2012 ACJC Prelim H2 Math Soln

7/29/2019 2012 ACJC Prelim H2 Math Soln

1/15

1

Anglo-Chinese Junior CollegeH2 Mathematics 9740

Qn 2012 Prelim Exam Paper 1 Solution

1(a)

2 2 2

2

2

2 1

1 2 4 2

4 7 2 4 7 4 7

1 2ln 4 7

2 2 3

1 2 2ln 4 7 tan

2 3 3

x xdx dx

x x x x x x

x x dx

x

xx x C

(b)

3 3

2 2

0 0

0

5

3 2

20

0

5522

2 2

3 3

22 2

0 3 3 5

2 2 40

3 5 15

a

a a

a

a

a x a xx a xdx x dx

a x

a x dx

aa

2

2 2

2 2 2

2

2

2 3 2 30

2 5 2 2 5 2 2 5 2

1 2

0(2 1)( 2)

1 2 0, (2 1)( 2) 0

0.5 2

, 0 0.5 2

x x x x

x x x x x x

x

x x

Since x x x

x or x

Since x x or x

3 y = f(x) = 2 lnx (lnx)2

y = 0 gives (lnx)(2 lnx) = 0

lnx = 0 or 2

x = 1 or e

2

[Note: May use GC to write e2

correct to 3 sf since

question did not ask for exact values.]

The line 0y cutsy = f(x) at 2 points.

Therefore f is not a one-one function.

Thus 1f does not exist.

dy/dx = (2/x) 2(lnx)/x = 0 lnx = 1 x = e (so maximum point is at (e, 1))

h1 exists when greatest possible a is e. [May write a = 2.72 (3 sf) as question did not ask for exact value.]

To find h1

, lety = h(x) so thatx = h1

(y) where 0


2/15

2

2

ln 1 1 ln 1 1y x x y but lnx 1 since 0


3/15

3

6

1

2

3

1

4 1

3

1 14

2 2 1 11

4

3

3 32

2 4 1 31

2

9

9 94

6 10 1 91

4

27

27 27 310

18 28 1 27 1 31

10

n

n n

u

u

u

u u

Let P(n) be the statement

1

1

3

1 3

n

n nu

for all 0n .

When 0n , LHS =0

1

4u RHS =

1

1

1

3 1311 3 4

13

LHS = RHS P(0) is true.

Assume that P(k) is true for some 0k , i.e.,1

1

3

1 3

k

k ku

.

To prove P(k+ 1) is true, i.e., 13

1 3

k

k ku

,

1

1

1

1

1

1

1 1 1 1 1

1

3LHS

2 1

33

1 3by assumption

32 1

1 3

3

3 3 31 3RHS2 3 1 3 2 3 1 3 1 3 3 1 3

1 3

kk

k

k

k

k

k

k

k k kk

k k k k k k

k

uu

u

Since P(0) is true and P(k) is true P(k+1) is true, by the Principle of Mathematical Induction,

we conclude that P(0), P(1), P(2), P(3), are all true. Hence P(n) is true for all integers 0n .

72

2 , tx t y e

0 x

y

(0, 1)

Either work out values yourself,Or use GC MODE SEQ to find thevalues.

Note that the answers in the GC TABLE arein decimals so you have to convert

them to fractions yourself, you should re-write the expression

into u(n) =

3u 1 1

2u 1 1

n

n

first so that

it is easier to key into the GC, you should use the BLUE u above

the 7 in your GC, not the green U.


4/15

4

2

2

x t

dx

dt

2

2

2

t

t

y e

dyte

dt

2

tdy

tedx

Equation of normal at 2

2 , pp e is 2 21

2p py e e x p

p

Equation of normal at1

2,Ce

is 1

2y e xe

At ,0A x :2

2 2 2

1 1 2 12 2

ex x

e e e

At 0,B y :21 1 1 2

2 2e

y e y ee e e

2 2

2

12 1 1 2OA e e

e eOB e

: 1: eOA OB

82 2 23 2x a a

y x ax a x a

so asymptotes are ,x a y x a

2

2

d 21

d

y a

x x a

Since 222 0 & 0 for a x a x ,

2

2

21 1 0

dy a

dx x a

2 2 23 2x a ay x a

x a x a

has no stationary points

(shown)

Graph of

2

2

2

f 1

a

y x x a

is

From graph of fy x , f is increasing forx a

OR

2

3

4f 0 0

ax x a x a

x a

x a

1y

x

y

x=a

y=x+a

0

(0,3a)

3,0a 3,0a

2 23x ay

x a

Note that the answers should

involvep only ortonly, notbothp and t.

