1. MOZ@C
3472/1
Additional Mathematics
Kertas 1
Oktober 2008
2 Jam
Name : ………………..……………
Form : ………………………..……
SEKOLAH BERASRAMA PENUH
BAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH / KLUSTER
KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN AKHIR TAHUN
TINGKATAN 4 2008
ADDITIONAL MATHEMATICS
Kertas 1
Dua jam
JANGAN BUKA KERTAS SOALAN INI
SEHINGGA DIBERITAHU
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 subpart of a question are shown in brackets.
9. A list of formulae is provided on pages 2 and 3.
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.
For examiner’s use only
Question
Total Marks
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
Marks
Obtained
2
4
4
4
2
2
3
3
3
4
3
3
4
3
4
2
3
3
4
3
2
3
4
4
4
TOTAL
80
Kertas soalan ini mengandungi 15 halaman bercetak
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[Lihat sebelah
SULIT

2. MOZ@C
SULIT
3472/1
The following formulae may be helpful in answering the questions. The symbols given are the ones
commonly used.
ALGEBRA
x=
−b ± b 2 − 4ac
2a
2
a m ÷ an = a m -
4
loga mn = log am + loga n
loga
7
am × an = a m + n
3
5
6
1
log a mn = n log a m
(am) n = a nm
m
= log am - loga n
n
n
8 logab =
log c b
log c a
CALCULUS
1
dy
dv
du
= u +v
dx
dx
dx
y = uv ,
u dx
=
,
v dy
2
y=
3
v
du
dv
−u
dx
dx ,
2
v
dy dy du
=
×
dx du dx
GEOMETRY
1 Distance =
( x1 − x 2 ) 2 + ( y1 − y 2 ) 2
2 Midpoint
y + y2 
 x1 + x 2
, 1

2 
 2
(x , y) = 
3472/1
3 A point dividing a segment of a line
 nx + mx 2 ny1 + my 2 
( x,y) =  1
,

m+n 
 m+n
4 Area of triangle
1
= ( x1 y 2 + x 2 y 3 + x3 y11 ) − ( x 2 y1 + x3 y 2 + x1 y 3 )
2
2007 Hak Cipta SBP
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[ Lihat sebelah
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3. MOZ@C
STATISTIC
1
x =
2
x =
3
σ =
4
σ=
x
N
 fx
f
 (x − x )
x
2
N
 f ( x − x)
f
5
I=
I=
2
=
−x
 fx
f
2
−x
2
Q1
×100
Q0
7
N
_2
1

2N−F
m = L+
C
 fm 




6
=
2
 w1 I1
 w1
TRIGONOMETRY
1 Arc length, s = r θ
2 Area of sector , L =
3
1 2
rθ
2
2
2
4 a = b +c
5
3472/1
a
b
c
=
=
sin A sin B sin C
2
- 2bc cosA
Area of triangle
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
=
1
absin C
2
3
[ Lihat sebelah
SULIT

4. MOZ@C
For
examiner’s
use only
Answer all questions.
1. Diagram 1 shows the linear function f .
f
f(x)
x
1
3
3
5
5
k
7
9
Set A
Set B
Diagram 1
(a) State the value of k.
(b) Using function notation, write a relation between set A and set B.
[ 2 marks]
Answer : (a) ……………………..
1
(b) ...…………………...
2
2.
f :x→
4x
,x≠h
x+3
g : x → 1 − 2x
The following information above refers to the functions f and g .
(a) State the value of h.
(b) Find the value of fg −1 (3) .
[ 4 marks ]
2
Answer : (a) …………………….
(b) ..............................
3472/1
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SMS MUZAFFAR SYAH , MELAKA
[ Lihat sebelah
SULIT
4

5. MOZ@C
For
examiner’s
use only
3.
Given the function f : x → 2 x − 1 and fg : x → 6 x + 1 . Find
(a) the function g(x)
(b) the value of x when gf ( x ) = 4 .
[ 4 marks ]
Answer : (a).........…………………
3
(b).....................................
4
4.
Given that the roots of the quadratic equation x 2 − 1 = 5 x − k are 3 and p .
Find the value of k and of p .
[4 marks]
Answer : k =.........………
4
p =....................
4
5.
The quadratic equation 2 x 2 − 2 x − p + 1 = 0 has two distinct roots. Find the range of
values of p .
[2 marks]
5
2
Answer : .............………
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
SULIT

