TRIAL ADDMATE SPM 2011 Selangor Paper 2 Answer

8/4/2019 TRIAL ADDMATE SPM 2011 Selangor Paper 2 Answer

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1

Program Peningkatan Prestasi Sains & Matematik 2011 (Trial)

Additional Mathematics Marking Scheme - Paper 2

Solution Marks Solution Marks

1

3

2

yx or 23 xy

4)23()23(222 xxxx

or 43

22

3

2 22

y

yy

y

021182 xx

or 0168 2 yy

)8(2

)2)(8(4)11()11(2

x

or)8(2

)16)(8(4)1()1(2

y

x = 1.159, 0.216,

y = 1.478, -1.353

P1

K1

K1

N1

N1

5

2a)

45)(2 xxxf

b)(i) xxxf 5)(2

4

= ]44

25)

2

5[(

2 x

=4

9)

2

5(

2 x

Minimum point )4

9,

2

5(

(ii) 0)4)(1(4)5( 2 p

4

9p

K1

N1

K1

N1

K1

N1

6

3 a) qxpxdx

dy

23 ,

xxqxpx 6623 22

3p = 6 @ 2q = -6

p = 2 ,

q = 3

r 23 )2(3)2(29

r= 5

b) 0)1(6 xx

x = 0 , x = 1

turning point (0,5)

turning point (1,4)

P1

K1

N1

N1

N1

K1

N1

N1

8

4 a)

Shape of graph (sin )

Max = 1 and Min = -1

2

11 cycles

Reflection about x -axis

b) 4 sin2

3x + 3x = 0

- sin2

3x =

4

3x

y =4

3x

Number of solution = 3

K1

(line)

P1

P1

P1

P1

N1

N17

2 x

1

2

y

-1

1.5


2/6

2

5 a) OLBOBL or ABOAOM or equivalent

yxBL 84

)812(2

112 yxxOM

=yx 46

b) )46( yxhON

)48(4 xykxON

)48(4)46( xykxyxh

4h = 8k, h = 2k

6h = 44k

6(2k) = 44k

2

1,

4

1 hk

K1

N1

N1

K1

K1

K1

N1N1

8

6(a) =

6251A ,

4

6252A ,

8

6253A

Since , then it is a

Geometric Progression

r =

(b)

4

11

625

S

=3

2500

P1

K1

N1

N1

K1

N1

6

7a) Area of )2)(6(2

1 k

2

0

2)2( dyy =

2

0

3

23

y

y

= 0)2(23

23

=

3

20

3

38= 63

3

20 k

k312 k= 4

b) V = 2

0

22)2( dyy

= 2

0

24)44( dyyy

2

0

35

43

45

y

yy

= 0)2(43)2(4

5)2(

35

= 15

376unit

3

K1

K1

K1

K1(+)

K1(=)

N1

P1

K1

K1

N1

10

8x 1 2 4 6 8 10

log10 y 0.84 0.99 1.29 1.57 1.87 2.16

Paksi betul, skala seragam & plot 1 titik

Plot semua titik betul

Garis lurus penyuaian terbaik

y = h kx

log10 y = log10 h + x log10k

i) log10k = 0.146

k= 1.4

( accept 1.351.43)

log 10h = 0.70

h = 5.012

ii ) log 10y = 1.7 , x = 6.8

N 1

K1

N1

N1

P1

K1

N1

K1

N1

N1

10


3/6

3

9 a)(i)

k = 13

(ii)

or equivalent

(b)(i)

E(-2, -5)

(ii) Area of triangle ODE

=

=

= 13 unit2

(c) 22 )4()6( yx

22)3()4( yx

K1

N1

K1

N1

K1

N1

K1

N1

K1

N1

10

10

a) 90,45 AOCABC 9)571.1(

2

1 2 r

r = 6 cm

b)

)571.1(6ACarc or equivalent= 9.426 cm

Perimeter = 2(6) + 9.426

= 21.426 cm

c)

radororBOC 713.42

3135

22

sec 834.84)713.4)(6(

2

1cmA tor

135sin)6(2

1)713.4)(6(

2

1 22segmentA

= 72.104

Area of shaded region = 2(72.104)

= 144.208 cm2

P1

K1

N1

K1

K1

N1

P1

K1

K1

N1

10

11a)(i)P(X 2)= P(x = 0) + P(x =1) + p(x = 2)

=60

06

52

53C

+

51

16

5

2

5

3C

+

42

26

5

2

5

3C

= 0.1792

(ii) Min = np = 1205

2= 48 @

Sisihan piawai = 28Min = 48 & Sisihan piawai = 48

b)(i)P( X > 50 ) = P( Z >9

3050 )

= P( Z > 2.22)

= 0.01321

(ii) P( X < t ) = 0.70

P( Z 9

30t ) = 0.3

52409

30t

t = 34. 716

K1

K1

N1

K1

N1

K1

N1

P1

K1

N110

12(a)9

35sin

6

sin

ADB

48.22DBA or '2922

(b) 48.57ADC

AC= 10.43 cm

(c) 48.2235180BAD = 122.52

BD = 13.23 cm

(d) Area =

= 45.53 cm2

K1

N1

P1

K1

N1

K1

K1N1

K1

N1

10


4/6

4

13

a) 125100168

x

or 10090

99y

210RMx y = 110

c) m + n = 35

115

100

4011010512025125

nm

8m + 7n = 265

8(35n) + 7n = 265

m = 20

n = 15

d) 115100

120I

= 138

K1

N1

N1

P1

K1

K1

N1

N1

K1

N110

14a) 200 500 x 700 yx

2002 yx

b) Graf (satu garis betul)

(semua garis betul)(rantau betul)

c)(i) Kek lapis maksimum = 450

(ii) (500, 200)

k = 0.8x + 0.6y

= 0.8(500)+0.6(200)

= RM 520

N1

N1

N1

K1

N1

N1

N1

N1

K1

N1

10

15 a) 622 ttv

16,0

cmsvt

b) 14 tdt

dv= 0

4

1t

6

4

1

4

12

2

min

v

=1

8

49 cms

c) 062,02 ttv

0)2)(32( tt

2t

d) dttts )62(2

s = cttt

6

23

223

ttt

s 623

223

N1

K1

K1

N1

K1

N1

K1

15d) )2(62

)2(

3

)2(2,2

23

st

=3

26 or

)4(62

)4(

3

)4(2,4

3

st

= s3

32

Total distance travelled

=3

32

3

262

= 28 cm

K1

K1

N1

____

102

3 2


5/6

5

Graph for Question 14

100

200

300

400

y

500

600

700 8000

100

200

300

400

500

600

700

800

R

500,200

x


6/6

6

Question 8

2 4 6 x

0.2

0.4

0.6

0.8

1.0

1.2

2.0

1.8

1.4

1.6

log10 y Graph log10 y against x

8

x

x

x

x

x

x

0

2.2

10

C = 0.70

x = 6.8

TRIAL ADDMATE SPM 2011 Selangor Paper 2 Answer

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