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SIXTH TERM

EXAMINATION PAPERS

WORKED PROBLEMS

VOLUME 1


2/17


A FEW WORDSA FEW WORDSA FEW WORDSA FEW WORDS

The amount of Sixth Term Examination Papers (STEP) resources on the World Wide

Web are far and few, hence my motivation to produce this volume to prepare the student

more adequately for this exceptionally demanding examination standard. All solutions were

personally (painstakingly if I may add) drafted by myself- a process which took me several

months. Not to mention numerous hours spent on the careful selection of questions. Kindly

note that the worked problems presented here belong to difficulty level III; the actual

question texts are not reproduced in consideration of copyright matters. It is my sincere hope

that you will find this compilation a worthy study-aid. Good luck and god bless.

Humbly,

Frederick Koh

BEng (Hons)

BSc (Hons)

Note: This is a short version of the preview; for a longer version head over to

http://www.facebook.com/whitegroupmathematics(Click on FOR FANS ONLY tab).

The full version of this volume will be out later in 2013.


3/17


STEP III, 2000 Question 2

SOLUTIONS :

For the substitution ,cos2 =x

sinsin == dxd

dxd

When ,2=x2

= ; ,

2

3=x

3

=

ddxx

x)(sin

cos1

cos1

3

1 2

3

2

12

2

3 +

=

d)(sincos1

cos1

cos1

cos12

3

+

=

( ) ( )

dd )(sinsin

cos1)(sin

cos1

cos1 2

3

2

3

2

=

=

=2

3

cos1

d [ ]

==

2

3

31

2sin 2

3

12

3

6+=

(shown)

The appropriate substitution would be ,cos22

+=

abbax

where

sin2

sin2

=

=

abdx

ab

d

dxd

When ,2

baqx

+==

2

= ; ,

4

3 bapx

+==

3

=

Then

dab

abba

b

aabba

dxxb

axq

p

+

+

+

=

sin2

cos22

cos22

2

3

2

1


4/17


dab

abab

abab

+

= sin2cos

22

cos22

2

3

dab

abab

abab

abab

abab

+

= sin2cos

22

cos22

cos22

cos22

2

3

dab

abab

abab

= sin2cos22

cos22

2

3

2

22

dab

ab

abab

= sin2sin

2

cos22

2

3

=

2

3

cos1

2

ab

d [ ]

=

=

2

3

3

1

22

sin

2

2

3

abab

+

= 1

2

3

62

ab

+

=

6

633

2

ab

( )12

633)( +=

ab (shown)


SOLUTIONS :

iiei

2

3

2

11

3

2sin

3

2cos111 3

2

2 +

=++=+=+

==+=

i

ei3

2

3

2

1

(shown)


5/17


Im

B

3

2

q p

3

0 Re

C r

== OAOCAC pr

=AN ( ) ( )

prprepri

== 23

22

)1( rpprpANOAON==+=

(shown)

== OAOBAB pq 2

=LB ( ) ( )

pqpqpqepqi

=== 3232 [Q ]13 == ie

pqpqpqqLBOBOL +=++=++== 122 (shown)

== OBOCBC 2 qr

=BM ( ) ( )

qrqreqri

=

=

1232

qrqrqrqBMOBOM ===+= )( 22 (shown)

Argument of complex number =N 222 )11(arg)1(arg rprp =

( ) [ ]222 )(argarg rprp +==


6/17


( )33

5

3

2)arg()1arg(arg 22

==+=+==

Hence, NB is a continuous straight line passing through the origin.

(By understanding the above functions on the fact that the chords ONandOB are inclined at thesame angle to the horizontal as well as the vertical, though proving the horizontal part indirectly also

implies the vertical part is true)

Argument of complex number =L ( )3

arg)arg(

==+ pq

Hence, LC is a continuous straight line passing through the origin.

Argument of complex number = = )arg( qr

Hence, MA is a continuous flat horizontal line(lying on the real axis) passing through the origin.

