2009CEmock-Maths2 Q&A OscarTam

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2009 MATH Paper 2

Beacon College Pre-mock 311.15 am 12.45 pm (1 1 2 hours) Subject Code 180

By Oscar Tam

1.

Read carefully the instructions on the Answer Sheet and insert the information required (including the Subject Code) in the spaces provided. When told to open this book, you should check that all the questions are there. Look for the words END OF PAPER after the last question. All questions carry equal marks. ANSWER ALL QUESTIONS. You should mark all your answers on the Answer Sheet. You should mark only ONE answer for each question. If you mark more than one answer, you will receive NO MARKS for that question. No marks will be deducted for wrong answers.

2.

3. 4. 5.

6.

2009- CE MATH 2

Reference material, not for sale 1

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FORMULAS FOR REFERENCE SPHERE Surface area

= 4r 2 = 4r 24 3 r 3 4 = r 3 3 == 2rh = 2rh

Volume

CYLINDER

Area of curved surface Volume

= r 2 h = r 2 h= rl = rl

CONE

Area of curved surface Volume

1 = r 2 h 3 1 2 = r h 3= base area height =

PRISM PYRAMID

Volume Volume

1 = base area height 3 1 = 3

2009- CE MATH 2

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There are 36 questions in Section A and 18 questions in Section B. 36 18 The diagrams in this paper are not necessarily drawn to scale. Choose the best answer for each question. Section A

(1) If m and n are positive integers, then m n

(m + n )n (n + m )m

(m + n )n (n + m )m

=

=

A. B. C. D.1 x 1 1 x

m n (m + n )nm

(m + n ) mn

1

(2)

=

A. B. C. D.

x 1+ x x 1 x 1 x 1 x x +1

2009- CE MATH 2


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(3) If y =

x+2 , then x = x+3 x+2 y= x = x+3

A. B. C. D.

1 y+2 3y 2 y 1 2 3y y 1 2 3y y +1

(4) Solve x + 3 >

3x + 4 , where x is a positive integer. 2 3x + 4 x+3> x 2 A. B. C. D. 1 only 1 2 only 2 0, 1 0, 1, 2

(5) Let x km/h be the speed of the current. The speed of a boat in still water is 12 km/h. The boat travels 51 km upstream and 25 km downstream in a total time of 7 hours. Then x km/h 12 km/h 51 km 25 km 7 A. B. C. D. 51 25 12 x 12 + x 51 25 + 12 x 12 + x 51 25 12 + x 12 x 51 25 + 12 + x 12 x=7 =7 =7 =7

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(6) Which of the following may represent the graph of y = x 2 + ax a 2 , where a > 0 ? y = x 2 + ax a 2 a > 0 A.y

B.y

O

x O

x

C.y

D.y

O

x

O

x

(7) The length of a rectangle is 4 times its width. If the perimeter of the rectangle is 30 cm, find its area. 4 30 cm A. B. C. D. 24 cm 2 28 cm 2 36 cm 2 64 cm 2

2009- CE MATH 2


(8) If f ( x ) =

1 2 x + x , then f ( x 1) = 2 1 f ( x ) = x 2 + x f ( x 1) = 2

(

)

(

)

A. B. C. D.

x2 x 2 2 x x 1 2 2 x + x 1 2 2 x + x +1 2

(9) In the figure, the 1st pattern consists of 3 dots. For any positive integer n, the (n + 1)th pattern is formed by adding (n + 3) dots the nth pattern. Find the number of dots in the 8th pattern. 1 n (n + 1) n (n + 3) 8

...

A. B. C. D.

42 52 53 63

(10) The dimensions of a study room are 6.7 m 9.2 m corrected to 1 decimal place. Let A m 2 be the actual area of the study room. Then 6.7 m 9.2 m 1 A m 2 A. B. C. D.6.65 9.15 A < 6.75 9.25 6.68 9.18 A < 6.72 9.22 6.65 9.25 A < 6.75 9.15 79 A < 89

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(11) P sold an article to Q at a profit of 20%. Q sold it to R at a profit of 15%. R paid $38000 more than it cost P. How much did P gain? P Q 20%Q R 15% R P $38000 P A. B. C. D. $14000 $16000 $18000 $20000

(12) A sum of money is deposited at a rate of 4% per annum for 2 years. If the difference between the compound interest (compounded annually) and simple interest is $4650, find the sum of money correct to the nearest ten. 4% () $4650 A. B. C. D. $2817750 $2906250 $2975400 $3185250

