Math Prelims Soln

8/8/2019 Math Prelims Soln

1/97


2/97

3. (i) 1

st Term, 000102

2002

=

=a

Common ratio, 169

43

2

=

=r

169

1

00010

=

S

22 cm00072cm808717000160 = d nT n

846.31350

150

50)1(

=+ x x

7> x or 11< x

Substitute x with xe ,7> xe or 11< xe (N.A.)

7ln> x or 1.95 (3 s.f.)

2.

(a)(i) Normal of p1 =

=

312

11

2x

111

Direction of l1 =

111

Angle between l1 and normal of p1 =

.

128)111)(914(

11

1

31

2

cos 1 =++++


11/97

3.25 + 2.5 = 0 . (2) = 2, = 2.6

=

1

2 of Normal 3 b p

=

01

5.0 of vectorDirection 2l

1;001

5.0

1

2==

bb .

islinecontainingplaneof Eqn 24 l p

r. 10

1

12

.

5.2

075.3

1

12

=

=

263

61

610

andbetweenDistance 32 == pl

3. (i) Since f g D R = , fg does not exist.

(ii) fh exists if ),0(g R . Hence 40)3ln( >> x x hence least a = 4

)3ln(2)3ln(:fh

+ x

x x a , 4for > x

(iii) y

x

x x y

2+= y = x

)22,2(

y

x

x x y

2+= y = x

)22,2(


12/97

Since the line y = c (candidates should indicate a horizontal line where >c22 k , and f is 1-1 for x > 2 ,the least value of 2=k .

4. (a) Since y = x and 1=dxdy

,

RHS x

x x LHS =

+== 2

22

21

(b) ux y =

dxdu

xudxdy +=

)(2

222

ux x xu x

dxdu

xu+=+

u

u

dx

du xu

2

1 2+=+

uuu

dxdu

x

uuu

dxdu

x

221

21

22

2

+=

+

=

)(2

1 2shown

uu

dxdu

x

=

= dx xduuu 1

12

2

=

dx x

duuu 1

12

2

C xu += ln)1ln( 2

x A

x y

x A

u x

Au

=

=

=

2

2

2

2

1

1

1


13/97

Since entire population is wiped out by the disease eventually, asHence, C = 0, D = 0.

t ae x 2= a represents the initial population of the fish (in thousands).

Section B 5. (a)(i) No. of ways = 6! = 720

(ii) No. of ways = 2 6 = 64(iii) No. of ways = 63CCCCCC 6

65

64

63

62

61

6 =+++++

(b) Case 1 : Daen and Vera not in the groupNo. of ways = 565

8 =C Case 2 : Daen and Vera in the groupNo. of ways = 563

8 =C Case 3 : Vera in the group without DaenNo. of ways = 704

8 =C

Total no. of ways = 56 + 56 + 70 = 182

t e x 22 =

t e x 2=


14/97

6. (i)0.02 Deformed

Supplier A p0.98 Not deformed

0.03 Deformed1 p

Supplier B

0.97 Not deformed

(ii)31= p

P(fish-ball is deformed) = )1(03.002.0 p p +

752308.0

3)2(03.0

302.0

=

=

+=

(iii)

)deformedisitP(

)deformedisandbysuppliedP()deformedisit|bysuppliedP()f(

B

B p

=

=

p

p

p p p

p p p

=

+

=

+=

3

)1(3)2)(()3)(1(

)3)(1()02.0)(()03.0)(1(

)03.0)(1(

2

2

)3(6

)3()1)(1()3)(1(

3)('f

p

p p p

p

=

=


15/97

7. (i) Let X be the no. of times the particular residential area is flooded in 4 mon

34

~ oP X

P( X 2) = 0.3849400 = 0.385

(ii) Let Y be the no. of 4-month periods, out of 12, in which the particular resiarea is flooded at least twice.P( X 2) = 0.3849400

Y ~ B(12, 0.3849400)

P(Y 5) = 0.704(iii) Let X be the no. of times the particular residential area is flooded in 5 yea

( )20~ oP X

Since = 20 > 10,( )20,20~ N X approximately

P( X 11) = P( X 10.5)= 0.983

(iv) Let Y be the no. of years, out of 40, in which there are at most 3 floodings Y ~ B(40, 0.433470)

Since n = 40 > 30, np = 17.3388 > 5, nq = 22.6612 > 5,

Y ~ N (17.3388, 9.82295) approximately

P(Y n) < 0.8 P(Y n 0.5) < 0.8 P(Y n 0.5) > 0.2 n 0.5 > 14.7010 n > 15.2010

Least n = 16

8. Let r.v. A be the mass of a snapper fish and r.v. B be the mass of a pomfret fishA ~ N(1, 0.1 2); B ~ N(0.6, 0.05 2)(a)(i) A 1 + A2 + A3 + B1 +B2 ~ N(4.2, 0.035)P[A A A B B 4 5] 0 0544


16/97


17/97

(ii) In this case, s 2 =n

n 1 [sample variance] =9

9-1 [0.2002] = 0.045 or

9200

test-statistic, T =

x 0sn

=

x 14

0.0459

Since complaint not valid, do not reject H 0, p-value > 0.04

x 140.045

9

< invT(1 0.04,8)

x 14

0.0459

< 2.004151525

x < 14.14171491

x < 14.14 (to 2 d.p.)

11. (i) r = 0.906

(ii)

PP

% of graduates, x

% earning more than$5000, y


18/97

(v) 4.6384ln)3.22(4.35 =+= y

(vi) Since x does not lie within the data range, extrapolation will make the estiunreliable.


19/97

Dunman High School2010 Year 6 H2 Mathematics (9740) Preliminary Examination Paper

Suggested SolutionsQn Suggested Solution1 1

2

2 2

2

1 1(1 ) 1

2 2 2

11 ...

2 2 2 2

1 1 1 1 1...

2 2 4 4 81 3 3

...2 4 8

= + +

= + +

= + + +

= + +

x x x

x

x x x

x x x x

x x

Valid values of x: 1 2 22

< < < x

x

2Let Pn be the proposition

1

1(2 1)(2 3) 3(2 3)

=

=+ + +

n

r

nr r n

for n

When n = 1:

LHS =1

1

1 1 1(2 1)(2 3) (2(1) 1)(2(1) 3) 15=

= =+ + + +

r r r

RHS =1

3(2(1

1Since LHS RHS, P is true.=

Assume Pk is true for some ,k +

i.e.1

1,

(2 1)(2 3) 3(2 3)

k

r

k r r k =

=+ + +

to prove1

P +k

is true,

i.e.1

1

1 1.

