2004 Hints and Answers

2005 Hints and Answers

2006 Hints and Answers
2007 Examiners' Report
2007 Solutions
2008 Examiners' Report
2008 Solutions
2009 Examiners' Report
2009 Solutions
2010 Examiners' Report
2010 Solutions
2011 Examiners' Report
2011 Solutions
2012 Examiners' Report
2012 Solutions
2013 Examiners' Report
2013 Solutions
2014 Examiners' Report
2014 Solutions
2015 Examiners' Report
2015 Mark Scheme
2015 Solutions
2016 Examiners' Report
2016 Mark Scheme
2016 Solutions
STEP mark schemes 2017

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 STEP

Mathematics
STEP 9465, 9470, 9475

STEP Hints and Answers

June 2005

STEP 2005

Oxford Cambridge and RSA Examinations 38

OCR (Oxford, Cambridge and RSA Examinations) is a unitary awarding body,
established by the University of Cambridge Local Examinations Syndicate and the
RSA Examinations Board in January 1998. OCR provides a full range of GCSE, A
level, GNVQ, Key Skills and other qualifications for schools and colleges in the
United Kingdom, including those previously provided by MEG and OCEAC. It is also
responsible for developing new syllabuses to meet national requirements and the
needs of students and teachers.

This mark scheme is published as an aid to teachers and students, to indicate the
requirements of the examination. It shows the basis on which marks were awarded
by Examiners. It does not indicate the details of the discussions which took place at
an Examiners’ meeting before marking commenced.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the relevant
knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question papers and
the Report on the Examination.

OCR will not enter into any discussion or correspondence in connection with this
mark scheme.

© OCR 2005

Any enquiries about publications should be addressed to:

OCR Publications
PO Box 5050
Annersley
NOTTINGHAM
NG15 0DL

Telephone: 0870 870 6622

Facsimile: 0870 870 6621
E-mail: publications@ocr.org.uk

39
 CONTENTS

STEP Mathematics (9465, 9470, 9475)

HINTS AND ANSWERS

Unit Content Page

9465 STEP Mathematics I 1

9470 STEP Mathematics II 15

9475 STEP Mathematics III 23

REPORT ON THE COMPONENTS

Unit Content Page

9465 STEP Mathematics I 40

9470 STEP Mathematics II 42

9475 STEP Mathematics III 53

* Grade Thresholds 56

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Step I Hints and Answers June 2005

Step I, Hints and Answers

June 2005

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Step I Hints and Answers June 2005

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Step I Hints and Answers June 2005

14
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Step II Hints and Answers June 2005

Step II, Hints and Answers

June 2005

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Step II Hints and Answers June 2005

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Step II Hints and Answers June 2005

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Step II Hints and Answers June 2005

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Step III Hints and Answers June 2005

Step III, Hints and Answers

June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Step III Hints and Answers June 2005

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Report on the Components taken in June

Report on the Components

June 2005

39
80
Report on the Components taken in June

9465 - Mathematics 1

General comments

This paper was found to be more straightforward than last year’s, with the exception of the
questions on Probability and Statistics. The Mechanics questions (in particular, 9 and 10) were
more popular than previously. Inaccurate algebraic manipulation remains the biggest obstacle to
candidates’ success: at this level, the fluent, confident and correct handling of mathematical
symbols is necessary and is expected. Many good starts to questions soon became unstuck after a
simple slip. There was little evidence that candidates were prepared to check their working, doing
so would have improved many candidates’ overall mark.

The weakness of many candidates’ integration was striking, and somewhat alarming.

Comments on specific questions

1 This was a popular question, and most candidates were familiar with the underlying
principle that, if there are n symbols of which a are of one type and b are of another type
etc, then there are n! ÷ (a! × b! × …) distinct rearrangements of the n symbols. It was
important to enumerate systematically the combinations totalling 39, to avoid counting
possibilities more than once.

2 This was a popular question, but was one which required careful algebraic manipulation. It
was pleasing that most candidates saw how to use the statement that “(1, 0) lies on the line
PQ” to deduce that pq = –1. Proving that PSQR was a rectangle was rarely done in full:
most candidates proved necessary conditions (e.g. there were two interior right angles)
rather than sufficient conditions (e.g. there were three interior right angles in a quadrilateral,
hence there were four). Considering the lengths of the sides without considering at least
one interior angle did not remove the possibility that the quadrilateral was a (non-
rectangular) parallelogram.

3 Most candidates who tackled this question knew what to do, but did not express clearly the
necessary reasoning. In part (i) the solution “x = √ab or x = –√ab” did not show that the
given equation had “two distinct real solutions”; there was needed an explicit statement that
since a and b were either both positive or both negative then ab > 0, hence √ab was real.
Similarly, in part (ii) most candidates did not explain why c2 ≠ 0.

In such questions, candidates are reminded of the need to explain clearly each component
of the result they have been given.

4 Part (a) was usually well done, though quite a few candidates did not justify the negative
value of sin θ. Arithmetical errors marred many evaluations of cos 3θ. The given identity
was not found difficult to prove, and most candidates saw that in part (b) they were being
asked to solve 2x3 – 33x2 – 6x + 11 = 0. Unfortunately, very few were able to make further
progress: substituting (correctly) x = ½ was rarely seen. Most of those who made progress
remembered to explain which of the three values of x was the value of tan θ.

5 Neither integral in this question was at all difficult, so it caused some concern to see poor
implementation of routine techniques such as integration by substitution or by parts. In part
(i) not every candidate linked the two cases together. In part (ii) m = 0 was often asserted to
be a special case before the integration had been performed. Even after that, the terms m +

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Report on the Components taken in June

1 and m + 2 in the denominator of the answer did not always prompt candidates to consider
m = –1 and also m = –2 as special cases.

6 Part (i) was well done. Those who attempted part (ii) derived the appropriate equations, but
then found it hard to proceed: a common error was to assert that if ax2 + bx + cy2 = d was to
be the same as x2 + 14x + y2 = 51, then a = 1, b = 14, c = 1 and d = 51, rather than the
correct deduction that b ÷ a = 14, c ÷ a = 1 and d ÷ a = 51.

7 This was not a popular question, though part (i) was usually well done. Part (ii) required the
factorisation of r2 – 1, and those who saw this usually simplified the product. Very few
solutions to part (iii) were seen: the replacing of cot θ with cos θ ÷ sin θ and the
simplification of the two terms into a single fraction was very rarely seen. Those who
reached this stage did not all recognise that cos A sin B + cos B sin A can be simplified
further.

8 Most candidates who attempted this question did so confidently and largely successfully. It
was pleasing to see that they understood how to use the hint implicit in the first result they
derived. However, a lot of solutions were flawed by the omission of the constant of
integration.

9 Many candidates attempted this question, and made good progress. Commonly seen was
the incorrect statement that ½ T sin θ = µR (derived by resolving horizontally on the rod).
There was no statement in the question that the rod was about to slip, hence it was wrong
to assert that the frictional force equalled µR. Full marks were not awarded unless the
candidate was careful to state that F ≤ µR.

10 This was the most popular Mechanics question, and was often well done. Clearly labelled
diagrams would have helped both candidates and examiners. Candidates are encouraged
to simplify answers as fully as possible: the results in parts (i) and (ii) were not always
reduced to their simplest forms. It was not necessary to do so to achieve full marks, but at
this level candidates should expect to give answers as neatly as possible.

11 The Mechanics tested by this question was not demanding, but candidates found that
solving the resulting equations was taxing. Great care was needed in part (iii). Many
solutions began “v = kr so 2 cos 2t = k × sin 2t, and –2 sin t = k × 2 cos t”. The subsequent
deduction that 2 cot 2t = k = – tan t should not have been made, without considering
whether sin 2t or cos t equalled zero.

12 This question was very poorly answered. In part (a) almost every candidate assumed that
hat-wearing and pipe-smoking were independent, and so multiplied together the given
probabilities. A handful of attempts at part (b) were seen. It was intended that this question
be tackled with Venn diagrams rather than tree diagrams: candidates seemed utterly
unfamiliar with these.

13 Very few attempts at this question were seen, which was surprising since, with the aid of
sketch graphs and a tree diagram, it was probably a lot more straightforward than question
12.

14 No successful attempts at this question were seen. Those who started it usually failed to
realise that they had been given the cumulative distribution function rather than the
probability density function.

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9475 - MATHEMATICS III

Report for Publication to Centres

General Comments

Almost all candidates this year chose to answer questions 2, 3, 5, 6 and 7, together with question 1
or 4. Many, usually the weaker candidates, tackled more than six questions, though most of these
were usually incomplete, or even barely started. Some good candidates concentrated on very full
and thorough answers to fewer than six questions. Few attempts were made at the Mechanics
questions and even fewer at the Probability and Statistics questions.

Some very impressive work was again seen from the best candidates, and this year there were far
fewer candidates who were essentially unable to answer any of the questions. Those doing less
well are typically candidates who can do effectively what they are explicitly told to but are unable to
make progress where the method is not provided by the question or those who can see what do,
but who are hampered by poor technical skills, especially in algebra, but also in trigonometry or
calculus.

Lack of clarity about the direction of implications is often a weakness, even of good candidates: for
⎛π ⎞ π
instance (in question 1) cos B = sin ⎜ − B ⎟ so sin A = cos B ⇒ A = − B , but cos is an even
⎝2 ⎠ 2
π
function so A = ± B . Many are not clear what is required to “show” a result: either sufficient
2
working to indicate the steps taken or a written explanation of what is going on are essential. For
instance, it was common to see (in question 2)
⎛ ⎞ ⎛ ⎞
c2 2
c2 2 2

(
d 2
)
x + y 2 = 2x ⎜1 − ( )
⎟ so d x 2 + y 2 = 2 ⎜ 1 − ⎟ + 8c x
⎜
( ) ⎟ ⎜
( ) ( )
⎟ x 2 + a2 3
2
dx x 2
+ a 2 dx 2 x 2
+ a 2
2

⎝ ⎠ ⎝ ⎠
where candidates were asked to show the latter result, which is on the question paper: of course, it
is easy to reconstruct the steps in the differentiation, but that is what the question is asking for.

Few candidates seem prepared to check their work, or to go back to look for obvious errors. It was
common to see the plaintive remark “I must have made a slip somewhere” at the end of a
derivation that failed to give the expected answer, where the error was a simple as the
mistranscribing from one line to the next of a negative sign as a positive sign.

Comments on Specific Questions

1 This question was surprisingly unpopular, given that it was the first question on the paper –
perhaps the sight of unfamiliar trigonometric graphs put candidates off. The first two parts
π
were very poorly done: for the first result almost all only showed that A = ( 4n + 1) ± B was
2
sufficient for sin A = cos B, while in the second many looked for turning points but assumed
without comment that the maximum of the modulus of a function must occur at the maximum
of the function. Almost all could then use these results to show that sin ( sin x ) = cos ( cos x )
had no solutions. The first two graphs were usually correctly sketched, though many had
cusps on one or other curve, but the graph of y = sin(2 sin x ) was very often attempted

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Report on the Components taken in June

π
without further calculation and almost always then had a maximum at x = instead of a
2
minimum there and maxima either side of this. Candidates should be aware that graph
sketches in STEP papers will often require considerable working, such as determining
turning points and their nature, even if this is not explicitly indicated in the question.

2 This question was attempted by almost all candidates and most managed the early parts
successfully, though many used the expression 2 + 2 ( y ′ ) + 2yy ′′ for the second derivative of
2

x 2 + y 2 , which made this part much harder than necessary. Very few than achieved full
marks for determining the closest points to the origin on the curve, noticing correctly the
existence or otherwise of two turning points of x 2 + y 2 other than at x = 0, and showing
clearly which points were minima, under the two conditions on a and c.

3 Almost all candidates attempted this question, and could complete the algebra correctly;
most could also solve the quartic equation in the last part using one of the ideas from the
earlier parts. Relatively few, however, understood what was meant by a necessary condition
on a, b and c and simply gave formulae for these in terms of p, r and s, while virtually no-one,
even among those who found that a3 + 8c = 4ab was necessary, could establish the
sufficiency of this condition.

4 This question was one of the less popular Pure questions. There seemed to be a fairly sharp
divide between the many candidates who dealt very effectively with the induction and the
many who were unsure what the induction hypothesis was or what was required for the base
case or who went round in circles with the recurrences. In the last part, most were able to
find the required conditions successfully.

5 Almost all attempted this question and found the first two parts straightforward. Most (apart
from the significant number who did not understand the phrase ‘common tangent’) could also
identify the discriminant condition for there to be only one tangent. Disappointingly few,
however, could link this accurately to the condition for the two curves to touch. The last part
was almost always poorly done, with candidates either continuing to use the discriminant,
which is not appropriate in this case, since the equation is linear, or being unable to state
clearly what the condition is for a linear equation to have exactly one solution.

6 This question was tackled by most candidates. Almost all could do the first part; most could
b
show that u + is a root of the equation, by a variety of methods, and relate their results to
u
the final equation. Fewer could convincingly establish the quadratic satisfied by the other two
roots of the cubic, and startlingly few could accurately solve this quadratic to get roots in
terms of ω with, in particular, very many sign errors.