Reconciling all 3 observations gives the fact that lines LC, A and NBmeet at the origin. (shown)

rqprqppqrLC ++=++== )(|||| [Q ]1|| = (shown)

rqprqpqrpMA ++=++== |)(||| (shown)

( ) rqprqprpqNB ++=++== |)(|||||2222

[Q

]1||

2

= (shown)


SOLUTIONS :

Let nPbe the proposition that 213 = nnn aaa , where2

2

11

+=n

nn

a

aa , 110 == aa and 2n

For ,2P 2133 012 === aaa and 21

111 2

0

2

12 =

+=

+=

a

aa

Since LHS=RHS, 2P is true.

Assume ,1kP kP are both true for some positive integer value ofkwhere ,2k

ie 321 3 = kkk aaa and 213 = kkk aaa


7/17


Considering :1+kP( )

1

2

221

2

1

1

2

21

1

2

1

691311

+

++=

+=

+=

k

kkkk

k

kk

k

k

ka

aaaa

a

aa

a

aa

21

1

2

2 691

++

= kkk

k aa

a

a

213 69 += kkk aaa

223 6)(3 ++= kkkk aaaa

23 33 += kkk aaa

( )3233 = kkk aaa

13 = kk aa

,1kP kP are true 1+kP is true.

Since 2P is true, by mathematical induction, nP is true for all .2n (shown)

The difference equation is given by ,03 21 =+ nnn aaa 2n

Auxiliary equation is2

530132

==+ mmm

Hence, general solution is given by ,2

53

2

53nn

n BAa

+

+= 1n

Recognising that +=+

+=+

12

511

2

53,

and11

2

5111

5121

2)15(1

253 =+=+=

=

, where

251

+= ,

The general solution can therefore be rewritten as ( )n

n

n BAa

++=

1

11

When as ,1=n ( ) ( ) )1(1

11

1111

++=

++==

BABAa


8/17


When as ,2=n ( ) ( ) )2(1

11

112

2

2

2

2

2

++=

++==

BABAa

:)2(1

)1( +

( ) 21

11 2

2

+

=

++

A

21

211 2 +

=

+++

A

211

1 2 +=

+++A

( )

+=

+++

111

1 22A

( )

+=

+

+

111

1 2A

( )

+=

+

+ 11

1 2A

21

1

+=A

:)2()1()1( ++

2)1(11

22

++=

+

B

=

++

1

1212

22

B

=

++ 1

121

12

B

=

++ 1

111

2B

( ) ( )

=

+++ 11

11

2B

11

12

=

+

B


9/17


11

2

2

=

+

B

2

2

1

+

=B

Substituting these into the general solution gives

( )n

n

na

+++

+=

1

11

1

12

2

2

( ) )3(1

11

1 22

++

+

n

n

Before proceeding, certain results must be made available to simplify (3):

+=+

=+

=

+= 1

2

53

4

526

2

512

2

5

1

51

2

5

1

55

2

2

531

1

1

12

=

+=

+=

++

=+

Hence, (3) becomes

( ) ( )[ ]nnna 2225

1 +=

( ) ( )[ ]nn

222

51 +=

[ ]nn 2225

1 +=

[ ]nn 21125

1 +=

5

)12(12 +=

nn

(shown)

( )( )( ) ( ) ( )

2

2

2

1

1

1

111 ==+

=

+

+=


10/17



SOLUTIONS :

(i) When the sphere intersects the line, ( ) ( ) 2ambmb =++

( ) ( ) 222 ammmbbb =++

( ) ( ) ( ) 02 22 =++ abbmbmm

Where we have a quadratic equation in .

