Jawapan Pelangi Analysis Tingkatan 4 & 5 Matematik Tambahan (Dwibahasa)

Apabila/When t = 10, (c) 2t – 10 = 8
–312(6160)323
s = – 4(10)2 – 20(10) 2t = 18
= m
t = 9

Sesaran/Displacement, s
1
Apabila/When t = 12, = 3 (9)3 – 5(9)2 + 28(9)

s = 1 (12)3 – 4(12)2 – 20(12) = 90 m
3

= –240 m 4. (a) a = 8 – 2t

t = 10 t = 12 t=0 Halaju/Velocity, v = ∫ a dt

= ∫ (8 – 2t) dt

–266 2 m –240 m 0m = 8t – t2 + c
3
Diberi/Given t = 2, v = 16,
Jumlah jarak yang dilalui
1 6 = 8(2) – (2)2 + c
Total distance travelled 16 = 16 – 4 + c
c = 4
  = 2 2
266 3 + 266 3 – 240 Halaju awal/Initial velocity
= 8(0) – (0)2 + 4
= 293 1 m = 4 m s–1
3

3. (a) s = 1 t3 – 5t2 + nt (b) Apabila halaju adalah maksimum,
3
ds
Halaju/Velocity, v = dt = t2 – 10t + n When the velocity is maximum,

Pecutan/Acceleration, a = dv = 2t – 10 dv = a = 0.
dt dt

Pada halaju minimum, 8 – 2t = 0
t = 4
At minimum velocity,

dv = 0 Halaju maksimum/Maximum velocity, v
dt = 8(4) – (4)2 + 4

2t – 10 = 0 = 20 m s–1

t = 5 (c) Jarak dilalui dalam saat ke-3,

Apabila/When t = 5, Distance travelled in the 3rd second,
halaju/velocity = 3 m s–1
s = ∫23(8t – t2 + 4) dt
(5)2 – 10(5) + n = 3
  = 1 3
25 – 50 + n = 3 4t2 – 3 t3 + 4t 2

n = 28   = 1
3
(b) t2 – 10t + 28 = 12 4(3)2 – (3)3 + 4(3)
t2 – 10t + 16 = 0
(t – 2)(t – 8) = 0   1
t = 2, t = 8 – 4(2)2 – 3 (2)3 + 4(2)

= 17 2 m
3

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5. (a) Halaju/Velocity, v = ∫ a dt (c) Apabila zarah berhenti seketika, v = 0

= ∫ (4t + 1) dt dan t = 2.

= 2 t2 + t + c When particle stops instantaneously, v = 0

Diberi/Given t = 0, v = –10. and t = 2.

Apabila/When t = 0, s = 0

2(0)2 + 0 + c = −10 Apabila/When t = 2,

c = −10 s = 2 (2)3 + 1 (2)2 – 10(2)
3 2
v = 2t2 + t – 10 2
3
Sesaran/Displacement, = –12 m

s = ∫ v dt Apabila/When t = 5,

= ∫ (2t2 + t – 10) dt s = 2 (5)3 + 1 (5)2 – 10(5)
2  3 2
3 1 5
= t3 + 2 t2 – 10t + c = 45 6 m

Apabila/When t = 0, s = 0, t=2 t=0 t=5

maka/therefore c = 0.

Fungsi sesaran bagi zarah itu ialah –12 2 m 0m 45 5 m
3 6
Displacement function of the particle is

s = 2 t3 + 1 t2 – 10t Jumlah jarak yang dilalui dalam 5 saat
3 2
pertama/Total distance travelled in the first

(b) Apabila zarah bergerak ke arah kiri, 5 seconds

When the particle moves towards the left, = 12 2 + 12 2 + 45 5
3 3 6
v  0
1
2t2 + t – 10  0 = 71 6 m
(2t + 5)(t – 2)  0
6. (a) Sesaran/Displacement, s
Apabila/When 2t2 + t – 10 = 0,
= ∫ v dt
(2t + 5)(t – 2) = 0
= ∫ (–pt2 + 2t + 15) dt
t = − 52 , t = 2
= –  3p t3 + t2 + 15t + c

5 2 t Apabila/When t = 0, s = 0,
2 maka/therefore c = 0.
–

Diberi/Given t = 3, s = 45
− 52  t  2
–   3p(3)3 + (3)2 + 15(3) = 45
–9p + 9 + 45 = 45

t  0, maka julat masa apabila 9p = 9
zarah itu bergerak ke arah kiri ialah
0  t  2. p = 1

t  0, therefore the time interval when (b) v = −t2 + 2t + 15 dv
the particle moves towards the left is dt
0  t  2. Pecutan/Acceleration, a = = –2t + 2

Apabila/When t = 7, a = –2(7) + 2

= –12 m s–2

© Penerbitan Pelangi Sdn. Bhd. 102

(c) Apabila zarah itu berhenti seketika, Jumlah jarak yang dilalui dalam 9 saat
When the particle stops instantaneously,
pertama/Total distance travelled in the
v = 0.
first 9 seconds

–t2 + 2t + 15 = 0 = 58 1 + 58 1 + 27
t2 – 2t – 15 = 0 3 3
(t + 3)(t – 5) = 0 2
t = –3, t = 5 = 143 3 m

t > 0, maka/therefore t = 5 7. (a) v = kt – t2 dv
dt
Apabila/When t = 5 Pecutan/Acceleration, a = = k – 2t

s = – 31 (5)3 + (5)2 + 15(5) Halaju adalah maksimum apabila t = 3.