7 This question was popular and almost all could obtain the general result quoted, with most
being able to go on to use this result on the given integrals. Many could then complete the
⌠ du
integration for (i), but very few knew how to tackle the integral ⎮ in (ii). Candidates
⌡ u u +1
(perhaps because this was the most recent technique they had learnt) almost all saw this as
an opportunity to substitute u = sinh2t or u = tan2t, which will work in principle, but are not as

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simple to execute correctly as u = t2 – 1. Many perfectly satisfactory methods of integration

not based on the general result given were also seen.

8 This was easily the least popular Pure question, though still attempted more frequently than
any of the applied questions. Solutions getting beyond the first couple of parts successfully
were extremely rare, with many not recognising |a – c| as the radius of the circle, and hardly
any being able to use the result 2aa* = ac* + ca* to show that the conditions for B and B’ to
lie on the circle were equivalent, let alone the converse.

9 This was the most popular Mechanics question and most of those who attempted it seemed
to know what to do, though those who had a formulaic approach to Newton’s Law of
difference of final velocities
restitution, writing it as −e = , frequently made at least one sign
difference of initial velocities
error in using this equation, and there were many who did not make a consistent decision
about the sign convention for the final velocities, for instance by drawing a diagram.
However, virtually all were defeated by the algebra required – it was very common indeed,
( )
for instance, to see the expression (1 − e ) miscopied as 1 − e 2 at some point in the
2

calculation, or vice versa. Unfortunately, the problem was unforgiving about this, and
incorrect early results made it very difficult to complete the question successfully.

10 There were a reasonable number of attempts at this question, with some good efficient
solutions, but with most candidates giving up when they did not get the required result in (i).
This was usually because they had included the work done by friction on only one disc, or
because they had not realised that the extension in the band is twice the distance apart of
the centres, or because they assumed that there would be no elastic potential energy stored
in the band if the two discs were in contact, or some combination of these.

11 This question was rarely attempted. There were a few good solutions, but most could only
tackle the first part – distinguishing the two equilibria by their stability (for example by finding
the value of sin θ and cos θ at each) was found difficult, and consequently the last part was
inaccessible.

12 There were very few attempts at this question and few of these progressed beyond the first
part, the result E ⎡⎣Y 2 ⎤⎦ = Var [Y ] − E [Y ] not being recognised as useful.
2

13 This was the only Probability and Statistics question attempted by more than a handful of
candidates, but was still only tackled by a small minority. In the first part, most could derive a
correct expression for the probability that the player wins exactly £r, but hardly any could
either sum the series for the expectation or, alternatively, spot the connection with a
Geometric random variable. Many did not then proceed to the second part, but those who did
often found it more straightforward.

14 There were hardly any attempts at this question: most of these could successfully get to the
expected value of V, but were unable to use the asterisked result to find the density function.

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Report on the Components taken in June

STEP Mathematics (9465/9470/9475)

June 2005 Assessment Session

Unit Threshold Marks

Unit Maximum S 1 2 3 U
Mark
9465 120 96 80 62 45 0
9470 120 89 64 49 31 0
9475 120 80 59 47 35 0

The cumulative percentage of candidates achieving each grade was as follows:

Unit S 1 2 3 U
9465 11.3 27.5 47.9 70.3 100.0
9470 15.0 45.0 65.5 86.9 100.0
9475 14.7 40.5 61.9 81.6 100.0

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101
 STEP

Mathematics
STEP 9465, 9470, 9475

STEP Hints and Answers

June 2006

STEP2006

Oxford Cambridge and RSA Examinations 102

OCR (Oxford, Cambridge and RSA Examinations) is a unitary awarding body,
established by the University of Cambridge Local Examinations Syndicate and the
RSA Examinations Board in January 1998. OCR provides a full range of GCSE, A
level, GNVQ, Key Skills and other qualifications for schools and colleges in the
United Kingdom, including those previously provided by MEG and OCEAC. It is also
responsible for developing new syllabuses to meet national requirements and the
needs of students and teachers.

This mark scheme is published as an aid to teachers and students, to indicate the
requirements of the examination. It shows the basis on which marks were awarded
by Examiners. It does not indicate the details of the discussions which took place at
an Examiners’ meeting before marking commenced.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the relevant
knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question papers and
the Report on the Examination.

OCR will not enter into any discussion or correspondence in connection with this
mark scheme.

© OCR 2006

Any enquiries about publications should be addressed to:

OCR Publications
PO Box 5050
Annersley
NOTTINGHAM
NG15 0DL

Telephone: 0870 870 6622

Facsimile: 0870 870 6621
E-mail: publications@ocr.org.uk

103
 CONTENTS

STEP Mathematics (9465, 9470, 9475)

HINTS AND ANSWERS

Unit Content Page

9465 STEP Mathematics I 1

9470 STEP Mathematics II 13

9475 STEP Mathematics III 21

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Step I, Hints and Answers
June 2006

1
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Step I Hints and Answers June 2006

2
107
Step I Hints and Answers June 2006

3
108
Step I Hints and Answers June 2006

4
109
Step I Hints and Answers June 2006

5
110
Step I Hints and Answers June 2006

6
111
Step I Hints and Answers June 2006

7
112
Step I Hints and Answers June 2006

8
113
Step I Hints and Answers June 2006

9
114
Step I Hints and Answers June 2006

10
115
Step I Hints and Answers June 2006

11
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Step I Hints and Answers June 2006

12
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Step II, Hints and Answers
June 2006

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Step II Hints and Answers June 2006

STEP MATHEMATICS PAPER 2 (9470) June 2006

HINTS AND ANSWERS

Q1 If you read through at least part (i) of the question, you will see that it is necessary to
work with u1, u2, u3 and u5 (and hence, presumably – as the sequence is defined
recursively – with u4 also). Although it is not the only way to go about the problem, it
makes sense to work each of these terms out first. Each will be an expression
involving k and should ideally be simplified as you go. Thus, u1 = 2 gives u2 = k –
2
36 k  18k  36
18, u3 = k – = , etc. Then, for (a), u2 = 2; for (b), u3 = 2; and,
k  18 k  18
for (c), u5 = 2. Each result leads to a polynomial equation (of increasing orders) to be
solved. Finally, you need to remember that, in the case of (c) for instance, of the four
solutions given by the resulting equation, two of them must have arisen already in
parts (a) and (b) – you’ll see why if you think about it for a moment. Ideally, you
would see this beforehand, and then this fact will help you factorise the quartic
polynomial by the factor theorem.

A simple line of reasoning can be employed to establish the first result in (ii) without
36 36
the need for a formal inductive proof. If un t 2, then un + 1 = 37 – t 37 – =
un 2
19 > 2. Since u1 = 2, it follows that all terms of the sequence are t 2. In fact, most of
them are much bigger than this. Then, for the final part of the question, the informal
observation that, eventually, all terms effectively become equal is all that is required.
Setting un + 1 = un = l (say) leads to a quadratic, with two roots, one of which is
obviously less than 2 and can therefore be rejected.

Answers: (i) k = (a) 20; (b) 0; (c) r 6 2 . (ii) 36.

Q2 The formula books give a series for ex. Setting x = 1 then gives you e as the limit of
an infinite sum of positive terms, and the sum of the first four of these will then
provide a lower bound to its value.

In the next part, you (again) can provide a perfectly sound argument for the required
result without having to resort to a formally inductive one (although one would be
perfectly valid, of course). Noting firstly that 4! = 24 > 16 = 24, (n + 4)! consists of
the product of 4! and n positive integers, each greater than 2; while 2n + 4 consists of
16 and a further n factors of 2. Since each term in the first number is greater than the
corresponding term in the second, the result follows. [Alternatively, 4! > 24 and n! >
2n Ÿ (n + 1)! = (n + 1) u n! > 2 u n! (since n > 4) > 2 u 2n (by hypothesis) = 2n + 1,
and proof follows by induction.] Now, adding the terms in the expansion for e
beyond the cubed one, and noting that each is less than a corresponding power of 12
8
using the result just established, gives e < 3 + the sum-to-infinity of a convergent
GP.

There are two common methods for showing that a stationary value of a curve is a
max. or a min. One involves the second derivative evaluated at the point in question.

14
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Step II Hints and Answers June 2006

There are several drawbacks involved with this approach. One is that you have to
differentiate twice (which is ok with simple functions). A second is that you need to
know the exact value(s) of the variable being substituted (which isn’t the case here).
d2 y
Another is that the sign of doesn’t necessarily tell you what is happening to the
dx 2
4 d2 y
curve. (Think of the graph of y = x , which has = 0 at the origin, yet the
dx 2
stationary point here is a minimum!)

Thus, it is the other approach that you are clearly intended to use on this occasion.
dy
This examines the sign of slightly to each side of the point in question. When x =
dx
1
2
, using e < 67
24
shows …..; at x = 1, using e > 83 shows …..; and at x = 54 , we can
use any suitable bound for e, such as e < 3 for instance, to show that …..

Finally, since the answers are given in the question, it is important to state carefully
the reasoning that supports these answers.

1
Q3 If you fail to notice that = 5 – 24 , then this question is going to be a bit
5  24
of a non-starter for you. The idea of conjugates, from the use of the difference of two
squares, should be a familiar one. As is the binomial theorem, which you can now use
to expand both (5 + 24 )4 and (5 – 24 )4. When you do this, you will see that
all the 24 bits cancel out, to leave you with an integer. For the next part, some
fairly simple inequality observations, such as
2
20. 25 < 24 < 25 Ÿ 4.5 < 24 < 5 and 2 u100 = 200 < 208 = 11 u 19 Ÿ <
19
11
100
help to establish the required results. It follows that 0.14 < (5 – 24 )4 < 0.114 and
the difference between the integer and (5 + 24 )4 is this small number, which lies
between …..

For part (ii), it is simply necessary to mimic the work of part (i) but in a general
1 2
setting, again starting with the key observations that N  N  1 and
N  N 2 1

that the binomial expansions for N  N

2
1   N 
k
2
N 1 
k
will lead to the
cancelling of all surd terms, to give an integer, M say. Now N  N 2  1  
k
is positive,
  k
and the reciprocal of a number > 1, so N  N 2  1 o 0+ as k o f. Also,
1 1
2N – 1
2 < N  N 2  1 < 2N Ÿ > N  N 2 1 > .
2N  1 2N
2

15
120
Step II Hints and Answers June 2006


Thus N  N 2  1 
k
M  N 2
N 1  differs from an integer (M) by less than
k

k
§ 1 ·
¨ ¸ = (2N – 1
)–k.
¨ 2N  1 ¸ 2
© 2¹

Answers: (i) 9601.9999

Q4 Using the given substitution, the initial result is established by splitting the integral
ʌ
into its two parts, and then making the simple observation that ³ x f(sinx ) dx =
0
ʌ

³ t f(sint ) dt .
0

This result is now used directly in (i), along with a substitution (such as c = cos x).
The resulting integration can be avoided by referring to your formula book, or done
by using partial fractions. In (ii), the integral can be split into two; one from 0 to S,
the second from S to 2S. The first of these is just (i)’s integral, and the second can be
determined by using a substitution such as y = x – S (the key here is that the limits
will then match those of the initial result, which you should be looking to make use of
as much as possible). In part (iii), the use of the double-angle formula for sin 2x
gives an integral involving sines and cosines, but this must also count as a function of
sin x, since cos x = 1  sin 2 x . Thus the initial result may be applied here also. Once
again, the substitution c = cos x reduces the integration to a standard one.

Answers: (i) 1
4 S ln 3 ; (ii)  12 S ln 3 ; (iii) S ln 4
3 .

Q5 The crucial observation here is that the integer-part (or INT or “floor”) function is a
whole number. Thus, when drawing the graphs, the two curves must coincide at the
left-hand (integer) endpoints of each unit range, with the second curve slowly falling
behind in the first instance, and remaining at the integer level in the second. Note that
the curves with the INT function-bits in them will jump at integer values, and you
should not therefore join them up at the right-hand ends (to form a continuous curve).

The easiest approach in (i) is not to consider ³y 1 dx – ³y 2 dx (i.e. separately), but

rather ³ y 1  y 2  dx. This gives a multiple of x – [x] to consider at each step, and this
1
simply gives a series of “unit” right-angled triangles of area 2 to be summed.

In (ii), several possible approaches can be used, depending upon how you approached
(i). If you again focus on the difference in area across a representative integer range,
then you end up having to sum k + 116 from k = 1 to k = n – 1. Otherwise, there is
some integration (for the continuous curves) and some summation (for the integer-
part lines) to be done, which may require the use of standard summation results for
¦ k and ¦ k 2 .

16
121
Step II Hints and Answers June 2006

3
Answers: (i) 2
n(n – 1) .

§a· § x·
¨ ¸ ¨ ¸
Q6 The two vectors to be used are clearly ¨ b ¸ and ¨ y ¸ . The inequality arises when you
¨c¸ ¨z¸
© ¹ © ¹
2
note that cos T d 1. The statement is an equality (equation) when cos T = r1, in
which case the two vectors must be parallel, so that one is a (non-zero) multiple of the
other. [The question cites an example of a result widely known as the Cauchy-
Schwarz Inequality.] The equality case of the inequality is then used in the two
following parts; simply in (i) – since we must have y = z = ….., from which it
follows that x = 12 this. In (ii), you should check that this is indeed an equality case
of the inequality when the two vectors are ….. and ….. The parallel condition (one
being a multiple of the other) now gives p, q and r in terms of some parameter (say
O), and you can substitute them into the linear equation (of the two given this is
clearly the more straightforward one to use), find O, and then deduce p, q and r; these
values actually being unique.

