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SULIT 3472/1
SEKTOR SEKOLAH BERASRAMA PENUHKEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2007
This question paper consists of 13 printed pages
Kod Pemeriksa
QuestionFull
Marks
Marks
Obtained
1 2
2 3
3 44 2
5 3
6 3
7 2
8 3
9 4
10 3
11 3
12 4
13 4
14 4
15 3
16 4
17 3
18 4
19 3
20 3
21 4
22 2
23 3
24 3
25 4
TOTAL 80
Name : ..
Form : ..
3472/1
Additional Mathematics
Paper 1
Oct 2007
2 hours
ADDITIONAL MATHEMATICS
Paper 1
Two hours
DO NOT OPEN THIS QUESTION PAPER
UNTIL INSTRUCTED TO DO SO
1 This question paper consists of 25 questions.
2. Answer all questions.
3. Give only one answer for each question.
4. Write your answers clearly in the spaces provided in
the question paper.
5. Show your working. It may help you to get marks.
6. If you wish to change your answer, cross out the work
that you have done. Then write down the new
answer.
7. The diagrams in the questions provided are not
drawn to scale unless stated.
8. The marks allocated for each question and sub-part
of a question are shown in brackets.
9. A list of formulae is provided on pages 2 to 3.
10. A booklet of four-figure mathematical tables is
provided.
.
11 You may use a non-programmable scientificcalculator.
12 This question paper must be handed in at the end of
the examination .
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SULIT 3472/22
The following formulae may be helpful in answering the questions. The symbols given are the ones
commonly used.
ALGEBRA
12
42
b b acxa
2 am an = a m + n
3 am an = a m n
4 (am
)n
= anm
5 loga mn = log am + loga n
6 loga n
m
= log am loga n
7 log a mn
= n log a m
8 logab =a
b
c
c
loglog
9 Tn = a + (n 1)d
10 Sn = ])1(2[2
dnan
11 Tn = arn 1
12 Sn =r
ra
r
rann
1
)1(
1
)1(, (r 1)
13r
aS
1
, r
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SULIT 3472/1
STATISTICS
1 Arc length, s = r
2 Area of sector , L =21
2r
3 sin2A + cos
2A = 1
4 sec2A = 1 + tan
2A
5 cosec2
A = 1 + cot2
A
6 sin 2A = 2 sinA cosA
7 cos 2A = cos2A sin
2A
= 2 cos2A 1
= 1 2 sin2A
8 tan 2A =A
A2
tan1
tan2
TRIGONOMETRY
9 sin (A B) = sinA cosB cosA sinB
10 cos (A B) = cosA cosB sinA sinB
11 tan (A B) =BtanAtan
BtanAtan
1
12C
c
B
b
A
a
sinsinsin
13 a2
= b2
+ c2
2bc cos A
14 Area of triangle = Cab sin2
1
1 x =N
x
2 x = f
fx
3 =N
xx 2)(= 2
2
xN
x
4 =
f
xxf2
)(= 2
2
xf
fx
5 m = Cf
FN
Lm
2
1
6 1
0
100Q
IQ
71
11
w
IwI
8
)!(
!
rn
nPr
n
9!)!(
!
rrn
nC
r
n
10 P(A B)=P(A)+P(B)-P(A B)
11 p (X=r) = rnrrn qpC
, p + q = 1
12 Mean = np
13 npq
14 z =
x
3
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SULIT 4 3472/1
Answer all questions.
1 Diagram 1 shows the relations between two sets of number.
(a) State the object of 9,
(b) Represent the relation by a set of ordered pairs.
[ 2 marks ]
Answer: (a) ..
(b) ...
2 Given the function 13 xx:f and 59 xx:gf . Find
(a) )(1 xf ,
(b) g (x)
[ 3 marks]
Answer: (a) ..
(b) ...
3
2
2
1
1
3
5
9
8
7
DIAGRAM 1
For
xaminers
use only
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SULIT 5 3472/1
3 Diagram 2 shows the function kx,x
baxx:h where a , b and kare constants.
