SULIT 3472/2 Additional Mathematics Kertas 2 Ogos 2016 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2016 PERCUBAAN SIJIL PELAJARAN MALAYSIA ADDITIONAL MATHEMATICS Kertas 2 PERATURAN PEMARKAHAN Peraturan pemarkahan ini mengandungi 12 halaman bercetak
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SULIT 3472/2 Kertas 2 2016 BAHAGIAN PENGURUSAN … · 2016-09-02 · ADDITIONAL MATHEMATICS [P2] TRIAL SPM 2016 3 Number Solution and Marking Scheme Sub Marks Full Marks 2 2x 4y 24
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SULIT
3472/2
Additional Mathematics
Kertas 2
Ogos
2016
BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH
DAN SEKOLAH KECEMERLANGAN
PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2016
PERCUBAAN SIJIL PELAJARAN MALAYSIA
ADDITIONAL MATHEMATICS
Kertas 2
PERATURAN PEMARKAHAN
Peraturan pemarkahan ini mengandungi 12 halaman bercetak
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2016
2
Number Solution and Marking Scheme Sub
Marks
Full
Marks
1 (a)
1st method
2
2
4
pxpSee
p
pq
2
44 2See
p
pq
pxp)x(f
422
K1
K1
N1
2nd
method
px
2
4See
qpp
pp
f
24
222
p
pq
pxp)x(f
422
K1
K1
N1
3rd
method
2
2
4
pxpSee
qp
p
2
2
4See
p
pq
pxp)x(f
422
K1
K1
N1
(b)
10
2
1
24
42
q
p
p
pqor
p
K1
N1
N1
6
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2016
3
Number Solution and Marking Scheme Sub
Marks Full
Marks
2 2442 yx or 2223yxyx K1
yx 212 or 2
12 xy
P1
22*2*
3212212 yyyy or K1
2
2
2*
2
123
2
12
xx
xx
02 yy or 0128 xx K1
,0y 2y
N1
,12x 8x
N1
Area / Luas = 48 cm2 N1 7
3 (a) L0 = 35.5 or F = 16 or fm = 10 P1
510
16255.35
K1
40 N1
(b)
Mid point : 28, 33, 38, 43, 48, 53, 58
3914610511
)58(1)53(4)48(6)43()38(10)33(5)28(11
x
x
x = 8
Bil pekerja yang berpindah : 2 orang
P1
K1
N1
N1
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2016
4
Number Solution and Marking Scheme Sub
Marks Full
Marks
4 (a)
2 , 23
2
, 2
3
2
3
2
or 2 ,
9
8,
3
4 or r =
3
2
T5 =
15
3
22
= 81
32
K1
N1
(b)
n = 26
S26 =
3
21
3
212
26
= 5.9998 // 6
P1
K1
N1
(c) (c) S =
3
21
2
= 6
K1
N1
7
5 (a) i) APCACP or ABCACP
K1
xy 46 N1
(ii) CBCR
3
1
xy 126
3
1
xy 42
N1
(b) ADCACD
yxyxyh 366*)46( K1
Compare
yyh 96 or xxh 64 K1
2
3h N1 6
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2016
5
Number Solution and Marking Scheme Sub
Marks Full
Marks
6 (a) 2coscot or 1 cos 2 2sin
sin
2coscot dan 1 cos 2 2sin
sin
= sin 2 ( RHS)
P1
N1
(b)
1sin 2
2 or 210 º or 330 º
0 0 0105 ,165 ,285 ,345o
P1
N1
(c)
y
1
2
3
2 2
P1
P1
P1
P1
8
Shape
of sin graph
2 cycles for 20 x
Modulus
Shifted
1
0
3
2
1
0
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2016
6
Number Solution and Marking Scheme Sub
Marks Full
Marks
7 (a)i) orXPorXPorXP )12()11()10(
K1
012
12
12111
11
12210
10
12 )25.0()75.0()25.0()75.0()25.0()75.0( CorCorC
orXPXPXP )12()11()10(
K1
012
12
12111
11
12210
10
12 )25.0()75.0()25.0()75.0()25.0()75.0( CCC
= 0.3907
N1
(ii)
75.025.012
25.22
K1
N1
(b) i)
See 40
520515
0.45026
K1
N1
(ii)
See 800
480
See 0.253 or – 0.253
88.509k
P1
K1
N1
10
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2016
7
Number Solution and Marking Scheme Sub
Marks
Full
Marks
8 (a)
212 6
2x x
(2,4)A
K1
N1
10
(b) i)
2
0
3
1 26
x
xA
442
12 A
21 AA
33.13//3
40
K1
K1
K1
N1
(ii)
4
22 4y dy
42
24y y
2 24 4(4) 2 4(2)
4
P1
K1
K1
N1
9 (a)
x 3.16 4.47 5.48 6.32 7.07 7.75
N1
(b)
Refer to graph
Correct axes, uniform scale and one point correctly plotted
All points correctly plotted
Line of best fit
K1
K1
N1
(c) i)
ii)
bxay 5
Use 5a = gradient
a = 0.196
Use b = y – intercept
b = 2.1
x = 36
P1
K1
N1
K1
N1
N1
10
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2016
8
Number Solution and Marking Scheme Sub
Marks Full
Marks
10 (a)
i)
2m
)4(29 xy
172 xy
K1
N1
(ii)
2
6172
xx
)1,8(B
K1
N1
(iii) 0
5
2)8(3
x or 3
5
2)1(3
y
)9,12( D
K1
N1
(iv)
area BOD = 1901
81208
2
1
= 72122
1
= 30 unit2
K1
N1
(b)
PBPA
2222 )1()8()9()4( yxyx
42 yx
K1
N1
10
ADDITIONAL MATHEMATICS [P2]
TRIAL SPM 2016
9
Number Solution and Marking Scheme Sub Marks Full
Marks
11 (a)
∠COD =
20
16cos 1
COD 280360
θ = 0.7529 rad
K1
K1
N1
(b) i)
s = (5) (0.7529)
Perimeter = (5) (0.7529) + 10+6+12+20
= 55.529 // 55.53
K1
K1
N1
ii)
Find area of sector OAB or sector OAM or sector OCE
)7529.0()10(2
1 2 or )6436.0()10(2
1 2 or )6436.0()16(2
1 2
Find area of AOB or OCD
13.43sin)10(2
1 2 or )12)(16(2
1
Use the right combination to find area of shaded region
Area of shaded region = 49.23 (accept 49.23 – 49.26)