Answers: x = Oa , y = Ob and z = Oc ; (i) x = 7 ; (ii) p = 24 , q = 6 , r = 1.

Q7 This is a reasonably routine question to begin with. The general gradient to the curve
can be found by differentiating either implicitly or parametrically. Finding the
gradient and equation of line AP is also standard enough; as is setting y = b in order
§ (1  k ) a ·
to find the coordinates of Q: ¨ , b ¸ . The equation of line PQ follows a similar
© (1  k ) ¹
§  (1  k 2 )b · b (1  k 2 )
line of working, to get y = ¨¨ ¸¸ x  . If you are not familiar with the
© 2 ka ¹ 2k
t = tan 12 -angle identities, the next part should still not prove too taxing, as you should
be able to quote, or derive (from the formula for tan(A + B) in the formula books),
the formula for tan 2A soon enough; and the widely known, “Pythagorean”, identity
cosec2A = 1 + cot2A will help you sort out the gradient and intercept of PQ to show
that the two forms of this line are indeed the same when k = tan( 12 D).

A sketch of the ellipse, though not explicitly asked-for, should be made (at least once)
so that you can draw on the lines PQ in the cases k = 0 and k = 1.

Answers: Yes; PQ is the vertical tangent to the ellipse.

Yes; PQ is the horizontal tangent to the ellipse.

Q8 I’m afraid that this question involves but a single idea: namely, that of intersecting
lines. The first two parts are simple “bookwork” tasks, requiring nothing more than
an explanation of the vector form of a line equation as r = p.v. of any point on the
line + some scalar multiple of any vector (such as y – x, in this case) parallel to the
line; then the basic observation that CB || OA Ÿ CB = Oa to justify the second
result.

17
122
Step II Hints and Answers June 2006

Thereafter, it is simply a case, with (admittedly) increasingly complicated looking

position vectors coming into play, of equating a’s and c’s in pairs of lines to find out
the position vector of the point where they intersect. If the final part is to be answered
numerically, then the parameter O must cancel somewhere before the end.

§ 1 · § 2Ȝ · § 2 ·
Answers: (ii) d = ¨ ¸c ; e = 1
3
a; f=c+ 1
2
Oa ; g = ¨ ¸a + ¨ ¸c ; h
©1 Ȝ ¹ © 2  3Ȝ ¹ © 2  3Ȝ ¹
2
= a.
5
Thus OH : HA = 2 : 3 (as H lies two-fifths of the way along the line OA).

Q9 The most important thing you can do on a question like this, is to draw a good,
decent-sized diagram first, marking on it all the relevant forces. In fact, since a lot of
extra forces come into play in the second part of the question, a completely new
diagram here is pretty much essential. It is also helpful to have the painter, P, in a
general position on the ladder; say, a distance xa from its base up along it. [Note that
xa is so much better than x, so that – since all distances are now multiples of a – these
will cancel in the moments equation and make things look simpler.] Now resolve
twice and take moments (easier about the base of the ladder), and use the Friction
Law (in its inequality form, since we don’t need to know when it attains its
maximum). And then sort out the remaining algebra. On this occasion, it is not
unreasonable to assume that P is at the top of the ladder when slipping is most likely,
and go from there.

In (ii), the extra forces involved are the weight of the table, the reaction forces
between its legs and the ground and the reaction of the ladder’s base on the table
(previously ignored when the ladder was on the ground). The standard approach now
is to assume that the system is rotationally stable and see when slipping occurs; then
to assume that the system is translationally stable and see when tilting occurs. Again,
this involves resolving twice and taking moments; using the Friction Law – with
equilibrium broken when one of the reactions between table and ground is zero – and
deciding which, if any, happens first.

Answers: Table slips on ground when P is distance 5a up the ladder. Table turns
about
edge furthest from the wall when P is distance 113 a up the ladder. Thus,
tilting
occurs first.

Q10 The first two collisions, between A and B and then between B and C, each require the
application of the principles of conservation of linear momentum (CLM) and
Newton’s experimental law of restitution (NEL or NLR). This will give the
intermediate and final velocities of B along with the final velocities of A and C
(although the latter is not needed anywhere) in terms of u. [It is simplest to take all
velocities in the same direction, along AB, so that “opposite” directions will then be
accounted for entirely (and consistently) by signs alone.] For a second collision
between A and B, VA > VB (irrespective of their signs!) and this leads to a quadratic
B

equation in k. Note that any negative solutions are inappropriate here.

18
123
Step II Hints and Answers June 2006

Using k = 1 (which presumably MUST lie in the range found previously), the
velocities of all particles can now be noted less algebraically. The time between
contacts is in two parts: the time for B to reach C, and then the time for A to catch up
with B (from its new position when B & C collide). After B leaves C, it is only the
relative speed of A and B that matters, and this simplifies the working considerably.

3
Answers: (i) 0 < k < 2
.

Q11 The equations of motion in the x- and y-directions can be found by integrating up
from accelerations, or by using the constant-acceleration formulae. Setting y = 0
gives t = 0 or t = ….. (as usual). Substituting this into the expression for x then
gives the distance OA.

x
In (i), the time when x = 0 must occur before the time found above. This gives an
inequality involving sine and cosine, which can be simplified to give the tangent of
the angle required.

In (ii), OB is just OA with T = 45o. Then OA is maximised either by calculus (a little

trickier here) or by using the double-angle formulae for sines and cosines and then
working with an expression of the form a cos 2T + b sin 2T + c, for which there is a
standard piece of work to yield the form R cos(2T – Ø) + c, which has an obvious
maximum of R + c (with R here being in terms of f and g).

For the very last part, f = g with T = 45o gives x = y for B’s motion, and the particle
moves up, and then down, a straight line inclined at 45o to the horizontal, to land at its
original point of projection.

§ g ·
Answers: (i) D = arctan ¨ ¸ ; (ii) answer as above.
©2f ¹

Q12 In (i), the probability that one wicket is taken is

p(A1 ˆ B0 ˆ C0) + p(A0 ˆ B1 ˆ C0) + p(A0 ˆ B0 ˆ C1),
each of which is a product of three terms from a binomial distribution. The
probability that it was Arthur who took the wicket is then the conditional probability
p (1,0,0)
.
p (1,0,0)  p (0,1,0)  p (0,0,1)
Although this looks a pretty ferocious creature with all its terms in it, in fact almost
all of them cancel in the fraction, and you are left with a few products to deal with
(most involving further cancellable terms).

§1 1 1·
Part (ii) is a “quickie” – 30 u ¨   ¸ – to point you towards the use of the
© 36 25 41 ¹
simple value of 3 in the next part. In (iii), since n is large and p is small, the Binomial
can be approximated by the Poisson; and p(W t 5) = 1 – {p0 + p1 + p2 + p3 + p4}.
From here, you can use either the approximation e3 = 20 (as given) and work with
Poisson terms directly, or just resort to the use of the Poisson tables in your formula
books.

19
124
Step II Hints and Answers June 2006

3
Answers: (i) 10
.

Q13 To be honest, this was more of a counting question than anything, at least to begin
with. Although it is possible to attack (i) by multiplying and adding various
probabilities, it is most easily approached by examining the 24 permutations of {1, 2,
3, 4} individually, and seeing what choice is made in each case. To make life easy for
yourself, be systematic in listing these possibilities.

This example should point you in the right direction, but don’t be tempted to just
write down the answer that you’ve spotted without any justification for how it arises
in the general case. To begin with, deal with what happens when the largest cone is
offered first; then the second-largest being first; then the third. By this stage it should
be easy to justify the general case as to what happens when the rth largest cone is the
first to be offered – then the largest is chosen if it appears first of the remaining (r –
1) cones that are bigger than the rth. With probability …..

7 4 1 2 1
Answers: (i) P4(2) = ; P4(3) = or ; P4(1) = or ;
24 24 6 24 12
n 1
1­0  1  1  1  .....  1 ½ or
(ii) ®
n¯ 2 3
¾
n  1¿
1
n ¦
r 1
1
r

1
Q14 For y = , y o f as x o 0 and y o 0 (+ve) as x o f are the obvious
x ln x
asymptotic tendencies of the graph. Since ln 1 = 0, there is also a discontinuity at x =
1, and you must decide what happens to the graph either side of this point.

1
For the rest of the question, its is essential to be able to integrate . This can be
x ln x
1
x
done either by the sneaky observation that it can be written in the form , so that
ln x
the numerator is exactly the derivative of the denominator – a standard log. integral
form – or by using a substitution such as u = ln x.

In (i) and (ii), it is now just a case of substituting in the limits and sorting out the log.
work. Having gained the answer for (ii), in log. form, the numerical approximation
arises from using the first few terms of the series, given in the formula books, for
ln(1 + x) with x = …..

In the very final part, a range is given that turns out to be outside the non-zero part of
the pdf. A little bit of work needs to be done to justify this, and then you can write
down the answer.

1 1
Answers: (i) O = 1
or – ; (iv) 0.
ln 2 ln 2

20
125
Step III, Hints and Answers
June 2006

21
126
Step III Hints and Answers June 2006

1 2 x ( x 2  5)
y
x2  4
2x
2x 
( x  2)( x  2)
Asymptotes are y 2 x , x r2 .

dy 2( x  2)( x  2)  4 x 2
2
dx ( x  2) 2 ( x  2) 2
(or equivalent).
Equation of the tangent at O is
5x
y .
2

(i) 3x( x 2  5) ( x 2  4)( x  3)

2 x( x 2  5) 2 x
œ  2 ( x z r2)
x2  4 3
2
y x  2 cuts the sketched curve in three points, so three roots.
3
(ii) 4 x( x 2  5) ( x 2  4)(5 x  2)
2 x( x 2  5) 5 x .
œ  1 ( x z r 2)
x2  4 2
5x
y  1 passes through the intersection of x 2 and y 2 x and is parallel
2
5x
to y so just one root.
2
(iii) 4 x 2 ( x 2  5)2 ( x 2  4)2 ( x 2  1)
2 x( x 2  5)
œ r ( x 2  1) ( x z r2)
x2  4
y r ( x 2  1) has two branches with asymptotes y r x , so there are six
roots.
2 (i) First “show” by change of variable T I (say).
Then

22
127
Step III Hints and Answers June 2006

S
2 cos 2 T S
2 cos 2 T
2I ³S 2 1  sin T sin 2D ³S 2 1  sin T sin 2D dT
d T 
S 2
2
³S 2 sec2 T  tan 2 T sin 2 2D dT
and next “show” follows.
(ii) S 1
J sec 2D ³S2 2
cos 2D sec2 T dT
2 1  (cos 2D tan T )
S 1
sec 2D ³S2 2
du (since cos 2D ! 0)
2 1 u

S sec 2D
(iii) I sin 2 2D  J cos2 2D S .
Result follows after use of (ii).
(iv) In this case, cos 2D  0 , so J S sec 2D .
1
Then I S cos ec2D
2
3 (i) tan x is an odd function.
Express both sides in terms of tan x .
From identity, substitute series and result follows by equating coefficients of
powers of x.
(ii) Show that cot x  tan x 2 cos ec2 x and follow same method.
(iii) Identity follows from 1  cot 2 x cos ec2 x .
Equate coefficients to show that all coefficients for even n are zero, and
1
a1 1, a3 .
3
4 Let x y and deduce first result.
2 f ( x) f (2 x)
Ÿ 2 f '( x) 2 f '(2 x)
Ÿ 2 f "( x) 4 f "(2 x)
then put x 0 to get f (0) 0, f ''(0) 0 .
Similarly all higher order derivatives are zero, so by Maclaurin the most
general function is cx , where c is a constant.
(i) Use properties of logs to show that G( x)  G( y) G( x  y).
Deduce that g ( x) ecx .
(ii) Show that H (u )  H (v) H (u  v)
so h( x) c ln x .
(iii) Let T ( x) t (tan x).
Deduce that t ( x) c arctan x .
5 There are essentially two different configurations, corresponding to clockwise
and anticlockwise arrangements of D , E , J taken in order.
1  3
In what follows, Z , the cube root of unity with modulus 1 and
2
2S
argument ; 1 Z  Z2 0 (*) is assumed.
3

23
128
Step III Hints and Answers June 2006

Then either E  J Z (J  D ) and E  J Z 2 (J  D ) expresses equality of

adjacent sides and the correct angle between them for each of the two cases;
by SAS this establishes an equilateral triangle.
These two are equivalent to [ E  J  Z (J  D )][ E  J  Z 2 (J  D )] 0 .
The required form is an expanded version of this, using (*).
NB It is essential to be clear that the argument works both ways.
If D , E , J are the roots of the equation given,
a D  E  J , b DE  EJ  JD , c DEJ .
Then a 2  3b D 2  E 2  J 2  DE  EJ  JD
so a 2  3b 0 is equivalent to the expression in the first part.
Result follows.
z o pw is an enlargement combined with rotation, so object and image are
similar. pw o pw  q is a translation so object and image are congruent.
Hence under the composition z o pw  q object and image are similar.
Result follows.
Aliter. Substitute z pw  q in the first equation, and simplify.
Compare coefficients to determine A and B in terms of a, b and c.
Then a 2  3b 0 Ÿ A2  3B 0 , so result follows.
6 x r cos T , y r sin T , r r (T )
dr
sin T  r cos T
dy dT
Ÿ
dx dr cos T  r sin T
dT
and result follows.

