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

Jawapan

Tingkatan 4 Nombor Indeks (b) (i)

BAB Index Numbers (115 × 4) + (108 × 1)

10 +4(1+112 × k) = 112.8
+ k

Masteri SPM 460 + 108 + 112k = 112.8
5+k

Kertas 2 568 + 112k = 564 + 112.8k

1. (a) 110 = Q2017 × 100 0.8k = 4
7
k = 5
Q2018
Q2017 = 7 × 110 (ii) 112.8 = 65 × 100
100
Q2018 = RM73.32
= RM7.70

(120 × 2) + (110 × 3) (c) 140 = I × 112 × 100
100 100
(b) + (90 × 1) + (p × 4) = 112
2+3+1+4 I = 125

240 + 330 + 90 + 4p = 112 3. (a) (i) Menokok/Increase 24%, m = 124
10
43.40
4p + 660 = 1 120 (ii) 124 = n × 100

4p = 460 n = 35

p = 115

Bertambah/Increase by 15% (124 × 10) + (116.4 × 20)

(c) (i) I– = Q2019 × Q2017 × 100 (b) –I = + (160 × 2)
Q2017 Q2015 10 + 20 + 2

= 118 × 112 × 100 = 3888
100 100 32

= 132.16 = 121.5

(ii) 132.16 = 80 × 100 (c) (i) 180 = 121.5 × –I × 100
Q2015 –I = 100 100

Q2015 = 80 × 100 148.15
132.16
(ii) 180 = Q2020 × 100
= RM60.53 120

16.20 Q2020 = RM216
12.00
2. (a) (i) h = × 100 RM85 000
RM216
= 135 = 393.5

(ii) 115 = Q2018 × 100 Bilangan maksimum beg = 393
12.00 Maximum number of bags

Q2018 = RM13.80

51 © Penerbitan Pelangi Sdn. Bhd.

4. (a) (i) 162 = h × 135 × 100 6. (a) 114 = 5.70 × 100
100 100 Q2016

h = 120 Q2016 = RM5.00 per kg

k = 130 × 90 × 100 (b)
100 100 (114 × 6) + (132 × 1)

k = 117 + (108 × 2) + (120 × m)
6 + 1+2+m
(ii) 108 = 5.40 × 100 = 116
Q2017
1032 + 120m
Q2017 = RM5.00 9 + m = 116

(150 × y) + (114 × 1) 1032 + 120m = 1044 + 116m

+ (162 × 4) + (108 × 7) 4m = 12

+ (117 × 6) m = 3
y+1+4+7+6
(b) = 126 (c) Untuk ayam/For chicken,

150y + 2220 = 126 I = 114 × 125 × 100 = 142.5
y + 18 100 100

150y + 2220 = 126(y + 18) Untuk udang/For prawn,

150y + 2220 = 126y + 2268 I = 120 × 90 × 100 = 108
100 100
24y = 48

y = 2 (142.5 × 6) + (132 × 1)

(c) 126 = Q2020 × 100 I– = + (108 × 2) + (108 × 3)
20 6+1+2+3

Q2020 = RM25.20 = 1527
12
Harga jualan/Selling price
= 127.25
= 25.20 + (25.20 × 65%)
Q2020 ×
= RM41.58 127.25 = 74 000 100

5. (a) 170 = Q2018 × 100 Q2020 = RM94 165
4
7.80
Q2018 = RM6.80 7. (a) 130 = Q2018 × 100

(110 × 20) + (130 × 50) Q2018 = RM6.00

(b) –I = + (80 × 10) + (170 × 20) (b) IW = 130 × 120 × 100 = 156
100 100 100

= 12 900 IX = 110
100

= 129 IY = 180 × 90 × 100 = 162
100 100
129 = 69 660 × 100
Q2014 IZ = 140 × 115 × 100 = 161
100 100
Q2014 = RM54 000

(c) –I = 129 × 140 × 100 = 180.6
100 100

Bertambah/Increase by 80.6%

© Penerbitan Pelangi Sdn. Bhd. 52

(156 × 150) + (110 × 300) Untuk durian/For durian,
I2019
(c) (i) I– = + (162 × 200) + (161 × 350) 115 = 100 × 125 × 100
150 + 300 + 200 + 350 100

= 145 150 I2019 = 92
1000
Untuk tembikai/For watermelon,
= 145.15 I2019
117 = 100 × 90 × 100
Q2020 100
(ii) 145.15 = 26 × 100
I2019 = 130

Q2020 = RM37.74 Untuk nanas/For pineapple,

Soalan KBAT 126 = I2019 × 105 × 100
100 100

k 120 I2019 = 120
100 100
1. (a) 126 = × × 100 (140 × 126) + (92 × 108)

k = 105 I– = + (130 × 72) + (120 × 54)
360
(b) x° + y° = 180°
y = 180 – x ......a = 43 416
360
Untuk durian/For durian,
180° – 72° = 108° = 120.6

(154 × x°) + (115 × 108°) 120.6 = Q2019 × 100
60 000

+ (117 × 72°) + (126 × y°) = 130.7 Q2019 = RM72 360
360°
2. (a) Indeks harga produk N pada 2020
154x + 126y + 20 844 = 47 052
berasaskan 2019,
154x + 126y = 26 208
Price index of product N in 2020 based on
77x + 63y = 13 104 ....b
2019,

Gantikan a ke dalam b, IN = 22.40 × 100
Substitute a into b, 14

77x + 63(180 – x) = 13 104 = 160

77x + 11 340 – 63x = 13 104 p = 160 × 90 × 100
100 100
14x = 1764

x = 126 = 144

y = 180 – 126
= 54

(c) Indeks harga pada 2019 berasaskan
2018:
Price index in 2019 based on 2018:

Untuk pisang/For banana,

154 = I2019 × 110 × 100
100 100

I2019 = 140

53 © Penerbitan Pelangi Sdn. Bhd.

(b) Indeks harga pada 2020 berasaskan 2019:
Price index in 2020 based on 2019:

Untuk M/For M,

135 = x + 100 × IM × 100
100 100

IM = 13 500
x + 100

Untuk N/For N,

IN = 160
 (120


× 200) + (108 × 600) + 13 500 × 400 + (160 × 800)
200 + 600 x + 100 = 130.9

+ 400 + 800

5 400 000 + 216 800 = 130.9(2000)
x + 100

5 400 000 = 45 000
x + 100

x + 100 = 120
x = 20

(c) 130.9 = 222 530 × 100
Q2019

Q2019 = RM170 000

© Penerbitan Pelangi Sdn. Bhd. 54

Jawapan

Tingkatan 5 Sukatan Membulat 3. Katakan/Let
AD = x cm, OD = 2x cm, OA = 3x cm
BAB Circular Measure

1 Luas kawasan berlorek/Area of shaded region

Masteri SPM = 32 cm2

21 (3x)2(0.8) – 1 (2x)2(0.8) = 32
2
Kertas 1
3.6x2 – 1.6x2 = 32

1. (a) 135o = 135o × π 2x2 = 32
180°
x2 = 16

= 0.75π rad x = 4

(b) Panjang lengkok PQ/Length of arc PQ sAB = 3(4)(0.8) = 9.6 cm
= 7(0.75π) sCD = 2(4)(0.8) = 6.4 cm
= 16.4955
Perimeter = 9.6 + 6.4 + 4 + 4
Perimeter = 16.4955 + 7 + 7
= 30.4955 cm = 24 cm

2. (a) sPQ = 28.8 – 9 – 9 4. PA = 1 OP = 1 r
2 2

= 10.8 cm s = jθ OA = r + 1 r = 3 r
2 2
∠POQ = 10.8 θ = s
9 j Perimeter = 60 cm

= 1.2 rad PA + sAB + BQ + sPQ = 60

(b) Katakan M = titik tengah PQ, 12 r + 3 r(3k) + 1 r + r(5k) = 60
2 2
Let M = midpoint of PQ, 19
2
sin 0.6 = PM rk + r = 60
9
19rk + 2r = 120
PM = 5.0818 cm
r(19k + 2) = 120

Panjang PQ/Length of PQ r = 120 2
= 2 × PM 19k +
= 2 × 5.0818
5. Dari 1:30 p.m. ke 2:05 p.m., jarum bergerak
= 10.1636 cm
From 1:30 p.m. to 2:05 p.m., the hand moved

Perimeter kawasan berlorek 31620° × 7 = 210°.
Perimeter of shaded region
θ = 210° × π
= 10.1636 + 10.8 180°
= 20.9636 cm
= 3.666 rad

Jarak yang dilalui oleh P/Distance covered by P
= panjang lengkok/length of arc
= 12(3.666)
= 43.992 cm

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6. (a) Katakan/Let ∠MON = θ, 9. (a)   sPR = x
rθ = x
j sin θ + j sin θ = 20
2 2 Luas/Area = 24 cm2
θ
2j sin 2 = 20 1 r2θ = 24
2
θ
2(12) sin 2 = 20 1 r(x) = 24
2 r =
θ 5 48
sin 2 = 6 x

θ = 0.9851 0θπ (b) (i) sPR = 3r
2 rθ = 3r
0  θ  π θ = 3 rad
2 2
θ = 1.9702

∠MON = 1.9702 rad (ii) Daripada/From rθ = x,

(b) Luas/Area 1 48 (3) = x
1 2 x x2 = 144
= 2 (12)2(1.9702) – (12)2 sin 1.9702

= 141.854 – 66.333 x = 12

= 75.521 cm2 r(3) = 12

7. (a) j + j + 4.5j = 39 r = 4

6.5j = 39 ∴ r = 4, x = 12

j = 6 cm 10. (a) 21 r2α = 18

sxy = 6(4.5) α = 36
= 27 cm r2

(b) Luas/Area = 1 (6)2(4.5) (b) Perimeter = r + r + rα
2  
= r + r + r 36
r2
= 81 cm2 36
r
= 2r +

8. (a) sPQ = 14(0.75) 11. (a) rx = 6
= 10.5 cm
6
sin 0.75 = PR  r = x
14
(b) ∠ADC = 90°
PR = 14 sin 0.75
  DC =
= 9.543 cm 10 2 – r2
x
OR = 142 – 9.5432
= 10.244 cm  =  10 2 –  6 2
x x
RQ = 14 – 10.244
= 3.756 cm  = 64
x2
Perimeter = 10.5 + 9.543 + 3.756
8
= 23.799 cm = x cm

(b) Luas/Area   Luas 1 8
1 1 ∆ACD/Area of ∆ACD = 2 (r) x
= 2 (14)2(0.75) – 2 (10.244)(9.543)

= 73.5 − 48.879

= 24.621 cm2

© Penerbitan Pelangi Sdn. Bhd. 56

Luas kawasan berlorek 2. (a) OC : CB = 3 : 1 ⇒ OC : OB = 3 : 4

Area of shaded region OC = 3 OB = 3 (16) = 12 cm
4 4
    =2 1 (r) 8 – 1 r2(x)
2 x 2 Katakan M = titik tengah OC

=  6  8  –  16 2 Let M = midpoint of OC
x x x
2 (x)

48 18 OM = 6 cm
x2 x
= – kos ∠AOM = OM
OA
48 – 18x
= x2 = 6
16

12. (a) Luas/Area = 1 (38)2(2) ∠AOM = 1.1864 rad
2
∴ ∠AOB = 1.1864 rad
= 1444 m2

(b) Katakan/Let BC = x m, (b) sAB = 16(1.1864)
maka/then OB = (38 − x) m = 18.9824 cm

Panjang pagar/Length of fence BC = 16 – 12
= 4 cm
= perimeter kawasan berlorek
 perimeter of shaded region Perimeter = 18.9824 + 4 + 16
= 38.9824 cm
= sAB + BC + sCD + AD
= (38 – x)(2) + x + 38(2) + x (c) Luas/Area = 1 (16)2(1.1864)
= 76 – 2x + x + 76 + x 2
= 152 m 1
– 2 (16)(12)sin 1.1864

Kertas 2 = 151.8592 − 88.9944

= 62.865 cm2

1. (a) 4BC = AB 3. (a) OT = 1 OQ = 1 (18) = 6 cm
3 3
BC = 1 (12)
4 1
OS = 2 OP
= 3 cm

AC = 12 + 3 = 1 (18)
= 15 cm 2

kos ∠CAD = 12 = 9 cm
15
∠CAD = 0.6435 rad kos ∠SOT = 6
9

(b) CD = 152 – 122 ∠SOT = 0.8411 rad
= 9 cm
∴ ∠POQ = 0.8411 rad
sBD = 12(0.6435) = 7.722 cm
(b) TS = 92 – 62 = 6.708 cm
Perimeter = 7.722 + 9 + 3 sPQ = 18(0.8411) = 15.140 cm
= 19.722 cm SP = OS = 9 cm
TQ = 2OT = 12 cm
(c) Luas/Area = 1 (12)(9) – 1 (12)2(0.6435)
2 2 Perimeter = TS + sPQ + SP + TQ
= 6.708 + 15.140 + 9 + 12
= 54 – 46.332 = 42.85 cm

= 7.668 cm2

57 © Penerbitan Pelangi Sdn. Bhd.