1 Velocity is maximum when t = 3.
3
= 58 m dv = 0
dt

Jarak dari O apabila zarah itu berhenti k – 2(3) = 0
1
seketika ialah 58 3 m. k = 6

The distance from O when the particle (b) v = 6t – t2
1 Apabila zarah bergerak ke arah kanan,
stops instantaneously is 58 3 m.
When particle moves towards the right,
(d) Apabila/When t = 0, s = 0 m v  0

Apabila/When t = 5, s = 58 1 m  6 t – t2  0
3 t(6 – t)  0

Apabila/When t = 9,

s = – 13 (9)3 + (9)2 + 15(9) t
= –27 m 06

t=9 t=0 t=5
0t6

–27 m 0 m 58 1 m
3

(c) Jarak yang dilalui dalam 9 saat pertama

Distance travelled in the first 9 seconds

= ∫06(6t – t2) dt + ∫69(6t – t2) dt 

    = 1 t3 6   3t2 1 t3 9
3t2 – 3 0 + – 3
 

6

        =1 1 1 1
3(6)2 – 3 (6)3 – 3(0)2 – 3 (0)3 +   3(9)2 – 3 (9)3 – 3(6)2 – 3 (6)3

 

= 36 + |–36|

= 72 m

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8. (a) Apabila/When v = 0, (b) Pada halaju maksimum/At maximum
–t2 + 3t + 18 = 0 dv
(–t – 3)(t – 6) = 0 velocity, dt = 0
t = –3,    t = 6
–2t + 3 = 0

t  0, maka/therefore t = 6 2t = 3

dv t = 1.5
dt
a = = –2t + 3 Halaju maksimum/Maximum velocity,

Apabila/When t = 6, a = –2(6) + 3 v = –(1.5)2 + 3(1.5) + 18

= –9 m s–2 = 20.25 m s–1

(c) Jumlah jarak yang dilalui dalam 8 saat pertama

Total distance travelled in the first 8 seconds

= ∫06(–t2 + 3t + 18) dt + ∫68(–t2 + 3t + 18) dt 

    = 3 6 – 31 t3 8
– 31 t3 + 2 t2 + 18t 0 +   + 3 t2 + 18t
2  

6

       = 3 3 3
– 31 (6)3 + 2 (6)2 + 18(6) – [0] +   – 13 (8)3 + 2 (8)2 + 18(8) – – 13 (6)3 + 2 (6)2 + 18(6)  

= 90 +  69 1 − 90 
3

= 90 +  −20 2  
3

= 110 2 m
3

Soalan KBAT

1. (a) s = 160t – 16t2 (b) Apabila/When t = 6,

Halaju batu/Velocity of stone, s = 160(6) – 16(6)2
= 384 m
v = d (160t – 16t2)
dt t = 5, maksimum / maximum s = 400

= 160 – 32t t = 6, s = 384

Apabila batu mencapai ketinggian

maksimum, v = 0.

When the stone reaches the maximum t = 0, s = 0

height, v = 0. Jarak yang dilalui oleh batu dalam
tempoh 6 saat pertama
160 – 32t = 0 Distance travelled by the stone in the first
6 seconds
32t = 160
= 400 + (400 – 384)
t = 5 = 416 m

Ketinggian maksimum,

Maximum height,
s = 160(5) – 16(5)2
= 400 m

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(c) Apabila batu berada 144 m dari (d) Apabila batu kembali semula ke
permukaan tanah, permukaan tanah,
When the stone is 144 m from the ground, When the stone falls back to the ground,

s = 144 m s = 0

160t – 16t2 = 144 160t – 16t2 = 0
16t2 – 160t + 144 = 0
16(t2 – 10t + 9) = 0 16t(10 – t) = 0
(t – 1)(t – 9) = 0
t = 1 atau/or t = 9 t = 0 atau/or t = 10

Batu berada 144 m dari permukaan Batu itu mula bergerak apabila t = 0.
tanah pada masa 1 saat dan sekali lagi The stone starts to move when t = 0.
pada masa 9 saat.
Stone is 144 m from the ground at time Maka, batu itu kembali semula ke
1 second and again at time 9 seconds. permukaan tanah pada masa 10 saat
selepas letupan.
Apabila/When t = 1, v = 160 – 32(1)
= 128 Therefore, the stone falls back to the ground
at time 10 seconds after the explosion.
Apabila/When t = 9, v = 160 – 32(9)
= –128

Halaju batu itu apabila t = 1 ialah
128 m s–1 dan halaju batu itu apabila
t = 9 ialah –128 m s–1.
Velocity of the stone when t = 1 is 128 m s−1
and velocity of the stone when t = 9 is
–128 m s−1.