(2 1)(2 3) 3(2 5)

k

r

k r r k

+

=

+=

+ + + 1

1LHS

k +

=


20/97

2

13(2 3) (2 3)(2 5)

(2 5) 33(2 3)(2 5)

2 5 33(2 3)(2 5)

(2 3)( 1)3(2 3)(2 5)

1RHS (shown)3(2 5)

k k k k

k k k k

k k k k

k k k k

k k

= ++ + +

+ +=

+ +

+ +=

+ +

+ +=

+ +

+= =

+

1P is true P is true+ k k

Since 1P is true, 1P is true P is true+k k , by mathematical inducti

for +n .1

33(2 3) 3 2

1 1as ,

3 63 2

=+ +

+

nn

n

n

n


21/97

3(i) 2 2

2

4 5 ( 2) 4 5

( 2) 1

=

n n n

n

+ = +

+

3(ii) 2 2

3

2 2

3

2 2

2 2

2 2

1 4 5

1 ( 2) 1

3 1 1 1

4 1 2 1

5 1 3 1

N

n N

n

n n n

n n

=

=

+ +

= + +

= + +

+ + +

+ + +

M M

M

2 2

2 2

2 2

2 2

( 2) 1 ( 4) 1

( 1) 1 ( 3) 11 ( 2) 1

1 ( 1) 1 5 2

N N

N N

N N

N N

+ + +

+ + +

+ + +

= + + +

M

3(iii) 2 2

2 2

2 2

1 ( 1) 1 5 3

2 1 ( 1) 2( 1) 1 (since >1)

( 1) ( 1 1)

1 (since >0)2 1

N N

N N N N N

N N

N N N

N

+ + +

< + + + + +

= + + +

= + +

= +

4(i) 0 0 31 1

3Area of d 1 d

4 R y x x x

= = + =

13 31 ( 1) y x x y= + =

( )

( )

1 43 3

2 22

43

3Area of d ( -1) d 1

4

31 1

4

bb b

S x y y y y

b

= = =

=


22/97

Volume required

=2 2 3 2

12 (1) ( 1) dk

b k x x

+ = 3. 53 (or 11.1) (unit cube)

5(i) If A, B and C are collinear, then

2 37 3 5 72 1 2

i.e. 2, 0, 4

AB BC

b b

a

a b

=

=

= = =

5(ii)If OA

is perpendicular to ,OB

then

0

23 7 0

2

i.e. 2 21 2 0

OA OB

b

a

b a

=

=

+ + =

2

2

2

2 33 5

1cos60

13 35

2(6 15 ) 13 35

31 168 1309 010 (nearest int.) or 4 (nearest int.)

20 (nearest int.) 6 (nearest int.)

a

a

a a

a a

a a

b b

=+

+ + = +

=

= =

= =

His claim is not necessarily true since points O, A, B and C may no

6(a)2

2

e 1d 3e (1 3e ) d

(1 3e ) 3

t t t

t t t = +

+


23/97

6(b)

( )( ) ( )2 2 2d tan 2 secd x x x x =

( )

( )

( )

3 2 2 2

2 2

2 2

1sec d 2 sec

2

1 tan21

tan l2

x x x x x

x x x

x x

=

=

=

6(c) 4 20

3 d x x x

3 42 2

0 3( 3) d ( 3) d x x x x x x= +

3 44 43 3

0 34 4 x x

x x

= +

27or 13.5

2=

7(i) No. The statement is not always true. It applies only for (polynomial) with real coefficients. 7(ii) 4 3+ i 0 z + = 4 3 i z =

5i

4 62e z

=

1 14 4

1 5 (12 5)i ( 2 ) i4 6 242 e 2 e

k k z

+= = , 0,1,2,3k =

1 1 1 14 4 4 4

5 7 19 17-i i i -i24 24 24 242 e or 2 e or 2 e or 2 e z

=

7(iii) Im

L

Re L

2 Z 3 Z

4 Z

O

1 Z


24/97

8(i) 236ON x=

( )

( )

2

2

12 12 2 36

22 6 36

A x x

x x

= +

= +

8(ii)( )2

2

d 1 22 36 2 6

d 2 36

A x x x

x x

= + +

( )

( )( )

2 2

2

2

2

2

72 2 12 236

4 18 3

364 6 3

36

x x x x

x x

x x x

x

=

=

+ =

dFor maximum , 0 : 0 3 cmd = > =

A A x x x

( ) ( ) ( )d d d 1 d 12 2d d d 10 d 20

xQR x x

t t t t = = = =

( ) ( )4 8 1dWhen 2, 32 4 2

d 32 A

x x

= = = =

2 1

d d d

d d d1

4 220

2cm s

5

A A x

t x t

=

=

=

P S

Q R

U T

2 x

2 x

6 6 O

N

6 6


25/97

9(a)(i)

ln(1 e )= + x y e 1 e = + y x

d d: e ed dd

ed

=

=

y x

x y

y x x

y x

2

2

2

2

d d d: e 1

d d d

d d d1 (shown)

d d d

=

=

x y y y x x x

y y y

x x x

9(a)(ii)

2

2

d 1 d 1When 0, ln2, ,

d 2 d 4 y y

x y x x

= = = =

2

2

11 4ln 2 ...2 21 1

ln 2 ...

2 8

= + + +

= + + +

x y x

x x

9(a)(iii)

2

2

2

22

2

2 2

2

ln(1 e ) ln 1 1 ...2

ln 2 ...2

ln 2 ln 1 ...

2 4

2 4ln 2 ...

2 4 2

1ln 2 ...

