KOMPIMatT DAERAH KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 1) / ANSWER FOR PAPER 2 (SET 1)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

14 (a) x + y ≤ 400
x ≤ 2y

4x + 3y ≥ 360

(b) Rujuk graf

(c) (i) 33

(ii) Jumlah maksimum
kutipan yuran sebulan
= 80(265) + 60(135)

= RM 29 300

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 1) / ANSWER FOR PAPER 2 (SET 1)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

15 (a) Sp = 4t + 3t2  t2 + c
2 3

Sp = 4(3) + 3(3)2  32 + c
2 3

Sp = 16.5
(b) Jarak C dari B/Distance C from B = 24 -16.5

= 7.5

v = 7.5 = 2.5
3

vq = - 2.5

(c) ( t – 4)(t + 1) = 0 ; t = 4

Sp = 4(4) + 3(4)2  42 atau/or Sp = 1832 atau/or Sq = 2.5  4
2 3

S = 1832 = 10 - 24
S = 423

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 1) / ANSWER FOR PAPER 2 (SET 1)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

16 (a) (i) v = 5t - 2t² + 12

(maximum velocity , a = 0)

5 – 4t = 0

t = 5
4

v = 121
8

(ii) 5t -2t² + 12 = 0

t=4

(b)

Shaping ∩
Passing any two points (0, 12), (0,4) or (5/4 , 121/8)

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 2 (SET 1) / ANSWER FOR PAPER 2 (SET 1)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

16 (c) 5t2 2t3 4
2 3 0
− =12t

5(4)2 − 2 4 3
2 3
=12(4) 0

136
3

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 1 (TINGKATAN 4) / CHAPTER 1 (FORM 4)
FUNGSI / FUNCTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
1 (i) MARKAH

g 2 = −2

h 2 = −1
a 2 +b= −2

2a + b = − 2

2 2 6 b= −1
−
6
4 − b=−1
4−b= −6

b = 10

2a + b = − 2

2a − 10 = − 2

2a = − 12
a = −6

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 1 (TINGKATAN 4) / CHAPTER 1 (FORM 4)
FUNGSI / FUNCTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
1 (ii) MARKAH

g y = −6y + 10
Biar y = g−1(x)

g(y) = x

−6y + 10 = x

6y = 10 − x
10 − x
y = 6

g−1 x = 10 − x
6

(iii) hg−1 x =h 10 − x
6
6
= 2 10 − x −10

6
6
= 10 −x −30

3
18
= −20 − x

= − x +1820, x ≠ −20

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 2 (TINGKATAN 4) / CHAPTER 2 (FORM 4)
FUNGSI KUADRATIK / QUADRATIC FUNCTION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

2 y = 5x – 1 ...... ①

y = 2x² + x + p ...... ②

Ganti/Substitute ① dalam/into ②.
5x – 1 = 2x² + x + p

2x² – 4x + p + 1 = 0

b² – 4ac < 0
(–4)² – 4(2)(p + 1) < 0

16 – 8p – 8 < 0

8 < 8p

p>1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 3 (TINGKATAN 4) / CHAPTER 3 (FORM 4)
SISTEM PERSAMAAN / SYSTEM OF EQUATIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

3 x – 2y + 3z = 3 …
–2x + 2y – pz = p …

y – 5z = –3 …

Hapuskan x daripada  dan 

Eliminate from  and  1
1
2 x  + : 2x – 4y + 6z + (–2x + 2y – pz) = 6 + p
6
–2y + (6 – p)z = 6 + p →
1
Hapuskan y daripada  dan  1

Eliminate y from  and  1
1
2 x  + : 2y – 10z + (–2y + (6 – p)z) = –6 + 6 + p

(– 4 – p)z = p

Sistem tiada penyelesaian apabila

System of linear equations does not have any solutions when

–4–p=0
p=–4

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 3 (TINGKATAN 4) / CHAPTER 3 (FORM 4)
SISTEM PERSAMAAN / SYSTEM OF EQUATIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

4 2x – 7y + 10 = 0
7y=10
x = 2 ----------①

-x2 − 4xy – y2 24x + 24y = 90 -----②

Ganti  dalam :