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Jawapan

Kertas Model SPM b2 – 4ac = (–2)2 – 4(1)(1)

Kertas 1 = 0 satu punca/one root

Bahagian A ∴ Hanya satu nilai x yang memuaskan
Only one value of x satisfies
1. (a) (i) f(x) = 4 1 x f  –1(x) = 9f(x).
–
2. (a) 3x2  14x + 5
Katakan/Let y = f(x)
3x2 – 14x – 5  0
f –1(y) = x
(3x + 1)(x – 5)  0
1
y = 4 – x Untuk/For (3x + 1)(x – 5) = 0,

y(4 – x) = 1 x = – 31 , x = 5

4y – xy = 1

xy = 4y – 1

x = 4y – 1 x
y 5
– 1 O
3

Maka/Hence, f  –1(x) = 4x – 1
x
Maka/Therefore, x  – 13 , x  5
  1
(ii) ff(x) = f  4 – x (b) y = px2 – px – 2x + 2 + p
= px2 + (–p – 2)x + (p + 2)
= 1
4 1
4 – – x Untuk/For y  0, b2 – 4ac  0

= 1 dan/and p  0.
4(4 – x) – 1
(–p – 2)2 – 4(p)(p + 2)  0

4–x p2 + 4p + 4 – 4p2 – 8p  0

= 1 –3p2 – 4p + 4  0
16 – 4x – 1
3p2 + 4p – 4  0
4–x
Untuk/For 3p2 + 4p – 4 = 0
= 4–x
15 – 4x (3p – 2)(p + 2) = 0
2
(b) f  –1(x) = 9f  (x) p = 3 , p = –2

4x – 1 = 94 1 x
x –

4x – 1 = 4 9 x –2 O 2 p
x – 3

(4x – 1)(4 – x) = 9x   

16x – 4x2 – 4 + x = 9x Maka/Therefore, p < –2, p > 2
3
4x2 – 8x + 4 = 0

x2 – 2x + 1 = 0 Oleh sebab/Since p > 0, p > 2
3

© Penerbitan Pelangi Sdn. Bhd. 106

3. (a) x + 4 = 2x – 1 – 1 5. 2p + q – 5 = 0 .........................a
x + 4 = 2x – 1 – 12 p2 + 3pq + 20 = 0 ....................b
x + 4 = 4(x – 1) – 4x – 1 + 1
4x – 1 = 4x – 4 + 1 – x – 4 Daripada/From a, q = 5 – 2p ................c

= 3x – 7 Gantikan/Substitute c ke dalam/into b,

p2 + 3p(5 – 2p) + 20 = 0

4x – 12 = (3x – 7)2 p2 + 15p – 6p2 + 20 = 0
16(x – 1) = 9x2 – 42x + 49
–5p2 + 15p + 20 = 0

16x – 16 = 9x2 – 42x + 49 p2 – 3p – 4 = 0

9x2 – 58x + 65 = 0 (p + 1)(p – 4) = 0

(x – 5)(9x – 13) = 0 p = –1, p = 4

x = 5, x = 13 ∴ p = –1, q = 5 − 2(−1)
9
=7
(b) logx 256 – logx 4 = 3
p = 4, q = 5 − 2(4)

logx 256 = 3 = −3
4
x3m
logx 64 = 3 6. y = x3m
64 = x 3 log3 y = h h
log3
3

x2 = 64 2 = log3 x3m – log3 h

x = 643 2 = 3m log3 x – log3 h
64
= 1 Kecerunan/Gradient = 3m
3

= 42 6 – 0 = 3m
4 – 3
= 16 6 = 3m

4. (a) Sn = n2 – 18n m = 2

T12 = S12 – S11 Di titik/At point (3, 0),
= [122 – 18(12)] – [112 – 18(11)] log3 y = 6 log3 x − log3 h
0 = 6(3) − log3 h
= –72 – (–77) log3 h = 18
h = 318
= 5 p q = 387 420 489
8 p
(b) =

p2 = 8q ....................a

8 + p + q = 168 7. (a) –h3––03 = 5–3
–3 – 0
p + q = 160
q = 160 – p ....................b
–6 = – 32
Gantikan/Substitute b ke dalam/into a, h

p2 = 8(160 – p) 6 = 2
h 3
p2 = 1 280 – 8p
h = 9
p2 + 8p – 1 280 = 0

(p – 32)(p + 40) = 0

p = 32, p = – 40

∴ p = 32, q = 160 – 32

= 128

p = –40, q = 160 – (–40)

= 200

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(b) mPQR = –3 – 3 (b) ~p = 4~i + 2~j
9–0 ~q = 6~i – 2~j
~r = 9~i + 7~j
= –6
9

= – 32 ~r = m~p + n~q
9 ~i + 7~j == 4mm(4~i~i++22m~j~)j + n6(n6~i~i––22n~j~j)
m = – 1 +
mPQR

= –  1 (9 – 4m – 6n)~i = (2m – 2n – 7)~j
– 23 ~i dan ~j ialah dua vektor bukan sifar

= 3 dan tidak selari, jadi
2
~ni oat npdar~jalalreel, two non-zero vectors and are
so
Persamaan garis lurus yang melalui P
dan berserenjang dengan garis lurus 9 – 4m – 6n = 0
PQR, 4m + 6n = 9 .............a