2 4 2 4

1 1ln 2 ... (verified)2 8

+ = + + + +

= + + +

= + + + +

+ = + + +

= + + +

= + +

x x x

x x

x x

x x

x x

x x x

x x

9(b) 10 tan 3 cos 2 = x x ( )22

10 3 12

= x

x


26/97

10(a)(i) e cos , sin cos , 0 4

x y

= = +

d de (cos sin ), cos sin ,d d

x y = =

-d d d / ed d d y y x x

= =

At (e cos ,sin cos ), + the equation of the tangent is

-( sin cos ) e ( e cos ), y x =

Set ,6

=

at (63e 3 1

,2 2

+) , the equation of the tangent is

-3 1( ) e

2 2 = y

-6 1e

2

= + y x

10(a)(ii)

Area under the curve C is

40

2 240

40

(sin cos ) e (cos sin ) d

e (cos sin ) d

e cos 2 d ( )

0.68 (2d.p.)

A

shown

= +

=

=

=

10(b)(i) y

f '( ) y x=

O x

-2


27/97

10(b)(ii)

11(a) 11 2

2 11 2

1 2 1

1 1 2 11

0 1, (since 0)2

ar

a r r a

a

a

a a r

=

= =

Therefore given the above initial condition, Bobs model corresponds to solution curv(I) in part (ii).

Therefore in Bobs model, the volume of water approaches infinity in the long run (norealistic) whereas in Andys model, the volume of water reasonably diminishes to zelong run/after some time.

Thus, Andys model is more appropriate than Bobs model.

C

t

C

V 1(I)4

C >

1(II)

4

C


33/97

(ii) 1

1

2

= =

1 2d n n

Set z=0,2 4 10

3 8

1, 3

x y

x y

x y

+ =

+ =

= =

11 1

3 1 , .

0 2

l

= +

1 1: r a + d =

Alternative

1

2 4 10

3 8

Let ,2 4 10

3 8

1 , 3 ,2 2

1 1

: r 3 1 ,20 2

x y z

x y z

z t x y t

x y t

t t x y

t l

+ + =

+ + =

= + =

+ =

= + =

= + =

(iii) Since the point with co-ordinates (6,m.5) lies on the first plane,

1 1

6 2

4 10

5 1

12 4 5 10

D

m

m

=

=

+ + =

a d

7

.4

m =

(iv) 2 2


34/97

SECTION B

5(i) To obtain a quota sample of size 80:Identify and categorise the parents into mutually exclusive sub-groups accordieducation levels . Set a quota, i.e. a target number of respondents for each grototal adds up to 80.Poll respondents on a first-come-first-serve basis, say, when the parents arrive at schomorning with their children, until the number for each category is filled.

(ii) Stratified sampling is more representative in terms of the proportion of parents educaqualifications in each category.

(iii) 42080 142400 =

6(i)

From GC, 0.860r =

(ii) 260.56From GC,regression line 37.612 xt

= +

ie, ( )37.6 , 261 3 sig figa b= =

Suggested model between x and1

t is a better fit with | r |= 0.930 > | r |= 0.860

model between x and t .

(iii) 260.5637.612 78.95.0

x = + =

5t = lies outside the data range of t , thus model may not be valid and estimate noreliable.

x

t


35/97

7(i) Let J be the event where Mylo wears a jacket and T be the event where Mylo we

( ) 0.6

( ) 0.6( )

( )0.6

0.2( ) 0.6 0.2 0.12

P T J

P T J P J

P T J

P T J

=

=

=

= = (ii)

[ ]( )

( ) '

1 ( )

1 ( ) ( ) ( )

1 0.4 0.2 0.12

1 0.48

0.52

P T J

P T J

P T P J P T J

=

= +

= +

=

=

(iii)

Let n J be the nth

day where Mylo wears a jacket.

3 1 3 2

3 1 3 2 1 2

Required Probability

( ) ( )( ) ( ) ( )

(0 2)(0 6)(0 4) (0 8)(0 2)(0 4)

P J J P J J P J J P J J P J J

+ =

+ +

+

' J

J

J

J

J

' J

0.2

0.8

0.4

0.6

0.2

0.8

0.2

J

J

J

' J

' J

' J

' J

0.8

' J

0.4

0.6

0.6

0.4

0.2

0.8

Mon Tue Wed


36/97

8(i)(a)

Number of ways 10!(5) 18144000= =

(b)S S S S S S

( )( )( )Number of ways

= 6! 5! 3 259200=

(ii)(a) ( )( )

Number of ways= 4! 5! 2880=

(ii)(b)

Case One: 8 Questions (4M and 4S) between B and KB K

Case Two: 8 Questions (3M and 5S) between B and KB K

Case Three: 9 Questions between B and KB K

Pure Mathematics Questions Statistics Questions

( )( ) ( )( ) ( )( )

Number of ways

4 5= 3! 5! 4! 4! 5! 4!

3 4

2880 2880 28808640

+ +

= + +

=

4! ( )5

4!4

( )4

3!3

5!

4! 5!

or

=(5!)6 5 4 3 (3!) 259200 =


37/97

9(i)2

Let 100

50 , 4008

u x

u u

=

= =

Unbiased estimate of population mean:100 4.16667 100 104.17 104 (3 s.f .) x u= + = + =

Unbiased estimate of population variance:2

2 1 50400811 12

s

=

= 345.42 345 (3s.f.)

(ii) To test H 0: 115 =

against H 1: 115 < One-tail test at 5% level ( = 0.05)

Use t -test since 2 is unknown and sample size of 12 is small

under H 0, T =115

345.42/12

X ~ t (11).

From GC, p-value = 0.0342

Since p-value =0.0342< 0.05, there is sufficient evidence to reject H 0 at significance and conclude that the mean IQ score is less than 115, hence the manuclaim is disputable.

(iii) (a) The IQ score of customers is normally distributed.

(b) For 2-tailed test, p-value =2(0.0324) =0.0684 > 0.05. H 0 will not bconclusion would be different.

10 (i)

Let X be the number of unsolicited text messages received in a day.

( )5Po 7P( 2) 0.125 (3 s.f.)

X

X = =

(ii) Let Y be the number of unsolicited text messages or phone calls received in a week.( )Po 8

P( 10) 0.816 (3 s.f.) (shown)

Y

Y =

(iii) Let W be the number of weeks where receives more than 10 unsolicited text messagephone calls in a week out of 10 weeks

( )P 16T


38/97

( )Po 16Since =16>10, N(16,16) approximately.

P( 20) P( 19.5) (apply c.c.)

