trial addmate spm 2011 selangor paper 2 answer
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8/4/2019 TRIAL ADDMATE SPM 2011 Selangor Paper 2 Answer
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Program Peningkatan Prestasi Sains & Matematik 2011 (Trial)
Additional Mathematics Marking Scheme - Paper 2
Solution Marks Solution Marks
1
3
2
yx or 23 xy
4)23()23(222 xxxx
or 43
22
3
2 22
y
yy
y
021182 xx
or 0168 2 yy
)8(2
)2)(8(4)11()11(2
x
or)8(2
)16)(8(4)1()1(2
y
x = 1.159, 0.216,
y = 1.478, -1.353
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5
2a)
45)(2 xxxf
b)(i) xxxf 5)(2
4
= ]44
25)
2
5[(
2 x
=4
9)
2
5(
2 x
Minimum point )4
9,
2
5(
(ii) 0)4)(1(4)5( 2 p
4
9p
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6
3 a) qxpxdx
dy
23 ,
xxqxpx 6623 22
3p = 6 @ 2q = -6
p = 2 ,
q = 3
r 23 )2(3)2(29
r= 5
b) 0)1(6 xx
x = 0 , x = 1
turning point (0,5)
turning point (1,4)
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4 a)
Shape of graph (sin )
Max = 1 and Min = -1
2
11 cycles
Reflection about x -axis
b) 4 sin2
3x + 3x = 0
- sin2
3x =
4
3x
y =4
3x
Number of solution = 3
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(line)
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2 x
1
2
y
-1
1.5
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8/4/2019 TRIAL ADDMATE SPM 2011 Selangor Paper 2 Answer
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5 a) OLBOBL or ABOAOM or equivalent
yxBL 84
)812(2
112 yxxOM
=yx 46
b) )46( yxhON
)48(4 xykxON
)48(4)46( xykxyxh
4h = 8k, h = 2k
6h = 44k
6(2k) = 44k
2
1,
4
1 hk
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6(a) =
6251A ,
4
6252A ,
8
6253A
Since , then it is a
Geometric Progression
r =
(b)
4
11
625
S
=3
2500
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7a) Area of )2)(6(2
1 k
2
0
2)2( dyy =
2
0
3
23
y
y
= 0)2(23
23
=
3
20
3
38= 63
3
20 k
k312 k= 4
b) V = 2
0
22)2( dyy
= 2
0
24)44( dyyy
2
0
35
43
45
y
yy
= 0)2(43)2(4
5)2(
35
= 15
376unit
3
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8x 1 2 4 6 8 10
log10 y 0.84 0.99 1.29 1.57 1.87 2.16
Paksi betul, skala seragam & plot 1 titik
Plot semua titik betul
Garis lurus penyuaian terbaik
y = h kx
log10 y = log10 h + x log10k
i) log10k = 0.146
k= 1.4
( accept 1.351.43)
log 10h = 0.70
h = 5.012
ii ) log 10y = 1.7 , x = 6.8
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8/4/2019 TRIAL ADDMATE SPM 2011 Selangor Paper 2 Answer
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9 a)(i)
k = 13
(ii)
or equivalent
(b)(i)
E(-2, -5)
(ii) Area of triangle ODE
=
=
= 13 unit2
(c) 22 )4()6( yx
22)3()4( yx
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10
10
a) 90,45 AOCABC 9)571.1(
2
1 2 r
r = 6 cm
b)
)571.1(6ACarc or equivalent= 9.426 cm
Perimeter = 2(6) + 9.426
= 21.426 cm
c)
radororBOC 713.42
3135
22
sec 834.84)713.4)(6(
2
1cmA tor
135sin)6(2
1)713.4)(6(
2
1 22segmentA
= 72.104
Area of shaded region = 2(72.104)
= 144.208 cm2
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11a)(i)P(X 2)= P(x = 0) + P(x =1) + p(x = 2)
=60
06
52
53C
+
51
16
5
2
5
3C
+
42
26
5
2
5
3C
= 0.1792
(ii) Min = np = 1205
2= 48 @
Sisihan piawai = 28Min = 48 & Sisihan piawai = 48
b)(i)P( X > 50 ) = P( Z >9
3050 )
= P( Z > 2.22)
= 0.01321
(ii) P( X < t ) = 0.70
P( Z 9
30t ) = 0.3
52409
30t
t = 34. 716
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N110
12(a)9
35sin
6
sin
ADB
48.22DBA or '2922
(b) 48.57ADC
AC= 10.43 cm
(c) 48.2235180BAD = 122.52
BD = 13.23 cm
(d) Area =
= 45.53 cm2
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8/4/2019 TRIAL ADDMATE SPM 2011 Selangor Paper 2 Answer
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13
a) 125100168
x
or 10090
99y
210RMx y = 110
c) m + n = 35
115
100
4011010512025125
nm
8m + 7n = 265
8(35n) + 7n = 265
m = 20
n = 15
d) 115100
120I
= 138
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14a) 200 500 x 700 yx
2002 yx
b) Graf (satu garis betul)
(semua garis betul)(rantau betul)
c)(i) Kek lapis maksimum = 450
(ii) (500, 200)
k = 0.8x + 0.6y
= 0.8(500)+0.6(200)
= RM 520
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15 a) 622 ttv
16,0
cmsvt
b) 14 tdt
dv= 0
4
1t
6
4
1
4
12
2
min
v
=1
8
49 cms
c) 062,02 ttv
0)2)(32( tt
2t
d) dttts )62(2
s = cttt
6
23
223
ttt
s 623
223
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15d) )2(62
)2(
3
)2(2,2
23
st
=3
26 or
)4(62
)4(
3
)4(2,4
3
st
= s3
32
Total distance travelled
=3
32
3
262
= 28 cm
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3 2
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8/4/2019 TRIAL ADDMATE SPM 2011 Selangor Paper 2 Answer
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Graph for Question 14
100
200
300
400
y
500
600
700 8000
100
200
300
400
500
600
700
800
R
500,200
x
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8/4/2019 TRIAL ADDMATE SPM 2011 Selangor Paper 2 Answer
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Question 8
2 4 6 x
0.2
0.4
0.6
0.8
1.0
1.2
2.0
1.8
1.4
1.6
log10 y Graph log10 y against x
8
x
x
x
x
x
x
0
2.2
10
C = 0.70
x = 6.8