skema trial spm 2010 paper 2

SULIT3472/2

Additional MathematicsPaper 2Sept 2010

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA

SEKOLAH MENENGAH MALAYSIA (PKPSM) CAWANGAN MELAKA

DENGAN KERJASAMA

JABATAN PELAJARAN 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 = m2 +m-2 or

m2+m+6+4m=2 or 4m+m2+m-2+8=2 or

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

P1

K1K1N1N1

52.(a)

(b)

(c)

m2 = 2y -5= 2(x+1) or equivalenty = 2x +7

Solve simultaneous equations:

E( -2, 3)

C( -3, 1)

K1K1N1

K1

N1

K1N1

73.(a)

(b)

Shape of sin xGraph of sin x shifted up 1 unitMiximum = 4, minimum = -2

1½ cycles in

or equivalent

Sketch the straight line involves x and yNumber of solutions = 4

P1P1P1

P1

N1

K1N1

74.(a) Gradient of tangent = 5

kx2-x = 5 substitute x = 1 to kx2-x = 5

K1K1K1

x

4

1

-2

02

2

3

3

y

(b)

k = 6

or equivalent

N1

K1

K1

N1 75.(a)(i)

(ii)

(b)

L = 20.5 F = 57 fQ3 = 28

= 26.93

=19.4

= 11.30

= 126.02

P1

K1

N1

K1

K1

N1

K1

N18

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

P1K1N1

K1N1

N16

7.(a)

(b)

(c)

(i)

(ii)

log10x 0.15 0.28 0.40 0.51 0.61 0.74log10y 0.52 0.72 0.90 1.08 1.20 1.42

Refer to the appendixOne point correctly plotted with uniform scales6 points correctly plotted with uniform scalesLine of best fit

-5k =1.52

log10p = 0.29

N1N1

K1N1N1

P1

K1N1

K1

N110

8(a)(i)

(ii)

(b)

(c)

or equivalent

½ x 8 x h = 20h = 5

K1

N1

N1

P1P1

K1

N1

N1

K1N1

109(a)

(b)

(c)

(d)

or equivalent

OQ = 6

// 0.2618 rad

Area of sector ROS =

=32.99 or 10.5π

Perimeter of shaded region =

=20.52

K1N1

K1

N1

P1

K1

N1

K1K1N1

1010(a)

(b)

(c)

Use b2 – 4ac = 0(-4)2 – 4(1)k =0k = 4Solve equation : (x – 2)(x – 2) = 0A(2, 2)

=

K1N1K1N1

K1

K1

N1

(d)=

= 8.1π

K1

K1

N110

11(a)(i)

(ii)

(b)(i)

(ii)

= 0.2151 – P(x=0) – P(x=1) or P(x=2)+P(x=3)+P(x+4)+ …………….+P(x=10)

=

= 0.9983

= 0.1587 // 0.15866

P(x >342) = 1 – P(z > 1.6) or other valid methodNumber of cakes = 0.9452 x 1500 = 1417 // 1418

K1N1P1

K1N1

K1

N1

K1K1N1

10

12(a)

(b)

(c)

(d)

Use

x = 135, y = 80, z = 140

m = 6

P2005 = 493.50

= 106.38

K1

N2,1,0

K1K1

N1

K1

N1

K1

N110

13 (a)

(b)

(c)

= 9.22

= 47.80° // 47°48 or 92.2°

QR = 12.44

K1N1

K1

N1P1N1

K1

N1

K1N1

10

14.(a)

(b)

(c)(i)

(ii)

x + y ≤ 40 or equivalent4x + 3y ≥36 or equivalentx ≤ 2y or equivalent

Refer to the appendixDraw correctly at least one straight line involves x and yDraw correctly all the three straight linesRegion R shaded correctly

y = 6min passengers = 11substitute any point in the region R into 40x + 30ySubstitute ( 26, 14) to 40x + 30y and maximum fares = 1460

N1N1N1

K1N1N1

K1N1K1N1

1015(a) (b)

(c)

(d)

18 ms-1

2t -9 = 0

=

Use sketch quadratic graph // number lines // others valid method fort2-9t + 18 < 0

3 < t < 6

P1K1

K1

N1

K1

N1

K1

3 6

Substitute t = 3 or t = 5 into

S3 = 22.5

S5 =

Total distance traveled

=

OR

=

K1

K1

N1

K1K1

K1

N110

7(a)

14(b)

R

x+y=40

x＝2y

4x+3y=36