Substitute  into 
2
− 7y − 10 − 4 7y −10 y − y2 + 24 7y − 10 + 24y = 90
2 2 2

− 49y2 − 140y + 100 −2 7y2 − 10y − y2 + 12 7y − 10 + 24y − 90 = 0
4
-49y2 + 140y – 100 – 56y2 + 80y – 4y2 + 336y – 480 + 96y – 360 = 0

-109y2 + 652y – 940 = 0

109y2 - 652y + 940 = 0

a = 109, b = -652, c = 940

y = − −652 ± −652 2 4(109)(940)

−

2(109)

y = 3.56 atau/or y = 2.42

Apabila/ When y = 3.56, x = 7.46

Apabila/ When y = 2.42, x = 3.47

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 4 (TINGKATAN 4) / CHAPTER 4 (FORM 4)
INDEKS, SURD DAN LOGARITMA / INDICES, SURDS AND LOGARITHMS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

5 (a) ( 2 + 3 ) ( 2 + 3 )
=
(2− 3) (2+ 3 )

= 4+3+4 3
4−3

=7+4 3

(b) log2m − log2n = 3
log24
2log2m − log2 n = 6
log2mn² = 6
m2
n = 64

m=8 n

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 4 (TINGKATAN 4) / CHAPTER 4 (FORM 4)
INDEKS, SURD DAN LOGARITMA / INDICES,SURDS AND LOGARITHMS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

6 (a) 2 x 3x = 18
3 3−

3 3x − 3x = 18
2 3x = 18
3x = 9
3x = 81 = 34
x=4

(b) = log2 7.5
log2 4
= log2 3 + log2 5 − log2 2
2
p + q 1
= 2 −

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 5 (TINGKATAN 4) / CHAPTER 5 (FORM 4)
JANJANG / PROGRESSION

SOALAN PENGIRAAN MARKAH JUMLAH
7 (a) MARKAH

(b) ar3÷ ar2

r = 8 = 1 , a = 216
24 3

216

1 − ( 1 )
3

= 324

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

8 n 2 4 + n −1 12 = 580
2
12n2 − 4n − 1160 = 0
3n2 − n − 290 = 0

(3n + 29)(n − 10) = 0

n = 10

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 5 (TINGKATAN 4) / CHAPTER 5 (FORM 4)
JANJANG / PROGRESSION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

9 a = 4 000, d = –250, Tn= 1 000

4 000 + (n – 1) (–250) = 1 000 1

4 000 – 250n + 250 = 1 000

250n = 3 250

n = 13 15

13 (4 000 + 1 000) = 32 500 1
2 1
1
Total cost = 32 500 × RM0.50

RM16 250

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 6 (TINGKATAN 4) / CHAPTER 6 (FORM 4)
HUKUM LINEAR / LINEAR LAW

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

10 (a) hy = kx2 + hk
k
y = h x2 + k

pintasan-y/y-intercept = 6
k=6

Kecerunan/Gradient = 2
k
h=2
k
h= 2

h = 6
2
h=3

(b) y =5x + a
x
Pada/at (−1, 0)
0
−1 = 5( − 1)+a

a=5

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 6 (TINGKATAN 4) / CHAPTER 6 (FORM 4)
HUKUM LINEAR / LINEAR LAW

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

11 (i) y = 4x − 6 ............................❶

y = x3 − x − 2................❷
gantikan ❶kedalam ❷/Subs ❶ into ❷
= x3 − x − 2
x3 − 5x+ 4=0
(x - 4)(x - 1) = 0
x = 1 atau x = 4
apabila/when x = 1
y = 4(1)-6

= -2
Apabila /when x = 4
y = 4(4) - 6

= 10
Koordinat/ Coordinate P(1, -2). Q(4, 10)

(ii) Titiktengah/ Midpoint PQ = 1+4 , (−2) +10
2 2
5 8
= 2 , 2

= 5 ,4
2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 8 (TINGKATAN 4) / CHAPTER 8 (FORM 4)
VEKTOR / VECTOR

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
12 (a) −3
5
-3i + 5j ( Jangan Terima / Don’t Accept)

