Skema Add Math P2 Trial Spm Zon A Kuching 2008

3472/2 Matematik Tambahan Kertas 2 2 ½ jam Sept 2008

SEKOLAH MENENGAH ZON A KUCHING LEMBAGA PEPERIKSAAN PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2008

MATEMATIK TAMBAHAN Kertas 2 Dua jam tiga puluh minit

JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU

MARKING SCHEME

Skema Pemarkahan ini mengandungi 15 halaman bercetak

2

ADDITIONAL MATHEMATICS MARKING SCHEME TRIAL ZON A KUCHING 2007 – PAPER 2 QUESTION NO.

1

SOLUTION

p  1 m @ m  1 p

MARKS

5

P1

*  1  m   2m 2  *  1  m  m  8 2

@

Eliminate p or m

K1

p 2  2*  1  p   p  1  *  1  p    8

m

2

  1 

@ p

K1

Solve the quadratic equation by using quadratic formula @ completing the square

 1  4  2   7  2  2

  3 

2

 3  4  2   6  2  2 2

m  2.138, m  1.637 @ N1 p  1.137, p  2.637 OR p  1.137, p  2.637 N1 @ m  2.138, m  1.637 2 (a)

K1

Note : OW-1 if the working of solving quadratic equation is not shown.

5

dS = 8 r or dr

dS dS dr =  dt dr dt

K1 8 (3)  0.2 N1

4.8 @ 15.08

3

3 QUESTION NO.

(b)

SOLUTION

MARKS

P1

4

dy d2y  2 x  3 and 2 dx dx 2

2 + (2x  3) 2 + 14x  11 = x  3x + 2 K1 (3x  1)(x + 2) = 0

x=

3 (a)

1 , 2 3

x  2.9 or



N1 7 fx 2  291 291  (2.9) 2 30

  1 1358

(b) (i)

(ii)

K1

3

P1

K1

Use the formula

N1

Or equivalent

2(2.9)  c  2.8

K1

c=3

N1

21.1358* = 22716*

N1

3

6

4 QUESTION NO. 4 (a)

SOLUTION

cos 

3 or 4

MARKS

P1

= 4141 @ 4125

2

N1

07227 rad 7937 or 12(07227)

or

8672

3

K1

(b)

7937 + 12(07227) + 3 K1

19609

(c)

N1

1 1 (12) 2  0 7227  @  9  7 937 2 2 1 1 (12) 2  0 7227    9  7 937 2 2

K1

K1 N1

1631 5 (a)

3

K1

4a(2)9 = 10 240 a=5

8

6

N1

(b)

a = 1, r = 4(both correct) 410  1 4 1 

or

410  1 44  1  4 1 4 1

 349 440

K1

44  1 4 1

K1

K1 N1

6

5 QUESTION NO.

6 (a) (i) (ii)

SOLUTION

CQ =

(b) (i)

P1

BP = 3 y − 4 x CQ =

2 CB 3

or

3

K1

Equivalent

8 x − 4y 3

N1

 1  BR =   BP  m + 1

P1

(

 1  BR =   − 4x + 3y  m + 1

(

MARKS

)

5 P1

)

2 3  1    − 4 x + 3 y = (− x + y) m + 1 3 4   −4 2 =− m +1 3

m=5

@

3 1 = m +1 2

K1

K1 N1 8

N1

N1

6 QUESTION NO. 7

(a)

SOLUTION

MARKS

y 2.88 2.30 1. 92 1.64 1.44

10

xy 2.88 4.6 5.76 6.56 7.2

N1

All values of xy correct (accept correct to 2 decimal places)

xy = −dy + c

P1

Refer to the graph.

K1

Plot xy against y

(b)(i) (ii)

N1

(iii)

N1

6 points mark correctly

Line of best fit

10 m = −c

K1

7

p = 11⋅5 ± 0⋅2

QUESTION NO.

y = 2⋅16

K1

2⋅315

N1

N1

SOLUTION

d = 3 ± 0⋅2

N1

MARKS

8

8 (a)

sin x cos 2 x + cos x sin 2 x

1

K1

sin x 1 − 2 sin 2 x + cos x 2 sin x cos x

K1

1 2 sin x cos x K1 1 or sin 2 x

cosec 2x

N1

y

(b) (i) & (ii)

6

3

1 x 2

1

Shape of sine curve

P1

Amplitude of 2 and 1 period

P1

0 Translation    2

P1

2x 1 

N1

y

Draw the straight line correctly Number of solutions = 2

K1 N1

10

9

QUESTION NO.

