SULIT

3472/2

Additional

Mathematics

Paper 2

Sept

2010

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA

SEKOLAH MENENGAH MALAYSIA (PKPSM) CAWANGAN MELAKA

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2010

ADDITIONAL MATHEMATICS

Paper 2

MARKING SCHEME

This marking scheme consists of 13 printed pages

Question Mark Scheme of Paper 2 Trial SPM 2010 Melaka Sub

Marks

Total

Marks

1. n= -6 – 4m or n = m

2 +m-2

or

4

6 nm

−−=

m2+m+6+4m=2 or 4m+m

2+m-2+8=2 or 2)

4

6(

2)

4

6( =−

−−+

−−n

nn

(m+4)(m+1)=0 or (n-10)(n+2)=0 or use formula/completing the square

m= -4, -1 or n=10, -2

n=10, -2 or m= -4, -1

P1

K1

K1

N1

N1

5

2.(a)

(b)

(c)

m2 = 2

y -5= 2(x+1) or equivalent

y = 2x +7

Solve simultaneous equations: 22

172 +−=+ xx

E( -2, 3)

)3,2()2

5,

2

)1(( −=

+−+ yx

C( -3, 1)

K1

K1

N1

K1

N1

K1

N1

7

3.(a)

(b)

Shape of sin x

Graph of sin x shifted up 1 unit

Miximum = 4, minimum = -2

1½ cycles in π2

30 ≤≤ x

12

+= xyπ

or equivalent

Sketch the straight line involves x and y

Number of solutions = 4

P1

P1

P1

P1

N1

K1

N1

7

4.(a)

Gradient of tangent = 5

kx2-x = 5

substitute x = 1 to kx2-x = 5

K1

K1

K1

x π x

4 x

1

-2

0 2

π

2

3π

3

y

(b)

k = 6

cxx

y +−=23

6 23

c+−=−2

)1(

3

)1(62

23

2

7

22

23

−−=x

xy or equivalent

N1

K1

K1

N1

7

5.(a)(i)

(ii)

(b)

L = 20.5 F = 57 fQ3 = 28

1028

57)100(4

3

5.203

−

+=Q

= 26.93

100

)5.45(6)5.35(9)5.25(28)5.15(32)5.5(25 ++++=x

=19.4

222222

4.19100

)5.45(6)5.35(9)5.25(28)5.15(32)5.5(25−

++++=σ

= 11.30

222

2

98

20401940

98

204050415 ）（ −−−

−−=σ

= 126.02

P1

K1

N1

K1

K1

N1

K1

N1

8

6. (a)

(b)

d = 2πr or 2πr, 4πr, 6πr

d = 4πr - 2πr = 6πr - 4πr

2πr = 2πr

T10 = 4 + 9(4) or T10 = 2πr + 9 (2πr )

= 40 = 20 π (r)

Circumference = 2π(40) = 20 π (4)

= 80π = 80 π

P1

K1

N1

K1

N1

N1

6

7.(a)

(b)

(c)

(i)

(ii)

log10x 0.15 0.28 0.40 0.51 0.61 0.74

log10y 0.52 0.72 0.90 1.08 1.20 1.42

Refer to the appendix

One point correctly plotted with uniform scales

6 points correctly plotted with uniform scales

Line of best fit

xkpy 101010 log5loglog −=

-5k =1.52

35.025.0 −↔−=k

log10p = 0.29

N1

N1

K1

N1

N1

P1

K1

N1

K1

N1

0.29.1 ↔=p

10

8(a)(i)

(ii)

(b)

(c)

OBAOAB += or equivalent

yx 36 +−=

xyOD2

1

4

9+=

)2

1

4

9( xymOC +=

)36( yxnBC +−=

)36()2

1

4

9(3 yxnxymy −−+=

5

6=m

10

1=n

½ x 8 x h = 20

h = 5

K1

N1

N1

P1

P1

K1

N1

N1

K1

N1

10

9(a)

(b)

(c)

(d)

