1. SULIT 1449/2
1449/2
Mathematics NAMA :
Kertas 2 TINGKATAN :
Mei 2010
1
2 jam
2
UNIT PEPERIKSAAN & PENILAIAN
SMK TUNKU BESAR BURHANUDDIN
71550 SERI MENANTI, N. SEMBILAN
PEPERIKSAAN PERTENGAHAN TAHUN
TINGKATAN 5 2010
Pemeriksa
MATHEMATICS Markah Markah
Bahagian Soalan
Kertas 2 Penuh Diperoleh
1 4
Dua jam tiga puluh minit
2 4
3 3
JANGAN BUKA KERTAS SOALAN INI 4 3
SEHINGGA DIBERITAHU
5 5
1.Kertas soalan ini mengandungi dua bahagian : A 6 6
Bahagian A dan Bahagian B. Jawab semua
soalan daripada Bahagian A dan empat soalan 7 5
dalam Bahagian B. 8 5
2.Jawapan hendaklah ditulis dengan jelas dalam 9 6
ruang yang disediakan dalam kertas soalan. 10 6
Tunjukkan langkah-langkah penting. Ini boleh
membantu anda untuk mendapatkan markah. 11 5
12 12
3.Rajah yang mengiringi soalan tidak dilukis
mengikut skala kecuali dinyatakan. 13 12
B 14 12
4.Satu senarai rumus disediakan di halaman 2 & 3
15 12
5.Anda dibenarkan menggunakan kalkulator
saintifik yang tidak boleh diprogram.
Jumlah
Kertas soalan ini mengandungi 14 halaman bercetak
1449/2@2010 [Lihat sebelah
Di Sediakan Oleh Di Semak Oleh
SULIT
______________________
(Sunita binti Mukhtar )
Panatia Matematik SMKTBB

2. Untuk
SULIT Kegunaan
Pemeriksa
Rumus-rumus berikut boleh membantu anda menjawab soalan. Simbol-simbol yang diberi adalah yang
biasa digunakan.
PERKAITAN
1. a m × a n = a m +n .
2. a m ÷ a n = a m −n
3. (a )
m n
= a mn
1  d −b
4. A−1 =  
ad − bc  − c
 a 

n( A)
5. P ( A) =
n( S )
6. P ( A' ) = 1 − P ( A)
7. Jarak = ( x1 − x2 ) 2 + ( y1 − y2 ) 2
8. Titik Tengah
( x, y ) =  x1 + x2 , y1 + y2 
 
 2 2 
jarak yang dilalui
9. Purata laju = masa yang diambil
hasil tambah nilai data
10. Min = bilangan data
hasil tambah( nilai titik tengah × ke ker apan )
11. Min = hasil tambah ke ker apan
12. Teorem Pithagoras
c 2 = a 2 + b2
y2 − y1
13. m=
x2 − x1
p int asan − y
14. m =−
p int asan − x
BENTUK DAN RUANG
1
1. Luas trapezium = × hasil tambah dua sisi selari × tinggi
2
2 [Lihat sebelah
1449/2 SULIT

3. SULIT
Untuk
Kegunaan
Pemeriksa 2. Lilitan bulatan = πd = 2πj
3. Luas bulatan = πj2
4. Luas permukaan melengkung silinder = 2πjt
5. Luas permukaan sfera = 4πj2
6. Isipadu prisma tegak = luas keratan rentas × panjang
7. Isipadu silinder = πj2t
1 2
8. Isipadu kon = πj t
3
4 3
9. Isipadu sfera = πj
3
1
10. Isipadu piramid tegak = × luas tapak × tinggi
3
11. Hasil tambah sudut pedalaman poligon = ( n − 2 ) × 180 0
panjang lengkok sudut pusat
12. =
lili tan bula tan 360 0
luas sektor sudut pusat
13. =
luas bula tan 360 0
PA '
14. Faktor skala, k =
PA
15. Luas imej = k 2 × luas objek
Section A
[52 marks]
Answer all questions in this section
1449/2 3 SULIT

4. Untuk
SULIT Kegunaan
Pemeriksa
1 Using factorisation, solve the following quadratic equation :
h2 = 9( h – 1)
2
[4 marks]
Answer:
2 Calculate the value of v dan w taht satisfy the following simultaneous linear equations :
v + 2 w = 1 and 4v − 3w = −18
[4 marks]
Answer :
3 Venn Diagram in diagram 1(a) and 1(b) shows set ξ, P, Q dan R . In diagram provided in the
answer spances. shade
(a) P ∩ (Q ∪ R)′ (b) P ′ ∩ (Q ∩ R)
Jawapan :
(a) (b)
ξ ξ
P P P P
Q Q
R R
Diagram 1(a) Diagram 1(b) [3 marks]
4 The diagram shows a solid formed by joining a cone with hemisfere
The height of the con is 7cm, Diameter of the cone and the hemisfere is 6 cm
4 [Lihat sebelah
1449/2 SULIT