0.191 (3 s.f.)

T

T

T T

= >

=

11(a)

2N(2, 2 )Y X

P( 3) 0.4

3 2P( ) 0.4

2

From GC,1

= 0.253352

= 2.7910

Y X

Z

> =

> =

2 21 2

21 2

1 2

Var( ) 2 2.7910 3.9471

(8,3.9471 )P(8 12) 0.345 (3 s.f.)

X X

X X N X X

+ = =

+

< + < =

b(i) Let X min be the amount of time spent by a student online each day.

( )1 2 60

221 2 60

E( ) 60 120 7200

Var( ) 60 45 90 15

X X X

X X X

+ + = =

+ + = =

L

L

Since n=60 is large, by Central Limit Theorem,

( )21 2 60 7200, 90 15 X X X N

+ + L approximately.

1 2 60P( 7000)

0.717 (3 s.f.)

X X X + +

=

L

(ii) Since n=60 is large, by Central Limit Theorem,245

120,60

X N

approximately.

P( 120 5) X <


39/97

1 Solve the inequality

2

42 ( 2)

x x x

,

giving your answer in an exact form.

Hence solve 2e 4

e 2 (e 2)

x

x x

+ +.

2 The sequence of numbers nu , where n = 0, 1, 2, 3, , is such t

1

1

( 2)2 1

nn

n

n uu

u n

+=

+ +.

Proof by induction that, for 0,n 2

2 1n

nu

n

+=

.

3 The functions f and g are defined as follows:( ) ,212:f 2 x x a 1< x ,

( )g : ln , x x a+a 1> x .

(a) Define1

f

in a similar form.

(b) State the value of a such that the range of g is (0, ) .

(c) Show that the composite function gf exists, and find the rangyour answer in terms of a .

4 A curve is defined by the parametric equations

21t

xt

=+

, 21t

yt

=

, where 1, 1.t

(i) Show that the tangent to the curve at any point with parameter


40/97

5 Robert took a study loan of $100 000 from a bank on 1 st Januarycharges an annual interest rate of 10% on the outstanding loan at thyear. After his graduation, Robert pays the bank $ x at the beginninThe first payment is made on 1 st January 2014. Let nu denote theRobert at the end of nth year after 2013, where 0+n .

(i) Find 0u .

(ii) Show that nu = 01.1 (1.1 1)n nu kx , where k is a constant to b

(iii) Given that Robert owes the bank less than $1000 at the end ofminimum value of x, giving your answer to the nearest dollar.

6 (a) Find 21

d3 4

t t .

(b) Use the substitution 5 xu = to find ( )25 cos 5 d x x x .

7 It is given that the function ( )f y x= has the Maclaurins series 1

satisfies ( ) ( )2 2d1 1d y

x b y x

+ = + , where a and b are real constants.

(i) Show that b = 2 and find the value of a .

(ii) Find the series expansion of f( )4 x

x+in ascending powers

including the term in 2 x .

(iii) State the equation of the normal to the curvef( )4 x

y x

=+

at x


41/97

8 (i) Express 4( 1) ( 2)

r r r r

+in the form

1 2 A B C

r r r + +

+.

(ii) Hence find2

4( 1) ( 2)

n

r

r r r r =

+ .

Give a reason why the series is convergent, and state its limit.

(iii) Use your answer to part (ii) to find2

3

( 1)( 3)

n

r

r

r r r =

+ + .

9 On a single Argand diagram, sketch the loci given by(i)

21 i 2 z ,

(ii) 1

arg123 i

z +

+ ,

(iii) 1 z z> .

Hence, or otherwise, find the range of values of i z and arg ( z

10 A file is downloaded at r kilobytes per second from the internet vconnection. The rate of change of r is proportional to the differenceconstant. The initial value of r is 348. If r is 43, it remains at this co

(i) Show thatd

( 43)d

r k r

t = .

(ii) Hence obtain an expression for r in terms of k and t .

The total amount of data downloaded, I kilobytes, in time t second

dd I

r t

= .

(iii) Given that there is no data downloaded initially find I in term


42/97

11

The diagram above shows part of the structure of a modern art museuMarcus, with a horizontal base OAB and vertical wall OADC . Pvectors i, j , k are such that i and k are parallel to OA and OC resp

The walls of the museum BCD and ABD can be described resequations

15 36

6

=

r and

14 5 10 4 00 0 4

= + +

r , where

(i) Write down the distance of A from O .

(ii) Find the vector equation of the intersection line of the two w

ABD .

(iii) Marcus wishes to repaint the inner wall ABD . Find the area o

Suppose Marcus wishes to divide the structure into two by adding athat it intersects with the walls BCD and ABD at a line. This described by the equation 2 7 x y z + = , where , .

(iv) Find the values of and .

(v) Another designer, Jenny, wishes to construct another partidescribed by the equation 2 7x y z + = where

B

k

O

C

D

i

j

A


43/97

12 The curves 1C and 2C have equations2 2 2( 2) (1 ) x a y = and

1 2,<


44/97


45/97

Since P(0) is true & P(k) is true P(k + 1) is also true, hence by minduction P( n) is true for all { }0 .n +

3

(a)

(b)

(c)

x

( )

7,2

2

1

2

1:f

1221

21

122

212

1

2

>+


46/97

(ii)

32

2

d 1d 1 y t x t

+=

Equation of tangent:

( ) ( ) ( ) ( )( )

32

2 2 2

3 3 2 22 2 2 2

32 3

11 1 1

1 1 1 1

1 4

t t t y x

t t t

t y t x t t t t

t x t

+ =

+

= + + +

= +

When12

t = ,d

27d y x

=

Let be the acute angle between the two lines.

