1. MARKING SCHEME
ADDITIONAL MATHEMATICS PAPER 2
SPM TRIAL EXAMINATION 2010
N0. SOLUTION MARKS
1 x = 10 − 2 y P1
K1 Eliminate x
y 2 + (10 − 2 y ) y = 24
y 2 − 10 y + 24 = 0 K1 Solve quadratic
( y − 4) ( y − 6) = 0 equation
y=4 or y=6
N1
x=2 or x = −2
N1
5
2
(a) k=6 P1
(b) Mid point 23 , 28 , 33 , 38 , 43 P1
(i) Mean
=
∑ fx =
1 × 23 + 4 × 28 + 7 × 33 + 5 × 38 + 3 × 43
K1 Use formula and
calculate
∑f 1+ 4 + 7 + 5 + 3
685
= = 34.25 N1
20
(ii) Varian
=
∑ fx2 − x 2
∑f K1 Use formula and
1 × 232 + 4 × 282 + 7 × 332 + 5 × 382 + 3 × 432 calculate
= − 34.252
20
24055
= − 34.252
20 N1
= 29.69
(iii) Median , m
1  1  K1 Use formula and
 2N −F   2 (20) − 5  calculate
= L+  C = 30.5 +  5
 fm   7 
    N1
= 34.07
8
2

2. N0. SOLUTION MARKS
3 1
y = x3 − x 2 + 2
3
(a) dy
= x2 − 2 x = 3 K1 Equate and solve
dx
x2 − 2 x − 3 = 0 quadratic
equation
( x + 1) ( x − 3) = 0
x = −1 , 3
2
x = −1 y=
3
x=3 y=2
 2
 −1,  and ( 3, 2 ) N1 N1
 3
(b) Equation of normals :
1
mnormal = −
3 K1 Use mnormal to form
2 1 1 equations
y− = − ( x + 1) y−2=− ( x − 3)
3 3 3
1 1 1 N1 N1
y=− x+ or equivalent y = − x+3 or equivalent
3 3 3
6
4
y
(a)
P1 Modulus sine
shape correct.
2 y = 3sin 2 x − 1
P1 Amplitude = 3
[ Maximum = 2
1
and Minimum =
-1]
O π π 3π 2π x P1 Two full cycle in
-1 2 2 0 ≤ x ≤ 2π π
3x
y = 1−
-2
2π P1 Shift down the
graph
3

3. N0. SOLUTION MARKS
4
3x
(b) 3sin 2 x − 1 = 1 −
2π
or N1 For equation
3x
y = 1−
2π
3x
Draw the straight line y = 1− K1 Sketch the
2π straight line
Number of solutions = 5. N1
7
5
(a) Common ratio, r=4 N1
(b) 1 K1
A6 = π ( 32 ) 1
T6 = ar = π ( 4 )
2 5 5
4 OR 4
N1
= 256π = 256π
(c)
S6 − S 2 K1 Use S6 or S2
1
(
π 46 − 1
1
) (
π 42 − 1 ) K1 Use S6 - S2
= 4 −4
4 −1 4 −1
N1
= 341.25π −1.25π
= 340π
6
4

4. N0. SOLUTION MARKS
6
(a) K1 for using vector
(i) uuu uuu uuu
r r r triangle for a(i) or
OD = OC + CD
a(ii)
= 6a + 12b
% % N1
(ii) uuu uuu uuu
r r r
AB = OB − OA
1 uuu uuu
r r
= OD − OA
2
= 3a + 6b − 3a
% % % N1
= 6b
%
OR
uuu 1 uuu
r r
AB = CD = 6b [ K1 N1 ]
2 %
(b)
uuur
AE
uuu uuu
r r
= AB + BE K1 for using vector
uuu
r
 1 uuu 
r triangle and BE
= 6b + h  OD 
% 2 
= 6b + h ( 3a + 6b )
% % %
a + kb = 3ha + ( 6 + 6h ) b
K1
% % % %
k = 6 + 6h K1 for equating
3h = 1 coefficients
1
1 = 6 + 6  correctly
h=  3
3
=8 N1 N1
8
5

5. N0. SOLUTION MARKS
7
(a)
x 1 2 3 4 5 6
N1 6 correct
log10 y 0.65 0.87 1.08 1.30 1.52 1.74
values of log y
log10 y
(b) K1 Plot log10 y vs x
Correct axes &
uniform scale
N1 6 points plotted
correctly
N1 Line of best-fit
0.43
0 x
(c) log10 y = ( k log10 A ) x + log10 A P1
(i) x = 2.6 N1
(ii) y-intercept = log10 y K1
A = 2.69 N1
gradient = k log10 A K1
gradient
k=
log10 A *
= 0.51 N1
10
6

6. N0. SOLUTION MARKS
8
(a) P(5, 1 ) P1
Q(1, 0 ) P1
(b) 1
∫ (4y + 1) dy ∫ x dy
2
A= K1 use
0
K1 correct limit
 4y3  1
=  + y
 3  0
K1 integrate
correctly
7
= OR equivalent
3 N1
5
x −1
(c) V= π∫ dx K1 integrate
4
1
π ∫ y 2 dx
K1 correct limit
π  x2 5
=  − x
4 2  1
K1 integrate
correctly
= 2π N1
10
7