(13) If 2 x 2 + px + 4 (1 x )(2 + qx ) + 2 , then 2 x 2 + px + 4 (1 x )(2 + qx ) + 2 A. B. C. D.

p = 4 , q = 2 p = 4 , q = 2 p = 4 , q = 2 p = 4, q = 2

(14) If ab < 0 and 5a 2 + 28ab 12b 2 = 0 , then a : b = ab < 0 5a 2 + 28ab 12b 2 = 0 a : b = A. B. C. D.1 : 6 6 :1 2 : 5 5:2

2009- CE MATH 2


(15) x and y are two variables. The table below shows some values of x and their corresponding values of y. xy x y

x y

4 36

6 16

12 4

24 1

Which of the following may be a relation between x and y? x y A. B. C. D.x y 1

x x

y

1 y 1 x 2 y

(16) A boat sailed 730 km due east from P to Q, and then sailed 520 km due south to R. Find the bearing of P from R, correct to the nearest 0.1. P 730 km Q 520 km R R P 0.1 A. B. C. D. N54.5W N35.5W S54.5E S35.5E

(17) In the figure, OAB is a sector with centre O. Given that the radius of the sector is r cm and the perimeter of the sector is 3r cm, find correct to the nearest degree. OAB O r cm 3r cm A. B. C. D. 29 30 57 60B

O r cm A

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(18) In the figure, the total surface area of the right circular cone is A. B. C. D. 135 cm 2 180 cm 2 216 cm 2 261 cm 2

15 cm 12 cm

(19) The figure shows a frustum of a right circular cone. The radii of the upper face and the base are 3 cm and 5 cm respectively. If the height is 8 cm, find the volume. 3 cm 5 cm 8 cm A. B. C. D. 1384 cm 3 15 128 cm 3 392 cm 3 3 136 cm 33 cm

8 cm

5 cm

(20) In the figure, BAC = BAC = A. B. C. D. 32 60 74 116B

A 32

D

C

2009- CE MATH 2


(21) In the figure, AB // DC. Which of the following must be true? AB // DC Area of ABE = Area of CBE ABE = CBE II. ABE ~ CDE III. DAE = CBE I. A. B. C. D. II only II I and II only I II II and III only II III I, II and III III III

A

B

E

D

C

(22)

1 cos A = sin A tan A A. B. C. D.1 sin A sin A cos A sin A

(23) For 0 90 , the greatest value of 0 90

1 31+ sin 2

is

1 31+sin2

A. B. C. D.

1 3 1 9 1 3

(24) For 0 90 , how many roots does the equation tan = tan 3 have? 0 90 tan = tan 3 A. B. C. D. 0 1 2 3

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(25) In the figure, P and Q are two houses along a straight road running east to west. R is another house such that the true bearing of R from P and Q are 028 and 298 respectively. Find the distance from R to the road. P Q R P Q R 028 298 R A. B. C. D.

PQ cos 28 sin 62 PQ cos 28 sin 28 PQ tan 62 sin 62 PQ tan 62 cos 62

R N N

P

Q

E

(26) Which of the following plane figures does NOT have rotational symmetry? I. II. III.

A. B. C. D.

I only I II only II I and II only I II II and III only II III

2009- CE MATH 2

Reference material, not for sale11

(27) In the figure, which of the following must be true?

O

I.

The area bounded by the figure remain unchanged after rotating 90 clockwisely through O. O 90 II. The figure has reflectional symmetry. III. The figure has 2-fold rotational symmetry. 2 I only I II only II I and III only I III I, II and III III III

A. B. C. D.

(28) In the figure, ABC is a straight line. If BC : AC = 1 : 3 , then x = ABC BC : AC = 1 : 3 x = D A. B. C. D. 15 30 45 6030 10 8 5A B x E

4C

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(29) In the figure, BCDE is a square and ABE is an equilateral triangle. Find EAC . BCDE ABE EAC A. B. C. D. 35 40 45 60A B C

E

D

(30) If the point (3,4 ) is rotated anti-clockwisely about the origin through 270, then the coordinates of its image are (3,4 ) 270 A. B. C. D.

( 4,3) ( 4, 3) (4,3) (4, 3)

(31) In the figure, O is the pole. If OP = 5 , then the polar coordinates of P are O OP = 5 P A. B. C. D.