( )

1

1

tan 27 87.879

tan 1 45

42.9

A

B

A B

= =

= =

= =

Alternative Solution

( )( )1

27 1 26tan1 27 1 28

26tan 42.9

28

= =+

= =

A B

Note: = A - B tan A = 27 , tan B = 1


47/97

2 23 0

3 3 20

1.1 1.1 1.1 (12 ) 1.1(12 ) 12

1.1 1.1 (12 ) 1.1 (12 ) 1.1(12 )

= =

u u x x x

u x x x

::

( )( )

( )

10

10

0

0

1.1 1.1 (12 ) 1.1 (12 ) ... 1.1(12 )

1.1 12 1.1 1.1 ... 1.1

1.1 1.1 11.1 12

0.1

1.1 132 1.1 1

=

= + + +

=

=

n n nn

n n n

n

n

n n

u u x x x

u x

u x

u x

(iii) n = 7 at end of 20207 7

01.1 132 (1.1 1) 1000u x < $2270.30 x >

Least x to the nearest dollar = $2271

6(a)2

2

2

2

1 d3 41 1

d344

1 1d

432

31 1 2ln4 3 3

2

3 2 3ln12 2 3

t t

t t

t

t

t C

t

t C t

=

=

= +

+

= ++

(b)

d5

x

x

u

u =


48/97

2

2

2

5 cos (5 ) d

1cos d

ln 51

cos dln 5

1(1 cos 2 ) d

2ln51 sin 2

2 ln 5 2

1 sin 2(5 )5

2 ln 5 2

x x

x x

x

u u uu

u u

u u

uu C

C

=

=

= +

= + +

= + +

7i

( ) ( )2 2

dWhen 0, 1, 4.

dd

1 1dd

(2) 4d

2 (Shown)

y x y

x y

x b y

x y

b x

b

= = =

+ = +

= =

=

( )2

22

2

2

d d d1 2 2 2

d d d

d16d

168

2!

y y y x x y

x x x

y x

a

+ + =

=

= =

ii 12

12 2

2 2

f( )(4 )

1(1 4 8 ...)(1 )2 4

1 3( )( )1 1 2 2(1 4 8 ...)(1 ( )( ) ( ) ...)

2 2 4 2! 4

+

= + + + +

= + + + + + +

x x

x x x

x x x x


49/97

1 162 31

y x=

8(i) 4( 1) ( 2) 1 2

r A B C r r r r r r

= + +

+ +

4 ( 2) ( 1)( 2) ( 1)r Ar r B r r C r r = + + + + 1, 2, 1 A B C = = =

(ii)

2

2

4( 1) ( 2)

1 2 11 21

1 14

1 2 12 3 51 2 13 4 61 2 14 5 7

1 2 14 3 1

1 2 13 2

1 2 12 1 1

1 2 11 2

1 2 1 1 1 2 12 3 3 1 21 1 1 16 1 2

n

r

n

r

r r r r

r r r

n n n

n n n

n n n

n n n

n n n n

n n n

=

=

+

= + +

= +

+ +

+ +

+ +

+

+ +

+ +

+ + +

+ + +

= + + + ++ +

= + ++ +

M

(iii) 1 1 1lim 0

+ + = hence the series in (ii) converges


50/97

(iv)

2

3( 1)( 3)

n

r

r r r r =

+ +

1 3 ( 1) 3...(2)(3)(5) ( 1)( )( 2) ( 1)( 3)n n

n n n n n n

= + + + + + +

1

2

4 2( 1) ( 2) (1)(2)(4)

n

r

r r r r

+

=

=

+ 1 1 1 1 16 1 2 3 12n n n

= + + + + +

1 1 1 1

12 1 2 3n n n= + +

+ + +

9.

21 i 2 z

(1 i) 2 z +

1arg

123 i z +

+

arg( 1)12 6 4

z

+ + =

Method 1:1 7

24 2

QR = = ;7 1

24 4

PQ = + =

Method 2: QR is the perpendicular bisector, so PQ = 2 (radius)i 2z >

- 1 x

O

y

(1,1)P

Q

1

1 R


51/97

Given 43r = whend

0d

r t

= ,

0 (43 )Since 0, then 43

k a

k a

=

=

d( 43)

dr

k r t

= (shown)

(ii)

1

1d d

43ln 43

r k t r r kt C

=

= +

1

1

43 e

43 e where e

kt C

C kt

r

r A A

+ =

= + =

When 0t = , 348r = .305 A = .43 305e kt r = +

(iii)

2

d

(43 305e ) d

30543 e

kt

kt

I r t

t

t C k

=

= +

= + +

When 0t = , 0 I = .

2305

C k

=

30543 (e 1)kt I t k

= +

(iv) Given 5700 I = and 90t = ,90

90

90

3055700 43(90) (e 1)

3051830 (e 1)

6 e 1

k

k

k

k

k

k

= +

=

=

Solving using GC,0.167k = or 0k = (NA)


52/97

iii 4 12+


53/97

iii 4 128 8 0

8(Reason: j is zero.)

48

0

12 14 2

0 0 08 0 8

12 4 80 8 88 0 8

+

= + = =

=

= =

= =

uuuv uuuv

%

uuuv

uuuv

uuuv

OD OD

OB

AD

BD

1Area =2

8 2 1 1 41

8 0 8 1 0 8 52

8 8 1 4 1

8 42 51.8 (3 s.f.)

= = =

= =

uuuv uuuv ABD BD AD

iv 2(4) 7( 8) (0)2(12) 7(0) 8

64, 5

+ =

+ =

= =

OR

The 3 planes intersect at the line4 18 1 ,

0 1r

= +

%

2 1

4 2


54/97

4 28 . 7 8 56 64

0 564

= + = =

v Their partitions are parallel to each other.

There is no intersection point.12 Ellipse

(a) 2 2 2( 2) (1 ) x a y = 2

22

( 2)1

x y

a

+ =

Method 1:Sequence of transformations:

1) Scale // to x-axis by factor a .2) Translate in the positive x-direction by 2 units.

Method 2:Sequence of transformations:

1) Translate in the positive x-direction by 2a

units.

2) Scale // to x-axis by factor a .

(bi) 1x


55/97

(bi)2

2

11 4

( )

4( 1)

3

+

+

x x x

x x

x

x

x

y

(bii)Sub

2 41

=

+

x y

xinto 2 2 2( 2) (1 ) x a y = :

222 2 4( 2) 1

1

= +

x x a

x

( ) ( ) ( )22 22 2 2 21 ( 2) 1 4 + = + x x a x a x --- (*)(shown)

Hence the x-coordinate of the points of intersection of 1C and 2C satisfy equation (*).

(b)(iii)

From (ii), number of intersection points between 1C and 2C gives t

real roots of the equation (*).