Find
(a) the value of k
(b) the values ofa and of b
[ 4 marks ]
Answer: (a) .........
(b)..............................................
4 Form the quadratic equation which has the roots 2 and41 . Give your answer in
the form ax2
+ bx + c = 0, where a, b and c are constants.
[2 marks]
Answer: .........
5 Given that the quadratic equation (p + 1)x2
= 3x 4 has two equal roots.Find the value of p.
[3 marks]
Answer: ..
5
3
4
x
9
62
1
h (x)
DIAGRAM 2
Fo
examin
use o
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SULIT 6 3472/1
6 Given the roots of 3x2
4x + 5 = 0 are and . Form a quadratic equation with roots
13 and 3 +1.
[3 marks]
Answer: .............................
7 Find the range of values ofx for which 2x2
+ 5x 12 < 0.
[2 marks]
Answer: ...
8
Answer: (a) .............................
(b) ....................................
(c) ....................................
For
xaminers
use only
3
6
2
7
y
x
DIAGRAM 3
4 ( 6, 4 )
Diagram 3 shows the graph of the
function y = (x p)2 -5, where p isa constant.
Find
(a) the value ofp,
(b) the equation of the axis of
symmetry,
(c) the coordinates of the
minimum point.
[ 3 marks]
3
8
O
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SULIT 7 3472/1
9 The graph of the quadratic function f(x) = x2
4mx +2 7m touches the x axis at
two points. Find the range of values ofm.
[4 marks]
Answer: ..........
10 Express 12 n 3 ( 2n ) + 7( 12 n ) in the simplest form.[3 marks]
Answer: ........
11 Solve the equation 24 x =x8
2.
[3 marks]
Answer: ........
For
examin
use on
10
9
11
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SULIT 9 3472/1
15 Given that P ( 3 , 6 ) and Q ( 9 , 8 ). Find the equation of a straight line with gradient
3
2that passes through the midpoint ofPQ.
[3 marks]
Answer: ........................................
16 A point P moves such that its distance from two fixed points, A(3 ,2) and B ( 1, 4), arein the ratio of PA : PB = 4 : 3. Find the equation of locus P.
.
[4 marks]
Answer: .....................................
17 A set of data consists of four numbers. The sum of the numbers, x 12kand the sum
of the squares of the numbers, 2x = 100. Express the variance in terms ofk.[3 marks]
Answer: ..........
For
examin
use on
17
16
15
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SULIT 10 3472/1
18 A set of positive integers 1, m 1, 6 and 8 is in ascending order.
Find the value ofm if
(a) the mode is 1,
(b) the mean is 4,
(c) the median is 5. [4 marks]
Answer: (a) ........
(b) ....................................
(c) ....................................
19 Diagram 4 shows a semicircle OPQR with centre O.
Given that the radius is 3 cm and the perimeter of sector QOR is 7.55 cm . Find the value
of.
[3 marks]
Answer: ..........
For
xaminers
use only
Q
P RO
DIAGRAM 4
3
19
4
18
3 cm
rad
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SULIT 11 3472/1
20 Diagram 5 shows a sector OPQ with centre O.
Given that the area of the sector OPQ is 64 cm2
and OPQ = 2 rad. Find the lengthof the arc PQ.
(Use = 3.142)
[3 marks]
Answer: ..........
21 Diagram 6 shows a sector MTKR with centre T.
Given that the radius of the sector is 5 cm and R is the midpoint of MK.
Find the perimeter of the shaded region.
[Use 3.142 ][4 marks]
Fo
examin
use o
OP
2 rad
DIAGRAM 5
2
20
T
M K
5 cm74
DIAGRAM 6
R
4 cm
Q
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SULIT 12 3472/1
Answer: ..........
22 Given that2
31
1
)x()x(f
, find f (x).
[ 2 marks]
Answer: ..........
23 The curve y = 2x2
5x + 1 has a gradient 3 at the point Q . Find the coordinates of
the point Q.
