1. The document contains the mark scheme and solutions for an Additional Mathematics exam paper with 10 questions. 2. Question 1 involves solving a pair of simultaneous equations to find the values of x and y. Question 2 involves finding the maximum value and graph of a quadratic function. 3. Question 3 examines areas of shapes in a geometric progression. The final area is found to be 17066 cm^2. 4. Subsequent questions cover topics such as function graphs, probability distributions, logarithmic graphs, trigonometry, and simultaneous equations. 5. The last few questions deal with topics like probability, normal distribution, and kinematic equations. Overall the document provides a comprehensive breakdown of the marking scheme

1. 1
SPM TRIAL EXAM 2010
MARK SCHEME ADDITIONAL MATHEMATICS PAPER 2
SECTION A (40 MARKS)
No. Mark Scheme Total
Marks
1 x = 1− 2y P1
2(1 − 2 y ) + y + (1 − 2 y )( y ) = 5
2 2
K1
7y2 − 7y − 3 = 0
− ( − 7) ± ( − 7 ) 2 − 4( 7 )( − 3)
y= K1
2( 7 )
y = 1.324 , − 0.324
N1
x = −1.648 , 1.648
N1
OR
1− x
y=
2
P1
1− x  1− x 
2x 2 +   + x =5
 2   2  K1
7 x 2 − 19 = 0
− ( 0) ± ( 0) 2 − 4( 7 )( − 19)
x=
2( 7 )
K1
x = −1.648 , 1.648
N1
y = 1.324 , − 0.324
N1
5

2. 2
2 (a)
f ( x ) = −( x 2 − 4 x − 21)
K1
 2
−4 −4
2

= − x 2 − 4 x +   −  − 21

  2   2  

N1
= −( x − 2) + 25
2
(b) Max Value = 25 N1
(c) f ( x)
25 (2,25)
21
x
-3 2 7
Shape graph N1
Max point N1
f ( x ) intercept or point (0,21) N1
d) f ( x ) = ( x − 2) 2 − 25 N1
7
3 1 1
a) List of Areas ; xy, xy, xy K1
4 16
1
T2 ÷ T1 = T3 ÷ T2 =
4
1
This is Geometric Progression and r = N1
4
n −1
1 25
b) 12800 ×   =
4 512

3. 3
1
n −1
1 K1
  =
4 262144
n −1 9
1 1
  = 
4 4
n −1 = 9 K1
n = 10
12800 N1
S∞ =
(c) 1
1−
4 K1
2 N1
= 17066 cm 2
3
7
4 a)
4 cos 2 − 1 − 1 K1
4 cos 2 − 2
2( 2 cos 2 − 1)
N1
2 cos 2θ
b) i)
2
1
π 2π
-1
-2
P1
- shape of cos graph P1
- amplitude (max = 2 and min = -2) P1
- 2 periodic/cycle in 0 ≤ θ ≤ 2π
θ K1
b) ii) y = 1 − (equation of straight line)
π
Number of solution = 4 (without any mistake done) N1
7

4. 4
5 a)
Score 0–9 10 – 19 20 – 29 30 – 39 40 – 49
Number 3 4 9 9 10 N1
 1 
 ( 35) − 7 
4
b) Q1 = 19.5 +  10 P1
 9 
 
  K1
= 21.44
3 
 ( 35) − 25 
Q3 = 39.5 +  4 10
 10  K1
 
 
= 40.75
Interquatile range
= 40.75 − 21.44 K1
= 19.31
N1
6
6 (a) OQ = OA + AQ K1
OQ = (1 − m ) a + m b N1
~ ~
(b) (
PO + OQ = n PO + OR ) K1
4
OQ = (1 − n ) a + 3n b N1
5 ~ ~
(c)
4 4  K1
(i)  − n  = 1 − m or 3n = m
5 5 
3 1
m= ,n= N1
11 11 N1
8 3 N1
(ii) OQ = a+ b
11 ~ 11 ~
8

5. 5
7 2
(a)(i) Area = ∫ ( 2 y − y ) dy
2
K1
0
2
 y3 
=  y2 − 
 3 0
4 2 N1
= unit
3
1 2
(ii) Area region P = ∫ y dy + ∫ ( 2 y − y ) dy
2
K1
0 1
2
1   y3 
=  × 1× 1  +  y 2 −  K1
2   3 1
7 2 N1
= unit
6
4 7 1 2
(b) Area region Q = − = unit
3 6 6 K1
7 1
= :
6 6
N1
=7:1
1
(c) Volume = π ∫ ( 2 y − y ) dy
2 2
0 K1
1
 4 y3 y5 
=π  − y4 +  K1
 3 5 0
8
= π unit
3
15 N1
10
8 x 0.000 0.707 1.000 1.414 1.732 N1
1
log10 y 1.000 1.330 1.477 1.672 1.826 N1
P1
(a) P1
P1
P1
Using the correct, uniform scale and axes
All points plotted correctly