6. MOZ@C
6. Diagram 2 shows the graph of the function y = ( x + 3) 2 − p where p is a constant and
( k + 1,−4) is a minimum point.
y
For
examiner’s
use only
5
x
(k + 1,−4)
Diagram 2
Find
a) the value of k.
b) the value of p.
2 marks ]
Answer : (a) ................................
6
(b) .................................
2
___________________________________________________________________________
7Find the range of values of x for which 3 ≤ x (2 x − 5) .
[3 marks]
7
Answer : ..................................
3
8.
Solve the equation 5 x + 2 − 5 x +1 = 4 .
[3 marks]
8
Answer : …….……………….
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[ Lihat sebelah
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3

7. MOZ@C
9.
Given log p 3 = x and log p 2 = y , express log8 12 in terms of x and y.
For
examiner’s
use only
[3 marks]
9
Answer : ..................................
3
10.
Solve the equation log 3 (3 x + 2) − log 3 ( x − 1) = 0
[4 marks]
10
Answer : ...................................
4
11.
The points A(t , t ) , P ( r ,2) and B (−9,−2) are on a straight line. P divides AB
internally in the ratio of 3 : 4 . Find the value of t and of r .
[3 marks]
11
Answer : t = ..............................
r = ..............................
3
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
SULIT

8. MOZ@C
12.
For
examiner’s
use only
Diagram 3 shows a straight line PQ with equation 4 x + 3 y − 12 = 0 .
y
Q(0,h)
P(k,0)
O
x
Diagram 3
Find
(a) the value of h and of k
(b) the equation of PQ in intercept form.
[3 marks]
Answer : (a) ....……………...………..
(b) .........................................
12
3
13.
Find the equation of the straight line that passes through a point P ( −3,1) and is
x y
+ = 1.
perpendicular to the straight line
5 4
[4 marks]
13
Answer: …...…………..….......
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[ Lihat sebelah
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4

9. MOZ@C
For
examiner’s
use only
14. The coordinates of point P and Q are (−3,1) and (6,10) respectively. The point X moves
such that XP : XQ = 2 : 3 . Find the equation of the locus of X .
[3 marks]
14
Answer: …...….………..….....................
3
___________________________________________________________________________
15.
Table 1 shows the distribution of the weight of 40 pupils in form 4 Alpha.
Weight (kg)
31 – 35
36 – 40
41 – 45
46 – 50
51 – 55
56 – 60
61 – 65
Number of pupils
7
4
8
7
6
4
4
Table 1
(a) Find the range of the weight.
(b) Without drawing an ogive, calculate the median of the distribution of weight.
[4 marks]
15
Answer : (a)
4
…………………….
(b) ….……………….....
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
SULIT

10. MOZ@C
For
examiner’s
use only
16. The mean of ten numbers is m . The sum of squares of the number is k and the
standard deviation is 4. Express k in terms of m .
[2 marks]
16
Answer : .…………………
2
17.
Diagram 4 shows a circle with centre O.
A

O
B
Diagram 4
The length of the minor arc AB is 3.9275 cm and the angle of the major sector AOB is
315o. Using π = 3.142 , find
(a) the value of θ , in radians,
(Give your answer correct to four significant figures.)
(b) the length, in cm, of the radius of the circle.
[3 marks]
17
Answer:
……..…….…………...
___________________________________________________________________________
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[ Lihat sebelah
SULIT
3

11. MOZ@C
For
examiner’s
use only
Diagram 5 shows a sector OTV of a circle, centre O.
T
7 cm
50°
O
V
Diagram 5
Find the perimeter of the shaded region.
[3 marks]
18
Answer: ...………………………
3
___________________________________________________________________________
19.
Diagram 6 shows a sector of a circle OPQ with centre O and OPR is a right angle
triangle.
P
5 cm
O
R 1 cm Q
Diagram 6
Find the area, in cm2, of the shaded region.
[4 marks]
19
4
Answer:………………………
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
SULIT

12. MOZ@C
SULIT
20.
3472/1
For
examiner’s
use only
Evaluate the following limits,
(a) lim
x →1
(b)
1
x +1
x2 − 9
x →3 x − 3
lim
[ 3 marks ]
Answer: (a) ....…………..….......
20
(b) .................................
3
___________________________________________________________________________
21. The straight line y = −2 x + 1 is the tangent to the curve y = x 2 − 4 x at the point P.
Find the x-coordinate of the point P .
.
[2 marks]
21
Answer: ……………………..
2
22.
Given that y = px 2 −
dy
4
q
= x + 2 , where p and q are constants , find the
and
dx
x
x
value of p and of q .
[3 marks]
22
3
Answer: ……………………..
3472/1
2008 Hak Cipta SBP
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[ Lihat Sebelah
SULIT

