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SULIT 3472/2 Additional Mathematics Paper 2 Sept / Okt 2010 PEJABAT PELAJARAN WILAYAH PERSEKUTUAN PUTRAJAYA KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5 2010 ADDITIONAL MATHEMATICS Paper 2 MARKING SCHEME This marking scheme consists of 11 printed pages

# SPM Trials 2010 - Add Maths P2 - Putrajaya (Scheme)

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PEJABAT PELAJARAN WILAYAH PERSEKUTUAN PUTRAJAYA KEMENTERIAN PELAJARAN MALAYSIA

PEPERIKSAAN PERCUBAAN SPM TINGKATAN 5

2010

Paper 2

MARKING SCHEME

This marking scheme consists of 11 printed pages

SKEMA PERMARKAHAN ADDITIONAL MATHEMATICS KERTAS 2 PEPERIKSAAN PERCUBAAN SPM 2010

Number Solution and marking scheme Sub Marks

Full Marks

1

@

@

@

and and

and and

P1

K1

K1N1

N1

5

2 (a)

( b)

y=-5(x2 + )

Turning point = (

= 2 – 3k

k=

K1

N1

K1

N1

2

Number Solution and marking scheme Sub Marks

Full Marks

6

(c)

N1 shape

All correct

N1

3 (a)

(b)

x, x + y , x+2y, a = x , d = y given T = 200 x + 5y = 200 ……………………….(1)

and S = 2520

……………(2)

Solve (1) and (2) y = 20 , x = 100

T of Encik Ali = T of Encik Tan 100 + (n-1)20 = 200 + (n – 1)10 100 + 20n – 20 = 200 + 10n – 10 10n = 110 n = 11.

K1 ( either (1) or (2)

K1N1N1

K1K1

N1 7

4 (a)

K1

N1

3

(

3

-54

-9

-20

Number Solution and marking scheme Sub Marks

Full Marks

( b )

( c )

Sketch straight line

*to get N1 for no. of solution all marks for graph must 5 marks

P1

P1

P1

N1

K1

N1 8

5 (a)

(b )

The mean =

= 40.25 kg

(20) = 5th observation

First quartile’s class = 35 – 39

= 36.5

(20) = 15th observation

Third quartile’s class = 40 – 44

Third

= 43.875 The interquartile range = 43.875 – 36.5 = 7.375 kg

K1N1

P1 (34.5 or 39.5 )

K1

K1

K1N1

7

6 ( a) Gunakan

K1N1

7

(b) (i)

(ii)

Gunakan

K1N1

K1

K1

4

Shape of sin graph

No of solution = 2

Amplitude (max=4, min=0 )

1 periodic /cycle

Number Solution and marking scheme Sub Marks

Full Marks

N1

7 (a)

b (i)

( ii )

<4)

m = 6

equation of the curve,

K1

K1

K1

K1N1

K1

N1

K1

K1

N110

5

Number Solution and marking scheme Sub Marks

Full Marks

8 a)

b) Lihat graf

c) i)

ii)

iii)

x 15 20 25 30 35 40lg y -0.82 -0.42 -0.022 0.37 0.77 1.17

Using the correct axes and uniform scale6 points plotted correctly *Line of best fit

log10 y = x log10 c –k or –k log10 10y-intercept, c = -k or c = –k log10 10

- 2.00 = - k

k = 2.00

m = log10 c c = 1.2

x = 37.5

N1

N1N1N1

P1

K1N1

K1N1

N1 10

9 (a)

( b)

(c )

( d )

mPQ=

mAC=mPC=

or equivalent

m for perpendicular bisector of AC =

(x - 4)

5y = 3x - 2

KQ = 6

= 6

x2 + y2 - 16x - 8y + 44 = 0

K1

K1

K1

N1

K1

K1N1

K1K1N1

10

10. (a)

(b)Area of

N1N1

K1

K1

N1

6

Number Solution and marking scheme Sub Marks

Full Marks

10

( c ) K1

N1

K1 ( either one )

K1 N1

11 a) (i)

( ii )

P(X>3) = P(X=4) + P(X=5) + P(X=6)

=0.98415

P1K1

N1

K1N1

10

(b) i)

ii)

P(X>54) → P(Z>

= 0.2266

P(X<m) = 0.08

Z =

P(Z< ) = 0.08 From standard normal table,

P(Z<-1.406) = 0.08

Therefore, = -1.406

m = 28.13

K1

N1

K1

K1

N112 (a)

( b )

The initial velocity = 0 ms-1

K1N1

Limit K1

K1K1

7

0.08

z O

Number Solution and marking scheme Sub Marks

Full Marks

(c )

=

=

= 14.67 m Distance travelled during the third second = 14.67 m

When v < 0,

The object moves to the left when t > 4

K1

N1

K1

K1

N1

10

13(a) i)

(ii)

= 116.670 BD = 10.80 cm

cos BCD =

= -0.9221 BCD = 157º14’

K1

N1

K1

N1

10

8

Number Solution and marking scheme Sub Marks

Full Marks

( b )

i)

ii)

C

D’ B

BDC = 14º31’ DBC = 180º - 157º14’ - 14º31’ = 8º15’ BD’C = 180º - 14º31’ = 165º29’ BCD’= 180º - 165º29’ - 8º15’ = 6º16’

The area of ΔBCD’

= 1.528 cm2

N1

K1

K1

K1

K1

N1

1014. (a)

(b)Lihat graf

(c ) i)

ii)

Draw correctly at least one straight line

from the *inequalities which involves and

Draw correctly all the three *straight lines Note : Accept dotted lines

N1N1N1

K1

K1N1

N1

K1N1

9

Number Solution and marking scheme Sub Marks

Full Marks

Use for point in the shaded region

Maximum point (33, 67)

RM 9 320

N1

15 (a)

(b )

(c ) (i)

( ii)

Use or b/20 x 100 = 125

,

b = 130

: 143

; RM28.60

K1

N1 N1

P1K1N1

K1 N1

K1 N1

10

10

11

100

90

80

70

60

50

40

30

20

10

R(33, 67)

75

50

NO 14

R

y = 30

x + y = 100

y = 2x

y

12

5 10 15 20 25 30

x

-2.0

-1.5

-1.0

-0.5

0

0.5

1.0

1.5

y10log

x

x

x

x

x

x

No.8(b)

35 40

-2.5

10 20 30 40 50 60 70 80 90 100

x

13