1

(a) y-intercept = 2 pintasan-y (b) m = 34 – 12.5

12 – 4 = 21.58 = 2.689

y = 2.689x + 2

2 (a)

(b) m = 7.0 – 11.0

10 – 2

= –48 = – 1

2

c = 12

y = – 12 x + 12

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3 (a)

(b) m = 6.2 – 1.710 – 1 = 4.5

9 = 12

Q = 1

2 P + 1.2

4 (a)

(b) When x = 3.5, y = 58 Apabila x = 3.5, y = 58

5 (a)

(b) When Q = 68, P = 50 Apabila Q = 68, P = 50 Whe P = 70, Q = 76 Apabila P = 70, Q = 76

6 (a)

20 40 60 80 100 120

x

60

50

40

30

20

10

y

0

(b) y = ax + b Y = mX + c

m = 50 – 18100 – 20

= 8280

= 0.4 c = 9 a = m = 0.4 b = c = 9

7 (a)

(b) q = ap + b Y = mX + c

m = 49 – 175 – 1

= 324 = 8

c = 9 a = m = 8 b = c = 9

8 (a) y = a(5)– bx

log10 y = (– blog10 5) 1

x + log10 a

Y = log10 y

m = – blog10 5

X = 1x , c = log10 a

(b) log10 y 2.28 2.27 2.25 2.21 2.201x 0.042 0.038 0.036 0.031 0.029

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0.01 0.02 0.03 0.04 00.050.035

2.30

2.28

2.26

2.24

2.22

2.20

2.18

2.16

2.14

2.12

2.10

2.08

2.06

2.04

2.02

2.00

1.98

log10y

1x

(c) m = 2.24 – 2.200.035 – 0.029 = – b

log 5

= 0.040.006 = 6.667

b = 6.667–log10 5

= –9.538

y-intercept / pintasan-y = log10 a log10 a = 2.004 a = 100.925

9 (a) x 1 2 3 4 5 6

yx 7 15 23 31 38 46

(b) (i) ax + by = x2

byx = x – a

yx = 1

b x – ab

Y = mX + c

m = 46 – 76 – 1 = 7.8

1b = m = 7.8

b = 0.128 c = –1 = – a

b a = 0.128

(ii) When / Apabila x = 7, yx = 54

y = 54 × 7 = 378

10 (a) y = pxq

log10 y = log10 p + q log10 x log10 y = q log10 x + log10 p, Y = log10 y, m = q, X = log10 x, c = log10 p,

log10 x 0.30 0.48 0.60 0.70 0.78

log10 y 1.05 1.32 1.51 1.65 1.77

(b) m = 1.65 – 1.050.70 – 0.30 = 0.6

0.4 = 1.5

q = m = 1.5 From the graph, Y-intercept = 0.6 Daripada graf, pintasan-Y = 0.6 Therefore, c = log10 p = 0.6 Oleh itu, c = log10 p = 0.6 log10 p = 0.6 p = 100.6 = 3.98

11 (a)

x 3 4 5 6 7

y√x 0.03 0.09 0.14 0.20 0.26

(b) (i) y = a√x + b

√x y√x = ax + b

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m = 0.20 – 0.096 – 4

= 0.112 = 0.005

a = m = 0.055 b = –0.14 (ii) y = 0.10

12

x 2 4 8 10 15

y2 0.7 1.0 1.3 1.6 1.8

m = 1.8 – 1.015 – 4 = 0.8

11 = 455

b = m = 455

a = c = 0.7

13

log10 q 1.38 1.68 1.98 2.28 2.58

p 1 2 3 4 5

q = abp

log10 q = log10 a + p log10 b

m = 2.58 – 1.685 – 2 = 0.9

3 = 0.3

log10 b = m = 0.3 b = 1.995 log10 a = c = 1.08 a = 12.023

14

x 0.30 0.48 0.60 0.78 0.85

xy 0.72 0.93 1.08 1.29 1.38

y = p + qx

xy = px + q

m = 1.29 – 0.720.78 – 0.30 = 0.57

0.48 = 1.19

p = m = 1.19 q = c = 0.36

15

x2 2.25 4.00 6.25 12.25 16.00

log10 y 0.58 0.72 0.89 1.37 1.66

y = abx2

log10 y = log10 a + x2 log10 b

m = 1.66 – 0.7216.00 – 4.00 = 0.94

12 = 0.0783

log10 b = m = 0.0783 b = 1.198 log10 a = c = 0.4 a = 2.51

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PAPER 1 KERTAS 1

1 ∫7x 3 dx = 7x 3 + 1

3 + 1 + c

= 7x 4

4 + c

= 74 x 4 + c

2 ∫ – 5x 2 dx = ∫ – 5x –2 dx

= –5x –1

–1 + c

= 5x –1 + c

= 5x + c

3 ∫(x 4 + 3x 2 – 9) dx = x 5

5 + 3x 3

3 – 9x + c

= 15 x 5 + x 3 – 9x + c

4 ∫(x –4 – 4x –3 + 5x –2 – 6) dx = x –3

–3 – 4x –2

–2 + 5x –1

–1 – 6x + c

= – 13x 3 + 2

x 2 – 5x – 6x + c

5 dy

dx = 4x – 7

dy = 4x – 7 dx y = ∫(4x – 7) dx = 2x 2 – 7x + c …… Substitute (2, –3) into Gantikan (2, –3) dalam –3 = 2(2)2 – 7(2) + c –3 = –6 + c c = 3 y = 2x 2 – 7x + 3

6 f '(x) = 3x 3 – 5x y = f(x) = ∫(3x 3 – 5x) dx

= 34 x 4 – 5

2 x 2 + c

Passes through (1, 7) Melalui (1, 7)

7 = 34 – 5

2 + c

c = 354

y = 34 x 4 – 5

2 x 2 + 354

7 dydx = (3x + 2

x )2

y = ∫(9x 2 + 12 + 4x 2 )dx

= 3x 3 + 12x – 4x + c ……


1 = 3(2)3 + 12(2) – 42 + c

1 = 24 + 24 – 2 + c c = –45 y = 3x 3 + 12x – 4

x – 45

8 (a) ∫(7x 3 + 2) dx = 74 x 4 + 2x + c

= px 4 + 2x + c

p = 74

(b) 74 x 4 + 2x + c = 14

Substitute x = 2 Gantikan x = 2

74 (2)4 + 2(2) + c = 14

28 + 4 + c = 14 c = –18

9 Let / Biar U = 4x + 3

dUdx = 4 dx =

dU4

∫U 5dU

4 or / atau ∫(4x + 3)5dx = (4x + 3)5 + 1

(5 + 1)(4) + c

= 14 ∫U5 dU = (4x + 3)6

24 + c

= 14 [ U 6

6 ] + c

= U 6

24 + c

= (4x + 3)6

24 + c

10 Let / Biar

U = 2x + 5, dUdx = 2, dx =

dU2 or / atau

∫ 3(2x + 5)4 dx = 3∫(2x + 5)–4 dx

= 3∫U–4(dU2 )

= 32 ∫U

–4 dU

= 32 (U –3

–3 ) + c

∫ 3(2x + 5)4 dx = 3∫(2x + 5)–4 dx

= 3[ (2x + 5)–3

2(–3) ] + c

= (2x + 5)–3

–2 + c = – 1

2(2x + 5)3 + c

= – 12 U –3 + c

= – 12U 3 + c

= – 1

2(2x + 5)3 + c

11 Let / Biar U = 7x – 10, dU

dx = 7, dx = dU7 or / atau

∫2(7x – 10)7 dx = 2∫U 7 dU7

∫2(7x – 10)7 dx = 2(7x – 10)8

7(8) + c

= (7x – 10)8

28 + c

= 27 ∫U

7 dU

= 27 ( U 8

8 ) + c

= U 8

28 + c = (7x – 10)8

28 + c

12 Let / Biar U = 4 – 3x, dU

dx = –3, dx = – dU3 or / atau

∫ 8(4 – 3x)6 dx = ∫ 8

U 6 (–

dU3 )

∫ 8(4 – 3x)6 dx = ∫8(4 – 3x)–6 dx

= 8(4 – 3x)–5

(–3)(–5) + c

= 8

15(4 – 3x)5 + c

= – 83 ∫U

–6 dU

= – 83 (U –5

–5 ) + c

= 8

15U 5 + c

= 8

15(4 – 3x)5 + c

13

2

∫1 (2x2 – 5x) dx = [ 2x3

3 – 5x2

2 ] 2

1 = [ 2

3 (2)3 – 52 (2)2] – [ 2

3 (1)3 – 52 (1)2]

= (16

13 – 10) – ( 23 – 5

2 ) = –2 56

14 1

∫–1(x 3 – 3x 2 + x – 3) dx = [ x4

4 – x 3 + x2

2 – 3x] 1

–1

= [ 14

4 – 13 + 12

2 – 3(1)] – [ (–1)4

4 – (–1)3 + (–1)2

2 – 3(–1)] = ( 1

4 – 1 + 12 + 3) – ( 1

4 + 1 + 12 + 3)

= – 13

4 – 194 = –8

AM F5•J(61-75).indd 69 05/05/2010 15:34:28


15 3

∫05(2x + 7)–3 dx = [ 5(2x + 7)–2

–2(2) ] 3

0

= [– 54 (2x + 7)–2] 3

0

= (– 54[2(3) + 7]2) – (– 5

4[2(0) + 7)]2 ) = – 5

676 + 5196

= 1508 281

16 5

5

∫1g(x) dx = 5[ 2

∫1g(x) dx +

5

∫2g(x) dx]

= 5[6 + 10] = 80

17

1

∫3g(x) dx –

1

∫3

12 dx = –

3

∫1g(x) dx – [ 1

2 x] 1

3 = –6 – [ 1

2 (1) – 12 (3)]

= –6 – ( 12 – 3

2 ) = –6 – (–1) = –5

18 3

∫1 f(x) dx =

3

∫0 f(x) dx –

1

∫0 f(x) dx

= 4 – 1 = 3

19 3

∫–2[3x + kg(x)] dx =

3

∫–23x dx +

3

∫–2k g(x) dx

= [ 3x2

2 ] 3

–2+ k

3∫–2

g(x) dx

= [ 3(3)2

2 – 3(–2)2

2 ] + k(2)

= ( 272 – 12

2 ) + 2k

= 152 + 2k = 6

2k = – 3

2

k = – 34

20 Area of shaded region = 0

∫–2 y dx +

4

∫0 y dx

Luas rantau berlorek

| 0

∫–2(x 2 + 3x) dx| +

4

∫0(x 2 + 3x) dx = |[ x3

3 + 32 x 2] 0

–2 | + [ x3

3 + 32 x 2] 4

0 = |0 – [(–2)3

3 + 32 (–2)2]| + [ 43

3 + 32 (42) – 0]

= |– 10

3 | + 1363

= 10

3 + 1363

= 48 23 unit2


∫–2x dy

Luas rantau berlorek =

3

∫–2

13 y2 dy

= [19 y3] 3

–2 = 19 (3)3 – 19 (–2)3

= 3 – (– 89 )

= 359 unit2


∫0x(3 – x) dx –

2

∫0x dx

Luas rantau berlorek =

2

∫0(3x – x 2) dx –

2

∫0x dx

= [32 x 2 – 13 x3] 2

0 – [1

2 x 2] 2

0 = [3

2 (2)2 – 13 (2)3 – 0] – [12 (2)2 – 0]

= 6 – 83 – 2 = 43 unit2

23 Volume of shaded region = 3

∫0π(x 2 – 3x)2 dx

Isi padu rantau berlorek = π

3

∫0(x 4 – 6x 3 + 9x 2) dx

= π[15 x 5 – 64 x 4 + 3x 3] 3

0 = π(1

5 (3)5 – 64 (3)4 + 3(3)3 – 0) = π(243

5 – 2432 + 81)

= 8 110 π unit3

24 Volume of shaded region = π 5

∫2( 3y )2

dy Isi padu rantau berlorek = π

5

∫2

9y2 dy

= π[– 9

y ] 5

2 = π(– 9

5 + 92 )

= 2 7

10 π unit3

25 Volume of shaded region = π 1

∫0 y1

2 dx + π 2

∫1 y2

2 dx Isi padu rantau berlorek = π

1

∫0 x4 dx + π

2

∫1(4 – x)2 dx

= π[ x5

5 ] 1

0+ π[ (4 – x)3

3(–1) ] 2

1 = π[ 1

5 – 0] + π[(4 – 2)3

–3 – (4 – 1)3

–3 ] = 1

5 π + π(– 83 + 9)

= ( 15 + 19

3 )π = 6 8

15 π unit3

PAPER 2 KERTAS 2 1 ∫ 4

x2 dx = – 4x + c

2 = – 41 + c

2 = –4 + c c = 6

2 (a) y = x …… y = 6x – x 2 …… Substitute to Gantikan ke dalam x = 6x – x 2

x 2 – 6x + x = 0 x 2 – 5x = 0 x(x – 5) = 0 x = 0 or / atau x – 5 = 0 x = 5 y = x, y = 5 P = (5, 5)

(b) Area of shaded region = 5

∫0(6x – x 2)dx –

5

∫0x dx

Luas rantau berlorek = [3x 2 – 1

3 x 3 ] 5

0– [ 1

2 x 2 ] 5

0 = (75 – 125

3 – 0) – ( 252 – 0)

= 100

3 – 252

= 20 5

6 unit2

3 (a) x 2+ 5 = y …… y = 7 – x …… Substitute to Gantikan ke dalam x 2+ 5 = 7 – x x 2 + x – 2 = 0 (x + 2)(x – 1) = 0 x = –2, x = 1 From / Dari , y = x 2+ 5 and / dan y = (1)2 + 5 = 6

AM F5•J(61-75).indd 70 05/05/2010 15:34:29

71

P = (1, 6)


∫0(x 2 + 5) dx + 1

2 (7 – 1)(6) Luas rantau berlorek = [ 1

3 x 3 + 5x] 1

0+ 1

2 (6)(6) = ( 1

3 + 5 – 0) + 12 (36)

= 163 + 18

= 23 13 unit2

4 (a) x = (y – 2)2

= y 2 – 4y + 4 …… y = x …… Substitute to Gantikan ke dalam y = y 2 – 4y + 4 y 2 – 5y + 4 = 0 (y – 1)(y – 4) = 0 y = 1, and / dan y = 4 x = 1 x = 4 P = (1, 1), Q = (4, 4)


∫1y dy –

4

∫1(y 2 – 4y + 4) dy

Luas rantau berlorek = [ 1

2 y 2] 4

1– [ 1

3 y 3 – 2y 2+ 4y] 4

1

= (8 – 12 ) – [(64

3 – 32 + 16) – ( 13 – 2 + 4)]

= 152 – (16

3 – 73 )

= 92 unit2

5 When / Apabila y = 0, x(x – 4)(x + 3) = 0 x = 0, x = 4, x, –3

Area of shaded region = | 0

∫–3x(x – 4)(x + 3) dx|+

4

∫0x(x – 4)(x + 3) dx

Luas rantau berlorek

= | 0

∫–3(x 3 – x 2 – 12x) dx| +

4

∫0(x 3 – x 2 – 12x) dx

= |[ 14 x 4 – 1

3 x 3 – 6x 2] 0

–3 | + [ 14 x 4 – 1

3 x 3 – 6x 2] 4

0 = |0 – (81

4 + 273 – 54)| + (64 – 64

3 – 96 – 0) = 24 3

4 + 53 13

= 78 112 unit2

6 (a) y = 6 – x 2 …… y = 5x …… Substitute to Gantikan ke dalam 5x = 6 – x 2

x 2+5x – 6 = 0 (x – 1)(x + 6) = 0 x = 1 or / atau x = –6 When / Apabila x = 1, y = 5(1) = 5 When Apabila y = 0, 6 – x2 = 0 x 2 = 6 x = ±√6 A = (1,5) and / dan B = (√6, 0)

(b) Volume of shaded region = 13 πr2h +

√6

∫ 1πy2 dx Isi padu kawasan berlorek = 1

3 π(5)2(1) + π √6

∫ 1 (6 – x 2)2 dx

= 253 π + π

√6

∫ 1 (6 – x 2)2 dx = 25

3 π + [36x – 4x3 + 15 x 5] √6

1 ) = 25

3 π + (36√6 – 4(√6)3 + 15 (√6)5 – 32 1

5 ) = π(25

3 + 14.8302) = 23.1635π unit3

7 (a) At P when x = 0 Titik P apabila x = 0 y = 9 – 02 = 9 Point Q when y = 0 Titik Q apabila y = 0 0 = 9 – x 2

x 2 = 9 x = ±3 P = (0, 9), Q = (3, 0)

(b) Volume of shaded region = π 9

∫0(√9 – y)2 dy – 1

3 π (3)2(3) Isi padu kawasan berlorek = π

9

∫0(9 – y) dy – 9π

= π[9y – y 2

2 ] 9

0 – 9π

= π (81 – 812 – 0) – 9π

= 3112 π unit3

8 (a) x + 2y = 6 , x = 6 – 2y …… y2 = x – 3 …… Substitute to Gantikan ke dalam y2 = 6 – 2y – 3 y2 + 2y – 3 = 0 (y + 3)(y – 1) = 0 y = –3 or / atau y = 1 When / Apabila y = 1, x = 6 – 2 = 4 P = (4, 1) From , when x = 0, Daripada , apabila x = 0, 6 – 2y = 0 y = 3 Q = (0, 3)


∫0 x2 dy + 1

3 πr2h Isi padu rantau berlorek = π

1

∫0(y 2 + 3)2 dy + 1

3 π(4)2(3 – 1)