(5, 45) (5, 135) ( 5, 225) (5, 225)P

O 45

x

2009- CE MATH 2


(32) Which of the following straight lines will cut with 3x 2 y + 7 = 0 ? 3x 2 y + 7 = 0 I. 6x + 4 y 5 = 0 2 x + y =1 II. 3 III. 6 y = 9 x + 1 A. B. C. D. I only I II only II I and II only I II II and III only II III

(33) Let A = ( 5, 10) and B = (16,5) . If the line joining A and B cuts the x-axis at C, then C= A = ( 5, 10) B = (16,5) A B x C C = A. B. C. D.

(8, 0) (0, 8) (9, 0) (0, 9)

(34) Two fair dice are thrown. What is the probability of getting a total of 10 or 12? 10 12 A. B. C. D. 1 6 1 9 1 12 1 15

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(35) The box-and-whisker diagram shows the distributions of ages of a group of teenagers joining a swimming class.

ages 20 21 22 23 24 25 26

Which of the following must be true? I. Range = 6 = 6 II. Mean = 24 = 24 III. 50% of the teenagers are above 25. 50% 25 A. B. C. D. I only I II only II I and II only I II I and III only I III

2009- CE MATH 2


(36) The stem-and-leaf diagram below shows the time spent by 50 students on study last week. 50 Boys () Leaf (1 hour) (1 ) 7 5 4 3 1 1 1 9 7 7 4 2 9 8 7 3 9 7 3 6 6 4 4 2 2 Stem (10 hours) (10 )0 1 2 3 4

Girls () Leaf (1 hour) (1 ) 3 5 6 6 2 4 6 8 1 3 5 6 2 4 5 5 6 6 7 7

7 9 8 8 8

Which interval has the highest frequency spent on studying? A. B. C. D. 0 9 hours 0 9 10 19 hours 10 19 20 29 hours 20 29 30 39 hours 30 39

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Section B

(37) Which of the following is a factor of x 3 + x 2 14 x 24 ? x 3 + x 2 14 x 24 A. B. C. D.

(x + 3)(x 2) (x 3)(x 4) (x + 3)(x 4) (x + 3)(x + 4)

(38) Let f ( x ) = x 2 2 x + 9 and g ( x ) = 2 x 2 8 x + 3 . Find the maximum value of f (x ) + 2 g (x ) . f ( x ) = x 2 2 x + 9 g ( x ) = 2 x 2 8 x + 3 f ( x ) + 2 g ( x ) A. B. C. D.

14 24 30 42

(39) If the graph of y = 3 x 2 + 7 x + 2 is given, which of the following straight line should be added on the graph in order to solve the equation 3x 2 + 7 x 8 = 0 ? y = 3x 2 + 7 x + 2 3x 2 + 7 x 8 = 0 A. B. C. D.

y = 10 y = 6 y=6 y = 10

(40) Solve log x 2 = 2 log 6 . log x 2 = 2 log 6 A. B. C. D. 6 6 0 or 6 06 6 or 6 6 6

2009- CE MATH 2


(41) Which of the following can be the graph of y = 3 x + 5 ? y = 3 x + 5 A.y

B.

y

8 5 O x 5 y O x

C.

D.

y

O 3 5 x

x

6 5 O

(42) In the figure, is the angle between the diagonals BE and CH of the rectangular block. Find correct to the nearest degree. BE CH A. B. C. D. 60 90 114 120B C

A O

D

10

G

F 6

H

18

E

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(43) In the figure, OAB is a sector of a circle with radius 20 cm. If the area of the shaded region is 40.8 cm 2 , find the length of AC correct to 2 significant figures. OAB 20 cm 40.8 cm 2 AC 2 B

A. B. C. D.

2.9 cm 8.5 cm 11 cm 17 cm

20 cm

48 O C A

(44) The figure shows the graphs of y = a cos bx and y = c cos dx , where a, b, c and d are positive constants. a b + c + d = y = a cos bx y = c cos dx abc d y ab+c+d = A. B. C. D. 0 3 2 3 2 15 42

y = a cos bx

1

O

180

x

1 2

y = c cos dx

(45) Convert the decimal number 2 7 + 2 5 + 2 4 + 1 to binary number. 2 7 + 2 5 + 2 4 + 1 A. B. C. D.10110012 10110000 2 101100012 101100112

2009- CE MATH 2


(46) Which of the following systems of inequalities has its solution represented by the shaded region in the figure? 3 x 2 y 0 x + y 12 x0 3x 2 y 0 x + y 12 y0 3x 2 y 0 x + y 12 y0 3x 2 y 0 x + y 12 y0

A.

y

3x 2 y = 0

B.

