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SPM TRIAL EXAM 2012 Marking Scheme
Additional Mathematics Paper 2 SECTION A
Question Important Steps Marks 1 y = 3x – 4 1
5x2 – 4x(3x – 4) + (3x – 4)2 = 9 or 2x2 – 8x + 7 = 0
1
)2(2)7)(2(4)8()8( 2 −−±−−
=x 1
x = 2.707, 1.293 1 y = 3(2.707) – 4 , y = 3(1.293) – 4 = 4.121 = – 0.121
1
TOTAL 5
2 (a) Change base of logarithm : 4
22
log (1 2 )log 4log (1 2 ) xx −
− = or
equivalent 1
Use n log x = log x n : 2 log 2 ( 1 – 2x ) = log2 ( 1 – 2x )2 1 Solve : (2x + 5 ) = ( 1 – 2x ) 2 1 x = 22
1 ,− 1
x = 21− 1
(b) 31 1
TOTAL 6
3 (a) – 8 1
(b)
Use Tn = a + (n – 1 ) d : – 8 + ( 22 – 1 ) ( 3 ) 1 Use Sn = ])([ dnan 122 −+ : 2 [2( 8) ( 1)(3)]n n− + − 1 Solve : ])()()([ 31822 −+− nn = 55 1 n = 10 1
(c) T13 1 28 1
TOTAL 7
4 (a)
x = 3 1
(b) 5
59105.9 ⎟⎠
⎞⎜⎝
⎛ −+=median
1
Median = 10.5 1
(c) All midpoints are correct. 1
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1120220
20)22(2)17(4)12(5)7(6)2(3
==++++
=x 1
∑ =++++= 3150)22(2)17(4)12(5)7(6)2(3 222222fx or Σf (x- x )2 = 730
1
22
20220
203150
⎟⎠
⎞⎜⎝
⎛−=σ or 207302 =σ 1
= 36.5 1 TOTAL 8
5
(a) y
xyπ2
=
x
Shape 1
Max/min 1
One period
1
Complete from 0 to
2π 1
(b) Equation xy
π2
= 1
Straight line xyπ2
= 1
2 solutions 1 TOTAL 7
6
(a)
32QSm = 1
)6(231 −=− xy 1
823
−= xy 1
(b) Q(0, − 8) 1
√[(x − 6)2 + (y − 1)2 ] or √[(6 − 0)2 + (1+8)2] 1
x2 − 12x + 36 + y2 − 2y + 1 = 117 1 x2 + y2 − 12x − 2y − 80 = 0 1
TOTAL 7
π 2π π 2
3π 2
3
-3
O
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SECTION B
7 (a)
x 1 1.5 2 2.5 3 3.5 4
log10 y 0.04 0.18 0.28 0.40 0.52 0.62 0.76
1
(b) Plot log10 y against x [ correct axes and uniform scales ] 1 All 7 points plotted correctly 1 Line of best fit 1
(c) (i) 3.72 1
(ii) 22101010 hk xy loglog )(log −= 1
Use :log210 hc −= 202
10 .log −=− h 1 h = 2.5 1
(iii) Use :log210 km = 2402
10 .log =k 1 k = 3.0 1
TOTAL 10
8 (a)
∫−= dxxy 2 1
cxy +−
=22 2
1
92 +−= xy 1
(b)
3 2
0( 9)x dx− +∫
1
33
0
93x x
⎡ ⎤−+⎢ ⎥
⎣ ⎦
1
−)10)(10(21
3 2
0( 9)x dx− +∫ or 50 –
3 2
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(π 1
92
0
92yyπ
⎡ ⎤−⎢ ⎥
⎣ ⎦
1
812
orπ equivalent 1
TOTAL 10
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9 (a)
Angle AOB = 6.5/5 1 = 1.3 rad. 1 Angle POQ = 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 arc PQ = 6 × 0.8667 = 5.2002 1 Perimeter = 3.811 + 5.2002 + 1 + (6−3.2367) 1 = 12.77 cm 1
(c) Area of sector OPQ = ½ × 62 × 0.8667 1 Area of shaded region = 15.60 − ½ (3.811)(3.2367) 1 = 9.433 1