5. SULIT
Untuk
Kegunaan
Pemeriksa
Diagram 2
22
Using π = , Calculate the volume, in cm3 , of the solid
7
Answer :
5.
y
K
M( -3, 7)
L
N
x
In Diagram 2, O is the origin. Straight line KL is parallel to straight line MN. The equation of
straight line KL is 2x + y = 4
The points L and N lie on the y- axis
Find
(a) The equation of straight line MN
1449/2 5 SULIT

6. Untuk
SULIT Kegunaan
Pemeriksa
(b) The x –intercept of the straight line MN
[5 marks]
Jawapan : (a)
(b)
(c)
6
3 4  1 0
(a) M is a 2 X 2 matrix where M  =
÷  ÷ find the matrix of M
,
2 5  0 1
(b) Write the following simultaneous linear equations as matrix equation :
3 p + 4q = 5
2 p + 5q = 8
Hence, using matrix method, calculate the value of p and q
[6 marks]
Answer :
(a)
(b)
.
7. Diagram 4 shows the speed-time graph of a particle for a period of 18 seconds
6 [Lihat sebelah
1449/2 SULIT

7. SULIT
Untuk
Kegunaan Speed(ms/s)
Pemeriksa
22
u
4
6 10 18 Time(s)
Diagram 4
a) State the the length time, in seconds, the particle moves with uniform speed
b) Calculate the rate of change, in ms -2, in the first 6 seconds
c) Calculate the value of u, if the average speed of the particle in the last 12 seconds
[5marks]
Answer :
a)
b)
c)
8 Diagram 5 shows the distance – time of the journey of a car from Pulau Pinang to Johor Bahru over a
period of t hours.
Distance (km)
800
300
Time (hour)
0 4 2 t
(a) State the length of time, that car is stationary
(b) State the distance in km, from Johor Bahru when the car is stationary
(c) Calculate the car average speed, in km h-1, during the 6 first hours
(d) If the average speed of the car for the whole journey is 80km/h
Calculate the value of t
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8. Untuk
SULIT Kegunaan
Pemeriksa
[5 marks]
Answer :
(a)
(b)
(c
d)
9.
Diagram 6
Diagram 6,shows OKJH with the centre O and a circle FGH with centre M.
F is the midpoint of OH and OK = 28
22
Using π = , calculate
7
(a) The perimeter in cm of the whole diagram
(b) The area cm of the shaded region
[6 marks]
Answer
a)
8 [Lihat sebelah
1449/2 SULIT

9. SULIT
Untuk
Kegunaan
Pemeriksa
b)
10 Diagram 7 shows three numbered cards in box P and four numbered in box Q
5 15 19 3 6 15 17
Box P Box Q
A card is picked at random from each of the boxes
By Listing the possible outcomes, find the probability that
a) a card with a number which is a multiple of 2 or a multiple of 5 are picked
b) a card with a number which is a multiple of 2 and a multiple of 5 are picked
[ 5 marks ]
Answer :
a)
b)
1449/2 9 SULIT

10. Untuk
SULIT Kegunaan
Pemeriksa
11 (a) ( I ) Complete the following mathematical sentence in the answer space by
10 5
( ii ) Complete the following statement the quantifier “all” or “ some” to make it a true
statement .
--------------------------- multiples of 3 are divisible by 2
[2 marks ]
(b) Complete the conclusion in the following argument
Premise 1 : If x is a factor of 12, then x is a factor of 24
Prermise 2 : x is not a factor of 24
Conclution : _____________________________________
[1 marks]
(c) Write down two implications based on the following sentence :
3m > 15 if and only if m > 5
[2 marks]
Answer :
(a) ( I )_____________________________________________________________________
( ii ) ____________________________________________________________________
(b)Conclusion:____________________________________________________________________
(c)Implication 1 : ______________________________________________________________
Implication 2 :______________________________________________________________
Section B
10 [Lihat sebelah
1449/2 SULIT