13. MOZ@C
SULIT
For
examiner’s
use only
23.
Given the curve
y = x+
3472/1
1
.
2x2
(a) Find the coordinates of the turning point.
(b) Hence, determine whether it is a maximum or a minimum point.
[4 marks]
Answer: (a)…...…………..….......
23
(b)....................................
4
___________________________________________________________________________
24. A cylinder has a fixed height of 10 cm and a radius of 5 cm. If the radius decreases by
0.05 cm, find (in terms of π )
(a) the approximate change in the volume of the cylinder,
(b) the final volume of the cylinder.
[4 marks]
24
Answer: (a)……………………………
4
(b)……………………………
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
SULIT

14. MOZ@C
SULIT
25. Given y =
3472/1
For
examiner’s
use only
d2y
( x − 2) 2
when x = −1 .
, find the value of
dx 2
x
[4 marks]
25
Answer: …...…………..….......
END OF QUESTION PAPER
3472/1
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[ Lihat Sebelah
SULIT
4

15. MOZ@C
PEPERIKSAAN AKHIR TAHUN TINGKATAN 4 2008
MARK SCHEME KERTAS 1
No.
1
Solution and mark scheme
(a) k = 7
(b) f : x → x + 2
or
Sub marks
1
f ( x) = x + 2
1
(a) h = - 3
1
(b) - 2
2
3
B2 :
g −1 ( x) =
1− x
2
2
(a) g ( x) = 3 x + 1
B1 :
4
2 g ( x) − 1 = 6 x + 1
2
(b) x = 1
B1 :
4
4
4(−1)
−1+ 3
B1 :
3
Full marks
2
3(2 x − 1) + 1 = 4
p = 2, k = 7 (both)
4
4
2
2
k = -4
1
2
p=4
1
B3 :
p + 3 = 5 and 3p = k – 1
B2 :
p+3=5
or
3p = k – 1
B1 : x 2 − 5 x + k − 1 = 0
5
p>
1
2
B1 :
6
(−2) 2 − 4(2)(− p + 1) > 0
2
SMS MUZAFFAR SYAH , MELAKA

16. MOZ@C
7
8
4
3
5 x = 5 −1
5 x × 5 2 − 5 x × 51 = 4
x + 2y
3y
B2 :
B1 :
x=−
log p 3 + log p 2 + log p 2
3 log p 2
log p (3 × 2 × 2)
log p 2 3
3
2
B3 :
3x + 2 = x − 1
B2 :
3x + 2
= 30
x −1
B1 :
11
4
3
x = -1
B1 :
10
3
2 x 2 − 5x − 3 ≥ 0
B2 :
9
3
(2 x + 1)( x − 3) ≥ 0
B1 :
3
3
B2 :
3
3
1
(both)
2
x ≥ 3, x ≤ −
log 3
3x + 2
=0
x −1
r = -1
B2 :
t=5
B1 :
2=
− 6 + 4t
4t − 27
or r =
7
7
3
SMS MUZAFFAR SYAH , MELAKA

17. MOZ@C
a) h = 4
1
k=3
12
1
b)
x y
+ =1
3 4
y=
5
19
x+
or equivalent
4
4
3
1
4
4
3
3
(a) 30
1
4
(b) 46.21
3
13
B3 :
B2 :
m2 =
B1 :
14
( y − 1) =
m1 = −
5
( x + 3)
4
5
4
4
5
5 x 2 + 5 y 2 + 102 x + 62 y − 454 = 0
B2 :
B1 :
15
3 ( x + 3) 2 + ( y − 1) 2 = 2 ( x − 6) 2 + ( y − 10) 2
3XP = 2XQ
B2 :
B1 :
16
 40

 2 − 19 
m = 45.5 + 
5 (all values correct)
 7 




45.5 ,19 , 7 , 5 (at least two are correct)
2
k = 160 + 10m 2
B1 :
16 =
k
− m2
10
4
SMS MUZAFFAR SYAH , MELAKA
2

18. MOZ@C
a) 0.7855
1
b) r = 5
17
2
B1 :
18
3
3.9275 = r (0.7855)
Perimeter = 12.0263
3
3
4
4
1
2
1
3
(b) 6
2
B2 :
B1 :
19
length PQ = 2r sin 25 0 = 5.9167
arc PQ = r = 7 (0.8728) = 6.1096
Area = 2.045
B3 :
B2 :
 = 0.6436 rad
B1 :
20
1 2
1
(5 )(0.6436) − (4)(3)
2
2
tan  =
(a)
( x + 3)( x − 3)
x →3
x−3
B1 :
21
3
4
lim
2
B1 :
22
1
and q = 4
2
3
dy
= 2x − 4
dx
p=
2
3
x =1
1
or q = 4
2
B2 :
p=
B1 :
dy
q
= 2 px + 2
dx
x
5
SMS MUZAFFAR SYAH , MELAKA

