Jpwp Mark Scheme Spm 2008 Trial Add Maths P1

SULIT

3472/1(SP)

3472/1 Matematik Tambahan Kertas 1 Ogos 2008 2 jam

JABATAN PELAJARAN WILAYAH PERSEKUTUAN KUALA LUMPUR

PEPERIKSAAN PERCUBAAN SIJIL PELAJARAN MALAYSIA 2008

SKEMA PEMARKAHAN

ADDITIONAL MATHEMATICS Paper 1

3472/1(SP)

[Lihat sebelah SULIT

2

SULIT Question 1

Solution (a) 8,10 (b) 4

2

f

1

( x) 

1 3 x h h

1 2 k 6

2

1

1,1

3

1 1 1 3

(p - 1)x2 - 8x - 4 = 0 (-8)2 – 4(p – 1)(– 4) > 0 p > –3

1 1 1

x2 – 4x – 21  0 (x – 7)(x + 3)  0 or

5

Full Mark

1

h

4

Sub Mark 1

gf ( x )  g (3x  2) 1  (3 x  2) = 3x  2 3  3x 2 = , x 3x  2 3

3

3472/1(SP)

3

1 1

-3 x  -3 , 6

7

(a)

7 x 7

h = –2 k=6

1 1 1

1 (b) c =  (0  2) 2  6 2 c=4 log 10 5 2 x 1  log 10 8 x

1 1

(2x+1)log 10 5  x log 10 8

1

0.699 (2 x +1) = 0.9031x

1

x =

4

 0.699 0.4949

= 1.412 3472/1(SP)

3

1

3 SULIT

3

SULIT 8

log 3 x 

4 3 0 log 3 x

3472/1(SP)

1

log 3 x 2  3 log 3 x  4  0 Let y = log 3 x y2  3y  4  0 ( y  4)( y  1)  0 or (log3 x + 4) (log3 x - 1) = 0 y = - 4 or y = 1 log 3 x  4 or log 3 x  1 1 x , x  3 (Both) 81 9

1 4

1 1

2

10

1

a) ar = 108 or 1 r= 3 1 2 b ) a ( ) = 108 3 a = 972 12 r= or x 1

4

ar = 12

1 4 1 r=

12 4x  4 = x 1 12 2 4 x + 8 x – 140 = 0 ( x – 5 ) ( x + 7) = 0 x = 5, x = −7

4x  4 12

1

1 (Both)

6 6 [a  8]  39 OR [2a  5d ]  39 and a + 5d =8 2 2

11

1

1 1

4

1, 1 3 1

a=5 12

(a)

log4 y = (blog4 A) x + log4 A 1 or log4 A = 2 A = 16

(b)

blog4A =

1

2–1 . =  0 – (–2)

b = 

1 1

4 [Lihat sebelah

3472/1(SP)

SULIT

4

SULIT 13

1

 6    (2)  1  a a  12

14

(a)

(b)

1

1 6 4 k 6 2  8  4  14  8 1 = 24  56  8k  32  4k  84 2 1 =  36  4k 2 = 18  2k

(a)

18  2k  0

16

4

1

~

  3 (a) OP     4 (b) Magnitude of OP = 5

3472/1(SP)

1

1

uur uuur uuur TR  TQ  QR 3 uuur uuur  SQ  QR 5 3  (15 p  10 q )  10q ~ ~ 5 %  9 p  4q % %

 3   1  3   5 or   5  4    4   5

1

1

uuur uuur uuur QS  QP  PS =  15 p 10 q ~

(b)

2

Area =

k = 9

15

3472/1(SP)

1 1 1

4

1

1

1

3

SULIT

5

SULIT 17

3472/1(SP)

(a) 1  t 2

1

(b) tan(900  x )  cot x t = 1 t2

1

18

1

POR = 0.7854 rad PO = 42  42  32  5.6569 QR = 5.6569  4 = 1.6569

or

Perimeter = ( 0.7854  32 ) + 4 + 1.6569 = 10.1 or 10.0998

19

Gradient of tangent = -3 1 k  1  3 2 k  8

3

1

1 1

3

1 1 1

3

h( x)  3(2 x  1)3 20

h( x)  9(2 x  1) 4 (2) 18 h' (1)  [2(1)  1]4 h(1)  18

1 1 1

3

k

 4 x2   2  6 x   24  2 ( 2k2 + 6k ) – ( 8 + 12 ) = –24 k 2 +3k +2 = 0 ( k +1) k + 2 ) = 0 k = –1 , k = –2 (Both)

21

22

(a) New interquartile range =

1 1 1

1 8  2 4

(b) New standard deviation  25  =

1

1 4

5 or 1.25 4

4

1

1 1

3

[Lihat sebelah 3472/1(SP)

SULIT

6

SULIT 23

3472/1(SP)

(a)

2 1  7 3 2  21

1

(b) 2 2 1 5    7 3 3 7 9  21

24

1

1

n = 10, p = 0.3, q = 0.7, r = 2

1

P(X = 2) = 10C2 (0.3)2 (0.7)10  2

1

= 0.2335

1

3

3

25

1  0.6103  0.3897

1

From the table, a  0.28

1

2

END OF MARK SCHEME

3472/1(SP)

SULIT