= π 1

∫0(y 4 + 6y2 + 9) dy + 1

3 π(16)(2) = π[ y5

5 + 2y3 + 9y] 1

0 + 32π

3

= π( 15 + 2 + 9 – 0) + 32π

3

= 56π5 + 32π

3

= 211315π unit3

9 dydx = 3x – 2

dy = (3x – 2) dx

y = ∫(3x – 2) dx

= 32 x 2 – 2x + c

y = 6 when x = –4 y = 6 apabila x = –4

6 = 32 (–4)2 – 2(–4) + c

6 = 24 + 8 + c c = –26 y = 3

2 x 2 – 2x – 26

10 (a) dydx = a

x3 + 4

4 = a8 + 4

a = 0

(b) dydx = 4

dy = 4 dx

y = ∫4 dx = 4x + c 6 = 4(2) + c c = 6 – 8

AM F5•J(61-75).indd 71 05/05/2010 15:34:30


= –2 y = 4x – 2 When / Apabila x = 3, y = 4(3) –2 = 12 – 2 = 10

11 (a) 1

∫–3(3f(x) – 4x) dx =

1

∫–33f(x) – 1

∫–34x dx

= 3 1

∫–3 f(x) dx –

1

∫–34x dx

= 3(6) – [2x 2] 1

–3 = 18 – [2 – (18)] = 18 – (–16) = 34

(b) 1

∫–3

f(x) + 43 dx =

1

∫–3

f(x)3 dx +

1

∫–3

43 dx

= 13

1

∫–3 f(x) dx + [ 4

3 x] 1

–3 = 1

3 (6) + [ 43 – (–4)]

= 2 + 163 = 22

3

(c) –3

∫ 1 (4 – 2f(x)) dx = –3

∫ 1 4 dx – –3

∫ 1 2f(x) dx

= [4x] –3

1 – [–2 1

∫–3 f(x) dx]

= (–12 – 4) + 2 1

∫–3 f(x) dx

= –16 + 2(6) = –4

12 (a) Area of shaded region = 4

∫2

4x 2 dx

Luas rantau berlorek = [– 4

x ] 4

2

= –1 – (–2)

= 1 unit 2


∫2( 4x 2 )2

dx Isi padu kawasan berlorek = 16π

4

∫2

1x 4 dx

= 16π[– 13x 3 ] 4

2 = 16

3 π[– 164 + 1

8 ] = 7

12 π unit 3

13 (a) Given / Diberi y = x2 + 3 x = √y – 3

Area of shaded region = 9

∫4(y – 3)

12 dy

Luas rantau berlorek = 2

3 [(y – 3)32 ] 9

4 = 2

3 [(9 – 3)32 – (4 – 3)]3

2

= 23 [6√6 – 1]

= 9.1313 unit2


∫4y – 3 dy

Isi padu kawasan berlorek = π[ y 2

2 – 3y] 9

4 = π[ 81

2 – 27 – ( 162 – 12)]

= 17 12 π unit3

14 (a) Area of shaded region = 1

∫0(2x2 + 4) dx + 1

2 (8 – 1)(6) Luas rantau berlorek = [2x3

3 + 4x] 1

0+ 1

2 (7)(6)

= ( 23 + 4 – 0) + 21

= 25 23 unit2

(b) Volume generated = 1

∫0 πy 2 dx + 1

3 πr2h Isi padu yang dijanakan

= π 1

∫0(2x2 + 4)2 dx + 1

3 π(6)2(8 – 1)

= π 1

∫0(4x4 + 16x2 + 16) dx + 7

3 π(36) = π[ 4x5

5 + 16x3

3 + 16x] 1

0+ 84 π

= [( 45 + 16

3 + 16) – 0] π + 84 π

= 106 215π unit3

15 (a) Given / Diberi y = x + 9 …… and / dan y = (x – 3)2 …… Substitute into Gantikan ke dalam x + 9 = x2 – 6x + 9 x 2 – 7x = 0 x(x – 7) = 0 x = 0 or / atau x = 7 When / Apabila x = 7, y = 7 + 9 = 16 When / Apabila x = 0, y = 9 M = (0, 9), N = (7, 16) Area of K / Luas K =

7

∫0(x + 9) dx –

7

∫0(x – 3)2 dx

= [ x 2

2 + 9x] 7

0– [ (x – 3)3

3 ] 7

0

= [(492 + 63) – 0] – [ 43

3 – (–3)3

3 ] = 175

2 – 913

= 57 16 unit2

(b) From y = (x – 3)2, when y = 0, x = 3 Daripada y = (x – 3)2, apabila y = 0, x = 3


∫0 πy2 dx

Luas rantau berlorek = π

3

∫0(x – 3)4 dx

= π[(x – 3)5

5 ] 3

0 = π[0 – (–3)5

5 ] = 243

5 π unit 3


PAPER 1 KERTAS 1

1 (a) OA = (129 )

(b) |OA| = √ 122 + 92 = 15 Unit vector in direction OA: Vektor unit dalam arah OA :

= OA|OA|

= 12 i~ + 9 j~

15 = 45

i~ + 35 j~

2 PR = PQ + QR = (–8i~ + 10 j~) + (–12 i~ – 8 j~) = –20 i~ + 2 j~

3 (a) OA = (86)

(b) AB = AO + OB = (–8 i~ – 6 j~) + (–10 j~) = –8 i~ – 16 j~

4 (a) PQ = λ QR 5a~ – 3b~ = λ[3a~ – (3 + k)b~] 5 = 3λ –3 = λ(–3 – k) λ = 5

3 –3 = 53 (–3 – k)

– 95 = –3 – k

k = – 65

(b) PQ = 53 QR

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73

PQ : QR = 5 : 3

5 OD = 34 OB

= 34 ( OC + CB )

= 34 (7 y~ + 15 x~)

6 (a) 2a~ – b~ = 2( 68 ) – ( 3

4 ) = (12

16) – ( 34 ) = ( 9

12) (b) The unit vector in direction of 2a~ – b~ Unit vektor dalam arah 2 a~ – b~

= 2a~ – b~|2a~ – b~ |

= ( 912)

√92 + 122

= 115 ( 9

12)

7 (a) h + 5 = 0 h = – 5 (b) k – 2 = 0 k = 2

8 (a) QR = QP + PR = –3a~ + 8b~

(b) PT = PR + RT

= PR + 14 RQ

= 8b~ + 14 (3a~ – 8b~)

= 8b~ + 3

4a~ – 2b~

= 34

a~ + 6b~

9 v~ = λw~ 2p i~ – 8 j~ = –3λ i~ + λ(p – 4) j~ 2p = –3λ λ = 2p

–3 –8 = (p – 4)λ

–8 = (p – 4)( 2p–3 )

–8 = 2p2 – 8p–3

24 = 2p2 – 8p 2p 2 – 8p – 24 = 0 p 2 – 4p – 12 = 0 (p – 6)(p + 2) = 0 p = 6 or / atau p = –2

10 PQ = –2a~ + 2b~

11 RP = RQ + QP = (3i~ + 7 j~) + (–2 i~ – 5 j~) = i~ + 2 j~ Unit vector in direction of RP Vektor unit dalam arah RP

= i~ + 2 j~√12 + 22

= i~ + 2 j~√5

12

PQ = 4a~ – 2b~ compared to PQ = ha~ + kb~ PQ = 4a~ – 2b~ dibandingkan dengan PQ = ha~ + kb~ h = 4 , k = –2

13 |v~| = 2|u~| √ m2 + 82 = 2√ 32 + (–4)2 m2 + 64 = 4(25) m2 = 36 m = ±√36 m = 6, m = –6

14 (a) PQ = PO + OQ = 9 i~ – 5 j~ + 3 i~ + k j~ = 12 i~ + (k – 5) j~

(b) QR = QO + OR = –3 i~ – k j~ + 2 i~ – 5 j~ = – i~ – (k + 5) j~

15 (t + 2)u~ + 9v~ = 0 …… 3u~ + (t – 4)v~ = 0 ……

÷ : t + 23 = 9

t – 4

(t + 2)(t – 4) = (9)(3) t 2 – 2t – 8 = 27 t 2 – 2t – 35 = 0 (t – 7)(t + 5) = 0 t = 7 or / atau t = –5

16 AB = AO + OB 9 i~ + k j~ = (–3 i~ – 4 j~) + (h i~ – k j~) = (–3 + h) i~ + (–4 – k) j~ Compare / Bandingkan, –3 + h = 9 or / atau –4 – k = k h = 12 –4 = 2k k = –2

17 (a) BE = BC + CA2

= –25a~ + 3b~

2

(b) CE = CB + BE

= 25a~ + ( –25a~ + 3b~ 2 )

= 25a~ + 3b~ 2

18 AC = –[ 13 (6a~)] = –2a~

CB = CO + OB = –4a~ + 4b~

CE = 14 CB = 1

4 (–4a~ + 4b~ ) = –a~ + b~

AE = AC + CE = –2a~ + (–a~ + b~) = –3a~ + b~

19 CA = CB + BA = (–2 i~ – 6 j~) + (– i~ + 2 j~) = –3 i~ – 4 j~ Unit vector in direction of CA Vektor unit dalam arah CA

= –3 i~ – 4 j~

√ (–3)2 + (–4)2

= –3 i~ – 4 j~

5

AM F5•J(61-75).indd 73 05/05/2010 15:34:34


20 AE = AB + BC + CD + DE = 4 r~ – 2q~ + 2p~ – 8 r~ = –4 r~ – 2q~ + 2p~

QE = 12 AE

= 12 (4 r~ – 2q~ + 2p~ – 8 r~)

= 2 r~ – q~ + p~ – 4 r~ = p~ – q~ – 2 r~

QD = QE + ED = p~ – q~ – 2 r~ + 8 r~ = p~ – q~ + 6 r~

21 AB = DC (ABCD parallelogram) (ABCD segi empat selari)

XB = 12 DB

= 12 (6a~ + 2b~ )

= 3a~ + b~

YB = 13 AB

= 2a~

YX = YB + BX = 2a~ – 3a~ – b~ = –a~ – b~

22 (a) BC = i~ – 4 j~

(b) |BC | = √ 12 + (–4)2 = √17 unit

(c) CA = –3 i~ + j~ |CA| = √ (–3)2 + 12 = √10 Unit vector in the direction CA :Vektor unit dalam arah CA :

CA|CA |

= 1√10

(–3 i~ + j~)

23 CB = CA + AB

= (–3–4) + ( 1

2 ) = (–2

–2) CD = CB + BD = (–2

–2) + (–4–4)

= (–6–6)

24 (a)

r~

y

xq~

(5, 4)

(1, –3)

(b) r~ + q~ = ( 54 ) + ( 1

–3) = ( 61 )

| r~ + q~| = √(62) 2 + (1) 2

| r~ + q~| = √37 unit

25 (a) FC (b) 0~

PAPER 2 KERTAS 2

1 AB = AC + CB = a~ + b~ AD = 1

3 AB

= 13 a~ + 1

3 b~ DC = DA + AC = – 1

3 a~ – 13 b~ +a~

= 23 a~ – 1

3 b~

= 13 (2a~ –b~)

FA = FC + CA = – 1

3 b~ – a~

= – ( 13 b~ + a~)

2 AC = AB + BC = a~ + 13 b~

Given / Diberi AC = 13 AE

AE =3a~ + b~ CE = 2

3 (3a~ + b~)

= 2a~ + 23 b~

DE = DC + CE = – 2

3 b~ + 2a~ + 23 b~

= 2a~ DE = 2a~ = 2AB . Proven / Terbukti

3 CA = CB + BA = 4a~ + b~ – b~ = 4a~

CE CA

= a~

4a~ = 14

CE : CA = 1 : 4 BE = 1

3 BD BC + CE = 1

3 BD –3a~ – b~ = 1

3 BD

BD = –9a~ – 3b~

CD = CB + BD = 4a~ + b~ + (–9a~ – 3b~)

= –5a~ – 2b~

4 EB = EA + AB

= 12 CA + AB

= 12 (b~ – a~) + a~

= 12 b~ + 1

2 a~

DC = DB + BC

= 2 EB + BC

= 2( 12 b~ – 1

2 a ~) + (–b~) = a~ = AB parallel / selari

5 AC = AE + EC = –a~ + b~ AD = AC + CD =(–a~ + b~) – 2

3 b~

= –a~ + 13 b~

AF = 12 AD

AM F5•J(61-75).indd 74 05/05/2010 15:34:36

75

= – 12 a~ + 1

6 b~

EF = EA + AF

= a~ – 12 a~ + 1

6 b~

= 12 a~ + 1

6 b~

6 BD = BA + AC + CD = (–2 i~ + 2 j~) + (– i~ – j~) + (2 i~ – j~) = – i~ | BD| = 1 unit

7 u~ + v~ = –4 i~ + (1 – p) j~ |u~ + v~ | = 5

√ 16 + (1 – p)2 = 5 (1 – p)2 = 25 – 16 = 32 (1 – p) = ±3 p = –2 or / atau p = 4 8 x~ = u~ – v~ = 4 i~ + j~– 2 i~ – 3 j~ = 2 i~ – 2 j~ | x~ | = √ (2) 2 + (– 2)2

| x~ | = √8 = 2√2Unit vector in direction of x~ Vektor unit dalam arah x~

x~

| x~ | = 2 i~ – 2 j~

2√2 = 1

√2 ( i~ – j~)

9 Au~ – B v~ = 4 i~ + 3 j~ A(3 i~ + 5 j~)– B(2 i~ + 7 j~) = 4 i~ + 3 j~ (3A – 2B) i~ + (5A – 7B) j~ = 4 i~ + 3 j~ 3A – 2B = 4 B = 3A – 4

2 5A – 7B = 3 5A = 21A – 28

2 + 3

10A – 21A + 28 = 6 11A = 22 A = 2 B = 6 – 4

2 = 1

10 AB = 2( AD + DX ) = 2(– i~ + 4 j~) = –2 i~ + 8 j~ AB = DC parallelogram / segi empat selari

Hence / Seterusnya, CD = 2 i~ – 8 j~ | CD| = √ (2) 2 + (– 8)2

| CD| = √68 = 2√17

11 (a) AR = AB + BR

= AB + 13 BD

= AB + 13 ( BA + AD)

= x~ + 13 (– x~ – 3 y~) = 2

3 x~– y~

PR = PA + AR = y~ + 2

3 x~ – y~ = 23 x~

(b) AR = h AC

= h( AD + DC ) 2

3 x~ – y~ = h(–3 y~ + k x~ – y~) 2

3 x~ – y~ = hk x~ – 4h y~

h = 14 , k = 8

3

12 (a) AC = 2 i~ – j~

(b) BA = 3 i~ + j~ | BA | = √ 3 2 + 12

| BA | = √10

Unit vector in direction of BA Vektor unit dalam arah BA

BA|BA |

= 3

√10 i~ + 1

√10 j~

13 (a) DE = DA + AE

= a~ – 12 AB

= a~ + 12 (–3a~ + 6b~)

= – 12 a~ + 3b~

CE = CD + DE

= 2a~ + (– 12 a~ + 3b~)

= 32 a~ + 3b~

(b) FE = 13 CE

= 12 a~ + b~

| FE | = 2 + (1)2

| FE | = √52

14 (a) (i) PA = PO + OA = –4 y~ + 3

5 OQ = –4 y~ + 3

5 (5 x~) = 3 x~ – 4 y~

(ii) OB = OP + PB = 4 y~ + 1

3 PQ = 4 y~ + 1

3 ( PO + OQ )

= 4 y~ + 13 (–4 y~ + 5 x~)

= 4 y~ – 43 y~ + 5

3 x~

= 53 x~ + 8

3 y~

(b) OC = h OB = h( 5

3 x~ + 83 y~)

= 5h

3 x~ + 8h3 y~

PC = k PA PO + OC = k(3 x~ + 4 y~) OC = 3k x~ – 4k y~ + 4 y~

5h3 x~ + 8h

3 y~ = 3k x~ + (4 – 4k) y~

5h3 = 3k 8h

3 = 4 – 4k

5h = 9k 83 ( 9k

5 ) = 4 – 4k

h = 9k5 24k = 20 – 20k

44k = 20

k = 511

h = 95 ( 5

11 ) = 9

11

12√

AM F5•J(61-75).indd 75 05/05/2010 15:34:37


(c) | QP| = √ (5(5)2) + (4(3))2 = √769 unit

15 (a) AC = AB + BC

= 3 y~ + 710 AD

= 3 y~ + 710 (5 x~)

= 3 y~ + 72 x~

(b) (i) AF = AE + EF

= 25 AD + m3 AB

= 25 (5 x~) + m3 (3 y~)

= 2 x~ + m y~

(ii) AF = k AC

2 x~ + m y~ = k(3 y~ + 72 x~)

2 = 7k2 m = 3k

k = 47

m = 3( 47 )

= 127


PAPER 1 KERTAS 1 1 420° = 360° + 60°

x

y

60°

420°

Angle 420° lies in quadrant I. Sudut 420° terletak di dalam sukuan I.

2

--210°x

y

Angle –210° lies in quadrant II. Sudut –210° terletak di dalam sukuan II.

3 206 π = 10

3 π = 2π + 4π3

x

y

43

10 3

Angle 206 π lies in quadrant III.

Sudut 206π terletak di dalam sukuan III.

4

x

y

-- 85

Angle – 85π lies in quadrant I.

Sudut – 85π terletak di dalam sukuan I.

5

x

y

--0.8

Angle –0.8π lies in quadrant III. Sudut –0.8π terletak di dalam sukuan III.