From the graphs, there are 2 points of intersection between 1C andreal roots .

1= x

1= y x

2

2 41

=

+

x y

x

4 2

22

( 2)1

x y

a

+ =

2 a 2 a+ 2

HCI Prelim H2 Mathematics P2 Solutions

2010 HCI H2 Mathematics Preliminary Examination Paper 2 Solution


56/97

2010 HCI H2 Mathematics Preliminary Examination Paper 2 SolutionQn Solutions1 Surface area of the tin and lid

=2

2 2 10 x xy x + + = 400

2200 5 x x y

x

=

Volume of the container

( )

22

3 2

200 5

200 5

x x x

x x x x

=

=

( )2d 200 3 10dV

x x x

=

d 200

d 3V

x x

= = or 10 x = (rejected)

( )2

2

d6 10 0

dV

x x

= < when203

x =

V is maximum when203

x = .

When203

x = ,553

y =

(or x = 6.67, y = 18.3).

2(i) 5 32 0 z = 5 32 z = e i0 = 32e i2k 2 i

52ek

z

= where 0, 1, 2.k = (ii)

The highest power in the equation5

2 132

ww

+ =

is four since the terms

canceled out. Hence the equation has only four roots.

51

2 32w

+ =

2 i512 2

k z e

w

+ = =


=2 4

2 2cos 2cos 4

+


57/97

= 2 2cos 2cos 45 5

+

=2 4

4 cos cos 2 .5 5

+

Or use GC,1 2 3 4

1 1 1 1w w w w

+ + + = 10.

3 2nS an bn c= + +

1 1 100U S a b c= = + + =

2 4 2 190S a b c= + + =

10 100 10 360 100 90 550S a b c= + + = + + = Using GC,

5a = , 105b = , 0c = Thus 25 105nS n n= +

1n n nU S S =

( ) ( )( )225 105 5 1 105 1n n n n= + + 110 10 n=

( )1

110 10 110 10 10n nU U

n n

= +

10 (a constant)= Hence sequence is an AP.4i 0 1 1

2 (1 ) 0 2

0

0 0 0

0 (1 ) 2 2 2

2

0 1 1

2 2 2 2 4

2 3

= + =

= + = +

= = + +

uuuv

uu uv

uu uv

OX

t t

OY

t t t t

XY

t t t t t


1 1


58/97

( )1 2 2 2 XBt t t

= =

uuur

0 0

2 2

2 2

BY

t t

= =

uuur

1 0 1

2 2 2 2 4

2 3

XY XB BY

t t t t t

= + = + =

uuur uuur uuur

ii Suppose O , X , Y are collinear.Then

1 0

2 2 2

2

1 0 1 (Out of range)

=

= +

= =

uuuv uuuv

OX k OY

k

t t t

Thus O , X , Y are not collinear.

iii 1 0

2 2 2

2

OX OY

t t t

= +

uuuv uuuv

= (4 4 + t 2 2 t 2)= 0

2

2 2

4 1 10 (reject) or = +

4 2 2 2t t t

+ = =+ +

For all t \{0}, 0 < < 1.Thus XOY can be 90 when 0t .


iv 1 uu uv


59/97

2 4

3

projection vector

1 4 4

2 4 . 1 . 1

3 0 0

174

4 4 2 41170

42

117

0

XY

t t

t t

=

+

+ =

+

=

=

5(i) ( ) ( )2 26 2 3 x y x + + = +

( ) ( ) ( )( )

2 2 22 3 6

9 2 3

y x x

x

+ = +

=

(ii)

(iii)


When 7 y = , 6 x = .


60/97

Method 1: Using d x y ( ) ( )

( ) ( )

2

2 2

2 9 2 3

2 23 13

2 18 2 9

y x

y y x

+ =

+ += + = +

( )

( ) ( )

27

1

73

1

2

213 d 2(6)

2 9

1 ( 2)3 12

2 27

121 27 3 1 12

2

10 units

y R y

y y

+ = + +

= +

= + + =

Method 2: Using d y x ( ) ( )22 9 2 3

2 3 2 3 [ 2 3 2 3 N.A.]

y x

y x y x

+ =

= + =

( )6

2

632

2

4(7) 2 3 2 3 d

28 2 (2 3)

R x x

x x

= +

= +

= { 28 [(12 + 27) (4 + 1)]}= 10 units 2

(iv) Volume required= vol. of cylinder (vol. generated by curve from y = 2 to y = 1)

( )6

22

2(7) (4) 2 3 2 3 d x x = +

= 196 92 = 327 unit 3 (3 s.f.)

6 Use random sampling method to select a sample from each class The number


The flight is not fully booked so the chosen seat could be empty.


61/97

OR The passenger may ignore the questionnaire.

It is not appropriate to use simple random sampling as passengers from differenhave different opinions on the service. The number of passengers in the first clsmall, so the passengers from the first class may not be chosen at all using the sampling method.

7(i)No. of ways =

10!12600

4!3!2!=

7(ii) Case 1: The 2 blue tiles and 1 yellow tile are in the 4th row with the 4 thgreen.No. of ways= no. of ways with B, B, Y, G in 4th row + no. of ways with B, B, Y, R in 4th r

=4! 6! 4! 6!

32402! 2!2! 2! 2!2!2!

+ =

Case 2: The 2 blue tiles and 1 yellow tile are in the third row.

No. of ways =3! 7!

18902 2!2!2! =

Total no. of ways = 3240 + 1890 3!2!

3!2!

4!2!

3!2!

4!2! 3!

= 5130 108 216 = 4806

7(iii) No. of ways such that less than 3 yellow tiles are in the fourth row

= 12600 4 3C 7!

4!2!

=12600 420 =12180

7 lastpart

No. of ways

=6!

3!2! 7C4 = 2100

8(i) 2010 10(9000)9201,

10 x

+= =

( )( )( )222 90001 5071479000

9 10 9

xs x

= =

H0 : 9000 = H1 : 9000 >

T t St t9000

(9) x

t

=


8(ii) H0: 9000 = vs H 1: 9000 >


62/97

8( ) H0: 9000 vs H 1: 9000

Under H 0, X ~ N(9000,25 2

10 ) = N(9000, 62.5).

Test Statistic = X 9000

62.5~ N(0, 1).