C.

O

x

x + y = 12

D.

(47) Three numbers are in arithmetic sequence. It is known that their sum is 69 and their product is 11339. What is the value of the smallest number? 69 11339 A. B. C. D. (48) 17 21 23 27

7 7 7 =2 3

A. B. C. D.

7 7

7 49.

3 2

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(49) Two dices are thrown, and the sum of the numbers shown is 8. What is the probability that these two numbers are both odd? 8 A. B. C. D. 1 18 5 36 2 5 1 4

(50) Which of the following changes can not affect the value of the inter-quartile range of a group of data? A. B. C. D. Dividing each datum of the group by 3. 3 Subtracting each datum of the group by 2. 2 Deleting one of the data in the group. Adding the datum 0 to the group. 0

(51) In the figure, PA and PB touch the circle at A and B respectively. C is a point on the circumference. If APB = x , then ACB = PA PB A B C APB = x ACB = A. B. C. D.x 2 xC A

x 2 180 x90 x B P

2009- CE MATH 2


(52) In the figure, O is the in-centre of PQR . Which of the following must be true? O PQR

OQ bisects Q . OQ Q II. PO produced bisects QR. PO QR III. Area of PQR = 2 Area of the circle. PQR = 2 I. A. B. C. D. I only I I and II only I II I and III only I III II and III only II IIIO Q

P

R

(53) The equation of a circle is 2 x 2 + 2 y 2 + 4 x 6 y 15 = 0 . Which of the following must be true? 2 x 2 + 2 y 2 + 4 x 6 y 15 = 0 The circle cuts the y-axis at two distinct points. y II. The centre lies in the fourth quadrant. III. The origin lies outside the circle. I. A. B. C. D. I only I I and II only I II I and III only I III I, II and III III III

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(54) M and N are two fixed points in a coordinate plane. P is a variable point such that the area of PMN is 7. What is the locus of P? M N P PMN 7 P A. B. C. D. a circle a parabola a straight line a pair of straight lines

END OF PAPER ~ ~

2009- CE MATH 2


Solution(1) B.

(m + n )nm . (m + n )n (n + m )m= (m + n )nm

.

(2) C.

1 . x 11 = x 1 x 1 1 x x 1 = . x 1 1 x

(3) C.

2 3y . y 1

y =y ( x + 3) yx + 3 y x( y 1) x

x+2 x+3 = x+2 = x+2 = 2 3y 2 3y = . y 1

(4) A.

1 only. 3x + 4 2 2 ( x + 3) > 3 x + 4 2 x + 6 > 3x + 4 x < 2.x+3 >

x = 1.(5) B. 51 25 + = 7. 12 x 12 + x

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(6) D.y

O

x

As the graph has a negative y-intercept a 2 , the answers (B) and (C) are not true. y a 2 (B) (C) Equation of the axis of symmetry: x =

a 2 < 0.

(D) is the answer. (D) 36 cm 2 .

(7) C.

Let x cm and y cm be the width and length of the rectangle respectively. x cm y cm

y = 4x 2 x + 2 y = 30Put (1) into (2), (1) (2)

(1) . (2)

2 x + 2(4 x ) = 30 10 x = 30 x = 3.

y = 4(3) = 12 .

2009- CE MATH 2


Required area () = xy = 3(12 ) = 36 cm 2 .

(8) A.

x2 x . 2f ( x 1) =

1 (x 1)2 + (x 1) 2 1 2 = x 2 x + 1 + ( x 1) 2 x2 x = . 2

[

]

[(

)

]

(9) B.

52. Required number () = 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 52 .

(10) A.

6.65 9.15 A < 6.75 9.25 .

Absolute maximum error () =

0.1 m 2 = 0.05 m.

6.65 9.15 A < 6.75 9.25 .(11) D. $20000.

Let $C be the cost of the article of P. P $C

Cost of the article of Q ( Q ) = $C (1 + 20% ) = $1.2C

and

R paid (R ) = $1.2C (1 + 15% ) = $1.38C . 1.38C C = 38000 C = 100000 .

