KOMPIMatT DAERAH KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 9 (TINGKATAN 4) / CHAPTER 9 (FORM 4)
PENYELESAIAN SEGITIGA/ TRIANGLES SOLUTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
16 1
(a) (i) KM2 = 142 + 4.52 −2(14)(4.5) cos 53°
KM2 = 140.4213

KM = 11.85 1
1
(ii) sin ∡LMK = sin 98°
4.1 11.85

sin ∡LMK sin 98 x 4.1
= 11.85

∡LMK = 20.04° 1
10
∡LKM = 180 – 98 − 20.04°
1
= 61.96°
(b) (i) K

L 4.1c K’
m
1

M

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 9 (TINGKATAN 4) / CHAPTER 9 (FORM 4)
PENYELESAIAN SEGITIGA/ TRIANGLES SOLUTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
16
(b) (ii) ∡ K’LM = 180° − 118.04 − 20.04

= 41.92 1
1
K′M = 4.1 1
sin 41.92 sin 20.04

K’M = 7.993

Luas segitiga K’LM = 1 (4.1)(7.993) sin 118.04
2

= 14.46 cm2 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 9 (TINGKATAN 4) / CHAPTER 9 (FORM 4)
PENYELESAIAN SEGITIGA/ TRIANGLES SOLUTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
17 (a) I16/14= (136 ×10)+(142 ×35)+(160×45)+(110 ×10) 1
100 1 10
1
= 14630 = 146.3 1
100
1
136 x 1.1 @ 142 x 1.1

149.6 @ 156.2

(b) (i) Katakan y ialah no indeks komoditi kayu yang berubah /

Let y is the index number that changed

(149.6 ×10) + (156.2×35) + (y×45) + (110 ×10) = 156.23
100

8063 + 45y = 15623

y = 168 1

∴ peratus peningkatan nombor indeks komoditi kayu ialah / 2
1
the percentage of increase of timber commodity index number is 1

168−160 ×100=5 %
160

(ii) Q18 ×100 = 156.23
400

Q18 = 624.92 juta / million

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 1 (TINGKATAN 5) / CHAPTER 1 (FORM 5)
SUKATAN MEMBULAT / CIRCULAR MEASURE

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

1 (a)  AOB 1
OA2 = OP2+ PA2 14
OA2 = 602+ 452
OA2= 75 cm 1
1
 POA= BOR

tan  POA= 45
60

 POA = 36.87

 BOA = 90−2(36.87)

= 16.26 × 3.142
180

= 0.2838 rad

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 1 (TINGKATAN 5) / CHAPTER 1 (FORM 5)
SUKATAN MEMBULAT / CIRCULAR MEASURE

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

1 (b) Perimeter/Perimeter = OA + AB + BO

= 75 + 75 × 0.2838  + 75 1
1
= 171.29 cm 1

Luas / Area AOB = 1 (752)(0.2838) 1
2

=  799. 19 cm2

(c) Katakan jejari dasar topi/let the base radius 6
of hat = j cm

lilitan dasar topi/circumference of hat
base = panjang lengkok AB/length of arc
AB

2πj = 21.285 1
1
j= 3.3908
2π

j = 3.39 cm

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 1 (TINGKATAN 5) / CHAPTER 1 (FORM 5)
SUKATAN MEMBULAT / CIRCULAR MEASURE

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

2 (a) Sudut binaan kolam /Pool construction angle

= 2 ×2π
3

= 4 π rad 1
3 14

(b) Luas / Area = 90 m2

1 × j2× 4 π = 90
2 3

j = 6.55 m

s = 6.55 × 4 π 1
3

= 8.733π m atau/or 27.44 m

Perimeter = 27.44 + 6.55 + 6.5

= 40.54 m 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 1 (TINGKATAN 5) / CHAPTER 1 (FORM 5)
SUKATAN MEMBULAT / CIRCULAR MEASURE

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

2 (c) Sudut sektor kolam kanak – kanak/Angle of the

children's pool sector

= 1 × 4 π 1
3 3

= 4 π atau/or 80o
9

Luas tangga kolam/Pool stairs area

= 1 × 6.552 × 4 π  1 ×6.55 ×6.55 ×sin 80o 16
2 9 2 1
1
= 29.96 – 21.13
1
= 8.83 m2 1

(d) Kos memasang Jubin/The cost of installing tile

= 8.83 ×125
1.5

= RM735.83

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 2 (TINGKATAN 5) / CHAPTER 2 (FORM 5)
PEMBEZAAN / DIFFERENTIATION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
1
3 (a) 3y + x + y + x + 2y + 2x = 16 1

6y + 4x = 16  3y + 2x = 8  y = 8 − 2 x
3 3

Luas / Area ; A = 3xy + 2xy

A = 5xy 1

A = 5x 8 − 2 x 1
3 3

= 40 x − 10 x2
3 3

dA = 40 − 20 x 1
dx 3 3 1

A maksimum / maximum , dA = 0 ; 40 − 20 x=0 10
dx 3 3

20 x =  40
3 3

x=2 1

y= 8 − 2 2 1
3 3

y = 4
3

(b) Luas maksimum = 5xy

=5 2 4 1
3

= 40 m2 1
3

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 2 (TINGKATAN 5) / CHAPTER 2 (FORM 5)
PEMBEZAAN / DIFFERENTIATION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

