Trial SPM Perlis 2010

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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


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


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


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


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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