9 (a) (i)

SOLUTION

p=

30 = 0.3 100

q = 0.7

K1

= 0.1585

(b)(i)

N1

P[X = 0] + P[X = 1] + P[X = 2] 12 1 11 12 2 10 = (0⋅7)12 + C1 (0.3) (0.7) + C2 (0.3) (0.7)

K1

= 0⋅2528

N1

42 − 40   P Z >  5  

P  X  m   0.15

5

K1

= 0.3446

(ii)

5

P1

∴ P[ X = 5] = 12C5 (0.3)5 (0.7)7

(ii)

MARKS

N1

K1

m  40  1.036 5 m  45.180

K1

N1

10

10

QUESTION NO.

10 (a)

SOLUTION

mBC = −

1 2

mAD × −

or

y – 7 = 2(x – 7)

or

1 = −1 2

or

MARKS

mAD = 2

7 = 2(7) + c

N1

1 y= − x+8 2

P1

1 2x – 7 = − x + 8 2

or

equivalent

x + 7(2) =6 1+ 2

N1

or

y + 7(2) =5 1+ 2

E(4, 1)

(d)

3 K1

D(6, 5)

(c)

3

K1

y = 2x − 7

(b)

K1

( x − 12) 2 + ( y − 2) 2 = 5

x² + y² – 24x – 4y + 123 = 0

K1

2

N1

K1

2

N1

10

11

QUESTION NO. 11 (a)

SOLUTION

MARKS

N1

x = 0, 5 125  3

6 K1

5





2 1  (5) 2 (5)    x 2  4 x dx 3 4

 x5 x3    2 x 4  16  3   5

K1

5

K1 4

  55   45   16 16    2(5) 4  (5)3    2(4) 4  (4)3     5  3 3   5    

34

(b) (i)

K1

2  15

= 2px – q

N1

or

5 = 2p – q @ equivalent

or

2=pq

4

K1 N1

p=3,q=1

Gradient of normal = − (ii)

5y + x = 11

or

1 5

equivalent

K1 K1

N1

10

12

QUESTION NO. 12 (a) (b)

SOLUTION

Vo  3

MARKS

P1

1

K1 Use v = 0

4t 2  8t  3  0

2

(2t  3)(2t  1)  0 t

(c)

3 1 , 2, 2

N1

3

4  [ t 3  4t 2  3t ] 12 3 2

Integrate

4 3 3 3 4 1 1 1  [ ( ) 3  4( ) 2  3( )]  [ ( )  4( ) 2  3( )] 3 2 2 2 3 2 2 2 2  m 3

(d)

a  8t  8

 15ms 1

v dt

K1

3

K1

N1

4

K1

t  3s V3  4(3)2  8(3)  3



N1 K1 N1

10 QUESTION

SOLUTION

MARKS

13 NO. 13 (a)

52  62  72 cos BCD  2(5)(6) ∠BCD = 78⋅46° @ 78°28′

N1

sin CAE sin 78 46o  8 10

(b)

*

3

K1

CAE = 5161 @ 5137

N1 N1

*AEC  49 93o AC  102  82  2(10)(8) cos 49 93o

(c)

AC = 7⋅8105

2

K1

N1

1 1 o Area of ∆BDE =   5  6  sin 78 46 3 2

(d)

2

K1

1

K1

K1

3

Use of the formula

1

= 4⋅8989

N1 (a)

2

ab sin C

10

x  y  60 1 y x 2

I. II.

4 x  3 y  360

III. (b)

3

Refer to the graph, 1 or 2 straight lines correct 3 st. lines correct Correct shaded area

Answer for question 14

y

100 30 40 20

10

N1

(c) (i) (10, 50)

N1 Max balance after purchase = RM[3 600  1900] N1

20

30

40

= RM 1 700

50

60

70

80

14

K1 N1 N1

90

N1

K1

K1

80

N1

40(10) + 30(50) @ 1900 10

70 60 50

(10, 50)

15

x

15 (a)

Use of formula I 

Q1  100 Q0

5

K1

p = 109.1 q = 200

N 2, 1, 0

r = 486

(b)

I

=

130  6  109.1 5  125  3  135  4 18

2240.5 18

K1

N1

= 124.5

(c)

Monthly expenditure for Year 2007 = 1986 x  100  128 1986 RM2542.08

K1

K1

K1

N1

10 Answer for question 7

xy

16 12

(a)

11

y

2.88

2.30

xy

2.88

4.6

1.9 2 5.76

1.64

1.44

6.56

7.2

10

9

8

×

7

× 6 × 5 × 4

3

×

2

1

0

0⋅4

0.88

1⋅2

1.6

2⋅0

2.4

2.8

3.2

y