Level of significance = 1%P( Z > 2.326347877 ) = 0.01At the 1% significance level, reject 0H if z 2.326347877.

z = x 9000

62.5 2.326347877

x 9018.391395 = 9020 .

Assumptions: The standard deviation of the life span remains unchanged afteprocess.

9Firstpart

~ N(190, 576) X

1 20 21 300.001( ... ) 0.001(2)( ... ) ~ N(0, 0.03456)T X X X X = + + + +

P( | | 0.15) P( 0.15 0.15)T T = = 0.580

OR

A = X 1 +... + X 20 2( X 21 +... + X 30) ~ N(0, 34560)

0.15P( | | ) P( 150 150)

0.001 A A =

= 0.5809(i) Let Y be the r.v. denoting the mass of a randomly chosen apple from Mark's or

2~ N( , 30 )Y


63/97


11(i)L t X b th d ti g th b f ll i i k H X


64/97

Let X be the r.v. denoting the number of callins in a week. Hence ~ X

We are looking for the such that P( X 9) = 0.701

From graph, the value of 32.5 = (to 3 sig.fig).

The condition is that the rate of callins received by the centre is constant thromonth / the callin occurs randomly / The callins occur in a month are indepanother

11(ii) Let Y be the r.v. denoting the number of callins in a week.~ Po(32.5)Y

Since the mean is bigger than 10, hence~ (32.5, 32.5)Y N approximately.

c.c(25 40) (25.5 40.5)P Y P Y < < < = 0.810

11(iii)

Let S be the r.v. denoting the number of successful cases out of the n pegroup.

3~ ( , )

20S B n

Since the number of groups concerned, which is 70, is large, therefore by apply3 3

13 20 20~ ( , )

20 70

nS N n

approximately.

EITHERn ( 4)P S

( 9)P X


( 4) 0.3P S < < 4 0.15 n


65/97

4 0.15( ) 0.3

0.1275

704 0.15

0.52440.1275

70

0.12754 0.15 ( 0.5244 )

70

0.12750.15 (0.5244 ) 4 070

5.23912 or 5.0899(reject)

27.45

nP Z

n

n

n

n n

n n

n n

n

<

> <

>

Least n = 28.

12(i) Location F should be omitted as the road distance cannot be smaller than the stdistance, indicating that it is an incorrect data entry.

From the scatter diagram, another location that should be omitted is location Houtlier based on the scatter diagram.

12(ii) The suitable regression line is the regression x on y:0.3936554 0.81702935 x y= +

When y = 20.0,

16.7 x = km

12(iii)

s


66/97


67/97

2

2

232

03

x

x

>

>


68/97

2 203 3

2 2or

3 3

x x

x x

+ >

< >

3 ( )

2

12 3 sin 602

V Base Area height

x h

=

=

(Note: some students might use Pythagoras Thm or Trigo. to find base area.)

2

2

1 32 3

2 2

8(shown)

x h

h x

=

=

Total cost of constructing prism,( ) ( ) ( )

( ) ( )2

1 edges 2 3 2 triangles 2 3 rectangles

1 33 6 2 3 2 2 3

2 2

C

h x x xh

= + +

= + + +

2

2 2

22

2 1 2

8 1 3 83 6 2 3 2 2 32 2

24 486 3

3 6 48 24

x x x x x

x x x x x x x x

= + + +

= + + +

= + + +

2 3d 6 6 48 48 0d

C x x x

x

= + =

2 3

d0

d6 6 48 48 0

C x

x x x

=

+ =

Minimum cost C is $54.

( ) ( )2

3 4

2

d6 96 2 144 2 0

dC

x

= + + >


69/97

4(a) 23 4w i= +

Let w x iy= + 2( ) 3 4 x iy i+ = +

2 22 3 4 x xyi y i+ = + 2 2 3 x y = ----- (1)

2 4 xy = ----- (2)

From eq (2): 2 y x

=

Sub into eq (1):2

2 4 22 3 3 4 0 x x x x

= =

Solving, we get 2 x = , 1 y = Hence (2 )w i= +

(b)Let 4 16 z =

14

4

4 ( 2 )

( 2 )

16

16

2 , 2, 1,0,1

i

i k

i k

z e

z e

z e k

+

+

=

=

= =

3 34 4 4 42 , 2 , 2 , 2i i i i z e e e e

=

1 z 2 z

3 z 4 z

Re( z)

Im( z)

2

4

O

(ii) y = f( x) y


70/97

(iii)1gf ( ) ( 3) 2 3 4 = + = + + x g x x

(b)(i)

(ii)

h(16) + h(25) = h(4) + h(1)= (12 6) + (12 1) = 17

6(i)

2 2t

y = f -1( x) y = f -1f( x)

6-6 0 12

12

3

0 x

x

y

( )cos 02

t t = =

When2

t =

, 12

y = +

. When2

t =

, 12

y =

.


71/97

1 12 2

y

+

(ii) 1 cos x t = ; sin y t t = + d d d 1 cosd d d sin y y t t x t x t

+= =

When2

t = , 1 cos 1 0 1

2

= = = x

, sin 1

2 2 2 y

= + = +

1 cosd 1 02 1d 1sin

2

y x

+

+ = = =

Equation of line l ,

( )1 1 12

1 12

2

y x

y x

y x

+ =

=

= +

(iii)When x = 1, 1

2 y = +

or 1

2 (by symmetry)

One of the points of intersection is the origin. From the graph in part (i), there

is another point of intersection when1

2m +

or 1

2m

7(i) 2 4 82 6

x y z

x z

+ =

+ =

From G.C, x = 6

2 z, y = 1 + 1.25 z, z = z

vector equation of l:6 2

1 1.25 ,0 1

+

r =


72/97

190.8 1

20 1916 1

1 2020

n

n

n A

= =


73/97

0.4n R n=

From GC, An = Rn when 32.4n (or tabulate it to get n = 33)The robot will overtake the athlete after 33 pulls.

(ii) 16 A = (Note: n A is an increasing sequence.) He will never reach the top.

(iii) [ ]2 2 ( 1)( 0.02)22 0.02 ( 1)

+

=

n

x nnx n n

Robot will reach the top after16

400.4

= pulls.

