1. MARKING SCHEME
ADDITIONAL MATHEMATICS PAPER 2
SPM TRIAL EXAMINATION 2009
N0. SOLUTION MARKS
1 y 3x 2 P1
3 2 x (3 x 2 1) K1 Eliminate y
0 6x2 6x 3
2x2 2x 1 0 K1 Solve quadratic
2 2
( 2) 4(2)( 1) equation
x
2(2)
N1
x 1.366 or x 0.366
y 3(1.366) 2 y 3( 0.366) 2
2.098 3.098 N1
5
2 y 2x 3 K1 Use equation or
(a) 3 3 gradient
P ( h, 3) 3 2h 3 2 h 3
0 h
h 3 OR k 3 N1
Q (4, k ) k 2 4 3 2 k 11
4 0
k 11 N1
1 11 0 1 1
(b) m1 2 m2 P1 For m2
2 2
4 x 2
y 11 1
1
x 4 2 OR 11 2 x
2 K1 Use equation or
1
QR : y x 13 x 26 gradient
2
y 0 x 26
x-intercept = 26 N1
(c) Area
1 3 4 26 3
2 3 11 0 3
K1 Use formula and
1 find the area
33 78 12 286
2 triangle
1
385
2
192.5 N1
8
2

2. N0. SOLUTION MARKS
3
(a)
y
P1 Negative sine
shape correct.
y 2 sin 2 x
3
P1 Amplitude = 1
[ Maximum = 3
and Minimum =
1]
2
P1 Two full cycle in
0 x 2
x
1 y 2
2
P1 Shift up the graph
O 3 x
2
2 2
x
(b) 2 sin 2 x 2
2
or N1 For equation
x
y 2
2
x
Draw the straight line y 2 K1 Sketch the
2 straight line
Number of solutions = 4. N1
7
3

3. N0. SOLUTION MARKS
4
dy 2
(a) 1 1 1 2 N1
dx
(b)
y ( x3 x 2 )dx
K1 Integrate gradient
x 4 x3
y c function
4 3
1 1 K1 Substitute ( 1,1)
1 c
4 3 into equation y
5
c
12
x4 x3 5
y N1
4 3 12
(c)
dy
x2 x 1 0
dx
x 0@1
d2y d2y
3x 2 2 x K1 Find
dx 2 dx 2
d2y
x 1 1 0
dx 2
1 1 5 1
y
4 3 12 3
1
1, Minimum point. N1 N1
3
7
4

4. N0. SOLUTION MARKS
5 K1 Use formula of
(a) mean and/or
(i) standard deviation
x
11 x 550 N1
50
(ii)
x2
112 8
50
x2 50(64 121)
N1
9250
K1 Find new mean
(b) and/or new
(i) variance
New mean = 11 1.8 5 24.8
N1
(ii) 2 N1
New variance = (8 1.8) 207.36
6
6 l r
(a)
a 5
P1 Value of a and/or d
d 3
5 n 13 104 K1 Use Tn = 104
n 1 33
n 34 N1
34
(b) S34 2 5 33 3 K1 Find S34
2
1853
5821cm
N1
58.21m
Total cost
RM 4 58.21 K1 RM4 S34
RM 232.8
RM 233 N1
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5

5. N0. SOLUTION MARKS
7 1 1 1
(a) P1
y x
1
1.04 0.83 0.71 0.50 0.45 0.33 N1
y
1
0.67 0.5 0.4 0.25 0.2 0.1 N1
x
1 K1 for correct axes
y and scale
N1 for all points
plotted correctly
N1 for line of
best-fit
0.2
1
0 x
(b)
(i) 1 K1 for y-intercept
= y-intercept
=5 N1
(ii) K1 for gradient
= gradient
N1
= – 6.25
10
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6. N0. SOLUTION MARKS
8
(a) ( x – 2 )2 = 4 – x K1 Solve for x
Q(3, 1 ) N1
3
(b) A= (4 x ) ( x 2 ) 2 dx K1 use
0
3 ( y2 y1 ) dx
2
= ( 3x x ) dx
0
K1 integrate
3 correctly
3x 2 x3
=
2 3 0
K1 correct limit
9
=
2 N1
Note : If use area of trapezium and y dx , give marks accordingly.
2
(c) V= ( x 2 ) 4 dx K1 integrate
0
y 2 dx
2
( x 2) 5
= K1 integrate
5 0
correctly
K1 correct limit
32
=
5 N1
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7

