Spm add math 2009 paper 1extra222

SPM ADD MATH 2010 Paper 1 1
SIJIL PELAJARAN
MALAYSIA 2009
ADDITIONAL MATHEMATICS
Paper 1
3472/1
2 hours

Diagram 1 shows the relation between set X and set Y in
the graph form.
1
State
(a) the objects of q,
(b) the codomain of the relation.
( 2 , q )
( 2 , s )
( 4 , p )
( 4 , r )
( 6 , q )
( 2 , q )
( 6 , q )
(a)the objects of q
2 , 6
(b) the codomain of the relation.
{ p , q , r , s }

Given the functions g:x  2x - 3and h:x  4x, find
(a) hg(x),
(b) the value of x if hg(x) = ½ g(x).
2
h g (x )
= h ( 2x - 3 )
= 4 ( 2x - 3 )
= 8x - 12
(a) (b) hg(x) = ½ g(x)
8x - 12 = ½ ( 2x - 3 )
16x - 24 = 2x -3
14x = 21
x = 3/2

3 Given the function g : x  3x -1, find
(a) g(2),
(b) the value of p when g-1
(p) = 11.
g : x  3x -1
(a) g(2)
= 3( 2 ) -1
(b) g-1
(p) = 11
= 5
3(11) -1 = p
p = 33-1
= 32

4 The quadratic equation x2
+ x = 2px - p2
, where p is a
constant, has two different roots.
Find the range of values of p.
x2
+ x = 2px - p2
x2
+ x -2px + p2
= 0
x2
+ x( 1 -2p) + p2
= 0
b2
- 4ac > 0
( 1-2p )2
- 4(1)(p2
) > 0
1 – 4p + 4p2
- 4p2
> 0
1 – 4p > 0
4p < 1
p < ¼

5 Diagram shows the graph of a quadratic function
f(x) = - (x + p)2
+ q, where p and q are constants.
State
(a)the value of p,
(b) the equation of the axis of symmetry.
f(x) = - (x + p)2
+ q
x = -3
x + 3 = 0
x + 3 = x + p
p = 3
(a)
(b) x = -3

6 The quadratic function f(x) = -x2
+ 4x + a2
, where a is a
constant, has maximum value 8.
Find the values of a.
f(x) = -x2
+ 4x + a2
= -( x2
- 4x ) + a2
= -[ x2
- 4x + ( -2 )2
- ( -2 )2
] + a2
= -( x - 2 )2
+ ( -2 )2
+ a2
( -2 )2
+ a2
= 8
a2
= 4
a = ± 2

7 Given 3n - 3
x 27 n
= 243, find the value of n.
3n - 3
x 27 n
= 243
3n - 3
x ( 33
) n
= 35
3n - 3
x 33n
= 35
3n – 3 + 3n
= 35
n – 3 + 3n = 5
4n = 8
n = 2

8 Given that log8 p - log2 q = 0, express p in terms of q.
log8 p - log2 q = 0
log8 p = log2 q
q
p
2
2
2
log
8log
log
=
log2 p = 3log2 q
log2 p = log2 q3
p = q3

9 Given the geometric progression -5, 10/3 , - 20/9 ,...,
find the sum to infinity of the progression.
10 20
5 , , ,...
3 9
− −
a = -5
10
3
5
r =
−
2
3
r = −
5
2
1
3
S∞
−
=
 
− − ÷
 
5
5
3
−
=
3= −

10 Diagram 10 shows three square cards.
The perimeters of the cards form an arithmetic
progression. The terms of the progression are in
ascending order.
(a)Write down the first three terms of the progression.
(b)Find the common difference of the progression.
(a) 4(3) , 4(5), 4(7)
= 12 , 20, 28
(b) d = 20 - 12
= 8

11 The first three terms of a geometric progression are x, 6,
12. Find
(a)the value of x,
(b) the sum from the fourth term to the ninth term.
(a) x , 6 , 12
6
126
=
x
6
2
x
=
6
2
x =
= 3
(b)
a = 3 r = 2
( )
12
123 9
9
−
−
=S = 3( 29
– 1 )
= 1533
( )3
3
3 2 1
2 1
S
−
=
−
= 3( 23
– 1 )
= 21
4 9 1533 21S −> = − = 1512

12 Diagram 12 shows a sector BOC of a circle with centre 0.
It is given that ∠BOC= 1.42 radians, AD = 8 cm and 0A =
AB = OD = DC = 5cm.
Find
(a) the length, in cm, of arc BC,
(b) the area, in cm2
, of the coloured region.
(a) arc BC
= 10 ( 1.42 )
= 14.2
(b) the area
= ½ (10)2
( 1.42 )
- ½ (8) (3)
= 71 - 12
= 59

Given that a = 13 i + j and b = 7 i – k j, find
(a) a - b in the form x i + y j,
(b) the values of k if | a - b | = 10.
13
a = 13 i + j , b = 7 i – k j,
(a) a - b
= 13 i + j – ( 7 i – k j )
= 6 i +(1+k) j
(b) | a - b | = 10
( ) 1016
22
=++ k
36 + 1 + 2k + k2
= 100
k2
+ 2k – 63 = 0
( k +9 ) ( k -7 ) = 0
k = -9 , 7

