SKEMA JAWAPAN
PEPERIKSAAN PERCUBAAN KERTAS 1
2017
SOALAN 1
BIL JAWAPAN SUB MARK MARKAH
PENUH
1
1)32(log:B1
5)32(:B2
5
1
xx
xx
2
5x
3 3
2 1
3
1
2
x
x
2log
x
x
3
B2
B1
3
3
2527:1
25133:2
213
logloglog
loglog
aaa
aa
aB
B
nm
3
4 3x – 3y
B2 : y
x
3
2
1
2
3
)3(
B1 : m = 9x atau n = 33
y
3 3
5. 1
B2 :
3
13
loglog
5
5
B1 :
3 3
SOALAN 2
1
1
7
7
7
7
1
7
777
7log
)753(log
7
1log
105log:B1
7log
7log5log3log:B2
1
1:B3
atau
kh
1kh
4
4
2
kh
hk
2
12
3log2log2
5log2log3log2
55
555
5log2log2log 55
2
5 or 3log2log 5
2
5 or
5log2log3log 12122
12
12log
90log
5
5 or 2 log 5 3 or 2 log52
4
B3
B2
B1
4
3 (a)
p
1
(b) 12
25
p
p
13log
23log2
5
m
m
m
m
m
m
9log
243log 2
1
3
B2
B1
4
4
2
23 xy
B3 : 2
log8loglog2 222 bc
B2: 4log
8loglog
2
2
2
2 bc
B1: bc 8loglog 4
2
4 @ 4log
8log
2
2
2b
c
4 4
5
q
pq
1
B3 : log2 3(1 – p) = pq
B2 : log2 3 = p(log2 4 + log2 3)
B1 : 12log
3log
2
2 = p
4 4
SOALAN 3
1 15
B2 : )2)(1()5)(1()42( nxnx
B1 : )5)(1()42( nx atau
)2)(1( nx
3 3
2 n = 42
87)2)(1(5 n
d = 2
3
B2
B1
3
3 1256
)168(2))8(4)200(2(2
55 S
)8(42005 T
3
B2
B1
3
4 a) k = 2h + 1
k -3h = (h + 2) – k
b) 9 – 6h
2
B1
1
3
5 a) d = 6
b) 5
B1 :
)5(283)6(283 )5()6(
22
1
2
3
SOALAN 4
1 7 1 3
4
B1 : 282 T atau 7
28
2
2
6
1
3
11
9
1
3
1r
3
B2
B1
3
3 3
2
6r dan p = 6
32
6r atau p = 6
p
p
p
ppp
18
2
36
6,0,6
2
3
B2
B1
3
4 x1 = 3 dan x2 = –1 B2 : (x – 3)(x + 1) = 0
B1 : 3x2 – 6x – 9 = 0
3
5 -24
B2 :
4
31
6
B1 : a = - 6 , 4
3r
3 3
SOALAN 5
1 (a) 5
B1 : )2(
)4(
)4(
)2(
x
x
x
x
2 4
(b)
2
243
B1 : a = T1 = 81 ...seen atau
3
11
81
2
2 (a)
3
2
5
4
27
8 or equivalent
2
3
1
2
T
T
T
T
(b) 3
4or equivalent
3
21
9
4
3 a) x = 6
B1 : 12
2412
x
b) 6096
B1 :
12
16
12
16 22310
1
1
4
4 atau 5.5
B3 :
2
111
2T
B2 : a = 11
B1 : 22
2
11
a
4 4
5 a) a = 8 , - 2
B1 :4
224
a
a
a
a
b) 16
2187 atau 136.6875
B1 :
2
38
9 3
16T
1
1
4
SOALAN 6
1 y = x3
3
3
B2 : 4 = 4 + c atau 12 = 12 + c atau c = 0
B1 : m = 1
2 p = 2 and 1q
p = 2 or 1q
04
)5(3
p or 55 q
qpxxy 52 2
3
B2
B1
3
3 s = 3 and t = 5
7 = 2s + 1 or t = 2(2) + 1
3
B1
3
4 r = 1, p = 0.001 (BOTH)
r = 1 or p = 0.001 (either one)
=-3 or r = or
3
5 m =
2
3 dan n = 3
B1 : –2 = m
3
y = –2x
2 + 3x
2
1
3
SOALAN 7
BIL JAWAPAN SUB MARK MARKAH
PENUH
1 2 rad
B3 : 0)2)(2(
4
4
B2 : 1002
40
2
12
B1 : 40 rrr
2 (a) 842.1
5.65
(b) 23.025
)842.1()5(
2
1 2 * (candidate’s from a)
2
B1
2
B1
4
3 (a) 10.8 cm
(b) 19.44 cm2
)142.3
1809.0tan(612
2
19.012
2
1 2
6
9.0tanh
1
3
B2
B1
4