Label intersections with axesin coordinate form (see

question).


5/15

5

9

: or2

k k

or ,2

10(i)

1 1 a b a b

a ba b a b a b

(ii)

2500

1

2500

1

2500

1

1

4 3 4 1

4 3 4 1

4 3 4 1

14 1 4 3

4

5

1

4

r

r

r

r r

r r

r r

r r

1

9

5

13 9

9997 9993

10001 9997

10001 1

4

(iii) 2500 2501

0 12500

1

1 1

4 1 4 5 4 3 4 11 1

4 3 4 1 10001 10005

10001 1 1

4 10001 10005

10000 1 1

4 10001 10005

124.75

10001 10005

24

r r

r

r r r r

r r

2

1

2

2

O6

2

R

3

Re(z)

Im(z)

3

3,1

Shade the regionR lightly to make itclear to the marker which is the

correct answer.

The rest of the circle and lines should

be in dashes.

Replace 10001 with a number for which you can find the root easily and the result is LESS than the previous

value


6/15

6

11

(i)

2

a cOM

2

2

2

cos60

2 2 2

0.5

2

0.5 3( )

2 4Shown

a c ca c c a c c cLength of Projection

c c c

a c cNote: a c

c

c c cc

c

11

(ii)

2

1 1 3sin 60

4 4 8

3

83

8

Area of OMC

k

a c a c c c Note: a c

c

11

(iii)5

2OD c [Note: You may want to sketch a diagram to obtain this result.]

sin 605 5 5 3 5 3

2 2 4 4

5 3

4

Shortest OMC OD

t

c a c aa c a= c

a a a a

12 2

2 4

d2

d dGiven , diff. w.r.t. to getd

yt yt

y w tw ttt t

22

2 3 3

4 4 2 2

3 2 3 2 3 2 2

22 2

2 2

2 2 2 2

dyt yt

dw y y ydtt w t wt w t t t dt t t t t

dy dyt t y t y yt yt y y

dt dt

2

11

2

1 1 11

2 2 2 2

1 1ln ln 2

2 2

1 ln ln 22 2 2

dy dt y y

dy dy dt y y y y

y y t c

y yt c t by y


7/15

7

2

2 2

2

2

2

2 2

e e2

1 2 e

2 e

1 e

1 2 e1 e

t b

t t

t

t

t

t

yA where A

y

y Ae A

Ay

A

Awt A

WhenA=0, the solution is the t-axis.

13Common ratio r= tan. For S to exist, 1r , i.e. 1 tan 1

4 4

For in this range,

2 3 11 tan tan tan1 tan

1 3 3

1 tan 2

21 tan

3 3

2tan 1

3 3

1 3tan

3 3

1 3 3 3tan

3 3 3 3

2 3tan6

1tan

3

6

Hence4 6

.

14(a)

2 2

2 2 2

2 2

2 2

2 2

22

3 2 Re

3 2

3 4

3 6 9 0

12 3 0

3

11 4 0

3

11

4 12

z z

x y x

x y x

x x y

x x y

x y

x y

Note:x > 0 since 3 2Rez z

(-1, 0) (3, 0)

x

y3( 1)y x 3( 1)y x

2 2( 1)1

4 12

x y

t

w

A>0

You may want to sketch the graph ofy = tanx

in your GC to find .

Use Y1 = tanX, Y2 = 1, Y3 = 1and adjust WINDOW to

Xmin = /2Xmax = /2

Xscl = /4


8/15

8

14(b) 2 2 3 iw 24 arg

3w w

i2 /3 i2 /32 2

4 4 4 cos sin3 3

nn n n n n nw e e i

2is real sin 0

3

3 , even, or 3 ,2

n nw

n m m m n k k

*5050 50 50

50 50

100 100 100

100

*

100 100 100 1004 cos sin 4 cos sin

3 3 3 3

2 32 2 sin 2 2 2 3

3 2

2 3

w w w w

i i

i i i

k

Qn 2012 Prelim Exam Paper 2 Solution

1 (i)

(ii)

2.5 3 4

3 0.5 7

2 0.5 5

, , ,

2.5 4 3

0.5 3 7 0

1.5 5 0

, :

k

k k

k k

Equating components

k

k

k

Solving Equations k

i j k

---(1)

---(2)

---(3)

(1) (2), (3) 2 . 1 2. ,

2.5 (2) 2 7

2, 2, 2 3 2 8

2 2 2 6

7,8,6

When k

r

Intersection point is

2 Let y f x 1sin 2 1

2

2

ln sin 2

1 2diff. w.r.t. : 1 4 2

1 4

xy e y x

dy dyx x y

y dx dxx

22

22

4diff. w.r.t. : 1 4 2

1 4

x dy d y dyx x

dx dxdxx

Use exponential form forindices.Use Cartesian ortrigonometric

forreal and imaginary parts.


