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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