1
3472/1
Matematik
Tambahan
Kertas 1
2 jam
Ogos 2012
BAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2012
PERCUBAAN SIJIL PELAJARAN MALAYSIA
ADDITIONAL MATHEMATICS
Paper 1
Skema Pemarkahan ini mengandungi 6 halaman bercetak
MARKING SCHEME
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Muat turun (percum
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uat turun (percuma) skem
a lain di
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PERATURAN PEMARKAHAN- KERTAS 1
No. Solution and Mark Scheme Sub
Marks
Total
Marks
1(a)
(b)
25
4
1
1
2
2(a) 2 1 3
(b) 4
B1: 12 3OR 2 [use (2) ].
3 2
x yk y
x y
2
3(a)
(b)
3x
B1: 3
5
)(
5
xxf
1
2
1
3
4 1m
B2: 0)3)(2(4)6( 2 m
B1: 0362 2 xxm
3
3
5 12 p
B2 : 0)1)(2( pp
B1: 0232 pp
3
3
6(a)
(b)
(c)
1x
1
(1, 4)
1
1
1
3
7 2x
B2: xx 652
1)32(2 or xx 6532
B1: xx
652
1)32(2
333
3 3
or -2 -1
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Muat turun (percum
a) skema lain di : M
uat turun (percuma) skem
a lain di
3
8
3625 n or 1
3625n
B3: 625
3
1
n
m or equivalent
B2: 4log
3
15
n
m or equivalent
B1: 125log
log
5
5 m (for change base)
4
4
9 h = 2 and k = 11 [both]
B2: h = 2 or k = 11
B1: - 7 + 3d = 20 OR - 7 + 3(20 – k) = 20
OR - 7 + 3( h – (- 7)) = 20
3
3
10
(a)
(b)
2
3r
3
10
B1 :
1
22
1 (* )3
S
1
2
3
11 75 55h
B2 : 10
[2(3 1) 9( 1)]2
h h
B1: 3 1a h or 1d h
3 3 w
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Muat turun (percum
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uat turun (percuma) skem
a lain di w
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m w
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mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
4
12
(a)
(b)
3p
q
11:
yB pq p
x x
or 3pq
5
2
1
3
13 3 5
2 4y x or 4 6 5y x or equivalent
B3 : 3 3
12 2
y x
B2 : 2
3
2m or
3,1
2
B1 : 1
2
3m OR (3,0) and (0,2)S T
4 4
14
2 2 4 4 92 0x y x y
B2 : 2 2 2 2( 2) ( 2) (6 ( 2)) ( 4 2)x y
B1 : PS = PQ OR 2 2(6 ( 2)) ( 4 2)
3
3
15(a)
(b)
5 12i j
5 12 55 12 1or or
1213 13 13 13
i ji j
B1 : |OR |=13
1
2
3
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Muat turun (percum
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uat turun (percuma) skem
a lain di
5
16
4
3k
B2: 034 k
B1: (4 3) (4 2)k i k j
3
3
17(a)
(b)
0.9506 rad / 0.9505 rad / 0.9507 rad
15.36 or 15.355
B1 : arc OS = 5 (*0.9506) or PS = 3.602
1
2
3
18 x =15°,75°,195°,255°
B2 : 0 0 0 02 30 ,150 ,390 ,510x or sin 2x = 1
2
B1 : 2(2sin cos ) 1x x
3 3
19(a)
(b)
80 32x
5.2x or 2
5x
B1 : 03280 x
1
2
3
20
2
54 xy or equivalent
B2 : 142
3 xy or equivalent
B1 : 4 or ( 1) 3dy dy
dx dx
3
3
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Muat turun (percum
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uat turun (percuma) skem
a lain di w
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mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
6
21(a)
(b)
10
3
B2 : 3
1
5 7
2 2hx
B1:
3 3
1 1
( ) 7
2 2
f xhdx dx
1
3
4
22(a)
(b)
3
1.648 or 1.6475
B2: 2163 51( )
20 20
or equivalent
B1 : 2 2 2 2 2
251 2(0) 3(1) 2(2) 8(3) 5(4)or
20 20x
1
3
4
23(a)
(b)
252
66
B1 : 4 6 4 6
3 2 4 1or 60 OR or 6C C C C
1
2
3
24(a)
(b)
1
10
B1 : 3 1 2
5 4 3
19
20
B1 : 3 1 1
15 4 3
2
2
4
25(a)
(b)
0.1741
49.69
B2 : 45
0.9385
k
B1 : z = 0.938
1
3
4
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Muat turun (percum
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uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
1
3472/2
Matematik
Tambahan
Kertas 2
Ogos 2012
2 ½ jam
BAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN
KEMENTERIAN PELAJARAN MALAYSIA
PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 2012
PERCUBAAN SIJIL PELAJARAN MALAYSIA
ADDITIONAL MATHEMATICS
Paper 2
Skema Pemarkahan ini mengandungi 10 halaman bercetak
MARKING SCHEME
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Muat turun (percum
