KOMPIMatT DAERAH KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
3 (a) 3x2 − 6x − 7=0 1
1 5
(b) α + β = −b = 2 1
a 1
c −7 1
αβ = a = 3

(c) b2−4 3 −7

=120 >0
2 punca nyata dan berbeza

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
4 1 MARKAH
(a) y = 2x – 8
xy = 384 15
x (2x – 8) – 384 1
x2 – 4x – 192 = 0 1
(x + 12)(x – 16) = 0
x = 16 1
y = 24

(b) Perimeter = 24(2) + 16(2) = 80
Kos = 15(5) + 1(5) = 80

Kos = RM 130(5) + RM 11.50(5) = RM
707.50

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
5 1 MARKAH

(a) x2 – 8x + 8 2 8 2
2 − 2 −8

(x2 – 8x + 16)(−16 − 8)

(x – 4)2 – 24 1

(b) q = − 8 18
Bila x – 4 = 0, maka x = 4 3
Titik minimum (4, -8)
1
(c) bentuk graf
pintasan-y 1
titik minimum

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
1
6 m = 1
2

y − −2 = 1 x−2 1
2

y= 1 x − 3 atau x = 2y + 6 1
2

2y + 6 + y 2y + 6 − y = 15 1
7
3y + 6 y + 6 =15
3y2 + 24y + 21=0

y + 7 y + 1 =0 1

y = − 7 atau − 1

x=4 atau −8

A 4 , − 1 , B −8 , − 7 atau B 4 , − 1 , A − 8, − 7) 1

Titik tengah= 4−8 , −1 − 7
2 2

= −2 , − 4 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

7 Katakan; 1
Bilangan pear = x biji Bilangan oren = y bijiBilangan epal = z biji 1

x + y + z = 22 …………………( 1 ) 17
1
2(x) + 1.8(y) + 2.2(z) = 43.20 ………………………( 2 )
1
y = 2z ……………………….(3) 1
1
Gantikan ( 3 ) ke dalam ( 1 )
x + 2z + z = 22

X = 22 – 3z ……………………….( 4 )

Gantikan ( 3 ) dan ( 4 ) ke dalam ( 2 )
2(22 – 3z) + 1.8(2z) + 2.2(z) = 43.2

44 – 6z + 3.6z + 2.2.z = 43.2
0.2z = 0.8
z=4

y = 2(4) = 8
x = 22 – 3(4) = 10

Bilangan Pear = 10 biji
Bilangan Oren = 8
Bilangan Epal = 4

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
8 (a) ( p + q ) ln e = 3 1
q=3–p 1 9
2
(b) 2 + 4 7x = 6(7x) 1

(c) log2 a − log2 b + log2 22 1

= log2 4a 1
b

(d) ( 2x − 2)2 − ( x + 1 )2

2x − 2 = x + 2 x + 1

x−2 x−3= 0

x − 3 ( x +1) = 0 1
1
x =3 x=−1

x=9 x= 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
9 1 MARKAH
a = 25 d = − 2
1 10
25 + n − 1 −2 =1
1
n =13 terdapat 13 lapisan.
1
a=1 d=2
1
S12 = 12 [2(1) + (11)(2)] = 144
2 1
1
Purata sehari = 144
8 1
1
= 18 tin 1

Berbaki 1 lapisan paling bawah = 25 tin

18 x 3=54
Tidak cukup untuk 3 hari lagi

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

10 (a) H1= 0.85Ho = 0.852Ho 1
0.85nHo 1
H2= 0.85H1
hence, Hn = m ke
gantikan n = 7 dan Ho = 2 dalam persamaan.
(b) Hn = 0.85nHo

H7 = 0.857 2 1
= 0.6412 m 1
17
= 64 cm
1
(c) jadi, 0.805.n8HH5nno<<<000..5.55HHoo

lmoga1su0kk0a.n85long1<0,log10 0.5
n log10 0.85 < log10 0.5
n −0.0706 < −0.3010

n > 4.26 1

∴n = 5,

ketinggian bola akan kurang daripada separuh

ketinggian asalnya untuk pertama kali selepas

melantun sebanyak lima kali.

