Are all the questions of a similar level of diculty?
No. Not surprisingly, many of the
original STEP questions were more suitable for examinations than for private study, and I did not
hesitate to adapt a question where I saw the need or where I saw an op
PREFACE
I have set out below the answers to the questions you will want to ask before getting down to work.
If I have forgotten anything, let me know and I will include it in the next of what I hope will be
many editions.
Is this booklet suitable for me?
CONTENTS
Contents
i
Preface
iii
Problems
Odd-numbered pages
Solutions
Even-numbered pages
Each problem has been given a title (not used later on), to remind you what it is about
if you have already looked at it, followed by a rough indication of the mathe
Question 22
The function f satises the condition f (x) > 0 for a
a suitable change of variable, prove that
b
a
x
b, and g is the inverse of f. By making
f(x) dx = b a
g(y ) dy,
(1)
where = f(a) and = f(b). Interpret this formula geometrically, by means o
Solution to Question 21
g(t)
The rst integral in equation (1)
in the question is equal to the
area under the lower curve in the
gure and the dierence between
the two integrals is equal to the
area between the curves, both areas being obviously positive.
f
Question 21
Show by means of a sketch that if 0
f (t)
g (t) for 0
x
0
t
x
f (t) dt
0
g (t) dt.
1
10
0
(1)
0
Starting from the inequality 0 cos t 1, prove that if 0
1
that 1 2 x2 cos x 1. Deduce that
1
1800
x then
x
dx
(2 + cos x)2
x
1
2
then 0
sin x
x and
Solution to Question 20
(i) Since we are dealing with integers, a < b is the same as a
b 1. Therefore, bc
as c > 0.
Similarly, d < c implies that ad a(c 1), as a > 0, i.e. ad a(c 1).
(a + 1)c,
Putting these together gives
bc ad
(a + 1)c a(c 1) i.e. bc ad
Question 20
Let a, b, c, d, p and q be positive integers. Prove that
(i) If a < b and d < c, then bc ad a + c;
a
c
(ii) If
< p < , then (bc ad)p a + c;
b
d
p
c
a+c
a
< < , then p
and q
(iii) If
b
q
d
bc ad
b+d
.
bc ad
Find all fractions with denominators
Solution to Question 19
(i) We need to show that (C1 )T CT = I, because multiplying this on the right by (CT )1 gives the
required formula. Start from CC1 = I and take the transpose, using the given formula for the
transpose of a product:
CC1 = I = (CC1 )
Question 19
Let A and C be n n matrices, and denote the n n unit matrix by I and the n n zero matrix
by O.
(i) Show that if C is non-singular then (CT )1 = (C1 )T , where CT denotes the transpose of C.
(ii) Now assume that the matrix (I A) is non-singular
Solution to Question 18
The general solution can be written in the form
x = A cos kt + B sin kt ,
where A and B are arbitrary constants.
At time t after the rope becomes taut, let y be the displacement of the trailer from its original
position and x be th
Question 18
Write down the general solution of the equation
x = k 2 x,
where k is a constant.
A truck is towing a trailer of mass m across level ground by means of an elastic tow-rope of natural
length l and modulus of elasticity . At rst the rope is slac
Solution to Question 17
(i) The inverse of f(x) = ax is f 1 (y ) = y/a, since then f 1 f(x) = x as required. Substituting into
Lagranges formula gives
f 1 (y ) = y +
1
1 dn1
[y ay ]n
n! dy n1
= y+
= y+
1
1
= y+y
(1 a)n
1 dn1 n
y
n! dy n1
(1 a)n y
(1 a)
,
Question 17
If y = f(x), the inverse of f is given by Lagranges identity:
f 1 (y ) = y +
1
1 dn1
n
y f(y ) ,
n1
n! dy
when this series converges.
(i) Verify Lagranges identity when f(x) = ax.