7. N0. SOLUTION MARKS
9
(a) 4 K1 Use ratio of
cos ∠ POQ =
10 trigonometry or
equivalent
∠ POQ = 1.16 rad.
N1
(b)
( 2π – 1.16 ) rad P1
PQ = 10 ( 2π – 1.16 ) K1 Use s = rθ
= 51.24 cm N1
(c)
P1
10 2 − 4 2
= 9.17 cm
1 K1
Area of trapezium POQR = ( 6 + 10 ) × 9.17 *
2
= 73.36 cm2
1 K1 Use formula
Area of sector POQ = (10) 2 (1.16)
2 1
A = r 2θ
= 58 cm2 2
Area of shaded region
= 73.36 – 58 K1
= 15.36 cm2 N1
10
8

8. N0. SOLUTION MARKS
10.
(a) Equation of str. line PQR :
1 K1
m= −
2
1 N1
y= − x+1
2
(b) 1 K1 solving
2x + 6 = − x+1
2 simultaneous
equation
P( –2, 2) N1
(c) 1( x) + 2(−2) 1( y ) + 2(2) K1 Use the ratio
=0 or =1
1+ 2 1+ 2 rule
R( 4, –1) N1
(d)
(i) 1 K1 Use distance
( x − 4) 2 + ( y + 1) 2 = ( x + 2) 2 + ( y − 2) 2
2 formula
4 [ x2 – 8x + 16 + y2 + 2y +1 ] = x2 + 4x + 4 + y2 – 4y +4
x2 + y2 – 12x + 4y + 15 = 0 N1
(ii) Substitute x = 0, y2 + 4y + 15 = 0 K1 Substitute x = 0
b2 – 4ac = (4)2 – 4(1)(15) 2
and use b – 4ac
= – 44 < 0 to make a
conclusion
⇒ No real root for y,
⇒ The locus does not intercept the y-axis. N1 if b2 – 4ac = -44
9

9. 10
N0. SOLUTION MARKS
11
(a)
µ = 80, σ = 12
65 − 80 X −µ
P ( X ≥ 65 ) = P ( Z ≥ ) K1 Use Z =
12 σ
= P ( Z ≥ − 1.25 )
= 1 – 0.1056 K1 Use 1 – Q(Z)
= 0.8944 N1
(b) 0.1056 × 4000 K1
= 422 or 423 N1
(c) 200 P1
= 0.05
4000
Q( Z ) = 0.05
Z = 1.645 K1 Find value of Z
m − 80 m−µ
= − 1.645 K1 Use
12 σ
K1 Use negative
value
m = 60.26 g N1
10
10

10. N0. SOLUTION MARKS
12
(a) - 5 ms-1 N1
(b) v<0 K1
2
t - 4t - 5 < 0 K1
(t – 5) (t +1) < 0
0<t<5 N1
v
(c) 8
7 7
6
P1 (for shape )
4
2
-5
0
0 2 5 6 10 15
t
-2
2 6
-4
P1 min(2,-9) , (6,7)
-5 -5
-6 &(0,-5) must be
-8 seen
-9 -9
-10
-12
(d) Total dis tan ce
5 6 K1 for
= ∫ vdt ∫
+ vdt 5 6
0 5
∫
0
and ∫ 5
5 6
t3  t3 
=  − 2t 2 − 5 t  +  − 2 t 2 − 5 t 
3
 0
 3
 5

K1 (for Integration;
either one)
 5 3    216  5
3 
=  − 2 (5 ) 2 − 5 (5 )  − ( 0 )  +   2( 36 ) − 30  −  − 2 (5 ) 2 − 5 (5 )  
 3


   3
    3


 K1 (for use and
summation)
1  1 
= −33 +  −30 − (−33 ) 
3  3 
2
= 36 m
3 N1
11

11. 10
N0. SOLUTION MARKS
13
P09
(a) × 100 = 125 K1
60
N1
P09 = RM 75
(b) (125 × 4 ) + (120 m) + (80 × 5 ) + 150 m + 450 K1
(i) 120 =
12 + 2 m
K1 (use formula)
1440 + 240 m = 1350 + 270 m
N1
m = 3
(ii)
100
P07 = RM 30 × K1
120
N1
= RM 25
(c)
120 + (120 × 0.15) = 138 K1
K1
(125 × 4 ) + (138 × 3) + (80 × 5 ) + (150 × 6 )
I 10 / 07 =
18
N1
= 123
12

12. 10
N0. SOLUTION MARKS
14
(a) y ≥ 200 N1
x+y ≤ 800 N1
4x + y ≤ 1400 N1
(b) y
1000
4x + y = 1400
900
800
700
600 (200,600)
500
400
R
300
y = 200
200
100
x + y = 800
x
100 200 300 400 500 600 700 800 900
• At least one straight line is drawn correctly from inequalities K1
involving x and y.
N1
• All the three straight lines are drawn correctly
• Region is correctly shaded N1
(c)
(i)
650 N1
(ii)
Maximum point (200, 600) N1
Maximum profit = 20(200) + 6(600) K1
= RM 7600 N1
13

13. 10
N0. SOLUTION MARKS
15
(a) TQ2 = 92 + 62 – 2(9)(6)cos56o K1
TQ = 7.524 cm N1
(b) sin ∠QTR sin 56 0 K1
=
6 7.524
∠QTR = 41o 23’ N1
(c) 1 K1
42.28 = ( RS )(6 )sin 56 o
2
RS = 17
ST = 17 − 9 (or ST + 9 in formula of area) K1
= 8 cm N1
(d) 1
Area ∆ QTR = (9)(6) sin 56 0
2 K1
2
= 22.38 cm
Area of quadrilateral PQTS = 2(42.28) – 22.38 K1
= 62.18 cm2 N1
10
END OF MARKING SCHEME
14