P gained (P ) = $(1.2C C ) = $0.2C

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= $0.2(100000 ) = $20000 .

(12) B.

$2906250.

Let $P be the principal. $P Compound interest () = $ P(1 + 4% ) $ P = $0.0816 P2

and Simple interest () = $

P 4 2 100 = $0.08 P

p = 4 , q = 2 .

0.0816 P 0.08 P = 4650 P = 2906250 .

(13) A.

2 x 2 + px + 4 (1 x )(2 + qx ) + 2 (2 + qx 2 x qx 2 ) + 2 qx 2 + (q 2)x + 4 .

2 = q . p = q 2 p = 4 , q = 2 .(14) B. 6 :1.

5a 2 + 28ab 12b 2 = 0 (5a 2b )(a + 6b ) = 0 5a = 2b (rejected) or () a = 6b a : b = 6 : 1 . (15) B.

x

1

y

.

2009- CE MATH 2


x y x yx y xy

4 36 2 3 24 144 5184

6 16 3 2 24 96 1536

12 4 6 24 48 192

24 1 24 24 24 24

xy 2 x1 .

y

(16) A.

N54.5W.730 km

N

In PQR , PQR tan = 730 520 = 54.5.

P

Q

520 km

Required bearing () = N54.5W.

R

(17) C.

57.

AB = 3r r r = r cm.

2r

=r 360 = 57 .

(18) C.

216 cm 2 .

Let r cm be the base radius of the circular cone. r cm

r 2 + 12 2 = 15 2 r = 9.Total surface area () = (9 ) + (9 )(15) = 216 cm 2 .2

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(19) C.

392 cm 3 . 3

A

ACB ~ AED (AAA),

x 3 = x +8 5 x = 12.

x cm C 3 cm B

Required volume () 1 1 2 2 = (5) (12 + 8) (3) (12 ) 3 3 392 cm 3 = 3 32.DAC =

8 cm

E

5 cm

D

(20) A.

180 32 2 = 74 .

ACB = 74 . BAC = 180 74 74 = 32 .

(21) A.

II only.

ABE = CDE BAE = DCE ABE ~ CDE . II is true.I and III are not true in general. I III (22) B.sin A .

(alt. s, AB // DC) (alt. s, AB // DC) (AAA)

(AB // DC) (AB // DC)

1 cos A 1 cos A = sin A tan A sin A sin A cos A 1 cos 2 A = sin A sin A

2009- CE MATH 2


1 cos 2 A sin A sin 2 A = sin A = sin A .

=

(23) A.

1 . 3 Greatest value () = 11+ 0

3 1 = . 3

(24) B.

1.tan 3 tan + tan tan tan 2 + 1 tan

(

)

= tan 3 =0 =0 =0 = 0 .

There is only one root. PQ cos 28 sin 28 .PRQ = 28 + 62 = 90.28 N R 28 62 N 62 S Q 298 E

(25) B.

In PQR , PQR cos 62 = PR PQ PR = PQ cos 62.

62 P

In PRS , PRS sin 62 =

RS PR RS = PQ cos 62 sin 62 = PQ sin 28 cos 28 .

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(26) C. (27) C. (28) B.

I and II only. I and III only. 30.BC : AC = 1 : 3 AB : BC = 1 : 2 .

DAB ~ ECB . (3 sides prop.) () x = 30 . (29) C. 45.

In ABC , ABC ABC = 60 + 90 = 150 .

BAC =

180 150 2 = 15 .

EAC = 60 15 = 45 .

(30) A. (31) D. (32) C.

( 4,3) . (5, 225)I and II only.

3 x 2 y + 7 = 0 and 6 x + 4 y 5 = 0 are non-parallel. They will cut each other. 3 x 2 y + 7 = 0 6 x + 4 y 5 = 0 I is true. 3 x 2 y + 7 = 0 and 2 x + y = 1 are non-parallel. They will cut each other. 3

2 3 x 2 y + 7 = 0 x + y = 1 3 II is true.

2009- CE MATH 2


3 x 2 y + 7 = 0 and 6 y = 9 x + 1 are parallel and non-overlapping. They will not cut each other. 3 x 2 y + 7 = 0 6 y = 9 x + 1 III is not true. (33) C.

(9, 0) .

