This document contains the marking scheme for the Additional Mathematics trial SPM 2009 paper 1. It provides the full workings and marks for each question. The key points assessed include algebraic manipulation, logarithmic and trigonometric functions, vectors, and statistics such as variance. In total there are 22 questions on topics commonly found in Additional Mathematics exams.

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
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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

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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
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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
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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
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