TOTAL 10
10 (a) (i)
RTPRPT +=
)148(21 xyRT +−=
1
)148(218 xyyPT +−+=
= yx 47 +
1
(ii)
RS RP PS= +uuur uuur uuur
)14(318 xy +−=
1
yx 8314
−= 1
(b) (i) yhxhPM 47 += 1
(ii) RMPRPM +=
)8314(8 yxky −+=
1
= ykxk )88(314
−+
1
(c) 147
3h k= or hk 488 =−
1
21
=h 1
43
=k 1
TOTAL 10
11 (a)(i)
377
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)8 1
= 0.09956 1
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(b) (i)
6.0 7.2 8.1 7.2( )1.2 1.2
P z− −< < OR ( 1 0.75)P z− < < 1
= 2266.01587.01 −− 1
= 0.6147 1
(ii) 8.0
6048
2.12.7
==⎟⎠
⎞⎜⎝
⎛ −>tzp 1
842.02.12.7
−=−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)
1.4t −0.3t2 + 0.5 = 0 1 (3t +1)(t − 5) = 0 or using quadratic formula 1 t = 5 1
(c)
s = 2(1.4 0.3 0.5)t t dt− +∫ = 0.7t2 − 0.1t 3 + 0.5t + c integrate 1 At t = 0, s = 0 ⇒ c = 0 finding c
or ∫ ∫ +−++−5
0
10
5
22 5.03.04.15.03.04.1 dtttdttt limits √ 1
When t = 5, s = 7.5 m, when t = 10, s = −25 m or substitute t=0, 5, 10 in [0.7t2 − 0.1t 3 + 0.5t] 1
Total distance = 7.5 x 2 + 25 or 7.5 + |−25 − 7.5| 1 = 40 m 1
TOTAL 10
13 (a) 158 100
130x = × 1
= 121.54 1
(b)
150 100 107.14140
A⇒ × = 1
121.54 100 90.03135
B⇒ × = 1
120 100 109.09110
C⇒ × = 1
123 100 102.5120
D⇒ × = 1
(c) 12
)25.102()209.109()503.90()314.107( ×+×+×+×=I 1
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= 99.56 1
(d) 56.99
100120
2013 ×=I 1
= 119.47 1 TOTAL 10
14 (a) (i)
sin sin3012 7ACB∠ °
= 1
∠ACB = 59° 1
(ii) 2 2 24 11.47 12cos2(4)(11.47)
AKB + −∠ = 1
cos ∠AKB = 0.0388 ∠AKB = 87.78° or 87°47’ 1
(iii) ∠ABC = 91° 1
Area rABC = ½ (7)(12) sin 91° or Area of rAKB = ½ (4)(11.47) sin 87.78°
1
Area of quadrilateral = Area rABC + Area of rAKB = 41.99 + 22.92
1
= 64.91 cm2 1
(b)(i)
1
(ii) ∠A’C’B’ = 121° 1 TOTAL 10
15 (a)(i)
x + y ≤ 10 or equivalent 1 y – x ≤ 4 or equivalent 1 x ≤ 2y or equivalent 1
(b) Draw correctly one straight line from the inequalities 1 Draw correctly two more straight line from the inequalities 1 Region R correctly shaded 1
(c)(i) Maximum point ( 3 , 7 ) 1 RM [ 10(3) + 25(7) ] = RM 205 1
(ii) Minimum point (2 , 6 ) 1 RM [ 10(2) + 25(6) ] = RM 170 1
TOTAL 10
B’
A’ C’
12 cm 7 cm
30o
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GRAPH FOR QUESTION 7
0 0.5 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 FOR QUESTION 15
y
0 1
1
x 2 3 4 5 6 7 8 9
2
3
4
5
6
7
8
9
10 y – x = 4 = 2y
x + y = 10
x = 2y
R
( 3 , 7 )
10x + 25y = 150 ( 2 , 6 )
y = 6
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