Let () C = (c, 0 ) . Slope () of AB 10 ( 5) 5 16 5 7 c C = (9, 0 ) . (34) B. 1 . 9 (1, 1) (2, 1) (3, 1) (4, 1) (5, 1) (6, 1) (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4) (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) 4 36 1 = . 9 (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6)= Slope () of AC 10 0 = 5c 10 = 5c = 9.

Required probability () =

(35) A.

I only. Range () = 26 20 = 6.

I is true. Mean could not be determined from the box-and-whisker diagram. II is not true.

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Only 25% of the teenagers are above 25. 25% 25 III is not true. (36) A. (37) C. 0 9 hours.

(x + 3)(x 4) .

Let () f ( x ) = x 3 + x 2 14 x 24 .f ( 2 ) = ( 2) + ( 2) 14( 2) 24 = 0.3 2

x + 2 is a factor () of f ( x ) . By long division, f ( x ) = ( x + 2 ) x 2 x 12 = ( x + 2 )( x + 3)( x 4 ) .

(

)

(C) is the answer. (38) D. 42.f (x ) + 2 g (x ) = x 2 2 x + 9 + 2 2 x 2 8 x + 3

(

)

= 3x 18 x + 15 = 3 x 2 + 6 x + 3 2 + 15 + 3 3 22

= 3( x + 3) + 42 .2

(

)

( )

Maximum value () = 42 . (39) D. y = 10 . 3x 2 + 7 x 8 = 0 3x 2 + 7 x = 8 3x 2 + 7 x + 2 = 10 . The straight line y = 10 should be added. y = 10 6 or 6 .

(40) D.

2009- CE MATH 2


log x 2 = 2 log x = x2 = x = 6 or () (41) C.y

2 log 6 log 6 2 62 x = 6 .

6 5 O x

(42) C.

114.F

In HEF , HEF

HF = 18 2 + 6 2 = 360 .H 18

6

E

In CHF , CHF CH =

C

(

360

)

2

+ 10 2H

10

= 460 .360F

OH = OE = In OHE , OHE

460 . 2

460 460 460 460 18 = 2 + 2 2 2 2 cos 2

2

2

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cos =

47 115 = 114 .

(43) A.

2.9 cm. (20) 2

48 1 (20)(OC )sin 48 = 40.8 360 2 167.55 7.4314OC = 40.8 OC = 17.056 cm.AC = 20 17.056 = 2.9 cm.

(44) A.

0.

From the figure, a = 2 , b = 4 , c = 1 and d = 1 .

a b + c + d = 2 4 +1+1 = 0. 101100012 . 3x 2 y 0 x + y 12 . y0

(45) C.

(46) C.

(47) A.

17.

Let a d , a and a + d be the three numbers, where d > 0 . a d a a + d d > 0

(a d ) + a + (a + d )

= 69 3a = 69 a = 23

and

(a d )(a )(a + d ) (23 d )(23)(23 + d ) (23 d )(23 + d )

= 11339 = 11339 = 493

529 d 2 = 493 d 2 = 36Reference material, not for sale35

2009- CE MATH 2

d = 6.

Smallest number () = 23 6 = 17 . 7.1 1 1

(48) B.

7 7 7

= 72 74 78 = 72=7 = 7.1 1 1 + + + 4 8

1 2 1 1 2

(49) C.

2 . 5

Possible outcomes (): (2, 6), (3, 5), (4, 4), (5, 3), (6, 2). Favourable outcomes (): (3, 5), (5, 3). Required probability () = (50) B. (51) C. 2 . 5

Subtracting each datum of the group by 2. 90 x . 2

PA = PB . (tangent properties) () PAB = 180 x 2 x = 90 . 2 ( sum of ) ( )

ACB = 90 (52) A. (53) A. I only. I only.

x . ( in alt. segment) () 2

Put x = 0 into the circle, x = 0

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2 y 2 6 y 15 = 0 . = ( 6 ) 4(2 )( 15)2

= 156 > 0.

The circle cuts the y-axis at two distinct points. y

I is true. 2 ( 3) Centre () = , 2 2 3 = 1, . 2 II is not true. Radius () ==

( 1)2 + 3 43 . 4

15 2 2

2

Let d be the distance between the centre of the circle and the origin. d d ==

( 1 0)2 + 3 0 2

2

13 4 < Radius ()

The origin lies inside the circle.

III is not true. (54) D. a pair of straight lines.

2009- CE MATH 2


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Locus of P P P N M

2009- CE MATH 2

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2009CEmock-Maths2 Q&A OscarTam