19. MOZ@C
23
 3
(a) 1, 
 2
B1 :
2
1 − x −3 = 0
d2y
= 3 > 0 , minimum point
(b)
dx 2
B1 :
24
(b)
B1 :
25
2
d2y
= 3x −4
dx 2
2
(a) − 5π
B1 :
4
4
dV
= 20π (5) or 100 π
dr
2
245π
Vnew = 10π (5) 2 − 5π
4
−8
B3 :
x 2 (2 x) − ( x 2 − 4)(2 x)
d2y
= 8 x −3 or
x4
dx 2
B2 :
2 x ( x − 2) − ( x − 2) 2
dy
= 1 − 4 x − 2 or
dx
x2
B1 :
y = x − 4 + 4 x −1
END OF MARK SCHEME
6
SMS MUZAFFAR SYAH , MELAKA
4

20. MOZ@C
3472/2
Additional
Mathematics
Kertas 2
2 ½ jam
OKT 2008
SEKTOR SEKOLAH BERASRAMA PENUH
BAHAGIAN PENGURUSAN
SEKOLAH BERASRAMA PENUH / KLUSTER
KEMENTERIAN PELAJARAN MALAYSIA
PEPERIKSAAN AKHIR TAHUN
TINGKATAN 4 2008
ADDITIONAL MATHEMATICS
Kertas 2
Dua jam tiga puluh minit
JANGAN BUKA KERTAS SOALAN INI SEHINGGA DIBERITAHU
1. This question paper consists of three sections : Section A, Section B and Section C.
2. Answer all question in Section A , four questions from Section B and two questions from
Section C.
3. Give only one answer / solution to each question..
4. Show your working. It may help you to get marks.
5. The diagram in the questions provided are not drawn to scale unless stated.
6. The marks allocated for each question and sub-part of a question are shown in brackets..
7. A list of formulae is provided on pages 2 to 3.
8. A booklet of four-figure mathematical tables is provided.
9. You may use a non-programmable scientific calculator.
Kertas soalan ini mengandungi 13 halaman bercetak
3472/2
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[ Lihat sebelah
SULIT

21. 2
MOZ@C
The following formulae may be helpful in answering the questions. The symbols given are the
ones commonly used.
ALGEBRA
1
5
− b ± b 2 − 4ac
2a
x=
log a mn = log a m + log a n
m
= log a m − log a n
n
log a m n = n log a m
log b x
log a x =
log b a
2
a m × a n = a m+ n
6
3
a m ÷ a n = a m−n
7
4
(a )
8
m n
= a mn
log a
KALKULUS (CALCULUS)
1
y = uv ,
dy
dv
du
=u +v
dx
dx
dx
u dy
=
,
v dx
2 y=
v
3
dy dy du
=
×
dx du dx
du
dv
−u
dx
dx
2
v
STATISTIK (STATISTICS)
1
x=
2 x=
x
5 I=
N
 fx
f
1N −F
C
6 M = L+ 2
 f

m


 (x − x )
2
3 σ =
Q1
× 100
Q0
N
= =
x
N
2
− x2
−
7 I =
W I
W
i i
i
4 σ =
3472/2
 f (x − x )
f
2
=
 fx
f
2
− x2
2008 Hak Cipta SBP
SULIT
SMS MUZAFFAR SYAH , MELAKA

22. 3
MOZ@C
GEOMETRI (GEOMETRY)
1.
3.
Jarak (Distance)
2
Titik yang membahagi suatu
tembereng garis
2
(x1 − x2 ) + ( y1 − y2 )
2.
(A point dividing a segment of a line)
 nx + mx 2 ny1 + my 2 
( x, y ) =  1
,