6 sin � = 45

7 tan � = 815

8 tan 150° = sin 150°cos 150°

= 0.5–0.866

= –0.5774

9 sin 240° = –sin (240° – 180°) = –sin 60° = –0.8660

10 cos 240° = –cos (240° – 180°) = –cos 60° = –0.5

11

-- 2

2

90° 180° 270° 360°x

y

O

y = 2 sin 2x

12

3

90° 180° 270° 360°x

y

O

y = 3|cos 2 x|

13

0° 90° 180° 270° 360°x

yy = 5 tan x

14

1

2

0 90° 180° 270° 360°x

y

y = sin x + 1

15

-- 1

-- 2

-- 3

-- 4

0 90° 180° 270° 360°x

y

y = 2 cos x -- 2

AM F5•J(76-88).indd 76 05/05/2010 15:35:27

77

16 cot2 x – tan2 x = (cosec2 x – 1) – (sec2 x – 1) = cosec2 x – 1 – sec2 x + 1 = cosec2 x – sec2 x

17 1 – tan2 x1 + tan2 x = 1 – (sec2 x – 1)

sec2 x

= 2 – sec2 xsec2 x

= 2sec2 x – sec2 x

sec2 x

= 2sec2 x – 1

= 2 cos2 x – 1

18 2 cos2 � + sin � – 2 = 0 2(1 – sin2 �) + sin � – 2 = 0 2 – 2 sin2 � + sin � – 2 = 0 –2 sin2 � + sin � = 0 sin �(–2 sin � + 1) = 0 sin � = 0 or / atau sin � = 1

2 � = 0°, 180°, 360°, � = 30°, 150° � = 0°, 30°, 150°, 180°, 360°

19 cos xsin x + 2 cos x = 0

cos x + 2 cos x sin x = 0 cos x(1 + 2 sin x) = 0 cos x = 0 or / atau 1 + 2sin x = 0 x = 90°, 270° sin x = – 1

2 x = 210°, 330° x = 90°, 210°, 270°, 330°

20 3 cos 2x = 8 sin x – 5 3(1 – 2 sin2 x) = 8 sin x – 5 3 – 6sin2 x = 8 sin x – 5 –6sin2 x – 8 sin x + 8 = 0 3sin2 x + 4 sin x – 4 = 0 (3sin x – 2)(sin x + 2) = 0 sin x = 2

3 or / atau sin x = –2 (No solution)

x = 41.81°, 138.19° x = 41.81°, 138.19°

21 sin 72° cos 42° – cos 72° sin 42° = sin(72° – 42°) = sin 30° = 0.5

22 cos 55° cos 35° – sin 55° sin 35° = cos(55° + 35°) = cos 90° = 0

23 2 sin 45° cos 45° = sin 2(45°) = sin 90° = 1

24 sin 2�1 + cos 2�

= sin 2�1 + (2 cos2 � – 1) = sin 2�

2 cos2 �

= 2 sin � cos �2 cos2 �

= sin �cos �

= tan �

25 3 cos 2x + cos x = 2 3(2 cos2 x – 1) + cos x – 2 = 0 6 cos2 x + cos x – 5 = 0 (6 cos x – 5)(cos x + 1) = 0 6 cos x – 5 = 0 or / atau cos x + 1 = 0 cos x = 5

6 cos x = – 1 x = 33°33', 326°27' x = 180° x = 33°33', 180°, 326°27'

PAPER 2 KERTAS 2 1 (a)

320°x

y

Angle 320° lies in quadrant IV. Sudut 320° terletak di dalam sukuan IV. (b) 375° = 360° + 15°

15°

375°x

y

Angle 375° lies in quadrant I. Sudut 375° terletak di dalam sukuan I.

(c) 500° = 360° + 140°

140°

500°x

y

Angle 500° lies in quadrant II. Sudut 500° terletak di dalam sukuan II.

2 (a)

--350°x

y

Angle –350° lies in quadrant I. Sudut –350° terletak di dalam sukuan I.

(b) –430° = 360° + (–70°)

--70°

360°x

y

Angle –430° lies in quadrant IV. Sudut –430° terletak di dalam sukuan IV.

(c) – 94 π = –2π + (– π4 )

x

y

-- 4

--2

Angle – 94π rad lies in quadrant IV.

Sudut – 94π rad terletak di dalam sukuan IV.

3 (a)

x

y

--0.7

Angle –0.7π rad lies in quadrant III. Sudut –0.7π rad terletak di dalam sukuan III.

AM F5•J(76-88).indd 77 05/05/2010 15:35:39


(b) 103 π = 2π + 4

3 π

x

y

43

103

Angle 103 π rad lies in quadrant III.

Sudut 103π rad terletak di dalam sukuan III.

(c) 2.4π = 2π + 0.4π

x

y

0.4

2.4

Angle 2.4π rad lies in quadrant I. Sudut 2.4π rad terletak di dalam sukuan I.

4 (a) cos 54° = sin(90° – 54°) = sin 36° = 0.5878

(b) cot 54° = tan(90° – 54°) = tan 36° = 0.7265

5 (a) tan A = 724

xOA

y

24

257

(b) sec A =

1cos A

= 1– 24

25

= – 2524

(c) cosec A = 1

sin A = 1

– 725

= – 257

6 (a) sec 155° = 1cos 155°

= 1–cos (180° – 155°)

= 1–cos 25°

= 1–0.9063

= –1.1034

(b) cot ( 53 π) = cot ( 5

3 × 180°) = cot 300°

= 1tan 300°

= 1–tan (360° – 300°)

= 1–tan 60°

= 1–1.7321

= –0.5773

(c) cosec 260° = 1sin 260°

= 1–sin (260° – 180°)

= 1–sin 80°

= 1–0.9848

= –1.0154

7 (a) and (b)

-- 2

-- 1

2

y = cos 2x

y = 2 sin 2x

1

0 90° 180° 270° 360°x

y

(c) Number of solutions = 4 Bilangan penyelesaian = 4

8 (a) and (b)

2

3y = |3 cos 2x|

1

0x

y

22

y = 2 -- x 2

3 2

2 – |3 cos 2x| = x2π

|3 cos 2x| = 2 – x2π

y = 2 – x2π

When , x = 0, y = –2, Apabila, x = 2π, y = 1 Number of solutions = 8 Bilangan penyelesaian = 8

9 (a) and (b)

2

5

y = 2 -- 3 sin x

y = -- 2 cos x

1

0x

y

22

--2

--13 2

–2 cos x + 3 sin x = 2 –2 cos x = 2 – 3 sin x y = 2 – 3 sin x When x = 0, y = –2, Apabila x = 2π, y = 2 Number of solutions = 2 Bilangan penyelesaian = 2

10 (a) 1 – 2 sin2 xsin x + cos x = sin2 x + cos2 x – 2 sin2 x

sin x + cos x

= cos2 x – sin2 xsin x + cos x

= (cos x – sin x)(cos x + sin x)(sin x + cos x)

= cos x – sin x

(b) 1 – tan2 x1 + tan2 x = 1 – (sec2 x – 1)

sec2 x = 2sec2 x – sec2 x

sec2 x

= 2sec2 x – 1 = 2 cos2 x – 1

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79

11 (a) 2 sin2 x + cos x – 2 = 0 2(1 – cos2 x) + cos x – 2 = 0 2 – 2 cos2 x + cos x – 2 = 0 2 cos2 x – cos x = 0 cos x(2 cos x – 1) = 0 cos x = 0 or / atau 2 cos x – 1 = 0 x = 90°, 270° cos x = 1

2 x = 60°, 300° x = 60°, 90°, 270°, 300°

(b) sin2 x cos x = 2 cos2 x – 2 sin2 x cos x = 2(1 – sin2 x) – 2 sin2 x cos x + 2 sin2 x = 0 sin2 x(cos x + 2) = 0 sin2 x = 0 or / atau cos x = –2 ( No solution) x = 0 ( Tiada penyelesaian) x = 180°, 360° x = 0°, 180°, 360°

12 (a) 3 cos 2x = 8 sin x – 5 3(1 – 2 sin2 x) = 8 sin x – 5 – 6 sin2 x – 8 sin x + 8 = 0 3 sin2 x + 4 sin x – 4 = 0 (3 sin x – 2)(sin x + 2) = 0 sin x = 2

3 or / atau cos x = – 2( No solution) / ( Tiada penyelesaian)

x = 41.81°, 138.19°

(b) 15 sin2 x = sin x + 4 sin 30° 15 sin2 x = sin x + 2 15 sin2 x – sin x – 2 = 0 (5 sin + 2)(3 sin x – 1) = 0 sin x = – 2

5 or / atau cos x = 13

x = 203.58°, 336.42° x = 19.47°, 160.53° x = 19.47°, 160.53°, 203.58°, 336.42°

13 (a) 2 tan x2 – sec2 x = 2 tan x

2 – (1 + tan2 x) = 2 tan x1 – tan2 x

= tan x + tan x1 – tan x tan x = tan 2x

(b) 1 + cos 2�1 – cos 2�

= 1 + (2cos2 � – 1)1 – (1 – 2sin2 �)

= 2 cos2 �2 cos2 �

= cot2 � = cosec2 � – 1

14 (a) cot x + 2 cos x = 0 cos x

sin x + 2 cos x = 0

cos x + 2 cos x sin x = 0 cos x(1 + 2 sin x) = 0 cos x = 0 or / atau 1 + 2 sin x = 0 x = 90°, 270° sin x = – 1

2 x = 210°, 330° x = 90°, 210°, 270°, 330°

(b) 2 sin 2x – sin x = 0 2(2 sin x cos x) – sin x = 0 4 sin x cos x – sin x = 0 sin x(4 cos x – 1) = 0 sin x = 0 or / atau 4 cos x = 1 x = 0°, 180°, 360° cos x = – 1

4 x = 75°31', 284°29' x = 0°, 75°31', 180°, 284°29', 360°

15 (a) 2 sin 2x – sin x = 0 2(2 sin x cos x) – sin x = 0 4 sin x cos x – sin x = 0 sin x(4 cos x – 1) = 0 sin x = 0 or / atau 4 cos x – 1 = 0 x = 0°, 180°, 360° 4 cos x = 1 cos x = 1

4 x = 75°31', 284°29' x = 0°, 75°31', 180°, 284°29', 360°

(b) sin2 x – 2 sin x = cos 2x sin2 x – 2 sin x = 1 – 2 sin2 x sin2 x – 2 sin x – 1 + 2 sin2 x = 0 3 sin2 x – 2 sin x – 1 = 0 (sin x – 1)(3 sin x + 1) = 0 sin x – 1 = 0 or / atau 3 sin x + 1 sin = 1 sin x = – 1

3 x = 90° x = 199°28', 340°32' x = 90°, 199°28', 340°32'

MID-TERM EXAMINATIONPEPERIKSAAN PERTENGAHAN PENGGAL

PAPER 1 KERTAS 1 1 In arithmetic progressions, Dalam janjang aritmetik, Tn + 1 = Tn – 1n – 1

(1 – 3h) – (5k – 4) = 7k – (1 – 3h) 1 – 3h – 5k + 4 = 7k – 1 + 3h –6h = 12k – 6 h = 1 – 2k

2 (a) In arithmetic progressions, Dalam janjang aritmetik, Tn = a + (n – 1)d Given Diberi T3 = 33 = a + 2d a = 33 – 2d ……

T6 = 51 = a + 5d a = 51 – 5d …… : 33 – 2d = 51 – 5d 3d = 18 d = 6 a = 51 – 5(6) = 21

(b) Tn = a + (n – 1)d 69 = 21 + (n – 1)6 6(n – 1) = 48 n – 1 = 8 n = 9 The number of terms is 9. Bilangan sebutan ialah 9.

3 (a) Given Diberi Sn = 33 = n2 (19 + 9n)

a = S1 = 12 (19 + 9)

= 12 (28)

= 14 T2 = S2 – S1 = 2

2 [19 + 9(2)] – 14

= 37 – 14 = 23 d = T2 – T1

= 23 – 14 = 9

(b) Sum from T10 to T15

Hasil tambah T10 hingga T15

= S15 – S9

= 152 [19 + 9(15)] – 9

2 [19 + 9(9)]

= 2 310 – 9002

= 705

4 T1 = a = 27 r = T2

T1 = – 18

27 = – 2

3 Let the number of terms = n Biar bilangan sebutan = n Tn = arn – 1

256243 = 27(– 2

3 )n – 1

(– 23 )n – 1

= 2566 561

= (– 23 )8

n – 1 = 8 n = 9

AM F5•J(76-88).indd 79 05/05/2010 15:35:44


5 a = T1 = 81 r = T2

T1 = 27

81 = 1

3

S� = a1 – r

= 811 – 1

3 = 81 × 3

2 = 121.5

6 a = 4, r = 64 = 3

2 Let the number of terms = n Biar bilangan sebutan = n Given Diberi Tn < 103 4( 3

2 )n – 1 < 103

( 32 )n – 1

< 25.75

(n – 1) log 1.5 < log 25.75 0.1761(n – 1) < 1.411 n – 1 < 8.011 n < 9.011 n = 9

7 0.545454 … = 0.54 + 0.0054 + 0.000054 + … = T1 + T2 + T3 + … = S�

a = 0.54, r = 0.00540.54

= 0.01 S� = a

1 – r = 0.54

1 – 0.01 = 0.54

0.99 = 6

11 0.545454 … = 6

11

8 Given Diberi y = x

7x – 2

1y = 7x – 2

x = 7x

x – 2x = –2( 1

x ) + 7

Y-intercept pintasan-Y = a = 7

Gradient kecerunan = 7 – b0 – 4 = –2

7 – b = –2(–4) = 8 b = –1

9 Gradient Kecerunan = 3 – 1

3 – 2

= 2 Y – 1 = 2(X – 2) Y = 2X – 3 log y = 2 log x – 3 = log x2 – log 103

= log ( x2

103) y = x2

103

10 Given Diberi y = 2x + 3x

xy = 2x2 + 3 Compare with Bandingkan dengan Y = 2X + c Y = xy and dan X = x2

11 6

∫3(7g(x) – kx2)dx =

6

∫37g(x) dx –

6

∫3kx 2 dx

21 = 7 6

∫3g(x) dx – k

6

∫3x 2 dx

= 7(12) – k[ x3

3 ] 6

3

= 84 – k(72 – 9) = 84 – 63k 63k = 84 – 21 = 63 k = 1

12 (a) Given Diberi

1

∫–3[ f(x) – 2x] dx = 14

1

∫–3 f(x) dx –

1

∫–32x dx = 14

1

∫–3 f(x) dx = 14 + [x2] 1

–3

= 14 + (1 – 9) = 6

1

∫–3[2 f(x)] dx = 2 × 6

= 12

(b) 4

∫–3[2 f(x) – 3] dx = 2

1

∫–3 f(x) dx + 2

4

∫1 f(x) dx –

4

∫–33 dx

= 2(6) + 2(7) – [3x] 4–3

= 26 – (12 + 9) = 5

13 Given gradient Diberi kecerunan

= dydx = 12x2 + 6x – 1

y = ∫(12x2 + 6x – 1) dx = 4x3 + 3x2 – x + c Equation passes through (1, 11). Persamaan melalui titik (1, 11). 11 = 4 + 3 – 1 + c c = 11 – 6 = 5 Equation of curve: Persamaan lengkung: y = 4x3 + 3x2 – x + 5

14 dydx = 2x(6x2 – 3x + 1)

= 12x3 – 6x2 + 2x

y = ∫(12x3 – 6x2 + 2x) dx = 3x4 – 2x3 + x2 + c The curve passes through (2, 30) Lengkung melalui titik (2, 30) 30 = 3(24) – 2(23) + (22) + c = 48 – 16 + 4 + c c = 30 – 36 = –6 Equation of the curve: Persamaan bagi lengkung: y = 3x4 – 2x3 + x2 – 6

15 Given y = –x2 + 3x + 4 Diberi When x = 3 and / dan y = k,� Apabila k = –32 + 3(3) + 4 = 4 When y = 0, Apabila y = 0, y = –x2 + 3x + 4 = 0 x2 – 3x – 4 = 0 (x – 4)(x + 1) = 0 x = 4 or/atau x = –1 Coordinate of B = (–1, 0) Koordinat B = (–1, 0)


∫–1 y dx – 1

2 (4)(k) Luas kawasan berlorek =

3

∫–1(–x2 + 3x + 4) dx – 2(4)

= [– x3

3 – 3x2

2 + 4x] 3

1– 8

= 332 + 13

6 – 8

= 563 – 8 = 32

3

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81

16 a~ = –2 i~ + 5 j~, b~ = 3 i~ + 2 j~ a~ + 2b~ = (–2 i~ + 5 j~) + 2(3 i~ + 2 j~) = –2 i~ + 5 j~ + 6 i~ + 4 j~ = 4 i~ + 9 j~ |a~ + 2b~| = √ 42 + 92 = √97 Unit vector Vector unit = 1

√97 (4 i~ + 9 j~)

= 4

√97 i~ + 9

√97 j~

17

OR = 5 i~ + 4 j~ , OS = – i~ – 5 j~ RT : TS = 1 : 2, TR = 1

3 SR

(a) SR = SO + OR = i~ + 5 j~ + 5 i~ + 4 j~ = 6 i~ + 9 j~ TR = 1

3 (6 i~ + 9 j~) = 2 i~ + 3 j~

(b) OT = OS + ST = OS + 2 TR = – i~ – 5 j~ + 2(2 i~ + 3 j~) = – i~ – 5 j~ + 4 i~ + 6 j~ = 3 i~ + j~

18 OA = 6 i~ + 2 j~ , OB = 3 i~ + 5 j~ AB = OC = AO + OB = –6 i~ – 2 j~ + 3 i~ + 5 j~ = –3 i~ + 3 j~ AC = AO + OC = –6 i~ – 2 j~ – 3 i~ + 3 j~ = –9 i~ + j~

19 (a) Since X, Y and Z are collinear, Oleh kerana X, Y dan Z adalah segaris,

XY = m YZ . 5a~ – 2b~ = 3ma~ + m(4 – k)b~ Compare both sides: Bandingkan kedua-dua belah: 3m = 5 m = 5

3 and dan –2 = m(4 – k) = 5

3 (4 – k)

4 – k = – 65

k = 265

(b) XY = m YZ = 5

3 YZ

3 XY = 5 YZ

XYYZ

= 53

XY : YZ = 5 : 3

20 (a) c~ = 2a~ + 3b~ = 2(–2 i~ + 3 j~) + 3(3 i~ – 7 j~) = –4 i~ + 6 j~ + 9 i~ – 21 j~ = 5 i~ – 15 j~

(b) |c~| = √ 52 + (–15)2 = √250 = 5√10 Unit vector, Vektor unit,

| c~ | = c~

|c~| 1

5√10 = (5 i~ – 15 j~)

