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WHITE GROUP MATHEMATICS (PREVIEW COPY) Page 1
WG
AAAA WHITE GROUP MATHEMATICS PUBLICATIONWHITE GROUP MATHEMATICS PUBLICATIONWHITE GROUP MATHEMATICS PUBLICATIONWHITE GROUP MATHEMATICS PUBLICATION
(www.whitegroupmaths.com)
SIXTH TERM
EXAMINATION PAPERS
WORKED PROBLEMS
VOLUME 1
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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.
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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
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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)
STEP III, 2000 Question 3
SOLUTIONS :
iiei
2
3
2
11
3
2sin
3
2cos111 3
2
2 +
=++=+=+
==+=
i
ei3
2
3
2
1
(shown)
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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 +==
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( )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)
STEP III, 2000 Question 8
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
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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
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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
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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 ==+
=
+
+=
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STEP III, 1998 Question 8
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)
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(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)
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STEP III, 1998 Question 5
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 ,
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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)
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STEP III, 1999 Question 1
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):
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( ) 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
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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)
STEP III, 1999 Question 3
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
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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)