m+n 
 m+n
Titik tengah (Midpoint)
 x + x y + y2 
( x, y ) =  1 2 , 1

2 
 2
4.
Luas segitiga (Area of triangle) =
1
(x1 y 2 + x2 y3 + x3 y1 ) − (x 2 y1 + x3 y 2 + x1 y3 )
2
TRIGONOMETRI (TRIGONOMETRY)
1. Panjang lengkok, s = j
Arc length, s = r
1
2. Luas Sektor, L = j 2 
2
1
Area of sector, A = r 2 
2
3.
4.
a
b
c
=
=
sin A sin B sin C
a 2 = b 2 + c 2 − 2bckos A
a 2 = b 2 + c 2 − 2bccos A
5.
Luas segitiga (Area of triangle)
= 1 ab sin C
2
3472/2
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[ Lihat sebelah
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23. 4
MOZ@C
SECTION A
[40 marks]
Answer all questions in this section
1.
Solve the simultaneous equations
5 x + 2 y = −4
x 2 − 3 x + 2 y = 16
[5 marks]
2.
Express f (x) = 1 – 6x + 2x2 in the form f (x) = m(x + n)2 + k , where m, n and k are
constants.
(a)
State the values of m, n and k.
[3 marks]
(b)
Find the maximum or minimum point.
[1 marks]
(c)
2
Sketch the graph of f (x) = 1 – 6x + 2x .
[2 marks]
3.
(a)
The straight line y = 1 – tx is a tangent to the curve y2 – 3y + 3x – x2 = 0.
Find the possible value of t.
[3 marks]
(b)
Given p and q are the roots of the quadratic equations 2x2 + 7x = m – 5,
where pq = 3 and m is a constant. Calculate the values of m, p and q.
[4 marks]
4
(a)
Given that 92 x. 27 y = 1 , find the value of x when y = −4
[2 marks]
(b)
Solve the equation 3x.6 x = 183 x− 2
[2 marks]
(c)
Solve the equation log 2 ( x − 2) = 2 + 2 log 4 (4 − x)
[3 marks]
3472/2
2008 Hak Cipta SBP
SULIT
SMS MUZAFFAR SYAH , MELAKA

24. 5
MOZ@C
5
(a)
A set of N numbers have a mean of 8 and standard deviation of 2.121. Given that
the sum of the numbers,  x , is 64. Find
(i) the value of N
(ii) the sum of the squares of the numbers.
[4 marks]
(b)
If each of the numbers is divided by h and is added by k uniformly , the new mean
and standard deviation of the set are 5 and 1.0605 respectively.
Find the value of h and k.
[4 marks]
6.
The gradient of the tangent to the curve y = px − qx 2 , where p and q are constants, at
the point (1 , 4) is 2.
Find
(a) the value of p and q.
[4 marks]
(b) the equation of normal to the curve at the point ( 2, 4)
[3 marks]
3472/2
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[ Lihat sebelah
SULIT

25. 6
MOZ@C
SECTION B
Answer four questions in this section
7
Diagram 1 shows function g maps x to y and function h maps z to y.
1
Given g (x) =
, x ≠ m and h (z) = 1 + 4z.
2x −1
g
x
h
y
z
a
b
Diagram 1
(a)
State the value of m.
(b)
Find
[1 marks]
(i)
(ii)
gh -1(x)
h g -1(1).
[6 marks]
(c)
Find the value of
(i)
(ii)
a
b
[3 marks]
3472/2
2008 Hak Cipta SBP
SULIT
SMS MUZAFFAR SYAH , MELAKA

26. 7
MOZ@C
8.
Diagram 2 shows a quadrilateral PQRS with vertices R (−2,5) and S (−1,1) .
Q
R(−2, 5)
S(−1, 1)
O
P
Diagram 2
Given the equation of PQ is 4 y = 7 x − 14 . Find ,
(a)
the equation of QR
[4 marks]
(b)
the coordinates of Q
(c)
the coordinates of P
(d)
the area of quadrilateral PQRS
[2 marks]
[1 marks]
[3 marks]
3472/2
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[ Lihat sebelah
SULIT

27. 8
MOZ@C
9
Total Time (hours)
Number of students
5 – 14
12
15 – 24
17
25 – 34
26
35 – 44
31
45 – 54
16
55 – 64
10
65 – 74
8
Table 1
Table 1 shows the total time spent on watching television by 120 students for a period of
3 weeks. Calculate ,
(a)
the mean,
(b)
the standard deviation,
[2 marks]
(c)
the third quartile,
[3 marks]
[5 marks]
of the distribution
3472/2
2008 Hak Cipta SBP
SULIT
SMS MUZAFFAR SYAH , MELAKA

28. 9
MOZ@C
10. Diagram 3 shows a circle ABCF with radius 6 cm and centre O.
D
A
B
O
F
C
E
Diagram 3
Given that ∠ODB = 30o , EBD is the tangent to the circle and OD = OE = 12 cm.
Find ,
(a)
the length of BD
(b)
the area of shaded region
(c)
the perimeter of the whole diagram
[3 marks]
[3 marks]
[4 marks]
3472/2
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[ Lihat sebelah
SULIT

