SPM TRIAL EXAM 2010 MARK SCHEME ADDITIONAL MATHEMATICS PAPER 2

SECTION A (40 MARKS)

No. Mark Scheme Total Marks

1

yx 21−=

( ) ( )( ) 521212 22 =−++− yyyy

0377 2 =−− yy

( ) ( ) ( )( )( )72

37477 2 −−−±−−=y

324.0,324.1 −=y

648.1,648.1−=x OR

21 xy −

=

52

12

12 2 =

−

+

−

+xxxx

0197 2 =−x

( ) ( ) ( )( )

( )72197400 2 −−±−

=x

648.1,648.1−=x

324.0,324.1 −=y

P1

K1

K1

N1

N1

P1

K1

K1

N1

N1

5

2

(a) ( ) ( )

−

−−

−+−−=

−−−=

2124

244

21422

2

2

xx

xxxf

( ) 252 2 +−−= x (b) Max Value = 25 (c) ( )xf 25 (2,25) 21 x -3 2 7 Shape graph Max point ( )xf intercept or point (0,21) d) ( ) ( ) 252 2 −−= xxf

K1

N1

N1

N1 N1 N1

N1

7

3 a) List of Areas ; xyxyxy

161,

41,

41

2312 =÷=÷ TTTT

This is Geometric Progression and 41

=r

b) 51225

4112800

1

=

×

−n

K1

N1

262144

141 1

=

−n

91

41

41

=

−n

91 =−n 10=n

(c)

411

12800

−=∞S

= 3217066 cm 2

K1

K1

N1

K1

N1

7

4 a) 11cos4 2 −−

2cos4 2− ( )1cos22 2−

θ2cos2 b) i) 2 1 π π2 -1 -2

- shape of cos graph - amplitude (max = 2 and min = -2) - 2 periodic/cycle in πθ 20 ≤≤

b) ii) πθ

−= 1y (equation of straight line)

Number of solution = 4 (without any mistake done)

K1

N1

P1 P1 P1

K1

N1

7

5 a) Score 0 – 9 10 – 19 20 – 29 30 – 39 40 – 49 Number 3 4 9 9 10

b) ( )

109

73541

5.191

−+=Q

= 21.44

( )10

10

253543

5.393

−+=Q

= 40.75 Interquatile range

44.2175.40 −= = 19.31

N1

P1

K1

K1

K1

N1

6

6 (a) AQOAOQ +=

( )~~

1 bmamOQ +−=

(b) ( )ORPOnOQPO +=+

( )~~

3154 bnanOQ +−=

(c)

(i) mn −=

− 1

54

54 or mn =3

111,

113

== nm

(ii) ~~ 11

3118 baOQ +=

K1

N1

K1

N1

K1

N1 N1

N1

8

7

(a)(i) ( )2

2

0

2Area y y dy= −∫

= 23

2

03yy

−

= 243

unit

(ii) ( )1 2

2

0 1

2Area region P y dy y y dy= + −∫ ∫

23

2

1

1 1 12 3

yy = × × + −

= 276

unit

(b) 24 7 13 6 6

Area region Q unit= − =

7 1:6 6

=

= 7 : 1

(c) ( )1

22

0

2Volume y y dyπ= −∫

13 5

4

0

43 5y yyπ

= − +

= 38 15

unitπ

K1

N1

K1

K1

N1

K1

N1

K1

K1

N1

10

8

(a) Using the correct, uniform scale and axes All points plotted correctly Line of best fit

(b) 10 10 101log log log3

y x p k= +

x 0.000 0.7071 1.000 1.414 1.732

log10 y 1.000 1.330 1.477 1.672 1.826

N1

N1

P1 P1 P1

P1

(i) 10 log∗ =use c k k = 10.0

(ii) 101.83 1.0 1* 0.47977 log1.73 0 3

use m p−= = =

−

27.5p =

K1 N1

K1 N1

10

9

(a) 126

COD π ∠ =

1 1.0473

radπ= =

(b) (i) 1 20103 3

Arc ABC orπ π π = − =

2 2 120 10 20cos6

Length AC or radπ = −

20 120cos 38.2673 6

Perimeter cmπ π= + =

(ii) ( )21 2 210 sin2 3 3

Area of shaded region π π = −

= 61.432cm2

(c) 16

CDE CAD radπ∠ = ∠ = ( alternate segments )