6. 6
Line of best fit K1
N1
1
(b) log10 y = x log10 p + log10 k K1
3
(i) use ∗ c = log10 k N1
k = 10.0
1.83 − 1.0 1
(ii) use * m = = 0.47977 = log10 p
1.73 − 0 3
p = 27.5
10
9
1  K1
(a) ∠COD = 2  π 
6 
1 N1
= π = 1.047 rad
3
 1  20 K1
(b) (i) Arc ABC = 10  π − π  or = π
 3  3
1  K1
Length AC = 202 − 102 or 20 cos  π rad 
6 
20 1 N1
Perimeter = π + 20 cos π = 38.267cm
3 6
(ii) Area of shaded region =
1
2
( ) 2
3
2 
102  π − sin π 
3 
K1
= 61.432cm2
N1
1
(c) ∠CDE = ∠CAD = π rad ( alternate segments ) K1
6
Area =
1
2
( ) 1 
102  π 
6 
K1
N1
= 26.183cm2

7. 7
10
10 (a) T ( 4, 2 ) P1
6+ x 6+ y
= 4, =2 K1
2 2
S ( 2, −2 ) N1
(b) y − 2 = 2 ( x − 4 ) K1 K1
y = 2x − 6 N1
3 x + 24 3 y + 24 K1
(c) = 2 or = −2
7 7
 10 38  N1
U − ,− 
 3 3 
K1
( x − 2) + ( y + 2) = 2 ( x − 4) + ( y − 2)
2 2 2 2
(d)
N1
3 x 2 + 3 y 2 − 28 x − 20 y + 72 = 0
10
11 (a) (i) P ( X = 0 ) = C0 (0.6) (0.4) or P ( X = 1) = C1 (0.6) (0.4) K1
10 0 10 10 1 9
P ( X ≥ 2) = 1 − [ P ( X = 0) + P ( X = 1) ]
10 0 10 10 1 9
= 1 ─ C0 (0.6) (0.4) ─ C1 (0.6) (0.4) K1
= 0.9983 N1
2
(ii) 800 × K1
5
N1
= 320
(b)(i) P ( −0.417 ≤ z ≤ 1.25 ) K1
=1 − 0.3383 − 0.1057
= 0.556 N1
(ii) P ( X > t ) = 0.7977
Z = −0.833 P1

8. 8
t − 4.5 K1
−0.833 =
1.2
t = 3.5004 N1
10
Sub Total
No Mark Scheme
Marks Mark
12a i) 1 K1 3
(14) (5) sin θ = 21
2
θ = 36.87° or 36° 52 '
∠ BAC = 180° − 36.87° K1
= 143.13° or 143° 8'
N1
ii) BC 2 = 142 + 52 − 2(14)(5) cos 143.13° K1 2
BC 2 = 333
BC = 18.25 cm N1
iii) sin θ sin 143.13° K1 2
=
5 18.25
θ = 9.46° or 9° 28' N1
b i) A'
14 cm N1 1
5 cm
B' C'
ii) ∠ ACB = 180° − 143.13° − 9.46° K1 2
= 27.41°
∠ A ' C ' B ' = 180° − 27.41°
= 152.59° or 152° 35' N1 10

9. 9
Sub Total
No Mark Scheme
Marks Mark
13 a) 4.55 n 3
m= × 100 or × 100 = 112 K1
3.50 4
m = 130 n = RM 4.48 N1 N1
b) 110(70) + * 130( x) + 120( x + 1) + 112(2) K1 2
= 116.5
7 + x + x +1+ 2
x=3 N1
c i) See 140 P1 3
x (116.5)
= 140 K1
100
x = 120.17 / 120.2 N1
ii) x K1 2
× 100 = 140
25
x = RM 35 N1 10

10. 10
Sub Total
No Mark Scheme
Marks Mark
15 a) v 0 = − 30 ms −1 N1 1
b) − 3t 2 + 21t − 30 > 0 K1 2
( t − 5)( t − 2 ) < 0
2<t<5 N1
c) a = − 6t + 21 K1 3
a 5 = − 6(5) + 21 K1
a 5 = − 9 ms − 2 N1
d) − 3t 3 21t 2 K1 4
S = + − 30t
3 2
21t 2
S = − t3 + − 30t
2
21(3) 2 K1
S 3 = − (3) +
3
− 30(3) = − 22.5 or
2
21(5) 2
S 5 = − (5)3 + − 30(5) = −12.5
2
Total distance = − 22.5 + (− 22.5) − ( −12.5) K1
= 32.5 m N1 10

11. 11
Answer for question 14
(a) I
I. N1
y I
II. N1
I
III.
N1
(b) Refer to the graph,
1 graph correct K1
3 graphs correct N1
90 Correct area N1
(
(c) i)
N1
80 ii) k = 10x + 20y
max point ( 20,50 ) N1
70 Max fees = 10(20) + 20(50)
K1
= RM 1,200
(20,50) N1
10
60
50
40
30
20
100 10 20 30 40 50 60 70 80 x

12. 12
log10 y Answer for question 8
2.0
1.9
X
1.8
1.7
X
1.6
1.5
X
1.4
X
1.3
1.2
1.1
1.0 X 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8
0
x