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
11 MARKAH

(a) n=25, d = 3, T2 = -5 1
d = T2 – T1 1
3 = – 5 – T1
T1 = – 8 17
1
(b) T22 = Sn n
−8 + 21 3 2 1
= 2 −8 + n−1 3 1
1
3n2 − 19n − 110=0

( n – 10 )( 3n + 11 ) = 0

n = 10 , n = − 11
3

n = 10

(c) T13 = − 8 + ( 25 – 1 )(3)
= 64

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
12 MARKAH
(a) T8 = 0.2 + (8 – 1)(0.05) 1
= 0.55 1

= 0.55 X RM 2.50

= RM 1.38 1
1
(b) S10 = 10 2 0.2 + 10 − 1 0.05
2 1
= 4.25 1 10

= 4.25 × RM 2.50 1
1
= RM 10.63 1

(c) = 522.50 = 209 1
2.50

n (2 0.2 + n−1 0.05 =209
2
n2+7n − 8360=0

n − 88 n + 95 =0

n = 88, n = − 95

n = 88

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
13
(a) First semicircle , a = 6π
d = 4π 1
1
6π + (n – 1)(4π) = 46π
1
n = 11 1
46π is the 11th semicircle.
7
(b) n 12π+ n − 1 4π =390π 1
2
4πn2 + 8πn − 780π=0 1
n2 + 2n − 195=0 1

n − 13 n +15 =0

n = 13 , n = − 15

n =13

T13= 6π + 13 − 1 4π
= 54π

Panjang OC = 54 cm

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
14 (a)
1.00 0.50 0.33 0.25 0.20 0.17
y 4.50 3.00 2.50 2.25 2.10 2.00 1

(b)

Semua titik
diplot dengan
betul [2]

3 atau 4 titik 6
diplot dengan
betul [1]

Paksi betul dan
mengikut skala
[1]

Garis lurus
penyuaian
terbaik [1]

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

(c) (i) y = 1 a + 2b 1 1
2 x

1 a = pintasan−y = 1.5 1
2 15
a = 3.00 1
1
(ii) 2b = kecerunan = 3.0 − 1.5 = 3
0.5 − 0

b = 1.5

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
1
15 (a) log10x 0 0.48 0.70 0.85 0.95

log10 y 0.16 0.48 0.63 0.72 0.80

(b) Semua titik diplot

dengan betul
[2]

3 atau 4 titik diplot 6
dengan betul [1]

Paksi betul dan
mengikut skala
[1]

Garis lurus
penyuaian terbaik
[1]

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

c (i) log10y = n log10x + log103 1
+ m+1
m 1

log103 = pintasan−y = 0.15 1
m+1 15

m = 2.18 1
1
(ii) n 1 = kecerunan = 0.78 − 0.15
m+ 0.93 − 0

n = 2.15

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
16
(a) y = 4 x − 1 ; m1= 4
3 2 3

4y = 48 + kx
k k
y = 12 + 4 x ; m2 = 4 2
1
m1 m2 = 4 x k = -1
3 4 1
1
k = -3
2
(b) 4( 4 x − 1 ) = 48 – 3x 1
3 2
1
x = 6 , y = 7.5 3

Q = ( 6, 7.5 ) 1
1
(c) Let R = ( x, y ) 1

6 1 + x (3) , 7.5 1 + y (3) = ( 0, 12 ) 2
1+3 1+3 1

6 + 3x =0 7.5 + 3y = 12
4 4

R = ( -2, 13.5 )

(d) y – 13.5 = 4 [ x – (−2) ]
3

y = 4 x + 55
3 6

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
17
(i) Koordinat-y/y- coordinates of S = 1 – (6-1)

S = -4 1

S(3, -4) 1

(ii) Persamaan garis QS/Equation of QS: x = 3 1

(iii) Luas sisiempat PQRS/Area of quadrilateral PQRS 1
1 1
= 2x 2 2+9 x5

= 55 units2

(iv) Koordinat T/Coordinates of T = ( 3,1) 1
10
tan ∡TRQ = 5
6 1
∡TRQ=39.81° or 39°48′

