spm trial 2012 addmath a perak
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SPM TRIAL EXAM 2012
Marking Scheme
Additional mathematics Paper I
Number Solution and marking scheme
Sub
Marks
Full
Marks
1 (a) {p, r, s}
(b) {a , b, c, d}(c) Many to One
1
11
3
2(a)
2
7
B1: 2x 5 = 2 atau f(x) =2
5+x
2 2
3 (a)5(b) 42 +x
B2: ( ) 7653 =+ xxf B1 : ( ) 53 + xf
1
3
4
4
p = 7 , q = 10 ( both )
B2: p = 7 or q = 10
B1:33
7 p= or33
10 q= or 01073 2 = xx
or 3(1)2
+p(1)+q = 0, 3(10/3)2
+p(10/3) + q = 0
3 3
5 h =2 andk= 3
B2 : h = 2 atau k= 3
B1 : 1
3
=h
or 922 =++ kh
3 3
6
45 x
B2: ( ) 0)5(4 + xx or
B1 : x2 +x 20 0
3 3
4-5
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2
7
pr
12+ or
pr
rp +2
B3:mm 3
3
2
2
log
3log2log
2log +
B2: log m2 + 2logmmm 3
2
3
2
2
log
3log
log
2log+3or
B1 : log 2 + log32
or 2logmmlog
2log3 or or
mlog
3log
4 4
8 158 += px
B2: )32(43 += px
B1: 32 or )32(42 +p
3 3
9 a) 3b) 360
B1: [ ]18
2(3) (18 1)22
+
1
2
3
10 13=a , 4=d (both)
B2: 13=a or 4=d
B1: 52 =+ da or 157 =+ da
3 3
11 (a) r=x
2
(b) 31
B1:2
2
18
1
x
x
=
1
2
3
12a) qpx
x
y+= 2
b) 2=p , 8=q B2 : 2=p or 8=q
B1:42
04
=p or q+= )4)(2(0 or 4 =p(2) + q
1
3
4
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13 2,12
B2: 110
)1(3
4
73=
hh or form equation using Pythagoras
theorem
B1:4
73
hor
10
)1(3
h of find the lengths of AB, BC and AC using
distance formula
3 3
14(a)
w
1
(b) 212 ww B1:
21 w or cos2w
1
2
3
15 26.57 ,116.57 , 206.57 , 296.57 B3: 26.57 and 116.57 B2: )2)(tan1tan2( + xx
B1: 02tan3tan2 2 =+ xx
4 4
16(a)13(b)k = 13
B1:
+
++25
112 kor jik )25()112( ++++
1
2
3
17 (a) 4a + 4b(b) ba 42 +
B1: )44(6 baa ++
12
3
18 8
B2: 9.42)3()3()3(3.13.1 =+++++ rrrrrr
B1: r3.1 or )3(3.1 r
3 3
19
5
12
B2:
+0
12
)2(32
2
B1:
2
0
2 1
32
+x
x
3 3
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20 122 + aa
B2: ))4()4(()( 22 ++ aa
B1: xx +2
3 3
21 k= 8
B3: (k+1)(k8) = 0
B2:
22 2 22 5 2 56
3 3
k k+ + + + =
B1:3
52 kx
++= or 2222 52 kx ++=
4 4
22 1 2( , )3 3
B2 : x =1
3, y = 2()(32)
B1 : 12x 4 = 0
3 3
23 (a) 1(b)i) 5040
ii) 288
B1: 2 3! 4!
