SPM TRIAL EXAM 2013Marking Scheme

Additional Mathematics Paper 2SECTION A

Question Important Steps Marks1 y = 3x – 4 P1

5x2 – 4x(3x – 4) + (3x – 4)2 = 9 Eliminate x or y

or 2x2 – 8x + 7 = 0 K1

solve the quadratic eq. by quadratic

formula or completing the square

K1

x = 2.707, 1.293 N1y = 3(2.707) – 4 , y = 3(1.293) – 4= 4.121 = – 0.121 N1

TOTAL 5Question Important Steps Marks

2(a)

(b)

a=5 , d=8 K1

77 = 5+ (n-1) 8 K1n=10 N1

a = 5 , d = 8

= 410 The total cost = RM 90.20

K1K1

N1N1

TOTAL 7

3Shape of cos graph

Amplitude=31.5 periods

Modulus

Straight line graph drawn, seen/ implied

Number of solution =7TOTAL

11111117

Question Important Steps Marks

Note:OW-1 if the working of solving quadratic equation is not shown

4(a)

(b)

( c )

mPQ = or mQR =

P1

K1

3y = 2x +16 or N1

Solve simultaneous equation between QR and PRR(4,8)

Area =

= 22

K1N1

K1

N1

TOTAL 7Question Important Steps Marks

5(a)

(b)

h + k = 11 P1

K1

h = 5 N1 k = 6 N1

Refer to the given histogram.Correct axes, consistent scale and draw one block correctly.Draw all the blocks correctlyTry to find the mode correctlyMode = 44

P1P1K1N1

TOTAL 8Question Important Steps Marks

6(a)

(b)

log3(3x+6) – 4log9x2+4log3x

=log3(3x+6) – +4log3xK1

=log3(3x+6) – + log3x4 K1

= log3(3x+6) – + log3x4 K1

=

=

= 2

N1

K1

x = 1 N1

TOTAL 6Question Important Steps Marks

7(a)

(b)

(a ) x 1 2 3 4 5 6 N1

lo g 1 0 y 0.2553 0.4314 0 .6 0 7 5 0.7839 0.9595 1.136 N1

Using the correct, uniform scale and axes P1

2 d.p

N1N1

Using the correct, uniform scale and axes P1 All points plotted correctly

P1 Line of best fit

P1

P1P1P1

(b ) lo g 1 0 y lo g 1 0 k x lo g 1 0 p (or implied)

use *m = log10 k K 1

P1

use *m = log10 k k 1.501

u se * c lo g 1 0 p p 1.202

K1N1

K1N1

TOTAL 10Question Important Steps Marks

8(a)(i)

(ii)

(iii)

(b)(i)

(ii)

N1

K1 = N1

= =

= h( ) =

= =

6 = 9h h=

3k – 3 = 3h

N1

N1

K1

N1

K1N1

3k – 3 = - 2

k=

N1TOTAL 10

Question Important Steps Marks9(a)

(b)

(c)

K1N1

Arc SU = 20(1.231)=24.62 N1(ST)2=602 - 202

ST=56.57

Perimeter = 40+24.62+56.57 = 121.19

Area of Area sector of SOU

=319.5

N1

K1N1

K1K1K1

N1TOTAL 10

Question Important Steps Marks10(a)

(b)

-2x + 5 = 5 - x2 K1x(x - 2) = 0 for solving quadratic equation K1(2,1) N1

A= use

=

= integrate correctly

= sub. the limit correctly

=

Note: if use area of trapezium and , give mark accordingly

K1

K1

K1

N1

c

(b)

correct limit K1

integrate correctly K1

= N1

TOTAL 10Question Important Steps Marks

11 (a)

(b)

(i) K1

= 0.2786 N1

(ii) P(X OR K1 P(X=2) +P(X=3)+P(X=4)+P(X=5)+P(X+6)+P(X=7)+P(X=8) =1- (for using

)= 0.9964

(i) P(X<155)= P (for using z-score formula)

= 0.3385

(ii) P(X > k) = 0.7

P =0.7 (for using z-score formula)

(accept -0.525)

k = 153.7

K1N1

K1N1

K1

K1

N1

TOTAL 10

Question Important Steps Marks12(a)

K1

N1

(b)(i)

(ii)

(c )

K1K1

=121.25

Seen 137.5 and 135

=130

N1

K1N1

K1

K1N1

TOTAL 10Question Important Steps Marks

13(a)

(b)

N1

K1

AC – 10.78 cm N1

K1

(c )

K1

BC = 15.21

K1

- 5.07 cm

DC = 15.21-5.07 = 10.14

Area of

= 35.86 cm2

or equivalent

N1

K1

K1N1

TOTAL 10Question Important Steps Marks

14(a) or equivalent N1

or equivalent N1

or equivalent N1

(b)

(c )

One straight line drawn correctly K1All straight line drawn correctlyRegion R shaded

(i) Using straight line x = 6 ymin=8

(ii) k = 16(7)+12(10) = RM232

K1N1

K1N1

K1N1

TOTAL 10

Question Important Steps Marks15(a)

(b)

(c )

K1

N1

v > 0K1N1

P1

(d)

Use (for limits)

= (for integration)

= (substitution)

= 8

P1

K1

K1

P1

N1

TOTAL 10

For minimum shape

Passing through points (0,8) and (4,0)

SULIT 11

Soalan 7(a) lo g 1 0 y

1.2

1.1

1.0

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0 1

2 3 4 5 6

x