T
Gradient of the normal is tan t , say. Then we have
2
dr
 r tan T
d T 2t
t  , tan T
dr
tan T  r 1 t2
dT
This reduces to

24
129
Step III Hints and Answers June 2006

dr
rt
dT
sin T2
Ÿ ln r ³ cos T2 dT
ª Tº
2 ln « c cos »
¬ 2¼
2 T
Ÿ 2
1  r cos T (using 1  cos T 2 cos 2 )
cr 2
This corresponds to the standard equation of a parabola in polars.
7 (i) Express sinhx in terms of exponentials, factorise and solve to get
u e x or u e x (or  cosh x r sinh x ).
dy
Use both of these as equal to and integration to get alternative solutions
dx
y e r x  c.
From the given conditions the particular integral is
y 1  e x .
(ii) Solve the quadratic as before to get either
1 r cosh y
u (or equivalent)
sinh y
dx sinh y
Ÿ
dy 1 r cosh y
Ÿ x ln(cosh y  1)  c1
or x  ln(cosh y  1)  c2
Only the first can satisfy the conditions x 0, y 0 ; then we have
2
x ln
1  cosh y
Ÿ cosh y 2e  x  1
This is undefined for x ! 0 .
For x o f Ÿ cosh y o f , and there will be two branches, corresponding
to y o rf , as cosh is an even function.
So x o f Ÿ cosh y o f Ÿ y o f Ÿ e y 4e  x Ÿ y  x  ln 4
in one case, and similarly y x  ln 4 in the other.

25
130
Step III Hints and Answers June 2006

8 Use (iv) with f ( x) { 1, g ( x) { 1 to show that '1 0 .

Use (iii) with O { k , f ( x) { 1 to show that 'k 0 .
By (iv), (i) 'x 2 2 x ; ditto 'x3 3x 2 .
Now show 'kx n knx n 1 by induction.
Initial step is 'k 0 ; inductive hypothesis is that 'kx N kNx N 1 .
Use (iii) and (iv) with hypothesis to show that 'kx N 1 k ( N  1) x N .
Now express any Pk ( x) , a polynomial of degree k, as a sum of such powers,
and so use (ii) to establish required result.
9 Take O as the zero level for potential energy. Then
PE of bead at B is mgy ; PE of particle at P is mgr  mgl .
For perpetual equilibrium, the PE must have the same value in any position,
in particular its value at H; result follows.
Express equation shown in polar coordinates to get
2h
r
1  sin T
x
Differentiate and make T the subject so
x
x r (1  sin T ) 2
T  .
2h cos T
These two expressions give the desired result.
By conservation of energy if PE is constant so is KE. Hence KE in a general
position is equal to the initial value. That gives
2 2
§ x· x
V2 ¨ rT ¸  2 r
© ¹
x x
Speed of the particle at P is r . Use the expressions for V 2 , T to derive the

required result.
10 Use conservation of angular momentum for the first result.
Use conservation of energy to derive
k 2  a2 2
v2 2
:  (k 2  r 2 )Z 2
k
dr
and so by use of the first result and v 
dt
second result follows.
dT dr dr dT
Now use Z and and the two displayed result to derive
dt dT dt dt
the third.
The suggested substitution transforms the third displayed equation to
du
1 u2 .
dT
Invert and integrate to get the desired result.
k
Hence r .
sinh(T  D )
As T o f, r o 0  , but r 0 is impossible.

26
131
Step III Hints and Answers June 2006

11

The equations of motion are

T  ( M  m) g ( M  m)a1
Mg  T Ma1
mg
Ÿ a1
2M  m
Now consider relative motion of the tile with acceleration ( g  a1 ) .
If the time of the first stage is t1 , s ut  12 at 2 gives
(2M  m)h
t1
Mg
and then for the absolute motion of the tile v u  at gives the required final
velocity.

The middle diagram shows the situation before the impact and the third after.
The forces acting on the left-hand system (lift plus tile) are exactly the same
as those on the right, so the changes in momentum must be equal in the first
stage of the motion. Thus given that all is stationary initially
( M  m)v2  mv1 Mv2
m
Ÿ v2 v1 D v1 (*), say.
2M  m
In the collision, the equality of impulsive tensions given means that the
change in momentum on one side equals change in momentum on the other.
Hence we have
 Mw2  Mv2  mw1  mv1  ( M  m)( w2  v2 )
Ÿ w2  v2 D ( w1  v1 )
Thus from the two last equations
w2 D w1 (**) .
Newton’s experimental law and the two asterisked equations give
w1 ev1 .

27
132
Step III Hints and Answers June 2006

Then the change of energy in the collision

2 2 2 2
2 (2M  m)(v2  w2 )  2 m(v1  w1 )
1 1

simplifies to the required expression when the above relations are substituted.
Loss of energy of a tile dropping to the floor of a fixed lift and bouncing
would be just the same.
12 Model each tourist as trial with success probability ½. If X is the number of
potential passengers X Bin(1024, 1 2) , ie N (512,162 ) approximately.
Lost profit corresponds to X ! 480. Hence if L is the loss, we have
32
E[ L ] ¦ kpr ( X
k 1
480  k )  32 pr ( X ! 512)
32
1 § k ·
¦ k. 16 .I ¨© 2  16 ¸¹  16
k 1

32 x § x·
|³ I ¨ 2  ¸dx  16
0 16 © 16 ¹
x
32 1 ( x  32) 2
³0 16 ˜ 2S ˜ exp 512 dx  16
Now use substitution to show that this evaluates to
16
2S
 e 2  1  32) (2) .

In the course of year the expectation is 50 times that figure, so that is the
maximum tolerable licence fee.
13

There are three cases to consider: (i) both on the circumference, (ii) P1 on the
diameter and P2 on the circumference, and (iii) vice versa.
For case (i), if P1 lies in the arc (D , D  GD ) P2 lies in the arc (T , T  GT ) ,
GT 1
with probability , the area is r sin T . The expected area given P1
S 2 2
1
lies in the arc (D , D  GD ) is by integration .
S 2
For case (ii), if P1 lies in (r , r  G r ) and P2 lies in the arc (T , T  GT ) , with
GT 1
probability , the area is r sin T . The expected area given P1 lies in
S 2 2
r
(r , r  G r ) from O is by integration .
S 2
Case (iii) is essentially the same as case (ii).
Thus the expected area is
S 1 1 1 r 1
³0 S  2 S  2
. d D  2 ³1 S  2. S  2 dr
where the first integral corresponds to case (i) and the second to (ii) and (iii).

28
133
Step III Hints and Answers June 2006

This evaluates to the answer given.

14 E[aX 1  bX 2 ] aE[ X 1 ]  bE[ X 2 ]
E[ X 1 X 2 ] E[ X 1 ]E[ X 2 ]
E[ P ] 2P1  2P2
E[ P 2 ] 4 E[ X 12 ]  8E[ X 1 ]E[ X 2 ]  4 E[ X 22 ]
E[ X 12 ] P12  V 12
var[ P ] 4(V 12  V 22 )
The standard deviation is the square root of that expression.
E[ A] P1P2
E[ A2 ] P12 P 22
var[ A] V 12 P22  V 22 P12  V 12V 22
Again the standard deviation is the square root.
Now find
cov[ P, A] 2P2V 12  2P1V 22
This is not zero (as independence would imply) with given conditions.
Similarly
cov[ Z , A] 2V 12 P2  2V 22 P1  D ( P12V 22  P22V 12  V 12V 22 )
That too is non-zero when D is not the excluded value.
We consider the exceptional case with the given information.
8
We have P1 P2 2, V 12 V 22 1, D .
9
Only three values of A are possible - 1, 3 and 9 - and they correspond to
unique values of Z. Dependence can be shown by considering, for example,
§ 28 · 1 § 28 ·
pr ¨ Z ¸ , pr ¨ Z A 3¸ 0 .
© 9 ¹ 4 © 9 ¹

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June 2006

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REPORT ON THE COMPONENTS

Unit Content Page

9465 STEP Mathematics I 5

9470 STEP Mathematics II 7

9475 STEP Mathematics III 13

* Grade Thresholds 15

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9465 – Mathematics I

General comments

This paper was found to be more difficult than last year’s; somewhat worryingly, this was perhaps
because the paper placed a greater emphasis on algebraic and numerical manipulation than
previously. At this level, the fluent, confident and correct handling of mathematical symbols (and
numbers) is necessary and is expected; many good starts to questions soon became unstuck after a
simple slip. The applied questions appeared to be beyond many candidates; it has been suggested
that this reflects the reduction of the amount of applied mathematics in single maths A-level.

There were of course some excellent scripts, but the examiners were left with the overall feeling
that some candidates were not ready to sit the examination. The use of past papers to ensure
adequate preparation is strongly recommended. A student’s first exposure to STEP questions can be
a daunting, demanding experience; it is a shame if that takes place during a public examination on
which so much rides.

Comments on individual questions

1 Some candidates could not square correctly a three digit number; of those who did, not all
recognised that 1842 – 33127 = 272. This was intended to be a straightforward “warm-up”
question, but it was not to be found to be so.

2 The key word in this question was “prove”, but most candidates assumed that the goat
would graze the maximum area if it were tethered to a corner of the barn, and the minimum
if tethered to the middle of a side. This unjustified assumption severely reduced the awarded
marks. Candidates are advised to ensure they understand what, at this level, is required by an
instruction to prove a result.

3 Parts (i) and (ii) were reasonably well done, but very few successful attempts to (iii) were
seen. Most candidates had not realised that they were supposed to be thinking about the
graph of y = x3 + px + q, in particular its stationary points and its y-intercept.

4 The two graphs were well drawn, although sometimes the horizontal scale was in degrees.
The area formula was often derived correctly, but the subsequent differentiation often
contained a major error, most commonly a failure to apply the chain rule when
differentiating tan( / n). Most candidates found it hard to construct a coherent argument
(using part (ii)) about the ratio of the polygon to the circumcircle.

5 This integration question was tackled much more successfully than last year’s. It was
particularly pleasing to see how many candidates were able to cope with the unusual partial
fractions that arise in part (ii); some imaginative methods were seen.

6 This was a popular, straightforward question, which was often answered very well.
However, algebraic errors still occurred – even when expanding
(3a + 4b)2.

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7 Only a few candidates saw the connection between the two halves of part (i), and therefore
most evaluations of the definite integral failed to remove the modulus of the integrand
correctly. Interestingly, part (ii) was often found more straightforward.

8 This was a well-answered question (using either vectors or the cosine rule), although some
candidates tried to derive the result about d by using the vector equation of the plane ABC:
the instruction “hence” required the use of the first two parts of the question. Candidates
should ensure that they understand the distinction between “hence” and “hence or
otherwise”.

9 It was not thought that this would be a difficult question, but many candidates were unable
to model correctly the motion of three connected particles. A common error was to consider
xg – yg, the resultant force acting on the whole system, but to divide it by x + y rather than x
+ y + 4 when calculating the acceleration.

10 Very few attempts at this question were seen, and those that did rarely progressed beyond
the first paragraph.

11 Hardly any attempts at this question were seen, but those candidates who did tackle it were
usually able to produce a mostly accurate solution. It was remarkable how few diagrams, not
to mention labelled diagrams, were seen: it is always much easier for both the candidate and
the examiner if symbols are clearly defined in a diagram.

12 Many different (and correct) arguments were seen to the first part: candidates’ careful
analysis of the different possibilities was encouraging. Unfortunately, hardly any candidates
recognised that the second part of the question was asking for a conditional probability. This
prompted the examiners’ concern that candidates were too reliant on a verbal clue such as
“given that”, and found it very hard to identify the inherently conditional structure of an
event such as was described in this question.

13 Very few attempts at this question were seen, although it was not expected to be popular
since it was known that some candidates would not have studied the Poisson distribution.
However, knowledge of it (and the Normal distribution) remains in the published
specification, and so candidates may wish to ensure that they are familiar with both of these.

14 Part (i) was well answered by most of those who attempted it. Solutions to part (ii) often
began with a correct product of fractions, but it was surprising how often factorials were
employed in an (unsuccessful) attempt to simplify an expression that cancelled down very
easily to 1/(n + 1). The implicit fact that n + 1 • r was not often realised, leading to “n = 0”
as the modal answer.

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9470 – Mathematics II

General comments

This was an accessible paper, with up to half the marks on each question available to candidates
of a suitable potential, The candidature represented the usual range of mathematical talents, with a
goodly number of truly outstanding students, many more who were able to show insight and flair on
some of the questions they attempted, and (sadly) a significant number of students for whom the
experience was not to prove a particularly profitable one. Of the total entry of nearly 700, around
40% were awarded grade 1’s (or better), while only around 20% received an unclassified grade.

Really able candidates generally produced solid attempts at six questions, while the weaker brethren
were often to be found scratching around at bits and pieces of several questions, with little of
substance being produced. In general, few candidates submitted serious attempts at more than six
questions – a practice that is not to be encouraged, as it uses valuable examination time to little or
no avail. It is, therefore, important for candidates to spend a few minutes at some stage of the
examination deciding upon their optimal selection of questions to attempt.

As a rule, question 1 is intended to be accessible to all takers, with question 2 usually similarly
constructed. In the event, at least one – and usually both – of these two questions were among
candidates’ chosen questions. Of the remaining selections, the majority of candidates supplied
attempts at the questions in Section A (Pure Maths) only. There were relatively few attempts at the
Applied Maths questions in Sections B & C, with Mechanics proving by far the more popular of the
two options. Question 10, in particular, was relatively popular. Overall, there were remarkably few
efforts submitted to the Statistics questions in Section C, although several of these were of
exceptional quality.