4 (a) θ = 0.842 rad
kos θ = 15
10
(b) 126.708 cm2
Area = 88)18.11)(10(
2
1)841.0()15(
2
1 2
2
B1
2
B1
4
5 (a) 1.75 rad
(b) 46. 5 rad
SOALAN 8
BIL JAWAPAN SUB MARK MARKAH
PENUH
1 (a)
15
17
(b) 221
21
13
5
17
15
13
12
17
8 B1
1
2
3
2 (a) tan A =
12
5
(b) sin (A – B) = 65
56
5
3
13
12
5
4
13
5
1
2
B1
3
3 θ = 45o, 63.43
o, 206.57
o, 225
o
θ = 63.43o, 206.57
o OR θ = 45
o, 225
o
(tan θ - 2)(tan θ - 1) = 0
3
B2
B1
3
4
7
17 atau 2.4286
B2 :
)1)(5
12(1
15
12
atau tan (112.62 – 45)
B1 : tan = 5
12 atau tan 45 = 1
3 3
5 (a)
p
1
1
2
3
(b)p
p21
21 p seen
B1
SOALAN 9
BIL JAWAPAN SUB MARK MARKAH
PENUH
1 x = 0o, 30
o, 150
o, 180
o, 360
o (all correct)
B3 : 0o and 30
o (both)
B2 : sin x (-2sinx +1 )= 0
B1 : (1-2sin2x) + sin x - 1 = 0
4 4
2 0, 120, 180, 240, 360
B3 : 0, 180, 360 atau 120, 240
B2 : sin x(2 kos x + 1) = 0
B1 : 2 sin x kos x + sin x = 0
4 4
3 x =18.435o, 45
o,198.435
o, 225
o
B3 : 18.435o, 198.435
o or 45
o, 225
o
B2 : (3 tan x – 1 ) ( tan x – 1 ) = 0
3 tan2 x – 4 tanx +1 =0
B1 : 3 ( 1 + tan2x)- 4 tan x -2 = 0
4 4
4 x = 68o12’, 90
o, 248
o12’ , 270
o
B3 : cos x = 0 and x = 90o,270
o
OR
4 4
B3 : tan x = 2
5 and x = 68
o12’, 248
o12’
B2 : cos x (2 sin x - 5 cos x) = 0
B1 : 5 sin2x = 5 - 2sin x cos x
5 x = 35o 16’, 144
o44’ , 215
o16’, 324
o44’
B3 : 3
1sin x
B2 : 3(1 - 2sin2 x) = 1
B1 : 3 (cos2 x - sin2 x) = 1
4 4
SOALAN 10
BIL JAWAPAN SUB MARK MARKAH
PENUH
1 (a) 36.6
B1: (16.8 x 2) + 3
2
1
3
(b) 7
2 6
B2 : 22
10
1060a
a
B1 : Set 1 : x2 = 24 + 6a
2 atau Set 2 : x
2 = 36 + 4a
2
dengan a = min
3 3
3 a) 12/3 = 4
b)13.2665
1
2
3
4 m = 11
B1 : (m - 11)(m + 4) = 0
B1 : 143
7
3
25422
mm
3 3
5 1A and 323.1B
B1 : 22 2
8
462
8
40 BA OR
(b) Set A is to be preferred as the standard deviation of
set A is smaller than set B
2
1
3
SOALAN 11
BIL JAWAPAN SUB
MARK
MARKAH
PENUH
1 (a) 252 1
3 (b) 66
B1 : 4C3 ×
6C2 atau 60 atau
4C4 ×
6C1 atau 6
2
2 277 200
3 3
B2: 2
5
3
8
4
12 CCC
3 6270
B2 : 4C3 x
11C4 +
4C2 x
11C5 +
4C1 x
11C6 +
4C0 x
11C7
B1 : 4C3 x
11C4 atau
4C2 x
11C5 atau
4C1 x
11C6 atau
4C0 x
11C7
3
3
4 (a) 210 1
3 (b) 30
B1: 2
2
2
5
2
3 CCC
2
5 11
B2: 4
4
3
4
2
4 CCC
B1: seenCorCorC ...4
4
3
4
2
4
3
3
SOALAN 12
BIL JAWAPAN SUB
MARK
MARKAH
PENUH
1
(a) 10
3
1
3
(b) 10
1
B1 : n (M N) = 2 atau {3, 15}
2
2
15
1 or an equivalent single fraction
B2: 6
2
5
2
6
3
B1: 6
3 or
5
2 or
6
2
3
3
3 6
B2: 24x
x x
24
22 +
24
24
x x
24
2 =
4
1
B1: 24x
x x
24
22 atau