(v) Panjang PQ /Length of PQ Panjang PQ /Length of PQ =

= (3+2 )2+(6−1 )2 (9−3 )2+(1−6 )2 1
or = 61 units
= 50 units

Perimeter sisiempat PQRS / Perimeter of quadrilateral 1
PQRS = 2  ( 50 + 61)

= 29.76 units 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
18
(a) 2y = −x − 6 , x = − 1 x – 3 , mBD = − 1
2 2

− 1 x mMB = −1
2

mMB = 2 1
y – 9 = 2 ( x + 4)
1
y = 2x + 17 17
1
(b) 2(2x + 17) = -x – 6 1

2x + 17 = −x−6 1
2 1

4x + 34 = −x −6 , x = −8

B ( −8, 1)

(c) B(−8, 1) , S( 0,−3), D(x, y) , BS = 2
SD 3

3 −8 +2x =0 atau/ or 3 1 +2y = −3
5 5

D (12, −9)

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
19
(a) PR = 1
PS 3
3PR=PS 1

3 x − 4 2+ y − 2 2 = x+2 2+ y − 10 2 1

9 x2 − 8x+16+y2 − 4y+4 = x2+4x+4+y2 − 20y+100 1
18
9x2 − 72x+144+9y2 − 36y+36 = x2+4x+4+y2 − 20y+100 1
8x2 − 76x+76+8y2 − 16y = 0 1
2x2 − 19x+2y2 − 4y+19 = 0
(b) Q 1, 1 1
LHS = 2(1)2 − 19(1)+2 1 2 − 4(1)+19
LHS = RHS. Q(1, 1) lies on the locus of P 1
(c) x = 0, 2y2 − 4y − 19=0
b2 − 4ac = −4 2 − 4 2 19
= − 136
b2 − 4ac < 0 ; Not real root.

Lokus bagi P tidak bersilang dengan paksi-y. / Locus of P
does not intersects the y-axis

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
20 (a) mRQ= – k 1
2 1 7
1
k= 4 1

(b) mRS = 3 1
7 1

RS tidak selari dengan OP/ RS is not parallel 1
to OP.

(c) m = 1
2

c = –34

y = 21x – 3
4

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
21 (a) a = 8 1
1 7
10 − 5
mAB = mCD = 8 − 3 = 1 1

D = (8,b) 9− b = 1 1
11− 8 1

b=6

(b) 9−5 1
mAC = 11 − 3 = 2

mBE = −2

10 − y = −2 1
8−9 1

y=8

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
22
(a) Titik tengah PR/midpoint of PR = 1+7 , 3 +7
2 2

= (4,5)

Kecerunan PR/Gradient of PR = 7− 3
7−1
3
=2

Kecerunan normal PR/Normal gradient of PR = − 2
3

Persamaan pembahagi dua sama serenjang PR/

Equation of the perpendicular bisector of PR:
2
y−5= − 3 (x − 4) 14
1
3x + 2y − 22 = 0
1
(b) Q(0 , y) ; 2y − 22 = 0

y = 11 Q ( 0, 11)

titik tengah QS = titik tengah PR/Midpoint QS= midpoint PR

x + 0 , y + 11 = (4,5)
2 2

x=8 , y = −1 S (8, − 1) 1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
22
(c) Luas PQRS/Area of PQRS

= 1 1 0 7 8 1 1
2 3 11 7 −1 3

= 1 11 − 7 + 24 − (77+ 56 − 1)
2

= 52 unit2 1

QR = (7 − 0)2 + (7 − 11)2 4

= 65 1
Biar jarak tegak P dari QR/Let’s perpendicular distance P 1

from QR = h

65  h = 52
h = 6.4498 unit

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

23 A(-2,-4), B(5, -2), C(p, q), D(-1, 4)

(a) −2+p = −1+5
2 2

-2 + p = 4
p=6

−4 + q = −2 + 4
2 2

-4 + q = 2

q=6

Maka/ therefore C(6,6)