(b) −3 1 = −7
5− 2 3

RT = RO + OT atau setara / or equivalent

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

13 Diberi CD selari dengan AB/ Given CD is parallel to AB

∴ CD = kAB
m 10
7 =k 14 1
m
7 = 10k 1 3
14k 1
7 = 14k 1
1 2
k = 2

m = 10k m=10 1 m=5
2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 8 (TINGKATAN 4) / CHAPTER 8 (FORM 4)
VEKTOR / VECTOR

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
14
(a) OA = −2 , OB = 3 , OC = 8
3 5 7
2 3 5
AB = −3 + 5 = 2 2
5
BC = −3 + 8 = 5
−5 7 2 2

AB = BC 1

Oleh itu, segaris/ Therefore, collinear

(b) OP= 5
11
2 5 7
AP= −3 + 11 = 8 1
3
BP= −3 + 5 = 2
−5 11 6 1
−8 5 −3
CP= −7 + 11 = 4 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

1 (a) ∠AOB = ∠α ; sin α = 15
17

α = 61.93° 1
1
= 61.93°× 3.142 1
180°
7
= 1.081 rad
1
(b) Perimeter kawasan tidak berlorek/ Perimeter of non- 1
shaded region = AC + CB + BA
OA2 =172 − 152 1
1
OA = 8 cm

AC = 8 1.081

= 8.648 cm

= 8.648 + 17 − 8  + 15

= 32.648 cm atau 32.65 cm

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
2

Nisbah jejari/Radial ratio : 1 : 2 1 6
3.95 m : 7.9 m
1
(a) Perimeter landskap/Perimeter landscape = 3.95 m x 2π 2
1
= 7.9π atau 24.82 m
(b) Luas/Area = π × 7.92 − π × 3.952 1

= 46.81π m2 atau 147.07 m2
(c) Kos mencukupi kerana luas landskap hanyalah 49.02

Tmh2e cost is sufficient because the area of the landscape is

only 49.02 m2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

3 y = 5x2+1  ❶ [1m]
y+δy = 5 x+δx 2+1

=5 x2 + 2xδx +  δx 2 +1

=5x2 + 10xδx + 5 δx 2 + 1  ❷ [1m]

❷❶ : δy = 10xδx + 5 δx 2

δy =10x + 5δx [1m]
δx [1m]

Maka, dy = had ∂y
dx δx→0 ∂x

= had 10x + 5δx
δx→0

= 10x + 5 0

= 10x

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

4 V= 3 h h2 − 6
5

V= 3 h3 − 18 h 1
5 5

dV  = 9 h2 − 18
dh 5 5

dV  = 16 4
dt 1

dh  = dh  dV ; dh = 1 × 16 1
dt dV dt dt 95h2 − 158 1

= 9 1 − 158 × 16
5 (3)2

=1.2698 cms−1 or 1.27 cms−1 or 1.270 cms−1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

5 (a) 5 (3f x dx + 2x) dx
0

=3 5 f x dx + 5 2x dx
0 0

=3 6  + 2x2 5 1
2 0 1

= 18 + 52 − 02

= 18 + 25

= 43 1 6

(b) k ( f x  + 3 dx ) =29 1
0 1

k f x dx + k 3 dx =29
0 0

5 f x dx + k f x dx +  3x 0k = 29
0 5

6 + 2 +  3 k − 3 0  = 29

8 + 3k = 29

3k = 21

k=7 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 4 (TINGKATAN 5) / CHAPTER 4 (FORM 5)
PILIHATUR DAN GABUNGAN / PERMUTATION AND COMBINATION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

6 (a) Bilangan cara = 8 P3 1
= 336
1
(b) Bilangan cara = 7 P4 × 2
1
= 1680
1

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

7 (a) Bilangan cara = 14 C 4 1
= 1001
1

(b) Bilangan cara = 6 C 2  8 C 2 + 6 C 3  8 C 1 + 6 C 4  8 C 0 1

= 595 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 5 (TINGKATAN 5 ) / CHAPTER 5 ( FORM 5)
TABURAN KEBARANGKALIAN KERTAS 1 / PROBABILITY DISTRIBUTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