= 1

√10 i~ – 3

√10 j~

21 sin2 x – sin x = 3 cos2 x = 3(1 – sin2 x) = 3 – 3sin2 x sin2 x + 3sin2 x – sin x – 3 = 0 4 sin2 x – sin x – 3 = 0 (4 sin x + 3)(sin x – 1) = 0 4 sin x + 3 = 0 or/atau sin x – 1 = 0 sin x = – 3

4 sin x = 1

x = 90°, 228.59°, 311.41°

22 4 cot x sin x = 3 sin x – 3 sin 2x + 2 4(cos x

sin x ) sin x = 3 sin x – 3(2 sin x cos x) + 2

4 cos x = 3 sin x – 6 sin x cos x + 2 6 sin x cos x + 4 cos x – 3 sin x – 2 = 0 (3 sin x + 2)(2 cos x – 1) = 0 3 sin x + 2 = 0 or/atau 2 cos x – 1 = 0 sin x = – 2

3 cos x = 12

x = 221.81°, 318.19° x = 60°, 300° x = 60°, 221.81°, 300°,318.19°

23 8 sin2 � 2

– 6 sin � 2

– 6 sin � + 9 cos � 2

= 0

8 sin2 � 2

– 6 sin � 2

– 12 sin � 2

cos � 2

+ 9 cos � 2

= 0

8 sin2 �

2

cos � 2

– 6 sin �

2

cos � 2

– 12 sin � 2

+ 9 = 0

8 tan � 2

sin � 2

– 6 tan � 2

– 12 sin � 2

+ 9 = 0

(2 tan � 2

– 3)(4 sin � 2

– 3) = 0

2 tan � 2

– 3 = 0 or/atau 4 sin � 2

– 3 = 0

tan � 2

= 1.5 sin � 2

= 0.75

� 2

= 56.31°, 236.3° � 2

= 48.59°, 131.41°

� = 112.62°, 472.6° � = 97.18°, 262.82° � = 97.18°, 112.62°, 262.82°

24 Let Biar � – 90° = x 12 cos x – 4 cot x + 9 = 3

sin x 12 cos x sin x – 4 cot x sin x + 9 sin x = 3 12 cos x sin x – 4(cos x

sin x ) sin x + 9 sin x – 3 = 0

12 cos x sin x – 4 cos x + 9 sin x – 3 = 0 (4 cos x + 3)(3 sin x – 1) = 0 (4 cos x + 3) = 0 or/atau 3 sin x – 1 = 0

cos x = –0.75 sin x = 0.333 � – 90° = 138.59°, 221.41° � – 90° = 19.47°, 160.53° � = 228.59°, 311.41° � = 109.47°, 250.53° 180° � � � 360°, � = 228.59°, 250.53°, 311.41°

25

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(a) 1 + sec2 � = 1 + 1cos2 �

= 1 + ( 2p3 )2

= 1 + (4p2

9 ) = 9 + 4p2

9

(b) sin � – cot � = sin � – 1tan �

= √ 4p2 – 9

2p – 3

√ 4p2 – 9

= 4p2 – 9

2p√ 4p2 – 9 – 3(2p)

2p√ 4p2 – 9

= 4p2 – 9 – 6p

2p√ 4p2 – 9 = 4p2 – 6p – 9

2p√ (2p – 3)(2p + 3)

PAPER 2 KERTAS 2 1 (a) T 1 = 1

2 (4p)(2p + 1)

= 2p(2p + 1) = 4p2 + 2p d = T 2 – T 1

= 12 (4p)(2p + 3) – (4p2 + 2p)

= 2p(2p + 3) – 4p2 – 2p = 4p2 + 6p – 4p2 – 2p = 4p

(b) T 9 = a + 8d 38 = (4p2 + 2p) + 8(4p) = 4p2 + 2p + 32p 4p2 + 34p – 38 = 0 2p2 + 17p – 19 = 0 (2p + 19)(p – 1) = 0 p > 0, p = 1

(c) a = 4p2 + 2p = 4 + 2 = 6 d = 4p = 4 Sum of areas from T6 to T10 = S10 – S5 Jumlah luas dari T6 hingga T10

= 102 [2(6) + 9(4)] – 5

2 [2(6) + 4(4)]

= 5(48) – 52 (28)

= 170 cm2

2 (a) a = RM19 800, r = 1.05 Annual salary for year 2010 = T8

Gaji tahunan untuk tahun 2010 = T8

T8 = ar7

= 19 800(1.05)7 = RM27 861

(b) Total saving Jumlah simpanan = T1 × 0.3 + T2 × 0.3 + … + Tn × 0.3 = Sn × 0.3 = 19 800(1.05n – 1)

1.05 – 1 × 0.3

= 118 800(1.05n – 1) Minimum value of n: Nilai minimun n: 118 800(1.05n – 1) � 31 500 1.05n � 1.2652 n(log 1.05) � log 1.2652 n � 0.1026

0.0212 n � 4.84 n integer, n = 5

(c) 2005 n = 3, 2009 n = 7 Total salary = S7 – S2

Jumlah gaji = a(r7 – 1)r – 1 – a(r2 – 1)

r – 1 = 19 800(1.057 – 1) – 19 800(1.052 – 1)

1.05 – 1 = 8 060.584 – 2 029.5

0.05 = RM120 621.68 ≈ RM120 622

3 (a) (i) Given Diberi AP : PB = 3 : 2 PB = 2

5 AB

PB = PC + CB = –3 y~ + x~ + 3 y~ = x~ AB = 5

2 PB

= 52

x~

(ii) Given Diberi AC = 3RC RC = 1

3 AC

AC = AB + BC = 5

2x~ – x~ – 3 y~

= 32

x~ – 3 y~ RC = 1

3 ( 32

x~ – 3 y~) = 1

2x~ – y~

(b) AP : PB = 3 : 2 AP = 3

5 AB

PC = PA + AC = 3

5 BA + AC

= 35 (– 5

2x~) + 3

2x~ – 3 y~

= – 32

x~ + 32

x~ – 3 y~ = – 3 y~ QC = 5

9 (– 3 y~)

= – 53 y~

RQ = RC + CQ = 1

2x~ – y~ + 5

3 y~ = 1

2x~ + 2

3 y~ AC = 3RC, AR = 2RC AR = 2( 1

2x~ – y~) = x~ – 2 y~

RB = RA + AB = –x~ – 2 y~ + 5

2x~

= 32

x~ + 2 y~ = 3( 1

2x~ + 2

3 y~) = 3RQ

Since RB = 3RQ with common point R, thus R, Q and B are collinear Oleh kerana RB = 3 RQ dengan titik sepunya R, maka R, Q dan B adalah segaris

(c) Area of triangle ABC = 12 × | AB | × | PC |

Luas segi tiga ABC | AB | = | 5

2 x~| = 5

2 |x~| = 5

2 (2) = 5

| PC | = |–3 y~ | = 3| y~ | = 3(3) = 9

Area Luas = 12 (5)(9) = 22.5 unit2

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4 (a) (i) Given Diberi PB = 2 i~ + j~ AB = 3 PB = 3(2 i~ + j~) = 6 i~ + 3 j~ (ii) AC = AB + BC = 6 i~ + 3 j~ + i~ – 5 j~ = 7 i~ – 2 j~

(b) ED = 23 AB

= 23 (6 i~ + 3 j~)

= 4 i~ + 2 j~ EC = EA + AC = i~ + 6 i~ + 7 i~ – 2 j~ = 8 i~ + 4 j~ = 2(4 i~ + 2 j~)

= 2 ED

Since EC = 2ED, thus E, D and C are collinear. Oleh kerana EC = 2 ED , maka E, D dan C adalah segaris

(c) Let Biar DB = r~ AD = AE + ED = – i~ – 6 j~ + 4 i~ + 2 j~ = 3 i~ – 4 j~ r~ = DA + AB = –3 i~ + 4 j~ + 6 i~ + 3 j~ = 3 i~ + 7 j~ | r~| = √ 32 + 72

= √58 Vector unit, Unit vektor,

| r~ | = r~

| r~| = 1

√58 (3 i~ + 7 j~)

5 (a) Right hand side Sebelah kanan

= 1sin x cos x – cot x

= 1sin x cos x – cos x

sin x = 1 – cos2 x

sin x cos x

= sin2 xsin x cos x

= sin xcos x

= tan x = Left hand side Sebelah kiri

(b)

π2 sin x cos x – π2cot x + x – π = 0

π2 sin x cos x – π2cot x = –x + π

12( 1

sin x cos x – cot x) = – 1π

x + 1

12 tan x = – 1

π x + 1

Draw y Lukis y = – 1π

x + 1 Number of solutions = 3 Bilangan penyelesaian = 3

6 (a) Left hand side Sebelah kiri = cosec2 x + 2 cos2 x – cot2 x – 2 = cosec2 x + 2 cos2 x – (cosec2 x – 1) – 2 = cosec2 x + 2 cos2 x – cosec2 x + 1 – 2 = 2 cos2 x – 1 = cos 2x = Right hand side Sebelah kanan

(b) (i)

(ii) 4π cosec2 x + 8π cos2 x – 4π cot2 x – 8π – 3x + 4π = 0

4π cosec2 x + 8π cos2 x – 4π cot2 x – 8π = 3x + 4π

cosec2 x + 2 cos2 x – cot2 x – 2 = 34π x – 1

Draw line y = 34π x – 1

Lukis garis y Number of solutions = 4 Bilangan penyelesaian = 4

7 (a)

x 1 2 3 4 5 6yx 2.50 4.55 6.50 8.40 10.56 12.50

x

yx

13

12

11

10

9

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6

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(b) Given Diberi y = abx2 + 2b2x y

x = abx + 2b2

Gradient = ab, Y-intercept = 2b2

Kecerunan = ab, pintasan Y = 2b2

From the graph, gradient = 12.5 – 2.56 – 3 = 10

5 Daripada graf, kecerunan ab = 2 …… Y-intercept = 2b2 = 0.5 pintasan Y b2 = 1

4 b = 1

2 ……

: a( 12 ) = 2

a = 4

8 (a)

x2 + 1 2.00 3.25 5.00 7.25 10.00 13.25

log10 y –0.12 0.10 0.40 0.80 1.28 1.86

(b)

log10 y

x 2 + 1

2.0

1.8

1.6

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0

–0.2

–0.4

–0.6

2 4 6 8 10 12 14

(2.7, 0)

–0.47

8.1

(10, 1.28)

(c) (i) When Apabila y = 8.9, log10 8.9 = 0.95 From the graph, when log10 8.9 = 0.95, Daripada graf, apabila log10 8.9 = 0.95, x2 + 1 = 8.1 x2 = 7.1 x = ±√7.1 = ±2.665 (ii) Given Diberi y = mnx2 + 1

log y = logn(x2 + 1) log m Gradient = log n = 1.28 – 0

10 – 2.7 Kecerunan = 0.175 n = 1.5 (iii) Y-intercept = log m = –0.47 Pintasan-Y m = 0.3388

9 (a) From Dari y = 12x – 36, When Apabila y = 0, 12x – 36 = 0 12x = 36 x = 3

Coordinate of points S = (3, 0) Koordinat titik S When Apabila x = 5, y = 12(5) – 36 = 60 – 36 = 24

Coordinate of points T = (5, 24) Koordinat titik T

(b) Area of shaded region Luas kawasan berlorek = 1

2 (5 – 3)(24) – 5

∫3 2k

3 (x – 3)3 dx

12 = 24 – 2k3 [ (x – 3)4

4 ] 5

3

2k3 (4 – 0) = 12

2k3 = 3, k = 9

2

(c) Equation of curve: y = 3(x – 3)3

Persamaan lengkung

Volume generated = Volume of cone –

5

∫3 πy2 dx

Isi padu janaan Isi padu kon

= 13 π(242)(2) – π

5

∫3 9(x – 3)6 dx

= 384π – 9π[ (x – 3)7

7 ] 5

3

= 384π – 9π( 27

7 – 0) = 384π – 1 152π

7 = 219 3

7 π unit3

10 (a) Let K = coordinate of x Biar K = koordinat x y = –x + 7 …… y = √2x + 1 y2 = 2x + 1 …… : (–x + 7)2 = 2x + 1 x2 – 14x + 49 = 2x + 1 x2 – 16x + 48 = 0 (x – 12)(x – 4) = 0 x = 12 or / atau 4 Since / oleh kerana k < 5, k = 4

(b) From y = –x + 7, when y = 0, x = 7 Daripada y = –x + 7, apabila y = 0, x = 7 y = –x + 7 intersects with x-axis at x = 7

y = –x + 7 bersilang dengan paksi-x pada x = 7 Area of shaded region Luas kawasan berlorek

= 4

∫0√(2x + 1) dx + 1

2 (7 – 4)(3) =

4

∫0(2x + 1)

12 dx + 9

2 = [ (2x + 1)

32

2( 32 ) ] 4

0

+ 92

= 26

3 + 92

= 13 16 unit2

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85

(c) Volume generated = 4

∫0 πy2 dx + Volume of cone

Isip adu janaan Isip adu kon

= π 4

∫0(2x + 1) dx + 1

3 π(32)3

= π[x 2 + x] 40 + 9π

= π(20 – 0) 9π = 29π unit3

11 (a) (i) DC = DA + AC = –(–4x~ + 10y~) + 4x~ + 8 y~ = 4x~ – 10y~ + 4x~ + 8 y~ = 8x~ – 2y~

(ii) AM = 12 AC

= 12 (4x~ + 8y~)

= 2x~ + 4y~ MD = MA + AD = –(2x~ + 4y~) + (–4x~ + 10y~) = –2x~ – 4y~ – 4x~ + 10y~ = –6x~ + 6y~

(b) DB = DA + AB 10x~ – 6y~ = –(–4x~ + 10y~) + ax~ + b y~ = 4x~ – 10y~ + ax~ + b y~ = (4 + a)x~ + (b – 10)y~ 4 + a = 10 and dan b – 10 = –6 a = 6 b = 4

(c) MN = MA + AN = 1

2 CA + 12 AB

= 12 ( CA + AB )

CB = CA + AB Since CB = 2 MN , therefore MN is parallel to CB . Oleh kerana CB = 2MN, maka MN selari dengan CB .

(d) NC = NA + AC

= – 12 (6x~ + 4y~) + 4x~ + 8 y~

= –3x~ – 2y~ + 4x~ + 8 y~ = x~ + 6y~ Let Biar

NC = r~

| r~| = √ 12 + 62

= √37 Unit vector, Vektor unit,

| r~ | = r| r~|

= x~ + 6y~√37

= 1√37

(x~ + 6y~)


PAPER 1 KERTAS 1 1 5! = 120

2 9!2! = 362 880

2 = 181 440

3 11!2! 2! 2! = 39 916 800

2(2)(2)

= 4 989 600

4 Number of arrangements = 8! Bilangan susunan = 40 320

5 6P4 = 360

6 9P6 = 9!

(9 – 6)! = 9!

3! = 362 880

6 = 60 480

7 Number of arrangements Bilangan susunan = 4C1 × 5C1 × 4C1 × 3C1

= 240 or / atau 4 × 5P3 = 240

8 First digit = 1: Digit pertama

Number of 5-digit even number less than 3 000 can be formed Bilangan nombor genap 5 digit kurang daripada 3 000 yang dapat dibentuk = 1 × 4C1 × 3C1 × 2C1 = 24 First digit = 2: Digit pertama

Number of 5-digit even number less than 3 000 can be formed Bilangan nombor genap 5 digit kurang daripada 3 000 yang dapat dibentuk = 1 × 4C1 × 3C1 × 3C1 = 36 Total / Jumlah = 24 + 36 = 60

9 Last digit = 4: Digit akhir Number of 5-digit even number less than 6 000 can be formed Bilangan nombor genap 5 digit kurang daripada 6 000 yang dapat dibentuk = 4C1 × 4C1 × 3C1 × 2C1 × 1 = 96 Last digit = 6,8: Digit akhir Number of 5-digit even number less than 6 000 can be formed Bilangan nombor genap 5 digit kurang daripada 6 000 yang dapat dibentuk = 2(3C1 × 4C1 × 3C1 × 2C1 × 1) = 144 Total / Jumlah = 96 + 144 = 240

10 4-digit numbers: = 3C1 × 4C1 × 3C1 × 2C1

Nombor 4 digit: = 72 5-digit numbers: = 5! Nombor 5 digit: = 120 Total / Jumlah = 72 + 120 = 192

11 Arrangements of vowels: 3! Pilihatur huruf vokal

Total arrangements = 3! × 6! Jumlah pilihatur = 4 320

12

5 6P2 4

Number of consonants: T, R, N, G, L = 5 Bilangan konsonan

Number of arrangements = 5 × 6P2 × 4 Bilangan pilihatur = 5 × 30 × 4 = 600

13 3P2 × 5! = 6 × 120 = 720

14 (a) Number of codes = 7P4

Bilangan kod = 840

(b) Number of codes = 6P3 × 5 Bilangan kod = 600

15 (a) Number of different 4-digit numbers = 6P4

Bilangan nombor 4 digit yang berlainan = 360

(b) Number of odd numbers = 5P3 × 4 Bilangan nombor ganjil = 240

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16 Number of ways Bilangan cara

20C4 = 20!16! 4!

= 4 845

17 Number of ways Bilangan cara 10C6 = 10C4

= 10!4! 6!

= 210

18 Number of ways Bilangan cara 3C2 × 6C3 = 3 × 20 = 60

19 Number of ways Bilangan cara (6C3 × 5C2) + (6C4 × 5C1) + (6C5 × 5C0) = 200 + 75 + 6 = 281

20 Number of ways Bilangan cara 1C1 × 5C3 = 1 × 10 = 10

21 Number of ways Bilangan cara 5C5 = 1

22 Number of ways Bilangan cara 1C1 × 1C1 × 5C3 = 1 × 1 × 10 = 10

23 Number of ways Bilangan cara

9C4 = 9!4! 5!