OC = 6i + 6j

(b) −1 + 5, −2 + 4
2 2
M= = (2,1)

OM = 2i + j

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
24
(a) ma˷ – mb˷ = kc˷

m 2 −n 1 =k 3
−3 2 −1

2m − n = 3k
−3m − 2n −k

2m – n = 3k ---------- ❶ 4
-3m – 2n = -k -------- ❷

k = 3m + 2n
2m – n = 3(3m + 2n)
2m – n = 9m + 6n

-7m = 7n

m=-n

(b) ma˷ – mb˷ = - c˷

m 2 −n 1 =− 3
−3 2 −1

2m − n = −3
−3m − 2n 1

2m – n = -3 ---------- ❶ x 2
-3m – 2n = 1 -------- ❷
4m – 2n = 6

-7m = 7

m=-1
Ganti m = -1 dalam ❶

n=1

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
24
(c) Magnitud vector unit/magnitude of unit vector =1 4
3
Kecerunan garis lurus/gradient of straight line = −

(2m – n)2 + (−3m – 2n) 2 = 1-------- ❶
−3m – 2n 4
2m –n =− 3 -------- ❷

3(-3m – 2n) = -4(2m − n)

−9m – 6n = −8m + 4n

m = - 10n

Ganti m = −10n dalam ❶

[2(-10n) – n)]2 + [-3(-10n) -2n] 2 = 1

(-21n)2 + (28n)2 = 1

1225n2 = 1 1
35
n = ±

m = −10 (315) = −27 m = −10 (− 315) = 2
7

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

25 OB = 10 ˷y OA = 6 x˷
OB = AO + OD = − 6x˷ + 5˷y
OB = BO + OC = −10y˷ + 4x˷

(a) (i) OE = OA + AE = OA + h AD

= 6x˷ + h(−6x˷ + 5y˷ )

= (6 − 6h)˷x + 5h˷y

(ii) OE = OB + BE = OB + k BC

= 10˷y + k(-10˷y + 4k˷x)

= (10 – 10k)˷y + 4kx˷

6 – 6h = 4k ------- ❶

10 – 10k = 5h -------- ❷ ÷ 5

2 – 2k = h

Ganti/substitude h = 2 – 2k dalam ❶

k = 3
4

Ganti/subsitude k = 3 dalam h = 2 -2k
4

PANIhT=IA12 MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
25
(b) OD = 5 x 1 = 5 unit, OA = 6 x 1.5 = 9 unit

Menggunakan petua kos/using cosine rule
(i) AD2 = 52 + 92 – 2(5)(9)cos120º

AD = 12.288 unit @ ( 151)

(ii) Menggunakan petua sin/using sin rule

sinAOD = sin 120°
5 151

sinAOD = sin 120°
151

AOD = 20.63º

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 9 (TINGKATAN 4) / CHAPTER 9 (FORM 4)
PENYELESAIAN SEGI TIGA/ SOLUTION OF TRIANGLES

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
26
(a) (i) QS2 = (10.4)2 + (9.8)2 – 2(10.4)(9.8)kos/cos 82º

QS2 = 204.2 – 28.369

QS2 = 175.831

QS = 13.26 cm

(ii) sin  PSQ = 9.8 sin 82°
13.26

sin  PSQ = 0.7319

 PSQ = 47º 2´

(ii) Luas/ Area △QRS = 1 (13.26)(10.6) sin 43°.
2
= 47.93 cm2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 9 (TINGKATAN 4) / CHAPTER 9 (FORM 4)
PENYELESAIAN SEGI TIGA/ SOLUTION OF TRIANGLES

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
27
(a) (i) PSR= 180° – 110° = 70°

PR2 = 82 + 92 – 2(8)(9)kos/cos 70°

PR2. = 95.749

PR = 9.785 cm

(ii) sin QPR = 4sin110°
9.785

sin QPR = 0.3841
QPR = 22o 35’

PRQ = 180o– 110o – 22o 35’

PRQ = 47o 24’

