1. PEPERIKSAAN BERSAMA
SEKOLAH-SEKOLAH MENENGAH KEBANGSAAN KELOMPOK M6
PPD UTARA SEBERANG PERAI PULAU PINANG
___________________________________________________
PEPERIKSAAN PERCUBAAN BERSAMA
SPM 2009
MATEMATIK TAMBAHAN
Kertas 1
SKEMA
PEMARKAHAN
________________________________________________________________
Kertas soalan ini mengandungi 6 halaman bercetak
www.cikgurohaiza.com

2. No Marking Scheme Marks
1 (a) 9 1
1
(b) {5,8,11}
2 5( 5x – 3 ) - 3 1
= 25x −18 1
3 − 7 5x 1
f 1( ) =
x
f −1 (2) = − 1
7 5(2)
2
f −1 ( ) = 3
2 1
4 a) 4x -11x-3=0
2
(4x+1)(x-3 ) =0 1
X=-1/4 or x=3 1
b) b -4ac =0
2
p -4(2)(q)=0
2
1
p -8q=0
2
q = p /8 2
1
5 f(x) = x -6x+7
2
X -6x+(-6/2) -(-6/2) +7
2 2 2 1
(x-3) -2 2 1
Compare with f(x) = (x-h) +k 2
h=3 and k=-2 1
6 2x > 5x+3
2
2x -5x-3>0
2
(2x+1)(x-3)>0 1
1
x>3 or x<-1/2 1

3. www.cikgurohaiza.com

4. 7 log (x − x+
2
3) + 2 = log ( 6) 2
log (x − =
2
3) + 2log 2 log (x + 6)
2 2
1
log (x − 3) + log 2 = log ( x + 6) 2 1
2 2 2
log 4(x − 3) = log (x + 6)
2 2
1
4(x − 3) = x + 6
x− =x+6
4 12
x= 1
3 18
8 x=6
x − x+3 = 0.
216 36
x=x+3 1
216 36
6 x=6 ( x+3)
3 2 1
3x = 2( x + 3)
1
x+
3x = 2 6
9 x=6 1
d = 5x+y – ( x + y ) 1
= 4x
10 a = 90 , d = -3 1
90 + ( n – 1 )( -3 ) = 36
1
n = 17
1
11 a
= 27 1
1− 1
3
a = 18
12 a ) log y = log a + 3 log x
10 10 10
b) i. log a = 2 10
log y = 2 + 3log x
10 10
+ h
ii. 6 = 2 3
4

5. 1
1
1
h= 3 1
13 a. ) y x = 3 − x 1
1
b. ) gradient = -1
1
y-intercept = 3
www.cikgurohaiza.com

6. 14 PA=2PB
( )( 4) 2
=2 ( )( ) 2
1, 1
x−3 + y− x + 2 + y −1
x + 3y + 2 = 1
3 2 22x − 5 0
15 (a) PQ = PO + OQ
= (3i + -4j) + (6i + -7j)
9i − 11j 1
(b) PQ = 9 +11 =
2 2
202 1
Unit vector = 1(9i − 11j) 1
202
16 (a) QS = + 1
QP PS
= -4 a + 6b 1
(b) PA = +
PQ QA
1
2
= 4 a + 3 QS
= 4 a +2( -4 a + 6b )
3
4 1
= a + 4b
3
17 tan A − 2 sec A =
2
2
(sec A – 1) – 2 sec A – 2 = 0
2
1
Sec A – 2 sec A – 3 = 0
2
1
(sec A + 1) ( sec A – 3) = 0
Sec A + 1 = 0 , sec A – 3 = 0
1 co sA

7. cos A
=− 1 1=3 1 1
cos A = - 1 cos A = 3
1
A = 180 ,
o
A = 70.53 , 289.47
o o
www.cikgurohaiza.com

8. 18 (a)
1
(2 j )θ = 1(j )(π
2 2
)
2 2 2 1
θ = 0.3927
(b) s = rθ
= 9 = 3.534 cm
1
π
8
Perimeter of sector POQ = 9 + 9 +3.534 = 21.534 cm 1
19 When x=3,2kx-8 =4 1
6k=12
K=2 1
20 y = 4x -7x +1
3 2
dy = 12x – 14x 2
dx
When x=2, dy = 12(2)2-14(2)
dx
=20 1
δy =2.1-2.0 =0.1
δ dy δ
Using y= × x
dx
0.1= 20 x δx
δx = 0.005
23
21
- 2m + [7(3) − 7(4)] = 17
-2m – 7 = 17
m = -12
22
1020 −

9. 1
(11) 2
1 1
Variance, σ = 2
6 1,1
= 170 –
121 1,1 1
= 49 1
(a) P = 120
5
5 1
(b) P – (2 x P ) =
5
5
4
4
120 – 48
1,1
= 72
1
www.cikgurohaiza.com