9/15

9

22 2

2

22

2

22

2

4 1 4 2 1 4

1 4 4 2 2

1 4 4 4 (shown)

dy d y dyx x x

dx dxdx

d y dyx x y

dxdx

d y dyx x y

dxdx

3 2 2

2

3 2 2diff. w.r.t. : 1 4 8 4 4 4

d y d y d y dy dyx x x x

dx dxdx dx dx

Whenx = 0, 0 1, ' 0 2, '' 0 4, ''' 0 16f f f f

2 38

1 2 2 ...3

f x x x x

1sin 2 16

2 3

6

sin 26

1 122 4

1 1 8 1 21 2 2 1

4 4 3 4 3

xe e x

x x

e

3i

ii

a)

1 sin

diff. w.r.t. : cos

x u

dxx u

du

22

2

2 2

2 1

2 2 1 sin 1 sin cos

2 2sin 1 2sin sin cos

cos 2 11 sin cos cos

2

1sin 2 1 2sin cos 1 12 sin 1

4 2 4 2 2 2

x x dx u u u du

u u u u du

uu u du u du du

xu u uu c u c x x x c

1 1 1 192 2 9

0 0 0 0

12 10 2

2 1

0

Area =2 2 2 =2 2 2

1 1 1 192 2 2 sin 1 2 2

2 2 2 10 2 10 4 5 2

R x x x dx x x dx x x x dx x xdx

x x x xx x x x


10/15

10

b)

21 1 2

2 9

0 0

Vol. of revolution formed when R is rotated completely about -axis

2 2 2 14.995

x

x x x dx x x dx

4 (i)

4(ii)

Method 1:

1 1

: 1 20 3

rline l

Since (1,1,0) lies on2

p ,

1 1

1 1 1 1 2

0 1

a a a

Method 2:

Since direction vector of line lis to normal of2

p ,

1 1

2 0 1 2 3 0 2

3 1

a a a

Method 3:1 1 1

1 2 1 1 2

1 1 1

a

a a a

a

Method 1: Normal method

1

1 0 1

: 1 2 : 0 1

1 1 1

0 11

0 1 2 1 13

1 1

p and line AN

r r

0 1 11 1

0 1 13 3

1 1 4

ON

Method 2: Projection Method

1 1 1 1 1 1 11 1 1 1 1 1 1 1. 1 1 1 1 1 1 1

3 33 3 3 31 1 1 1 1 1 1

AN AB

1 0 11 1

1 0 13 3

1 1 4

AN ON OA

ON AN OA


11/15

11

4

(iii)Let the acute angle between planes

1p and

2p be .

1 1

1 2

1 1 1 2 1 2cos

1 1 3 6 18

1 2

1 1

61.9

4

(iv) Distance from Origin top12 2

1 3

1

1

Distance from Origin top3 ( 0)2 12

2

2

b b b

23

12 3

23

12 3

4 6 10

b

b

b

4

(v)

(vi)

1p ,

2p and

4p meet in a line l.

Hence (1,1,0) lies on4

p .

1

1 2

0 3

2

c

d

c d

Normal of4

p is perpendicular to line l.

1

2 2 0

3 3

4 9 0

5

7

c

c

c

d

5 , 7c d


12/15

12

5 Randomly choose 10 programmers, 1 secretary and 1 section head for the sample.

Stratified sampling guarantees a representative sample of each group (i.e. programmers,

secretaries and section heads) in the population.

CANNOT accept any of the following answers:

allows the opinions of different strata to be considered separately.

accurate

unbiased

6

(a)

(i)

(ii)

(b)

4 boys and 3 girls

_ B _ B _ B _ B _ 4! 5P3 = 24 60

= 1440

Or 4! (5C3 3!) = 24 60 = 1440

G BBBB GG type so 4! 4! = 576

Case (i): 3 boys and 3 girls4C3 (3 1)! 3!= 4 2 6 = 48

Case (ii): 4 boys and 2 girls

(4 1)! 3C2

4P2

= 6 3 12

= 216

Or (4 1)! 3C2

4C2 2! = 216

Total number of arrangements = 48 + 216 = 264

7 Plot ofy againstx.

Highesty = 6.3; Lowesty = 2.7

Highestx = 12.3; Lowestx = 4.4

Correlation coefficient = r= 0.263

A linear model is not appropriate as the scatter diagram

shows that the points are not close to a straight line and

the value ofris quite close to 0.