Hence, if the two distinct points of intersection occur, then

( ) ( )( ) 042 22 > abbmmmb

( ) ( )( ) 044 22 > abbmmmb

( ) ( )( ) 022 > abbmmmb

Since m is a unit vector, then 1|| 2== mmm and the inequality is further reduced to

( ) ( ) 022 > abbmb

( ) 022 >+ abbmb

( )22 mbbba > (shown)

When only one point of intersection occurs, then ( ) ( )( ) 042 22 = abbmmmb

and similar further simplification would give rise to ( )22 mbbba = (shown)

For this situation, mb

mm

mb=

=

2

2

and ( )mmbbp += (substituting the above expression forinto the line

mbr += gives point P)

( )mmbbp =

( )[ ] mmmbbmp =

( )[ ] mmmbbmp =

( )( ) 0=== mbmbmmmbmbmp (shown)


11/17


(ii) If the circle is tangential to the plane, then only one point of intersection exists. If we let this

point have position vectorq , we have andq = .

Equation of plane is lnr = ; since q clearly lies on the plane,

lnq =

Substituting the above equation into this gives ( ) lndan =+

lndnna =+ )(

lnda =+

la = (shown)

(Note: nd =0 because the normal of the tangential plane would be perpendicular to the

radius vector through the centre of the second circle)

(iii) Equation of second sphere is given by2)()( adrdr =

When the two spheres intersect, rrdrdr = )()(

rrdddrrr =+ 2

( )dddr =2

1

Let a possible point of intersection(out of the infinite number of points) have position vector = c

In order for the circles to intersect in a perpendicular configuration,

( ) )3(0 = dcc and

( ) )4(2

1= dddc

,

From (3), we have dccc = ; substituting this into (4) gives ( )ddcc =2

1

Since this point of intersection also lies on the first circle,2acc =

and therefore ( ) 22 22

1addadd == (shown)


12/17



SOLUTIONS :

IM =

=

=

=

10

01

10

01

01

10

01

102

(shown)

...................!4

1

!3

1

!2

1)(

!

1)exp( 443322

0

+++++==

=

MMMMIMr

M r

r

...........!4

1

!3

1

!2

1 432 +++= IMIMI

+++

++= ............

!7!5!3............

!6!4!21

753642

MI

sincos MI += (replacing both series by their Maclaurins equivalents)

=

+

=

cossin

sincos

0sin

sin0

cos0

0cos (shown)

)exp( M represents a transformation matrix which serves to rotate any given set of coordinates( inthe yx plane) by an angle of about the origin in the clockwise direction. (shown)

=

=

00

00

00

10

00

102N (null matrix)

====

00

00..........543 NNN

=

+

=+==

= 10

1

00

0

10

01)(

!

1)exp(

0

sssNIsN

rsN r

r

)exp(sN represents a transformation matrix which serves to shear any given set of coordinates

parallel to the x axis such that the original x coordinate 1x is adjusted to 11 syx + while leaving the

coordinate 1y unchanged. (shown)

When )exp(sN )exp( M = )exp( M )exp(sN ,


13/17


10

1 s

cossin

sincos=

cossin

sincos

10

1 s

=

+

cossin

sincossincos ss

+

cossinsin

sincoscos

s

s

By observation, , n= where n (shown)

The sequence of transformations (rotation and shearing) is commutative only IF the angle of rotation

is an integer multiple of . (shown)

By considering integration by parts and chain recurrence, show that

( )!1!!

)1(),(1

0++

== baba

dtttbaIba

for ,0a 0b

( Note: I have modified this question substantially from its original version.)

SOLUTION :

dttta

bt

a

tdtttbaI bab

aba 1

1

0

1

1

0

11

0

)1(1

)1()1(

)1(),( ++

+

+

+==

)1,1(1

)1(1

1

1

0

1 ++

=+

= + baIab

dttta

b ba

)3,3(3

2

2

1

1)2,2(

2

1

1+

+

+

+=+

+

+= baI

a

b

a

b

a

bbaI

a

b

a

b

)0,(.1

..........3

2

2

1

1baI

baa

b

a

b

a

b+

+

+

+

+=

)0,(1

..........3

2

2

1

1!