On a more technical note, many solutions to those questions which were not already quite structured
suffered a lamentable lack of clearly directed working. Large numbers of candidates would benefit
considerably from the odd comment to indicate the direction that their working was taking. This
was especially the case in questions 3, 5, 10 and 13, where it was often very difficult for examiners
to decide what candidates were attempting to do, and where they had gone wrong, without any clear
indication as to what they themselves thought they were doing.

Comments on individual questions

1 Almost all candidates attempted this question and most managed at least some measure of
success; although the high level of algebra required to see matters through to a successful
conclusion proved to be a decisive factor in whether attempts got much over half-marks. A
minority of candidates worked with un and un + r (for the appropriate r’s) and thereby made
the algebra rather harder for themselves; whereas it had been intended that they should work
with u1 (with the given value of 2) and the appropriate ur in order to determine periodicity.
The other major problem arose when candidates worked backwards from (say) u5 towards
u1, rather than forwards. This often generated nested sets of bracketed expressions of the
form
36
u5 = k –
36
k
k  k 36.....

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which only the hardiest were able to unravel successfully; while a forwards approach would
have found each of u2, u3, … successively as much simpler (rational) terms.

Another common error arose when candidates failed to note that, if k = 20 gives a constant
sequence, then, for a sequence of period 2, the answers “k = 20 and 0” can’t both be correct.
Similarly, for a sequence of period 4, the values 0 and 20 should appear as possible
solutions when equating u5 to u1, but should be discounted. Whilst many candidates noted
these points – and some shrewdly used their existence to help factorise the arising quartic
equation in k – it is still clearly the case that a large proportion of A-level students, even the
better ones, are happy to assume that any solution to an equation they end up having to solve
is valid, irrespective of the context of the underlying problem or the logic of their work (viz.
necessary and/or sufficient conditions).

Although only the most basic of arguments was required to establish that un t 2 at the
beginning of part (ii), it was clear that most candidates were really not comfortable handling
inequalities, and lacked practice in constructing reasonable mathematical arguments. Far too
many failed to work generally at all, and simply showed that the first few terms were greater
than or equal to 2, concluding with a waffle-y “etc., etc., etc.” sort of argument. In the very
last part, it was important to appreciate that a limit is approached when successive terms
effectively become the same. No formal work beyond this simple idea was required, and the
resulting quadratic gave two solutions, only one of which was greater than 2. Rather a lot of
candidates were happy with this idea and rattled it through very quickly.

2 This question was the second most popular on the paper (in terms of the number of attempts)
and really sorted out those who were comfortable with inequalities from those that weren’t.
Those who were generally scored very high marks on the question; even those who weren’t
generally managed several bits and pieces to get around half-marks on it.

Once again, there was an informal (possibly induction-type) proof required for the second
bit of the question, although this was handled slightly more capably than the easier one in
Q1, possibly because so many candidates seemed happier to effectively produce a formally
inductive line of reasoning. Most candidates then picked up on the purpose of this bit in
helping create a convergent GP to sum, which helped establish the next inequality for e.

The differentiation proved undemanding, and most candidates managed to realise that the
minimum and maximum points referred to would be established by considering the sign of
dy
at x = 12 , 1 and 54 . Rather fewer were entirely happy to use the given bounds on e to
dx
help them do so, with many going off to lengthier (although often equally correct) workings-
out. (In the final part, the use of e < 3 would have done the trick.) Those candidates who
used approximations rather than inequalities were missing the point, as were those who tried
d2 y
to use without actually knowing the exact values of x which they could use in it.
d x2

3 A lot of candidates made a faltering start to this question before moving on to pastures
greener. This was usually occasioned by a realisation that life was going to be very tough
1
here – which it was if they failed to appreciate that = 5 – 24 . Those who saw
5  24
this early on generally made their way to at least the first 8 marks. Although there are other

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ways to go about the first part, the use of the binomial theorem, with the 24 -bits all
cancelling out, establishes that the given expression is indeed an integer (without necessarily
having to find out which). The three modest inequalities that followed were easily
established with just a modicum of care. However, it was again the case that candidates’
lack of comfort with inequalities once more prevented a convincing conclusion to (i) since
most candidates resorted to approximation: showing that N | 9601.9999 is NOT the same
as showing that, because N lies between … and … , it is actually equal to it (to four
decimal places). Sadly, most candidates did not seem to understand such a difference in
logical terms.

For part (ii), it was necessary only to mimic the work of part (i) but in a general setting.
Most candidates attempting this question were happy to leave it at this point; of those who
continued, many picked up two or three marks – only a handful actually polished it off
properly.

4 Another difficult start again put most candidates off this question at the outset (if not before)
and there were relatively few efforts at it. Most of these were pretty decent and scored well.
The use of the initial result in (i) was straightforward, provided one is prepared to spot a
decent substitution (such as c = cos x). The formula books then helped bypass the
integration required. In (ii), the given integral splits into the answer to (i) + a second
integral, which must be considered separately. A simple linear substitution helped here,
although quite a few candidates incorrectly assumed a result over the interval (S, 2S) similar
to the given one could just be assumed to hold. This was often the case in (iii) also, although
fewer candidates tried such a move: the sin(2x) forcing them to consider more sensible
approaches, such as (again) a linear substitution (after using the double angle formula for
sine).

5 Despite the introduction of a non-standard function – often called the floor or the INT
function – this was a popular question to attempt. As mentioned earlier, finding the areas
required candidates to structure their working and, since there are several ways to break up
the bits of the process, a teensy-weensy bit of explanation would have been greatly
appreciated by the examiners. The easiest approach to the area in (i) is to work straightaway
with the difference (y1 – y2) which immediately gives a whole load of “unit triangles” to
sum. Attempts varied from excellent-and-concise all the way down to scrambled-heap-of-
integrations-and summations. Part (ii) was handled similarly, although it is strange to say
that – despite the slightly greater degree of care needed with the various bits and pieces –
there were slightly more correct answers arrived at here.

6 In hindsight, it might have been more generous to have included an “or otherwise” option to
the very opening part of this question, as many candidates – particularly overseas ones –
preferred an algebraic approach to obtaining the given result, rather than the vector one
asked-for. It does, however, illustrate a pretty important examination point: namely, that if
you don’t actually answer the question that has been asked, you may not actually get any
marks for your time and effort! These candidates reduced the given inequality to

(bx – ay)2 + (cy – bz)2 + (az – cx)2 t 0,

and this represents some pretty decent mathematics. It is also very easy to deduce when
equality holds in the result from this alternative statement. Such candidates were able to
get the remaining sixteen marks on the question, however.

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Part (i) didn’t actually require candidates to use the given result to solve this quadratic
equation, but those who did were guided towards the helpful notion of considering the
equality case of the given result, which was intended to help them approach part (ii). [The
question cites an example of a result widely known as the Cauchy-Schwarz Inequality.]

Overseas candidates apart, this was not a very popular question at all. Those who attempted
it generally did quite well, and a surprisingly high proportion of them saw it through right to
the end.

7 This proved to be a relatively popular choice of question, usually being pretty well-done, at
least up to the point where trig. identities came into play, and often all the way through. It is
suspected that the principal reasons for this were that the question had a fairly routine start,
and then developed in a fairly straightforward A-level manner thereafter.

Most attempts established the opening result easily enough, and also managed to acquire Q’s
coordinates without much difficulty, and usually the equation of the line PQ also. A
common shortfall at the next stage was not so much the introduction of the trig., which
clearly put some candidates off, but rather the use of the trig. to show that the two lines were
the same when these identities were used. A very surprising number of candidates seemed
content to suggest that the two forms of the line were the same on the basis of their
gradients only.

Those who got as far as the last part usually handled it very capably, showing that the two
cases led to PQ being the vertical and horizontal tangents (respectively) to the ellipse.

8 Clearly vectors weren’t a popular choice for candidates, as there were very few attempts
made at this question. The first six marks, however, are gifts and almost all attemptees
gained these. Thereafter, it is simply a case, with (admittedly) increasingly complicated
looking position vectors coming into play, of equating a’s and c’s in pairs of lines to find
out the position vector of the point where they intersect. Candidates’ efforts tailed off fairly
uniformly as the question progressed, and examiners cannot recall anyone actually getting to
the end and finding h (the p.v. of H) correctly, although there were several attempts that
gained all but the final two marks.

9 These leaning-ladder questions are actually pretty standard, and it was disappointing to see
so few attempts made at this one. More disappointing still was the lack of a decent diagram
from which candidates might have been able to extract some support for their working.
Similar dismay was evoked by the widespread inability, on the part of almost all candidates,
to be able to say what mechanical principles they were attempting to use at any stage of their
working. Of the relatively small number of attempts seen, most suffered from at least one of
these deficiencies. Consequently, although there were many partially or totally successful
attempts at (i), the number of even half-decent attempts at (ii) were very few. The extra
forces that needed to be considered in (ii) were either overlooked completely, or were
missing from (i)’s diagram that candidates were trying to re-use.

The other painfully obvious shortfall here lay in candidates’ dislike of using the Friction
Law in its more general, inequality, statement rather than in the equality case given by
limiting equilibrium. Such a shortfall was overlooked, even when it wasn’t explained
correctly (although it contributed substantially to problems in part (ii), when working was to

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be found). Those making a stab at (i) usually managed to make correct statements from
resolving and taking moments, although arguments putting everything together and
explaining why the ladder was stable were often less than entirely satisfactory.

10 The most popular of the three Mechanics questions, and generally the best done. Even so,
marking was often made unnecessarily difficult by candidates’ failure to explain what was
going on and/or simplify their working at suitable stages in the proceedings. Setting up and
finding the post-collision velocities of the various particles was relatively straightforward –
although the algebra did prove too demanding for quite a few candidates – and most
attempts correctly indicated the condition required to give a second collision between A and
B. The number of unsuccessful attempts to solve the resulting quadratic was a surprise –
most presumably faltering due to the lack of a unit x2 term! – as was the number who
preferred to use the quadratic formula rather than factorisation.

Problems generally arose here in part (ii), where a lack of explanation was a big problem.
Those candidates who simply work out times and distances, without saying what they are
supposed to be, do themselves no favours, as it is very difficult for the examiners to give
credit to the working until a coherent strategy has emerged. Any error, no matter how small
– and especially those made by candidates working “in their heads” – can render it almost
impossible to spot such a strategy and reward it. On a more fundamental level, part (ii)
should have opened up with the statement of the three relevant velocities, given in terms of
u, using k = 1. Most efforts made mistakes because this simple task was left until much later
on in the working, and some candidates even insisted on working with a general k
throughout.

11 It was felt by examiners that this was the nicest (and easiest) of the three Mechanics
questions, yet it drew very few serious attempts from the candidature. Most serious efforts
coped very easily with the first two parts. Thereafter, it was often the case that maximising
OA proved to be a greater difficulty than it should have done, despite the fact that the option
to use calculus was available (although much less concise an approach than using a
trigonometric one). There had been concerns that, for the final part, candidates might not
grasp what was going on but, happily, this proved not to be the case and several candidates
spotted the significance of having f = g and described the resulting motion adequately.

12 This was the least popular of the Statistics questions, even amongst the relatively small
number of attempts at any Section C questions. Of those seen, examiners can recall only two
which got the answer of 103 in (i). This was due to the almost total lack of appreciation that
the result “1 wicket taken” required three probabilities.

The clear guidance towards the use of a Poisson distribution in (ii) and (iii) was, however,
picked up by candidates. The calculation of the ensuing probabilities, either directly or via
tables, was actually very straightforward, and candidates coped very easily when they
ventured this far.

13 To be honest, this was more of a counting question than anything, at least to begin with, and
several candidates picked up relatively large amounts of marks for very little working.
Whilst several attacked (i) by multiplying and adding various probabilities, it was possibly
most easily approached by looking at the 24 permutations of {1, 2, 3, 4} individually. Those
candidates who adopted a mix-‘n’-match approach without explanation often got themselves
into a bit of a muddle, but still picked up several of the marks available here.

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The example provided by (i) was intended to help direct candidates’ thinking in (ii) as well
as give them with a non-trivial case to use as a check. Of the attempts received, many
explained things very poorly, even when they arrived at the correct expression. Sadly, rather
too many seemed to deduce the (correct) answer on the basis of (i)’s example alone, and
seemed unable to grasp that anything needed to be explained or justified.

14 This was a relatively popular choice of question, perhaps partially because it started off with
a couple of bits of Pure Maths: namely, curve-sketching and integration. Strangely, though,
very few sketches were fully correct, even when followed-through by “reciprocating” a
correct sketch of y = x ln x.

1
Further progress was going to be impossible without integrating , and some attempts
x ln x
fell at this hurdle. Pleasingly, several candidates spotted the log. form immediately, while
many others correctly used the substitution u = ln x, or equivalent.

Thereafter, it was a routine statistical exercise in some respects. However, the log. work
required to simplify matters in (i) proved beyond rather too many candidates – whereas it
proved much less of a difficulty in (ii). Only a few candidates realised that there was a
standard series expansion ready to hand for ln( 43 ), and those that did generally only went
up to the cubed term, which was a shame as the given answer arose from using the next one
as well.