14 Diagram 14 shows a triangle PQR.
Given and point S lies on QR such
that QS : SR = 2 : 1, express in terms of a, and b


= -3 a + 6 b
SR RP= +
uur uuur
( )
1
3 6 6
3
a b b= − + −
2 6a b b= − + −
4a b= − −

15 Diagram shows a straight line AC.
The point B lies on AC such that AB : BC = 3 : 1.
Find the coordinates of B.

( ) ( ) ( ) ( )3 4 1 2 3 0 1 3
,
3 1 3 1
B
 + − +
=  ÷
+ + 
12 2 0 3
,
4 4
− + 
=  ÷
 
10 3
,
4 4
 
=  ÷
 
5 3
,
2 4
 
=  ÷
 

16 Solve the equation 3 sin x cos x - cos x = 0 for 00
≤ x ≤
3600
.
3 sin x cos x - cos x = 0
cos x ( 3 sin x – 1 ) = 0
cos x = 0 3 sin x – 1 = 0
x = 90o
, 270o sin x = 1/3
x = 19.47o
, 160.53o

17 It is given that sin A = 5/13 and cos B = 4/5 where A is
an obtuse angle and B is an acute angle.
Find
(a) tan A,
(b) cos(A - B).
A5
13 B
5
4
12
(a) tan A
5
12
= −
3
(b) cos( A – B )
= cosA cosB – sinA sinB
12 4 5 3
13 5 13 5
     
= − + ÷ ÷  ÷ ÷
     
48 15
65 65
= − +
33
65
= −

18
Given that and∫ =
m
dxxf
5
6)( [ ]
5
( ) 2 14
m
f x dx+ =∫
find the value of m.
∫ =
m
dxxf
5
6)([ ]
5
( ) 2 14
m
f x dx+ =∫
5 5
( ) 2 14
m m
f x dx dx+ =∫ ∫
5
6 2 14
m
dx+ =∫
[ ]5
2 8
m
x =
2 ( m – 5 ) = 8
m – 5 = 4
m = 9

19 The gradient function of a curve is = kx - 6, where k
is a constant.
It is given that the curve has a turning point at (2, 1).
Find
(a)the value of k,
(b) the equation of the curve.
dy
dx
6
dy
kx
dx
= −(a)
( )0 2 6k= −
2 6k =
3k =
3 6
dy
x
dx
= −(b)
2
3
6
2
x
y x c= − +
( )
( )
2
3 2
1 6 2
2
c= − +
7c =
2
3
6 7
2
x
y x= − +

20 A block of ice in the form of a cube with sides x cm, melts
at a rate of 972 cm3
per minute. Find the rate of change of
x at the instant when x = 12cm.
3
V x=
2
3
dV
x
dx
=
( )
2
3 12=
432=
dV dV dx
dt dx dt
= •
972 432
dx
dt
= •
972
432
dx
dt
=
2.25 /cm s=

21 Diagram shows part of the curve y = f(x) which passes
through the points (h, 0) and (4, 7).
Given that the area of the coloured region is 22 unit2
,
find the value of h∫4
f(x)dx.
∫
4
)(
h
dxxf
= 4(7) - 22
= 6

22 There are 4 different Science books and 3 different
Mathematics books on a shelf.
Calculate the number of different ways to arrange all the
books in a row if
(a) no condition is imposed,
(b) all the Mathematics books are next to each other.
S1 S2 S3 S4 M1 M2 M3
(a) 7
7p
7!=
5040=
(b)
S1 S2 S3 S4M1 M2 M3
5 3
5 3p p×
5! 3!= ×
120 6= ×
720=

23 The probability that a student is a librarian is 0.2. Three
students are chosen at random.
Find the probability that
(a) all three are librarians,
(b) only one of them is a librarian.
P( x = 3 )(a)
= 1 × ( 0.2)3
× 1
= 0.008
X ~ Bin ( 3 , 0.2 )
= 3
C3 (0.2)3
(0.8)0
P( x = 1 )(a)
= 3 × ( 0.2) × 0.64
= 0.384
= 3
C1 (0.2)1
(0.8)2

24 A set of 12 numbers x1, x2, ... , x12 , has a variance of 40
and it is given that ∑x2
= 1 080.
Find
(a)the mean,
(b) the value of ∑ x.
(a) 2
2
2
x
N
x
−=
∑σ
21080
40
12
x= −
2
40 90 x= −
2
50x =
7.071x =
(b)
12
x
x =
∑
12(7.071)x =∑
84.853=

25 The masses of apples in a stall have a normal
distribution with a mean of 200 g and a standard
deviation of 30 g.
(a) Find the mass, in g, of an apple whose z-score is 0.5.
(b) If an apple is chosen at random, find the probability
that the apple has a mass of at least 194g.
X ~ N ( 200 , 302
)
(a) X
Z
µ
σ
−
=
200
0.5
30
X −
=
15 200X= −
215X =
(b) ( 194)P X >
194 200
( )
30
P Z
−
= >
( 0.2)P Z= > −
1 ( 0.2)P Z= − >
1 0.4207= −
0.5793=

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