Athlete must reach top by 39 th pull.

2

2(19) 0.02(19)(18) ( 19 0.02) 16

39 0.02(19) 160.595

x x

x

x

+

The minimum value of x is 0.60 (2 d.p.).

9(i)1

1

1

tan

2

tan

2

tan

d ed

1d e

d dd 1

e

x

x

x

y x

x y

x x x x

y C

=+

=+

= +

When x = 0, y = 1 01 e 0C C = + = Thus

1tane x y

=

(ii)

( )

1

tan2 2

2

d ed 1 1

d1

d

x y y x x x

y x y

x

= =+ +

+ =

(iii) ( )

( )

22

2

3 2 22

d d1 (2 1) 0

ddDifferentiating w.r.t. ,

d d d d1 2 (2 1) 2 0

y y x x

x x x

y y y yx x x

+ + =

+ + + + =


74/97

( )

( )

3 2 2

3 22

3 2

1 2 (2 1) 2 0dd d d

d d d1 (4 1) 2 0

dd d

x x x x x x x

y y y x x

x x x + + + =

When x = 0, y = 1 (given)2 3

2 3d d d

1, 1, 1d d d y y y x x x

= = =

Thus Maclaurin series is

2 31 112 6

y x x x= + + + L

(iv)

(a)

( )

11 2

2

2 3 2

2

tantane e (1 )

(1 )

1 1= 1 ... 1 2 3 ...

2 6

3= 1 ...2

x x x

x

x x x x x

x x

= ++

+ + + + +

+ +

(b) e12 tan x x+ = e 2 x e

1tan x

=2

2 3(2 ) 1 11 2 ... 1 ...2! 2 6 x

x x x x + + + + + +

= 29

1 3 ...2

x x+ + +

10(a) ( ) ( )1

Let P( ) be the statement ! 1 ! 1, for alln

r

n r r n n+

=

= +

Proving P(1)

( )

( )

1

1

! 1

1 1 ! 1 1r

LHS r r

RHS

=

= =

= + =

P(1) is true

A i h P( k) i f i i i k

( ) ( ) ( )

( ) ( )( ) ( )( )

1 ! 1 ! 1 1

1 ! 1 1 1

1 ! 2 1

2 ! 1

k k k

k k

k k

k

= + + + + = + + +

= + +

= +


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( )2 ! 1k P(k)true P(k 1)is true +

Since P(1) is true,

and P(k) is true P(k 1) is true

by Mathematical Induction, P(n) is true for all n +

+

(b)(i)

(ii)

(iii)

2, 1, 3 = = = (by GC)

1

3 2

lim lim

1 5 8 5

n nn n x x L

L L L

+

= =

= + + (or any statement to the same effect)

( )

3 2

3 2

3 2 2

3 2

1 5 8 5

1 5 8 5

3 3 1 5 8 52 5 6 0

L L L

L L L

L L L L L

L L L

+ = +

+ = +

+ + + = +

+ =

As 3 22 5 6 0 L L L + = , hence L is a root of the equation 3 22 5 y x x x= Therefore, L = , or

( )

1

23

23

32

2 3 2

3 2

1 5 8 5

5 8 5 1

5 8 5 1

5 8 5 3 3 1

2 5 6 0

n n

n n n

n n n

n n n

n n n n n

n n n

x x

x x x

x x x

x x x

x x x x x

x x x

+ .

14

0

2(sin cos ) 2(cos sin )d

sin cos

+ =

+

1 14 4 (cos sin )

2 d 2 d

=


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0 0

4 40 0

sin cos

2[ ] 2[ln(sin cos )]

ln 2 (or 2 ln 2 )2 2

+

= +

=

(b)(i)2

2( 1)

t t +

= 2( 1) ( 1) A B

t t +

+ +

Solving, we have A = 2, B = 2 (ii) Using substitution t = 2 1 x ,2 1

,2

d 1 1d 2 1

t x

t x t x

+=

= =

When x =1, t = 1

x = 5, t = 3Hence

( )25 3

11 1 2

3

21

3

21

3

21

31

1 1d ( ) d

2 1

2d

2 1

2 d( 1)

1 12 ( ) d from ( )

1 ( 1)

12[ln( 1) ]

1

12ln22

t x t t

x x t

t t

t t

t t t

t t t

t t

+=

+ +

=+ +

=+

= + +

= + ++

=

b i

Solutions to SAJC H2 Maths PRELIM EXAM 2010

Paper 2

Qn Solutions

1 (i)


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1 (i)

Length of projection = AC uuur

= 2 2

2

1

2

2 1 2

AG

+ +

uuur

= 2 2

1 2

2 1

4 2

2 1 2

+ +

= 2 2 83

+ += 4 units

(ii)2 2

1

2 2

AC

= =

uuur

4 AC =uuur

2 2 24 4 16 + + =

43

=

24

13

2

AC

=

uuur

By ratio theorem,2 3

5 AG AC

AI +

=

uuur uuuruur

2 8

4 4

8 8

510 5

1 28 4

5 516 8

AI

+ =

= =

uur

1

1 24

2 13

4 2sin 60.8

1 4 16 4

= =

+ + o or 1.06 (in radians)


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2 (i) x + y = 100d d

(100 ) (100 )d d x x

x k xt t

=

1.9 = k (100 5) k = 0.02d

0.02(100 )d

x x

t =

(ii) d0.02(100 )

d x

xt

=

1d 0.02 d

100ln 100 0.02

=

= +

x t x x t C

When t = 0, x = 5, ln(95) = C

So,

( )ln 100 0.02 ln(95)

ln 100 ln(95) 0.02 since 100

1001n 0.0295

=

= 0.01, we do not reject H 0. There isinsufficient evidence, at 1% level, that the pH level differs from 8.5.

(ii) The pH level of the water in the tanks is assumed to be normallydistributed.

(b) (i) 8.31 x = ,2

21 ( 15.2)

232.2 2.9079 80s

= = (3 s.f.)

(ii) By CLT,2.9027

~ (8.31, ) X N n

approximately

P( 8.2 X < ) < 0.38.2 8.31

0.52442.9027

0.11 0.8934

8.1218

65.96

n

n

n

n

<

<

>

>

Least value of n is 66.


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Math Prelims Soln

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