1
12
4
24(a)
15
4
(b) 35
B1 :3 2 2 1 3 1 2 2
1 or5 3 5 3 5 3 5 3
+ +
1
2
3
25(a) 53 (b)
625
144
B1:
32
2
5
5
2
5
3
C
1
2
3
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1
SPM TRIAL EXAM 2012
Marking Scheme
Additional Mathematics Paper 2SECTION A
Question Important Steps Marks
1 y = 3x 4 1
5x2 4x(3x 4) + (3x 4)2
or 2x= 9
21
8x + 7 = 0
)2(2
)7)(2(4)8()8( 2 =x
1
x = 2.707, 1.293 1y = 3(2.707) 4 , y = 3(1.293) 4
= 4.121 = 0.121 1TOTAL 5
2(a)
Change base of logarithm :4
2
2
log (1 2 )
log 4log (1 2 )
xx
= or
equivalent
1
Use n logx = logx n : 2 log 2 ( 1 2x ) = log2 ( 1 2x ) 12
Solve : (2x + 5 ) = ( 1 2x ) 12
x = 221 , 1
x = 21
1(b)
31 1
TOTAL 6
3(a)
8 1
(b)Use Tn 1= a + (n 1 ) d : 8 + ( 22 1 ) ( 3 )
Use Sn ])([ dnan 122 += : 2 [ 2 ( 8) ( 1) (3) ]n n + 1
Solve : ])()()([ 31822
+ nn = 55 1
n = 10 1
(c)T 11328 1
TOTAL 7
4
(a)x = 3
1
(b)5
5
9105.9
+=median
1
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Median = 10.5 1
(c) All midpoints are correct. 1
11
20
220
20
)22(2)17(4)12(5)7(6)2(3==
++++=x 1
=++++= 3150)22(2)17(4)12(5)7(6)2(3222222fx
or f(x-x )21
= 7302
2
20
220
20
3150
= or
20
7302 = 1
= 36.5 1
TOTAL 8
5(a)
y
xy
2
=
x
Shape1
Max/min1
Oneperiod
1
Completefrom 0 to
2
1(b)
Equation xy
2= 1
Straight line xy
2= 1
2 solutions 1
TOTAL 7
6
(a)
3
2
QSm = 1
)6(2
31 = xy 1
82
3= xy 1
(b) Q(0, 8) 1
[(x 6)2 + (y 1)2 [(6 0)] or 2 + (1+8)2 1]
x2 12x + 36 +y2 1 2y + 1 = 117
x2
+y2
1 12x 2y 80 = 0TOTAL 7
22
32
3
-3
O
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SECTION B
7(a)
x 1 1.5 2 2.5 3 3.5 4log10 0.04y 0.18 0.28 0.40 0.52 0.62 0.76
1
(b)Plot log10 1y againstx [ correct axes and uniform scales ]All 7 points plotted correctly 1Line of best fit 1
(c) (i) 3.72 1
(ii)22101010 hk xy
loglog)(log = 1
Use :log210 hc = 20
210 .
log=
h 1
h = 2.5 1(iii)
Use :log210 km = 240
210 .
log=
k 1
k = 3.0 1TOTAL 10
8(a)
= dxxy 2 1
cx
y +
=2
2 2 1
92 += xy 1
(b)3
2
0( 9)x dx + 1
33
0
93
xx
+
1
)10)(10(2
1
32
0( 9)x dx + or 50
32
0( 9)x dx +
or )10)(10(21
33
0
93x x +
or 50
33
0
93x x +
1
= 32 1
(c) Volume = dyy)9( 1
92
0
92
yy
1
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81
2or equivalent
1
TOTAL 10
9(a)
Angle AOB = 6.5/5 1= 1.3 rad. 1
AnglePOQ = 0.8667 rad. 1(b) MN= 5 sin(0.8667 rad.) = 3.811 cm
or ON = 5 cos( 0.8667 rad.) = 3.2367 cm
1
Length of arcPQ = 6 0.8667 = 5.2002 1
Perimeter = 3.811 + 5.2002 + 1 + (63.2367) 1
= 12.77 cm 1(c) Area of sectorOPQ = 62 1 0.8667
Area of shaded region = 15.60 (3.811)(3.2367) 1
= 9.433 1TOTAL 10