Equation of the straight line that passes 2m – 2n – 7 = 0
through P and is perpendicular to straight 2m – 2n = 7 .............b
line PQR,
b × 2, 4m – 4n = 14 .............c

y – 5 = 3 a – c, 10n = – 5
x + 3 2 n = – 12

2y – 10 = 3x + 9

2y = 3x + 19 Gantikan nilai n ke dalam b,
3 19
y = 2 x + 2 Substitute the value of n into b,

    8. (a) ~a =–2,~b = 1   2m – 2 – 21 = 7
–1 –5
2m + 1 = 7

    –2 1 2m = 6
~a – ~b = –1 – –5
m = 3

  –3   t2 – 9
= 4 9. (a) had t–3
t → 3

 
  –3 had (t – 3)(t + 3)
~rˆ = 1 4 = t–3  
 (–3)2 + (4)2 t → 3

  = 1 –3 = had (t + 3)
5 4
t → 3

= 6

  = – 53  lim t2 – 9
4 t–3
t → 3

 =
lim (t – 3)(t + 3)
t–3
5 t → 3

= –  53 ~i + 4 ~j = lt i→m 3(t + 3)
5
=6

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(b) f(x) = (x2 – 3)4 11. (a) 9C6 × 5C2 = 840
2 – 3x
(b) 9C4 × 5C4 = 630
(2 – 3x)[4(x2 – 3)3(2x)]
(c) 9C5 × 5C3 + 9C6 × 5C2 + 9C7 × 5C1
f ʹ(x) = – (x2 – 3)4(–3) + 9C8 × 5C0
(2 – 3x)2
= 1 260 + 840 + 180 + 9
(2 – 3x)[8x(x2 – 3)3] + 3(x2 – 3)4 = 2 289
(2 – 3x)2
= 12. (a) p = 0.24, q = 0.76

f ʹ(0) = (2 – 0)(0) + 3(0 – 3)4 P(X  2) = 1 – P(X = 0) – P(X = 1)
(2 – 0)2
= 1 – 8C0(0.24)0(0.76)8

= 243  – 8C1(0.24)1(0.76)7
4
= 1 – 0.1113 – 0.2812

10. (a) (i) ∫2 3f(x)dx = 3 ∫2 f(x)dx = 0.6075

4 4 (b) P(X  1)  0.8

= –3 ∫4 f(x)dx 1 – P(X = 0)  0.8

2

= –3(12) 1 – nC0(0.24)0(0.76)n  0.8

= –36 1 – (0.76)n  0.8

(ii) ∫24 [2f(x) – 7x]dx (0.76)n  0.2

= ∫4 2f(x) dx – ∫4 7x dx log10 0.76n  log10 0.2

2 2 n log10 0.76  log10 0.2

  4 7x2 4
∫= 2 2 f(x) dx – 22 n  log10 0.2
log10 0.76
  7(42) 7(22)
= 2(12) – 2 – 2 n  5.865

= 24 – (56 – 14) n = 6

= 24 – 42 Bahagian B

= –18 13. (a) tan 2x = 7 ,
24
(b) ddxy = (2x – 3)3
90°  x  180°
y = ∫ (2x – 3)3 dx ⇒ 180° < 2x  360°

= (2x – 3)4 + c tan 2x  0
2(4) ⇒ Sudut 2x berada dalam sukuan ketiga
3)4
= (2x – + c Angle 2x is in the third quadrant
8
kos 2x  0 / cos 2x  0
x = 1, y = –2

–2 = [2(1) – 3]4 +c 25 7
8
(–1)4 2x
–2 = 8 + c 24

–2 = 1 + c ∴ kos 2x = – 2254 / cos 2x = – 2254
8

c = –2 – 1 Oleh sebab kos 2x = 2 kos2 x − 1,
8 Since cos 2x = 2 cos2 x − 1,
17
= – 8

∴ y = (2x – 3)4 – 17
8 8

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2 kos2 x – 1 = – 2254 / 2 cos2 x – 1 = – 2245

2 kos2 x = 1 / 2 cos2 x = 1
25 25

kos2 x = 1 / cos2 x = 1
50 50

(b) 11 tan y – kot y = sek y / 11 tan y − cot y = sec y

11 sin y – kos y = 1 y / 11 sin y – cos y = 1 y
kos y sin y kos cos y sin y cos

11 sin2­ y – kos2 y – 1 y = 0 / 11 sin2­ y – cos2 y – 1 y = 0
sin y kos y kos sin y cos y cos

 1

kos
11 sin2­ y – kos2 y  1
sin y
cos y
–1 = 0 / 11 sin2­ y – cos2 y – 1 = 0
sin y
y

    ko1s y
11 sin2­ y – (1 – sin2 y) – sin y = 0 / 1 y 11 sin2­ y – (1 – sin2 y) – sin y = 0
sin y sin y cos sin y sin y

    ko1s y
11 sin2­ y – 1 + sin2 y – sin y = 0 / 1 11 sin2­ y – 1 + sin2 y – sin y = 0
sin y cos y sin y

