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ADD MATHEMATICS PAPER 1,2010 Mark Scheme
No Answers Sub-marks
Total mark
1
a) 5 b) 3
1 1
2
2 -2
B1: 3
5)(
xxf
3 3
3 a) 3kx +8
a) 12
5
B1: 3k = 4
5
1 2
3
4 2.766 and – 1.266 B2:
)2(2
)7)(2(4)3()3( 2
B1: 0732 2 xx
3 3
5 a) 5)4( 2 x b) 11 B1: 5)40( 2
1 2
3
6 5
3
1 x
B2: B1: (3x + 4)(x – 5) < 0 Or roots
31 and 5
3 3
7 – 3 B2: 3x +1 = 2(x – 1)
B1: 21
)5(5 )1(413 xx or
3 3
8 n
m
22
5
B3:
2
log2log2log5 qp
B2: change base and 1 law of log B1: use 1 law of log
4 4
9 27
B2: 2
3
9x
B1: 3
2
x seen
3 3
No Answers Sub-
mark Total mark
10 a)3
2
B1: 729
8
18
1 5 arorar
b) 25.04
1or
B1: 12
1a
2 2
4
11 a) 15 b) 217
B2: )4)(17()19(22
7 or
)4(5)5(22
6)4(12)5(2
2
13
B1: 197 T or )4(12)5(22
13
or )4(5)5(22
6
1 3
4
12 xxy 42 2
B2: 42 xx
y
B1: m = 2
3 3
13 2
3 or – 1.5
B2: 15
6
4
5
p
B1: p4
5 or
5
6 seen
3 3
14 a) ji 34 or
3
4
B1: OBAOAB b) 5
2 1
3
15 3
10
B3: 10k = 3m B2: 2m and 35 k
B1: RSPQ seen
4 4
16 a)
k
1
b) 21 k B1: xx sinsincoscos
1 2
3
31 5
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No Answers Sub-mark
Total mark
17 a) 0.3948
B1: 13
12cos POR
b) 11.13 B1: 3948.013
2 2
4
18 25
3
2 3 xxy
B2: c )3(5)3(3
21 3
B1: xx 53
2 3
3 3
19 24x – 16 B1: – 6 or – 2 seen
2 2
20 3
1
B2: 12x – 4 = 0 B1: 12x – 4
3 3
21 4 B2: 4k – 2k = 8 B1: 5 seen
3 3
22 8 B2: 0)87(2 2 kk or
87(2 2 kk
B!: 22
3
7
3
29
kk
3 3
23 a) 7
3
B1: 7
2
3
1or seen
b) 21
2
B1: 7
2
3
1
2 2
4
24 a) 3 B1: p = 0.05
2
b) 2.85 B1: 95.005.060
2 4
25 a) 65.15
B1: 95.03
68
b) – 0.7 B1: 0.242
2 2
4
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ANSWER TRIAL PAPER 2, 2010 SECTION A
1. 3
333
xyoryx 1M
Substitute x or y into eqn. 2. 1M 10633 2 yy or
103
36
2
x
x or
equivalent. 0736 2 yy or
012272 2 xx
)6(2
)7)(6(433 2 y 1M
Or
)2(2
)12)(2(4)9()9( 2 x
y = 0.859, – 1.359 1M x = 5.577 // 5.576, – 1.077 // – 1.076 1M 2. (a) 54 kkxy 1M
(b) 82
25 2 kdk
dA 1M
082
25 2 k 1M
4
5,
4
5 kk 1M
32
2
25 kdk
Ad 1M
A minimum, 4
5k 1M
3. a) 4 + (25 – 1)x = 40 1M x = 1.50 1M
b) )50.1)(125()4(22
25 1M
= 550 1M
c) )5.1)(1()4(22
nn
1M
nnn
30)5.1)(1()4(22
1M
67.35n 1M N = 36 1M
4. (a)
sec
tan 1M
1
cos
cos
sin 2 1M
(b) (i)
graph cos 1M, Max 2 and min – 2 1M
2
11 cycle 1M
(ii) 22
x
y 1M,
Draw straight line graph 1M No. of solution = 3 1M
5. (a) k
k
242
5.3410359 1M
2.41242
5.3410359
k
k 1M
k = 58 1M (b) 5934652 fx 1M
2)2.41(300
593465 1M
= 16.76 1M
6. (a) 1128
xy
or 83
2 xy or
equivalent 1M
(b)
31
)8(1)0(3,
31