(ii) Show that one root of the equation x 1 x3 =
4
x=
0
3
4
is
Solution to Question 16
To obtain equation (4), we just subtract equation (3) from equation (2) and factorise:
b c = (y 2 zx) (z 2 xy ) = (y 2 z 2 ) + (xy xz ) = (y z )(y + z ) + (y z )x = (y z )(x + y + z )
By symmetry, we can write down equations for c
Question 16
Given that x, y and z satisfy
x2 yz =
y 2 zx =
z 2 xy
=
a,
b,
(1)
(2)
c,
(3)
where a, b and c are real, not all equal, and a + b + c > 0, show that
b c = (y z )(x + y + z ) .
(4)
By considering
(b c)2 + (c a)2 + (a b)2 ,
or otherwise, show tha
Solution to Question 15
1
For 0
tan < 1 , we have tan < 1,
4
so that tank is close to zero (i.e. much
smaller than 1) when k is large. This is
illustrated in the gure which shows three
cases: for k = 1 the graph is mildly curved;
for larger k the graph hu
Question 15
Give rough sketches of the function tank for 0
1
4
Show that for any positive integer n
/4
0
tan2n+1 d = (1)n
and deduce that
ln 2 =
Show similarly that
1
2
in the two cases k = 1 and k 1.
(1)m
m=1 2m
n
ln 2 +
(1)m
.
m
m=1
,
()
()
(1)m
=
.
Solution to Question 14
We can isolate z (i.e. eliminate x and y ) in one go because of the special form of the equations:
(1) a (2) gives
=
z (1 ab) = a2 a
a(a 1)
z=
1 ab
provided ab = 1.
Note the condition ab = 1; we have to take into account all possib
Question 14
Find the simultaneous solution of the three linear equations
a2 x + ay + z
ax + y + bz
=
=
a2
1
(1)
(2)
a2 bx + y + bz
=
b
(3)
for all possible real values of a and b.
Discussion
The simplest way to tackle this is to use an elimination (or sub
Solution to Question 13
This problem can be solved using tree diagrams; see Question 8 for an example of this kind of
solution. A more sophisticated (but not necessarily better) method is to use Bayes theorem.
Bayes theorem, in its simplest form, states:
Question 13
My two friends, who shall remain nameless, but whom I shall refer to as P and Q, both told me
this afternoon that there is a body in my fridge. Im not sure what to make of this, because P tells
the truth with a probability of p, while Q (indep
Solution to Question 12
To prove the trig. identity, we can use the formulae for cos(a + b), etc:
cos 3 = cos(2 + ) = cos 2 cos sin 2 sin
and then use double angle formulae:
= (cos2 sin2 ) cos (2 sin cos ) sin = cos3 3 sin2 cos ,
which turns into the giv
Question 12
Prove that cos 3 = 4 cos3 3 cos .
Show how the cubic equation
24x3 72x2 + 66x 19 = 0
()
can be reduced to the form
4z 3 3z = k
by means of the substitutions y = x + a and z = by , for suitable values of the constants a and b.
Hence nd the thre
Solution to Question 11
Making the suggested change of variable, we have
(x )( x)
sin2 + sin2 cos2 cos2
=
=
and
(cos2 1) + sin2 (1 sin2 ) cos2
=
( )2 sin2 cos2
dx
= 2 cos sin + 2 sin cos = 2( ) sin cos .
d
()
Now we need to work out the limits of the
Question 11
Show by means of the substitution x = cos2 + sin2 , or otherwise, that
1
(x )( x)
dx =
when < . What is the value of the integral when < ?
Using the substitution t = x1 , show that
b
a
dt =
ab
t (t a)(b t)
1
when 0 < a < b.
Evaluate, for 0 <
Solution to Question 10
a
The point represented by
a + (z a)ei
a + (z a)ei
is a rotation by anticlockwise about
the point represented by a of the point
represented by z .
z
There are a number of ways we could write the required conditions. One way (sugges