!baI

baa

b

a

b

a

b

a

a+

+

+

+

+=

+

+=+

+=

1

0)!(

!!)0,(

)!(

!!dtt

ba

babaI

ba

ba ba

( )!1!!

)1()!(!!

++=

+++=

baba

bababa (shown)


14/17



SOLUTIONS :

(i) =++= pakaakp 1

)1()1( 1 =++ pkka

( ) qakakakaaakq =++= ))(())(()( 11

qakaka =++ 2212

)2()1( 12 =++ qkka

rakaakr == ))(()( 1

)3(3 = ra

:)1()2( p

qa = (shown)

Substituting this into (3):33

3

rpqr

p

q==

033 = rpq (shown)

(ii) =++= pp

)1(=+ p

(where , and are generic roots of the equation)

qq=++=

( ) )2(=++ q

3

3

3

3

p

q

p

qr ===

)3(3

3

=

p

q

Substitute both (1) and (3) into (2):


15/17


( ) qpp

q=+

3

3

3233 )( qpppq =+

33323 4

qpppq =+

0332433 =+ qqppp

Let332433)( qqpppf +=

0322223332

4

3

3 =+=

+

=

qqpqpqq

p

qqp

p

qp

p

qp

p

qf

Hence,p

q= is a root of the equation 023 =+ rqxpxx (shown)

Substitutingp

q= into (3) gives

2

2

3

3

p

q

q

p

p

q=

= (shown)

An obvious relationship arising is =2

where ==

common ratio

Hence, the 3 roots are in geometric progression .( Note though that the sequencing of the roots

to form a proper GP in my solving context is in the following order: , , ) (shown)

(iii) Let the 3 roots be ,da a , da +

Then == ,3 pap ie3

pa =

( ) qdadadaaadaq =++++= ))(())(()(

qdaadaada =+++ 2222

qda = 223

qdp

=

22

93 [ Q

3

pa = ]

qpd = 3

2

2


16/17


rdaadar =+= ))()((

rada = 23

rqppp =

333

23

rpqpp

=+3927

33

= rppq

27

2

3

3

02729 3 = rppq

Necessary condition is given by 02729 3 = rppq (shown)


SOLUTIONS :

)(xfy =

0 1 2 x

n

11+

n

21+

n

31+

As can be observed from the above graph, total area of all rectangles for 21 x

=

+=

++

++

++

+=

n

m n

mf

nn

nf

nnf

nnf

nnf

n 11

11

1........

31

121

111

1


17/17


When ,n an infinitely large number of rectangles with negligible widths are considered,

expression therefore simply approaches towards the area under the graph of )(xfy = , ie dxxf2

1

)( .

(shown)

++

++

+=

++

++

+ nn

n

nn

n

nn

n

nnnnn nn

1............

2

1

1

1lim

1............

2

1

1

1lim

+

=

+

+

+

+

+

= =

m

mnn

nm

f

nnnn

nn

nn 1 1

11lim

1

11............

21

11

11

11lim

[ ] 2ln||ln1

2

1

2

1

=== xdxx(shown)

++

++

+=

++

++

+ 22

2

2

2

2

2

2222

1............

4

1

1

1lim............

41lim

nn

n

nn

n

nn

n

nnn

n

n

n

n

n

nn

+

+

++

+=

2

2

22 1

11............

41

11

11

11lim

n

nn

n

n

n

nn

+

+

+

+

+

+

+

+

+

+

+

=

2121

11............

22

122

1

11

21

121

1

11lim

222

n

n

n

nn

nn

n

nn

nn

)1(1)1(

1

22

12

1

2

2

1

2

+=

+= dxx

dxxx

Using the substitution ,tan1 =x (1) becomes

( )

24

0

2sec

1tan

1 +

d [ ]4

40

4

0

=== d

(shown)

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