The final twist, in part (iv), of giving a range that turned out to be outside the non-zero part
of the pdf, was twigged by slightly more than half of the candidates that got this far.

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9475 - Mathematics III

General comments

Candidates found the paper in general hard. Nevertheless they seemed aware of the importance of
depth and quality of answers rather than disposed to do lots of bits. There were disproportionately
small numbers of attempts on applied questions, particularly as regards the statistics section.
Questions 1 and 7 were very popular. It is pleasing to see that there is a core of candidates who are
well up to the challenge and who produce pleasing solutions, and many others who make sustained
attempts to rise to it.

Comments of individual questions

1 Many candidates attempted algebraic approaches to the later parts instead of relating the
problems to the graph.

2 This question required sustained and confident technical ability. Level of completion
reflected the levels of those abilities. The last part required a more subtle appreciation of
how integration by substitution works than is usual.

3 This question required a repeated and systematic use of trigonometric identities and their
derivation together with the technique of equating coefficients. It tested the ability of
candidates to work in a systematic and accurate fashion.

4 This question started with the use of function notation together with an appreciation of the
chain rule. This would be unfamiliar to many candidates but at the same time accessible to
those who had understood the chain rule well. The later parts required confident use of
function notation in a creative way.

5 There were many different approaches to this question; the key to their acceptability is the
extent to which they can be justified as sufficiently general. It is clearly not acceptable to
illustrate the result with one or more special cases only. It was also important to note the “if
and only if” phrase in the question. Those who managed to reach beyond the first proof
produced different approaches to the second part, particularly with a preference for the
algebraic approach.

6 The lack of structure in this question meant that there were again varied approaches. The
solution required some degree of confidence at interrelating elementary geometry with
calculus in the less familiar context of polar coordinates. Note that the question requires a
solution which starts with a geometrical property and finishes with a parabola, and not the
other way.

7 This was a very popular question, with many examples of sustained and correct work. The
solution was ambiguous at every stage with a heavy premium on balancing the options with
the appropriate correct choice at each stage. It tested candidates robustness very well.

8 This question was only accessible to those who were able to take seriously the required
justification in terms of the rules given at each stage of the solution. The last part can be
approached in several different variations of the same idea, and did require a careful

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construction of an inductive proof from the several elements of the first part. Again it was a
test of sustained thought.

9 The first part depends on the intuition that potential energy must be constant if the system is
in equilibrium in all positions. The remainder of the question can be attempted without that
by assuming the established result, but it makes its return in the complementary intuition
that kinetic energy must also be constant by the conservation law.

10 This question is reasonably routine to those candidates who are conversant with standard
rigid body results and are confident with integration..

11 The first part requires two separate equations of motion to establish the acceleration of the
connected parts; this is standard work. The second extends the usual situation to take in
impulsive tensions. The “given” assumption can be justified but is provided as a hint.

12 This question requires the build up of a suitable model with appropriate approximation.
This has to be followed through by converting a discrete sum into an approximate integral.

13 This question is another with no structuring. It requires an analysis that breaks the problem
into separate parts, which must be recombined at the end.

14 This was a long but straightforward question for those who could handle their definitions
with confidence.

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STEP Mathematics (9465/9470/9475)

June 2006 Assessment Series

Unit Threshold Marks

Unit Maximum S 1 2 3 U
Mark
9465 120 82 67 50 38 0
9470 120 85 60 49 31 0
9475 120 80 60 49 31 0

The cumulative percentage of candidates achieving each grade was as follows:

Unit S 1 2 3 U
9465 7.7 21.3 44.2 68.1 100.0
9470 12.9 41.4 57.2 82.4 100.0
9475 12.2 38.6 59.2 78.7 100.0

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STEP Mathematics I 2007: Report

General comments

There were significantly more candidates attempting this paper this year (an increase
of nearly 50%), but many found it to be very difficult and only achieved low scores.
In particular, the level of algebraic skill required by the questions was often lacking.
The examiners’ express their concern that this was the case despite a conscious effort
to make the paper more accessible than last year’s. At this level, the fluent, confident
and correct handling of mathematical symbols (and numbers) is necessary and is
expected; many good starts to questions soon became unstuck after a simple slip.
Graph sketching was usually poor: if future candidates wanted to improve one
particular skill, they would be well advised to develop this.

There were of course some excellent scripts, full of logical clarity and perceptive
insight. It was pleasing to note that the applied questions were more popular this year,
and many candidates scored well on at least one of these. It was however surprising
how rarely answers to questions such as 5, 9, 10, 11 and 12 began with a diagram.

However, the examiners were left with the overall feeling that some candidates had
not prepared themselves well for the examination. The use of past papers to ensure
adequate preparation is strongly recommended. A student’s first exposure to STEP
questions can be a daunting, demanding experience; it is a shame if that takes place
during a public examination on which so much rides.

Further, and fuller, discussion of the solutions to these questions can be found in the
Hints and Answers document.

Comments on specific questions

1 This question required little more than a clear head and some persistence:
candidates had either ample or very little of both, and thus most scores were
either high or very low. The examiners would like to stress that a solution to a
question such as this must be written out methodically and coherently: many
answers which began promisingly were soon hopelessly fragmented and
incoherent, leaving the candidate unable to regain his or her train of thought.
This was especially true when deriving the final expression given on the exam
paper. Examiners follow closely a candidate’s line of reasoning, and they have
to be certain that the candidate has constructed a complete argument, and that
he or she has not arrived at a printed result without full justification.

2 This was a popular question, and was usually well done. The argument at the
end was often incomplete, though: many candidates simply stated that t = 1 or
t = 2 without explaining why no other values were possible. To do so, use had
to be made of the fact that s and t have no common factor other than 1.

3 This was the most popular question on the paper, and many different methods
were seen. The intended method was to use the identities
cos4θ – sin4 θ ≡ cos 2 θ and cos4 θ + sin4 θ ≡ 1 – ½ sin2 2 θ to evaluate the
integrals of cos4 θ – sin4 θ and cos4 θ + sin4 θ, and hence be able to write down

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 separately the values of the integrals of cos4 θ and sin4 θ. A similar approach
works well for cos6 θ – sin6 θ and cos6 θ + sin6 θ. Other methods were, of
course, acceptable, and many candidates received high marks for this question.

4 This question was found to be very difficult. The initial factorisation was
beyond most candidates, even given the linear factor x + b + c. Anyone who
wants to read Mathematics at university must be able to factorise quickly
cubic expressions such as this one, and also x3 ± y3. The Hints and Answers
document discusses this in more detail.

Candidates who progressed to the second part of the question often deduced
that ak2 + bk + c = 0 and bk2 + ck + a = 0, but then tried to eliminate k; given
that the result they were asked to derive was still in terms of k, this was an
unwise strategy.

5 Only a few candidates made much progress with this question, even though it
only required GCSE Mathematics. Basic properties of triangles (for example,
the sine and cosine rule, and the location of the centroid, the circumcentre and
the incentre) are assumed knowledge at this level. It was surprising how many
candidates tried to answer this question without a diagram.

6 This was a popular, straightforward question, which was often answered well.
However, algebraic errors still occurred, for example when expanding (x – y)3.

7 Part (i) was well done by most of those who attempted this question, but many
then found it difficult to develop the strategy in part (ii). A certain amount of
trial and error is needed to complete the squares in an expression in terms of
both α and β, but the coefficients (in particular, 1α2, 1β2 and 26β2k2) do not
permit many possibilities. This question demanded some stamina, as
Mathematics at university level also does.

8 This question was answered poorly; many candidates were unable to sketch
the graphs correctly, even given the results derived earlier in the question. For
example, many graphs did not touch at (2, 8). Also, many graphs were drawn
with turning points, when a simple check of the derivative would have
revealed that there were none. In part (iii), the effect of the negative coefficient
of x3 was often ignored.

Graph sketching is a very important skill in all mathematical subjects – from

Economics to Engineering. STEP candidates are strongly advised to practise
this skill as much as possible.

9 This was a popular question, and was usually well done. Not many candidates
recognised that sin θ cos θ ≡ ½ sin 2 θ, which makes the final inequality easier
to obtain. Knowing identities “both ways” is important.

10 Only a few attempts at this question were seen, and those that did rarely made
much headway; worryingly, the accurate simplification of the solutions of
simple linear equations was found to be very difficult.

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11 Hardly any attempts at this question were seen. It was remarkable how few
diagrams were seen; it is always much easier for both the candidate and the
examiner if answers begin with a labelled diagram.

12 Very few tree diagrams were seen here, and hence very few correct solutions
were constructed; a clear tree diagram is invaluable when attempting a
complicated probability question such as part (ii). Most candidates identified
some (if not all) of the possible outcomes, but many mistakes were made (for
example, writing a denominator of N rather than N + 1 or N – 1).

The subsequent algebraic simplification was found to be very demanding.

Candidates would have probably made more progress if they had been more
willing to factorise groups of terms which had obvious common factors, rather
than (for example) attempting to write all the fractions with a common
denominator.

13 A lot of attempts at this question were seen, but conceptual errors undermined
many solutions. In particular, a lot of candidates seemed not to realise that
they were being asked to calculate conditional probabilities in parts (ii) to (vi).

14 Only a few attempts at this question were seen. Poor graph sketching limited
many candidates’ progress; the importance of the ability to sketch accurately
standard graphs such as y = xe–x cannot be overstated.

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STEP Mathematics II 2007: Report

General Remarks

Although the paper was by no means an easy one, it was generally found a more
accessible paper than last year’s, with most questions clearly offering candidates an
attackable starting-point. The candidature represented the usual range of mathematical
talents, with a pleasingly high number of truly outstanding students; many more who
were able to demonstrate a thorough grasp of the material in at least three questions;
and the few whose three-hour long experience was unlikely to have been a
particularly pleasant one. However, even for these candidates, many were able to
make some progress on at least two of the questions chosen.

Really able candidates generally produced solid attempts at five or six questions, and
quite a few produced outstanding efforts at up to eight questions. In general, it would
be best if centres persuaded candidates not to spend valuable time needlessly in this
way – it is a practice that is not to be encouraged, as it uses valuable examination
time to little or no avail. Weaker brethren were often to be found scratching around at
bits and pieces of several questions, with little of substance being produced on more
than a couple. It is an important examination skill – now more so than ever, with most
candidates now not having to employ such a skill on the modular papers which
constitute the bulk of their examination experience – for candidates to spend a few
minutes at some stage of the examination deciding upon their optimal selection of
questions to attempt.

As a rule, question 1 is intended to be accessible to all takers, with question 2 usually

similarly constructed. In the event, at least one – and usually both – of these two
questions were among candidates’ chosen questions. These, along with questions 3
and 6, were by far the most popularly chosen questions to attempt. The majority of
candidates only attempted questions in Section A (Pure Maths), and there were
relatively few attempts at the Applied Maths questions in Sections B & C, with
Mechanics proving the more popular of the two options.

It struck me that, generally, the working produced on the scripts this year was rather
better set-out, with a greater logical coherence to it, and this certainly helps the
markers identify what each candidate thinks they are doing. Sadly, this general
remark doesn’t apply to the working produced on the Mechanics questions, such as
they were. As last year, the presentation was usually appalling, with poorly labelled
diagrams, often with forces missing from them altogether, and little or no attempt to
state the principles that the candidates were attempting to apply.

Comments on responses to individual questions

SECTION A: PURE MATHEMATICS

Q1 Most candidates attempted this question and the majority coped fairly well
with the algebraic demands. Surprisingly, it was when the work went
numerical that candidates tended to let themselves down; poor arithmetic
providing the main difficulty. The final three marks available in (i) parts (a) and
(b) were the marks most frequently scorned, generally being lost by
candidates’ unwillingness or inability to simplify fractions and/or turn them into
decimals. In many cases, candidates had difficulty deciding on a suitable

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 value for k in (i) (b) and (ii). In (b), the value k = 50 was often selected, rather
than the intended value of – 4. Although this does lead to a similar set of
working, the ultimate approximation is relatively poor, and they lost the final
mark here. It is, rather more tellingly, indicative of the way in which many
modern A-level mathematicians have great difficulty in thinking only in terms
of positive integers! A small, but significant, number of candidates offered a
value of k that exceeded 100 (the denominator of the “x” term), and these
were penalised all four of the available marks for this part of the question, on
the not unreasonable grounds that they really should have appreciated the
k
general convergence condition | 100 | < 1 for binomial series of this kind.

Q2 This question was also a very popular one, although many candidates gave
up their attempt when the algebra started to get a little too tough for them,
which generally happened later if not sooner. With this in mind, it has to be
said that when candidates did get stuck at some stage of this question, the
principal cause was (again!) an unwillingness or inability to simplify algebraic
expressions before attempting to work with them. This was particularly
important when factorising otherwise lengthy expressions with lots of (q – p)s
involved in them.

The sketch required in (ii) was intended to be a gentle prod in the right
direction for later use in the question, and should have been four easy marks
for the taking. Strangely, however, it was often not very well attempted at all.
A surprising number of candidates couldn’t even manage to draw their cubic
through O; and many others seemed unable to make good use of the given
conditions, which – despite looking complicated – actually just ensured that all
the fun was going on in the first quadrant in an attempt to make life easy.
Even more surprising still was the number of sketches that had non-cubic-like
kinks, bumps and extra inflection points in them. I was particularly baffled by
this widespread lack of grasp as to what a cubic should actually look like! I
was equally baffled by the extraordinarily large body of candidates who failed
to do what the question explicitly told them they were required to do, by not
marking the point of inflection on their sketch and, in many cases, not even
attempting to describe the symmetry of it either.

