S ULIT 3472/1 Additional Mathematics Paper 1 Oct 2007 2 hours
3472/1 Name : ………………..…………… Form : ………………………..……
SEKTOR SEKOLAH BERASRAMA PENUH KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSA PEPERIKSAAN AN AKHIR TAHUN TAHUN TINGKATAN TINGKATAN 4 2007 Kod Pemeriksa ADDITIONAL MATHEMATICS Pap Paper 1 Two hours
Question
1
Full Marks 2
DO NOT OPEN THIS QUESTION PAPER UNTIL INSTRUCTED TO DO SO
2
3
3
4
4
2
5
3
6
3
7
2
8
3
9
4
1
This This questi question on paper paper consis consists ts 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.
10
3
11
3
5.
Show your working. It may help you to get marks.
12
4
6.
If you wish to change your answer, cross out the work that that you have have done done.. Then Then writ writee down down the new answer.
13
4
14
4
15
3
The The diag diagra rams ms in the the ques questi tion onss prov provid ided ed are are not not drawn to scale unless stated.
16
4
17
3
The marks allocated for each question and sub-part of a question are shown in brackets.
18
4
19
3
A list of formulae formulae is provided provided on pages 2 to 3.
20
3
21
4
22
2
23
3
24
3
25
4
7.
8.
9.
10. A booklet of four-figure mathematical tables is provided. . 11 You may use a non-programmable scientific calculator. 12 This question paper must be handed in at the end of the examination .
TOTAL
This question paper consists consists of 13 printed pages
80
Marks Obtained
2
SULIT
3472/2
The following following formulae may be helpful helpful in answering answering the questions. questions. The symbols symbols given are the ones commonly used.
ALGEBRA
b b 2 4ac
1
x
2
a
3
a
4
(a ) = a
5
loga mn = log am + loga n m loga = log am loga n n n log a m = n log a m
an = a m + n
m
an = a m n
m n
6 7
logab =
9
Tn = a + ( n 1)d
2a
m
log c b
8
log c a
n
[ 2a ( n 1)d ] 2 1 11 Tn = ar n n n a ( r 1) a(1 r ) 12 Sn = , (r 1) 1 r r 1 a 13 S , r <1 1 r 10
nm
Sn =
CALCULUS
1
2
y = uv ,
y
u
dx
dx
,
v
dy
dy
u
v
dv
v
dx
du dx
u v
2
du 4 Area under a curve
dx
b
y
=
dv
dx or
a
dx ,
b
=
x dy a
dy
3
dx
dy du
5 Volume generated
du
b
dx
=
y
2
dx or
2
dy
a b
=
x a
GEOMETRY
1 Distance =
2 2 ( x1 x 2 ) ( y1 y 2 )
2 Midpoint ( x , y) =
y y 2 x1 x 2 , 1 2 2
3
2 2 r x y
4
r ˆ
xi yj
5
A poin pointt divid dividing ing a seg segmen mentt of a line line nx mx2 ny1 my 2 ( x,y) = 1 , m n m n
6 Area Area of triangl trianglee 1 = ( x1 y 2 x 2 y 3 x3 y11 ) ( x 2 y1 x3 y 2 x1 y 3 ) 2
SULIT
3472/1
3 STATISTICS
1
2
3
4
x =
x =
x
fx f
8
n
N
=
x
2
f ( x x)
f
=
N
2
=
Q1 Q0
w1 I 1 w1 n!
Pr
n
9
( x x )
=
I
I
N
( n r )!
C r
2
x
f
fx 2
n!
( n r )!r !