(c) Luas kawasan berlorek (b) Luas/Area

Area of shaded region = 1 (12)2(0.723) + 1 (16)2(0.848)
2 2
= 1 (18)2(0.8411) – 1 (6)(6.708) 1
2 2 2
  − (10.583)(12)
= 136.258 – 20.124

= 116.13 cm2 = 52.056 + 108.544 − 63.498

4. (a) 8θ = 2.8 = 97.102 cm2
 θ = 0.35
6. (a) AC = AQ + QC
∠ROS = 0.35 rad = 130 + 130
= 260 m
(b) tan 50° = PQ
8 sin 1.2 sin ∠ABC
250 = 260
PQ = 8 tan 50o

= 9.534 cm sin ∠ABC = 0.9693

50° = 50° × 3.142 ∠ABC = 1.322 rad
180°
∠ACB = 3.142 – 1.322 – 1.2
= 0.8728 rad
= 0.62 rad

Luas kawasan berlorek sinA0B.62 = 250
sin 1.2
Area of shaded region

= 1 (8)2(0.35) + 1 (8)(9.534) AB = 155.85 m
2 2
1 BP = 155.85 – 130
– 2 (8)2(0.8728) = 25.85 m

= 11.2 + 38.136 − 27.9296 sPQ = 130(1.2)
= 156 m
= 21.4064 cm2

5. (a) sin 0.848 = 12 Perimeter kawasan berlorek
OP Perimeter of shaded region
12
OP = sin 0.848 = 25.85 + 156 + 130 + 250
= 561.85 m
= 16.0 cm
Jumlah kos pemagaran
sPQ = 16(0.848) Total cost of fencing
= 13.568 cm = 561.85 × RM30

OA = 162 – 122 = RM16 855.50
= 10.583 cm
(b) Luas/Area = 1 (260)(155.85) sin 1.2
AQ = 16 – 10.583 2
= 5.417 cm 1
– 2 (130)2(1.2)

∠OPA = 3.142 – 0.848 = 18 883.578 – 10 140
2
= 8 743.578 m2
= 0.723 rad

sAB = 12(0.723) Soalan KBAT
= 8.676 cm

BP = AP = 12 cm 1. (a) Perimeter = lilitan bulatan
 circumference of circle
Perimeter = 13.568 + 5.417 + 8.676 + 12
28π = 2πj
= 39.661 cm
j = 14 cm

© Penerbitan Pelangi Sdn. Bhd. 58

Kaedah Alternatif sPQ = 28.868(2.0947)
= 60.470 cm
Sudut di pusat dicangkum 30°
oleh setiap lengkok/Angle at Perimeter
centre subtended by each arc = 60.470 + 2(28.867) + 100
= 218.204 cm
= 180° − 2(30°) 120°
30°
= 120° (b) Luas kawasan berlorek
2π
= 3 rad Area of shaded region

Perimeter = 3 × panjang lengkok/length of arc = 1 (100)(28.868) – 1 (28.868)2(2.0947)
2 2
  28π = 3 × j 2π
3 = 1443.4 − 872.821
28π = 2πj
j = 14 cm = 570.58 cm2

(b) 3. (a) Katakan O ialah pusat semibulatan,
maka OP = 24 m.
14 cm Let O be the centre of the semicircle,
14 cm therefore OP = 24 m.
60°
60° PQ : QR = 2 : 1 ⇒ PQ : PR = 2 : 3

14 cm

Luas/Area PS = PQ S

= 3[luas sektor/area of sector − 2(luas = 2 PR 16 m
3
 segi tiga dalam sektor/area of triangle 16 m 24 m
2 P 24 m O
in sector)] = 3 (48)

       1 2π 1 π = 32 m
= 3 2 (14)2 3 – 2 2 (14)2sin 3

= 3(205.277 – 169.754) kos ∠SPQ = 16
24
= 106.569 cm2 ∠SPQ = 0.841 rad

2. (a) (i) OP = OB = jejari sektor (b) s = 32(0.841)
radius of sector = 26.91 m

∠AOB = 60° (c) ∠SOP = 3.142 – 2(0.841)
= 1.46 rad
tan 60° = AB
OB
Luas kawasan berlorek
tan 60° = 50 Area of shaded region
OB
= luas semibulatan/area of semicircle −
OB = 50 (luas sektor/area of sector SPQ
tan 60° + luas tembereng/area of segment PS)

= 28.868 cm

∴ OP = 28.868 cm = 21 (24)2(3.142) – [ 1 (32)2(0.841)
2
(ii) OA = 502 + 28.8682 1
= 57.735 cm + 2 (24)2(1.46 – sin 1.46)]

PA = QC = 57.735 – 28.868 = 904.896 − (430.592 + 134.246)

= 28.867 cm = 340.058 m2

120° = 120° × 3.142 Luas tembereng/Area of segment
180°
= 1 j 2θ – 1 j 2 sin θ
2 2
= 2.0947 rad
= 1 j 2(θ – sin θ)
2

59 © Penerbitan Pelangi Sdn. Bhd.

Jawapan

Tingkatan 5 Pembezaan (b) had 2 + 3x = 2 + 3(0) = – 32
x–3 0–3
BAB Differentiation x → 0

2 lim 2 + 3x = 2 + 3(0) = – 32
x–3 0–3
x → 0

Masteri SPM (c) had x2 – 4 = had (x – 2)(x + 2)
x+2 x+2
x → 0 x → 0

Kertas 1 = had (x – 2)

x → 0

1. = (0 − 2)
x −0.05 −0.01 −0.005 0 0.005 0.01 0.05
= −2
f(x) −2.15 −2.03 −2.015 −2 −1.985 −1.97 −1.85
lim x2 – 4 = lim (x – 2)(x + 2)
x+ 2 x+2
x → 0 x → 0

Daripada jadual nilai, apabila x menghampiri = lim (x – 2)

x → 0

sifar dari kiri dan kanan, f(x) menghampiri = (0 − 2)

nilai −2. Jadi, had f(x) = –2. = −2

x → 0
From the table of values, as x approaches zero
  4.  d p = – xp2 + k …a
from left and right, f(x) approaches a value of dx x + kx + 3

−2. So, lim f(x) = –2.   Diberi/Given  d p r
dx x 6
x → 0

2. (a) Apabila x menghampiri 0 dari kiri dan + kx + 3 = 2kx–2 +

kanan, f(x) menghampiri −4. = 2k + r  ... b
x2 6
∴ had f(x) = −4
Bandingkan a dan b,
x → 0

When x approaches 0 from left and right, Comparing a and b,

f(x) approaches −4. –p = 2k …......c

∴ lim f(x) = −4 r
6
x → 0

(b) Apabila x menghampiri 0 dari kiri dan dan/and k = …......d

kanan, f(x) menghampiri 9. Gantikan d ke dalam c,

∴ had f(x) = 9 Substitute d into c,

x → 0
When x approaches 0 from left and right,
–p = 2 r 
f(x) approaches 9. 6
lim
∴ f(x) = 9 –p = r
x → 0 3

(c) Apabila x menghampiri 0 dari kiri dan r = –3p

kanan, f(x) menghampiri 2. 1

∴ had f(x) = 2 5. (a) 3u23u   = 3u2 × u3

x → 0
When x approaches 0 from left and right,
7
f(x) approaches 2.
= 3u3

∴ lim f(x) = 2  d 3u23u   =  d 3u37 
du du
x → 0

3. (a) hx →ad 0 (x3 + 2x) = [03 + 2(0)] = 0 4
= 7u3
lim (x3 + 2x) = [03 + 2(0)] = 0
= 73u4
x → 0

© Penerbitan Pelangi Sdn. Bhd. 60

(b) r2ℎ = 72 8. y = px2 + x + 1

ℎ =  72  ddxy = 2px + 1
r2

 ddrh = −  1r434 Pada/At x = x0,  dy = k
dx
k = 2px0 + 1
Apabila/When r = 2,
k–1
 dh = −  12434 = −18 x0 = 2p
dr

6. y = (x2 + 1)n 9. y = x2 – 7x + k

 dy = n(x2 + 1)n – 1 . 2x  ddxy = 2x – 7
dx
= 2nx(x2 + 1)n – 1 ....a Kecerunan tangen/Gradient of tangent = 3
2x – 7 = 3
Diberi/Given  dy = px(x2 + 1)3q – 2 .....b x = 5
dx

Bandingkan a dan b, Gantikan x = 5 ke dalam persamaan tangen,

Comparing a and b, Substitute x = 5 into the equation of tangent,

2n = p .......c y = 3(5) − 13

dan/and n – 1 = 3q – 2 = 2

n = 3q – 1 .......d Koordinat titik/Coordinates of point A = (5, 2).

Gantikan d ke dalam c, Gantikan x = 5 dan y = 2 ke dalam persamaan

Substitute d into c, lengkung,

2(3q – 1) = p Substitute x = 5 and y = 2 into the equation of

6q – 2 = p the curve,

 p + 2 2 = 52 − 7(5) + k
6
q = k = 12

7. (a) y =  1  dan/and  u = 3x2 – 2x + 1 10. (a) x  −2, kecerunan/gradient +;
u3
x = −2, kecerunan/gradient = 0;
(b)  dduy = – u 34 ,  ddxu = 6x – 2 x  −2, kecerunan/gradient −
⇒ Q

Gunakan petua rantai, (b) x  0, kecerunan/gradient −;

Using chain rule, x = 0, kecerunan/gradient = 0;
x  0, kecerunan/gradient +
 ddxy =  dy ×  du ⇒P
du dx
(c) x  −2, kecerunan/gradient +;
= – u 34 × (6x – 2)
x = −2, kecerunan/gradient = 0;
= – (3x 32 (–6x2x– 2) x  −2, kecerunan/gradient +
+ 1)4 ⇒R

= – (3x 62 (–3x2x– 1)
+ 1)4

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11. r + h = 10 ∴ V = 4000 π ialah nilai maksimum.
h = 10 – r ....a 27

V = πr2h .......b V = 4000 π is the maximum value.
27
Gantikan a ke dalam b,
Substitute a into b, 12. A = x2

V = πr2(10 – r)  dA = 2x
= 10πr2 – πr3 dx

 ddVr = 20πr – 3πr2 Dengan petua rantai/With chain rule,

Untuk nilai maksimum dan minimum,  dA =  dA ×  dx
dt dx dt
For maximum and minimum values,
12 = 2x ×  dx
 ddVr = 0 dt

20πr – 3πr2 = 0  dx =  6 ....a
dt x
πr(20 – 3r) = 0
Apabila/When A = 4,
πr = 0  atau/or 20 – 3r = 0
r =  230 x2 = 4
r = 0
x = 2 (Abaikan/Ignore x = –2)

Gantikan r = 0 ke dalam a, Gantikan x = 2 ke dalam a,

Substitute r = 0 into a, Substitute x = 2 into a,

h = 10  dx  6
dt 2
Gantikan r =  20 ke dalam a, =
3
 20 = 3
3
Substitute r = into a, Maka, kadar perubahan x ialah 3 cm s−1

h = 10 –  20 =  10 apabila A = 4.
3 3 Therefore, the rate of change of x is 3 cm s−1

 dd2rV2 = 20π – 6πr when A = 4.
13. Katakan V ialah isi padu belon dan r ialah

Apabila/When r = 0, h = 10, jejari belon.

 dd2r V2 = 20π – 6π(0) Let V be the volume of the balloon and r be the
= 20π 0 radius of the balloon.