8 P = 0.45, q = 0.55

P(X = 0) = 0. 0145

nC0 (0.45)0 (0.55)n = 0.0145 1
n log 0.55 = log 0.0145 3

1

n = 7.18

n = 8 batang pokok hiasan 1

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

9 (a) a = 3, b = 2, c = 2

(b) Bilangan penyelesaian = 2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 6 (TINGKATAN 5) / CHAPTER 3
FUNGSI TRIGONOMETRIK / TRIGONOMETRIC FUNCTION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

10 8 (1 – sin2 x) + 2 sin x – 5 = 0

8 – 8 sin2 x + 2 sin x – 5 = 0

8 sin2 x – 2 sin x – 3 = 0

(4 sin x – 3)(2 sin x + 1) = 0

sin x = 3 atau sin x = – 1
2 2

sudut rujukan = 48.59° sudut rujukan = 30°

x = 48.59°, 131.41° x = 210°, 330°

x = 48.59°, 131.41°, 210°, 330°

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 6 (TINGKATAN 5) / CHAPTER 3
FUNGSI TRIGONOMETRIK / TRIGONOMETRIC FUNCTION

SOALAN P ENGIRAAN MARKAH JUMLAH
MARKAH

11 (a) Kot  = 1
q

(b) Kos (90 - ) = q
1 + q2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

BAB 6 (TINGKATAN 5) / CHAPTER 3
FUNGSI TRIGONOMETRIK / TRIGONOMETRIC FUNCTION

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

12 (a) kos²  = sin²  (1 + kot²  1)

= sin²  kos² 
sin² 

= kos²  (LHS)

(b) p2 1
1

θ

x

2
x2 = 1 2  p2 −1

x2 = 1 − p2 + 1

x = 2 − p2
 Kos  = p

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
13

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
14 I : x+y≤9
II : x ≥ 4y
III : 18x + 3y ≤ 80

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

15 (a) v = 0 ; 3t2 − 14t +15 = 0

(3t – 5)(t – 3) = 0

t = 5 , t = 3
3

(b) s = 3t2 −  14t +15 dt

= t3 − 7t2 + 15t + c

t = 0 , s = -10 ; (0)3 − 7(0)2 + 15(0) + c = -10
c = -10 ;  s = t3 − 7t2 + 15t −  10
t = 3 ; s = (3)3 − 7(3)2 + 15(3) −  10

s = -1

t = 5 ; s = ( 5 )3 − 7( 5 )2 + 15( 5 ) −  10
3 3 3 3

s = 5
27

5  (-1) = 32
27 27

(c) 3t2 − 14t +15  0

(3t – 5)(t – 3)  0

t  5 , t  3
3

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

16 (a) dv = 2pt + q ; a = 2pt + q
dt

t = 1 , a = 10 ;  10 = 2p (1) + q

q = 10 – 2p  ❶

t = 4 , v = 0 ;  0 = p(4)² + q(4)  q
0 = 16p + 4q  q  ❷

Ganti/Subtitute ❶ dalam/in ❷

0 = 16p + 4(10 – 2p)  (10 – 2p)

0 = 16p + 40 – 8p  10 + 2p

0 = 10p + 30

p=3

q = 10 – 2(-3)

q = 16

(b) v = pt² + qt  q  v = 3t² + 16t  16

3t² + 16t  16  0

3t²  16t + 16  0

(3t  4)(t  4)  0

t  4 atau/or t4
3

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

16 (c) s = 3t² + 16t 16dt

s = t³ + 8t²  16t + c
t = 0 , s = -10  -10 = (0)³ + 8(0)²  16(0) + c

c = -10

s = t³ + 8t²  16t  10

v = 0  3t² + 16t − 16 = 0

(-3t + 4)(t  4) = 0

t = 4 atau/or t=4
3

t = 4  s = (43 )³ + 8(34 )²  16(34 )  10
3

s =  526
27

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

JAWAPAN KERTAS 1 (SET 2) / ANSWER FOR PAPER 1 (SET 2)

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

17 (a) ds = 3t³  10t + 3
dt

v = 3t³  10t + 3

3t³  10t + 3 = 0

(3t  1)(t  3) = 0

t = 1 atau/or t=3
3

(b) v = 2t − 4 dt

= t² – 4t + c

t = 0 , v = 3  3 = (0)² – 4(0) + c

c=3

 t² – 4t + 3

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 1 (TINGKATAN 4) / CHAPTER 1 (FORM 4)
FUNGSI / FUNCTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
1
(a) x = 100, y = 212