2
10
P(A B)=P(A)+P(B)-P(A B)
11
p (X=r) =
12
Mean µ = np
13
14
z=
n
x 2
1 2 N F 5 m = L C f m 6
7
C r p r q n r , p + q = 1
npq
x
100 TRIGONOMETRY
9 sin ( A B) = sin A cos B cos A sin B
1 Arc length length,, s = r 1
2 Area Area of secto sectorr , L = 2
2
2
r 2
3 sin A + cos A = 1 2
10 cos ( A B) = cos A cos B 11 tan tan ( A B) =
2
sin A sin B
tan A tan B
1 tan A tan B
4 sec A = 1 + tan A 2
2
5 cose cosecc A = 1 + cot A
12
6 sin 2 A = 2 sin A cos A 2
2
7 cos 2 A = cos A – sin A 2 = 2 cos A – 1 2 = 1 – 2 sin A 8 tan 2 A =
2 tan A 2 1 tan A
a
sin A
2
b
sin B
2
c
sin C
2
13
a = b + c – 2bc cos A
14
Area Area of tria triang ngle le =
1 2
ab sin C
For examiner’s use only
SULIT
4
3472/1
Answer all questions. 1 Diagram Diagram 1 shows the relations relations between between two sets of number. number.
1
9
3
8
5
7
DIAGRAM 1 ( a)
State the object of 9,
(b) (b)
Repr Repres esen entt the the rela relatio tion n by a set set of orde ordere red d pair pairs. s. [ 2 marks ]
1
…………………….. ….. Answer : (a) ………………… (b) ……………………... ……………………...
2
2
Given the function f : x 3x 1 and gf : x 9 x 5 . Find Find (a) f 1 ( x) , (b) g ( x) [ 3 marks]
Answer : (a) ………………… …………………….. ….. 2
(b) ……………………... ……………………... 3
SULIT 3
5
3472/1
b Diagram Diagram 2 shows shows the function function h : x ax , x k where a , b and k are constants. x
x
For examiner’s use only
h (x)
9 2
6
1 DIAGRAM 2 Find (a) the value of k (b) the value valuess of a and of b [ 4 marks ]
...…………………… …………... …...... ... Answer : (a) ...……………
3
(b).............................................. 4
Form the quadratic equation which has the roots 2 and
1
4 2 the form form ax + bx + c = 0, where a, b and c are constants.
4
. Give your answer in [2 marks]
4
.........………………… Answer : .........…………………
5
2
Given that the quadratic equation ( p + 1) x = 3 x 4 has two two equal equal roots. roots. Find the value of p. 2
[3 marks] 5 3
Answer : ………… ……………… ………… …….. ..
For examiner’s use only
SULIT 6
6
3472/1
2
Given Given the roots roots of 3 x – 4 x + 5 = 0 are and . Form a quadratic equation with roots 3 1 and 3 +1. [3 marks]
6
Answer : ……............................. ……............................. 3
7
2
Find the range of values of x for which which 2 x + 5 x – 12 < 0. [2 marks]
7 Answer :
………… ……………… ………… ……….. …....
2
8
y
Diagram 3 shows the graph of the 2 function y = ( x p) -5, where p is a constant. Find
( 6, 4 )
4
x
O
(a) the value of p, (b) the equation of the axis of symmetry, (c) the coordina coordinates tes of the minimum point. [ 3 marks]
DIAGRAM 3 Answer : (a) …….............................
8
(b) .................................... .................................... 3
(c) .................................... ....................................
SULIT 9
7
3472/1 2
The graph graph of the quadratic quadratic function function f ( x) = x – 4 mx +2 – 7m touches the x – axis at two points. points. Find the the range of of values values of m. [4 marks]
For examiner’s use only
9 Answer : ……..……...………..... ……..……...……….....
4
the simple simplest st form. form. 10 Express 2 n 1 3 ( 2 ) + 7( 2 n 1 ) in the n
[3 marks]
10
……………...………..... Answer : ……………...……….....
11
Solve the equation 4 x 2 =
2 x
8
3
. [3 marks]
11
Answer : …...…………..………... …...…………..………...
3
For examiner’s use only
SULIT
8
3472/1
12 Solve the equation log 2 x log 8 x 2 .
[4 marks]
12 Answer : ……………...………..... ……………...………..... 4
13
Given log 2 m = x , and log 5 m = y, express log m 80 in terms of x and y. [4 marks]
13
…………...………..... Answer : …………...………..... 4
14 The following information refers to the equation of two straight lines, JK and RS which are perpendicular to each other.