10
(a) (i)RTPRPT +=
)148(2
1xyRT += 1
)148(2
18 xyyPT ++=
= yx 47 +
1
(ii)RS RP PS= +
)14(3
18 xy += 1
yx 83
14= 1
(b) (i) yhxhPM 47 += 1
(ii)RMPRPM +=
)83
14(8 yxky +=
1
= ykxk )88(3
14+ 1
(c)14
73
h k= or hk 488 = 1
2
1=h
1
4
3=k
1
TOTAL 10
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11(a)(i)
37
7
10 )55.0()45.0()7( CXP == 1
= 0.07460 1
(ii)
P(X= 0, 1, 3)= 10C0(0.45)0(0.55)10 + 10C1(0.45)1(0.55)9 + 10C2(0.45)2(0.55) 18
= 0.09956 1
(b) (i)
6.0 7.2 8.1 7.2( )
1.2 1.2P z
< < OR ( 1 0.75)P z < < 1
= 2266.01587.01 1
= 0.6147 1
(ii) 8.06048
2.12.7 ==
> tzp 1
842.02.1
2.7=
t 1
190.61896.6 ort = 1
TOTAL 10
SECTION C
12(a)
a = 1.4 0.6dv
tdt = 1
= 1.4 0.6(2)= 0.2 1
(b)1.4t0.3t2 1+ 0.5 = 0
(3t+1)(t 5) = 0 or using quadratic formula 1t= 5 1
(c)s = 2(1.4 0.3 0.5)t t dt + = 0.7t
2 0.1t3 1+ 0.5t+ c integrate
At t= 0,s = 0 c = 0 finding c
or +++
5
0
10
5
225.03.04.15.03.04.1 dtttdttt
limits
1
When t= 5,s = 7.5 m, when t= 10,s = 25 mor substitute t=0, 5, 10 in [0.7t2 0.1t3
1+ 0.5t]
Total distance = 7.5 x 2 + 25 or 7.5 + |25 7.5| 1= 40 m 1
TOTAL 10
13(a)
158100
130x = 1
= 121.54 1
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(b)
150100 107.14
140A = 1
121.54100 90.03
135B = 1
120 100 109.09110
C = 1
123100 102.5
120D = 1
(c)
12
)25.102()209.109()503.90()314.107( +++=I 1
= 99.56 1
(d)56.99
100
1202013
=I 1
= 119.47 1
TOTAL 10
14(a) (i)
sin sin 30
12 7
ACB =
1
ACB = 59 1
(ii) 2 2 24 11.47 12cos
2(4)(11.47)AKB
+ = 1
cos AKB = 0.0388AKB = 87.78 or 8747
1
(iii) ABC= 911
AreaABC= (7)(12) sin 91or Area ofAKB = (4)(11.47) sin 87.78
1
Area of quadrilateral= AreaABC+ Area ofAKB= 41.99 + 22.92
1
= 64.91 cm 12
(b)(i)
1
(ii) ACB = 121
1
TOTAL 10
B
AC
12 cm
7 cm
30o
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15(a)(i)
x +y 10 or equivalent 1y x 4 or equivalent 1x 2y or equivalent 1
(b)Draw correctly one straight line from the inequalities 1Draw correctly two more straight line from the inequalities 1RegionR correctly shaded 1
(c)(i)Maximum point ( 3 , 7 ) 1RM [ 10(3) + 25(7) ] = RM 205 1
(ii)Minimum point (2 , 6 ) 1RM [ 10(2) + 25(6) ] = RM 170 1
TOTAL 10
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GRAPH FORQUESTION 7
00.5 1 1.5 2 2.5 3 3.5 4
x
log10 y
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.1
0.2
( 0 , 0.2 )
( 4 , 0.76 )
3.2
0.57
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GRAPH FORQUESTION 15
01
1
x
y
2 3 4 5 6 7 8 9
2
3
4
5
6
7
8
9
10y x = 4
x +y = 10
x = 2y
R
( 3 , 7 )
10x + 25y = 150( 2 , 6 )
y = 6