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mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
2
No Solution and Mark Scheme Sub
Marks
Total
Marks
1
8 3
2
yx
OR
8 2
3
xy
P1
2 8 33 6 0
2
yy y
OR
28 2 8 2
3 6 03 3
x xx
K1
Replace a, b & c into formula K1
2( 24) ( 24) 4(7)( 12)
2(7)y
OR
2( 20) ( 20) 4(7)( 59)
2(7)x
0.443, 3.871y OR 4.664, 1.807x N1
4.664, 1.807x OR 0.443, 3.871y N1
5
5
2
(a)
kxy
kxxy
31)1(
32
2
2
1
431
k
k
(b)
3
3
6
(1,4)
3
3 -1
-Maximum shape P1
-*Maximum point K1
-Another 1 point y-intercept / x-intercept K1
K1
K1
N1
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Muat turun (percum
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uat turun (percuma) skem
a lain di
3
3(a) 16 ,8 ,4 ,...... OR
1
2r
P1
116 1
230.5
11
2
n
K1
4.416n K1
5n N1
4
7
(b) 64 ,16 ,4 ,...... OR
1
4r P1
64
11
4
S
K1
= 1
853 or 85.33 N1
3
4(a) (i) Change
13 6 0 2
3y x to y x or
1
3BCm
OR 3ABm K1
5 3 ( 6)y x OR any correct method K1
3 23y x N1
5
7
(ii) Use simultaneous equation to find point B
* 3 23y x and 1
3 6 0 23
y x or y x K1
B = 15 1
,2 2
N1
(b)
15 1 2( ) 3( 6) 2( ) 3(5)* , ,
2 2 5 5
x y
K1
D =39 25
,4 4
N1
2
Use r
raS
n
n
1
)1(
Use 1
aS
r
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Muat turun (percum
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m w
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mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
Muat turun (percum
a) skema lain di : M
uat turun (percuma) skem
a lain di
4
5(a)
(b)
Amplitude = 3 [ Maximum = 3 and Minimum = − 3 ] P1
Sine shape correct P1
Two full cycle in 0 x 2 P1
Negative sine shape correct(reflect) P1
4
7
53sin 2 2
xx
or
52
xy
N1
Draw the straight line 5
2x
y
K1
Number of solutions is 3 N1
3
6(a)
L = 79.5 OR F = 24 OR fm = 4 P1
3(36) 24
479.5 104
K1
87 N1
3
8
(b)
(i)
(44.5 4) (54.5 5) (64.5 6) (74.5 9) (84.5 4) (94.5 8)
36X
2602
36
= 72.28
5
y
3
–3
2
2
3 2 x
52
xy
O
3sin2y x
K1
N1
OR
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Muat turun (percum
a) skema lain di : M
uat turun (percuma) skem
a lain di
5
(ii)2 2 2 2 2 2(44.5) 4 (54.5) 5 (64.5) 6 (74.5) 9 (84.5) 4 (94.5) 8
K1
2197689(*72.28)
36
K1
16.34 N1
7 Rujuk Lampiran
8(a)
(b)
3 2
4 2y dx c
x x
2
2(4) , 2
( 1)c c
2
22y
x
2
2
5
21
5
1 1
22
22
1
2( 2) 2( 5)2( 2) 2( 5)
1 1
dxx
xx
3
3
10
(c )
33or 6.6
5
2
2
2
5
23 1
5
22 3 13 1
5 5
2( ) Volume ( 2)
4 84
3 1
4 5 8 54( 2) 8( 2)4 2 4 5
3 1 3 1
14.56
i dxx
x xx
4
K1
N1
K1
K1
N1
K1
K1
K1
N1
K1
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Muat turun (percum
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mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
6
9(a)
(b)
(c)
yx
yxxOD
yxAC
4
15
4
7
)57(4
37
57
3 7 157 5 5
4 4 4
21 7
4 4
3
15 155
4 4
3
x y h y k x y
k
k
h k
h
5
452
150
t
t
3
5
2
10
10(a)
(i) 6 410
66 0.3 0.7P X C
= 0.03676
(ii) 9 1 10 010 10
9 100.3 0.7 0.3 0.7C OR C
9 1 10 010 10
9 109 0.3 0.7 0.3 0.7P X C C
= 0.0001437
5
5
10
N1
K1
N1
K1
N1
K1
N1
K1
N1
2
3
4
155
4
15
3
4
7
4
21
4
15
4
7557
4
3
h
h
k
kk
yxkyhyx
K1
K1
N1
K1
N1
2
3
4
155
4
15
3
4
7
4
21
4
15
4
7557
4
3
h
h
k
kk
yxkyhyx
K1
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Muat turun (percum
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uat turun (percuma) skem
a lain di w
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m w
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mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
7
(b)
5.3
4548
5.3
45404840 )( ZPXPi
= 0.7278
( ) 0.7
450.524
3.5
43.166
ii P X m
m
m
11(a)
(b)
(c)
7
tan 1
0.78554
OR RQ PR cm
rad rad
2 2
7(1.571) 7(2.3565)
7 7 2(7)(7)(cos135 )o
OR
2 27 7 7(1.571) 7(2.3565) ( 7 7 2(7)(7)(cos135 )
54.4268
o
Perimeter
217
4
2 21 17 2.3565 7 sin135
2 2
o
2 2 21 1 17 7 2.3565 7 sin135
4 2 2
78.8996
oArea
2
4
4
10
K1
N1
N1
K1
K1
K1
N1
2
3
4
155
4
15
3
4
7
4
21
4
15
4
7557
4
3
h
h