29. 10
MOZ@C
11
Diagram 4 shows a rectangle PQRS and a parallelogram ABCD.
P
x cm
Q
B
x cm
C
6 cm
A
x cm
S
D
x cm
R
10 cm
Diagram 4
(a)
If L cm 2 is the area of the parallelogram,
(i)
Show that L = −2 x 2 + 16 x
(ii)
Find the value of x when L is maximum.
(iii) Find the maximum area of the parallelogram ABCD.
[7 marks]
(b)
3472/2
Given that the rate of change of x is 0.1 cms -1 , find the rate of change of L ,
in cms -1 , when x is 2 cm.
[3 marks]
2008 Hak Cipta SBP
SULIT
SMS MUZAFFAR SYAH , MELAKA

30. 11
MOZ@C
SECTION C
Answer two questions in this section
12. Diagram 5 shows two triangles ABC and ACD . BCD is a straight line.
C
B
55
D
o
9 cm
8 cm
Diagram 5
8 cm
A
Find ,
(a) ∠ADC
[3 marks]
(b)
the length of CD
(c)
the area of triangle ABD
[3 marks]
[4 marks]
13
A
B
50o
1040
C
100o
E
F
Diagram 6
Diagram 6 shows two triangles ABE and BCF , where ABC is a straight line.
Given that AE = 5 cm, BE = 7 cm, BC = 8 cm, CF = 9 cm, ∠BAE = 500 , ∠EBF = 1040
and ∠BCF = 1000. Calculate
(a)
∠AEB ,
[3 marks]
(b)
the length of BF.
[3 marks]
(c)
3472/2
if point E joint with F, find the area of the quadrilateral ACFE.
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[4 marks]
[ Lihat sebelah
SULIT

31. 12
MOZ@C
14
Table 2 shows the prices and the price indices of five types of food, H, I, J, K and L
represented the cost of food. Diagram 7 shows a percentage according to the food’s
pyramid.
Types of food
Price (RM) for the year
Price index for the year 2008
based on the year 2006
2006
2008
H
I
J
K
2.20
m
5.00
3.00
2.75
2.20
7.50
2.70
125
110
150
n
L
2.00
2.80
140
Table 2
L
I
10 %
K
20 %
J
25 %
H
40 %
Diagram 7
(a)
(b)
(c)
3472/2
Find the value of m and of n.
[3 marks]
Calculate the composite index for the cost of food in the year 2008 based on the
year 2006.
[3 marks]
The price of each food increases by 30% from the year 2008 to the year 2009.
Given that the cost of food in the year 2006 is RM80, calculate the corresponding
cost in the year 2009.
[4 marks]
2008 Hak Cipta SBP
SULIT
SMS MUZAFFAR SYAH , MELAKA

32. 13
MOZ@C
15
Item
Price Index for the year 2005
Based on the year 2003
Percentage of usage (%)
J
118
22
K
t
12
L
108
31
M
113
35
Table 3
Table 3 shows the price indices and percentage of usage of four items,
J, K, L and M, which are the main ingredients in the production of a brand of cake.
(a)
Calculate
(i)
the price of item M in the year 2003 if its price in the year 2005
was RM2.50.
(ii)
(b)
the price index of item J for the year 2005 based on the year 2001 if
its price index for the year 2003 based on the year 2001 is 108.
[5 marks]
The composite index of the cost of cake production for the year 2005
based on the year 2003 is 113. Calculate ,
(i)
the value of t
(ii)
the price of a cake in the year 2003 if its corresponding price in the year
2005 was RM 25,






[ 5 marks ]
END OF THE QUESTIONS
3472/2
2008 Hak Cipta SBP
SMS MUZAFFAR SYAH , MELAKA
[ Lihat sebelah
SULIT

33. 2
MOZ@C
SKEMA PERMARKAHAN MATEMATIK TAMBAHAN KERTAS 2
PEPERIKSAAN DIAGNOSTIK TINGKATAN 4, 2008
Number
1
Solution and mark scheme
Sub Marks
Full Marks
P1
−4 − 5 x
−4 − 2 y
y=
@x =
2
5
 −4 − 5 x 
x 2 − 3x + 2 
 = 16
 2 
K1
2
 −4 − 2 y 
 −4 − 2 y 
@
 − 3
 + 2 y = 16
 5 
 5 
( x − 10 )( x + 2 ) = 0 @ ( y + 27 )( y − 3 ) = 0
x = 10, −2 and y = −27,3
K1
N1, N1
5
2
5
(a)


2 x −
3
2
m=2,n=–
(b)
K1
7
 −
2
2
3
2
and k = –
7
N 0, 1, 2
2
3
7 3
(− , )
2 2
N1
1
P1
P1
2
y
(c)
1
x
o
•
7
( ,– )
3
2
2
Shape
Minimum point and y-intercept
6
SMS MUZAFFAR SYAH , MELAKA

