selangor spm 2010 trial add maths p1 maths/selangor/2010... · 6 13 (a) 100 120 1500 u m m = 1800...

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SULIT

3472/1 © 2010 Hak Cipta Jabatan Pelajaran Selangor SULIT

PROGRAM PENINGKATAN PRESTASI AKADEMIK 3472/1 SIJIL PELAJARAN MALAYSIA 2010 Matematik Tambahan Kertas 1

SKEMA JAWAPAN PEPERIKSAAN PERCUBAAN

SIJIL PELAJARAN MALAYSIA 2010

MATEMATIK TAMBAHAN

Kertas 1

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

Peraturan Pemarkahan ini mengandungi 4 halaman bercetak


SULIT 2 3472/1


1 (a)

23

t

(b) 0 ≤ f(x)≤13

1

2

6

p = – 2 and q = – 5

p = – 2 or q = – 5

(x + 2)2 – 5

3

B2 B1

3

2

2

(a)

211)4( f

2324

x

(b) 1k

21

23)2(2

k

B1

2

B1

7

2x

913 x

23 3333 xx

or equivalent

B2

B1

3

(a) 16x –5

4(4x – 1) – 1 (b) a =8 and b = 5 a =8 or b = 5

1

2 B1

4

2x2–11x– 6=0 (x– 6)(x +

21 ) = 0 or equivalent

2

B1

8 9

x = 243

5log 3 x

5log 22 3 x

23

y = 9 or G.P : 8, 12, 18 y2 – 8y – 9 = 0 or

32

13

yy

yy

3

B2

B1

3

B2

B1

5

2 < k < 6 (k – 2)(k – 6) < 0 k2–4(1)(2k – 3)< 0

3

B2

B1

10

n = 5

729)3(9 1 n

2

B1


SULIT 3 3472/1


11

–1

[ 8] – [9] S4 = [24-16] or equivalent OR S3 = [18 – 9] or equivalent

3

B2

B1

15

(a) 1 : 2

37

nmnm

(b) 4

2

B1

1

12

(a) 1

1cos2

p

(b) 212

pp

11

12

22 ppp or

equivalent

1 2 B1

16

8

bamba 9)2()64( ,

2

946

m

or equivalent

2

B1

13

h = 31 and k = –2

h = 31 or k = –2

231 kor

h

gradient = 2 or

h

kxy

11

4

B3

B2

B1

17

(a) ba 68 (b) ba 34

abbKH 86216

or equivalent

1

2

B1

14

(a) 3 (b) 05422 yyx 3)( 22 yx

1

2

B1

18

(a) 2. 5 radian

45.92)6.8(21 2

(b) 21.5cm 5.26.8

2 B1 2 B1


SULIT 4 3472/1


19

83

a and b = 3

8a + 2b = 3 or 8a + b = 0 4ax + b = 0

3 B2 B1

23

(a) 84 (b) 30 2

44

5 CC 2

44

5 CorC

1 3 B2 B1

20 (a)

23

p

2

12

p

(b) 4

2)(1

xpdxdy

2 B1 2 B1

24

(a) 185

(b) 1811

65

32 +

61

31

65

32 or

61

31

1 3 B2 B1

21

(a) – 4 (b) k = 2 0)2)(4( kk or

kk

2)2(

24

21 22

kxor

2

2

24

21

1 3 B2 B1

25

(a) k = 1.03 8485.01)( kzP (b) 198

1515.030

n

2 B1 2 B1

22

(a) m = 14 (b) std deviation = 4.598

2127

1156

1

2

B1


1

PROGRAM PENINGKATAN PRESTASI AKADEMIK 3472/2

SIJIL PELAJARAN MALAYSIA 2010 Matematik Tambahan Kertas 2

SKEMA JAWAPAN

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2010

MATEMATIK TAMBAHAN

Kertas 2

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

3472/2 ©2010 Hak Cipta Jabatan Pelajaran Selangor SULIT


2

Solution Marks Solution Marks

1 yx 34 or 3

4

xy

1)134( yy

or

1)1(3

4

x

x

0153 2 yy or

0732 xx

)3(2

)1)(3(4255 y

542.1,542.4 x

847.1,181.0 y

P1

K1

K1

N1

N1

___

5

2(a)

2

5

2

32 2 xxy

=

2

5

16

9)

4

3(2 2x

=8

49)

4

3(2 2 x

4

3

2

a or 1

88

49

b

2

3a

b = –41

(b) 02)3(2 2 xkx

0)2)(2(4)3( 2 k

0762 kk

0)7)(1( kk

7,1k

K1

K1

N1

N1

K1

K1

N1

___

7

3(a) y2 , y

2 + 2y, y

2 + 4y

yTTyTT 2,2 2312

Since yTTTT 22312 ,then A.P

(b)

i) T7 = 25 + 6(10)

= 85

ii) ]10)1()25(2[2

700 nn

0)14)(10( nn

10n

K1

N1

K1

N1

K1

K1

N1

7

4 (a) LHS = xx 44 sincos

= )sin)(cossin(cos 2222 xxxx

= 1( x2cos )

= x2cos

(b)

Shape of graph cos x

2 cycles

max = 2 and min = –2 within

20 x

2

3xy

Number of solutions = 3

K1

N1

K1

P1

P1

P1

N1

N1 __

8

y

Draw line

2

x 2

0

2

|


3

5 (a) 3 + p + 8 + q + 3 = 24

p + q = 10

(b)(i) p = 9

58

65.443 Q

= 48.25

Modal class = 40 – 44

7(a)