24
24
x x
24
2
3
3
4 (a)
10
3
B1: 2
1+ P(G) =
5
4
2
3
(b) 5
1
1
5 (a)
216
1 // 0.00463
1 3
(b) 72
25
B1: 6
1
6
5
6
5
6
5
6
1
6
5
6
5
6
5
6
1
2
SOALAN 13
BIL JAWAPAN SUB
MARK
MARKAH
PENUH
1 (a) 0.1573
B1: 0.5 – 0.3427
2
3
(b) 0.1573 1
2 (a) 60 2 3
B1: 5.012
54
X
(b) 30.85% 1
3 (a) 0.1977 1
3 (b) 31.75
B1: – 0.85 = 5
36x
2
4 (a) 0.9788 1
3 (b) 48
B1: 24
40
X
2
5 (a) 1.25
B1: 2.1
2.57.6 Z
2
3
(b) 0.1056 1
SOALAN 14
BIL JAWAPAN SUB
MARK
MARKAH
PENUH
1 (a) t = 1
B1: 23
24
t
2
4
(b) y = 2x – 4
B1: m2 = 2 atau 2 = 2(3) + c
2
2 (a) h = 4
B1:
(b) D (5, 9)
B1:
2
2 4
3 (10, 7)
B3: x = 10 atau y = 7
B2: 25
0
x atau 3
5
8
y
B1: AC : AB = 1 : 5 atau AC : CB = 1 : 4
4
4
4 9.7082
B3 : 22 )2(6()
2
16(
B2 : Koordinat titik pembahagi tembereng garis
dengan nisbah m : n; P(2
1, –2)
B1 :
31
)26(1)6(3,
31
)6(1)6(3
4
4
5 (a) h = 8
B1: 6
710
62
75
h
(b) (0, – 2)
B1: x = 0 atau y = – 2
atau
4
)1(6)3(2 atau
4
)1(7)3(5
2
2 4
SOALAN 15
BIL JAWAPAN SUB
MARK
MARKAH
PENUH
1 p = 1 dan q = ½
B2: p = 1 atau q = ½
B1: 3p + 4q – 5 = 0 atau p + 6q – 4 = 0
3
3
2 k =
5
6
B2 : 5 = 3m atau – 3 = – (3 + k) m
B1: 5a – 3b = m [3a – (3 + k)] b
3
3
3 h = 7
B2: 24 or 3 = )1(2
1h
B1:
4
)1
2
3 h
3
3
4 (a) – 3x
(b) 6x – 5y
B1 : QP = QR + RP
1
2 3
5 (a) 5w – 3u
(b) – 10w + 9u
B1: – 15w + 9u + 3u + 5w – 3u
1
2 3
SOALAN 16
BIL JAWAPAN SUB MARK MARKAH
PENUH
1 . a) = 5i – 8j
= 4i – 6j + i - 2j
PQ = PO + OQ
b) = 11i – 18j
= 15i -24j –(4i – 6j)
B1
B2
B1
3
OR = 15i – 24j –PO
PO + OR = 3(5i – 8j)
2 a) i – 4j
b) 17
4 ji
: 22 )4()1( atau 17
B1
2
B1
3
3 12.65
22 124
ji 124 or ji 124
3
B2
B1
3
4
53
27~~ji
53OC
~~~~34511 jiji
3
B2
B1
3
5 h =
7
8 k
–h + k = – + k and –8 = –7
AB =
8
khatau BC =
7
1 k
3
B2
B1
3
SOALAN 17
BIL JAWAPAN SUB MARK MARKAH
PENUH
1 = 2 18π = π h
r = 3 V = π h
V = π (6)
V = 2
= 4 πr
= x = x 2
=
2 r = 5
408 r
3.0
128
r
rdr
dA8
3
B2
B1
3
3 π1.2
3
3
)2.0()3(2
2.0dt
dratau r
dr
dA2
B2
B1
4 10
42
32
x or 4
2
23
2
2
23
xdx
dp
3
B2
B1
3
5
8
15
25
16h
2 = 9 x
4
1
dh
dV = 2
25
16h
3
B2
B1
3
SOALAN 18
BIL JAWAPAN SUB MARK MARKAH
PENUH
1 )(
8
1xf
)(4
1
2
1)(
2
1xfxdxxg
)(4
1)( xfdxxg
)()(4 xfdxxg
)(4)( xgxfdx
d
3
B2
B1
3
2
a) 2
2)(x
7
dx
dy
3
B2
3
2)2(
13222
x
xx
dx
dy
(b)c
2x
64x
B1
3 y =
3
3x+ x
2 + x + 4
1 = 3
)3( 3 +
2
)3(2 2 + (–3) + c
y = 3
3x +
2
2 2x + x + c
3
B2
B1
3
4
3
B2
B1
3
5 546 2 xxy
c = ‒ 5
c
)1(42
)1(62
2
cxx
y 42
6 2
y = dxx 46
46 xdx
dy
3
B2
B1
3
SOALAN 19