V = π(0)2(10) Maka/Then, V =  4 πr3
3
 dV
= 0 dr = 4πr2

∴ V = 00iisatlhaehmniinlaimi umminviamluuem. . dan/and δr = 5.8 – 6
V =
= –0.2 cm
Apabila/When r =  20 , h =  10 , Perubahan hampir bagi isi padu belon,
3 3

   dd2rV2 = 20π – 6π  20 Approximate change in the volume of the balloon,
3
= –20π  0 δV   dV × δr
dr
 4πr2 × (–0.2)
    V = π  20 2  10  4π(6)2 × (–0.2)
3 3
 –28.8π cm3
= 4000 π
27 Gantikan/Substitute r = 6

© Penerbitan Pelangi Sdn. Bhd. 62

14. (a) xy = x2 + 6 (b) Apabila/When x = 125, y = 3125 = 5
x2 + 6
y = x δx = 130 – 125 = 5

= x + 6  dy = 1
x dx 75

 ddxy = 1 – 6 Jadi/So,
x2
3x + δx  y + δy
Apabila/When x = 2,  y +  ddxy . δx

 ddxy = 1 – 6 1
(2)2 75
3125 + 5  5 + (5)
= – 12
 5.07

(b) δx = (2 + p) − 2 Maka, nilai hampir bagi 3130 ialah 5.07.
= p Therefore, the approximate value of 3130

Perubahan hampir dalam y, is 5.07.

Approximate change in y,

δy   dy × δx Kertas 2
dx
 – 12 p 1. (a) f(x) = 3
x

Nilai baru y/New value of y, f(x + δx) – f(x) = 3 – 3
y  y + δy x + δx x

     x + 6 = 3x – 3(x + δx)
x + – 21 p (x + δx)x

     2 6 = 3x – 3x – 3δx
2 + – 21 p (x + δx)x
+
–3δx
1 Gantikan/Substitute x = 2 = (x + δx)x
2
 5 – p Maka/Therefore,

15. (a) y = 3x f ʹ(x) = had  f(x + δx) – f(x)
δx
1 δx → 0

= x3 = had  –3δx
δx(x + δx)x
 ddxy = 1 x– 23 δx → 0
3
1 = had  –3
33x  2 (x + δx)x
= δx → 0

= – 3
x2

Apabila/When x = 125, (b) f ʹ(x) = – 31

 ddxy = 1 3 – 31
33125  2 x2
– =

= 1 x2 = 9
75
x = ±3

63 © Penerbitan Pelangi Sdn. Bhd.

2. (a) Katakan/Let y = 4x3 + 3x2 2x + 5y − 20 = 0
x
5y = −2x + 20
= 4x2 + 3x y = – 52 x + 4

 ddxy = 8x + 3 Maka, kecerunan tangen ialah – 52 .
Therefore,the gradient of the tangent is
(b) Katakan/Let y = (2x – 5)4 – 25 .

 ddxy = 4(2x – 5)3(2) Jadi/So, –10 = – 52
(x – 1)2
= 8(2x – 5)3
(x – 1)2 = 25
(c) Katakan/Let y = 6x2(1 – 3x)3
x – 1 = ±5
 ddx y =
6x2(3)(1 – 3x)2(–3) x – 1 = 5 atau/or x – 1 = –5
+ (1 – 3x)3(12x)
x = 6 x = –4
= –54x2(1 – 3x)2 + 12x(1 – 3x)3
Gantikan x = 6 dan x = –4 ke dalam a,
= (1 – 3x)2[–54x2 + 12x(1 – 3x)]
Substitute x = 6 and x = –4 into a,
= (1 – 3x)2(–54x2 + 12x – 36x2)
2(6) + 8 2(–4) + 8
= (1 – 3x)2(12x – 90x2) y = 6–1 y = –4 – 1

= 6x(1 – 3x)2(2 – 15x) = 4 = 0

(d) Katakan/Let y = 3x 5 Maka/Thus, A(6, 4), C(–4, 0).
2x –
(c) Persamaan tangen kepada lengkung
 ddxy = (2x – 5)(3) – 3x(2)
(2x – 5)2
pada titik A ialah

= 6x – 15 – 6x Equation of the tangent to the curve at
(2x – 5)2
point A is
y – 4 = – 25 (x – 6)
= – (2x1–5 5)2 y = – 25 x + 12
5
3. (a) y +4

A y = –  2 x + 32
5 5

CO Taxngen / Tangent Persamaan tangen kepada lengkung

Normal pada titik C ialah

Equation of the tangent to the curve at

point C is
y – 0 = – 25 [x – (–4)]
(b) y = 2x + 8 .........a
x–1
y = – 25 (x + 4)
 ddxy = (x – 1)(2) – (2x + 8)(1)
(x – 1)2
y = – 25 x – 8
= 2x – 2 – 2x – 8 5
(x – 1)2
Kecerunan normal/Gradient of normal
–10
= (x – 1)2   = –1
– 25
Tangen kepada lengkung pada A dan C
adalah selari dengan 2x + 5y − 20 = 0.
= 5
The tangents to the curve at A and C are 2
parallel to 2x + 5y − 20 = 0.

© Penerbitan Pelangi Sdn. Bhd. 64

Persamaan normal kepada lengkung (c) Gantikan x = q , y = –2 ke dalam
2
pada titik A ialah persamaan lengkung,

Equation of the normal to the curve at q
2
point A is 5 Substitute x = , y = –2 into the equation
2 of the curve,
y – 4 = (x – 6)
–2 q 2 k q 
y = 5 x – 15 + 4 –2 = 2 + 2 + 4
2
= – q22 q
y = 5 x – 11   –6 + k 2
2

Persamaan normal kepada lengkung Gantikan/Substitute k = 4p,

pada titik C ialah   –6 = – q22 + 4p q
2
Equation of the normal to the curve at –6 = – q22 + 2pq

point C is 5
2
y – 0 = [x – (–4)] –12 = –q2 + 4pq

y = 5 (x + 4) 4pq = q2 – 12
2
5 p = q2 – 12
y = 2 x + 10 4q

(d) Segi empat tepat. Kedua-dua tangen 5. y = 0.001x3 – 0.12x2 + 3.6x + 10
kepada lengkung di A dan C adalah
selari. Kedua-dua normal kepada  ddxy = 0.003x2 − 0.24x + 3.6
lengkung di A dan C adalah selari.
Tangen dan normal kepada lengkung Pada titik-titik paling tinggi dan paling
di A dan C adalah berserenjang antara
satu sama lain. rendah,

Rectangle. Both tangents to the curve at At the highest and lowest points,
A and C are parallel. Both normals to the
curve at A and C are parallel. The tangent  dy = 0
and normal to the curve at A and C are dx
perpendicular to each other.
0.003x2 − 0.24x + 3.6 = 0

3x2 − 240x + 3600 = 0

x2 − 80x + 1200 = 0

(x − 20)(x − 60) = 0

4. (a) A ialah titik maksimum kerana x = 20 atau/or x = 60
a = –2  0.
Apabila/When x = 20,
A is a maximum point because a = –2  0.
y = 0.001(20)3 − 0.12(20)2 + 3.6(20) + 10
(b) y = −2x2 + kx + 4 =42

 ddxy = –4x + k Apabila/When x = 60,

Pada titik pusingan A(p, q),  dy = 0. y = 0.001(60)3 − 0.12(60)2 + 3.6(60) + 10
dx =10

At the turning point A(p, q),  dy = 0. Beza jarak menegak
– 4p + k = 0 dx Difference in vertical distance

k = 4p = 42 – 10
= 32 m

65 © Penerbitan Pelangi Sdn. Bhd.

6. (a) y = −0.00002x3 + 0.006x2 (d) A = πr2
 ddxy = −0.00006x2 + 0.012x 314 = 3.14r2
r2 = 100
(b) −0.00006x2 + 0.012x = 0.45 r = 10 (abaikan/ignore r = −10)
0.00006x2 − 0.012x + 0.45 = 0
6x2 − 1200x + 45 000 = 0  dd Ar == 2π(10)
x2 − 200x + 7500 = 0 2(3.14)(10)
(x − 150)(x − 50) = 0
x = 150 atau/or x = 50 = 62.8

(c) Kecerunan/Gradient  0.45 Kadar perubahan A terhadap r apabila
–0.00006x2 + 0.012x  0.45 luas ialah 314 km2 ialah 62.8 km2 per
0.00006x2 – 0.012x + 0.45  0 km.
x2 – 200x + 7500  0
Rate of change of A with respect to r when
Jadi/So, 50  x  150 the area is 314 km2 is 62.8 km2 per km.

Apabila/When x = 50, 8. Isi padu kotak/Volume of box, V = (3x)(2x)(x)
y = −0.00002(50)3 + 0.006(50)2
= 12.5 = 6x3

Apabila/When x = 150,  ddVx = 18x2
y = −0.00002(150)3 + 0.006(150)2
= 67.5 Pertambahan kecil dalam x,

Julat tinggi bukit yang rumah tidak Small increase in x,
boleh dibina ialah
The range of height of the hill where houses δx = 1.5 × x
cannot be built is 100

12.5 m  y  67.5 m = 0.015x

Pertambahan hampir dalam isi padu,

Approximate increase in volume,

δV =  dV × δx
dx

= 18x2 × 0.015x

7. (a) A = πr2 = 0.27x3

Peratus pertambahan hampir dalam isi padu

(b)  dA = 2πr Approximate percentage increase in volume
dr
=  δV × 100%
V
(c) Apabila/When r = 2.5, 0.27x3
= 6x3 × 100%
 ddAr == 2π(2.5)
2(3.14)(2.5) = 4.5%

= 15.7

Kadar perubahan A terhadap r apabila

jejari ialah 2.5 km ialah 15.7 km2 per

km.

Rate of change of A with respect to r when

the radius is 2.5 km is 15.7 km2 per km.

© Penerbitan Pelangi Sdn. Bhd. 66

Soalan KBAT (c) δℎ = 1.95a − 2a

= −0.05a

1. (a) Daripada segi tiga serupa, Apabila/When h = 2a,  dA = –4πa
dh
From the similar triangles,  dA
dh
ar = h h a δA  × δh
–
 –4πa × (−0.05a)
r = ah  0.2πa2 cm2
h–a

= ah – a2 + a2 2. Katakan kadar perubahan isi padu air dalam
h –a
kon =  dV
= a(h – a) + a2 dt
h–a
Let the rate of change of the volume of water in

= a(h – a) + a2 the cone =  dV
h–a h–a dt
a2  ddVt = (10ℎ + 2) − 7
= a + h–a =10ℎ − 5

(b) A = πr2 Apabila/When ℎ = 5,
  =  ddVt = 10(5) − 5
π a2 2 = 45 cm3 s−1
h–a
a +

    ddhA = 2π a + a2 – (h a2
h–a – a)2
Katakan jejari permukaan air = r
  =
–2πa2 a + a2 Let the radius of water surface = r
(h – a)2 h–a  r
tan 30° = h .....a
 dh
Diberi/Given dt = –1.5 cm s−1 1 =  r
 3  h
Gunakan petua rantai,
h = 3r
Using the chain rule,

 dA =  dA ×  dh Isi padu air/Volume of water,
dt dh dt
  V = 1 πr2h
 ddAt = –2πa2 a + a2 × (−1.5) 3
(h – a)2 h–a 1
= 3 πr23r 

Gantikan/Substitute h = 2a,  3
3
   ddAt =
–2πa2 a + a2 a × (−1.5) = πr3
(2a – a)2 2a –
 ddVr = 3πr2
  =
–2πa2 a + a2 × (−1.5)
a2 a
Gunakan petua rantai/Using chain rule,
=−2π(2a) × (−1.5)
 ddVt =  dV  dr
=−4πa × (−1.5) dr × dt

= 6πa cm2 s−1 3πr2  dr
dt
45 = ×

 ddtr = 45
 3πr2

67 © Penerbitan Pelangi Sdn. Bhd.