k
kk
yxkyhyx
K1
K1
K1
N1
K1
K1
K1
N1
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Muat turun (percum
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uat turun (percuma) skem
a lain di w
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mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
8
No Solution and Mark Scheme Sub
Marks
Total
Marks
12(a)
a = 10 - 5t = 0
t = 2 s
cttv 2
2
510
c 2)0(2
5)0(1030
c = 30
302
510 2 ttv
30)2(2
5)2(10 2 v
= 40 ms- 1
4
10
(b)
302
510 2 ttv 0
2 6 0t t
0 6t
3
(c)
cttts 306
55 32
s = 0, t = 0 , c = 0
ttts 306
55 32
)6(30)6(6
5)6(5 32 s or )8(30)8(
6
5)8(5 32 s
= 180 = 133.33
Total distance = 180 + 33.133180
= 226.67 m
OR
3
Use v > 0
Integrate and
substitute t = 2
Use a = 0
Integrate a to
find v
K1
K1
N1
K1
N1
K1
K1
Integrate v dt K1
K1
N1
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Muat turun (percum
a) skema lain di : M
uat turun (percuma) skem
a lain di
9
67.226
67.46180
302
51030
2
510
8
6
2
6
0
2
dtttdttt
13(a)
12510055
10
P
P10 = RM 44
2
10
(b)
115534
58831404125110
hh
hh
*
h = 1
2
(c)
11510020
11 P
P07 = RM 23
2
(d)
I S = 110125100
88
110 *1 125 4 140 *1 3 110 5
1 4 4 5I
= 122.86
4
Integrate v
6
0
+ 8
6
K1
K1
N1
N1
K1
N1
K1
K1
N1
See 125 P1
K1
K1
N1
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Muat turun (percum
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a lain di w
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m w
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m w
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.mysch
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m w
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.mysch
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ren.co
mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
Muat turun (percum
a) skema lain di : M
uat turun (percuma) skem
a lain di
10
14 Rujuk Lampiran
15(a)
Using sine rule to find BAC .
sin sin30
27 14
oBAC
74.64oBAC
(obtuse) 180 74.64
105.36
o o
o
BAC
3
10
(b) 105.36 30 or 6o oDCB DC cm
Use cosine rule to find BD.
22 26 27 2 6 27 cos135.36
31.55
BD
BD
3
(c)
Use formula correctly to find area of triangle ABC or ACD.
180 30 105.36
44.64
o o o
o
ABC
22 227 14 2 27 14 cos 44.64
19.67
AC
AC
1Area (14)(27)sin 44.64
2
oABC or
1Area (6)(19.67)sin105.36
2
oACD
Use Area ABCD = sum of two areas
Area ABCD = 189.7 cm2 .
4
END OF MARKING SCHEME
N1
N1
K1
P1
K1
N1
K1
K1
K1
N1
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Muat turun (percum
a) skema lain di : M
uat turun (percuma) skem
a lain di
Pentaksiran Diagnostik Akademik SBP 2012 Paper 2 ADM
0.1 0.3 0.4 0.6
1
x
10
20
30
40
50
60
70
80
y
x
x
0.5
No.7(a)
0.7 0
x
0.2
x
x
Plot y against 1
x K1
(at least one point)
6 points plotted correctly K1
Line of best fit N1
1
x
0.1 0.2 0.25 0.4 0.5 0.8
y 62 54 50 38 29 4
12
ny m
m x
N1
P1
. 2 70i m
35m
N1
. 80m
iin
K1
2800n
N1
1. 0.37iii
x
K1
2.703x
N1
0.8
x
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Muat turun (percum
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mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
Answer for question 14
(a) I. 170x y
II. 80x y
III. 2 20y x
(b) Refer to the graph,
1 graph correct
3 graphs correct
Correct area
(c) max point ( 50,120 )
i) k = 100x + 80y
Max fees = 100(50) + 80(120)
= RM 14, 600
ii) 14080 y
10 20 30 40 50 60 70 0 80
20
40
140
120
100
160
180
80
60
(50,120)
10
N1
N1
N1
N1
N1
N1
K1
N1
K1
N1
y
x
ww
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Muat turun (percum
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uat turun (percuma) skem
a lain di w
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lchild
ren.co
m w
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mM
uat turun (percuma) skem
a lain di : Muat turun (percum
a) skema lain di
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