Straight line JK : y = (4+p)x+ k x
Express k in terms of p.
y
1 p k where p and k are constants.
Straight line RS :
[4 marks]
14 4
Answer : ………..……….......... ………..………..........
SULIT
9
3472/1
and Q ( 9 , 8 ). Find the equation equation of a straight straight line with gradient gradient 15 Given that P ( 3 , 6 ) and 2 that passes through the midpoint of PQ. 3 [3 marks]
For examiner’s use only
15 Answer : ........................................ ........................................
3
16 A point P moves such that its distance from two fixed points, A(3 ,2) and B ( 1, 4), are in the ratio of PA : PB = 4 : 3. Find the equation of locus P. .
[4 marks]
16
..................................... Answer : ..................................... 4
consists of four numbers. numbers. The sum of the numbers, numbers, 17 A set of data consists of the squares of the numbers,
x
2
x 12k and the sum
= 100. Express the variance in terms of k . [3 marks]
17 3
Answer :
For examiner’s use only
SULIT
10
3472/1
18 A set of positive integers 1, m – 1, 6 and 8 is in ascending ascending order. order. Find the value of m if
(a) the mode is 1, (b) the mean is 4, (c) the median is 5. [4 marks]
Answer : (a) .…………..……..…...
18
(b) .................................... .................................... (c) .................................... ....................................
4
19 Diagram 4 shows a semicircle OPQR with centre O. Q
3 cm rad P
O
R
DIAGRAM 4 Given Given that the radius is 3 cm and the perimeter perimeter of sector sector QOR is 7.55 cm . Find the value . of . [3 marks]
19 3
Answer :
…...……… …...…………..… …..……..… …..…... ...
SULIT
20
11
3472/1 For examiner’s use only
Diagram 5 shows a sector OPQ with centre O.
Q
2 rad P
O
DIAGRAM 5 Given that the area of the sector OPQ is 64 cm and OPQ = 2 rad. Find the length of the arc PQ. (Use = 3.142) 2
[3 marks]
20 Answer : …...…………..……..…... …...…………..……..…... 3
21 Diagram 6 shows a sector MTKR with centre T . T
74 5 cm 4 cm M
R
K
DIAGRAM 6
21
Given 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
[4 marks]
SULIT For examiner’s use only
12
3472/1 Answer : …...…………..……..…... …...…………..……..…...
22 Given that f ( x )
1 ( 1 3 x )
2
, find f (x). [ 2 marks]
22
…...…………..……..…... Answer : …...…………..……..…... 2 2
gradient 3 at the the point Q . Find the coordinates of 23 The curve y = 2 x – 5 x + 1 has a gradient the point point Q. [3 marks]
23 Answer : …...…………..……..…... …...…………..……..…... 3
24
Find the equation of the tangent to the curve y = x ( x – 4) at the point ( 3, 9 ). 2
[3 marks]
24 3
Answer : …...…………..……..…... …...…………..……..…...
SULIT
25
13
It is given that y = u – u and x = 2 u 3. Find 2
3472/1
dy dx
For examiner’s use only
in terms of x. [4 marks]
25
…...…………..……..…... Answer : …...…………..……..…...