4 (a) dx = 0.2 ; jejari / radius, r = 1.5x
dt

dr  = 1.5 1
dx

dr = dr  dx
dt dx dt

= 1.5 × 0.2 1
1
= 0.3 cms−1

(b) Luas / Area, A = 9x2 − π 1.5x 2 1
1
dA = 18x − 3πx
dx 10
1
dA = dA  dx = 18x − 3πx 0.2
dt dx dt 1
1
x = 4, dA = 18 4 − 3π 4 0.2
dt

= 6.8592 @ 6.86 cms−1

(c) δ = 4.01 − 4 = 0.01

δ = × δ


= 18 4 − 3 4 0.01   1

= 0.15444  @ 0.154 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 3 (TINGKATAN 5) / CHAPTER 3 (FORM 5)
PENGAMIRAN / INTEGRATION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
5 (a) Luas rantau berlorek / Area of the shaded region

= Luas rantau A + luas rantau B/ Area of region A + Area of

region B

 = 3 x2 − 2x − 3 dx + 5 x2 − 2x − 3 dx Nyata/State x =
03

3,

= x3 − 2x2 − 3x 3 + x3 − 2x2 − 3x 5
3 2 0 3 2 3

= 33 − 3 2−3 3 −0 + 53 − 5 2−3 5 − 33 − 3 2−3 3 5
3 3 3

= −9 + 125 − 25 − 15 − 9−9−9 y
3

= 59 unit2 y = x2 – 2x - 3
3

OA B x
x =5

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 3 (TINGKATAN 5) / CHAPTER 3 (FORM 5)
PENGAMIRAN / INTEGRATION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
5
(b) Isipadu janaan / Volume of revolution

= 5 πy2 dx
0

5
= π x2 − 2x − 3 2 dx

0

5 5
= π x4 − 4x3 − 2x2 + 12x + 9 dx

0

=π x5 − 4x4 − 2x3 + 12x2 +9x 5
5 4 3 2 0

=π x5 − x4 − 2x3 + 6x2 + 9x 5
5 3 0

=π (5)5 − 5 4 − 2(5)3 + 6(5)2 + 9(5) −0 y
5 3

= 335 π unit3 y = x2 – 2x - 3
3

O x
x =5

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 4 (TINGKATAN 5) / CHAPTER 4 (FORM 5)
PILIH ATUR DAN GABUNGAN / PERMUTATIONS AND COMBINATIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

6 (a) Bilangan cara / No. of ways = 6 C 2 6
= 15

(b) Bilangan cara / No. of ways = 6 C 3
= 20

(c) Bilangan cara / No. of ways = 6 C 4
= 15

SOALAN PENGIRAAN MARKA JUMLAH
H MARKAH
7 (a) Bilangan cara / No. of ways = 4 C 1  3 C 3
=4 6

(b) Bilangan cara / No. of ways = 1 C 1  3 C 3
=1

(c) Bilangan cara / No. of ways = 3 C 1  2 C 2  1 C 1
atau 4 – 1

=1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 5 (TINGKATAN 5) / CHAPTER 5 (FORM 5)
TABURAN KEBARANGKALIAN / PROBABILITY DISTRIBUTION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

8 (a) (i) np = 336 , npq = 126

336q = 126

q = 3
8

p = 5
8

(ii) n = 8

P(X > 6) = P (X = 7) + P(X = 8)
= 0.1118 + 0.0233

= 0.1351

(b) (i) μ = 60, σ = 15

P(x < 54) = P( Z < 54 − 60 )
15

= P(Z >  0.4)

= 0.3446

(ii) P( X > q ) = 0.25

P( Z < q − 60 ) = 0.25
15

q − 60 = 0.674
15

q = 70.11 kg

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 6 (TINGKATAN 5) / CHAPTER 6 (FORM 5)
FUNGSI TRIGONOMETRI/ TRIGONOMETRY FUNCTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

9 (a) LHS = sin θ + cos θ
cos θ sin θ

= sin2 θ +cos2 θ
cos θsin θ

= 1
cos θsin θ

= sec θcosec θ (shown)

(b)

sin shape 1

1 period and max=2 1

modulus 1
1
= 2
1

straight line

number of solutions = 1 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 6 (TINGKATAN 5) / CHAPTER 6 (FORM 5)
FUNGSI TRIGONOMETRI/ TRIGONOMETRY FUNCTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

10 (a) kos A = sin A
sin A kos A

kos2A − sin2A

sin A kos A

kos 2A ÷ sin 2 A 3
2 7

2 kos 2A

sin 2A

2 kot 2A

(b)

y = 2 − 2x 4
π

Bilangan Penyelesaian = 3

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 6 (TINGKATAN 5) / CHAPTER 6 (FORM 5)
FUNGSI TRIGONOMETRI/ TRIGONOMETRY FUNCTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