Luas permukaan membulat air, 3. dy
Circular area of water surface, dx

A = πr2 15

 ddAr = 2πr 10

5

–15 –10 –5 O 5 10 15 x
–5
Gunakan petua rantai/Using chain rule,
–10 y1 = fЈ(x)
 ddAt =  dA  dr
dr × dt –15

= 2πr × 45
 3πr2
90
=  3r

Daripada/From a, r = h tan 30°

Apabila/When h = 5, r = 5
 3

   ddAt =90

3 5
 3

= 18 cm2 s−1

© Penerbitan Pelangi Sdn. Bhd. 68

Jawapan

Tingkatan 5 Pengamiran Bandingkan/Compare

BAB Integration (1 – 2 + 5)n – 1 + c
n)(2x
3 k
+
Masteri SPM dengan/with (2x 5)p + c,
Kertas 1
p = n – 1 ....a dan/and k = 1 2 n
–

1 – n = 2
k
1. (a) ∫ (2x + 3) dx = 2x2 + 3x + c1
2 n = 1 – 2 ...b
k
= x2 + 3x + c1
Gantikan b ke dalam a,
(2x + 3)2
(b) ∫ (2x + 3) dx = 2(2) + c2 Substitute b into a,
2
1 p = 1 – k –1
4
= (2x + 3)2 + c2 p = – 2k

(c) x 2 + 3x + c1 = 1 (2x + 3)2 + c2 3. (a) Jika/If u = ax + b,
4

= 1 (4x2 + 12x + 9) + c2  ddxu = a
4
9
= x2 + 3x + 4 + c2 ∫ (ax + b)n dx = ∫ un ×  dx du
du
c1 = 9 + c2  1
4 = ∫ un × a du

∫ =  1 × ∫ un du
2. (a)  x2 – 2x – 15 dx a
x–5
=  1 × un + 1 +c
∫  (x – 5)(x + 3) a n+1
= x–5 dx
=  1 × (ax + b)n + 1 + c
= ∫ (x + 3) dx a n+1

= ∫ x dx + ∫ 3 dx (b)  (8x∫ a = 8, b = 3, n = −21
+ ∫ (8x + 3)–2 dx
= 1 x2 + 3x + c 3)2 dx =
2 =  1 (8x + 3)–2 + 1
8 × –2 + 1 + c
∫(b)
 (2x 4 5)n dx = 4 ∫ (2x + 5)–n dx =  1 × (8x + 3)–1 + c
+ 8 –1

 = 5)–n + 1
 (22x(+1 – n) = – 8(8x1+ 3) + c
4 +c

= (1 – 2 + 5)n – 1 + c 4. (a) Graf bagi f(x) ialah garis lurus (linear)
n)(2x dengan kecerunan = 3.

The graph of f(x) is a straight line (linear)
with gradient = 3.

69 © Penerbitan Pelangi Sdn. Bhd.

y1 = fЈ(x)

(b) y 7. (a) ∫ 8x3 dx = 2x4 + 5 + c1,

y = f(x) f(x) = 3x + 1 dengan keadaan c1 ialah pemalar.
where c1 is a constant.
1
= 2x4 + c,

– 1 O x dengan keadaan c ialah pemalar.
3 y1 = fЈ(x)
where c is a constant.

5. y y = 4x2 + 4 (c1 ialah pemalar, 5 + c1 juga ialah
y = 4x2 + 2 pemalar, maka kita gantikan 5 + c1
dengan c supaya keputusan kelihatan
y = 4x2 lebih mudah./c1 is a constant, 5 + c1 is also
a constant, then we replace 5 + c1 by c so
(0, 4) (1, 4) that the result looks simpler.)

(0, 2) x

O

y = 4x2 – 4 ∫ k 9
4
(0, –4) (b)    x2 dx = 48

0  

[Terima jawapan lain yang betul]   94 × x3 k
[Accept other correct answers] 3
= 48

0

6. (a) Apabila tangen kepada lengkung pada   3 x3 k = 48
4 0
titik (−2, 14) selari dengan paksi-x,
x30k = 64
When the tangent to the curve at the point

(−2, 14) is parallel to the x-axis, k 3 – 0 = 64

 ddxy = 0  di/at x = –2 k = 4

k(–2) – (–428)3 = 0 k = 16

k = 3 8. (a) ∫05 h(x) dx = ∫02 h(x) dx + ∫25 h(x) dx

(b)  ddxy = 48 =4+2
x3
3x – =6

∫  y =   3x – 48 dx ( b) ∫05 3h∫0(5x [)3hd(xx)++∫05k k] dx = 28
x3 dx = 28

= ∫ (3x – 48x–3) dx 3 ∫05 h(x) dx + [kx]05 = 28

= 3x2 – 48x–2 + c 3(6) + (5k − 0) = 28
2 –2
18 + 5k = 28

= 3x2 + 24 + c 5k = 10
2 x2
k = 2

Gantikan/Substitute x = –2 dan/and y = 14, 9. (a) ∫12 f(x) dx = 9

14 = 3(–2)2 + 24 + c [x3 + kx + c]12 = 9
2 (–2)2
(23 + 2k + c) − (13 + k + c) = 9

c = 2 7 + k = 9

Persamaan lengkung ialah k = 2
(b) ∫ f(x) dx = x3 + 2x + c
Equation of the curve is  d
dx (x3
y = 3x2 + 24 + 2 f(x) = + 2x + c)
2 x2
f(x) = 3x2 + 2

© Penerbitan Pelangi Sdn. Bhd. 70

10. (a) Luas rantau berlorek   13. (a) d dx1g(x) =f(x)
Area of shaded region 3
1
A =  ∫02 g(x) dx  ∫ab f(x) dx = 3 [g(x)]ab

(b) Luas rantau berlorek = 1 [g(b) – g(a)]
Area of shaded region 3
1
A + B = 24 – 8 = 3 (–16)
= 16 unit2
16
11. (a) Graf = –   3
Graph
y = f(x) y = g(x) = –5 1
 dy ✓✓ 3
dx  0
✓ Luas kawasan berlorek ialah 5 1 unit2.
✓ Area of the shaded region 3
 dy 1
dx = 0 is 5 3 unit2.

 dy  0 (b) f(x) = ∫(3x2 − 4x − 5) dx
dx
= 3x3 – 4x2 – 5x + c
3 2
 d 2y
dx2  0 = x3 − 2x2 − 5x + c

 d 2y Graf y = f(x) melalui titik (−1, 8),
dx2
= 0 Graph y = f(x) passes through point (−1, 8),

f(–1) = 8

 d 2y  0 (−1)3 – 2(−1)2 – 5(−1) + c = 8
dx2
c = 6

(b) ∫12 f(x) dx mempunyai nilai yang lebih Maka/Therefore, f(x) = x3 − 2x2 − 5x + 6.

besar. Graf y = f(x) berada di atas graf Kertas 2
y = g(x) antara x = 1 dan x = 2.
1. (a) y = x2(1 − 3x)4
∫12 f(x) dx has the larger value. The graph
Katakan/Let u = x2
of y = f(x) is above the graph of y = g(x)
between x = 1 and x = 2. maka/then  ddxu = 2x

12. (a) Rantau/Region B dan/and v = (1 – 3x)4
(b) Rantau/Regions A, B, C
(c) Rantau/Region A maka/then  ddxv = 4(1 − 3x)3(−3)
= −12(1 − 3x)3

 ddxy = u ddxv + v ddxu
= x2(−12)(1 − 3x)3 + (1 − 3x)4(2x)

= −12x2(1 − 3x)3 + 2x(1 − 3x)4

= −2x(1 − 3x)3[6x − (1 − 3x)]

= −2x(1 − 3x)3(9x − 1)

= −2x(9x − 1)(1 − 3x)3

∴ k = −2

71 © Penerbitan Pelangi Sdn. Bhd.

(b) ∫ −2x(9x − 1)(1 − 3x)3 dx = x2(1 − 3x)4 + c
∫ x(9x − 1)(1 − 3x)3 dx = − 21 x2(1 − 3x)4 + c
∫–11 x(9x − 1)(1 − 3x)3 dx = − 12 [x2(1 − 3x)4]−11

= − 21 {(12)[1 − 3(1)]4 − (−1)2[1 − 3(−1)]4}
= 120

2.  ddx 2y2 = 24x2 − 2 ( b)  dd M t == t(6 – t)
6t – t2
 ddx y = ∫ (24x2 − 2) dx
M = ∫ (6t – t2) dt

= 24x3 – 2x + c1 = 62t2 – t3 + c
3 3

= 8x3 − 2x + c1 Diberi/Given M = 0 apabila/when t = 0,

 d y maka/therefore, c = 0.
dx
Oleh sebab/Since = 5 apabila/when x = 1, M = 6t2 – t3
2 3
  5 = 8(1)3 − 2(1) + c1
Di akhir tindak balas, t = 6,
c1 = −1
At the end of reaction, t = 6,
 d y
∴ dx = 8x3 − 2x − 1 M = 6(6)2 – (6)3 = 36 g
− 2x − 2 3
y = ∫ (8x3 1) dx
Maka, jisim kuprum yang dihasilkan
y = 8x4 – 2x2 – x + c2 dalam tindak balas itu ialah 36 g.
4 2
y = 2x4 − x2 − x + c2 Therefore, the mass of copper produced in
the reaction is 36 g.
Titik (1, 4) terletak di atas lengkung,
Point (1, 4) lies on the curve, 4. (a) Koordinat/Coordinates of Q = (ℎ, ℎ2 + 2)

  4 = 2(1)4 − (1)2 − 1 + c2 Kecerunan tangen/Gradient of tangent PQ
c2 = 4
= ℎ2 +2 – k
h
Jadi, persamaan lengkung ialah

So, the equation of the curve is y = x2 + 2

y = 2x4 − x2 − x + 4 y = fЈ(x)  d y = 2x
1 dx

3. (a)  d M  0 Di titik/At point Q, x = h,
dt
t(6 − t)  0
∴  d y = 2h
dx
× (−1), t(t − 6)  0
ℎ2 +2 – k =
Maka/Therefore, h 2h

t ℎ2 + 2 – k = 2h2
06

k = 2 – h2

Maka, tempoh tindak balas kimia itu
ialah 6 saat.

Therefore, the duration of the reaction is
6 seconds.