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 9 (TINGKATAN 4) / CHAPTER 9 (FORM 4)
PENYELESAIAN SEGI TIGA/ SOLUTION OF TRIANGLES

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
28
(a) (i) Luas/Area △QRS = 14.41

1 x 6 x 7 sin QRS = 14.41
2

sin QRS = 14.41
21

sin QRS = 0.6862

QRS = 180o– 43o 19’ = 136o 41’

(ii) QS2 = 62 + 72 – 2(6)(7)kos/cos 136o 41’

QS2 = 146.12

QS = 12.088 cm

(iii) sin  RQS = 6sin136° 41′
12.088

sin RQS = 0.3452
RQS = 19o 54’

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 9 (TINGKATAN 4) / CHAPTER 9 (FORM 4)
PENYELESAIAN SEGI TIGA/ SOLUTION OF TRIANGLES

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
29
(a) 3.5 sin120°
(I) AC = sin 15°

AC = 11.71 cm
(ii) AE2 = 4.52 + 42 – 2(4.5)(4)kos/cos 50o

AE2 = 34.6835

AE = 5.89 cm

CE = 11.71 – 5.89 = 5.82 cm

(iii)  BAC = 180o – 120o – 15o = 45o

Luas/Area △ABC = 1 (3.5)(11.71)sin 45o
2

= 14.49 cm2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 10 (TINGKATAN 4) / CHAPTER 10 (FORM 4)
NOMBOR INDEKS/ INDEX NUMBER

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
30 (a) x × 100 =135 ⇒ x = 2.70 x = 2.70 1
2.00 y = 120 1 3
4.20 z = 3.40 1
y= 3.50 × 100 =120 1 3
1
4.93 × 100 = 145 ⇒ z = 3.40 2
z 1
2
(b) m =100 − 40 − 28 = 32 m = 32 1
Iҧ =133 1
Iҧ = 135 40 +120 32 +145(28) 1
100 1
13300
= 100

= 133

(c) (i) P2020 ×100=165⟹ P2020 = 165 P2018 = P2020 × P2018 ×100
P2014 P2014 100 P2014 P2014 P2020

P2020 ×100=133⟹ P2020 = 133 P2018 = 165 × 100 ×100
P2018 P2018 100 P2014 100 133

P2018 =124.06
P2014

(ii) P2020 ×100=165⟹ P2020 = 165 P2020= 165 ×20=33 sen
P2014 20 100 100

Bil maksΤMax no= 200×100 = 606.06 606
33
≈606

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 10 (TINGKATAN 4) / CHAPTER 10 (FORM 4)
NOMBOR INDEKS/ INDEX NUMBER

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

31 (a) 15 ×100 = 125 ⇒ x = 12
x
x =12 1
13
y × 100 = 115 ⇒ y = 28.75 y = 28.75 1
25 z = 120

18 × 100 = z ⇒ z = 120
15

(b) Iҧ = 125 40 +115 50 +110 n +120(65) 2
200 3
18550+110n
117.5 = 200 1

n = 45

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

BAB 10 (TINGKATAN 4) / CHAPTER 10 (FORM 4)
NOMBOR INDEKS/ INDEX NUMBER

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

31 (C)

Comic Price Index for the year
2007 based on 2005
P 120
Q 110
R 110
S 110

Iҧ = 120 40 +110 50 +110(45)+110(65) 1
200 4
Iҧ =112
1
I2007 = I2007 × I2005 ×100
I2002 I2005 I2002 1
I2007 112 117.5 1
I2002 = 100 × 133 ×100

I2007 =131.6
I2002

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

1 (a) sin ∠BAD= 8
12

∠BAD = 41.81° × 3.142
180°

= 0.7298 rad

(b) ∠AOB = 3.142 – 2(0.7298) = 1.6824 rad

Major ∠BOD = 1.6824 + 3.142 = 4.8244 rad

Perimeter seluruh rajah/Perimeter for the whole diagram
30°
= 4.8244(6) + 180°  × 3.142  × 8 + 8