34. 3
MOZ@C
Number
3
(a)
Solution and mark scheme
(1 – tx)2 – 3(1 – tx) + 3x – x2 = 0
(t2 – 1)x2 + (t + 3)x – 2 = 0
Full Marks
K1
(t + 3)2 – 4(t2 – 1)(–2) = 0
(3t + 1) (3t + 1) = 0
1
t=− .
3
(b)
Sub Marks
K1
N1
SOR or POR pq or p + q
POR: pq = 3
5−m
=3
2
m = −1.
3
K1
N1
.
SOR : p + q = −
7
2
3
q = − or − 2
2
3
p = −2or −
2
.
4
N1
N1
4
7
(a)
(32 ) 2 x ⋅ (33 ) y = 30
4x + 3 y = 0
K1
x=3
N1
18x = 183 x− 2
x = 3x − 2
K1
x =1
N1
2
(b)
(c)
log 2 ( x − 2) = 2 + 2
log 2 ( 4 − x )
log 2 4
Change base
2
K1
log2 ( x − 2) = log 2 4 + log2 (4 − x)
( x − 2) = 4(4 − x )
x = 3.6
K1
N1
3
7
SMS MUZAFFAR SYAH , MELAKA

35. 4
MOZ@C
Solution and mark scheme
Number
5
(a)
64
N
N =8
8=
i)
Sub Marks
Full Marks
K1
N1
ii)
2.121 =
x
2
8
 x = 547.99
− 82
K1
2
(b)
New mean =
N1
4
8
+k =5
h
or
New Stnd. Deviation =
2.121
= 1.0605
h
h =2
N1
8
+k =5
2
k =1
6
(a)
dy
= p − 2qx
……………
dx
2 = p − 2q (1)
Substitute ( 1, 4 ) in y = px − qx 2
4 = p – q …………
 –  ; q = 2 and p = 6
(b)
x = 2,
K1
dy
= 6 − 4( 2)
dx
K1
N1
4
8
K1
K1
K1
N1
4
K1
∴ m1 = −2
m2 =
1
2
y−4 =
y=
1
( x − 2)
2
1
x+3
2
K1
N1
3
7
SMS MUZAFFAR SYAH , MELAKA

36. 5
MOZ@C
Number
7
(a)
(b)
Solution and mark scheme
m=
(i)
1
.
2
P1
z −1
4
h −1 ( z ) =
gh −1 ( x) =
=
(ii)
Sub Marks
Full Marks
1
K1
1
 x −1
2
 −1
 4 
K1
2
, x ≠ 3.
x−3
N1
1 = g(x)
1
2x −1
x =1
1=
hg −1 (1) = h(1) = 1 + 4(1)
K1
K1
−1
hg (1) = 5
(c)
N1
6
(i)
1
a = g( )
4
1
a=
1
2( ) − 1
4
a = −2
K1
N1
(ii)
h (b ) = a
1 + 4b = −2
3
b=−
4
SMS MUZAFFAR SYAH , MELAKA
N1
3
10

37. 6
MOZ@C
Number
Solution and mark scheme
Sub Marks
(a)
(b)
M RS = −4
1
M QR =
4
1
y − 5 = ( x + 2 ) or other suitable method
4
4 y = x + 22 or equivalent
P1
Solve simultaneous equation
x + 22 = 7 x − 14
Q(6, 7)
K1
N1
2
P (2, 0)
8
N1
Full Marks
1
P1
K1
N1
4
(c)
(d)
Area of quadrilateral PQRS
=
1 2 6 −2 −1 2
2 0 7 5 1 0
K1
1
[2(7) + 6(5) + (−2)(1) + (−1)(0)] − [6(0) + (−2)7 + (−1)5 + 2(1)]
2
= 29.5
=
9
K1
N1
3
10
(a)
x=
12(9.5) + 17(19.5) + 26(29.5) + 31(39.5) + 16(49.5) + 10(59.5) + 8(69.5)
120
= 36.5
(b)
(c)
K1
N1
12(9.5) 2 + 17(19.5) 2 + ... + 8(69.5) 2
σ=
− (36.5) 2
120
= 266
= 16.31
L = 44.5 ; F = 86
3

 4 (120) − 86 
Q3 = 44.5 + 
10 K1 lower boundry 44.5
16




K1 using formula
= 44.5 + 2.5 = 47
SMS MUZAFFAR SYAH , MELAKA
2
K1
K1
N1
3
P1, P1
K1
K1
N1
5
10