212 = 1.8 (100) + k

212 = 180 + k

k = 32

(b) x = 32

y = 1.8(3.2) + 32

y = 57.6 + 32 6
y = 89.6 ° F
(c) Biar y = f‾¹ (x)

f(y) = x

1.8y + 32 = x

y = x − 32
1.8

f‾¹ (x) = x − 32
1.8

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 2 (TINGKATAN 4) / CHAPTER 2 (FORM 4)
FUNGSI KUADRATIK/ QUADRATIC FUNCTIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
2
(a) 3x² + 5x – 2 =0

(3x - 1) (x + 2) = 0

x = 1 atau −2
3

(b) hx² + kx + 3 = 0

Untuk punca yang sama , b² - 4ac = 0 6

k² − 4 h (3) = 0

k² − 12h = 0

h = k2
12

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 3 (TINGKATAN 4) / CHAPTER 3 (FORM 4)
SISTEM PERSAMAAN / SYSTEMS OF LINEAR EQUATIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

3 (a) x = 1 and y = m into kx2 – 4xy = y2

k(1)2 – 4(1)(m) = m2

k – 4m = m2 1

x = 1 and y = m into 13 – 5x – 4y = 0 1
13 – 5(1) – 4(m) = 0
8 – 4m = 0

m=2

k = m2 + 4m 1
k = 22 + 4(2) 1
k = 12
1
(b) 12x2 − 4xy − y3 = 0 → ❶ 6

y= 13−5x →❷
4

Ganti ❷ dalam ❶: 12x2 − 4x 13−5x − 13−5x 2
4 4 =0

247x2 − 78x − 169 = 0
19x2 − 16x − 13 = 0

19x+13 x−1 = 0

Apabila x= − 13 , y= 13−5 − 13 = 78
19 4 19 19

Titik persilangan yang satu lagi ialah − 13 , 78 1
19 19

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 3 (TINGKATAN 4) / CHAPTER 3 (FORM 4)
SISTEM PERSAMAAN / SYSTEMS OF LINEAR EQUATIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
4 (a) 60 x + 35 y + 40 z = 48 (50) →12x + 7 y + 8z = 480 satu persamaan 1
100 100 100 100 betul 1 7

30 x + 40 y + 30 z = 33 (50) → 3x + 4 y + 3z = 165 semua persamaan 1
100 100 100 100 1
1
10 x + 25 y + 30 z = 19 (50) → 2x + 5y + 6z = 190 betul 1
100 100 100 100 1

(b) 12x + 7 y +8z = 480 ...
3x + 4y + 3z =165 ...

2x + 5y + 6z =190 ...

12x + 7 y +8z − (12x +16y +12z) = 480 − 4(165)

9y + 4z =180 ...

12x + 7 y +8z − (12x + 30y + 36z) = 480 −1140

23y + 28z = 660 ...

9(23y + 28z) − 23(9y + 4z) = 9(660) − 23(180)

160z =1800
z =11.25

Ganti z = 11.25 dalam 

9y + 4(11.25) =180

y =15

Ganti z = 11.25 dan y = 15 dalam :

2x + 5(15) + 6(11.25) =190

x = 23.75

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 4 (TINGKATAN 4) / CHAPTER 4 (FORM 4)
INDEKS, SURD DAN LOGARITMA / INDICES, SURDS AND LOGARITHMS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
5 (a) 1000 1.05 5 1
1 8
= RM 1276.28 1
1
= RM 1276 1

(b) 1000 1.05 t ≥ 1500

1.05 t ≥ 1.5

t ≥ log10 1.5 1
log10 1.05
1
t ≥ 8.31 1

t = 9 tahun / year 2026

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 4 (TINGKATAN 4) / CHAPTER 4 (FORM 4)
INDEKS, SURD DAN LOGARITMA / INDICES, SURDS AND LOGARITHMS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
6 1
(a) 32(x+y)
33x