7. N0. SOLUTION MARKS
9
(i) K1 for using vector
QS = QP + PS
triangle
= – 2x + 3 y N1
~ ~
(ii)
PR = PS + SR
= x +2y N1
~ ~
(b)
QT = m QS
(i)
= m (– 2 x + 3 y )
~ ~
= – 2m x + 3m y N1
~ ~
(ii)
PT = n PR
=n(x +2y)
~ ~
= n x + 2n y N1
~ ~
(c)
PQ = PT + TQ
K1 for substituting
2 x = n x + 2n y + 2m x – 3m y & grouping into
~ ~ ~
~ ~ components
= (n + 2m) x + (2n – 3m) y
~ ~
K1 for equating
n + 2m = 2 coefficients
correctly
2n – 3m = 0
K1 for eliminating
m or n
4 N1
m=
7
6 N1
n=
7
10
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8. N0. SOLUTION MARKS
10
(a) 1 K1 Use correctly
cos POA
4 trigonometric
1 1 ratio
POA cos
4
= 75.52o @ 75o31"
= 1.318 rad. N1
(b) Arc PA = 20 ( 1.318 ) = 26.36 K1 Use s r
PQ2 = 802 202 K1 Use Pythagoras
PQ = 77.46 Theorem
Perimeter
= 60 + 26.36 + 77.46 K1
= 163.82 cm N1
(c) 1 K1 Use formula
Area OPQ = 20 77.46 774.6
2 1
A bh
2
1 2 K1 Use formula
Area sector POA = 20 1.318 263.6
2 1 2
A r
2
Area of shaded region
= 774.6 263.6 K1
= 511 cm2 N1
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9. N0. SOLUTION MARKS
11
(a) p = 0.4 q = 0.6 n=6 P1 Value of p and/or q
(i) AND p + q =1
P X 4
P X 4 P X 5 P X 6
6 4 2 6 5 1 6 6 0 K1 Use P(X = r)
C4 0.4 0.6 C5 0.4 0.6 C6 0.4 0.6
= n Cr prqn–r
0.13824 0.036864 0.004096
0.1792 N1
(ii) 2
npq 288
K1
288
n 1200 N1
0.4 0.6
(b) 62 8
(i)
P 60 X 68
X
60 62 68 62 K1 Use Z =
P Z
8 8
P 0.25 Z 0.75
1 P Z 0.25 P Z 0.75
K1
1 0.4013 0.2266
N1
0.3721
(ii)
P X 68 P Z 0.75 0.22663
K1
0.22663 3000 = 679.89
= 679 @ 680 N1
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10

10. N0. SOLUTION MARKS
-1
12 8 ms N1
(a)
dv K1
a 0
dt
2 – 2t = 1
1 N1
t
2
(b) 8 + 2t – t2 = 0 K1
(t – 4 ) (t + 2) = 0 K1
t=4 t = –2 (not accepted) N1
(for t = 4 only)
4 5 K1
(c) s v dt v dt 4 5
0 4 (for and )
0 4
4 5
2 t3 2 t3
= 8t t 8t t K1
3 3
0 4 (for integration)
64 125 64
= 32 16 (0 ) 40 25 32 16 K1
3 3 3
(for summation)
80 10
=
3 3
= 30 m N1
10
11

11. N0. SOLUTION MARKS
13 x
150 100 (or formula finding y /z)
(a) 0.4
x = RM 0.60 N1
y = 120 N1
z = RM 2.00 N1
(b) 45o P1
(150 x 30 ) (110 x90 ) (120 x75) (150 x 120) ( 120 x 45) K1
I
360
46800
=
360
= 130 N1
100 K1
P03 = (5000 )
(c) 130
= RM 3846.2 N1
K1
I 08 / 03 130 x 1.2 (or 130 + 130x0.2)
(d)
= 156 N1
10
12

12. N0. SOLUTION MARKS
14 40x + 20y 2000 or N1
(a) 2x + y 100
15x +30y 1200 or N1
x + 2y 80
y 3x N1
(b)
y
100 y = 3x
2x + y = 100
90
80
70
60 (20, 60)
50
40
(35, 30) y = 30
30
20
10
x + 2y = 80
x
10 20 30 40 50 60 70 80 90 100
At least one straight line is drawn correctly from inequalities K1
involving x and y.
N1
All the three straight lines are drawn correctly
N1
Region is correctly shaded
(c) 35 N1
N1
(d) Maximum point (20, 60)
K1
Maximum profit = 20(20) + 30(60)
N1
= RM 2200
10
13

13. N0. SOLUTION MARKS
2 2 2
15 11 = 9 + 7 – 2(9)(7)cos PQR K1
(a)
9 K1
cos PQR =
126
PQR = 85o 54’ N1
(b) PSR = 180o – 85o 54’
= 94o 6’ P1
'
sin PRS sin 94 0 6
K1
6 11
PRS = 32o 57’
RPS = 180o – 94o 6’ – 32o 57’ K1
= 52o 57’
RS 11
o
sin 52 o 57 ' sin 94 6 '
RS = 8.802 cm N1
1 1 o K1, K1
Area = (9)(7 )sin 85 o 54 ' (6 )(8.802)sin 94 6 '
2 2 (for using
(c) area= ½absinc
= 31.42 + 26.34 and summation)
= 57.76 N1
10
END OF MARKING SCHEME
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