)0(1)12(3
Either x or y correct 1M (– 9, 2) 1M c) Equation CD or gradient CD 1M
))9((2
32 xy or
x
mCD
9
02
Find x, when y = 0 or 2
3
9
02
x
and solve for x 1M
0,
3
23C 1M
– 2
2
2
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Section B
7. a) BCABAC or ABDADB 1M
yxAC 8 1M
yxDB 46 1M
b) (i) )46( yxkDH 1M
(ii) )8(4 yxmyDH 1M
)8(4 yxmy )46( yxk 1M
km 44 and km 68 1M Solve equations 1M
19
16k 1M,
19
12m 1M
8. a)
y10log
0.1206 0.301 0.4983 0.699 0.8633 1.061
1M b) qxpy 101010 logloglog 1M All points plotted 2M Line of best fit 1M c) (i) erceptyp intloglog 1010 1M p = 0.832 – 0.871 1M
(ii) gradientq 10log 1M q = 0.013 1M (iii) y = 6.03 1M
9. a) AR = 12 + 9 = 21 1M b) oSOQ 120 1M = 2.095 rad 1M
c) arc RQ = 180
6021 or
arc CQ = 095.29 1M
perimeter = 095.29 + 180
6021 +9+3 1M
= 52.85 1M d) Area of sector 1M Area of triangle 1M Area of shaded region =
2222 612)(12(2
1095.2)9(
2
1047.1)21(
2
1
1M = 83.66 1M
10. a) Solve equations 132 2 yxandxy 1M
1,2
1 yy or
2,4
5 xx 1M
Q (2, 1) 1M
Area of trapezium = )1(2
32
2
1
1M
yy
dyy 3)1(
1
0
2 1M
Area =
01
3
1
4
7 1M
= 12
5 // 0.417 1M
(c) V dxy 1
0
22 1
dxyyy
1
0
35
3
2
5
013
)1(2
5
1 35
867.1//15
28
11. (a) p = 0.75 and q = 0.25 1M Use nn
nC 1010 25.075.0 1M (i) P(X = 10) = 0.05631 1M (ii) P(X = 9) + P (X = 10) 1M = 0.244 1M
b) 8
174
xz 1M
(i) 0.6915 1M (ii) P 75.075.1 z 1M Finding correct area 1M 0.7333 1M
1M
1M
1M
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12. (a) 15,0 vt 1M
(b) 015183 2 tt and solve 1M t =1, t = 5 2M (c) 186 ta 1M = 6 1M (d) Shape minimum 1M 3 coordinates shown 1M
(e) Distance = 503 159 ttt 1M = 425 1M
240RM07 P
13. (a) (i) CD = 10 1M
(ii) oCE70cos
10 1M
CE = 3.42 1M (b) oAB 70cos)9)(12(2912 222 1M
29.12AB 1M
29.12
70sin
12
sin oABC
1M
9175.0sin ABC ooABC 57.66//56.66 1M
c) Area = 70sin)9)(12(2
1 or
Area = 70sin)42.3)(10(2
1 1M
Area ABED = 70sin)9)(12(2
1– 70sin)42.3)(10(
2
1
1M = 34.67 1M
14. (a) 1101004.1
54.1x
1151002
y
, y = 2.30 1M
10310018.6
z
, z = 6 1M
(b) (i)
360
)90(103)65(115)45(120)160(110 1M
4.110I 1M
(ii) 4.110100110
09 P
1M
44.12109 P (c) 8009/10 P 1M
100
08/0909/10 PP =
100
4.11080 1M
= 88.32 1M 15. (a) I : 80 yx 1M
II : yx 2 1M
III : 20 xy 1M (b)
One straight line drawn 1M The other two straight lines drawn 1M Region R 1M
(b) (i) 5010 x 1M (ii) yxp 4030 Finding maximum point (30, 50) Maximum profit = 30 (30) + 40 (50) 1M = 2900 1M
1M
15
1 5
1M
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