11. SULIT
Untuk
Kegunaan [48 marks]
Pemeriksa
Answer four questions from this section.
4
12 (a) Complete Table 1 in the answer space fo the equation y = −
x
x −4 −2.5 −1 -0.5 0.5 1 2 3.2 4
y 1 1.6 8 -8 -4 -1.25 -1
Table 1
[2 marks]
For this part of the question, use the graph paper provided on page 31. You may use a
flexible curve rule
(b) By using a scale of 2 cm to 1 unit on the x-axis and 2 cm kepada 2 unit on the-yaxis, draw the
4
graph y = − for −4 ≤ x ≤ 4 .
x
[5 marks]
( c ) From your graph, find
( i) the value of y when x = 1.8
(ii) the value of x when y = 3.4
[2 marks]
(a) Draw a suitable straight line on your graph to find all the value of x which satisfy the
4
equation y = − for −4 ≤ x ≤ 4
x
State these values of x [3 marks]
Answer :
(a)
x −4 −2.5 −1 -0.5 0.5 1 2 3.2 4
y 1 1.6 8 -8 -4 -1.25 -1
Table 1
(c) ( i ) y = ____________________
( ii) x = _____________________
(d) x = _______________________, _______________________
13 Diagram 7 shows rectangle D, E, F dan G.
1449/2 11 SULIT

12. Untuk
SULIT Kegunaan
Pemeriksa
y
7
D
6
5
G
4 E
3
2 F •
K
1
-2 -1 O 1 2 3 4 5 6 7 8 x
RAJAH 7
(a) Transformation P is a reflection about line x = 3and transformation R is clockwise rotation
of 900 about the point (4, 4). State the coordinate of image of point K under
(i) Transformation P
(ii) Transformation PR
[3 marks]
(b) Rectangle D is the image of rectangle of F under transformation V and rectangle E is the
image of rectangle D under transformation W.
Describe in full
(i) Transformation V
(ii) A single transformation which equavalent to transformation WV.
[5 marks]
(c) Rectangle G is the image of rectangle F under an enlargement
(i) State the centre coordinate of the centre of the enlargement
(ii) Given the area of rectangle F is 7 unit2, calculate the area of rectangle of G.
[4 marks]
Answer:
(a) (i)
(ii)
(b) (i)
(ii)
(c) (i)
(ii)
12 [Lihat sebelah
1449/2 SULIT

13. SULIT
Untuk
Kegunaan 14 The Data in diagram 8 shows the donations, in RM, of families to their children’s school welfared
Pemeriksa fund.
4 24 17 30 22 26 35 19
0
2 28 33 33 39 34 39 28
3
2 35 45 21 38 22 27 35
7
3 34 31 37 40 32 14 28
0
2 32 29 26 32 22 38 44
0
Diagram 8
(a) Using data in diagram 8 and a class interval of RM 5 , complete the table 2 in the answer space
[4 marks]
Answer :
Donations (RM) frequency Cumulative frequency
11 - 15
16 – 20
Table 2
(b) For this part of the questions, use the graph paper provided
By using a scale of 2 cm to RM 5 on the x- axis and 2 cm to 5 families on the y-axis, draw
an ogive based on the data
[6 marks]
(c) From your ogive in (b)
( i ) Find the third quartile
(ii ) hence, explain briefly the meaning of the third quartile
Answer :
(c)
(i)
(ii)
1449/2 13 SULIT

14. Untuk
SULIT Kegunaan
Pemeriksa
15 (a) Complete Table 1 in the answer space fo the equation y = 6 − x 3
x −3 −2.5 −2 -1 0 1 2 2.5
y 33 21.63 14 6 5 -9.63
Table 1
[2 marks]
For this part of the question, use the graph paper provided on page 31. You may use a
flexible curve rule
(c) By using a scale of 2 cm to 1 unit on the x-axis and 2 cm kepada 5 unit on the-yaxis, draw the
graph y = 6 − x 3 bagi nilai x dalam julat −3 ≤ x ≤ 2.5 .
[4 marks]
( c ) From your graph, find
( i) the value of y when x = 1.5
(ii) the value of x when y = 10
[2 marks]
(b) Draw a suitable straight line on your graph to find all the value of x which satisfy the
equation x 3 − 8 x − 6 = 0 for −3 ≤ x ≤ 2.5 .
State these values of x [4 marks]
Answer :
(a)
x −3 −2.5 −2 -1 0 1 2 2.5
y 33 21.63 14 6 5 -9.63
Table 1
(c) ( i ) y = ____________________
( ii) x = _____________________
(d) x = _________________________ ,_________________________
14 [Lihat sebelah
1449/2 SULIT

15. SULIT
Untuk
Kegunaan
Pemeriksa
1449/2 15 SULIT