= 28.946 + 4.189 + 8

= 41. 135 cm

(c) ∠BOD = 3.142 – 1.6824 = 1.4596 rad

Luas tembereng BED/Area of segment BED

= 1 (6)2 1.4596 − 1 (6)2 1.4596 ×180°
2 2 3.142

= 26.2728 – 17.8885
= 8.3843 cm2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

2 (a) (i) sin ∠AOC =  8
10

∠AOC = 53.13° × 3.142
180°

= 0.9274 rad

(ii) Jejari bulatan /Radius of the circle = 102 − 82

= 6 cm

(b) AC = 0.9274(6)
= 5.5644 cm

AB = 10 – 6 = 4 cm
Perimeter rantau ABC/Perimeter of region AB
= 5.5644 + 8 + 4

= 17.5644 cm

(c) Luas rantau ODB/Area of region ODB

= 1 3.142 (6)2 + 1 (6)(8)
2 2 2

= 28.278 + 24

= 52.278 cm2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

3 (a) tan ∠BOA= 12
5

∠BOA= 67.38° ×3.142
180°

= 1.176 rad

(b) OB = 122+52 = 13 cm

OQ = 4 13 = 10.4 cm
5

Perimeter kawasan berlorek/Perimeter of the shaded region

= 0.79(13) + 0.79(10.4) + 2(13 – 10.4)

= 10.27 + 8.216 + 5.2

= 23.686 cm

(c) Luas rantau ODB/Area of region ODB

= 1 (13)2 0.79 − 1 (10.4)2 0.79
2 2

= 66.755 – 42.7232

= 24.0318 cm2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

4 V = πj2h, j = 5, dV = p, dh = 2
dt dt
dV
V = 25πh , dh  = 25π

(a) dV = dV × dh
dt dh dt

p = 25π × 2 = 50π

(b) A = 2π 5 h

dA  = 10π
dh

dA  = dA × dh = 10π × 2 = 20π cm s-1
dt dh dt

(c) 1000π = 20 saat
50π

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

5 (a) P dan Q memintas paksi-x/P and Q cuts the x-axis,
maka/then y=0

x + 2 x − 1 x − 2 =0

x = − 2, x = 1 and x = 2
Oleh itu/Therefore , P ( 1 ,0 ) dan/and Q ( 2 , 0 )

(b) f x dx

1 f x dx − 2 f x dx
0 1

x + 2 x2 − 3x + 2 atau/or setara/equivalent

1 (x3 − x2 − 4x + 4 )dx+ 2 (x3 − x2 − 4x + 4 )dx
0 1

1 x3 − 1 x3 − 2x2 + 4x
4 3

1 11 atau/or − 7
12 12

2 1
2

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

6 (a) 9 − x 2 = 9 x+1 1
x − 24 x − 3 =0
1

3,6 1

(b) Luas Trapezium/Area of trapezium = 1 (3+9)(6) 1
2

Luas/ Area = 6 y2 − 1 dy= y3 − y 6 1
3 2 2 3

 Luas kawasan berlorek/Area of shaded region 19
1
= 36 – 4 1

= 32

(c) 3 y2 − 2 3 y4 − 2y2 + 1
0 9 dy atau/or π 0 81 9
v=π 1 dy

=π y5 − 2y3 + y 1 1
405 27 1

∗  integrate and sub the limit correctly

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

7 (a) 6  6 =36
(b) (i) 100C24  80C12 = 4.8062  1036
(ii) 24!  12! = 2.972  1032

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

8 (i) 10! = 2520
(2!5!3!)

(ii) 8! = 168
(2!5!)

(iii) 8! + 8! + 8! =784
(2!5!) (5!3!) (2!3!3!)