= 126

24 Number of ways Bilangan cara 4C2 × 6C3 × 8C5 = 6 × 20 × 56 = 6 720

25 (a) Number of ways Bilangan cara = 2! × 5! = 240

(b) Number of ways Bilangan cara = 8C4 × 5C2

= 70 × 10 = 700

PAPER 2 KERTAS 2 1 (a) Number of arrangements Bilangan susunan 6! = 720

(b) 5!2! = 120

20 = 60

(c) 9!2! 3! = 362 880

2(6)

= 30 240

2 (a) Number of arrangements = 5P4 = 120 Bilangan susunan

(b) Number of arrangements = 6P4 = 360 Bilangan susunan (c) Number of arrangements = 4P4 = 24 Bilangan susunan

3 (a) Number of arrangements = 5P3 × 3 = 180 Bilangan susunan (b) Number of arrangements = 5P3 × 3 = 180 Bilangan susunan (c) Number of arrangements = 4 × 5P3 = 240 Bilangan susunan

4 (a) Number of arrangements = 1P1 × 5P5 = 120 Bilangan susunan (b) Number of arrangements = 1P1 × 5P5 = 120 Bilangan susunan (c) Number of arrangements = 4P1 × 5P5 = 480 Bilangan susunan (d) Number of arrangements 5P5 × 2P1 = 120 × 2 = 240 Bilangan susunan

5 (a) Number of arrangements = 5P4 = 120 Bilangan susunan (b) Number of arrangements = 2 × 3 × 3P2 = 2 × 3 × 6 = 36 Bilangan susunan

6 (a) Number of arrangements = 4! × 3 × 2P2 = 144 Bilangan susunan (b) Number of arrangements = 4! × 2P2 = 48 Bilangan susunan (c) Number of arrangements = 6! = 720 Bilangan susunan

7 (a) Number of arrangements = 5P2 × 2! × 6! = 28 800 Bilangan susunan (b) Number of arrangements = 3! × 6! = 4 320 Bilangan susunan

8 (a) Number of arrangements = 9C5 × 6C2 = 126 × 15 = 1 890 Bilangan susunan (b) Number of arrangements = (9C7) + (9C6 × 6C1) + (9C5 × 6C2) + (9C4 × 6C3) Bilangan susunan = 36 + 504 + 1 890 + 2 520 = 4 950

9 (a) 35 (b) 1 (c) 72 (d) 32 400

10 (a) Number of arrangements = 22C10 = 646 646 Bilangan susunan (b) Number of arrangements / Bilangan susunan

= (10C7 × 12C3) + (10C8 × 12C2) + (10C9 × 12C1) + (10C10) = 26 400 + 2 970 + 120 + 1 = 29 491

11 (a) Number of arrangements = 7C4 = 35 Bilangan susunan

(b) Number of arrangements = 10C4 = 210 Bilangan susunan

(c) Number of arrangements = (3C2 × 11C2) + (3C3 × 11C1) Bilangan susunan = 165 + 11 = 176

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87

12 (a) Number of arrangements = (5C5 × 5C2) + (5C4 × 5C3) Bilangan susunan = 10 + 50 = 60

(b) Number of arrangements / Bilangan susunan

= (5C2 × 5C5) + (5C3 × 5C4) + (5C4 × 5C3) + (5C5 × 5C2) = 10 + 50 + 50 + 10 = 120

13 (a) Number of arrangements 13C5 = 1 287 Bilangan susunan

(b) Number of arrangements = 4C2 × 6C2 × 3C1

Bilangan susunan = 6 × 15 × 3 = 270

14 (a) Number of arrangements = 2C2 × 9C2

Bilangan susunan = 1 × 36 = 36

(b) Number of arrangements = 6C4 = 15 Bilangan susunan

15 (a) Number of arrangement = 5! = 120 Bilangan susunan (b) Number of arrangements Bilangan susunan

P L P L P 3C1

2C1 2C1

1C1 1C1

= 3C1 × 2C1 × 2C1 × 1C1 × 1C1

= 3 × 2 × 2 × 1 × 1 = 12 (c) Number of arrangements = 2! × 4 × 3! = 48 Bilangan susunan


PAPER 1 KERTAS 1 1 Numbers on dice = 1, 2, 3, 4, 5, 6 = 6 numbers Nombor pada dadu = 1, 2, 3, 4, 5, 6 = 6 nombor Odd numbers = 1, 3, 5 = 3 numbers Nombor ganjil = 1, 3, 5 = 3 nombor

P(odd number) = 36 = 1

2 P(nombor ganjil)

2 (a) Total balls = 15 Jumlah bola

P(yellow) = 515 = 1

3 P(kuning) (b) Total balls not blue = 9 Jumlah bola bukan biru

P(not blue) = 915 = 3

5 P(bukan biru)

3 (a) Total cards = 8 Jumlah kad

P(S) = 28 = 1

4

4 (a) Total cards = 10 Jumlah kad Total cards with even numbers = 6 Jumlah kad dengan nombor genap

P(even) = 610 = 3

5 P(genap)

5 (a) Total candies = 20 Jumlah gula

P(green) = 1020 = 1

2 P(hijau)

6 A = {odd numbers} {nombor ganjil} = {3, 5, 7, 9, 11, 15} B = {prime numbers} {nombor perdana} = {2, 3, 5, 7, 9, 11} A ∩ B = {3, 5, 7, 9, 11} P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 6

10 + 610 – 5

10 = 7

10

7 P(story and science) = 1030 = 1

3 P(cerita dan sains)

8 Odd numbers Nombor genap = {11, 13, 15, 17, 19, 21, 23, 25, 27, 29} Multiples of 3 Gandaan 3 = A = {15, 21, 27} Multiples of 5 Gandaan 5 = B = {15, 25} P(A ∩ B) = 1

10

9 P(odd number) = 36 = 1

2 P(nombor ganjil)

P(number 4) = 16 P(nombor 4)

P(odd number or number 4) = 12 + 1

6 = 23 P(nombor ganjil atau nombor 4)

10 P(head) = 12 P(kepala)

P(odd number) = 36 = 1

2 P(nombor ganjil)

P(head or odd number) = 12 + 1

2 = 1 P(kepala atau nombor ganjil)

11 P(yellow) = 310 P(kuning)

P(red) = 310 P(merah)

P(yellow or red) = 310 + 3

10 = 35

P(kuning atau merah)

12 P(black) = 38 P(hitam)

P(white) = 28 P(putih)

P(black or white) = 28 + 3

8 = 58 P(hitam atau putih)

13 P(A) = 49

P(T) = 29

P(A or T ) = 49 + 2

9 = 69 = 2

3 P(A atau T)

14 P(red) = 1016 = 5

8 P(merah)

15 Probability = P(swimming) + P(cycling) – P(both) Kebarangkalian = P(berenang) + P(berbasikal) – P(kedua-duanya)

= 3550 + 25

50 – 1550

= 4550

= 910

16 P(A) + P(D) = 1025 + 2

25

= 1225

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17 P(white) = 515 P(putih)

P(black) = 1015 P(hitam)

P(white and black) = P(white, black) + P(black, white) P(putih dan hitam) = P(putih, hitam) + P(hitam, putih)

= ( 515 × 10

15) + (1015 × 5

15) = 2( 1

3 × 23 )

= 49

18 P(Monday ∩ Tuesday) = P(Monday) × P(Tuesday) P(Isnin ∩ Selasa) P(Isnin) × P(Selasa)

= 34 × 2

5 = 6

20 = 3

10

19 P(A ∩ D) = P(A) × P(D)

= 35 × 1

6 = 3

30 = 1

10

20 P(A ∩ B ∩ C) = P(A) × P(B) × P(C)

= 15 × 3

4 × 23

= 660 = 1

10

21 P(all) = 23 × 4

6 × 24 = 16

72 = 29 P(semua)

22 P(all) = 1100 × 1

100 × 1100 = 1

1 000 000 P(semua)

23 Let R represent rotten durian Biar R mewakili durian yang rosak P(3 rotten durian) = P(R) × P(R) × P(R)

P(3 durian yang rosak) = 25 × 2

5 × 25 = 8

125

24 P(late for 3 days) = P(L) × P(L) × P(L) P(lewat 3 hari) = 0.1 × 0.1 × 0.14 = 0.001

25 Let S represent Joseph win in science quiz Biar S mewakili Joseph menang dalam kuiz sains Let M represent Joseph win in mathematic quiz Biar M mewakili Joseph menang dalam kuiz matematik

P(S) × P(M) = 15 × 2

7 = 235

PAPER 2 KERTAS 2 1 (a) Prime number = 2, 3, 7, 29 Nombor perdana

P(prime number) = 47 P(nombor perdana)

(b) Odd number = 3, 7, 29 Nombor ganjil

P(odd) = 37

P(ganjil)

(c) Two digit number = 12, 18, 29 Nombor dua digit

P(Two-digit) = 37

P(Dua digit)

2 (a) Even numbers = {2, 4, 6, 8, 10} Nombor genap

P(even) = 511 P(genap)

(a) Odd numbers = {1, 3, 5, 7, 9} Nombor ganjil

P(odd) = 511 P(ganjil)

3 (a) Total candies = 18 Jumlah gula

P(red candy) = 218 = 1

9 P(gula merah)

(b) P(green candy) = 618 = 1

3 P(gula hijau)

4 A = odd numbers nombor ganjil = {3, 7, 11, 15, 19} B = prime numbers nombor perdana = {3, 7, 11, 19} A ∩ B = {3, 7, 11, 19} (a) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 5

10 + 410 – 4

10 = 5

10 = 12

(b) P(A ∩ B) = 410 = 2

5

5 A = {5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31,33, 35} P3 = {9, 15, 21, 27, 35} P5 = {5, 15, 25, 35} P3 ∩ P5 = {15, 35} (a) P(P3 ∪ P5) = (P3) + (P5) – (P3 ∩ P5)

= 516 + 4

16 – 216 = 7

16

(b) P(P3 ∩ P5) = 2

16 = 18

6 (a) P(white or blue) = P(white) + P(blue) P(putih atau biru) = P(putih) + P(biru)

= 413 + 3

13 = 713

(b) P(white or black) = P(white) + P(black) P(putih atau hitam) = P(putih) + P(hitam)

= 413 + 6

13 = 1013

7 (a) Let B represent blue discs Biar B mewakili cakera padat biru

P(B) = 212 = 1

6 (b) P(not white) = P(red) + P(blue) P(bukan putih) = P(merah) + P(biru)

= 612 + 2

12 = 8

12 = 23

(c) P(white or red) = P(white) + P(red) P(putih atau merah) = P(putih) + P(merah)

= 412 + 6

12 = 10

12 = 56

8 (a) Numbers divisible by 3 = {3, 6, 9, 12, 15, 18, 21, 24, 27, 30} Nombor boleh dibahagi 3

P(divisible by 3) = 1030 = 1

3 P(boleh dibahagi 3)

(b) Numbers not divisible by 8 = {8, 16, 24} Nombor tak boleh dibahagi 8

P(not divisible by 8) = 1 – 330

P(tak boleh dibahagi 8) = 2730 = 9

10

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89

(c) Numbers less than 5 = {1, 2, 3, 4, …… 14} Nombor kurang daripada 5

P(N < 15) = 1430 = 7

15

9 (a) Male + Female = 1 050 Lelaki + Perempuan P(Female) = 1 – 0.460 = 0.54 P(Perempuan) Female students = 1 050 × 0.54 = 567 Pelajar perempuan (b) Let x = Total workers Biar x = Jumlah pekerja Part time = 54 workers Sambilan = 54 pekerja

= (100 – 55)100 × x

54 = 45100 × x

x = 120 workers pekerja

10 (a) P(pass both) P(lulus kedua-duanya)

= 35 × 4

5 = 12

25

(b) P(fail, pass) = (1 – 35 ) × 2

5 P(gagal, lulus) = 2

5 × 25

= 425

11 P(girls) = 49

P(perempuan)

P(boy) = 59

P(lelaki)

(a) P(all girls) = 49 × 4

9 × 49 =

64729

P(semua perempuan)

(b) P(2 girls) = P(B, G, G) + P(G, B, G) + P(G, G, B) P(2 perempuan) P(L, P, P) + P(P, L, P) + P(P, P, L)

= ( 59 × 4

9 × 49 ) + ( 4

9 × 59 × 4

9 ) + ( 49 × 4

9 × 59 )

= 3( 80729 )

= 80243

12 (a) From Bag A, PA (green) = 410 Dari Beg A, PA (hijau)

From Bag B, PB (green) = 716 Dari Beg B, PB (hijau)

P (green) = 410 × 7

16

P (hijau) = 7

40

(b) PA (green) × PB (yellow) = 410 × 9

16

PA (hijau) × PB (kuning) = 9

40

13 (a) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) Draw Seri = P(A ∩ B) = 1 – 0.2 – 0.5 = 0.3

(b) P(win) = PA(win) + PB(win) P(menang) = PA (menang) + PB (menang) = 0.2 + 0.5 = 0.7

14 (a) nP = 10 P(A or atau B) = P(A) + P(B) = 3

10 + 410

= 710

(b) nP = 10 P(B or atau C) = P(B) + P(C) = 4

10 + 310

= 710

15 (a) P(A) × P(B) × P(C)

= 23 × 5

6 × 14

= 1072

= 536

(b) P( 23 × 5

6 × 34 ) + P( 2

3 × 16 × 1

4 ) + P( 13 × 5

6 × 14 )

= 3072 + 2

72 + 572

= 3772


PAPER 1 KERTAS 1 1 p = 0.3, r = 2 q = 1 – 0.3 = 0.7, n = 5 P(X = 2) = 5C2(0.3)2(0.7)3

= 10(0.09)(0.343) = 0.3087

2 p = 0.4, r = 3 q = 1 – 04 = 0.6, n = 10 P(X = 3) = 10C3(0.4)3(0.6)7

= 120(0.064)(0.02799) = 0.2150

3 p = 16 , q = 1 – 1

6 = 56

n = 7, r = 2 P(X = 2) = 7C2 ( 1

6 )2( 56 )5

= 21 ( 136 )( 3 125

7 776 ) = 0.2344

4 p = 12 , q = 1

2 , n = 5, r = 3

P(X = 3) = 5C3( 12 )3( 1

2 )2 = 10( 1

8 )( 14 )

= 5

16 = 0.3125

5 p = 0.45, q = 0.55 n = 10, r = 4 P(X = 4) = 10C4(0.45)4(0.55)6

= 210(0.04101)(0.02768) = 0.2384

6 p = 30% = 0.30 q = 0.70 n = 10, r = 0, 1, 2 P(0, 1, 2) =P(X = 0) + P(X = 1) + P(X = 2) = 10C0(0.3)0(0.7)10 + 10C1(0.3)(0.7)9 + 10C2(0.3)2(0.7)8

= 0.02825 + 0.12106 + 0.23347 = 0.3828

7 p = 90% = 0.90, q = 0.10 n = 8, r = 7, 8 P(7 or 8 are in good condition) P(7 atau 8 dalam keadaan baik) = 8C7(0.9)7(0.1) + 8C8(0.9)8(0.1)0

= 8(0.4783)(0.1) + 0.43047 = 0.38264 + 0.43047 = 0.8131

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8 1 – p(0, 1 are not fresh) = 1 – [ 9C0(0.2)0(0.8)9 + 9C1(0.2)(0.8)8] 1 – p(0, 1 adalah tidak segar) = 1 – (0.13422 + 030199) = 0.5638

9 p = 45 = 0.8

q = 0.2 n = 4 r = 2, 3.4 P(2, 3 or 4 are rich) = 4C2(0.8)2(0.2)2 + 4C3(0.8)3(0.2) + 4C4(0.8)4(0.2)0

P(2, 3, atau 4 adalah kaya) = 0.1536 + 0.4096 + 0.4096 = 0.9728

10 = np = 300( 1

3 ) = 100 2 = npq = 300( 1

3 )( 23 ) = 200

3

= √ –200

3 = 8.165

11 p = 12 , q = 1

2 n = 500 = np = 500( 1

2 ) = 250

2 = npq = 250( 12 ) = 125

= √125 = 11.1803

12 p = 29 , q = 7

9 n = 150 = np = 150( 2

9 ) = 1003

2 = npq = 150( 29 )( 7

9 ) = 70027

= √ –70027 = 5.0918

13 P(Z 0.72) = 0.2358

14 P(Z 1.065) = 0.1434

15 (a) P(Z 1.38) = 1 – P(Z 1.38) = 1 – 0.08379 = 0.9162

(b) P(Z – 0.37) = P(Z 0.37) = 0.3557

16 (a) P(Z –0.6) = 1 – P(Z –0.6) = 1 – P(Z 0.6) = 1 – 0.2743 = 0.72587

(b) P(Z 0) = 0.500

17 (a) P(–2.31 Z 0.65) = 1 – P(Z 2.31) – P(Z 0.65) = 1 – 0.0104 – 0.2579 = 0.7317

(b) P(0.26 Z 2.11) = P(Z 0.26) – P(Z 2.11) = 0.3974 – 0.0174 = 0.38

18 (a) P(–1.3 Z 1.3) = 1 – P(Z 1.3) – P(Z 1.3) = 1 – 2P(Z 1.3) = 1 – 2(0.0968) = 1 – 0.1936 = 0.8064

(b) P(–2.61 Z –0.43) = P(Z 0.43) – P(Z 2.61) = 0.3336 – 0.00453 = 0.3291

19 Z = X – � �

= 12 – 104

= 24

= 0.5

20 � = 120 � = √30 g

P(X > 130) = P(Z > 130 – 120√30 )

= P(Z > 1.826) = 0.03393

21 � = 60 � = √5.25 P(X > 55) = P(Z > 55 – 60

√5.25 ) = P(Z > –2.1822) = 1-P (Z > 2.1822) = 0.9855

22 � = 60, � = 5 P(58 X 63) = P(58 – 60

5 Z 63 – 605 )

= P(–0.4 Z 0.6)