An apology has to be made at this point, since the region R in the question
was insufficiently clearly defined and there were, in fact, two possibilities.
Candidates were not penalised for choosing the “wrong” one at any stage of
the proceedings, although the choice of the “left-hand” R would have
prevented such candidates from using the short-cut for the following attempt
at the area. ALL scripts where candidates made the “wrong” choice were
passed to the Principal Examiner and given careful individual consideration.
Only about 25 candidates made such a choice: of these, over half had failed
to make any attempt at all at the area, and most of the rest had started work
on the area and, to all intents and purposes, given up immediately. Two more
had found the intended area anyway, despite their previous working (and
were not penalised for having switched regions), and (I think) only three had
pursued the “left-hand” area almost to a conclusion. Of course, they were
unable to get the given answer, but they did get 7 of the 8 marks available. In
each of these cases, it was fortunate (for us and them) that this was their last
question, so it was safe to say that they hadn’t been unduly penalised for time
in any way. It is, of course, impossible to say whether they might have seen
the intended short-cut approach. In this respect, however, it has to be said
that remarkably few candidates saw the symmetry approach anyhow. Partly, I

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 suspect, due to not having picked up the hint at the diagram stage (see
earlier)! On the plus side, for us, I imagine that the reference to the point of
inflection on the diagram had at least ensured that most candidates chose the
intended region R. Only 2 of the 25 or so candidates scored an overall mark
that fell just below a grade boundary, and both of these were given the benefit
of the doubt by the Chief Examiner.

Q3 This was another popular question, and was usually a good source of marks
for those candidates who attempted it. The first two parts were usually
successfully completed. In part (i) (b), candidates had to employ the t = tan ½
x substitution which seems to have fallen into disuse in recent years (due to
modularity!). Having said that, most candidates were able to make some
progress and, where they did fall down, it was generally due to a lack of
confidence in handling trigonometric identities. One of the advantages of
these last two parts to the question was that they could be done in one of two
directions, and many candidates were able to spot the connections and
exploit them satisfactorily. When errors arose, they were frequently due to a
lack of care with constants, and a correct final answer was not often to be
found as a result.

Q4 This question was a popular one for partial attempts; with most candidates
giving up towards the end of the introductory part and going elsewhere. It was
slightly surprising to see candidates being put off in this way, since the given
result made it perfectly possible to move successfully into the three following
cases. For those who did press on, many lost a mark for not verifying
(somewhere) that the chosen values of α, β and γ actually satisfied the
required condition. Then, in (ii), one of the two brackets was identically zero,
the significance of which was largely overlooked, with many candidates
offering again the same two solutions as had been found in part (i). In (iii), it
was important to note first A and B in terms of x, although some candidates
adopted a valid alternative approach by first collecting up the two 3x terms.

Q5 Although this was not a popular choice of question, those who attempted it
generally did rather well on it. Finding f2 and f3 was a routine algebraic slog,
and most attempts coped successfully with it. Spotting, and then exploiting,
the periodicity of the function was then a relatively easy matter. Pretty much
everyone used x = tan θ appropriately in (ii), with formal and informal
induction approaches evenly mixed. Some shrewder candidates identified the
two forms for the cases n = 1, 2, and 3 and then noted that the periodicity of
the tan function accounted for everything thereafter.

The final part of the question had intended to be a simple take on part (ii), but
with t = sin θ this time, so that 1 − t 2 = cos θ , and attempts at this part of
the question generally fell evenly into one of the two following camps: those
who gave up, and those who proceeded as intended. In all, I think there were
just three candidates who noticed the extra complication that can arise in this
case, with just two or three more following a separate line of enquiry without
realising the inherent dichotomy in the “powers” of the function g. A full
inspection of the function exposes the fact that gn takes different forms
depending upon which part of the domain of g is employed. This is because
the 1 − t 2 bit should actually be | cos θ |, and this leads to different
answers for g2 in the range ½ ≤ t ≤ 1 than in the rest of g’s domain, so that
candidates could get different answers from slightly different approaches.

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 With so few candidates expected to attempt this last part of the question, and
with the alternate route leading to a much easier answer (where the sequence
gn turns out to be periodic with period 2), it was considered to be a suitable
final part to the question. Candidates were not expected to take more than
one route, nor to comment on the potential for different answers. In the event,
none did the former, although a few gave a mention of the latter property.

Q6 This was one of the most popular questions on the paper, although the
number of completely successful attempts could be counted without having to
resort to toes! Part (i) was reasonably routine, although attempts at
simplification were often not very well done, and left many candidates having
to resort to “otherwise” approaches for integrating 3 + x 2 , which was a
great shame as they got no marks for ignoring the “hence” instruction in the
question. Treating the differential equation in part (ii) as a quadratic in dy/dx
proved an obstacle for many, but a lot of candidates seemed quite happy to
work with it as such and made good progress in the rest of the question. The
biggest hurdle to completely successful progress, however, once again lay in
candidates’ inability to simplify expressions at various stages, and sign and/or
constant errors proliferated.

Q7 Not very many candidates attempted this question, but those who did usually
found it to be relatively straightforward. It was only the very last part that
required much thought, and this was where most attempts lost a few marks. A
small number of efforts failed to get beyond part (ii); this was due to not
finding a suitable function to work with that gave what turns out to be the
Arithmetic Mean-Geometric Mean Inequality. This was a bit of a shame, since
the question actually gives the log. function at the very beginning, along with
the sine function, which is used in (i).

Q8 This is really just half of the (⇔) proof of Ceva’s Theorem. Several candidates
even recognised it as such. Of the remarkably small number of attempts
submitted, most fell down at some stage (again) by failing to be sufficiently
careful with signs/arithmetic/the modest amounts of algebra involved. It often
didn’t help those candidates who chose completely different symbols each
time they did a stage of the working.

SECTION B: MECHANICS

Q9 This was the least popular of the Mechanics questions, perhaps because it
commenced with a request for an explanation. As mentioned already, a
clearly labelled diagram or two would have been enormously helpful here!
The fact that there are only the two mechanical principles being employed
here should have made it an easy question, but efforts were generally very
poor.

Q10 This was the most popular of the three Mechanics questions, although most
efforts failed to get very far into it. The routine opening part, finding the
position of a centre of mass, probably accounts for its initial (relative)
popularity, but progress beyond this point was pitifully weak in most cases.
Resolving and taking moments frequently appeared, but often had to be
searched-for in amidst a sea of other statements, many of which were
incorrect, repetitive or just nonsensical. Very few candidates indeed grasped
the fact that the horizontal force P could be in either direction, and the given

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 answer was mostly fiddled, usually by simply placing modulus signs around
the answer.

Q11 This was almost as popular a question on Section B as Q10, being (in
principle, at least) a reasonably straightforward projectiles question. Whilst
many efforts were successful up to the final part, an awful lot of the attempts
foundered at the very outset by failing to do the simplest of tasks: namely,
noting exact values for sin θ and cos θ from tan θ = ½. It simply beggars
belief that serious candidates can proceed through quite a large part of a
question like this with expressions such as sin(arc tan ½) still in there! They
may as well just hang out a flag which says “I’m an incompetent
mathematician” on it! The three-dimensional aspect of the introduction was
enough to confound most candidates attempting this question, and they were
forced to resort to fiddling the given answer for the distance OP. Many
attempts picked up several marks here and there throughout the question
without producing anything particularly coherent, and few coped with the
hazards of the last part – largely, I suspect, due to the fact that they were
required to do some approximating!

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 SECTION C: STATISTICS

Q12 This was the least popular of the Statistics questions, with very few attempts
seen on it. Most of these tended to consist of muddled or unexplained
reasoning which led to the fiddling of (i)’s given result. Progress into (ii) was
either non-existent or sketchy as a consequence.

Q13 This is a lovely approach to a well-known problem, and employs a very handy
rational approximation to ln 2. Although it drew a small number of attempts,
many of these were partial attempts at best, and few were seen of a good
standard throughout. Disappointingly, several candidates arriving at the
correct quadratic equation in the third part didn’t seem to know how to go
about solving it. As was mentioned earlier, regarding the end of Q11, working
with approximations proved to be a particular obstacle for most candidates
who made it to the last part here.

Q14 This was the most popular of the Statistics questions, probably due to the
high pure mathematical content. The sketch introduction was intended to
ensure that candidates drew something which would remind them what
integrals they should be working with later on. As with Q2, it presented more
problems than should have been the case, with many candidates losing
marks for fairly trivial things which would have cost them dearly even on an
ordinary AS/A-level module paper. The integration for total probability was
generally done very well, although several candidates had often failed to gain
a and b in terms of k in a simplified form, or at all, and this rather hindered
them. In (iii), most candidates didn’t seem to feel that it was necessary to
justify which region of the function that the median lay in, often doing one
calculation after making an assumption about the matter. In general, it is
always best if candidates can justify their choices.

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STEP Mathematics III 2007: Report

Section A: Pure Mathematics

1. This question was popular. Many candidates did not simplify their first
expression into the symmetrical form which made it harder for them to spot the use of
the sums and products of roots results. A common slip was to make a 1 by default
which also obscured what was going on. Most struggled to take the given equation
requiring solution and produce a quartic equation in t ( tan ϑ ), some producing a
quartic equation in cosϑ , and somehow expecting to use the earlier results.

2. This question was popular though not well answered. Solutions to part (i)
were frequently unconvincing, though to part (ii) were quite good if they avoided
elementary errors in working. Part (iii) was less well attempted with some not
spotting to use integration, some stumbling over “+ c” and some not spotting the
value of x to substitute.

3. This question was popular. Many solutions to part (ii) were rambling and
lacked a sense of direction, even if correct. The induction in (iii) was frequently
incorrectly handled and a common error was to replace n by k/2. Part (iv) caused
difficulties.

4. This question was quite popular. A lot of attempts involved rambling

t
trigonometrical manipulations, and few spotted the standard differential of ln tan .
2
The curve sketch was often omitted or incorrect, and there was a lot of complicated
working using e.g. the equation of the normal etc. to find the centre of curvature.

5. This was frequently attempted, though lack of facility with hyperbolic

functions meant that few progressed beyond the first two differentials, and for those
going further, the working was not methodical enough to spot the factorial that would
emerge in the general result.

6. This was the least popular Pure question and very little success was achieved
by the few that attempted it. The first result was often obtained correctly by
expressing each of the four complex numbers in modulus-exponential form, but then
the perpendicularity was the stumbling block.

7. This was a very popular question. As the question led the candidates through
there were a number of unconvincing solutions to parts of the question, but overall it
was reasonably well handled.

8. This ranked alongside question 5 in popularity and success. Frequently, it was

calculation errors that obscured the path through part (i) and the two differences
between part (i) and part (ii) were enough to put most off the track for part (ii), even if
they had completed (i) successfully.

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Section B: Mechanics

9. This was little attempted. Some did struggle through to the solution of the
differential equation, but the appreciation of the three possible cases eluded them.

10. This was the most popular of the Mechanics questions, but less so than any but
question 6 of the Pure. Most managed to obtain the first two results correctly, but
then struggled to find the further result. The deduction for the largest R was rarely
spotted leading to some unnecessarily unwieldy calculus.

11. There were very few attempts at this question.

Section C: Probability and Statistics

12. There were some attempts at this question but they faltered when trying to find
the expectation of Y, even though some may have believed that they had obtained the
required result through false logic.

13. This was the most popular of the Probability and Statistics questions, ranking
alongside questions 5 and 8. The first two parts were competently handled, but most
got bogged down in the algebra of part (iii) through not having a clear strategy to
solve the equations.

14. There were few answers of any substance to this question.

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 STEP

Mathematics
STEP 9465, 9470, 9475

STEP Solutions

June 2007

STEP2007

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 CONTENTS

STEP Mathematics (9465, 9470, 9475)

Unit Content Page

9465 STEP Mathematics I

9470 STEP Mathematics II

9475 STEP Mathematics III

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June 2007

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June 2007

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STEP Examiners’
Report
2008

Mathematics
STEP 9465, 9470, 9475

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 Contents

Step Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 3
STEP Mathematics II 11
STEP Mathematics III 16
Grade Boundaries 18
Cumulative Percentages 18

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STEP I Examiners' Report June 2008

General Remarks
There were significantly more candidates attempting this paper this year (an increase
of nearly 25%), but many found it to be very difficult and only achieved low scores. The
mean score was significantly lower than last year, although a similar number of
candidates achieved very high marks. This may be, in part, due to the phenomenon of
students in the Lower Sixth (Year 12) being entered for the examination before
attempting papers II and III in the Upper Sixth. This is a questionable practice, as while
students have enough technical knowledge to answer the STEP I questions at this
stage, they often still lack the mathematical maturity to be able to apply their knowledge
to these challenging problems.

Again, a key difficulty experienced by most candidates was a lack of the algebraic skill
required by the questions. At this level, the fluent, confident and correct handling of
mathematical symbols (and numbers) is necessary and is expected; many students
were simply unable to progress on some questions because they did not know how to
handle the algebra.

There were of course some excellent scripts, full of logical clarity and perceptive
insight. It was also pleasing that one of the applied questions, question 13, attracted a
very large number of attempts.