© Penerbitan Pelangi Sdn. Bhd. 72

(b) Luas di bawah lengkung dari x = 0 Pada/At B(1, 0),

hingga x = h  ddx y = 6(1) − 6(1)2 − 4(1)3
= –4
Area under the curve from x = 0 to x = h

= ∫0h (x2 + 2) dx Oleh sebab  d y di x = –2 dan x = 1
dx
  = x3 + h
3 sama dengan kecerunan garis lurus AB,
2x
maka garis lurus AB ialah tangen
0

  = h3 + 2h −0 kepada lengkung di kedua-dua titik A
3
dan B.
= h3 + 2h  d y
3 Since dx at x = –2 and x = 1 is equal

Luas trapezium/Area of trapezium to the gradient of the straight line AB,
1
= 2 (k + ℎ2 + 2)(ℎ) therefore the straight line AB is a tangent

1 to the curve at both points A and B.
2
= (2 − ℎ2 + ℎ2 + 2)(ℎ) (c) y

A(–2, 12)

= 2ℎ

Luas rantau berlorek y = x2(3 – 2x – x2)
C
Area of shaded region

= luas di bawah lengkung – luas O B(1, 0) x

trapezium

area under the curve – area of trapezium Luas rantau yang dikehendaki
= h33 + 2h – 2ℎ Area of required region
= h33 unit2
= luas segi tiga ABC – luas di bawah
lengkung dari x = −2 hingga x = 1

5. (a) y = x2(3 − 2x − x2) area of triangle ABC − area under the
= 3x2 − 2x3 − x4
 ddx y = 6x − 6x2 − 4x3 curve from x = −2 to x = 1

=  21 [1 – (–2)](12) – ∫–12  (3x2 – 2x3 – x4) dx
  x5 1
(b) Kecerunan garis lurus AB, = 18 – 3x3 – 2x4 – 5 –2
3 4
Gradient of straight line AB,
  x4 x5 1
mAB = 12 – 0 = –4 = 18 – x3 – 2 – 5 –2
–2 – 1

Kecerunan tangen kepada lengkung,    13 – 14 – 15 – (–2)3 – (–2)4
= 18 – 2 5 2
Gradient of the tangent to the curve,
 d y  –
dx = 6x − 6x2 − 4x3 (–2)5
5

Pada/At A(–2, 12),   3 + 9 3
= 18 – 10 5
 d y
dx = 6(–2) − 6(–2)2 − 4(–2)3 = 8110 unit2

= –4

73 © Penerbitan Pelangi Sdn. Bhd.

6. (a)  ddx y = −4x + 10 Gantikan d ke dalam c,
Substitute d into c,
y = ∫(−4x + 10) dx
−4a3 + 30a2 − 3a(−2a2 + 10a) = 54
= − 42x2 + 10x + c
= −2x2 + 10x + c .....a −4a3 + 30a2 + 6a3 − 30a2 = 54

2a3 = 54

a3 = 27

Gantikan x = 0, y = 0 ke dalam a, a = 3
Substitute x = 0, y = 0 into a,
Gantikan a = 3 ke dalam d,
0 = −2(0)2 + 10(0) + c Substitute a = 3 into d,

c = 0 b = −2(3)2 + 10(3)
= −18 + 30
Maka, persamaan lengkung ialah = 12
So, the equation of the curve is
Jadi, koordinat titik P ialah (3, 12).
y = –2x2 + 10x ..........b Therefore, the coordinates of point P is
(3, 12).
(b) Luas di bawah lengkung dari x = 0

hingga x = a

Area under the curve from x = 0 to x = a (c) Apabila lengkung memotong paksi-x,
= ∫0a (–2x2 + 10x) dx
y y = 0.

  = –2x3 + 10x2 a P(a, b) When the curve cuts the x-axis, y = 0.
3 2 0
  = 0 = −2x2 + 10x
–2a3
3 + 5a2 – 0 2x2 – 10x = 0

–2a3 O x x2 − 5x = 0
3 Q
= + 5a2 x(x − 5) = 0

1 x = 0  atau/or  x = 5
2
Luas/Area of ∆OPQ = × a × b Isi padu janaan/Volume generated

= 1 ab = π ∫05 y2 dx
2 = π ∫05 (−2x2 + 10x)2 dx

Diberi luas yang dibatasi oleh lengkung = π ∫05 (4x4 − 40x3 + 100x2) dx
dan garis lurus OP
Given the area bounded by the curve and   = 5
the line OP π 4x5 – 40x4 + 100x3 0
5 4 3
= 9 unit2
  = 4x5 5
–2a3 1 π 5 – 10x4 + 100x3 0
3 2 3
+ 5a2 – ab = 9
  =
−4a3 + 30a2 − 3ab = 54 ........c π 4(5)5 – 10(5)4 + 100(5)3 –0
5 3

  =
Oleh sebab P(a, b) terletak di atas π 2500 − 6250 + 12 500
lengkung, gantikan x = a dan y = b ke 3
dalam b, 2
= 416 3 π unit3
Since P(a, b) lies on the curve, substitute
x = a and y = b into b,

b = −2a2 + 10a ........d

© Penerbitan Pelangi Sdn. Bhd. 74

7. (a) Diberi persamaan lengkung Apabila lengkung memotong paksi-y,
Given the equation of curve
x = 0 dan y = 1 (0)2­ + 1 = 1
y = px2 + q. ........a 2
When the curve cuts the y-axis, x = 0 and

Oleh sebab A(2, 3) terletak di atas y = 1 (0)­2 + 1 = 1
lengkung, 2

Since A(2, 3) lies on the curve, Isi padu janaan/Volume generated
3 = p(2)2 + q
4p + q = 3  ................b = π ∫13 x2 dy

Kecerunan normal/Gradient of normal PQ, = π ∫13 (2y – 2)dy

mPQ = 3–0 = – 12   = 2y2 – 2y 3
2–8 2 1

Maka, kecerunan tangen ialah 2 apabila = π{[(3)2 − 2(3)] − [(1)2 − 2(1)]}

x = 2. = 4π unit3

Therefore, the gradient of the tangent is 2 8. (a) a2 = 2a + 1

when x = 2. (b) π ∫0a [(2x + 1) − x2] dx

Daripada/From a,  d y = 2px (c) V = π∫0a [(2x + 1) − x2] dx
dx

Apabila/When x = 2,  d y = 2. = π∫0a(2x + 1 − x2) dx
dx
2 = 2p(2)   = a
π x2 + x – x3 0
p = 1 3
2
1    = π a3
Gantikan p = 2 ke dalam b, a2 + a – 3 −0

Substitute p = 1 into b,   = 1 a3 π unit3
2 a2 + a – 3
  4
1 + q = 3 9. (a) y B
2 q = 1
r
(b) Luas rantau berlorek y = r x A
Area of shaded region h

Oh x

  = ∫02  1 x2 + 1 dx + 1 (8 – 2)(3)
2 2
r
    =1 x3 2 Persamaan garis lurus OB ialah y= h x.
2 3 OB is y =
+x +9 Equation of straight line r x.
dx h
0 V = π ∫0h y2

  = 1 (2)3 + 2 – 0 + 9
6
∫   = π h r
1 h 2
3 x dx
= 12 unit2

0

(c) Daripada persamaan lengkung, ∫ = π h r2 x2 dx
0 h2
From the equation of curve, h
  = 0
y = 1 x2 + 1 πr2 x3
2 h2 3
12 x2 = y – 1
  = πr2 h3 –0
x2 = 2y – 2 h2 3

= 1 πr2h
3

75 © Penerbitan Pelangi Sdn. Bhd.

(b) Isi padu dijana lengkok AB Soalan KBAT
Volume generated by arc AB

 = π ∫–11 1 x2 + 1 2 dx 1. y2 = 4x
4
 = π ∫–11  Apabila/When y = 2,
1 x4 + 1 x2 + 1 dx 22 = 4x
16 2 x = 1
1
     = π1x5 + 1 x3 Diberi isi padu cawan/Given volume of cup,
16 5 2 3 +x

–1
1
 = π 1 x5 + 1 x3 V = 32π unit3
80 6 +x
π ∫1a  y2 dx = 32π y
–1 ∫1a 4x dx = 32 y2 = 4x

 = π1 (1)5 + 1 (1)3 + (1) [2x2]1a = 32 2
80 6 2(a2 − 12) = 32
O1
 –1 1 x
80 (–1)5 + 6 (–1)3 + (–1) a

283 a2 − 1 = 16
120
= π unit3 a2 = 17

Isi padu dijana lengkok BC a = 17
Volume generated by arc BC
Apabila/When x = 17, y2 = 417

  = π ∫19  5 x  2 dx y = 4.06
4
25 Maka, diameter bibir cawan
= 16 π ∫19  x dx Therefore, the diameter of rim of the cup

  = 25 π x2 9 = 2 × 4.06
16 2 1 = 8.12 unit/units

25
= 32 π –
(81 1)

= 125 π unit3
2

Isi padu pasu
Volume of the vase

= 283 π+ 125 π
120 2

= 64 103 π unit3
120

© Penerbitan Pelangi Sdn. Bhd. 76

Jawapan

Tingkatan 5 Pilih Atur dan Gabungan 7. (a) nC1 = n! = n
– 1)!1!
BAB Permutation and Combination (n

4 (b) (i) 7C4 = 35

Masteri SPM (ii) 5C1 × 7C3 = 5 × 35

= 175

Kertas 1 Kertas 2

1. (a) 6! = 720 1. (a) 18C4 = 3060
(b) 4P2 × 4! = 12 × 24 (b) 8! – (7! × 2!) = 30 240
= 288
2. (a) 12C5 = 792

2. (a) 8P5 = 6720 (b) 12C9 + C12 + C12 + C12
10 11 12
(b) 5 × 6P3 × 3
= 5 × 120 × 3 6P3 = 220 + 66 + 12 + 1
.................................
= 1800 = 299

5 cara 3 cara Soalan KBAT
5 ways 3 ways
(K, M, P ,T, R) (O, U, E)

3. (a) K, A, N secaman/identical. 1. 10 × 10 × 10 × 10 × 10 = 100 000
14! Kes dengan/Cases with 123:
2!2!4! = 908 107 200

(b) K, E, BARANG, K, A, L, I, A, N: 123 10 × 10
9 objek/objects; K, A secaman/identical.
123 10 × 10 Tiada kes berulang
9! No duplicate case
2!2!
= 90 720 1 2 3 10 × 10

4. (a) (8 – 1)! = 7! Bilangan kata laluan
Number of passcodes
= 5040
= 100 000 – (10 × 10) – (10 × 10) – (10 × 10)
(b) (5 – 1)! × 4! = 4! × 4!
= 24 × 24 = 99 700
= 576
2. (a) 72!! = 2520
5. (a) 8! = 40 320 5C3 ×
(b) 4C4 × 4C2 = 1 × 6 (b) 5! + 5C4 × 5! + 2! 5!
= 6

6. (a) 10C6 = 210 = 120 + 600 + 600
(b) (8C3 × 10C3) + (8C4 × 10C2) × (8C5 × 10C1)
+ 8C6 = 1320 5! : Tiada huruf E.
= (56 × 120) + (70 × 45) + (56 × 10) No letter E.
+ 28 5C4 × 5! : E + 4 huruf lain.
5C32×! 5! : 2E2EE+++43o3othhtheuerrrluelfettteltaersirns. .
= 6720 + 3150 + 560 + 28

= 10 458

77 © Penerbitan Pelangi Sdn. Bhd.