2

1

3

01

x

x = 1

L( 1, 0)

(b) trapeziumA = 2)1)(31(2

1 unit

2

A1 = 1

0

31

0

2 23

2

y

ydyy

= 23

1

= 3

7unit

2

Area = 3

12

3

7 unit

2

(c) V =

qq

xx

dxx2

2

2

22

)2(

)42(2

22

9 2

qq

0542 qq

( q – 5)(q + 1) = 0

q = 5 only

K1

N1

P1

K1

N1

P1

___

6

K1

N1

K1

K1

K1 (limit)

N1

K1

K1

KI

N1

10

6(a) i) OQOT2

1

= y2

1

ii) PSOPOS

= )(3

1OQPOx

= yxx3

1

3

1

= yx3

1

3

2

(b) PR =h PT

PO +OR = h PO( +OT )

yhxhyxkx2

1)

3

1

3

2(

hk

3

23 , hk

2

1

3

1

kk

3

2

3

23

4

3k

2

1h

N1

K1

N1

K1

K1

N1

N1

___

7

8( a)

(b) Refer to the graph

Correct & consistant scale and plot at

least 1 point correctly All 6 points plotted correctly

Line of best fit

(c) i) log10 y = x

2 log 10 a – log 10 b

13.7

5.098.0log10

a

a = 1.1918

ii) – log10 b = 0.425

b = 0.3758

N1

N1

K1 N1

N1

P1

K1

N1

K1 N1

10

x2 1 2.25 4 5..29 6.25 7.29

log10y 0.5 0.6 0.72 0.83 0.89 0.98


4

9(a) radBAO 621.02

9.1142.3

(b) 10

58.35cosAB

AB = 8.133 cm

BC = 20 – 8.133 = 11.867

cmSBE 21.6)9.1142.3(5 or

cmSCD 42.12)621.0(20

Perimeter = 10+ 6.21+ 12.42 + 11.867

= 40.497 cm

(c) 22

sec 2.124621.0202

1cmA tor

or

22 525.15242.152

1cm

22 83.118.108sin)5(2

1cmAtriangle

Area = 124.2 – 15.525 – 11.83

= 96.845 cm2

P1

K1

P1

K1

K1

N1

K1

K1

K1

N1

____

10

10 (a) 2

3,1

2

23

k

k

(b)

)14,9(

24

)3(2,0

4

)3(3

C

yx

(c) Midpoint (3, 6)

4

3

34

1

m

)3(4

36

xy

4

33

4

3

xy

(d) 22 )2()3( yx

22 )2()3( yx =3

4

9x2 + 9y

2 +54x + 36y + 101 = 0

N1

K1

N1

P1

K1

K1

N1

K1

K1

N1

10


5

11(a)

900

)5

4(720

5

4,

5

1

720144

n

n

pq

q

(b) X ~ N ( 28 , 32 )

i) P( 25 ≤ X ≤ 30 )

6667.01

3

2830

3

2825

ZP

ZP

= 1 – (0.1587 + 0.2525)

= 0.5888 or 0.589

(ii) P(X > m) = 0.1

846.31

282.13

28

1.03

28

m

m

mZP

(accept m = 32)

K1

N1N1

K1

N1

K1

K1

N1

K1

N1

___

10

12 (a)(i)

106cos)10)(15(21015 222AC

AC = 20.1914

(ii)

106sin

1914.20

sin

15

ACB

57.45ACB

(b)i)

43.13457.45180' ABC

(ii)

)]57.45(2180[106' ABC

= 17.14º

14.17sin)10)(15(2

1'ABCArea

= 22.1031

K1

N1

K1

N1

P1

K1N1

P1

K1

N1

____

10

A

B

A

C’

A


6

13 (a) 1201001500

m

m = 1800

OR

80100900

n

n =1125

(b)

100

3080451501512010130

100

122500

= 122.5

(c)(i) 100

5.122140

2007

2011

I

= 171.5

(c)(ii) 5.1711450

100

P

P = RM 2486.75

K1

N1

N1

K1

N1

N1

K1

N1

K1

N1

10

14 (a) xy 3

50 xy

1000 yx

(b) Refer to graph

Draw at least one line correctly

Draw all the lines correctly

Correct region

(c) (i) RM 720

(ii) (25, 75)

Using x + y = k on any point

in the region

RM 100

N1

N1

N1

K1

N1

N1

N1

N1

K1

N1

10

15 (a) when v = 0,

(2t + 3)(t – 2) = 0

t = 2

a = 4 t – 1

= 4(2) – 1

= 7 m s – 2

(b) a = 4t – 1 = 0

1

2

ms8

16

6)4

1()

4

1(2

4

1

v

v

st

K1

K1

N1

K1

K1

N1

15 (c)

dttts )62( 2

ttt

623

2 23

S2 = 3

28122

3

16

S3 = 2

1418

2

918

d = m6

512)]

2

14(

3

28[

3

28

K1

K1

K1

N1

10


7

Graph for Question 15

100

200

300

400

0

500

600

700

800

900

1000

y

100 200 300 400 500 600 700 800

x

x

R

R


8

Question 8

1 2 3 4 5 6 7 x2

0.1

0.2

0.3

0.4

0.5

0.6

1.0

0.9

0.7

0.8

log10 y Graph log10 y against x2

8

x

x

x

x

x

x

0


selangor spm 2010 trial add maths p1 maths/selangor/2010... · 6 13 (a) 100 120 1500 u m m = 1800...

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