[3 marks]
Answer: ..........
24 Find the equation of the tangent to the curve y = x2 ( x 4) at the point ( 3, 9 ).[3 marks]
Answer: ..........
For
aminers
se only
3
24
2
22
3
23
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SULIT 13 3472/1
25 It is given that y = u u2
and x = 2u 3. Finddx
dyin terms ofx.
[4 marks]
Answer: ..........
END OF THE QUESTION PAPER
Fo
exami
use o
25
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3472/1 2007 Copyright SBP Sector, Ministry of Education, Malaysia
SULIT
SULIT
3472/1
Additional
Mathematics
Paper 1
Okt
2007
SEKOLAH BERASRAMA PENUH
KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN AKHIR TAHUN
TINGKATAN 4
2007
ADDITIONAL MATHEMATICS
Paper 1
MARKING SCHEME
This marking scheme consists of 6 printed pages
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3
Number Solution and marking schemeSub
MarksFull
Marks
72
34 x
(2x 3)(x + 4 ) < 0 or2
B1
2
8 (a) p = 3
(b) x = 3
(c) ( 3 , - 5 )
1
1
1
3
92m ,
4
1m
(4m 1 )( m + 2 ) > 0 or
4m2
+ 7m 2 > 0
(-4m)2
4 (1)(2 7m) > 0
4
B3
B2
B1
4
10 5 ( 2n 1
)
2n
2
732
2n
21
3 ( 2n
) + 7 ( 2n
2- 1
)
3
B2
B1
3
11x =
5
3
2x + 4 = 1 3x
22(x+2)
3
B2
B1
3
-423
2
4
1
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4
Number Solution and marking schemeSub
MarksFull
Marks
12 x = 8
6
3
x
x
4
log x3
log x = 64 or logx3
log x = 6 log 2
82
2
log
xlogor 2 log22 or 2 log88
4
B3
B2
B1
4
13
yx
14
4 logm 2 + logm 5
logm 24 + logm 5
22
m
m
log
mlogmlog or
55
m
m
log
mlogmlog or
mlog
log
2
2 80 or
mlog
log
5
8 80 or log m( 24
x 5)
4
B3
B2
B1
4
14p
pk 4
14
p
kp
p
k and 4 + p
p
k or 4 + p
4
B3
B2
B1
4
15 3y = 2x + 15
y 7 =3
2( x 3 )
Mid point PQ = ( 3 , 7 )
3
B2
B1
3
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5
Number Solution and marking schemeSub
MarksFull
Marks
16 5x2
+ 5y2
+ 62x 68y + 49 = 0
9 [ x2
+ 6x + 9 + y2
4y + 4 ] = 4 [ x2
2x + 1 + y2
+ 8y + 16 ]
3 22 23 )y()x( = 4 22 41 )y()x(
3PA = 4PB
4
B3
B2
B1
4
17 25 9k 2
2100 12
4 4
k
n = 4
3
B2
B1
3
18 (a) 2
(b) 2
44
8611
)m(
(c) 5
1
2
B1
1
4
19 = 0.5167
5513 .
QR = 1.55
3
B2
B1
3
20 16
r = 8
642
1 2 r
3B2
B1
3
21 Perimeter = 12.4586
Perimeter = 6.4586 + 6
Arc MK = 5 ( 1. 2917)
1.2917 rad
4
B3
B2
B1
4
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6
Number Solution and marking schemeSub
MarksFull
Marks
22331
6
)x()x(f
)()x(f 3312 3
2
B1
2
23 Q ( 2, - 1)
4x 5 = 3
3dx
dyor 54 x
dx
dy
3
B2
B1
3
24 y = 3x 18
y + 9 = 3 ( x 3)
xxdx
dy83
2
3
B2
B1
3
25
2
2 x
dx
dy
dx
dy
2
2321
x
))(u(dx
dy
2
121
udx
dy21 or 2
du
dxor u =
2
3x
4
B3
B2
B1
4