3. (a) nC2 = nP2 nP2 = n! Maka, bilangan pepenjuru
2! (n – 2)!
n(n – 1)(n – 2)! Hence, the number of diagonals
(n – 2)! n(n – 1)
= n(n – 1) = = 2 – n
2×1
n(n – 1) = n(n –1) n2 – n 2n
2 2 2
= = –

(b) Bilangan sisi/Number of sides = n = n2 – 3n
2
Maka, bilangan bucu = n
Hence, number of vertices = n = n(n – 3)
2
Dua bucu disambung untuk membentuk
satu garis. Garis-garis yang dibentuk itu (c) Bilangan pepenjuru bagi poligon sekata
ialah pepenjuru dan sisi poligon sekata
itu. dengan 30 sisi
Two vertices are joined to form a line. Those
lines that are formed are the diagonals and Number of diagonals of a 30-sided regular
the sides of the regular polygon.
polygon

= 30(30 – 3)
2

Bilangan garis boleh dibentuk = 30(27)
2
Number of lines that can be formed
= 405
= nC2

= n(n – 1)
2

© Penerbitan Pelangi Sdn. Bhd. 78

Jawapan

Tingkatan 5 Taburan Kebarangkalian (b) P(X  2)

BAB Probability Distribution = 1 – P(X = 0) – P(X = 1)

5         0 8
= 1 – 8C0 2 1 – 8C1 2117
3 3 33

Masteri SPM = 1 – 0.0001524 – 0.002439

= 0.9974

Kertas 1 4. (a) P (X = 3) = 1
27
1. (a) h = 1 – 216 – 216 – 81 – 16 1
625 625 625 625 3C3p3q0 = 27
96
= 625 p3 = 1
27
(b) P(X  2) 1
p = 3
= 1 – P(X = 0) – P(X = 1)

= 1 – 16 – 96 (b) q = 1 – 1 = 2
625 625 3 3

= 0.8208 Bilangan bateri yang masih berfungsi

Kaedah Alternatif selepas satu tahun

P(X  2) Number of batteries which are still
= P(X = 2) + P(X = 3) + P(X = 4)
functioning after one year

= 216 + 216 + 81 = 15 × 2 = 10
625 625 625 3

= 0.8208 5. (a) h + k + P(X = 0) + P(X = 3) = 1

h + k + P(X  1) + P(X = 3) = 1

2. n = 10, p = 75% = 3 , q = 1 P(X = 3) + P(X < 1) = 1 – h – k
4 4
(b) P (X = 3) = 27
P(X > 7) 512

= P(X = 8) + P(X = 9) + P(X = 10) 3C3p3q0 = 27
512
            =
10C8  3 8 1 2 3 9 1 1 3 10 1 0
4 4 4 4 4 4
 + 10C9   + 10C10  p3 = 27
512
= 0.2816 + 0.1877 + 0.05631

= 0.5256 p = 3
8
3. (a) σ = npq
6. (a) (i) µ = 0 (ii) σ = 1
 = 8 2 1 2 
3 – 3 (b) Luas rantau berlorek

= 8 23  31  Area of shaded region

= P(–2σ  Z  0)

 = 16 = P(–2  Z  0)
9
= P(0  Z  2)

4 = 0.5 – P(Z  2)
3
= = 0.5 – 0.0228

= 0.4772

79 © Penerbitan Pelangi Sdn. Bhd.

7. (a) Z = X–µ 9. (a) µ = 13
σ
(b) P(6  X  20) = 1 – 0.21 – 0.21
1.5 = 30 – 20 = 0.58
σ
1.5σ = 10 10. (a) P(X  µ + 4) = 0.15

σ = 6.667   P Z 
µ+4–µ = 0.15
σ
(b) P(X > h) = 0.2420  P
4
  P Z  Z  σ = 0.15
h – 20 = 0.2420
6.667 4
σ = 1.037
h – 20 = 0.7
6.667 σ = 3.857

h – 20 = 4.6669 (b) 7:20 a.m. = 7:12 a.m. + 8 minit/minutes

h = 24.67 µ+8

8. (a) P(k  Z  0) = 0.3106  
P(X  µ + 8) = P Z  µ+8–µ
0.5 – P(Z  k) = 0.3106 3.857

0.5 – P(Z  –k) = 0.3106  =
8
P(Z  –k) = 0.1894 P Z  3.857

–k = 0.88 = P(Z  2.074)

k = –0.88 = 0.019

(b) σ2 = 9.61

σ = 9.61 = 3.1

45.36.1– µ = –0.88
45.6 – µ = –2.728
µ = 48.33

Kertas 2

1. (a) (i) np = 6
12p = 6
p = 0.5

(ii) P(X  3)
= P(X = 3) + P(X = 2) + P(X = 1) + P(X = 0)
= 12C3 (0.5)3(0.5)9 + 12C2 (0.5)2(0.5)10 + 12C1 (0.5)1(0.5)11 + 12C0 (0.5)0(0.5)12
= 0.05371 + 0.01611 + 0.002930 + 0.0002441
= 0.073

 (b) (i)
P(X > 7) = P Z  7–6 (ii) P(X  h) = 0.66
1.8
1 – P(X  h) = 0.66
= P(Z  0.5556)
P(X  h) = 0.34
= 0.2892   P Z 
h–6 = 0.34
1.8

h–6 = 0.413
1.8
h − 6 = 0.7434

h = 6.74

© Penerbitan Pelangi Sdn. Bhd. 80

2. (a) (i) P(X = 8) = 10C8(0.75)8(0.25)2

= 0.2816

(ii) p = 0.25, q = 0.75, n = 10

P(X  3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 10C0(0.25)0(0.75)10 + 10C1(0.25)1(0.75)9 + 10C2(0.25)2(0.75)8 + 10C3(0.25)3(0.75)7
= 0.05631 + 0.1877 + 0.2816 + 0.2503

= 0.7759

 (b) (i)  P Z  3 – 2.1 (ii) P(X  h) = 15 = 0.025
P(X 3) = 0.4 600

= P(Z > 2.25)   P Z  h – 2.1 = 0.025
0.4
= 0.0122   P Z
2.1 – h
 0.4 = 0.025

2.1 – h = 1.96
0.4

2.1 – h = 0.784

h = 1.316

3. (a) p = 2 = 0.4, q = 0.6, n = 12
5

P(X  9) = P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)

= 12C9(0.4)9(0.6)3 + 12C10(0.4)10(0.6)2 + 12C11(0.4)11(0.6)1 + 12C12(0.4)12(0.6)0
= 0.01246 + 0.002491 + 0.0003020 + 0.00001678

= 0.01527

 (b) (i) 4. (a) (i) P(X = 6) = 8C6(0.4)6(0.6)2
P(X  50) = P Z  50 – 60
12 = 0.04129

= P(Z  –0.8333) (ii) P(X  7) = 1 – P(X = 8)

= 1 – P(Z  –0.8333) = 1 – 8C8(0.4)8(0.6)0
= 1 – 0.0006554
= 1 – P(Z  0.8333)

= 1 – 0.2025 = 0.9993

= 0.7975  (b) (i)
P(X > 7.5) = P Z  7.5 – 6.2
Bilangan murid yang lulus ujian 1.5

itu/ Number of students who passed = P(Z > 0.8667)

the test = 0.7975 × 400 = 0.1930

= 319 (ii) P(X  h) = 0.7881

(ii) P(X  m) = 0.15  
PZ h – 6.2 = 0.7881
  P Z  1.5
m – 60 = 0.15   1
12 h – 6.2
m – 60 – PZ 1.5 = 0.7881
12 = 1.036
  1
m – 60 = 12.43 – PZ 6.2 – h = 0.7881
1.5
m = 72.43  
PZ 6.2 – h = 0.2119
Markah minimum untuk gred 1.5
cemerlang/Minimum mark for
excellent grade = 73 6.2 – h = 0.8
1.5

6.2 – h = 1.2

h = 5

81 © Penerbitan Pelangi Sdn. Bhd.

5. (a) (i) µ = np = 235

σ2 = npq
= np(1 – p)

∴ 141 = 235(1 – p)

141 = 235 – 235p

235p = 94

p = 0.4

(ii) P(X  7) = P(X = 8) + P(X = 9) + P(X = 10)

= 10C8(0.4)8(0.6)2 + 10C9(0.4)9(0.6)1 + 10C10(0.4)10(0.6)0

= 0.01062 + 0.001573 + 0.0001049

= 0.0123

 (b) (i) (ii) P(X > h) = 0.3
P(X < 152) = P Z  152 – 156
12   P Z 
h – 156 = 0.3
= P(Z < –0.3333) 12

= P(Z > 0.3333) h – 156 = 0.524
12
= 0.3696 h – 156 = 6.288

h = 162.3

6. (a) (i) 6C6 p6q0 = 0.046656 1
p6 = 0.046656
p = 60.046656 atau/or 0.0466566

= 0.6

(ii) P(X  4) = P(X = 5) + P(X = 6)

= 6C5(0.6)5(0.4)1 + 6C6(0.6)6(0.4)0
= 0.1866 + 0.04666


= 0.2333

(b) (i) P(X  M) = 0.283

  P Z 
M – 750 = 0.283
120

M – 750 = 0.574
120

M – 750 = 68.88

M = 818.9 g

  (ii) P(700
 X  780) = P 700 – 750  Z  780 – 750
120 120

= P(–0.4167  Z  0.25)

= 1 – P(Z  –0.4167) – P(Z  0.25)

= 1 – P(Z  0.4167) – P(Z  0.25)

= 1 – 0.3384 – 0.4013

= 0.2603

© Penerbitan Pelangi Sdn. Bhd. 82

7. (a) P(X = 2) = 6C2(0.3)2(0.7)4

= 0.3241

(b) (i) P(X > 3) = 0.188

  P Z 
3 – 2.5 = 0.188
h
3 – 2.5
h = 0.885

0.885h = 0.5

h = 0.565

  (ii) P(2
 X  3) = P 2 – 2.5  Z  3 – 2.5
0.565 0.565
= P(–0.885  Z  0.885)


= 1 – P(Z  –0.885) – P(Z  0.885)

= 1 – P(Z  0.885) – P(Z  0.885)

= 1 – 0.188 – 0.188

= 0.624

Bilangan durian yang mempunyai jisim antara 2 kg dan 3 kg
Number of durians that have masses between 2 kg and 3 kg

= 0.624 × 2558
 1596

8. (a) (i) µ = 3250

P(X  4230) = 0.1103

  P Z4230
 – 3250 = 0.1103
σ

 P Z  980 = 0.1103
σ

980 = 1.225
σ
σ = 800

  (ii) P(2500  X  3500) = P
2500 – 3250 Z 3500 – 3250
800 800

= P(–0.9375  Z  0.3125)

= 1 – P(Z  –0.9375) – P(Z  0.3125)

= 1 – P(Z  0.9375) – P(Z  0.3125)

= 1 – 0.1742 – 0.3772

= 0.4486

Bilangan pekerja yang menerima gaji antara RM2500 dan RM3500

Number of employees who receive salary between RM2500 and RM3500

= 0.4486 × 50
 22

83 © Penerbitan Pelangi Sdn. Bhd.

(b) P(X < k) = 0.2   2. (a) (i)
P(X  3.5) = P Z  3.5 – 3.2
  P Z  0.9
k – 3250 = 0.2
800 = P(Z  0.3333)
  P Z 
3250 – k = 0.2 = 0.3696
800
(ii) Katakan/Let
3250 – k = 0.842 jisim buah tembikai Gred II
800
mass of Grade II watermelons
3250 – k = 673.6 = h  X  3.5

k = 2576.4 jisim buah tembikai Gred III
mass of Grade III watermelons
Soalan KBAT =Xh

1. (a) P(memenangi hadiah/wins a prize) P(X < h) = 0.25

7 7 7   P Z 
= 10 × 10 × 10 h – 3.2 = 0.25
0.9
343  
= 1000 PZ 3.2 – h = 0.25
0.9
= 0.343
3.2 – h = 0.674
0.9
Bilangan permainan/Number of games
= 24 ÷ 4 3.2 – h = 0.6066
=6
h = 2.5934

P(X  1) = 1 – P(X = 0) Jisim minimum buah tembikai Gred
II ialah 2.5934 kg.
= 1 – 6C0(0.343)0(0.657)6 The minimum mass of a Grade II
watermelon is 2.5934 kg.
= 1 – 0.08043

= 0.9196

Ali mempunyai 91.96% peluang untuk (b) (i) P(X = 1) = 8P(X = 0)
memenangi sekurang-kurangnya satu
hadiah. Maka, Ali patut mencuba nC1(0.2)1(0.8)n – 1 = 8nC0(0.2)0(0.8)n
permainan itu. n(0.2)(0.8)n – 1 = 8(0.8)n

Ali has 91.96% chance of winning at least 0.2n = 0.8n
one prize. Hence, Ali should play the game. 8 0.8n – 1

= 0.8n – (n – 1)

(b) np = 3 = 0.8
n(0.343) = 3
n = 8.75 0.2n = 6.4

n = 32
(ii) σ = npq
Bilangan minimum permainan yang Ali
perlu cuba ialah 9. = (32)(0.2)(0.8)

The minimum number of games Ali needs = 5.12
to play is 9.
= 2.263

© Penerbitan Pelangi Sdn. Bhd. 84

Jawapan

Tingkatan 5 Fungsi Trigonometri

BAB Trigonometric Functions

6

Masteri SPM

Kertas 1
1. (a) tan θ = – 1p

(b) OA = p2 + 12
= p2 + 1

kosek (90o − θ) = sek θ    /  cosec (90o − θ) = sec θ

=  p2 + 1
p

2. y

12 13 x
x

–5 O

(a) sin x = 12
13

(b) kot (180o + x) = tan 1 + x)   /  cot (180o + x) = tan 1 + x)
(180° (180°

= 1 x x berada dalam sukuan II dan (180° + x) berada dalam sukuan IV dengan
tan sudut rujukan yang sama. Jadi, tan (180° + x) = tan x.
x lies in quadrant II and (180° + x) lies in quadrant IV with the same reference angle.
= 1 So, tan (180° + x) = tan x.
– 152

= – 152

85 © Penerbitan Pelangi Sdn. Bhd.

