Spb Addmaths Answer Spm 2009

SULIT 3472/1 Additional Mathematics Kertas 1 Peraturan Pemarkahan August 2009 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2009

PEPERIKSAAN PERCUBAAN SPM TAHUN 2009

ADDITIONAL MATHEMATICS KERTAS 1

PERATURAN PEMARKAHAN

UNTUK KEGUNAAN PEMERIKSA SAHAJA

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Question 1 (a)

1

Marks 1

1 (b)

3

1

2 (a)

Working / Solution

g 1

2(b)

2

4 , x  2 x2 4 4  , x  0 or x2 f ( x) x

f ( x) 

Total 2

4

B1

x  3

2

4

B1

4 2 x 3(a)

(b)

4

5

g ( x) 

6  5x 2

2

6  2x y 5 5 p= 2 6  5( x  2)  8  px 2 b = - 5 and c = - 2

4

B1 2 B1 3

b = - 5 or c = - 2

B2

5 2 ( x – 2) ( 3x + 1) = 0 OR x 2  x   0 3 3

B1

4 x5

3

2 2

( x  5)( x  4)  0 OR

4

5

x

B1 Must indicate the range correctly by shading or other method or 4

5

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Question 6

Working / Solution q  60 p  5 2

q  15(2 p ) 2  5

7

 6 p  3n or 5  q  15n 2  3

2

2( x  2)   4

8

9

5 2( x  2) or 1 x 1  x2 3  x  log   x 2

Marks 3

Total 3

B2 B1 3

3

B2 5 4 x OR 25 2 x

B1 3

3

B2 B1

2k  h  1 2h  k

4

2 log 5 3  log 5 2  log 5 5

4

B3

2 log 5 2  log 5 3 log 5 2 2  log 5 2  log 5 5 or log 5 2 2  log 5 3 or

B2

log12 3  log12 2  log12 5 2

log 5 90 or 2 log 5 3 or 2 log52 log 5 12 10

n = 42 5  (n  1)(2)  87 d=2

11

1 6

B1 3 B2 B1 3

1 9

B2

1 1 3 1 r 3

B1

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3

3

Question 12

13

Working / Solution p = 2 and q  1 p = 2 or q  1 3  (5) p or 5q  5 40 y  2 x  px 2  5q 3 9 y  x 4 2

P ( 0,8) or Q (-6,0) or m  PQ  

15

16

3 4

(10, 7)

 3  1  h )          2  4  7 i 2 j

3

B1 3

B2 B1 3

3

B2

3

3

~

B2 B1

11 i  5 j  4 i  3 j

17

3

B1

53 OC  53 ~ o

B2 B1

3

x  10 or y  7 x0 y 8  2 or 3 5 5 h=7 1 4  2 or 3 =  (1  h) 2

~

Total 4

B2

3 y  0   ( x  6) 4

14

Marks 4 B3

~

~

~

o

90 , 123.69 ,270o,303.69o

4

90o, 270o or 123.69o, 303.69o

B3

cos x(3 cos x  2 sin x)  0

B2

3 cos 2x + 2 sin x cosx = 0

B1

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4

Question 18 (a)

Working / Solution

  1.842 5  6.5

(b)

23.025 1 2 (5) (1.842) * (candidate’s  from a) 2

19

60

20

21

x 3 2(5  3 x)1 (3)  (5  3 x) 2 3 x 2

B2

2(5  3 x)(3) or 3 x 2 10 2  2     3  2   4 or  3  2   4 x  2    dp 2  3 2 dr x h=3

B1

B2

h h – =7 3 [2(3)  5] [2(2)  5]3

B2

23

  h ( with the correct l imit ) or x 32  3  (2 x  5)  2 a) m=5

3

4

3

3

B1 3

3

B1 2

2m  3  8  m  1 7 3

B1

b) 21 1 or an equivalent single fraction 15

1 3

3 2 2   6 5 6

B2

3 2 2 or or 6 5 6

B1

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Total

2 B1

3

3

22

Marks 2 B1

3

3

Question

Working / Solution

24(a)

14 or 2.8 5

24(b)

1.296

Marks 1

Total 3

2

2  2 7   1   or equivalent 5  5

B1

25 (a)

1.1 0.1357

2 B1

25(b)

61.2

2

70    *1.1 (candidate’s k) 8

“END OF MARKING SCHEME”

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B1

4

3472/2 Matematik Tambahan Kertas 2 2 ½ jam Ogos 2009

SEKOLAH BERASRAMA PENUH BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN KLUSTER KEMENTERIAN PELAJARAN MALAYSIA PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2009

ADDITIONAL MATHEMATICS Kertas 2 Dua jam tiga puluh minit

MARKING SCHEME

Skema Pemarkahan ini mengandungi 13 halaman bercetak

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2 QUESTION NO.