 1 sin2­ –  1

kos cos y
12 y – sin y 1 = 0 / 12 sin2­ y – sin y – 1 = 0
sin y sin y
y

1 =0 ⇒ tiada penyelesaian / 1 = 0 ⇒ no solution
kos y cos y

12 sin2­ y– sin y – 1 = 0
sin y

(4 sin y + 1)(3 sin y – 1) = 0

4 sin y + 1 = 0
sin y = – 14

y = 194°29ʹ, 345°31ʹ

3 sin y – 1 = 0

sin y = 1
3

y = 19°28ʹ, 160°32ʹ

∴ y = 19°28ʹ, 160°32ʹ, 194°29ʹ, 345°31ʹ

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14. (a) (i) 6y + x = 29 15. (a) (i) y = 2(x + 1)
y = 4, 4 = 2(x + 1)
6y = –x + 29 4 = 2x + 2
2x = 2
y = – 61 x + 29 x = 1
6

Kecerunan normal ∴ k = 1 y = 2(x + 1)
Gradient of normal (ii) y
= – 61
4 A (1, 4)
Kecerunan tangen
Gradient of tangent 2B y = (x – 2)(x – 5)
O1 x
= – –1 61 = 6

y = (x – 2)2 – 5 Luas (A + B)/Area of (A + B)

ddxy = 2(x – 2) = ∫1 y dx
= 2x – 4
0

= ∫1 (x – 2)(x – 5) dx

0

2x – 4 = 6 = ∫1 (x2 – 7x + 10) dx
2x = 10
x = 5 0

  = x3 – 7x2 + 10x 1
3 2 0

x = 5, y = (5 – 2)2 – 5   = 1 – 7 + 10 – (0)
3 2
=9–5
5
=4 = 6 6 unit2

Koordinat/Coordinates of P = (5, 4) Luas B/Area B = 1 (2 + 4)(1)
2
(ii) yx – 4 of
– 5
= 6 = 3 unit2

y – 4 = 6x – 30 Luas rantau berlorek A

y = 6x – 26 Area of shaded region A

(b) (i) PM + MQ = 50 = 6 5 – 3
x + MQ = 50 6
MQ = 50 – x
= 3 5 unit2
6
1
L = 2 (x)(MQ) (b) ∫h πx2 dy = 16π

0

= 1 x(50 – x) ∫h x2 dy = 16
2
0

= 25x – 1 x2 ∫ 0h (6 – y) dy = 16
2
  y2 h
dL 6y – 2
dx = 16
(ii) = 25 – x
0

25 – x = 0   6h –h2 – (0) = 16
2
h2
x = 25 6h – 2 = 16

L = 25(25) – 1 (25)2 12h – h2 = 32
2
h2 – 12h + 32 = 0
= 625 – 625
2 (h – 4)(h – 8) = 0

= 312.5 h = 4, h = 8

111 © Penerbitan Pelangi Sdn. Bhd.

Lengkung y = 6 − x2 memintas paksi-y Nisbah P(X  32) kepada
di y = 6 − (0)2 = 6 P(X  39)
Curve y = 6 − x2 intersects the y-axis at
y = 6 − (0)2 = 6 Ratio of P(X  32) to P(X  39)
= 0.0228 : 0.0668
∴h=4 = 1 : 2.9298
= 100 : 292.98

Kertas 2 ∴ n = 292.98
Bahagian A
(b) (i) P(0  Z  h) = 0.33

1. (a) LNM : LPM = 5 : 7 0.5 – P(Z  h) = 0.33

LNM 5 K P(Z  h) = 0.17
LPM 7
= h = 0.954

  KM6θ12 θ 1 θ (ii) X – 67
2 2.5

= 5 O = 0.954
7
L θ 6 cm M X – 67 = 2.385

112KM = 5 N X = 69.385
7
P 3. Harga tepung gandum/Price of wheat flour = x
5 Harga tepung sekoi/Price of millet flour = y
KM = 7 × 12 Harga tepung kuinoa/Price of quinoa flour = z

= 8.571 cm

(b) Luas/Area of KLNM = 50 cm2 4x + y + z = 58 ...........a
x + 3y + 2z = 83 .........b
  21 (8.571)2 1 θ = 50 2x + y = z ....................c
2 50

θ = 18.3655 Gantikan c ke dalam a,

= 2.7225 rad Substitute c into a,

1 4x + y + (2x + y) = 58
2
Luas/Area of OLPM = (6)2(2.7225) 4x + y + 2x + y = 58

= 49 cm2 6x + 2y = 58

3x + y = 29 ....................d

2. (a) (i) Diberi varians/Given variance = 4, Gantikan c ke dalam b,

sisihan piawai/standard deviation = 2 Substitute c into b,

  30 – 36 x + 3y + 2(2x + y) = 83
2
P(X  30) = P Z  x + 3y + 4x + 2y = 83

= P(Z  –3) 5x + 5y = 83 ....................e

= P(Z  3) d × 5, 15x + 5y = 145 ..................f

= 0.00135 f – e, 10x = 62

  (ii) P(X  P Z  32 – 36 x = 6.2
32) = 2
= P(Z  –2)
Gantikan x = 6.2 ke dalam d

= P(Z  2) Substitute x = 6.2 into d,

= 0.0228 3(6.2) + y = 29

  39 – 36 18.6 + y = 29
2
P(X  39) = P Z  y = 10.4

= P(Z  1.5)
= 0.0668

© Penerbitan Pelangi Sdn. Bhd. 112

Gantikan x = 6.2 dan y = 10.4 ke dalam c, (iii) Zarah P/Particle P:

Substitute x = 6.2 and y = 10.4 into c, S28 = 28 [2(162) + (28 – 1)(–6)]
2
2(6.2) + 10.4 = z

12.4 + 10.4 = z = 14(324 – 162)

z = 22.8 = 2 268 cm

∴ x = RM6.20, y = RM10.40, z = RM22.80 Zarah Q/Particle Q:

Jumlah harga yang perlu dibayar S28 = 28 [2(108) + (28 – 1)(–4)]
Total price that has to pay 2

= 5(6.20) + 2(10.40) + 22.80 = 14(216 − 108)
= RM74.60
= 1 512 cm

8 1 Q
16 2
4. (a) r = = , a = 16

16 1 512 cm

S∞ = 1 – 1 O 2 268 cm P
2


= 16 Jarak PQ/Distance of PQ
1
= 2 2682 + 1 5122
2 = 2 725.8 cm

= 32 4 4 12
12 12 12
(b) (i) Zarah P/Particle P: 5. (a) (i) = ×
162, 156, 150, …
= 412
a = 162, 12

d = 156 − 162 = 12
= –6 3

Tn = 0, 162 + (n – 1)(–6) = 0 = 4 × 3
3
162 – 6n + 6 = 0

6n = 168 = 23
3
n = 28

∴ p = 28 (ii) 3  + 1 + 2  – 1
3 – 2 3 + 2
(ii) Zarah Q/Particle Q:
y, y – 4, y – 8, … 3  + 13  + 2 

a = y, d = –4 = + 2  – 13  – 2 
3 – 2 3  + 2 
T28 = 0

y + (28 – 1)(–4) = 0 3 + 6 + 3 + 2 + 6 – 2
– 3 + 2
y + 27(–4) = 0
= 3 2 – 2 2
y = 108

= 1 + 26 + 22
3– 2

= 1 + 26 + 22

113 © Penerbitan Pelangi Sdn. Bhd.

(b) log3 (x + 2) – 4 log9 x3 + 3 log3 x 7. (a) y = 20 = 20x–4
x4
= log3 (x + 2) – 4 log3 x3 + log3 x3
log3 9 dy = –80x–5
dx
4 log3 x3 = – 8x05
= log3 (x + 2) – 2 + log3 x3

= log3 (x + 2) – 2 log3 x3 + log3 x3 Apabila/When x = 3, dy = – 8305
dx
= log3 (x + 2) – log3 x6 + log3 x3
= – 28403
= log3 (x + 2)(x3)
x6
dy
= log3 x+2 ∴ δy  dx × δx
x3
= – 28403 × (–0.02)
6. (a) x2 + 3(2x + h) = 0
 x2 + 6x + 3h = 0
= 0.006584

Hasil tambah punca/Sum of roots, Juga/Also, x = 3, y = 20 = 20
m + 3m = –6 34 81
4m = –6
m = – 32 Maka/So, 20  20 + 0.006584
(2.98)4 81
 0.2535


Hasil darab punca/Product of roots,  
dan/and 60  3 20
m(3m) = 3h (2.98)4 (2.98)4

3m2 = 3h  3(0.2535)

m2 = h  0.7605
  h = – 23 2
(b) dx = 0.03
9 dt
= 4

Jumlah luas permukaan = 224 cm2

(b) (m – 2) + (m + 5) = 2m + 3 Total surface area

= 2– 32  + 3 2(x2) + 4(3x)(x) = 224

2x2 + 12x2 = 224 3x

= –3 + 3 14x2 = 224

= 0 x2 = 16 x
x = 4 x
(m – 2)(m + 5) = m2 + 3m – 10
V = 3x(x)(x)
= – 32 2 + 3– 32  – 10
= 3x3

= 9 – 9 – 10 dV = 9x2
4 2 dx

= – 449 dV = dV × dx
dt dx dt


Persamaan kuadratik/Quadratic equation: = 9x2(0.03)

49 = 9(42)(0.03)
4
x2 – 0x – = 0 = 4.32 cm3 s–1

x2 – 49 = 0
4
4x2 – 49 = 0

© Penerbitan Pelangi Sdn. Bhd. 114

Bahagian B (d) B(9, 2), C(12, 0), D(13, 8)

8. (a) mAB = 8–2 Luas ∆BCD
0–9
= – 69 = 1   12 13 9 12
2 0 8 2 0

= – 32 = 1 |(96 + 26 + 0) – (0 + 72 + 24)|
2
1
Pintasan-y/y-intercept = 8 = 2 |122 – 96|

Persamaan garis lurus AB: = 13 unit2

Equation of straight line AB: 9. (a)
y = – 23 x + 8 X2 1 4 9 16 25 36
p2 4.24 10.89 22.66 38.19 59.44 83.72
Di pintasan-x/At x-intercept,
y = 0, – 23 x + 8 = 0 Graf p2 melawan X2
p2 Graph of p2 against X2