3. (a) (i) k = −1 (ii) p = 3
4

(b) 2 kos px = 1 2 cos px = 1
1
kos px = 1 cos px = 2
2
3 3
3 kos px = 2 3 cos px = 2

3 kos px − 1 = 3 − 1 3 cos px − 1 = 3 − 1
2 2

3 kos px – 1 = 1 3 cos px – 1 = 1
2 2

Garis lurus y = 1 bersilang dengan y = 3 kos px – 1 pada satu titik, maka bilangan
2
penyelesaian bagi 2 kos px = 1 untuk 0  x  2π ialah 1.
1
The straight line y = 2 intersects y = 3 cos px – 1 at one point, therefore the number of solutions for

2 cos px = 1 for 0  x  2π is 1.

4. (a) sek (180o – θ) = 1 θ) sec (180o – θ) = 1
kos (180° – cos (180° – θ)

= 1 = 1
–kos θ –cos θ

= –   1 = –   1
h h

(b) n = 7, r = 360° = 72o
5

5. (a) y = | n tan x kos2 x | y =  | n tan x cos2 x |

  =  n sin x (kos2 x)    =  n sin x (cos2 x) 
kos x cos x
=  | n sin x cos x |
= | n sin x kos x | =  n2 (2 sin x cos x)
=  n2 sin 2x 
=  n2 (2 sin x kos x) 
=  2n sin 2x 

Amplitud/Amplitude, m = n .
2

(b) 0 < k < m

6. y

p2 + 1 p

A x
O1

(a) sin (180o + A) = −sin A
= − p2 p+ 1

© Penerbitan Pelangi Sdn. Bhd. 86

(b) sek 2A = 1 1
kos 2A sec 2A = cos 2A

= 2 1 – 1 = 2 cos21A – 1
kos2 A

  = 1 2
1
2 –1
p2 + 1

= 1 –1
2

p2 + 1

= 2 p2 + 1 1)
– ( p2 +

= p2 + 1
1 – p2

7. y

45 x
θ

–3 O

(a) tan (315o – θ) = tan 315° – tan θ
1 + tan 315° tan θ

 = –1 – – 34
 1 + (–1) – 34
1
= 7

(b) kos θ = 2 kos2 1 θ – 1 cos θ = 2 cos2 1 θ – 1
2 2
 cos
 kos 1 cos θ + 1
1 θ = kos θ + 1 2 θ = 2
2 2
 = – 35 + 1
 = – 53 + 1 2
2
= 0.4472

= 0.4472

8. (a) tan x = 1 x  / tan x = 1
kot cot x

= 1
k

87 © Penerbitan Pelangi Sdn. Bhd.

(b) y

–k x x
O

–1
k2 + 1

kos 2x = 2 kos2 x – 1  /  cos 2x = 2 cos2 x – 1

 
= 2 –k 2 1
k2 + 1
–

= 2k2 –1
k2 + 1

= 2k2 – (k2 + 1)
k2 + 1

= k2 – 1
k2 + 1

9. sek2 x + 3 sek x = 10 sec2 x + 3 sec x = 10

sek2 x + 3 sek x – 10 = 0 sec2 x + 3 sec x – 10 = 0

(sek x − 2)(sek x + 5) = 0 (sec x − 2)(sec x + 5) = 0

sek x = 2  ,  sek x = −5 sec x = 2  ,  sec x = −5
1 – 51
kos x = 1 kos x = – 15 cos x = 2 cos x =
2

  x = 60o, 300o x = 101.54o, 258.46o

10. kot2 x − 3 kosek x + 1 = 0 cot2 x − 3 cosec x + 1 = 0

( kosek2 x – 1) – 3 kosek x + 1 = 0 (cosec2 x – 1) – 3 cosec x + 1 = 0

kosek2 x – 3 kosek x = 0 cosec2 x – 3 cosec x = 0

kosek x(kosek x – 3) = 0 cosec x(cosec x – 3) = 0

kosek x = 0, kosek x = 3 cosec x = 0, cosec x = 3

(tiada penyelesaian bagi kosek x = 0) (no solution for cosec x = 0)

∴ kosek x = 3 ∴ cosec x = 3

sin x = 1
3

x = 19.47o, 160.53o

Kertas 2 y = 4 – 2x
1. (a) y π

4 π 2π x

O y = 4 sin 3 x
–4 2

© Penerbitan Pelangi Sdn. Bhd. 88

(b) sin 3 x + x = 1
2 2π

sin 3 x = 1 – x
2 2π

4 sin 3 x = 4 – 2x
2 π

Graf ialah garis lurus/Graph is a straight line y=4 – 2x .
π
Bilangan penyelesaian/Number of solutions = 4

2. (a) y y = 3 – 5 kos 4x / y = 3 – 5 cos 4x

8

3 3x
π
y = x

O ππ
–2 2

(b) sin2 2x – kos2 2x = 3x – 3 sin2 2x – cos2 2x = 3x – 3
5π 5 5π 5

−(kos2 2x – sin2 2x) = 3x – 3 −(cos2 2x – sin2 2x) = 3x – 3
5π 5 5π 5

−kos 4x = 3x – 3 −cos 4x = 3x – 3
5π 5 5π 5
3
3 – kos 4x = 3x 5 – cos 4x = 3x
5 5π 5π

3x 3 – 5 cos 4x = 3x
π π
3 – 5 kos 4x =

3x
π
Graf ialah garis lurus/Graph is a straight line y= .

Bilangan penyelesaian/Number of solutions = 4

3. (a) y y = |tan 3 x|
4
2
y = 2π
x

1

O 2 π π 4 π x
3 3 2π

(b)  tan 3 x  – 2π = 0
4 x

  tan 3 x  = 2π
4 x

Graf ialah lengkung/Graph is the curve y = 2π .
x

x0 π 2π

y∞ 2 1

Bilangan penyelesaian/Number of solutions = 3

89 © Penerbitan Pelangi Sdn. Bhd.

   sin2 x cot x = sin2 x
4. (a) sin2 x kot x = sin2 x kos x cos x
sin x sin x

= sin x kos x = sin x cos x

= 1 (2 sin x kos x) = 1 (2 sin x cos x)
2 2

= 1 sin 2x = 1 sin 2x
2 2


(b) 8 sin2 x kot x = 1  /  8 sin2 x cot x = 1

  8 1 sin 2x = 1
2

4 sin 2x = 1

sin 2x = 1
4

2x = 14.48o, 165.52o, 374.48o, 525.52o
π
x = 7.24o, 82.76o, 187.24o, 262.76o

(c) (i) y y = 1 sin 2x
2
1
2 y = 1 – x
1 4 8π
4 3 π 2π x
ππ
O 2

2

– 1
2

(ii) 8π sin2 x cot x + x = 2π
8π sin2 x kot x + x = 2π
8π sin2 x cot x = 2π − x
8π sin2 x kot x = 2π − x x
1 8π
sin2 x kot x = 1 – x sin2 x cot x = 4 –
4 8π
1 1 x
1 sin 2x = 1 – x 2 sin 2x = 4 – 8π
2 4 8π

1 x
Garis lurus ialah/The straight line is y = 4 – 8π .

Bilangan penyelesaian/Number of solutions = 5

© Penerbitan Pelangi Sdn. Bhd. 90

Soalan KBAT

1. sin (x + y) = sin x kos y + kos x sin y / sin (x + y) = sin x cos y + cos x sin y

sin (x + y) = 3 + 1
5 4

sin (x + y) = 0.85

x + y = 58.21° ............a

sin (x − y) = sin x kos y − kos x sin y / sin (x – y) = sin x cos y – cos x sin y

sin (x − y) = 3 – 1
5 4

sin (x − y) = 0.35

x − y = 20.49° ............b

a + b, 2x = 78.70°

x = 39.35°

Gantikan x = 39.35° ke dalam a,

Substitute x = 39.35° into a,

39.35° + y = 58.21°

y = 18.86°

2. (a) sin 2x + π  + sin 2x – π 
3 3
= sin π π   = π π
2x kos 3 + kos 2x sin 3 sin 2x cos 3 – cos 2x sin 3

+ sin 2x kos π – kos 2x sin π    + sin 2x cos π – cos 2x sin π
3 3 3 3

= 2 sin 2x kos π = 2 sin 2x cos π
3 3

 = (2sin 2x) 1  = (2sin 2x) 1
2 2
= sin 2x
= sin 2x

(b) (i) sin  3 x + π  + sin  3 x – π  = 1
2 3 2 3 2

sin 3 x = 1
2 2

3 x = π rad, 5π rad, 13π rad, 17π rad π
2 6 6 6 6
π 5π 13π 17π π πx
x = 9 rad, 9 rad, 9 rad, 9 rad 2

(ii) y = sin 3x + π  + sin 3x – π  − 1 y
3 3 2
1
2

y = sin 3x − 1 O
2
– 1
2

y = sin 3x – 1
2
– 3
2

91 © Penerbitan Pelangi Sdn. Bhd.

Soalan KBAT dalam kod QR pada m.s. 221

1. (a) (i) 1 + sin x – kos x 1 + sin x – cos x

1 + sin x + kos x 1 + sin x + cos x

 = x x x  = x cos x – 1 – 2 sin2 x
1 + 2 sin 2 kos 2 – 1 – 2 sin2 2 1 + 2 sin 2 2 + 2
1 + 2 sin
1 + 2 sin x kos x + 2 kos2 x – 1 x cos x 2 cos2 x – 1
2 2 2 2 2 2

2 sin x kos x + 2 sin2 x = 2 sin x cos x + 2 sin2 x
2 2 2 2 sin 2 cos 2 + 2 cos2 2
= x x x x x x
2 sin 2 kos 2 + 2 kos2 2 2 2 2

 =2sin x kos x + sin x  = 2 sin x cos x + sin x
2 2 2 2 cos 2 2 2
x x x
x x x 2 sin 2 + cos 2
2 2 2
2 kos sin + kos sin x
2
sin x = x
2 cos 2
=
kos x x
2 = tan 2

= tan x
2


(ii) tan 22.5° = tan 45°
2

= 1 + sin 45° – kos 45°  /  1+ sin 45° – cos 45°
1 + sin 45° + kos 45° 1+ sin 45° + cos 45°

1 + 1 – 1
 2  2
=
1 1
1 +  2 +  2

= 1 2 2 22
 2  2 2
1 + 2 × =
 2
= 2

= 1
1 + 2

= 1 × 1 – 2
1 + 2 1 – 2

= 1 – 2
1–2

= 2 – 1

© Penerbitan Pelangi Sdn. Bhd. 92

π

(b) (i) y y = tan x + 3
2
3
2 y=2
O
π 2π x

(ii) 11 + sin x – kos x = –1  /  1 + sin x – cos x = –1
+ sin x + kos x 1 + sin x + cos x

tan x = −1
2
x
tan 2 + 3 = 2

Garis lurus ialah/The straight line is y = 2.
Bilangan penyelesaian/Number of solutions = 1

93 © Penerbitan Pelangi Sdn. Bhd.