3x 2 3y 2 1
3x 3
=

= r2s2 1
r3 8

= s2 1
r
1, 1
(b) b) log3 r = x and log3 s = y 1
1
log3 81+ log3 r3 − log3 s2 1

=4 + 3 log3 r − 2 log3 s

=4 + 3x − 2y

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 5 (TINGKATAN 4) / CHAPTER 5 (FORM 4)
JANJANG / PROGRESSIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
7 1
(a) y = 5103
567 y

y = 1701 1
1
(b) r = 1701
567

r=3 1
(c) a(2242) = 847
1
a= 7 9
(d) 7 ( 3n−1) > 10 000
1
(n−1) log103 > log10 (10 7000)
1

1

n > 7.612

n=8 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 5 (TINGKATAN 4) / CHAPTER 5 (FORM 4)
JANJANG / PROGRESSIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
8
47718 1
(a) = 1 − 48200
1
= 0.01 1
(b) = 48200(0.99) 1
(c) = 48200(0.99)6

= 45379.34 1 8
1
(d) Kadar susutnilai bagi kereta / depreciation rate of
car = 0.01

Kadar susutnilai bagi motosikal / depreciation rate 1
of motorbike = 0.03

motosikal menyusut nilai dengan lebih cepat / 1
motorbike depreciates more rapidly

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 5 (TINGKATAN 4) / CHAPTER 5 (FORM 4)
JANJANG / PROGRESSIONS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
1
9 31800 (1.0410 −1)
S10 =
1.04 −1

= 381 794.21 1
1
S10 = 27 600 (1.0810 −1)
1.08 −1

= 399 829.12 1
8

Rayyan patut memilih Syarikat Q/
Rayyan should choose Company Q

RM399 829.12 – RM381 794.21 = RM18 034.91 1
1
RM399 829.12 × 20 1
100

= RM79 965.82

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 6 (TINGKATAN 4) / CHAPTER 6 (FORM 4)
HUKUM LINEAR / LINEAR LAW

PENGIRAAN MARKAH JUMLAH
MARKAH

10 (a) x2 4.0 9.0 16.0 25.0 30.3 36.0 2
xy 9.4 16.5 26.4 39 46.2 54.6

(b) xy

9

2

Plot xy melawan x2 / Plot xy 1
against x2
1
5 titik diplot dgn betul / 5 1
points plotted correctly

garis lurus penyesuaian
terbaik / Line of best fit

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 6 (TINGKATAN 4) / CHAPTER 6 (FORM 4)
HUKUM LINEAR / LINEAR LAW

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
10
(c) y = mx + n
mx

xy = mx2 + n 1
m 1

Guna/using m = kecerunan/gradient atau n =c
m

Kecerunan graf /Gradient of the graph

m = 54.6 − 9.4
36.0 − 4.0

= 1.4125 1

n =pintasan−Y/Y−intercept
m

n = 4
1.4125

n = 5.65 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 7 (TINGKATAN 4) / CHAPTER 7 (FORM 4)
GEOMETRI KOORDINAT / GEOMETRY COORDINATE

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
11
(a) OK = ; OS =

SK = 2 + 2 =10 1
1
2 + 2 =100 1

(b) SR = 3 SR = 3 RK
RK 2 2

x = 3 σ, y = 2 θ
5 5

R= 3 σ, 2 θ 10
5 5 1

σ2 + θ2 =100

5 x 2+ 5 y 2 = 100
3 2

25 x2 + 25 y2 =100
9 4

100 x2 + 225 y2 =100
36 36

4x2 + 9y2 - 144 =0 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 7 (TINGKATAN 4) / CHAPTER 7 (FORM 4)
GEOMETRI KOORDINAT / GEOMETRY COORDINATE

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
11
(c) σ = 3, x = 3 σ
5

x = 3 3
5

= 9 or 1 4 1,1
5 5
1
σ2 + θ2 =100
32 + θ2 =100 1
θ2 =100 – 9 1

θ = 91

y = 2 θ
5
2
= 5 ( 91 )

= 2 91
5

Koordinat R / coordinates of R 1 4 , 2 91
5 5

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 8 (TINGKATAN 4) / CHAPTER 8 (FORM 4)
VEKTOR / VECTORS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
12
(a) (i) Guna/use AB + BC atau/or AD + DC 1