END OF THE QUESTION PAPER
4
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 marking scheme consists consists of 6 printed pages pages
3472/1
© 2007 Copyright SBP Sector, Ministry of Education, Malaysia
SULIT
Sub Marks
Full Marks
(a) 1 and 3
1
2
(b) (b)
1
Number 1
Solution and marking scheme
{ (1,9 (1,9 ,(3,9 ,(3,9 ),(5,9 ),(5,9)) }
2
(a) f 1( x )
3
1
(b) (b) g( x) x) = 3x 3x + 8
2
gff
3
x 1
1
B1
x 1 9 5 3
(a) k = 0 (b) a=4 and b = 2
1 3
a = 4 or b = 2 2a +
4
b
2
9 or a + b = 6
B1
2
4x +7x +7x - 2 = 0
p=
4
B2
2
or SOR SOR=
(x + 2)(4x - 1) = 0
5
3
7 4
& POR =
1
B1
2 3
7
16 (- 3) – 4 (p+1)(4) = 0
2
3
2
B2
2
(p+1)x + 3x + 4 = 0
6
B1
2 x 6 x 20 0
3 1 3 1 6
4 3
or
3
or 3 1( 3 1 ) 20 5
B2
3
B1 2
3
Number 7
Solution and marking scheme
4 x
Sub Marks
Full Marks
2
2
3 2
(2x (2x – 3)(x 3)(x + 4 ) < 0
or
-4
3
B1
2
8
9
(a) p = 3
1
(b) x = 3
1
(c) (c) ( 3 , - 5 )
1
m 2 , m
4
1
4 (4m (4m – 1 )( )( m + 2 ) > 0 or
2
1 4
2
4m + 7m – 2 > 0
5(2
n– 1
)
2 2 3 n
n
1
B3 B2
3
B2
2 n
n
-1
B1
)
3
3
5 2x + 4 = 1 – 3x
B2
x=
3
7
2 2 –3(2 )+7(2 2
11
4
B1
2
(-4m) – 4 (1)(2 – 7m) > 0
10
3
B1
2(x+2)
2
3
3
Number 12
Solution and marking scheme x = 8 x
64 3
4
3
log 2 x
x
B1 4
1 y
B2
4
logm 2 + logm 5 log m m
log 2 m
log 8 80 log 5 m
log m 2
4
B3
4 logm 2 + logm 5
k
B2
or 2 log22 or 2 log88
log 2 8
14
4
B3
log x – log log x = 64 or log log x – log x = 6 log 2
4
Full Marks
3
x
13
Sub Marks
or log 5 m
log m m log m 5
or log 2 80 or log 2 m
B1
4
or log m( 2 x 5)
4
p
4
4 p
k
B3
4 p 1 p
k p
k p
15
and 4 + p
B2
or 4 + p
B1
3y = 2x + 15 y–7=
2
3
B2
(x–3)
3 Mid point PQ = ( 3 , 7 )
B1
4
3
Number 16
Sub Marks 4
Solution and marking scheme 2
2
5x + 5y + 62x – 68y + 49 = 0 2
2
2
2
9 [ x + 6x + 9 + y – 4y + 4 ] = 4 [ x – 2x + 1 + y + 8y + 16 ] 2 2 2 2 3 ( x 3 ) ( y 2 ) = 4 ( x 1 ) ( y 4 )
3PA = 4PB
17
2
25 – 9k
12k 4 4
100
B1
(a) 2
1
(b) 2
2
4
4
B1
4
5
1
= 0.5167
3
(c)
3
B2
2
1 ( m 1) 6 8
19
B3 B2 B1
3
n= 4 18
Full Marks 4
3
3 1.55
B2
QR = 1.55
B1
20
16 r= 8 1 2 r 64 2
3 B2 B1
3
21
Perime imeter ter = 12.4 12.45 586
4
4
Peri Perime mete terr = 6.45 6.4586 86 + 6
B3
Arc MK = 5 ( 1. 2917) 2917)
B2
1.2917 rad
B1
5
Number 22
Solution and marking scheme
f ( x )
6
2
Q ( 2, - 1)
3
4x – 5 = 3
B2
dy dx
dy
3 or
dx
4x 5
y = 3 x – 18
dy dx dy
3
dy dx
B1
2 x
4
2
dx
x 3 2
1 2
B3
2 B2
dy dx dy dx
1
( 1 2u )( ) 2
1 2u or
3
B2
3 x 2 8 x
3
B1
y + 9 = 3 ( x – 3)
25
2
B1
24
Full Marks
( 1 3 x )3
f 2( 1 3 x ) 3 ( 3 )
23
Sub Marks
dx du
2 or u =
x 3
B1
2
6
4