7 Fory = ax2 + b, value ofr= 0.913

Fory = a lnx + b, value ofr= 0.971

Therefore, (b)y = a lnx + b is a better model as the value ofrin (b) is closer to 1 than for (a).Line of regression isy = 11.042 948 39 2.687 231 256 lnx

Wheny = 6.1, lnx = 1.839 420 548x = 6.29 (3 sf) Acceptx = 6.3 (1 dec place)

This estimate is valid since the value ofris close to 1 and the value ofyused in within therange of experimental data(4.5 y 6.3).Or, may say thatx comes from interpolationinstead of within the range of experimental data.

BB

BGG

G

B B

BGB

G

x4.4 12.

2.7

6.3


13/15

13

8(a) LetXbe the number of cars arriving at the jetty in 30 mins. X~ Po(6)

( ) 2 0.0446 ( ) 2 0.0620i P X ii P X 8(b) Let Ybe the number of cars arriving at the jetty in 20 mins. Y~ Po(4)

0.1

0.9

7

P Y k

P Y k

Least k

k P Y k

6 0.88933

7 0.948878 0.97864

9(a) LetXbe the time of journey from TownA to TownB.

2Let ,

1 160 70

4 20

60 1 70 1

4 20

60 700.67449 1. 64485

Solve: 53.0491 53

10.3054 10 (nearestminute)

X N

P X P X

P Z P Z

9(b) Let Ybe the time of journey from TownB to Town C.

80,9

0.02

86.1612Min 87mins

Last time of departure 10 33h

Y N

P Y k

kk

980,

20

80 2 78 82 0.997

Y N

P Y P Y

10 Let L be the length and B be the breadth of a tile

2 2

2 2

1 2 3 10

1 2 3 10

18.9,0.3 8.9,0.1

2 2

2(18.9) 2(8.9) 55.6

4(0.3 ) 4(0.1 ) 0.4

............ 10(55.6) 556

............ 10(0.4) 4

L N B N

Let P L B

E P

Var P

E P P P P

Var P P P P


14/15

14

10 Let Wbe the number of tiles with a red tint. W~ B(500, 0.6)

300 5 200 5, is large 300,120

0.95

0.05

10.05

2

1281.98

2

282.48

Greatest = 282

np and nq n W N approximately

P W k

P W k

P W k

k

k

k

11

0

1

0

0

To test : H : 27

H : 27 at 2%level

27Under H : 26

27

Test statistic = 1.22474

p-value = 0.1158 >0.02

Do not reject H

XT t

s

There is insufficient evidence at the 2% level of significance to conclude the mean foot length

of an 18 year old man of high intelligence is more than 27cm

Assumption: Assume the population of foot length of 18 year old man of high intelligence is

normally distributed

0

1

0

22

To test: H :

H : at 4% level

Under H by Central limit theorem, since is large

1, , (123.20) 2.08813660 59

k

k

n

sX N k approx where s

0To reject H : 2 0.04 0.02

60 60

2.05375 2.05375

60 60

26.6 26.62.05375 2.05375

2.088136 2.08813660 60

26.9831 26.21687

27.0 26.2

x k x kP Z P Z

s s

x k x kor

s s

k kor

k or k

k or k


15/15

15

12

(i)

LetXbe the r.v. number of shots, out of 20, that hit the target.X~ B(20, 0.25)

Most likely number of shots is 5.

12

(ii) P(X 5 |X 10) =

P 5 10

P 10

X

X

=

P 10 P 4

P 10

X X

X

=

0.5812163575

0.9960578583= 0.584 (3 sf)

12

(iii)

P(same number of shots each to hit target)

= P(1, 1) + P(2, 2) + P(3, 3) +

= 0.25

2

+ (0.75 0.25)

2

+ (0.75 0.75 0.25)

2

+

= 0.252 (1 + 0.752 + 0.754 + 0.756+ )

= 0.0625 2

1

1 0.75

= 0.0625 1

0.4375=

1

7or 0.143 (3 sf)

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2012 ACJC Prelim H2 Math Soln