Addmaths Johor Spm2008 P2 Answer

1449/2 Matematics Kertas 2 Sept. 2008

.JABATAN PELAJARAN NEGERI JOHOR

PEPERIKSAAN PERCUBAAN SPM TAIFL. 2008

SKEMA PEMARKAHAN

Pv]

MATEMATICS

Kertas 2 Duajam tiga puluh minit

144912

0 Hak Cipta .lPNJ 2008

SULIT

SKEMA PERMARKAHAN MATEMATIK TAMBAIUN KERTAS 2 PEPERIKSAAN PERCURAAN SPM TAHUN 2008

x=y-2

Sub Marks P1

y 2+ 2 y + ( y - 2 y = 9

K1

Solution and mark scheme

Number

I

Full Marks

2y2 - 2 y - 5 = 0 Y=

-(2)

+ J-40(-5)

N1 N1

x =0.158,-3.158 =

J(2x3 - dx)dr

KI

1

Kl

=-x4 -2x2 2 v = - 1x - 2 (b)

4

-2x

2

+C

+-7

M ,= -

N1

P1 P1

2 2y=x+3

N1

x-= -64 -=$

NI

8 ,

G x 2 = 580

/?

5

2

m, = 2(113-4(1) = -2 1

3(a)

KI

2(2)

6

P1

o = -= 2.9155

(b)

- ---2.9 1 55 on, -

4 = 0.7289

N1 K1 N1

5

6(a)

tan28 + 1 tan B sec2B tan B sec B cos ec8

N1

N1 for Sin shape

N1 for 1 periodic and max = 2 N1 for modulus N1 for

K1 for straight line Number of solutions = 3

Number

Solution and mark scheme

7

Sub Marks

Full Marks

NI NI logy = log a + x2 log h log a = 0.19(+0.0 1) a = 1.55 0.889 - 0.493 log b = - = 0.0792 9-4

P1 Kl N1 KI NI

Graph: K I for uniform scale on both axes N 1 for a1l correct points N I for best fit line 8(a>(i>

JG=Io m=8

(ii)

10 KI NI

K1 N1

(b) K1 L

= --(10!+l

3

l-j )

N1

K1 N1

(c>(i>

(i i)

I

K1 0,P and T are collinear.

N1

10

Number 9ja)

Solution and mark scheme sin LBEO = 0.5 :. LBEO = 30° = 0.5237rads

Sub Marks K1 N1

Perimeter of shaded region =AB+BD+AD = 5j1.572) + 1 O(0.5237) + 3.6602 =16.7522~~

Area of the shaded region = ABEO + sector AOB - sector BED

K1 for any correct formula used N1 for any correct answer

K 1 for all correct operation

Full Marks

1

-

-

Solution and mark scheme

--

-7Full Marks

--

Sub Marks

1 1 (a)

K1

N1

K1 for

I K1 for

ncrprqn-r

operation N1

K1 for x-P

z=-

LT

Nl

K1 for x-P

z=-LT

KI

NI

. . . p p -

Number

Solution and mark scheme

--Full Marks Sub Marks K1

v,,,, = 3(2)' - 12(2) + 9 = -3 m/s

K1 for formula I

21 100 = 140 15

I,, =--x

-

Solution and mark scheme

Number -

sin LQRP --sin 30'

9.5 5.8 LQRP = 125'1' LPRS = 180'-125'11= 54'59'

Sub -Marks K1

NI NI Kl NI

1 2

- (6.322)(1 1.2)sin LSPT

= 25

SPT = 44'55' LQPR = 24'59' Area APQS 1 1 = -(9.5)(5.8) sin 24'59'+-(5.9)(7.5)sin 54O59' 2 2 = 11.6359+18.12 = 29.7559cm2

KI N1

P1 K1

N1

Full Marks

Solution and mark scheme

Sub Marks

Full Marks

K1 N1

K1 Kl N1

I :100x+400y28000 II : 1 OOx + 200y I 9000

10

N1 Nl N1

III : x < y refer to the graph: K 1 for uniform scale on both axes K I for any one correct straight line drawn N 1 for correct region R

K1 K1 N1

(i) 16

PI

(ii) 1 2 x 4 9 =~ 180 ,(30,30) Maximum profit = 30(RM12) + 30(RM9) = RM630

N1 K1 for k=ax+by N1

10

~