1

SOLUTION

MARKS

y = 3 - 2x

P1

x (3 - 2x ) - 2(3 - 2x ) + 5 = 0

K1

2

2x - 7x + 1 = 0

- (- 7) ±

5

(- 7)2 - 4(2)(1)

K1

2(2)

x = 3.351 or 0.149

N1

y = - 3.702 or 2.702

N1

5 2 a)

b)

c)

7 ] 2 1 1 7 = 2[x 2 - 2x + ( (- 2))2 - ( (- 2)2 + ] 2 2 2 5 = 2[(x - 1)2 + ] 2 2 = 2(x - 1) + 5

f (x ) = 2[x 2 - 2x +

Minimum value = 5

K1

2 N1

N1

1

23 Shape – P1 Max point – P1 Other 2 points – P1

13 7

-

3

(1,5) -2 0 d)

f (x ) = - 2(x - 1)2 - 5

3

1

N1

7

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3

3 a)

36 , x , 20.25 x 20.25  36 x 3 x 4

(b)

(c)

3 T10  36  4  2.703

2

K1 N1 9

K1

2

N1

1  (0.75) n  36   140  1  0.75  n log 0.75  log 0.02778 n  12.46 n  13

K1

3

N1 N1

7 4 (a)

2



sin x cos x  cos x sin x 2 sin x cos x  sin 2 x  cos 2 x  sin 2 x

(b)

K1 2 N1

y 2 0.25

 2

2

x

Sine curve……………P1 1 period………………P1 Max/min value 2/-2…………P1 Sketcht straight line.….K1

x …………….N1 4 No. of solutions = 3…….N1

6

y

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8

4

5(a)

14.5 or 8 or 9 33+m

P1

1   4 (33  m)  9   5  15.125 8     m7

b)

K1 3 N1

Refer the graph paper Uniform scale, correct frequency and upper boundary Method to find the mode Mode = 22

3

K1 K1 N1

6 6 (a)

 3      (i ) AT  ( AO  OB ) or MT  MA  AT 4  3 AT  (12 a  4 b) ~ ~ 4   9 a  3b ~

(ii )

3 N1

~

  MW  hMT   MO  OW  h( a  3 b) ~

~

 1  (12 a )  kOB  h( a  3 b) ~ ~ ~ 3  4 a  k (4 b)  h a  3h b ~

7 (a)

N1

~

 2 MT  (12 a )  3 b  9 a ~ ~ 3 ~   a 3b ~

(b)

K1

 4  h h4 dy  2x dx 2x = 4 x =2

~

~

or

K1

3

~

4k  12 , k3

K1 N1

K1 N1 www.tutormuruli.blogspot.com

7

5 2

A1   x 2  3 dx

(b)

2

0

2

 x3  =   3x  3 0 8  =   6  0 3  26 2 = or 8 or 8.667 3 3

1 A2   2  (7  9) 2 = 16 A  16 

= 2

(c)

K1 K1

K1 5

26 3

K1

22 1 or 7 or 7.33 3 3

N1

V     x 2  3 dx 2

0

2

=   x 4  6 x 2  9 dx 0

3

2

 x 5 6 x3  =    9x 3 5 0

K1

 25  =    2(2)3  9(2)   0 5  2 202 = 40.4 or 40  or  5 5

K1 N1 10

8 (a)

log10 (x +1)

0.30

0.48

0.60

0.70

0.78

0.85

log10 y

0.70

0.81

0.89

0.95

1

1.04

N1 N1 2

(b)

refer to the graph paper

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6 (c)

log10 y = k log10 (x + 1) + log10 h (i) log10 h = 0.515 h = 3.27

P1 K1 N1 5

(ii) k =

1.04  0.7 0.85  0.3

K1 N1

= 0.6

10 9 (a)

(b) (i)

 10  EAD  tan 1    12   0.6948 rad

N1

COD  1.3896 rad

arc length  6(1.3896)  8.3376

(ii)

2

K1

K1 N1

AC 2  6 2  6 2  2(6)(6) cos 100.39 o

5

K1

 9.2187 Perimeter of shaded region  10  8.3376  2.4432  20.7808

(c)