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

9 (a) (i) ¹⁰C₇ (0.6)⁷ (0.4)³

= 0.215

1 P (x = 0) – P (x = 1) atau/or
P (x = 2) + P(x = 3) + P (x + 4) +……………+ P(x = 10)

(ii) 1 – ¹⁰Cₒ (0.6)⁰ (0.4)¹⁰  ¹⁰C₁ (0.6)¹ (0.4)⁹

= 0.9983 9

(b) (i) P (z < 345 - 350)
5

= 0.1587 or 0.15866

(ii) P(x>342 = 1 – P (z >1.6 ) or other valid method

Number of cakes = 0.9452 x 1500
= 1417 atau/or 1418

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

10 (a) n = 12, p= 0.8

(i) P (X = 10) = ¹²C₁ ₀ (0.8)¹º (0.2)²

= 0.2835

(ii) P (x < 10) = 1 – P (X = 11) – P (X= 12)

= 1 – ¹² C ₁ ₁ (0.8)¹¹ (0.2¹) – ¹² C₁ ₂ (0.8)¹²

(0.2)⁰

= 0.7251

(b) (i) μ = 1850, σ =60

P (X > 1935) = P (x – μ > 1935 – μ)
σ σ

=P ( Z > 1935 - 1850 ) 9
60

= P ( X > 1.417)

= 0.0783

(ii) P ( X . M ) = 0.8250

P (Z > m – 1850 ) = 0.8250
60

P ( Z . – 0.935 ) = 0.8250

m – 1850 = - 0.935
60

m = -1850 + 60 (-0.935)

PANITmIA= M- 19A0T6 EMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

11 (a) kosek2x + 2kos2 x – kot2 x – 2 = kot 2x

Sebelah kiri/LHS

= kosek2x + 2kos2 x – kot2 x – 2 Guna identiti asas/use 1
= kosek2x + 2kos2 x – (kosek2 x – 1)  2 basic identity
= kosek2x + 2kos2 x – ksekt2 x + 1  2 kot2 x = kosek2 x – 1

= 2kos2 x  1 1

= kos 2x 1
= Sebelah kanan/RHS

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
11 (b) (i)

11)
(b) (i)

Graf kosinus yang betul /Correct Cosine graph 1
2 kitaran/2 cycle 1
Amplitud/Amplitute (-1/2 hingga/to 3/2) 1
Graf dinaikkan sebanyak/Shifted graph - ½ 1

(b) (ii) 4kosek2x + 8kos2 x – 4kot2 x – 8 - 3x + 4 = 0 1
1
4kosek2x + 8kos2 x – 4kot2 x – 8 = 3x  4 1
kosek2x + 2kos2 x –kot2 x – 2 = π3x  1

y = π3x  1
Lukis garis/Draw the line ; y = π3x  1
Bilangan penyelesaian/Number of solutions = 3

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH

12 (a) 1 −  kot x =tan x
sin x kos x

Sebelah kiri/LHS

= sin x 1 x −  kot x
kos

= sin x 1 x − kos x 3
kos sin x 1

= 1- kos2 x 1
sin x kos x 1

= sin2 x x
sin x kos

= tan x

= Sebelah kanan/RHS

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
12
(b) 2 kos 1 x
sin x kos x sin 2
(i) − 2kot x = 1
2 x

2 tan x = kot 1 x
2

2 2tan 1 x = 1
2
1 1
1 − tan 2 x tan 2 x

4tan2 1 x = 1  tan2 1 x
2 2

5tan2 1 x = 1 3
2
1
tan2 1 x = 1 1
2 5 1

tan 1 x = ±  1
2 5

Sudut rujukan/Reference angle = 1 x = 24.09
2

1 x = 24.09, 155.91
2

x = 49.18 , 311.82

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021

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

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

SOALAN PENGIRAAN MARKAH JUMLAH
MARKAH
12 (b) (ii)

Graf tangen yang betul/Correct tangent graph 1
2 kitaran/2 cycle 4
Graf diturunkan sebanyak/Shifted graph 1
1
1
1

PANITIA MATEMATIK TAMBAHAN PPD 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

13 (a) x + y ≤ 1000
x ≥ 3y
5x + 4y ≥ 2800

(b) Rujuk graf

(c) (i) 450

(ii) Keuntungan
maksimum sebulan
= 10(750) + 8(250)

= RM9500

PANITIA MATEMATIK TAMBAHAN PPD KINTA UTARA 2021