2 x = 8 100
3
98

x = 12 80

∴ a = 12, b = 8

 (b) (9, 2) = + 60
m(12) + n(0) m(0) + n(8)
m+n , m n

  40
12m 8n
= m+n , m+n

12m 20
m+n
= 9 y
A(0, 8)
2 X2
12m = 9m + 9n m 0 10 20 30 40 42.25

3m = 9n B(9, 2) (b) ap2 = bX2 + 5
n
m = 9 O x   p2 = b X2 + 5
n 3 C(12, 0) a a

m = 3 Daripada graf, pintasan-Y = 2
n 1

m:n=3:1 From the graph,the Y-intercept = 2
a5 = 2
AB : BC = 3 : 1 a = 2.5

(c) mBD = 3 y
2 A(0, 8)
D(x, 8) Kecerunan graf/Gradient of graph
8x – 2 3
– 9 = 2 = 38 – 11
16 – 4
3x – 27 = 12 B(9, 2)

3x = 39 O C(12, 0) x = 2.25

x = 13 b = 2.25
∴ D(13, 8) a

2b.5 = 2.25

b = 5.625

115 © Penerbitan Pelangi Sdn. Bhd.

(c) Apabila/ When X = 6.5

X2 = 42.25

Daripada graf/From the graph,

X 2 = 42.25, p2 = 98
p = 98


= 9.899

10. (a) sin 2x + 10 kos2 x = 6 / sin 2x + 10 cos2 x = 6

2 sin x kos x + 10 kos2 x – 6 = 0 / 2 sin x cos x + 10 cos2 x − 6 = 0

sin x kos x + 5 kos2 x – 3 = 0 / sin x cos x + 5 cos2 x − 3 = 0

sinkoxsk2oxs x + 5 kos2 x – 3 x = 0 / sincoxsc2 oxs x + 5 cos2 x – 3 = 0
kos2 x kos2 cos2 x cos2
x

tan x + 5 – 3 sek2 x = 0 / tan x + 5 – 3 sec2 x = 0

tan x + 5 – 3(1 + tan2 x) = 0

tan x + 5 – 3 – 3 tan2 x = 0

tan x + 2 – 3 tan2 x = 0

3 tan2 x – tan x – 2 = 0

(3 tan x + 2)(tan x – 1) = 0

3 tan x + 2 = 0
tan x = – 32
x = 146°19ʹ, 326°19ʹ


tan x – 1 = 0
tan x = 1
x = 45°, 225°

∴ x = 45°, 146°19ʹ, 225°, 326°19ʹ

(b) sin x = kos x / sin x = cos x

ksoins x = 1 / sin x = 1
x cos x

tan x = 1

r2 – 1 = 1

r2 – 1 = 12

r2 = 2
r = ±2

(c) y 2 kos 2x + 4 x = 3 / 2 cos 2x + 4 x = 3
π π
3 4
2 y = 2 kos 2x 2 kos 2x = 3 – π x / 2 cos 2x = 3 – 4 x
1 y = 2 cos 2x π
3π
O 4 x 1 titik persilangan antara dua graf.
–1 π
–2 π
2 y = 3 – 4π x 1 intersection point between the two graphs.

∴ Bilangan penyelesaian = 1
 Number of solutions

© Penerbitan Pelangi Sdn. Bhd. 116

11. (a) (i) 9O:P = 3 P9:S (c) 9O:T = nO9:Q
5
  h~p + k~q = n~q
OPSP = 3
O9:P = 5  O9:S 1 – m ~p + 2 m~q = n~q
9 3
3    
9O:S = 8  O9:P 1– m ~p = n– 2 m ~q
9 3
8
= 3 1 – m = 0 n – 2 m = 0
9 3
8 ~p
3 m = 9 n – 2 (9) = 0
3
(ii) 9Q:S = Q9:O + 9O:S n – 6 = 0

= –~q + 8 ~p n = 6
3

= 8 ~p – ~q (d) ∆PQR dan ∆PQT mempunyai tinggi
3 yang sama, dengan tapak PR dan PT
masing-masing.
(b) QR : RS = 1 : 2
QR : QS = 1 : 3 PR : PT = 1 : 9

P9:T == 9P:O + 9O:T Maka, luas ∆PQR = 1
–~p + h~p + luas ∆PQT 9
k~q
∆PQR and ∆PQT have the same height,

mP9:R = –~p + h~p + k~q with bases PR and PT respectively.
+ 9Q:R = –~p + h~p + k~q
m9P:Q P R : PT = 1 : 9

  Hence, area of ∆PQR = 1
area of ∆PQT 9
1 Q9:S
m ~q – ~p + 3 = –~p + h~p + k~q Bahagian C

   1 8 ~p – ~q = –~p + h~p + k~q
m ~q – ~p + 3 3 12. (a) p = 100 − 10 − 30 − 35
  = 25
m ~q – ~p + 8 ~p – 1 ~q = –~p + h~p + k~q
9 3 (b) (i) 7Q02.07006 × 100 = 140
 
m – 19 ~p + 2 ~q = –~p + h~p + k~q 70.70 × 100
3 140
Q2006 =
– m9 ~p + 23m~q = –~p + h~p + k~q
= RM50.50

– m9 ~p + ~p – h~p = k~q – 2 m~q Harga bahan D pada tahun 2006
3 ialah RM50.50.
    The price of ingredient D in the year
– m9 + 1 – h ~p = k– 2 m ~q 2006 is RM50.50.
3

(i) –  m + 1 – h = 0 (ii) I = Q2021 × 100
9 h = 1 Q2001
– m
9 Q2021 Q2006
= Q2006 × Q2001 × 100
2
(ii) k – 3 m = 0 = 1.35 × 1.20 × 100

k = 2 m = 162
3

117 © Penerbitan Pelangi Sdn. Bhd.