–0.4 0.6

= 1 – P(Z 0.4) – P(Z 0.6) = 1 – 0.3446 – 0.2743 = 0.3811

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23 � = 3 000, � = 50 P(2 900 X 3 080) = P( 2 900 – 3 000

50 Z 3 080 – 3 000

50 ) = P(–2 Z 1.6)

–2 1.6

= 1 – P(Z 2) – P(Z 1.6) = 1 – 0.02275 – 0.0548 = 0.9225

24 P(0 < Z < a) = 0.5 – P(Z > a) P(Z > a) = 0.5 – 0.475 = 0.025

25 (a) P(Z a) = 0.025 from the table, a = 1.96 P(Z a) = 0.025 dari jadual, a = 1.96

(b) P(–a Z a) = 1 – 2p(Z a) = 1 – 2(0.025) = 1 – 0.05 = 0.95

PAPER 2 KERTAS 2 1 p = 3

5 , q = 25 , n = 7 (7 days a week)

(7 hari seminggu)

(a) P(X = 4) = 7C4( 35 )4( 2

5 )3 = 0.2903

(b) P(X 5) = P(X = 5) + P(X = 6) + P(X = 7)

= 7C5( 35 )5( 2

5 )2 + 7C6 ( 35 )6( 2

5 ) + 7C7 ( 35 )7( 2

5 )0 = 0.4199

(c) P(X < 5) = 1 – P(X 5) = 1 – 0.4199 = 0.5801

2 p = 0.8, q = 0.2, n = 5 (a) P(X = 3) = 5C3(0.8)3(0.2)2

= 0.2048

(b) P(X 3) = P(X = 3) + P(X = 4) + P(X = 5) = 5C3(0.8)3(0.2)2 + 5C4(0.8)4(0.2) + 5C5(0.8)5(0.2)0

= 0.9421

(c) P(X 3) = 1 – P(X > 3) = 1 – P(X = 4) – P(X = 5) = 1 – 5C4(0.8)4(0.2)1 + 5C5(0.8)5 (0.2)0

= 0.2627

3 p = 0.75, q = 0.25, n = 10 (a) P(X = 3) = 10C3(0.75)3(0.25)7

= 0.0031

(b) P(X 9) = P(X = 9) + P(X = 10) = 10C9(0.75)9(0.25) + 10C10(0.75)10(0.25)0

= 0.1877 + 0.0563 = 0.2440

4 p = 0.1, q = 0.9, n = 20 (a) P(X = 0) = 20C0(0.1)0(0.9)20

= 0.1216

(b) P(X = 2) = 20C2(0.1)2(0.9)18

= 0.2852

5 p = 0.40, q = 0.60, n = 10 (a) P(X 8) = P(X = 8) + P(X = 9) + P(X = 10) = 10C8(0.4)8(0.6)2 + 10C9(0.4)9(0.6) + 10C10(0.4)10(0.6)0

= 0.0123

6 p = 14 , q = 3

4 , n = 5

(a) P(X = 2) = 5C2( 14 )2( 3

4 )3 = 0.2637

(b) P(X 2) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

= 5C2( 14 )2( 3

4 )3 + 5C3( 1

4 )3( 34 )2 + 5C4( 1

4 )4( 34 )

+ 5C5( 14 )5( 3

4 )0 = 0.3672 or / atau

P(X 2) = 1 – P(X < 2) = 1 – P(X = 0) – P(X = 1)

= 1 – 5C0( 14 )0( 3

4 )5 – 5C1( 1

4 )1( 34 )4

= 0.3672

(c) P(X 4) = 1 – P(X > 4) = 1 – P(X = 5)

= 1 – 5C5( 14 )5( 3

4 )0 = 0.999

7 p = 0.70, q = 0.30, n = 5 (a) P(X � 3) = 1 – P(X > 3) = 1 – P(X = 4) + P(X = 5) = 1 – 5C4(0.7)4(0.3)1 – 5C5(0.7)5(0.3)0

= 0.4718

(b) P(X � 1) = 0.90 1 – P(X < 1) = 0.90 1 – P(X = 0) = 0.90 P(X = 0) = 0.10 nC0(0.70)0(0.30)n = 0.10 (0.30)n = 0.10 n = log 0.10

log 0.30

= 1.9125 n = 2

8 np = 20, √npq = 4 (a) npq = 16

npqnp = 16

20 q = 4

5 p = 1 – q

= 15

np = 20

n( 15 ) = 20

n = 100

(b) n = 10, p = 15 , q = 4

5 P(X 2) = 1 – P(X < 2) = 1 – P(X = 0) – P(X = 1)

=1 – 10C0 ( 15 )0( 4

5 )10 – 10C1 ( 15 )1( 4

5 )9 = 0.6242

9 P(Z > 2.12) = 0.017

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10 P(Z < 1.812) = 1 – P(1.812) = 1 – 0.035 = 0.965

11 P(1.742 < Z < 2.496) = P(1.742) – P(2.496) = 0.0408 – 0.0063

= 0.0345

12 � = 3 kg, �2 = 0.01 kg2

(a) P(X > 2.8) = P(Z > 2.8 –3.0√0.01

) = P(Z > –2) = 1 – P(Z > 2) = 1 –0.0228 = 0.9772

(b) P(2.9 > X > 3.2)

= ( 2.9 – 3.0√0.01

< Z < 3.2 – 3.0√0.01 )

= P(–1 < Z < 2) = 1 – P(Z > 1) – P(Z > 2) = 1 – 0.1587 – 0.0228 = 0.8185 Percentage of watermelon Peratusan tembikai = 0.8185 × 100% = 81.85%

13 � = 9.8 cm, P(X > 10.13) = 2841 000 = 0.284

(a) P(X > 10.13) = 0.284

P(Z > 10.13 – 9.18� ) = 0.284

P(Z > 0.33

� ) = 0.284

0.33� = 0.571

� = 0.5779

(b) P(X < 9.72) = P(Z < 9.72 –9.80.5779 )

= P(Z < –0.138) = 0.44512 Number of pen Bilangan pen = 0.44512 × 1 000 = 445.12 ≈ 445 pens

14 � = 1 500, � = 30

(a) (i) P(X > 1 532) = P(Z > 1 532 –1 50030

) = P(Z > 1.067) = 0.1 430

(ii) P(1 480 < X < 1 530) = P(1 480 – 1 50030

< Z < 1 530 – 1 50030 )

= P(–0.667 < Z < 1) = 1 – P(Z > 0.667) – P(Z > 1) = 1 – 0.2524 – 0.1587 = 0.5890

(b) P(X > n) = 0.05

P(Z > n – 1 50030 ) = 0.05

n – 1 50030 = 1.645

n = 1549.35 ≈ 1 549 Number of days = 1 549 days Bilangan hari = 1 549 hari

15 � = 55, � = 10, n = 500

(a) P(X 35) = P(Z 35 –5510 )

= P(Z –2) = 1 – P(Z 2) = 0.9773

(b) Number of students passing the examinations Bilangan pelajar yang lulus peperiksaan = 0.9773 × 500 = 488.65 ≈ 488 students pelajar

(c) Let a denotes the minimum marks for gred A. Bar a mewakili markah minimum bagi gred A. P(X a) = 0.13

P(Z > a – 5510 ) = 0.13

a – 5510 = 1.127

a = 66.27

The minimum marks for grade A is 66.27% Markah minimum bagi gred A adalah 66.27%.

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93


PAPER 1 KERTAS 1 1 s = t 2 – 2t + 2 t = 3, s = t 2 – 2t + 2 = (3)3 – 2(3) + 2 = 5

t = 4, s = t 2 – 2t + 2 = (4)2 – 2(4) + 2 = 10

2 s = 7t + t 2

t = 2, s = 7t + t 2

= 7(2) + (2)2

= 18

t = 5, s = 7t + t 2

= 7(5) + (5)2

= 60

3 s = 2t – t 2 + 4 s = 8, 8 = 2t – t 2 + 4 2t – t 2 = 4 t(2 – t ) = 4 t = 4, 2 – t = 4 t = 4, t = –2 (Not acceptable / Tidak diterima) t = 4 s

4 s = 2t2 – 3t + 2 s = 10, 10 = 2t2 – 3t + 2 2t 2 – 3t = 8 t(2t – 3 ) = 8 t = 8, 2 t = 11 t = 8, t = 11

2

t = 8 s , t = 112 s

5 t = 3, s = 8(3)2 – 2(3) = 72 – 6 = 66

t = 4, s = 8(4)2 – 2(4) = 128 – 8 = 120

Distance travelled Jarak yang dilalui = |120 – 66| = 54 m

6 t = 2, s = 6(2)2 – 2(2)2

= 12 – 8 = 4

t = 3, s = 6(3) – 2(3)2

= 18 – 18 = 0

Distance travelled Jarak yang dilalui = |0 – 4 | = 4 m

7 s = 3t – t 2, t = 1, s = 3(1) – (1)2

= 2

t = 4, s = 3(4) – (4) 2

= 12 – 16 = –4

Total distance travelled Jumlah jarak yang dilalui = | (2) – 0| + | (–4) – 2| = 2 + 6 = 8 m

s

t

8 s = t 2 – 2t, t = 1, s = (1)2 – 2(1)

= –1

t = 3, s = (3)2 – 2(3) = 9 – 6

= 3

Total distance travelled

s

t

Jumlah jarak yang dilalui = | (–1) – 0| + | (3 –(–1)| = 1 + 4 = 5 m

9 s = 2t – 3t2 + 4 v = ds

dt = 2 – 6t

t = 3, 2 – 6t = 2 – 6(3) = 2 – 18

= –16 m s –1

10 s = t 2 – 6t + 4 v = ds

dt = 2t – 6

t = 4, 2t – 6 = 2(4) – 6 = 8 – 6

= 2 m s–1

11 s = t 2 – 12t + 16 v = ds

dt = 2t – 12

v = 0, 2t = 12

t = 122

= 6 s

12 s = t 2 – 6t + 7 v = ds

dt = 2t – 6 = 0

2t = 6 t = 3 s

13 v = 6t – t 2 s = ∫v dt = ∫(6t – t 2) dt

= 6t 2

2 – t 3

3 + c

= 3t 2 – 13 t 3 + c

t = 0, s = 0 0 = 0 – 0 + c c = 0 s = 3t 2 – 13 t 3

When / Apabila t = 2, s = 3(2)2 – 13 (2)3

= 12 – 83

= 9 13 m

14 v = 6 – 4t s = ∫6 – 4t dt = 6t – 2t 2 + c

t = 0, s = 0, c = 0 s = 6t – 2t 2

When / Apabila t = 3 s = 6(3) – 2(3)2

= 18 – 18 = 0 m

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15 v = 3t 2 + 2t + 1 s = ∫(3t 2 + 2t + 1) dt

= 3t 3

3 + 2t 2

2 + t + c dt

= t 3 + t 2 + t + c t = 0, s = 0, c = 0 s = t 3 + t 2 + t

When / Apabila t = 2, s = t 3 + t 2 + t = 23 + 22 + 2 = 8 + 4 + 2 = 14 m

16 s = 2t + t 2 – t 3 – 4 v = ds

dt = 2 + 2t – 3t 2

v = 2 + 2t – 3t 2

When / Apabila t = 3, v = 2 + 2(3) – 3(3)2

= 2 + 6 – 27 = –19 m s–1

17 v = 6t 2 – 12t When the particle stops, Apabila zarah berhenti,

v = 6t 2 – 12t = 0 = t(6t – 12) = 0 t = 2 s

18 v = 9t – t2

a = dvdt = 9 – 2t

(a) t = 0, a = 9 – 2(0) = 9 m s–2

(b) t = 3, a = 9 – 2(3) = 3 m s–2

19 v = 2t 2 – 3t + 7

a = dvdt = 4t – 3

(a) t = 0, a = 4(0) – 3 = –3 m s–2

(b) t = 2, a = 4(2) – 3 = 8 – 3 = 5 m s–2

20 s = 2t 3 + 4t 2 – 3

v = ddt (2t 3 + 4t 2 – 3)

= 6t 2 + 8t

a = ddt (6t 2 + 8t)

= 12t + 8 (a) t = 0, a = 12(0) + 8

= 8 m s–2

(b) t = 2, a = 12(2) + 8 = 32 m s–2

21 s = 3t 3 + 2t 2 – 5t + 2 v = d

dt (3t 3 + 2t 2 – 5t + 2)

= 9t2 + 4t – 5

a = ddt (9t 2 + 4t – 5)

= 18t + 4 (a) t = 0, a = 18(0) + 4

= 4 m s–2

(b) t = 3, a = 18 (3) + 4 = 58 m s–2

22 a = 8t – 3 v = ∫(8t – 3) dt = 4t 2 – 3t + c t = 0, v = 5, c = 5

v = 4t 2 – 3t + 5 When / Apabila t = 2, v = 4(2)2 – 3(2) + 5 = 16 – 6 + 5 = 15 m s–1

23 a = 6 – 2t v = ∫(6 – 2t) dt = 6t – t 2 + c t = 0, v = 4, c = 4 v = 6t – t 2 + 4

s = ∫6t – t 2 + 4 dt

= 3t 2 – 13 t 3 + 4t + k

t = 0, s = 0, k = 0

s = 3t 2 – t 3

3 + 4t

When / Apabila t = 3, s = 3(3)2 – (3)3

3 + 4(3)

= 3(9) – 273 + 12

= 30 m

24 a = 8 – 4t v = ∫(8 – 4t) dt = 8t – 2t 2 + c t = 0, v = 8, c = 8 v = 8t – 2t 2 + 8 a = 0 8 – 4t = 0 4t = 8 t = 2 When / Apabila t = 2, v = 8t – 2t 2 + 8 = 8(2) – 2(2)2 + 8 = 16 m s –1

25 v = 2t(4 – t) v = 8t – 2t2 a = a = 8 – 4t a = 3, 8 – 4t = 3 4t = 5 t = 5

4 When / Apabila t = 5

4 , v = 2( 54 )(4 – 5

4 ) = 5

2 (16 – 54 )

= 52 (11

4 ) = 55

8 = 6.88 m s–1

PAPER 2 KERTAS 2 1 (a) t = 2s s = 2t 2 – t + 5 = 2(2)2 – (2) + 5 = 8 – 2 + 5 = 11 m

(b) s = 5m s = 2t2 – t + 5 2t2 – t + 5 = 5 2t2 – t = 0 t(2t – 1) = 0 t = 0, t = 1

2

t = 0.5 s

dvdt

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95

2 (a) t = 3s s = 6t 2 – t + 3 = 6(3)2 – (3) + 3 = 54 – 3 + 3 = 54 m

(b) s = 3m 6t 2 – t + 3 = 3 6t 2 – t = 0 t(6t – 1) = 0 t = 0, t = 1

6

t = 16 s

3 (a) t = 3, s = 8(3) – 3(3)2

= 24 – 27 = –3 t = 2, s = 8(2) – 3(2)2

= 16 – 12 = 4 Total distance travelled / Jumlah jarak yang dilalui = |–3 – 4 | = 7 m

(b) t = 7, s = 8(7) – 3(7)2

= 56 – 147 = –91 t = 6, s = 8(6) – 3(6)2

= 48 – 108 = –60 Total distance travelled / Jumlah jarak yang dilalui = |–91 – 60| = 151 m

4 (a) s = t 2 – 3t, t = 1, s = (1)2 – 3(1) = –2 t = 4, s = (4)2 – 3(4) = 16 – 12 = 4 Total distance travelled Jumlah jarak yang dilalui = | (–2) – 0| + |4 – (–2)| = 2 + 6 = 8 m

(b) s = t 2 – 3t t = 1, s = (1)2 – 3(1) = –2 t = 5, s = (5)2 – 3(5) = 25 – 15 = 10 Total distance travelled Jumlah jarak yang dilalui = | (–2) – 0| + |10 – (–2)| = 14 m

5 (a) t = 3, s = 8 – 4t 2

= 8 – 4(3)2

= 8 – 36 = –28 t = 4, s = 8 – 4(4)2

= 8 – 64 = –56 Total distance travelled Jumlah jarak yang dilalui = |–56 – 28| = 84 m

(b) t = 0, s = 8 – 4(0)2 = 8

t = 2, s = 8 – 4(2)2

= 8 – 16 = –8 Total distance travelled Jumlah jarak yang dilalui = |8 – 0| + |–8 – 0| = 16 m

6 (a) vp = 6 + 4t – 2t 2

ap = 4 – 4t

vp maximum when ap = 0 vp maksimum apabila ap = 0 4 – 4t = 0 t = 1

vp maximum = 6 + 4t – 2t 2

vp maksimum = 6 + 4(1) – 2(1)2

= 8 m s–1

(b) P at C when vp = 0 P di C apabila vp 6 + 4t – 2t 2 = 0 3 + 2t – t 2 = 0 (–t – 1)(t – 3) = 0 t – 3 = 0 t = 3

Distance of C from A = 3

∫0(6 + 4t – 2t 2) dt

Jarak C dari A = [6t + 2t 2 – 23 t 3] 3

0 = (6(3) + 2(3)2 – 2

3 (3)3) – 0

= 18 m

7 (a) v = 6t – 3t 2 + 1 When / Apabila t = 4,

v = 6(4) – 3(4)2 + 1 = 24 – 48 + 1 = –23 m s–1

Particle moves to the left of O. Zarah bergerak ke arah kiri O.