11 (a) (i)sin x+45° kos x+45°

= sin x kos 45 + kos xsin 4 5° kos x kos 45° − sin xsin 4 5°

= sin x 1 + kos x 1 kos x 1 −  sin x 1
22 2 2

= 1 1 sin x + kos x kos x − sin x 3
22

= 1 kos2x − sin2 x
2

1
= 2 kos 2x

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 6 (TINGKATAN 5) / CHAPTER 6 (FORM 5)
FUNGSI TRIGONOMETRI/ TRIGONOMETRY FUNCTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

11 (a) (ii) sin x + 45° kos x+ 45°  = − 1
3

1 kos 2x = − 1
2 3

kos 2x = −   2 3
3

∴ 2x = 131.81°, 228.19°, 491.81°, 588.19°

x = 65.91°, 114.10°, 245.91°, 294.10° 6
(b)

3

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 6 (TINGKATAN 5) / CHAPTER 6 (FORM 5)
FUNGSI TRIGONOMETRI/ TRIGONOMETRY FUNCTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

12 (a) kot x (1 – kos x) = 1 sin x x (1 – kos x) = sin 2
2 −kos

x

(b) Amplitud, a = 3; y maks= 4, y min = -2 3
Kala = 2π

Bilangan kitaran sehingga 2π = 1

(c) Bil. penyelesaian = 1

kot x (1 – kos x) = x - 1
2 3π 3

1 sin x x (1 – kos x) = x - 1 8
−kos 3π 3

3 sin = x - 1
π

3 sin + 1 = x 3
π

y = x
π

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 7 (TINGKATAN 5) / CHAPTER 7 (FORM 5)
PENGATURCARAAN LINEAR/ LINEAR PROGRAMMING

y

100

SOALAN PENGIRAAN MARKAH JUMLAH 90
MARKAH 80
13 (a) x + y ≥ 40 3 70 y = 2x
y ≤ 2x 3 10 60
3x + 2y ≤ 180 50 120x + 80y = 7200
4 40
(b) Rujuk Graf 30 R
20
(c) (i) Maksimum x + y = 40
= 53
Minimum x
= 30
10 20 30 40 50 60 70
(ii) RM3760

10

0
0

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 7 (TINGKATAN 5) / CHAPTER 7 (FORM 5)
PENGATURCARAAN LINEAR/ LINEAR PROGRAMMING

y

110

100 y = 10x

SOALAN PENGIRAAN MARKAH JUMLAH 90 R
MARKAH 80
14 (a) 3x + 2y ≤ 160 3 70 y = x + 10
y ≥ x + 10 3 10 60
x ≥ 110y 4 50
40
(b) Rujuk Graf 30

(c) (i) 76 orang

(ii) RM 744

20

3x + 2y = 160

10

0x

0 5 10 15 20 25 30 35 40 45

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 8 (TINGKATAN 5) / CHAPTER 7 (FORM 5)
KINEMATIK GERAKAN LINEAR/ KINEMATICS OF LINEAR MOTION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

15 s = 3 – 9t² + 24t + 5, 1
(a) t = 0 , v = 3(0) – 18(0) + 24
= 24

(b) t = 3 , v = 3(3²) – 18(3) = 24

= 27 – 54 + 24

=-3 1

(c) v = 0 ; 3t² - 18t + 24 = 0 7
3
t² - 6t + 8 = 0
(t – 2)(t – 4) = 0 t–4=0
t–2=0

t=2 t=4

(d) 3t² - 18t + 24 < 0 2
t² - 6t + 8 < 0
(t – 2)(t – 4) < 0

2<t<4

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 8 (TINGKATAN 5) / CHAPTER 7 (FORM 5)
KINEMATIK GERAKAN LINEAR/ KINEMATICS OF LINEAR MOTION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

16 (a) v = (12 −  4t) dt 3
v = 12t – 2t² + c
Apabila t = 0 , v = 0 maka c = 0 2
v = 12t – 2t²
Halaju maksimum
12 – 4t = 0
12 = 4t
t = 3s
v = 12(3) – 2(3²)
= 18 m −1

(b) 12t – 2t² = 0
2t(6 – t) = 0
12t – 2t² = 0
2t(t – 6) = 0
t = 0 atau 6

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 2) / ANSWER FOR PAPER 2 (SET 2)

BAB 8 (TINGKATAN 5) / CHAPTER 7 (FORM 5)
KINEMATIK GERAKAN LINEAR/ KINEMATICS OF LINEAR MOTION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

16 (c) s = 12t − 2t2 dt

s = 6t²  2 t2 + c 2
3

Apabila t = 0 dan s = 0 maka c = 0

s = = 6t²  2 t2
3

(d) Apabila Amin bergerak ke kiri , v < 0

bergerak ke kiri , v < 0
(t – 2)(t – 4) < 0

2<t<4

2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

Pejabat Pendidikan Daerah Kinta Utara

SEKTOR PEMBELAJARAN,
PEJABAT PENDIDIKAN DAERAH KINTA UTARA,

LEBUH CATOR, 30450 IPOH,
PERAK DARUL RIDZUAN.
TEL: 05-2544877
FAKS: 05-2419154


TAHUN 2021