135(10) + x(30) + 110(25) v  0

(c) (i) + 140(35) = 132 t2 – 6t + 5  0
100
(t – 1)(t – 5)  0
1350 + 30x + 2 750 + 4 900 =
100 132 Apabila/When v = 0,
t = 1 dan/and t = 5
9 000 + 30x = 132
100
v

9 000 + 30x = 13 200

30x = 4 200

x = 140 O1 t
5

(ii) Q5.230026 × 100 = 132
Q2006 =
5.32 × 100 Maka/So, 1  t  5
132
(ii) v = 0
= RM4.03
t2 – 6t + 5 = 0
13. (a) I : 12x + 18y  7 200
2x + 3y  1 200 (t – 1)(t – 5) = 0

II : x  2y t = 1, t = 5

y  1 x (iii) s = ∫ (t2 – 6t + 5) dt
2
= t3 – 3t2 + 5t + c
III : 3x + 7y  1 000 3

(b) y Apabila/When t = 0, s = 0 ⇒ c = 0

400 ∴ s = t3 – 3t2 + 5t
3
300 2x + 3y = 1200
Apabila/When s = 0,

200 y = 1 x t3 – 3t2 + 5t =0
2 3
3x + 7y = 1000
x = 600   tt2 =0
120 R 3 – 3t + 5

100

0 100 200 240 300 x t(t2 − 9t + 15) = 0
400 500 600

(c) (i) 600 Untuk/For t2 − 9t + 15 = 0,

(ii) Apabila/When y = 120, t = −(−9) ± (−9)2 − 4(1)(15)
x minimum = 240 2(1)

Keuntungan minimum = 9 ± 81 − 60
Minimum profit 2

= 3(240) + 7(120) = 9 ± 21
= RM1 560 2

14. (a) (i) a = 2t – 6 = 6.79, 2.21

v = ∫ (2t – 6) dt t = 0, t = 2.21, t = 6.79

= t2 – 6t + c Apabila/When t = 1,
1
Apabila/When t = 0, v = 5 ⇒ c = 5 s = 3 –3+5
∴ v = t2 – 6t + 5
= 7
3

© Penerbitan Pelangi Sdn. Bhd. 118

Apabila/When t = 5, Zarah itu kemudian bergerak ke
arah kiri dengan halaju semakin
s = 53 – 3(5)2 + 5(5) bertambah sehingga mencapai
3 halaju maksimum 4 m s–1 apabila
125 t = 3. Halajunya kemudian semakin
= 3 – 75 + 25 berkurang sehingga berhenti
seketika apabila t = 5.
= – 235 The particle starts to move from O
towards the right with a velocity of
Apabila/When t = 6, 5 m s–1. The velocity then decreases
until it stops instantaneously when
s = 63 – 3(6)2 + 5(6) t = 1. The particle then moves towards
3 the left with an increasing velocity
until the maximum velocity of 4 m s–1
= 216 – 108 + 30 when t = 3. The velocity then decreases
3 until it stops instantaneously when
s t = 5.
= –6

7

3

O 1 5 6 t

2.21 6.79

–6

– 25 15. (a) (i) B21C(1 =1)(1B13C5s)in×sin3283°8° = 35
3

Jumlah jarak dilalui dalam 6 saat
pertama/Total distance travelled in

the first 6 seconds = 10.34 cm

    =2 7 + 25 + 25 –6 (ii) AB2 = 112 + 10.342
3 3 3
– 2(11)(10.34) kos 38°
= 14 + 25 + 7 cos 38°
3 3 3
46 = 48.66
3
= AB = 48.66
= 6.98 cm
= 15 1 m
3
(iii) sin ∠ABC = sin 38°
(b) (i) a = dv = 0, 11 6.98
dt
sin ∠ABC = sin 38° × 11
2t – 6 = 0 6.98

2t = 6 = 0.9702
t = 3
∠ABC = 75°59ʹ

t = 3, v = 32 – 6(3) + 5 (b) (i) A

= –4 v 11 cm

6.98 cm

5 104°1Ј 38°

v = t2 – 6t + 5 M C

O1 3 5 t (ii) ∠MAC = 180° – 104°1ʹ – 38°

–4 = 37°59ʹ

(ii) Zarah itu mula bergerak dari Luas ∆ACM /Area of ∆ACM
O ke arah kanan dengan halaju
5 m s–1. Halajunya kemudian = 1 (6.98)(11) sin 37°59ʹ
semakin berkurang sehingga 2
berhenti seketika apabila t = 1.
= 23.63 cm2

119 © Penerbitan Pelangi Sdn. Bhd.