(b) s = ∫(6t – 3t2 + 1) dt

= ∫( 6t 3

2 – 3t 3

3 + t) dt

= 3t 2 – t 3 + t + c t = 0, s = 0, c = 0 s = 3t 2 – t 3 + t When / Apabila t = 3, s = 3t 2 – t 3 + t = 3(3)2 – 33 + 3 = 27 –27 + 3 = 3 m

8 (a) v = 6t 2 – 12t s = ∫(6t 2 – 12t ) dt s = 2t 3 – 6t 2 + c s = 0, t = 0, c = 0 hence maka s = 2t 3 – 6t 2 When particle stops 6t 2 – 12t = 0 Apabila zarah berhenti 6t(t – 2) = 0 t = 2 When / Apabila t = 2, s = 2t 3 – 6t 2 s = 2(2)3 – 6(2)2

= –8

(b) v = 6t 2 – 12t, s = 0 2t 3 – 6t 2 = 0 2t (t – 3) = 0 t = 3 When / Apabila t = 3, v = 6t 2 – 12t = 6(32) – 12(3) = 18 m s–1

9 (a) When / Apabila v = –3, 9 – 2t = –3 2t = 12 t = 6 s = ∫ 9 – 2t dt = 9t – t2 + c (b) When / Apabila t = 5, s = 5 5 = 9(5) – 52 + c c = –15

AM F5•J(89-106).indd 95 05/05/2010 15:37:57


s = 9t – t 2 – 15 When / Apabila t = 6, s = 9(6) – 62 – 15 = 3 When / Apabila t = 7, s = 9(7) – 72 – 15 = –1 Distance travelled = |3 – (– 1)| Jarak dilalui = 4 m

10 (a) Change direction, v = 0 Tukar arah 14 – 2t = 0 t = 7

s = ∫ 14 – 2t dt = 14t – t 2 + c t = 0 , s = 0, c = 0 s = 14t – t 2 When / Apabila t = 7, s = 14(7) – 72

= 49 m

(b) When / Apabila s = 0, 14t – t 2 = 0 t = 0, t = 14 v = 14 – 2(14) = –14 m s–1

11 (a) (i) t = 0 v = t 2 – 4t + 3 = 02 – 4(0) + 3 = 3

v = 3 m s–1

(ii) v < 0, t 2 – 4t + 3 < 0 (t – 1)(t – 3) < 0

1 < t < 3

(iii) v = t 2 – 4t + 3

a = dvdt

= 2t – 4 a > 0, 2t – 4 > 0 2t > 4 t > 2

(b)

(c) Total distance Jumlah jarak yang dilalui

= 1

∫0v dt + | 3

∫1v dt |

= 1

∫0(t 2 – 4t + 3) dt + | 3

∫1(t 2 – 4t + 3) dt |

= [ t 3

3 – 2t 2 + 3t] 1

0+ [ t 3

3 – 2t 2 + 3t] 3

1 = ( 1

3 – 2 + 3) + |( 273 – 18 + 9) – ( 1

3 –2 + 3)| = 4

3 + | 27 – 54 + 273 – 4

3 | = 4

3 + 43

= 83 m

12 (a) v = t 2 – 6t + 5 t = 0 , v = 5 v = 5 m s–1

(b) v = t 2 – 6t + 5

a = dvdt

= 2t – 6 a = 0 2t – 6 = 0 t = 3 When / Apabila t = 3, v = (3)2 – 6(3) + 5 = 9 – 18 + 5 = – 4 m s–1

v = – 4 m s–1

(c) v < 0, t 2 – 6t + 5 < 0 (t – 1)(t – 5) < 0 t = 1, t = 5 The range of values t = 1 < t < 5 Julai nilai t

(d) Total distance Jumlah jarak yang dilalui

= 1

∫0v dt + | 5

∫1v dt |

= 1

∫0(t 2 – 6t + 5) dt + | 5

∫1(t 2 – 6t + 5) dt |

= [ t 3

3 – 3t 2 + 5t] 1

0+ |[ t 3

3 – 3t 2 + 5t] 5

1| = ( 1

3 – 3 + 5) + |(1253 – 75 + 25) – ( 1

3 –3 + 5)| = 1 – 9 + 15

3 + |( 125 –225 + 753 ) – ( 1 – 9 + 15

3 )| = 7

3 + |(– 253 ) – ( 7

3 )| = 7

3 + |– 323 |

= 393

= 13 m

13 (a) v = 10 + 3t – t 2

v = 0, 10 + 3t – t 2 = 0 (t + 2)(t – 5) = 0 t = –2, t = 5 t = 5 seconds

a = dvdt

= 3 – 2t When / Apabila t = 5, a = 3 – 2(5) = –7 m s–2

(b) a = 0 3 – 2t = 0 t = 3

2 seconds

Maximun velocity, v = 10 + 3( 32 )

– ( 32 )2

Halaju maksimum,

= 10 + 92 – 9

4

= 12 14 m s–1

(c) s = ∫v dt

= 10t + 3t 3

2 – t3

3 + c

t = 0, s = 0, c = 0

s = 10t + 3t 2

2 – t3

3

When / Apabila t = 5, s = 10(5) + 3(5)2

2 – (5)3

3

= 50 + 752 – 125

2 = 45 56 m

When / Apabila t = 9, s = 10(9) + 3(9)2

2 – (9)3

3

= 90 + 2432 – 243 = –31 1

2 m

–31 m12 45 m5

60

s

Total distance in first 9 seconds Jumlah jarak dalam 9 saat pertama

= 2(45 56 ) +31 1

2 = 123 16 m

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97

14 (a) a = 25 – 3t2 When / Apabila t = 0, a = 25 – 3(0)2

= 25 m s–2

(b) v = ∫(25 – 3t 2) dt = 25t – t 3+ c When / Apabila t = 0, v = 4 maka c = 4 v = 25t – t3+ 4 When / Apabila t = 2, v = 25t – t3+ 4 = 25(2) – 23 + 4 = 46 m s–1

(c) When / Apabila v = 4, 25t – t 3 + 4 = 4 25t – t 3 = 0 t(25 – t 2) = 0 t = 0, or / atau t 2 = 25 t = ±5 t = 5

15 (a) (i) a = 4 – 2t v = ∫a dt = 4t – t 2 + c v = 12, t = 0, c = 12 v = 4t – t 2 + 12 a = 0, 4 – 2t = 0 2t = 4 t = 2 When / Apabila t = 2, v = 4(2) – (2)2 + 12 = 8 – 4 + 12 = 16 m s–1

(ii) v = 0, 4t – t 2 + 12 = 0 (–t + 6)(t + 2) = 0 t = 6, t = –2 (not accepted / Tidak diterima) t = Ps, P = 6 P = 6

(b) Total distance Jumlah jarak =

1

∫0v dt + | 6

∫1v dt |

= 1

∫0(4t – t 2 + 12) dt + | 6

∫1(4t – t 2 + 12) dt |

= [2t 2 – t3

3 + 12t] 1

0+ [2t 2 – t

3

3 + 12t] 6

1 = (2 – 1

3 + 12) + |(2(6)2 – (6)3

3 + 12(6)) – (2 – 13 + 12)|

= 6 – 1 + 363 + |( 216 – 216 + 216

3 ) – 6 – 1 + 363 |

= 413 + |72 – 41

3 | = 41

3 + 1753

= 72 m


PAPER 1 KERTAS 1 1

2

3

4

5

4

6

3x + 2y = 12

6

7

AM F5•J(89-106).indd 97 05/05/2010 15:37:59


8

y = x + 3

2x + y = 6

9

10

3

1

x + y = 3x = 3

312–

y = 2x + 1

11 x � 4 y � x + 2 x + 2y > 4

12 x � 5 y < 2x y � 1

3 x

13 y � 0 y < 2x x + y < 6 x + 2y � 8

14 y > 2 x + y < 6 3x + y � 6

15 x � 0 y � x 2y < x + 4

16 x + y � 95 x = total number of female teachers = jumlah bilangan guru perempuan y = total number of male teachers = jumlah bilangan guru lelaki

17 70x + 85y � 500 14x + 17y � 100 x = total number of shirts = jumlah bilangan baju y = total number of shoes = jumlah bilangan kasut

18 3x + y � 5 x = total number of chocolates = jumlah bilangan coklat y = total number of candies = jumlah bilangan gula-gula

19 x + y � 150 x � 100 x � 2y x = total number of adults = jumlah bilangan orang dewasa y = total number of children = jumlah bilangan kanak-kanak

20 (a) y � 3 (b) x + y � 12

21 x � 5 y � 7 280x + 420y � 5 000 14x + 21y � 250 180x + 240y � 4 000 9x + 12y � 200

22

x + y = 7

23

24

AM F5•J(89-106).indd 98 05/05/2010 15:38:00

99

25

x + y = 59y + 5x = 45

PAPER 2 KERTAS 2 1 (a)

y

(b)

(c)

(d)

2 (a) (i)

x = 0

(ii)

(b) (i) y < 4 y � 2x y > 1

2 x

(ii) x + y > 4

y � 13 x

x � 0 2x + y � 10

3 (a) P = (0, 3)

(b) 2y + 3x = 6 y = 0, x = 2 q = 2

(c) y < 5x + 3

2y + 3x > 6 y + 2x � 4

4 (a) x + y = 8 x = 3, y = 5 a = 5

(b) (i) x + y � 8

(ii) y � 8 y � 2x

5 (a) (i) x � 5

(ii) x + y � 3

(iii) 5y � 3x + 5

(b) 14

6 (a) & (b)

2 4 6 8 10 12

12

10

8

6

4

2

x + y = 2

3x + y = 12

x + 3y = 12

2x + 3y = 6

0 x

y

R

(c) Nilai min = 4 Nilai max = 15

AM F5•J(89-106).indd 99 05/05/2010 15:38:02


7 (a) (i) x � 2 3y + 2x > 16 y > 2x – 16

(b) x = 2, 2x + 6y = 34 2(2) + 6y = 34 6y = 30 y = 5

8 (a) y � 5 4y > 5x – 10 2y > –5x + 10

(b) y � 0 x � 0 y � x + 2 x + y � 6

9 x + y � 7 5x + 10y � 100 → x + 2y � 20

x � 2y

10 (a) (i) 50x + 55y � 5 000 10x + 11y � 1 000

(ii) 15x + 20y � 800 3x + 4y � 160

11 (a) x + y � 80

(b) x � 3y

(c) 3 000x + 2 700y � 30 000 10x + 9y � 100

12 (a) 150x + 60y � 900 5x + 2y � 30

60x + 40y � 480 3x + 2y � 24

(b)

1 2 3 4 5 6 7 8 9 10

12

11

10

9

8

7

6

5

4

3

2

1

5x + 2y = 30

3x + 2y = 24

0 x

y

(5, 4)

(c) (i) The number of lorries used is 5. Therefore, x = 5. From the graph, when x = 5, the maximum value y = 4 Therefore, the maximum number of vans used is 4. Bilangan lori yang digunakan ialah 5. Maka, x = 5. Daripada graf apabila x = 5, nilai maksimum y = 4 Maka maksimum van yang digunakan ialah 4 buah.

(ii) Cost of transportation k = 60x + 40y Let k = 480 thus, 60x + 40y = 480 3x + 2y = 24 From the graph, optimum point is (6, 0) Therefore, cost of transportation for 6 lorries = RM60 × 6 = RM360

Kos pengangkutan k = 60x + 40y andaikan k = 480 Maka, 60x + 40y = 480 3x + 2y = 24 Daripada graf, titik optimum ialah (6, 0) Maka kos pengangkutan bagi 6 buah lori = RM60 × 6 = RM360

13 (a) 60x + 20y � 720 3x + y � 36

30x + 40y � 360 3x + 4y � 36

xy � 1

3 3x � y

(b)

2 4 6 8 10 12

18

16

14

12

10

8

6

4

2

3x + y = 36

3x + 4y = 36

3x = y

0

R

x

y

14 (a) x + y � 100

y � 4x y � x + 5

(b)

10 20 30 40 50 60 70 80 90 100

100

90

80

70

60

50

40

30

20

10

y = x + 5

x + y = 100

y = 4x

R

x

y

15 (a) x + y � 8

x � 2y

y � 12 x

800x + 300y � 4 000 8x + 3y � 40

AM F5•J(89-106).indd 100 05/05/2010 15:38:04

101

(b)

1 2 3 4 5 6 7 8 9 10

8

7

6

5

4

3

2

1

8x + 3y = 40

x + y = 8

0

R

x

y

y = x12

SPM MODEL PAPERMODEL KERTAS SPM

PAPER 1 KERTAS 1

1 (a) 18 (b) f( y) = √ y – 2

2 (a) f(x) = 4x + 4 , fg(x) = f [g(x)] = 4

g(x) + 4

4g(x) + 4 = 7x – 2

g(x) + 4 = 47x – 2

g(x) = 4 – 4(7x – 2)7x – 2

= –28x + 127x – 2

g(x) 12 – 28x7x – 2

(b) Let / Biar f –1(x) = y Thus, Maka, f (y) = x 4

y + 4 = x

4x = y + 4

y = 4x – 4

f –1(x) = 4x – 4, f –1(5) = 4

5 – 4,

= – 165

3 (a) f(x) = 2x – 43x + 6 , f(6) = 2(6) – 4

3(6) + 6

= 824

= 13

(b) gf(6) = g( 1

3 ) = 6

13 k + 4 = 6

k = (6 – 4) × 3 = 6

4 (a) Let � and 3� be the roots of the equation. Biar � dan 3� menjadi punca bagi persamaan. x 2 – (� + 3�)x + �(3�) = x 2 – 8x + (3 + 2k) x 2 – (4�)x + 3�2 = x 2 – 8x + (3 + 2k) Compare both sides: Bandingkan kedua-dua belah: 4� = 8 � = 2 and / dan 3�2 = 3 + 2k 3 + 2k = 3(22) = 12 2k = 9 k = 9

2

5 (a) Let the roots be � and �. Biar punca-punca adalah � dan �. From / Dari y = x 2 – 4x – 3mx + 21 = 0, � + � = 4 + 3m …… and / dan �� = 21 ……

From / Dari y = –2(x – 5)2 + 2n, y = –2(x 2 –10x + 25) + 2n = 0 0 = –2x 2 + 20x – 50 + 2n = x 2 – 10x + ( 50 – 2n

2 ) � + � = 10 …… and / dan �� = 50 – 2n

2 …… = : 4 + 3m = 10 3m = 6 m = 2 = : 21 = 50 – 2n

2 2n = 50 – 42 = 8 n = 4

(b) From y = x 2 – 4x – 3mx + 21, since m = 2, therefore Dari y = x 2 – 4x – 3mx + 21, oleh kerana m = 2, maka y = x 2 – 10x + 21 = (x – 5) 2 – (–5) 2 + 21 = (x – 5) 2 – 4 Minimum point Titik minimum = (5, –4) From y = – 2(x – 5) 2 + 2n, since n = 4, therefore Dari y = – 2(x – 5) 2 + 2n, oleh kerana n = 4, maka y = – 2(x – 5)2 + 8 Maximum point Titik maksimum = (5, 8)

6 (x + 2)(2x – 1) < 5(2x – 1) 2x 2 + 4x – x – 2 < 10x –5 2x 2 + 3x – 2 < 10x – 5 2x 2 – 7x + 3 < 0 (2x – 1)(x – 3) < 0

1

2 < x < 3

7 5log 4x = 125 = 53

log 4x = 3 4x = 103 = 1 000 x = 250

8 log8 4p – log2 2q = 2 log2 4p

log2 8 = log2 2q + log2 22

log2 4plog2 23 = log2 2q + log2 4

13 (log2 4p) =log2(2q × 4)

log2(4p)13 = log2(8q)

(4p)13 = 8q

4p = (8q)3

= 512q3

p = 128q3

9 (a) Perimeter of circle / Perimeter bulatan = 2 r First 3 terms / 3 sebutan pertama = T1, T2, T3

= 2 (1), 2 (3), 2 (5) = 2 , 6 , 10

(b) Common difference / Beza sepunya = T2 – T1

= 6 – 2 = 4

10 (a) a = 3r, S = 3r1 – r = 6

3r = 6 – 6r 9r = 6 r = 2

3

�

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(b) T5 = ar4

= ( 23 × 3)( 2

3 )4

= 2( 16

81 ) = 32

81

11 T2 – T1 = T3 – T2

( h2 – 5) – (k + 5) = (7h –2k 2) – ( h

2 – 5) h

2 – 5 – k – 5 = 7h – 2k 2 – h2 + 5

2k 2 – k – 15 = 7h – h 6h = (2k + 5)(k – 3) h = 1

6 (2k + 5)(k – 3)

12 y 2 = 2x(m – nx)

y 2

x = 2m – 2nx

Y = –2nX + 2m ……

Substitute (0, 20) into : Gantikan (0, 20) kedalam : 20 = 2m – 2n(0) m = 10

Substitute (8, 4) into : Gantikan (8, 4) kedalam : 4 = 2m – 2n(8) = 20 – 16n (m = 10) 16n = 16 n = 1

13 (a) x-intercept = a = 6 and y-intercept = b = 8 Pintasan-x = a = 6 dan pintasan-y = b = 8

Equation Persamaan: x6 + y

8 = 1

(b) T = (0, 8), S = (6, 0), TM : MS = 3 : 2

M = (2(0) + 3(6)2 + 3 ,

2(8) + 3(0)2 + 3 )

= (185 , 16

5 )

14 Gradient of AB, MAB = – y-interceptx-intercept

Kecerunan AB, MAB = – pintasan-ypintasan-x

MAB = – ( 6–3 ) = 2

MBC = – 1MAB

= – 12

Equation of BC, y = – 12 x + 6

Persamaan BC, y = – 12 x + 6

At point C, y = 0 and x = k Pada titik C, y = 0 dan x = k

0 = – 12 (k) + 6

12 k = 6

k = 12

15 c~ = 10 i~ + 8 j~ = ka~ + tb~ = k(5 i~ – 2 i~) + t(–2 i~ + 4 j~) = (5k – 2t) i~ + (–2k + 4t) j~ Compare both sides, / Bandingkan kedua-dua belah, 5k – 2t = 10 …… and / dan –2k + 4t = 8 –k + 2t = 4 …… + , 4k = 14 k = 7 From , 5 – 2t = 10 Dari , 2t =

t = – 152

16 PQ = PO + OQ = –2a~ + 3b~

PM = 35 PQ

= 35 (–2a~ + 3b~)

= – 65

a~ + 95

b~ OM = OP + PM = 2a~ + (– 6

5a~ + 9

5b~ )

