09 trial melaka_s2

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SULIT

3472/2

Additional

Mathematics

Paper 2

Sept

2009

PERSIDANGAN KEBANGSAAN PENGETUA-PENGETUA
SEKOLAH MENENGAH MALAYSIA (PKPSM) CAWANGAN MELAKA
PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2009

ADDITIONAL MATHEMATICS

Paper 2

MARKING SCHEME

This marking scheme consists of 13 printed pages

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PAPER2

QUESTION FULL
WORKING MARKS
NUMBERS MARKS

1 y  2 x K1
x 2  2 x  (2  x)  4  0 K1
x  x6  0
2

K1
( x  2)( x  3)  0
x  2 , x  3 N1
y  4, y  1
N1
5

2 (a) 3x2 - 4x + 6 = 0 K1

4 K1
SoR = p + n = and PoR = pn = 2
3

New roots:

3(n  p )
SoR = =2
pn
K1
9 9
PoR = 
pn 2
N1
2 9
Eqn : x – 2x + = 0 or equivalent.
2

b) f(x) = 3 ( x  2) 2  4  9 ) K1

= 3 (x – 2)2 + 1 5 N1

Axis of symmetry x = 2 N1
7
Or Or

12 K1
X=
2(3)

f(2) = 15

f(x) = 3 (x – 2)2 + 1 5 K1

x=2 N1

3 1 1 sin x N1
a) LHS = . 
cos x sin x cos x


QUESTION FULL
WORKING MARKS
NUMBERS MARKS
1  sin 2 x
=
cos x sin x K1

cos 2 x cos x N1
= = = cot x = RHS
cos x sin x sin x

P1
b) Shape of graph of sin x
P1
Graph of sin 2x

1

0 
2
 3
2 8
2
1

x
y= K1
2

L1
Graph of the straight line

N1
Number of solutions = 4

4 a) T5 = 16T1 K1

ar4 = 16a K1

r4 = 16

r=2 N1

b) Circumference = 2π r = 88 cm K1

 25  1 
a + 2a + 4a + 8a + 16a = 88 or a
 2  1   88

  N1

31a = 88 K1

88 26 K1
a= = 2 cm 7
31 31 www.cikgurohaiza.com


QUESTION FULL
WORKING MARKS
NUMBERS MARKS

5 a) 44.5 , 54.5 , 64.5,74.5, (all seen) P1

44.5(12)  54.5(k )  64.5(30)  74.5(14) K1
60.25 =
12  k  30  14

24 = k
N1

b) L = 59.5 , F = 36 , fm = 30 , N = 80 , c = 10
K1
 80 
 2  36 
median = 59.5   10
 30  K1

 


= 60.83
N1 6

6    K1
a) i) QT  QP  PT
N1
=  9x  6 y

   2  
ii) RS  RQ  QS   PS  PT K1
3
N1
=  6x  2 y

   1  1 
b) VU  VP  PU   PS  PT K1
3 3
N1
=  3x

N1
h = -3 and k = 0
7



QUESTION FULL
WORKING MARKS
NUMBERS MARKS
7 K1
4  5x  x 2
x 2  5x  4  0
( x  4)( x  1)  0
x  4, x  1
N1

4

(b) Area   ( 5 x  x 2  4)dx
1

5x 2  x 3 K1
[  4 x]1
4

2 3
5 (4) 2  (4) 3 5 (1) 2  (1) 3
 [(  4(4))  (  4(1))]
2 3 2 3
8 11
 [ ]  [  
3 3
9

2 N1

1
K1
V   (4) (1)    y 2 dx
2

0
1
16    ( 5x  x 2 ) 2 dx
0

(c) 1
K1
16     ( 25x 2 10 x 3  x 4 ) dx
0

25 x 3 10 x 4 x 5 1
16    [   ]0
3 4 5
25 10 x
16    [   ] K1 10
3 4 5
181
  [16  ]
30
29
9 
30 N1



QUESTION FULL
WORKING MARKS
NUMBERS MARKS
8. x 10 15 20 25 30

y√x 27.4 40.0 53.4 66.7 79.4 N1

Graph

uniform scale

One point plotted wrongly and /All points plotted K1

correctly P1

Line of the best fit L1

10

B1

K1
y x  hx  k
N1
(c) (i) from graph
N1
79.4  27.4 K1
k
30  10
 2.6 N1

(ii) h = 1.5

(iii) y x = 22

Y = 7.778


QUESTION FULL
WORKING MARKS
NUMBERS MARKS
9 2
AEB  2   
3
  
(a)
 
1 or   K1 N1
2 3 6
3 K1
1

AEO  3
2
1
 
6 N1

K1
(b) Perimeter ACBO

K1
AE
ta n AOE 
AO
1
AE  5  tan  K1
3
 8.66 10
N1
arc ACB  r
1
 8.660  
3
 9.069
Perimeter  9.0699  10
 19.069