1 2 (6) (1.7524) 2

or

1 (6)(6) sin 100.39 0 2

Area of segment ABC 

K1 N1

3

K1

1 2 1 (6) (1.7524)  (6)(6) sin 100.39 0 2 2

K1

= 31.5432 – 17.7049 = 13.8383

N1

10 www.tutormuruli.blogspot.com

7 10

(a)(i) R(0, -4) (ii) x 

P1 2(0)  5(3) 2( 4)  3(11) and y  23 23

K1

Q(3, 5) (b) 14 

N1

1k 5 3 k 2 0 11 5 0

14 

1 11k  25  5k  33 2

6k – 8 = 28

K1

or 8 - 6k = -28

(d)

y

PS 

N1

2 x y x  4 or  1 3 6 4

 x  6

 x  6

2

2

N1

 y 2 or PQ 

 y2  2

 x  3

2

 x  3   y  5  2

 ( y  5)2

2

3 x 2  3 y 2  12 x  40 y  100  0

11

3

K1

k=6

(c)

3

K1 K1

4

N1

10

(a) (i) p = 0.8 , q = 0.2 P(X = 0) = 6C0 (0.8)0 (0.2) 6 or P  X  1  6C1 (0.8)1 (0.2)5 K1

P( X  2)  1   P( X  0)  P( X  1) 

= 1 - 6C0 (0.8)0 (0.2) 6 - 6C1 (0.8)1 (0.2)5 = 1 – 0.000064 – 0.001536 = 0.9984

K1 N1

(ii) 14  n(0.8)(0.2) n = 1225

K1 N1

(b)(i) P ( X  45)  0.2266 Z = -0.75

P1

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5

8 0.75 

  54

(ii)

12(a) (b)

Z

45   12

Z

or

42  54 ( in b(ii)) 12

N1

42  54 12

P  42  X  45   P(1  Z  0.75) = 0.2266 – 0.1587 K1 = 0.0679 N1 v=8 N1

5

10 1

K1

a = 2 – 2t = 0 2t = 2 t=1

3

K1

v = 8 + 2(1) – (1)2

N1

= 9 (c)

K1

v = 8 + 2t – t2 = 0 t2 – 2t – 8 = 0 K1

(t – 4) (t + 2) = 0

N1

t=4 (d)

2

s   (8  2t  t 2 ) dt s  8t  t 2 

t3 c 3

K1

t=0,s=0c=0 s  8t  t 2 

t3 3

t = 4 , s = 8( 4)  (4) 2 

43 80  3 3

or

K1

3

6 t = 6 , s = 8(6)  (6)   12 3

K1

80  80     12  3  3 

N1

2

Total distance = =

124 3

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4

10

9 13

(a)

P08  100  150 32

(i)

K1

P08 = RM 48

(ii)

5

K1

P05 100  110 P03

or

K1

P08 100  130 P05

I 08  03



P08 P05   100 P05 P03 130 110   100 100 100

K1

N1

= 143

(b) (i)

(ii)

115(40) + 150 (20) + 30x + 130 (10)

P1

115(40)  150(20  30 x  130(10)  122 100

K1

x = 110

N1

305 100 P05

K1

P05 = RM 250.00

N1

122 

5

10 14

(a)

cos BAC 

15 2  24 2  18 2 2  15  24

K1

2 = 0.6625  BAC = 48.51o

(b)

 AED = 180o – 48.51o – 60o www.tutormuruli.blogspot.com

N1

K1

10 = 71.49o

DE o

sin 48.51



N1 8 sin 71.49 o

K1

N1

DE = 6.319

(c)

area of ABC =

4

1  15  24 sin 48.51o 2

K1 N1

= 134.83

4 1  24  h  134.83 2

K1

h = 11.24

N1

10 15

. (a)

I : x + y  150 II : y 

1 x 2

3

N1

III : y – x  80

N1

(b) refer the graph paper

3 N1

(c) (i) x = 100 (ii) maximum point ( 35, 115)

N1

Profit = 3(35) + 5(115)

K1

= RM 680

N1

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4

10

11

Q5

frequency

16

14

12

10

8

6

4

2

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0

4.5

9.5

14.5

19.5

24.5

29.5

34.5 34. 34.

UpperBoundary

12 Q8 1.1 x x

1.0 x 0.9

x

x

0.8

x

0.7

correct axes and uniform scale

K1

all points plotted correctly

N1

line of best fit

N1

0.6

0.5

0.4

0.3

0.2

0.1

0.1

0.2

0.3

0.4

0.5

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0.6

0.7

0.8

0.9

13

Q 15

160

correct axes with uniform scale and one line correct( equation involved x and y) .

K1

all straight lines correct

N1

correct shaded region

N1

140

120 

100

80

60

R

40

20

20

40

60

80

100

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120

140

160