= 45

a~ + 95

b~

17 2 sin( + 60°) = cos( + 60°) 2(sin cos 60° + cos sin 60°) = cos cos 60° – sin sin 60°

2[sin ( 12 ) + cos (√ 3

2 )] = cos ( 12 ) – sin (√ 3

2 ) sin + √ 3 cos = 1

2 cos – √ 32 sin

2sin + 2√ 3 cos = cos – √ 3 sin 2sin + √ 3 sin = cos – 2√ 3 cos (2 + √ 3) sin = (1 – 2√ 3) cos

sin cos

= 1 – 2√ 32 + √ 3

tan = –2.46413.732

= –0.6603 = 146.57°, 326.57°

or / atau 2 sin( + 60°) = cos( + 60°)

sin ( + 60°)cos ( + 60°) = 1

2

tan( + 60°) = 0.5 + 60° = 26.57°, 206.57° = 146.57°, 326.57°

18 (a) s = r 6 = 8 = 0.75 rad

(b) OA : OB = 8 : OB = 2 : 3 2 OB = 24 OB = 12 AB = CD = 12 – 8 = 4 cm Length of arc BD = r Panjang lengkok BD = 12(0.75) = 9 cm

Perimeter of ABCD = AB + BD + CD + AC Perimeter ABCD = 4 + 9 + 4 + 6 = 23 cm

19 Let / Biar U = x 3 + 2 and / dan V = (1 – 2x)6

dUdx = 3x 2 dV

dx = 6(–2)(1 – 2x)5

= –12(1 – 2x)5

g'(x) = UdVdx + VdU

dx

= (x 3 + 2)(–12)(1 – 2x)5 + (1 – 2x)6(3x 2) = –12(x 3 + 2)(1 – 2x)5 + 3x 2(1 – 2x)6

g'(1) = –12(3)(–1)5 + 3(–1)6

= 36 + 3 = 39

20 Gradient of normal = m1, gradient of tangent = m2 Kecerunan garis normal = m1, kecerunan garis tangen = m2

m1 × m2 = –1 m2 = – 1

m1 = 4

y = 2x 2 + 8x – 3, dydx = 4x + 8 = 4

4x = –4 x = –1

Thus, Maka, y = 2(–1)2 + 8(–1) – 3 = –9 S = (–1, –9) Equation of tangent: Persamaan tangen: y + 9 = 4(x + 1) y = 4x – 5

( 72 ) 15

2

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103

21 y = x(x + 2)(x – 3) = x(x 2 – x – 6) = x 3 – x 2 – 6x Area of shaded region / Luas rantau berlorek

= 0

∫–2 y dx + | 3

∫0 y dx |

= 0

∫–2(x 3 – x 2 – 6x) dx + | 3

∫0(x 3 – x 2 – 6x) dx |

= [ x 4

4 – x 3

3 – 3x2] 0

–2 + |[ x 4

4 – x 3

3 – 3x 2] 3

0 | = 0 – ((–2)4

4 – (–2)3

3 – 3(–2)2) + |( 34

4 – 33

3 – 3(3)2) – 0| = (– 16

4 – 83 + 12) + (|81

4 – 9 – 27|) = 16

3 + 634 = 21 1

12 unit2

22 (a) Number of codes = 9P4

Bilangan kod = 3 024

(b) Number of codes = 4P3 × 4! Bilangan kod = 576

23 (a) Let A = families with durian trees. B = families with banana trees. C = families with no tree. Biar A = keluarga yang mempunyai pokok durian B = keluarga yang mempunyai pokok pisang. C = keluarga yang tidak mempunyai pokok.

P(A ∩ B) = k

52 = 926

k = 926 × 52 = 18

(b) P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

= 3252 + 29

52 – 926 = 43

52

24 (a) x = ∑xn

11 = ∑x8

∑x = 88

(b) New mean, Min baru x ,= 88 + 2y9 =12

88 + 2y = 108 2y = 20 y = 10

25 (a) P(0.365 < Z < k) = P(Z > 0.365) – P(Z > k) 0.156 = 0.3576 – P(Z > k) P(Z > k) = 0.2016 k = 0.836

(b) Z = X –

12 (0.836) = X – 85

2.8

X – 85 = 1.1704 X = 86.1704

PAPER 2 KERTAS 2SECTION A BAHAGIAN A1 2x + y = 3, y = 3 – 2x …… y 2 – 3x + 3 = 0 …… : (3 – 2x)2 – 3x + 3 = 0 9 – 12x + 4x 2 – 3x + 3 = 0 4x 2 – 15x + 12 = 0 x = 15 ± √ (–15)2 – 4(4)(12)

2(4)

= 15 ± √ 338

= 2.593 or / atau 1.157

When / Apabila x = 2.593, y = 3 – 2(2.593) = –2.186 When / Apabila x = 1.157, y = 3 – 2(1.157) = 0.686 x = 2.593, y = –2.186 or / atau x = 1.157, y = 0.686

2 (a) dydx = 3

x 2 – 3x

3t 2 – 3t = 0

t 3 = 1 t = 1

(b) dydx = 3

x 2 – 3x

d 2ydx2 = – 6

x 3 – 3

When / Apabila x = 1, d 2ydx2 = –6 – 3

= –9 < 0. So, the turning point (1, 7) is a maximum point. Maka, titik pusingan (1, 7) adalah titik maksimum.

(c) dydx = 3

x 2 – 3x

y = ∫ 3x 2 – 3x dx

= – 3x

– 32 x 2 + c

At point (1, 7), Pada titik (1, 7)

7 = –3 – 32 + c

c = 232

Equation of the curve: Persamaan lengkung:

y = – 3x

– 3x 2

2 + 23

2

3 (a) a = 2 minutes 15 seconds = 135 seconds, d = 10 seconds a = 2 minit 15 saat = 135 saat, d = 10 saat Tn = a + (n – 1)d, T11 = 135 + 10(10) = 235 seconds saat = 3 minutes 55 seconds 3 minit 55 saat

(b) Sn = n2 [20a + (n – 1) d]

S20 = 202 [2(135) + 19(10)]

= 4 600 seconds saat = 76 minutes 40 seconds 76 minit 40 saat

4 (a) Cumulative frequencyKekerapan longgokan

75

37

2.7

25

Mass (kg) / Jisim (kg)

100

90

80

70

60

50

40

30

20

10

01.05 1.55 2.05 2.55 3.05 3.55 4.05 4.55 5.05

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Mass (kg)Jisim (kg)

FrequencyKekerapan

Upper boundarySempadan atas

Comulative frequencyKekerapan longgokan

< 1.1 0 1.05 0

1.1 – 1.5 3 1.55 3

1.6 – 2.0 13 2.05 16

2.1 –2.5 15 2.55 31

2.6 – 3.0 20 3.05 51

3.1 – 3.5 27 3.55 78

3.6 – 4.0 16 4.05 94

4.1 – 4.5 4 4.55 98

4.6 – 5.0 2 5.05 100

(b) (i) From the ogive, Q1 = T25 = 2.35 Daripada ogif, Q2 = T75 = 3.50 Interquartie range = Q2 – Q1

Julat antara kuartil = 3.50 – 2.35 = 1.15 kg

(ii) Percentage of babies heavier than 2.7 kg Peratusan bayi yang lebih berat daripada 2.7 kg

= 100 – 37100 × 100% = 63%

5 (a) tan (cos 2 + 1) = tan (1 – 2 sin2 + 1)

= sin cos

(2 – 2 sin2 )

= 2 sin cos

(1 – sin2 )

= 2 sin cos

(cos2 )

= 2 sin cos = sin 2

(b) (i)

(ii) 1 – 3

2π – tan (cos 2 + 1) = 0

1– 3 2π

– sin 2 – 1 + 1 = 0

–sin 2 – 1 = 3 2π

– 2

Equation of straight line = y = 3 2π

– 2 Persamaan garis lurus Number of solutions = 3 Bilangan penyelesaian = 3

6 (a) P = (4, –1), OP = ( 4–1)

Q = (–5, 8), OQ = (–58 )

PQ = PO + OQ

= – ( 4–1) + (–5

8 ) = ( –4 – 5

1 + 8 ) = (–9

9 ) Let / Biar PQ = r~

Unit vector , r~ = r~

| r~| Vektor unit

= 1

√ (–9)2 + 92 (–9

9 )

= 19√2

(–99 ) = 1

√2 (–1

1 )

(b) (i) PQ = mQR –9 i~ + 9 j~ = m(OR – OQ ) = m[(k i~ + 2 j~ ) – (–5 i~ + 8 j~)] = m[(k + 5) i~ – 6 j~ ] = m(k + 5) i~ – 6m j~ –9 = m(k + 5) …… and / dan 9 = –6m m = – 3

2 ……

: –9 = – 32 (k + 5)

= – 32 k – 15

2

32 k = 3

2 k = 1

(ii) PQ = –9 i~ + 9 j~ |PQ | = √ (–9)2 + 92 = √162 PR = PO + OR = OR – OP = k i~ + 2 j~ – (4 i~ – j~) = k i~ + 2 j~ – 4 i~ + j~ = (k – 4) i~ + 3 j~ |PR | = √ (k – 4)2 + 32 = √ k2 – 8k + 16 + 9 = √ k2 – 8k + 25 |PQ | = |PR | : √162 = √ k2 – 8k + 25

k 2 – 8k + 25 – 162 = 0 k 2 – 8k – 137 = 0

k = 8 ± √ (–8)2 – 4(–137) 2

= 8 ± √612

2 = 8 ± 6√17 2

= 4 + 3√17 or / atau 4 – 3√17

SECTION B BAHAGIAN B 7 (a)

x – 1 1 2 3 4 5 6

log y 0.40 0.59 0.76 0.93 1.11 1.30

(b) y = ab x – 1

log y = log(ab x – 1) = (x – 1)log b + log a y-intercept, Persilangan-y, log a = 0.22 a = 1.66

Gradient, Kecerunan, log b = 1.3 – 0.46 – 1 = 0.18

b = 1.514

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105

8 (a) y = 3(x – 1)–2

dydx = –6(x – 1)–3

Gradient at point A: / Kecerunan pada titik A:

dydx = –6(2 – 1)–3 = –6

Equation of tangent: / Persamaan tangen: y – 3 = –6(x – 2) = –6x + 12 y = –6x + 15

At point B(x, 0), Pada titik B(x, 0), 0 = –6x + 15 x = 2.5 B = (2.5, 0)

(b) Area of shaded region: / Luas rantau berlorek:

= 2.5

∫23(x – 1)–2 dx – 1

2 (2.5 – 2)(3)

= [–3(x – 1)–1 ] 2.5

2– 1

2 (1.5)

= [– 3(x – 1) ] 2.5

2 – 0.75

= (– 31.5 ) + 3

1 – 0.75

= 0.25 unit2

(c) Volume generated: Isi padu janaan

= 2.5

∫2πy2 dx – Volume of cone with radius 3

Isi padu kon berjejari 3

= π 2.5

∫29 (x – 1)–4 dx – 1

3 (π)(32)(0.5)

= π[ –3(x – 1)3 ] 2.5

2– 3

2 π

= π[ – 3(1.5)3 + 3

1 ] – 32 π

= π(– 89 + 3 – 3

2 ) = 11

18 π unit3

9 (a) (i) QR : 2y + x – 16 = 0 2y = –x + 16 y = – 1

2 x + 8

MQR = – 12

MPQ = – 1MQR

= 2 PQ : y + 3 = 2(x + 3) = 2x + 6 y = 2x + 3

(ii) Equation of QR: 2y = –x + 16 …… Persamaan QR Equation of PQ: Persamaan PQ y = 2x + 3 …… : 2(2x + 3) = –x + 16 4x + 6 = –x + 16 5x = 16 – 6 = 10 x = 2 y = 2(2) + 3 = 4 + 3 = 7 Q = (2, 7)

(b) PQ : QS = 3 : 2

Q = ( nx1 + mx2

n + m , ny1 + my2

n + m ) (2, 7) = ( 2(–3) + 3x

2 + 3 , 2(–3) + 3y2 + 3 )

= ( –6 + 3x5 , –6 + 3y

5 ) –6 + 3x

5 = 2 and / dan –6 + 3y5 = 7

–6 + 3x = 10 –6 + 3y = 35 3x = 16 3y = 41 x = 16

3 y = 413

Therefore, Maka, S = ( 163 , 41

3 )

(c) Area of PQR = 12 | –3 2 6 –3

–3 7 5 –3 | Luas PQR = 1

2 | (–21 + 10 + 18) – (6 + 42 – 15) | = 1

2 |7 – 33| = 13 unit 2

10 (a) sin TOY = 69

= 0.667 TOY = 0.7298 rad XOY = 2 × 0.7298 = 1.4596 rad

(b) Length of arc XRY Panjang lengkok XRY = r = 9(1.4596) = 13.14

Length of arc XSY Panjang lengkok XSY

= 12 (2)(3.142)(6)

= 18.852

Perimeter of shaded region Perimeter rantau berlorek = 13.1364 + 18.852 = 31.99 cm

(c) Area of shaded region = Area of semicircle XTYS – Area of segment XTVR Luas kawasan berlorek = Luas semi bulatan XTYS – Luas tembereng XTYR

= πr12

2 – 12 r2

2( – sin )

= π(6)2

2 – 12 (9)2(1.4596 – sin 1.4596)

= 56.556 – 18.864 = 37.692 cm2

11 (a) � = 350 g, � = 75 g P(235 < x < 315) = P(235 – 350

75 < Z < 375 – 35075 )

= P(–1.533 < Z < 0.333) = 1 – P(Z > 1.533) – P(Z > 0.333) = 1 – 0.06264 – 0.36945 = 0.5679

(b) (i) P(235 < X < 375) = 0.5679 Number of papayas from sample Bilangan betik dari sampel = 600 × 0.5679 = 341 papayas betik

(ii) P(X > n) = 198600 = 0.33

P(Z > n – 35075 ) = 0.33

n – 350

75 = 0.44 n = 0.44(75) + 350 = 383

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SECTION C BAHAGIAN C 12 (a) Given / Diberi v = at – bt 2

The particle is at rest when t = 3 Zarah berhenti seketika pada t = 3 0 = 3a – 9b a = 3b …… Displacement, / Sesaran, s = ∫v dt = ∫at – bt 2 dt = at 2

2 – bt 3

3 + c

When / Apabila t = 0, s = 0 c = 0

Therefore, / Maka, s = at 2

2 – bt 3

3

When / Apabila t = 3, s = 18 18 = 9a

2 – 9b

36 = 9a – 18b …… : 36 = 9(3b) – 18b = 27b – 18b 9b = 36 b = 4 a = 3(4) = 12

(b) When a = 12 and b = 4, / Apabila a = 12 dan b = 4,

s = 6t 2 – 43 t 3

When particle returns to O, / Apabila zarah kembali kepada O, s = 0 0 = 6t 2 – 4

3 t 3

t 2(6 – 43 t) = 0

t = 0 or / atau 43 t = 6

t = 92

t = 0 not accepted, t = 92 s

t = 0 tidak diterima, t = 92 s

(ii)

When t = 3, s = 18 Apabila t = 3, s = 18 When t = 6, s = 6(62) – 4

3 (63) Apabila t = 6, = –72 Total distance Jumlah jarak = 18 + 18 + 72 = 108 m

13 (a) (i) I2009 = P2009

P2005 × 100

125 = 2.45x × 100

x = 1.96

(ii) I2009 = P2009

P2005 × 100

y = 5.204.00 × 100 = 130

(iii) I2009 = P2009

P2005 × 100

115 = z6 × 100

z = 6.9

(b)

FruitsBuah-

buahan

Price index for 2009 based on

2005 (I)Indeks harga bagi 2009 berdasarkan

2005 (I)

Weekly expenses

(RM)Perbelanjaan

mingguan (RM)

Weightage (W)

Pemberat (W)

IW

BananaPisang 125 150 15 1 875

OrangeOren 130 120 12 1 560

MangoMangga 125 130 13 1 625

WatermelonTembikai 140 90 9 1 260

MangosteenManggis 115 60 6 690

W = 55 IW = 7 010

I = IW

W = 7 01055 = 127. 45

(c) I = P2009

P2005 × 100

127.45 = P2009

2 050 × 100

P2009 = RM2 612.73

14 (a) I : x + y 450 II : x 3y y � x

3

III : 18x + 16y � 3 600 9x + 8y � 1 800

(b)

(225, 225)

450

400

350

300

250

200

150

100

50

9x + 8y = 720

9x + 8y = 1 800

y + x = 450

x

y = x13

50 100 150 200 250 300 350 400 450

110

R

y

(c) (i) When x = 110, yminimum = 100 Apabila x = 110, yminimum = 100 Minimum number of shirt B is 100 pieces Bilangan minimum baju B adalah 100 helai.

(ii) Maximum profit Keuntungan maksimum = 18x + 16y Draw the straight line Lukiskan garis lurus 18x + 16y = 1 440 9x + 8y = 720 Maximum point Titik maksimum = (225, 225) Maximum profit Keuntungan maksimum = 18(225) + 16(225) = RM7 650

15 (a) Area of triangle PQS Luas segitiga PQS

12 × 7 × 4 sin ∠SPQ = 12

sin ∠SPQ = 0.857 = 58.99°

(b) Cosine rule: Petua kosinus: SQ2 = 72 + 42 – 2 × 7 × 4 cos 58.99° = 49 + 16 –28.85 = 36.15 SQ = √36.15 = 6.012 cm

(c) Sine rule: / Petua sinus:

6.012sin ∠SRQ

= 8sin 75°

= 8.282 sin ∠SRQ = 0.7259 ∠SRQ = 46.54°

(d) ∠QRS = 180° – 75° – 46.54° = 58.46° Area of SQR / Luas SQR

= 12 (6.012)(8) sin 58.46 = 20.45 cm2

Area of quadrilateral PQRS / Luas sisi empat PQRS = Area of SQR + Area of PQS = Luas SQR + Luas PQS = 20.45 + 12 = 32.45 cm2

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IT Add Maths F5 Answer

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