AC = 4i + 5j 1
(i) Cari / Find AB 1

Vektor unit / unit vector= 2 + 2 1
1
8 8

(b) (i) AP = b + CB

= b + (a – b)

(guna / use AP = AC + CP )

untuk / for CB = a – b atau / or BC = b – a 1
=  a + (1 –  ) b 1
(ii) a + (1 –  ) b = a + (– 3a + 3b)

Guna / use AP = AB + BP 1

a = (1 – 3)a atau / or (1 –  ) b = 3b

banding pekali a atau b / compare the coefficient of a 1
or b 1

=–2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 8 (TINGKATAN 4) / CHAPTER 8 (FORM 4)
VEKTOR / VECTORS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

13 (a) (i) AP = AO + OP 1
1
AP = − OA + OP

AP = − 8a˷ + 4b˷

(ii) OQ = AO + AQ 1
1
OQ = 8a˷ + ¼ AB

= 8a˷ + ¼ ( AO + OB )

= 8a˷ + ¼ ( − 8a˷ + 4OP )
= 8a˷ + ¼ ( − 8a˷ + 4(4b˷))
= 8a˷ – 2a˷ + 4b˷
= 6a˷ + 4b˷

(b) (i) AR = hAP 1

= h (− 8a˷ + 4b˷ )
= − 8ha˷ + 4hb˷

(ii) RQ = kOQ

= k (6˷a + 4˷b)

= 6k˷a + 4kb˷ 1
PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 8 (TINGKATAN 4) / CHAPTER 8 (FORM 4)
VEKTOR / VECTORS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
13
(c) (i) AQ = AR + RQ

AQ = = − 8ha˷ + 4hb˷ + 6ka˷ + 4kb˷ 1

AO + OQ = − 8ha˷ + 4hb˷ + 6ka˷ + 4kb˷

- 28aa˷˷˷ + 6a˷ + 4˷b = − 8h˷a + 4h˷b + 6ka˷ + 4k˷b
- + 4b˷ = − 8h˷a + 4h˷b + 6ka˷ + 4kb˷

− 2 = 8h + 6k → − 1 = 4h + 3k → ❶ 1

4 = 4h + 4k → 1 = h + k → k = 1 – h → ❷

Ganti ❷ dalam ❶/substitute ❶ into ❷
− 1 = − 4h + 3(1 − h)
− 1 = − 4h + 3 − 3h)

− 4 = − 7h 1
1
h = 4
7

k = 1 - 4 = 3
7 7

˷˷

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 8 (TINGKATAN 4) / CHAPTER 8 (FORM 4)
VEKTOR / VECTORS

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
14
(a) ON =t(-4i+4j)+ (50i+20j) 2

=(50-4t) i + ( 20+4t) j 1

Bila Bus M menemu Bus N/When Bus M meets Bus N
OM = ON

∴ (6ti+8tj) = (50-4t) i + ( 20+4t) j 1

Banding pekali I / Compare the coefficient of i, 1
6t = 50 - 4t

10t = 50 1 10
t=5

(b) Banding pekali j / Compare the coefficient of j,

8t = 20 + 4t 1
4t = 20 ➔ t = 5 1
Kedua-dua t = 5, maka Bus M menemu Bus N
1
Both t = 5, therefore Bus M will meet up Bus N

Bus M akan menemu Bus N setelah 5 jam 1
berlepas dari stesyen.

Bus M will meet up Bus N after 5 hours of travelling.

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
15 1,1
(a) ∡DBA = 68⁰ ÷ 2 = 34⁰ (dibuktikan) 1
AB2 = 62 + 62 ‒ 2(6)(6) kos 112⁰

AB2 = 98.9717

AB = 9.95 cm 1

(b) ∡DBC = 100⁰ ‒ 68⁰ ‒ 34⁰ =78⁰ 1
10
∡ ABC = 78⁰ + 34⁰ = 112 ⁰
1
9.95 = 6 +CD 1
sin 34 sin 112⁰
2
∴ CD = 10.50 cm 1

(c) L△ABC = 1 (10.50 + 6)(9.95) sin 34⁰
2

L△ABC = 45.9 cm2 atau 45.90cm2 atau 45.903

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021