38. 7
MOZ@C
Number
10
Solution and mark scheme
Sub Marks
Full Marks
(a)
∠DOB = 600 or
DB
= Sin60°
12
DB
= tan 60 0
6
DB = 10.3923
(b)
K1
N1
3
DE = 2 ×10.3923
= 20.7846
N1
Area of shaded region
1
1
2
= ( 20.7846 )( 6 ) − ( 2.0944 )( 6 )
2
2
= 24.6546
(c)
P1
Major ∠AOC = 2400 or 4.1888 rad
S AFC = 6 ( 4.1888 )
= 25.13228
Perimeter = S AFC + AD + CE + DE
= 25.1328 + 6 + 6 + 20.7846
= 57.9174
SMS MUZAFFAR SYAH , MELAKA
K1
N1
3
P1
N1
K1
N1
4
10

39. 8
MOZ@C
Number
11
Solution and mark scheme
Sub Marks
Full Marks
(a)
(i)
 1  1

L = 10(6) − 2 x 2  − 2  (10 − x )(6 − x ) 
 2  2

= − 2 x 2 + 16 x
K1
N1
(ii)
dL
= −4 x + 16
dx
dL
= 0,
dx
− 4 x + 16 = 0
K1
x=4
(iii)
K1
N1
Lmax = −2(4) 2 + 16(4)
= 32
(b)
K1
N1
dx
= 0.1
dt
P1
dL
= ( −4(2) + 16) × 0.1
dt
= 0.8
7
K1
SMS MUZAFFAR SYAH , MELAKA
N1
3
10

40. 9
MOZ@C
Number
12
(a)
Solution and mark scheme
9
8
=
Sin∠BCA Sin55
0
K1
∠ADC = 67 09 ' or 67.15°
N1
N1
3
∠CAD = 180 − 2(6709')
= 450 42 ' or 45.7°
CD 2 = 82 + 82 − 2(8)(8) cos 450 42 '
CD = 6.213
(c)
Full Marks
0
∠BCA = 112 51' or ∠ACD = 67 9 '
(b)
Sub Marks
∠CAB = 1800 − 550 − 112051'
= 1209 '
1
1
( 9 )( 8 ) sin1209 '+ (8 )( 8 ) sin 450 42 '
2
2
= 30.48
Area =
K1
K1
N1
3
N1
K1, K1
N1
4
10
SMS MUZAFFAR SYAH , MELAKA

41. 10
MOZ@C
Number
13
Solution and mark scheme
Sub Marks
Full Marks
(a)
5
7
=
SinB Sin50°
∠ABE = 33.17°or 33°10'
K1
K1
∠AEB = 180 − 50 − 33.17°
= 96.83º or 96º 50’
N1
∠CBF = 42.83°or  42°50'
BF
9
=
sin100° sin 42.83°
P1
BF = 13.04 cm
(b)
N1
3
K1
3
OR equivalent
(c)
1
× 5 × 7 sin 96.83° or equivalent
2
= 17.38
Area AEB =
1
× 8 × 9sin100°
2
or equivalent
= 35.45
K1
Area BCF =
1
× 7 × 13.04sin104°
2
= 44.28
K1
Area BEF =
Area of quadrilateral ACFE = 100.08
SMS MUZAFFAR SYAH , MELAKA
K1
N1
4
10

42. 11
MOZ@C
Number
14
Solution and mark scheme
Sub Marks
Full Marks
(a)
2.20
× 100 = 110
m
m = 2.00
2.70
× 100 = n
3.00
n = 90
(b)
 IW = (125× 40) + (110 ×10) + (150 × 25) + (90 × 20) + (140 × 5)
100
W
12350
100
= 123.5
I 09 =
06
N1
N1
3
K1
K1
=
(c)
K1
N1
Q
130 123.5
×
×100OR  08 ×100 = 123.5
100 100
80
= 160.55Q08 = RM 98.80
Q
160.55
× RM 80  09 × 100 = 130
100
98.80
= RM 128.44.Q09 = RM 128.44
Q09 =
K1
K1
K1
N1
SMS MUZAFFAR SYAH , MELAKA
3
4
10

43. 12
MOZ@C
Number
15
(a)
Solution and marking scheme
Sub Marks
Full Marks
(i)
2.50
× 100 = 113
P03
P03 = RM 2.21
K1
N1
(ii)
P03
× 100 = 108
P01
P05
× 100 = 118
P03
 118  108 


 × 100 = 127.44
 100  100 
K1, K1
N1
5
(b)
−
118( 22) + t (12) + 108(31) + 113(35)
= 113
100
t = 116.75
(i) I =
(ii)
25
× 100 = 113
P03
P03 = RM 22.12
K1, K1
N1
K1
N1
5
10
END OF MARK SCHEME
SMS MUZAFFAR SYAH , MELAKA