( )21 1102 6

Area π =

= 26.183cm2

K1

N1

K1

K1

N1

K1

N1

K1

K1

N1

10

10

(a) ( )4, 2T

22

6,42

6=

+=

+ yx

( )2, 2S − (b) ( )2 2 4y x− = −

2 6y x= −

(c) 3 24 3 242 27 7

x y or + += = −

10 38,3 3

U − −

(d) ( ) ( ) ( ) ( )2 2 2 22 2 2 4 2x y x y− + + = − + − 2 23 3 28 20 72 0x y x y+ − − + =

P1

K1

N1

K1 K1

N1

K1

N1

K1

N1

10

11

(a) (i) ( ) 10 0 10

00 (0.6) (0.4)P X C= = or ( ) 10 1 911 (0.6) (0.4)P X C= =

[ ]( 2) 1 ( 0) ( 1)P X P X P X≥ = − = + =

= 1 ─ 10 0 100 (0.6) (0.4)C ─ 10 1 9

1(0.6) (0.4)C = 0.9983

(ii) 28005

×

= 320 (b)(i) ( )0.417 1.25P z− ≤ ≤ = 1057.03383.01 −− = 0.556 (ii) ( ) 0.7977P X t> = Z = 0.833−

4.50.8331.2

t −− =

3.5004t =

K1

K1 N1

K1 N1

K1

N1

P1

K1

N1

10

No Mark Scheme Sub Marks

Total Mark

12a i)

1 (14) (5)sin 212

θ =

36.87 36 52 'orθ = ° ° 180 36.87BAC∠ = ° − ° 143.13 143 8'or= ° °

K1

K1

N1

3

ii)

2 2 214 5 2(14)(5)cos 143.13BC = + − ° 2 333BC = 18.25BC cm=

K1

N1

2

iii)

sin sin 143.135 18.25θ °=

9.46 9 28'orθ = ° °

K1

N1

2

b i)

N1

1

ii)

180 143.13 9.46ACB∠ = ° − ° − ° 27.41= ° ' ' ' 180 27.41A C B∠ = ° − ° 152.59 152 35'or= ° °

K1

N1

2

14 cm 5 cm

'A

'B 'C

10

No Mark Scheme Sub Marks

Total Mark

13 a)

4.55 1003.50

m = × or 100 1124n× =

130m = RM 4.48n =

K1

N1 N1

3

b)

*110(70) 130( ) 120( 1) 112(2) 116.5

7 1 2x xx x

+ + + +=

+ + + +

3x =

K1

N1

2

c i)

See 140

(116.5) 140100

x=

120.17 / 120.2x =

P1

K1

N1

3

ii)

100 14025x× =

RM 35x =

K1

N1

2

10

No Mark Scheme Sub Marks

Total Mark

15 a)

10 30v ms −= −

N1

1

b)

23 21 30 0t t− + − > ( 5)( 2) 0t t− − < 2 5t< <

K1

N1

2

c)

6 21a t= − +

5 6(5) 21a = − +

25 9a ms−= −

K1

K1

N1

3

d)

3 23 21 30

3 2t tS t−

= + −

2

3 21 302tS t t= − + −

2

33

21(3)(3) 30(3) 22.52

S = − + − = − or

2

35

21(5)(5) 30(5) 12.52

S = − + − = −

Total distance = 22.5 ( 22.5) ( 12.5)− + − − − 32.5 m=

K1

K1

K1

N1

4

10

Answer for question 14

10

20

70

60

50

80

90

40

30

y

(20,50)

10 20 30 40 50 60 70 0 80 x

(a) I. 70x y+ ≤

II. 2x y≤ III. 2 10y x− ≤

(b) Refer to the graph, 1 graph correct 3 graphs correct Correct area (c) i) 15 40y≤ ≤ ii) k = 10x + 20y max point ( 20,50 ) Max fees = 10(20) + 20(50) = RM 1,200

10

N1

N1

N1

N1

K1

N1

N1 K1

N1

N1

10log y

x 0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8

1.0

1.2

1.4

1.1

1.5

1.3

1.6

1.7

1.8

1.9

2.0

X

X

X

X

X

Answer for question 8