SKEMA JAWAPAN PEPERIKSAAN PERCUBAAN KERTAS 1 2017 SOALAN 1 BIL JAWAPAN SUB MARK MARKAH PENUH 1 1 ) 3 2 ( log : B1 5 ) 3 2 ( : B2 5 1 x x x x 2 5 x 3 3 2 1 3 1 2 x x 2 log x x 3 B2 B1 3 3 25 27 : 1 25 1 3 3 : 2 2 1 3 log log log log log a a a a a a B B n m 3 4 3 x – 3y B2 : y x 3 2 1 2 3 ) 3 ( B1 : m = 9 x atau n = 3 3 y 3 3 5. 1 B2 : 3 1 3 log log 5 5 B1 : 3 3
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SKEMA JAWAPAN PEPERIKSAAN PERCUBAAN KERTAS 1 · skema jawapan peperiksaan percubaan kertas 1 2017 soalan 1 bil jawapan sub mark markah penuh 1 b1: log (2 3) 1 b2 : (2 3) 5 5 1 x x
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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