1230sin

OQ=° or equivalent

OQ = 6

θπ

62

=

=∠QOR rad12

πθ = // 0.2618 rad

radradROS 833.1/12

7/1051560180 π°=°−°−°=∠

Area of sector ROS = ROS∠××26

2

1

=32.99 or 10.5π

Perimeter of shaded region = )5.52sin6(2)12

7(6 °+π

=20.52

K1

N1

K1

N1

P1

K1

N1

K1K1

N1

10

10(a)

(b)

(c)

Use b2 – 4ac = 0

(-4)2 – 4(1)k =0

k = 4

Solve equation : (x – 2)(x – 2) = 0

A(2, 2) 3

2

32

32

3

−

xx

=

−

−

3

2

2

)2(3

3

3

2

)3(3 3232 －

K1

N1

K1

N1

K1

K1

N1

(d)

=6

7

3

0

543

54

6

3

9

+−

xxxπ

= 05

3

4

)3(6

3

)3(9 543

−

+−π

= 8.1π

K1

K1

N1

10

11(a)(i)

(ii)

(b)(i)

(ii)

377

10 )4.0()6.0(C

= 0.215

1 – P(x=0) – P(x=1) or P(x=2)+P(x=3)+P(x+4)+ …………….+P(x=10)

= 911

101000

10 )4.0()6.0()4.0()6.0(1 CC −−

= 0.9983

)5

350345(

−<zP

= 0.1587 // 0.15866

P(x >342) = 1 – P(z > 1.6) or other valid method

Number of cakes = 0.9452 x 1500

= 1417 // 1418

K1

N1

P1

K1

N1

K1

N1

K1

K1

N1

10

12(a)

(b)

(c)

(d)

Use 1000

1×=

P

PI

x = 135, y = 80, z = 140

141842

)8(140)4(125150)2(150=

+++

+++

m

m

m = 6

141100350

2005=×

P

P2005 = 493.50

100141

11502005/2006 ××=I

= 106.38

K1

N2,1,0

K1K1

N1

K1

N1

K1

N1

10

13 (a)

(b)

(c)

22 611 −=PQ

= 9.22

8

40sin

22.9

sin=

∠PRQ

= 47.80° // 47°48

PRQQPR ∠°°=∠ －－40180 or 92.2°

QR = 12.44

QVR∠−+= cos)10)(11(2101144.12 222 '2972//48.72 °°=∠QVR

4552

48.72sin10112

1 。＝ °××

K1

N1

K1

N1

P1

N1

K1

N1

K1

N1

10

14.(a)

(b)

(c)(i)

(ii)

x + y ≤ 40 or equivalent

4x + 3y ≥36 or equivalent

x ≤ 2y or equivalent

Refer to the appendix

Draw correctly at least one straight line involves x and y

Draw correctly all the three straight lines

Region R shaded correctly

y = 6

min passengers = 11

substitute any point in the region R into 40x + 30y

Substitute ( 26, 14) to 40x + 30y and maximum fares = 1460

N1

N1

N1

K1

N1

N1

K1

N1

K1

N1

10

15(a)

(b)

(c)

(d)

18 ms-1

2t -9 = 0

2

9=t

18)2

9(9)

2

9(

2max +−=V

= 4

12−

Use sketch quadratic graph // number lines // others valid method for

t2-9t + 18 < 0

3 < t < 6

ttt

s 182

9

3

23

+−=

P1

K1

K1

N1

K1

N1

K1

3 6

Substitute t = 3 or t = 5 into ttt

s 182

9

3

23

+−=

S3 = 22.5

S5 =

6

119

Total distance traveled

= )6

1195.22()05.22( −+−

6

525=

OR 5

3

233

0

23

182

9

318

2

9

3

+−+

+−

ttt

tt

+−−

+−+

−+−= )3(18

2

)3(9

3

3)5(18

2

)5(9

3

50)3(18

2

)3(9

3

3232323

=6

525

K1

K1

N1

K1K1

K1

N1

10

7(a)

14(b)

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

lg x

lg y

(0.15,0.52)

(0.28,0.72)

(0.40,0.90)

(0.51,1.08)

(0.61,1.20)

(0.74,1.42)

R

x+y=40

x＝2y

4x+3y=36