Jawapan

Tingkatan 5 Pengaturcaraan Linear 2. (a) I : 40x + 35y  2800
8x + 7y  560
BAB Linear Programming
II : y  3x
7

Masteri SPM (b) 3y  2x
2
y  3 x

Kertas 2 Bilangan kotak pen dakwat hitam

1. (a) I :  x + y  70 adalah sekurang-kurangnya 2 daripada
3
II : 5x + 9y  1200 bilangan kotak pen dakwat biru.

III : x  2 The number of boxes of black ink pens is
y 3 iantklepaesnts.32 of the number of boxes of blue

y  3 x
2
(c) y 8x + 7y = 5y6=0 3x
(b) y (19, 57)
60

140

y= 3 x 50
2
120

40 y = 2 x
3
100
R (64, 96) 5x + 9y = 1200 30 R

80 22
(0, 70)
20
60

40 10
x + y = 20
20 x + y = 70
x
O 10 20 30 40 50

5x + 3y = 120 x (d) (i) Apabila/When x = 32,
nilai minimum/minimum value of
O 20 40 60 80 y = 22

(c) Fungsi objektif/Objective function: Bilangan kotak minimum pen
5x + 3y = 120 dakwat hitam/Minimum number of
boxes of black ink pens = 22
Nilai minimum berlaku di (0, 70) dan
nilai maksimum berlaku di (64, 96). (ii) Fungsi objektif/Objective function:
Minimum value occurs at (0, 70) and x + y = 20
maximum value occurs at (64, 96).
Nilai maksimum berlaku di/
Jualan minimum/Minimum sales Maximum value occurs at (19, 57).
= 5(0) + 3(70) = RM210
Jumlah maksimum bilangan kotak
Jualan maksimum/Maximum sales
= 5(64) + 3(96) = RM608 pen/Maximum total number of boxes

∴ RM210  Jumlah jualan  RM608 of pens = 19 + 57
RM210  Total sales  RM608
= 76 kotak/boxes

© Penerbitan Pelangi Sdn. Bhd. 94

3. (a) I :  x + y  120 (b) y

II :  y  2 x 25 (13, 25)
5

III : 18x + 12y  420 20 R
3x + 2y  70
15
(b) y y = 2x 3x + 2y = 90

120 x + y = 120 10

3x + 5y = 110

100 5
5x + 8y = 80
80 x

O 5 10 12 15 20 25

60 R (c) (i) 12  Karpet jenis P  20
12  Type P carpets  20
2
40 (84, 36) y= 5 x (ii) Fungsi objektif/Objective function:
5x + 8y = 80
20
18x + 12y = 864 x Nilai maksimum berlaku di/
Maximum value occurs at (13, 25).
3x + 2y = 70

O 9 20 40 60 80

(c) (i) Apabila/When y = 22, Bilangan maksimum peserta
nilai minimum/minimum value of Maximum number of participants
x = 9
= 5(13) + 8(25)
Bilangan unit minimum pembesar = 265 peserta/participants
suara jenis P yang dijual
Minimum number of units of type P 5. (a) I :  x + y  15
loudspeakers sold
II : x  2y
= 9 unit/units
2y  x
(ii) Fungsi objektif/Objective function:
18x + 12y = 864 y  1 x
2
Nilai maksimum berlaku di/
Maximum value occurs at (84, 36). III : 72x + 36y  420

Keuntungan maksimum 6x + 3y  35
Maximum profit
(b) y
= 18(84) + 12(36)
= RM1944 14

4. (a) I : 1.5x + 2.5y  55 12

x + y = 15
10

15x + 25y  550 8 R
6x + 3y = 35
3x + 5y  110
6
II : 240x + 160y  7200 y= 1 x
2
24x + 16y  720 (10, 5)

3x + 2y  90 4
3 5x + 2y = 10
III : y  2x 2

x
O 2 4 6 8 10

95 © Penerbitan Pelangi Sdn. Bhd.

(c) (i) Apabila/When x = 5, (ii) Fungsi objektif/Objective function:
nilai minimum/minimum value of 1.6x + 1.2y = 24
y = 3 16x + 12y = 240
4x + 3y = 60
Bilangan minimum perjalanan ke
Q/Minimum number of trips to Q = 3 Nilai maksimum berlaku di/
Maximum value occurs at (48, 36).
(ii) Fungsi objektif/Objective function:
30x + 12y = 60 Keuntungan maksimum
5x + 2y = 10 Maximum profit

Nilai maksimum berlaku di/ = 1.6(48) + 1.2(36)
Maximum value occurs at (10, 5). = RM120

Keuntungan maksimum 7. (a) I : y  3 x
Maximum profit 2
= 30(10) + 12(5)
= RM360 II : y  1 x
3

III : 2x + y  120

6. (a) I : 2x + 4y  240 (b)
x + 2y  120 y
60
II : x + y  60 y = 3 x
2
III : y  17050x
50 (34, 51)

y  3 x 40
4
30
(b) y 2x + y = 120
20 R
60 5x + 6y = 120

x + 2y = 120 10 y= 1 x
50 3

R y = 3 x
4
40
x
34 (48, 36) O 10 20 2730 40 50 60

30

20 4x + 3y = 60 x + y = 60 (c) (i) Apabila/When y = 40,
10
nilai minimum/minimum value of
x x = 27
O 10 20 30 40 50
Bilangan minimum kipas jenis P/
(c) (i) Apabila/When x = 45, Minimum number of type P fans = 27

nilai minimum/minimum value of (ii) Fungsi objektif/Objective function:
y = 34 150x + 180y = 3600
15x + 18y = 360
Bilangan minimum bekas Q 5x + 6y = 120
Minimum number of containers Q
Nilai maksimum berlaku di/
= 34 Maximum value occurs at (34, 51).

Kos maksimum/Maximum cost
= 150(34) + 180(51)

= RM14 280

© Penerbitan Pelangi Sdn. Bhd. 96

8. (a) I : x + y  95 (b) y = x + 25
II : y  20 y (33, 57)
III : y  2x 60
50
(b) 40 x + y = 90
30
y y = 2x y = 1 x
x + y = 95 (32, 63) 2
R
70

60

20

50

40 10
R 3x + 7y = 105

30 x
O 10 20 30 40 50 60
20
y = 20 (c) (i) Nilai maksimum/Maximum value of
x = 60
10 5x + 6y = 120
x Bilangan maksimum jubin putih/
Maximum number of white tiles = 60
O 10 20 30 40 50 60 70 80
(ii) Fungsi objektif/Objective function:
(c) (i) 10 pelajar/students 1.2x + 2.8y = 42

(ii) Fungsi objektif/Objective function: 12x + 28y = 420
250x + 300y = 6000 3x + 7y = 105
25x + 30y = 600
5x + 6y = 120 Nilai maksimum berlaku di/
Maximum value occurs at (33, 57).
Nilai maksimum berlaku di/ Kos maksimum/Maximum cost
Maximum value occurs at (32, 63). = 1.2(33) + 2.8(57)
= RM199.20

Kutipan maksimum yuran 10. (a) I :  x + y  10
Maximum fee collection
II : x  2y
= 250(32) + 300(63)
= RM26 900 2y  x

9. (a) I : x + y  90 y  1 x
2

II : x  2y III : 750x + 400y  5000

2y  x 15x + 8y  100

y  1 x
2

III : y – x  25

y  x + 25

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(b) (b)
y y
10 60

9 15x + 8y = 100

8 50 y = x y = x – 15
7 (2.86, 7.14) 40 (36, 36)
30 R
6 20 x + y = 35 2x + 3y = 180
5R 2x + 3y = 172
4 x + y = 10
3
10

y = 1 x x
2 O 10 20 30 4043 50

2 (c) (i) Fungsi objektif/Objective function:
y=x

1 Kos maksimum berlaku di
5x + 3y = 15 Maximum cost occurs at
x
(36, 36).
O 123456
Kos maksimum/Maximum cost
(c) (i) Apabila/When x = 5, = 10(36) + 15(36)
= RM900
nilai minimum/minimum value of
y=3 Baki minimum peruntukan
Minimum balance of allocation
Bilangan minimum bas kecil/ = RM0
Minimum number of small buses = 3
(ii) Baki peruntukan/Balance of allocation
(ii) Fungsi objektif/Objective function: = RM900 – RM40
40x + 24y = 120 = RM860
5x + 3y = 15
10x + 15y  860
Nilai maksimum berlaku di 2x + 3y  172
Maximum value occurs at
(2.86, 7.14) ≈ (2, 7). Bilangan maksimum pen dakwat
hitam
Bilangan maksimum pekerja Maximum number of black ink pens
Maximum number of workers
= 40(2) + 24(7) = 43 dozen
= 248
2. (a) I : x + y  45
Soalan KBAT
II : 30x + 18y  1080
5x + 3y  180

1. (a) I :  x + y  35

II : 10x + 15y  900

2x + 3y  180

III : x – y  15

y  x – 15

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(b) y (ii) Apabila/When x = 10,

60 y = 5 x nilai minimum/minimum value of
3 y = 43
50
R Bilangan botol minimum krim N
43 yang perlu dijual
(18, 30) Minimum number of bottles of cream
40
N that should be sold
30 = 43

20 Jumlah jualan/Total sales
x + y = 45 = 30(10) + 18(43)

10 5x + 3y = 180 = RM1074

x Jumlah komisen/Total commission
O 10 20 30 40 50

(c) (i) x= 60 y = 9 × 1074
100 100

x = 3 y = RM96.66
5
5
y = 3 x

Bilangan botol minimum krim M
Minimum number of bottles of
cream M

= 18

Bilangan botol minimum krim N
Minimum number of bottles of
cream N

= 30

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Jawapan

Tingkatan 5 Kinematik Gerakan Linear (c) Pecutan/Acceleration, a = dv = 6t – 4
dt
BAB Kinematics of Linear Motion Apabila/When a = –1,

8

6t – 4 = –1

Masteri SPM 6t = 3
1
Kertas 2 t = 2

      s =1 3 – 2 1 2 – 6 1 = –3 3
2 2 2 8
1. (a) s = t3 – 2t2 − 6t

Halaju/Velocity, v = ds = 3t2 – 4t – 6 Zarah itu berada 3 3 m di sebelah kiri
dt O. 8

Apabila/When t = 2, v = 3(2)2 – 4(2) – 6 The particle is 3 3 m to the left of O.
8
= 12 – 8 – 6

= –2 m s−1 2. (a) s = 1 t3 + pt2 − 20t
3
Apabila/When t = 5, v = 3(5)2 – 4(5) – 6 ds
dt
= 75 – 20 – 6 Halaju/Velocity, v = = t2 + 2pt – 20

= 49 m s–1 Diberi/Given t = 2, v = –32,

Halaju seketika zarah itu pada saat ke-2 (2)2 + 2p(2) – 20 = –32
dan ke-5 masing-masing ialah –2 m s–1
dan 49 m s–1. 4p = –16

The instantaneous velocity of the particle p = –4
at the 2nd and 5th seconds are –2 m s–1 and
49 m s–1 respectively. (b) v = t2 – 8t – 20

Pecutan/Acceleration, a = dv = 2t – 8
dt
(b) Apabila/When v = 1 m s–1, Apabila/When t = 3,

3t2 – 4t – 6 = 1 a = 2(3) – 8

3t2 – 4t – 7 = 0 = –2 m s–2

(3t – 7)(t + 1) = 0 (c) Apabila zarah itu berhenti,

3t – 7 = 0, t + 1 = 0 When the particle stops,

t = 7 t = –1 v = 0
3
t2 – 8t – 20 = 0
7
t  0, maka/therefore t = 3 (t + 2)(t – 10) = 0

Halaju seketika zarah itu ialah 1 m s–1 t = –2, t = 10

apabila t = 7 . t  0, maka/therefore t = 10
3
The instantaneous velocity of the particle is Apabila/When t = 0,
7 1
1 m s–1 when t = 3 . s = 3 (0)3 – 4(0)2 – 20(0)

= 0 m

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