(c) Area of DACB

1 2 2
Area  (5) (   sin 120  ) K1 K1K1
2 3
25
 (1.2286 )
2
 15.3546 N1



QUESTION FULL
WORKING MARKS
NUMBERS MARKS
10 (a) Value of h

Equation of DC

2 2
y x
3 3
2
m AB  m DC 
3
0 1 2
 K1
h4 3
11
h
2
N1
(b) Equation of AD

1 3
m AD   
m AB 2
3 K1
( y  1)   ( x 4)
2
N1
3
y   x  5 @ 2 y   3 x  10
2 10

(c) Coordinate of point D

3y  2 x  2
2 y  3 x  10
9y 6 x 6
4 y  6 x  20 K1
13 x  26
K1
y2
x2 N1
(2,2)

(d) Lets P =( x, y)

AP = PC

( x  4) 2  ( y  1) 2  ( x  7) 2  ( y  9) 2 K1
x  8 x  16 )  ( y  2 y  1 )  ( x  14 x  49 )  ( y  18y  81 )
2 2 2 2

6 x  20 y 113  0 K1

N1



QUESTION FULL
WORKING MARKS
NUMBERS MARKS
1 1 P1
a) i) p = , q = , n = 10, r = 2
2 2
2 8
1 1 K1
P(x = 2) = 10C 2   
2 2
N1
= 0.04395

ii) 1 – P(x = 0) – P(x = 1)

0 10 1 9
1 1 1 1 K1
= 1 - 10C 0    - 10C1   
2 2 2 2

= 1 – 0.00098 – 0.0098

N1
= 0.9893
11

50  45
b) i) Z 
6

P(X > 50) = P Z  0.8333
K1
= 0.2025
N1

ii) P ( 0.8333< Z< 2.5 )
K1
= 0.1963
10
K1
6000 x 0.1963
N1
= 11777

a) V = 3 P1

b) v = 3 + 8t  3t2 = 0

(3t + 1) ( t – 3) = 0
K1
t =3
N1
12
(c) S = 3t  4t 2  t 3

S3 = 3(3) + 4(3)2 – (3)3 = 18 K1

S4 = 3(4) + 4(4)2 – (4)3 (either one)


QUESTION FULL
WORKING MARKS
NUMBERS MARKS
= 12 K1

18 + ( 18- 12) = 24 N1

Or Or

3 4

 3  8t  3t   3  8t  3t 2
2

0 3 K1

3 3 4
3t  4t 2  t  3t  4t 2  t 3
0 3

18 +  6
N1

24

dv K1
d)  8  6t  0
dt
10
4
t= s N1
3

v =3t2 – 6t – 6

3 3 K1
= 3( ) 2  6( )  6
2 2

 26
= - 8.667 ms-1 @ N1
3

4.20 P1
a) x = (100)  140
3.00

 150  P1
y = 0.80   = 1.20
 100 

13  100 
z = 0.60   = 0.75 P1
 80 

K1, K1
125(70)  140(60)  150(105)  80(125)
b) i) I 
360 10
= 119.17 N1


QUESTION FULL
WORKING MARKS
NUMBERS MARKS

 100  K1
ii) Cost in 2002 = 3600  
 119.17 

= 3020.89 N1

3600  360 110 K1
c) I  (100) or  119.17
3020.89 100
N1
= 131.09

14 a) K1
PQ 2  6.52  7.32  2(6.5)(7.3) cos1300
 42.25  53.29  2(6.5)(7.3)(0.6428) K1
 156.54
N1
PQ  156.54  12.51

P1

b)
P1

10

K1 K1



QUESTION FULL
WORKING MARKS
NUMBERS MARKS
K1
c) Area, PQR1
1
  6.5  7.3  sin 130 0
2
 18.17cm 2 K1

Area, QRR' N1
1
  6.5  6.5  sin 80 0
2
 20.80cm 2
Area, PQR'
 18.17  20.80
 38.97cm 2

15 a) x + 2y ≥ 30 P1

2x + 3y ≤ 100 P1

x ≥ 2y P1

b)Draw correctly all the three straight lines/ P2/ P1
Draw wrongly one straight line from the inequalities which involves x
and y
P1
Region shaded correctly

b) Line y = 10

X = 28 K1

(ii)k = 4x + 8y N1

= 4 ( 25) + 8(15) K1

= 220 N1



45

40

35

y = 640

30 2x + 3y = 100

25

20
2y = x

15

x + 2y = 30 (25, 12)

10

R
5 R

5 10 15 20 25 30 40 50


09 trial melaka_s2

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