BIL JAWAPAN SUB MARK MARKAH
PENUH
1 = 18
= 10 + (2(4) -2(0))
= 2 5 + [2x]4
0
24 4
0 0( ) 2f x dx dx
3
B2
B1
3
2 22
2
2
pp
)42(22
2
p
p
p
xx
2
2
22
3
B2
B1
3
3 5
12 – 7
seenataudxxfdxxg 7)()(4
2
4
2
3
B2
B1
3
4 h = 3
3]5)3(2[
h –
3]5)2(2[
h = 7
32
3
2
3)(
)52(xorimitlcorrectthewith
x
h
3
B2
B1
3
5 −209
−11(19)
66 + ∫b,6f(x) dx = 85
3
B2
B1
3
SOALAN 20
BIL JAWAPAN SUB MARK MARKAH
PENUH
1 (a) 1.............................1
(b) 2..................................9
1................0))(1(4)6( 2
q
Bq
1
B1
2
3
2 p = -16 and k = 5
2(3)2+3p+30 = 0
2x2 -16x + 30 = 0
(x - 3)(x - 5) = 0
3 (a) x − 3 or x 1
0)1)(3( xx or
0322 xx
3
B2
B1
3
4 x ≤ −4 or x ≥ 7 3 3
- 3 1
(x + 4)(x − 7) ≤ 0
B2
B1
5
3
B2
B1
3
SOALAN 21
BIL JAWAPAN SUB
MARK
MARKAH
PENUH
1
6)1( 2 x
26)10(4 2 aor
.6)1(2)( 2 xxf
B1
B2
3
2 h = −1 and k = −10
(x − 1)2 – 10
2(x2 − 2x + (−1)
2 − (−1)
2 − 4)
3
B2
B1
3 a) 1
1/2 -1/ 3
b)
c)
1
1
4 (a) 1)1( 2 xy
2)2
2()
2
2(2 222 xx
(c) ( −1,1)
2
B1
1
5 610 p
0)6)(10( pp or
0)8)(2(4)2( 2 pp
3
B2
B1
SOALAN 22
BIL JAWAPAN SUB MARK MARKA
H
PENUH
1 (-2)2- 4(1)(3 – k) >0
4 – 12 + 4k > 0
k > 2
B1
B2
3
2 b2 − 4ac < 0 : Tiada punca/ No real roots:
−12q2
(2pq)2 − 4(p
2 + 3)(q
2)
3
B2
B1
-10 6
3
B1
B2
3
4
15 xy , pxxy 22
pxxx 2215
0)1(42 2 pxx
Tiada punca nyata, 042 acb
0)1)(2(4)4( 2 p
088 p
1p
B1
B2
3
5
1)3(2 xmx
01)3(2 xmx
Dua punca nyata, 042 acb
0)1)(1(4))3(( 2 m
0469 2 mm
0562 mm
0)5)(1( mm
5,1 mm
B1
B2
3
SOALAN 23
BIL JAWAPAN SUB
MARK
MARKAH
PENUH
1
B1
B2
3
2
n(x) = 1 - x
x=2
1
B1
2
1
3
3
2,
23
32
m
m
m
2)1
(
)1
(3
m
mm
m
)1
(m
mg
3
B2
B1
4
g(x)=2x+k
g2(x) = 2(2x+k)+k
= 4x+3k
∴ @ sebarang satu nilai
h /k (Both)
B1
B2
3
5 (a) b
(b) a
b
3
5
a
b
b
53
1
2
B1
SOALAN 24
BIL JAWAPAN SUB MARK MARKA
H
PENUH
a)
b)
1
B1
2
2 (a) k = 1
(b) t = 4, t = -3
0122 tt
1
2
B1
3 1. f(-1) = 1
p-q-5 =1
p-q = 6 p-q=6 / p+q =-4
f(1) = -9 +
p+q-5=-9
p+q = -4
sebarang satu set
2p=2
p=2
∴1+q=-4 both
q=-5
B1
B2
3
4 a) 1
b) x = 22.5, 112.5
2x = 45°, 225°
1
2
B1
5 x =
4
3
5x + 3 = 5
3x
f –1
= 5
3x
3
B2
B1
SOALAN 25
BIL JAWAPAN SUB
MARK
MARKAH
PENUH
1 a) 5
b)
c) Hubungan satu ke banyak
1
1
1
2 (a) or
(b) many to one
c)
1
1
1
3 a) banyak-banyak
b)
c)
1
1
1
4 a) Kodomain = { a, b, c, d, e }
b) Satu kepada banyak
c) Julat = { a, b, d, e }
1
1
1
5 a) {(5, g), (7, h), (9, h), (9, i), (11, g), (11, h), (11, i)}
b) Many-to-many relation Hubungan banyak dengan banyak
c) {g, h, i}
1
1
1