However, the examiners were again left with the overall feeling that some candidates
had not prepared themselves well for the examination. The use of past papers and
other available resources to ensure adequate preparation is strongly recommended. A
student’s first exposure to STEP questions can be a daunting, demanding experience;
it is a shame if that takes place during a public examination on which so much rides.

Comments on individual questions

Q1 This question was primarily about logical thinking and structuring an argument.
While it was a very popular question, the marks were disappointing: only 30% of
candidates gained more than 6 marks.

Most candidates could describe vaguely what is meant by the term irrational, though
only a handful gave a precise, accurate definition. The popular offering of ‘a number
with an infinite decimal expansion’ was not acceptable.

It was pleasing to see, though, that the majority of candidates were capable of using
proof by contradiction to prove statements A and B, and they then went on to provide a
counterexample to statement C. A small number of very strong candidates justified their
counterexamples by proving that the numbers they presented were in fact irrational,
though any well-known irrational examples were given full marks without the need for
justification.

It is important to stress the difference between proving a statement and disproving one;
while a single (numerical) counterexample is adequate to disprove a statement, a proof
of the truth of a statement requires a general argument. Too many candidates wrote
things such as: ‘If pq = 3 , then p = 3 and q = 1, so . . . .’ Also, it is unknown what
an irrational number ‘looks like’, so the frequently occurring arguments such as ‘We

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STEP I Examiners' Report June 2008

know that e + π must be irrational because the numbers are not of the same form’
(when comparing this example to something like (1 − 2 ) + 2 ) are spurious.

Sadly, very few candidates made any significant progress on the main part of the
question. Several attempted (unsuccessfully) to prove that all four of the given numbers
are irrational. Others asserted that since π and e are both irrational, π + e must also be,
despite having just disproved statement C. A number of candidates successfully
showed that π + e and π - e cannot both be irrational by appealing to B, but then could
not see how to continue. The best attempts proceeded by using A and B repeatedly to
show that no pair of π ± e and π2 ± e2 could simultaneously be rational (that is, they
considered all six cases separately).

Q2 This was by far the most popular question on the paper, with about six out of
every seven candidates attempting it.

The very first part involving implicit differentiation was generally done very well with
most candidates scoring full marks for this part.

A majority of candidates then went on to successfully see how to apply this result to the
required integral, although a sizeable minority failed to understand that they were being
asked to perform a substitution. Some candidates resorted to the formula book and
quoted the standard integral ∫ 1/ x 2 + a 2 dx; however, this gained no credit as the
question explicitly said ‘hence’.

Having reached ∫ 1/ (t + b ) dt, the vast majority of candidates became unstuck. Firstly,
after integrating, some did not substitute back t = x + ··· to get an expression in terms of
x. The fundamental problem, though, was that the candidates were mostly unaware of
the need to use absolute values when integrating 1/x: almost everyone gave the
intermediate answer as ln(t + b)+ c rather than ln |t + b| + c. It turns out that in this
case, t + b is always positive so the absolute values may be replaced by parentheses,
but this requires explicit justification (which no-one gave).

This lack of appreciation of absolute values prevented all but the strongest candidates
from making a decent attempt at the last part of the question, the consideration of the
case c = b2 . Some candidates successfully substituted this in to the earlier result as
instructed, but many claimed that 2 x 2 + 2 x + b 2 = x + b . However, the correct expression
is |x + b|, which is x + b when this is positive, but −(x + b) when x + b< 0. Only the tiny
handful of candidates who appreciated this subtlety managed to correctly explain the
distinction between these two cases.

Q3 This was another popular question, although the scores were again fairly poor.

The proof of (*) was often done quite well. The main difficulties here arose because of a
lack of clarity in the logic; it is important to make clear where the starting point is and
what steps are being taken to move forward from there. A significant number of
candidates attempted to work backwards, and then divided by d − b or the like without
realising that this might be zero. Also, inequalities were multiplied without any regard to
the sign of the numbers under consideration; for example, while 0 > −1 and 1 > −2, it is
not true that 0 × 1 > (−1) × (−2).

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The beginning of part (i) was completed correctly by a majority of candidates. It is

important to stress again that if a question specifies “use (*)”, then this must be done to
gain any credit; no marks were given for the numerous answers which began with “as
(x − y)2 ≥ 0 for all x and y, we have x2 − 2xy + y2 ≥ 0” or similar.

The last part of (i) was often poorly tackled. It was sometimes interpreted to mean
“when x < 0” or other spurious cases, without understanding that the inequality had so
far only been shown in the case x ≥ y ≥ z. (Indeed, the intermediate result
z2 + xy ≥ xz + yz does not hold in the case x > z > y.) Many other candidates ignored
their preceding work and went on to prove the result from scratch using the inequality
(x − y)2 + (y − z)2 +(z − x)2 ≥ 0. Very few candidates explained the symmetry of the
situation.

Part (ii) was problematic because of the wording of the question. It turned out that there
is a very straightforward way to answer this part by making use of the results proved in
part (i). While this was not what was actually asked (“Show similarly . . . ”), it was felt
unfair to penalise candidates too harshly for taking this route. Thus they were awarded
partial credit and all such candidates were referred to the Chief Examiner for individual
consideration. Nonetheless, the attempts at this part, by whichever method, were
generally either close to perfect or non-starters.

Q4 The initial graph-sketching part of this question was designed to help candidates
solve the quadratic equation which was to come up later in the question. Whilst almost
2
all of the candidates successfully sketched y = sin x, the attempts at y = cos2x were
3
significantly poorer. Many candidates sketched curves with cusps at the x-axis,
presumably confusing cos2 x with | cos x|; others had curves which fell below the x-axis
in places. Perhaps few candidates had seen graphs of y = cos2x before or considered
1
that cos2x = (cos 2x + 1), making cos2x sinusoidal itself. Also, a large number of
2
candidates appeared to have spent a significant amount of time drawing beautiful and
accurate graphs on graph paper; it is important to appreciate the nature of a sketch as
a rough drawing which captures the essential features of a situation. In general, STEP
questions will not require accurate graphs, only accurate sketches.

The first derivative of f(x) was generally computed correctly, though a sizable proportion
of candidates failed to correctly apply the product rule to determine the second
derivative. Those candidates who obtained f’’(x) correctly generally realised that they
2
needed to solve the inequality cos2x ≥ sin x. Some appear to have guessed a value
3
of x which makes this an equality - this method is perfectly acceptable as long as some
justification of the claimed result is given (such as by explicitly substituting x = π / 6 into
the two sides). Most candidates who got this far correctly understood the connection
with the graph sketch and went on to give the correct intervals.

In part (ii), there was a lot of difficulty performing the differentiation. A number of
candidates made their life more difficult by substituting k = sin 2α before differentiating
g(x); this just made the expressions appear more complex and increased the likelihood
of error. Some candidates, for example, tried differentiating with respect to both x and α
simultaneously.

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Nevertheless, most candidates who were able to correctly compute g’’(x) went on to
solve the resulting trigonometric equation, finding the solution x = α, but many failed to
determine the second interval. A sketch of some sort would very likely have been
useful.

Q5 This was the least popular of the Pure Mathematics questions. There was a fair
amount of confusion as to the meaning of the summation, with the majority of attempts
at the n =1 case in part (i) failing to understand that the polynomial would be p(x)= x+a0
rather than just p(x)= x. A small number thought that the summation indicated a
geometric series, and proceeded to claim that p(x) = (1 − xn)/(1 − x) or other such
things.

Nonetheless, there were many good answers to the rest of part (i), with candidates
showing that they understood the statement of Chebyshev’s theorem. A small number
of strong candidates had a mature enough understanding of mathematics to use the
alternative method given in the sample solutions; most were content with finding the
maxima and minima. Some forgot to check the value of p(x) at the ends of the interval,
which was not penalised as long as they did not incorrectly assert that they had found
the value of M.

1
One recurrent incorrect assertion was that from the inequality p(x) ≥ , it necessarily
2
1
follows that M = , without showing that equality is obtained for some value of x.
2

There were few serious attempts at part (ii), but most of those achieved full marks or
very close to it. Several candidates had difficulty in explaining their reasoning: a sketch
would certainly have helped clarify why a maximum value of |p(x)| occurring in the
interval −1 <x< 1 necessarily forces this point to be a turning point.

Of the other attempts, many could not see the relevance of Chebyshev’s theorem to
this situation, or even if they did, then failed to divide the given polynomial by 64.
Arguments which did not invoke Chebyshev’s theorem were not given any credit (the
main alternative being to differentiate, then to find points where the derivative was
positive and negative, and use the intermediate value theorem to assert that there is a
point where the derivative must be zero).

Q6 This was another popular question which was gained a pleasing number of good
marks.

The sketch was generally done well. A significant number of candidates did not realised
that f(0) = 0 and f(1) = 1, so either had non-intersecting graphs or graphs which were
tangent to each other at the origin. A number of candidates sketched the graph of f(x)
for all real x, in spite of the question stating x ≥ 0; they were not penalised for this. Most
understood how the graphs of f(x) and g(x) related.

The determination of g(x) algebraically was performed correctly by a majority of

candidates. However, a disturbing number of candidates introduced absolute value
signs, writing g(x)= ln |(e−1)x+1|. Whilst technically correct in this range (and therefore
not penalised here), it is indicative of confusion about when absolute values are used

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with logs: when integrating 1/x (as in Question 2 above) they are required; when
inverting exponentiation they are not. A smaller number made very significant errors in
their handling of the logarithm function, writing such things as
ln(ex − x +1) = ln ex − ln x + ln 1.

The majority of candidates correctly integrated f(x). A small minority bizarrely asserted
⎛ 1⎞
that ∫01/ 2 f(x)dx = f ⎜ ⎟ - f(0)which was somewhat disturbing.
⎝2⎠

The integration of g(x) proved much more troublesome. Despite ∫ ln x dx being a

standard integral and explicitly mentioned in both the STEP Specification and A2
Mathematics specifications, the introduction of the linear function of x flummoxed most
candidates. Some differentiated instead of integrating, others just gave up. A small
number either attempted to use parts or to substitute, and a good proportion of such
attempts were successful. Some candidates confused differentiation with integration
during this process and tried to use a mixture of parts and the product rule.

Finally, of those who managed to reach this point, a decent number gave a very
1
convincing explanation of why ∫f + ∫g = k.
2

Q7 This was a reasonably popular question, tackled by about half of the candidates.
1
Most confidently showed that y = (y − 3 x) and went on to deduce the result for the
2
clockwise rotation. A small number of candidates lost marks here because their
presentation either failed to make clear which answer corresponded to which direction
of rotation, or the directions were reversed. Several candidates would have been
helped by including a sketch in their solution.

About two-thirds of the candidates were unable to progress beyond this point. Of those
who continued, the majority succeeded in finding h1, either by a direct argument or,
more usually, by using the earlier result as intended by the question. Despite the hint of
h1 being given with absolute value signs, a large number of candidates then claimed
that h2 = y rather than the correct |y|, suggesting that they do not understand what
absolute values mean and when they should be used.

Very few candidates correctly determined h3, the most common incorrect answer being
1
h3 = |y + 3 x| to parallel the answer for h1. Again, clear diagrams are essential if
2
marks are to be gained for questions such as these. There was also evidence of
confusion in the algebraic manipulation of absolute values, with some candidates
1 1
confusing |a−b| with |a|−|b|, thereby giving answers such as h3 = |y + 3 x|− 3
2 2

Only a handful of candidates made a significant attempt at the final part of the question,
and of those who did, the main difficulty stemmed from not appreciating that to prove
an “if and only if” statement, one has to prove the implication in both directions. The
sample solutions use the triangle inequality; it could equally and straightforwardly be
argued by considering all eight possible cases of where the point P might lie with
respect to each of the three sides of the triangle.

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Q8 This was another popular question, and many candidates achieved decent
marks on this question.

Many candidates were correctly able to differentiate (*), although a significant number
ran into difficulties with the ( y′ )2 term, where things such as 2 y′
d
( y′) = 2 y′. y′′ dy were
dx dx
common errors. Although the rest of (*) was usually differentiated correctly by these
candidates, since the rest of part (i) depended upon getting this first step correct, they
floundered from then on. Many of these candidates nevertheless went on to gain
additional marks by at least making a good start to part (ii).

Also, candidates must remember to read the question and to follow its guidance; the
question instructed them to differentiate (*), and those who tried rearranging it instead
got nowhere.

From y′′ = 0, most candidates deduced that y′ = m and substituted this back into (*) to
determine y = mx − m2 . However, when working like this, it is vital to check that the
purported y, call it ŷ say, satisfies d ŷ /dx = y′ , since y and y′ must be related by both
the given differential equation and also by y′ =dy/dx. There may be other arguments
which would allow one not to differentiate the obtained y, but these would have to be
given explicitly. The alternative method of determining that y = mx + c and then
substituting this into (*) was noticeably less common, but avoided this subtlety.

For the 2 y′′ = x case, similar comments again apply, although here it was concerning
1 2
how many candidates integrated to get y = x without including a constant of
4
integration.

In part (ii), few candidates succeeded in correctly differentiating the differential

d 2
equation. One of the most common errors was to claim that (y )=2 y′ . The few
dx
candidates who correctly differentiated the equation mostly applied the techniques from
part (i) to solve the equation successfully. Several fudged the solution of the resulting
equation (x2 − 1) y′ = xy by conveniently forgetting the absolute value signs when
integrating (as shown in the sample solutions), but this error was not penalised on this
occasion.

Q9 This was an unpopular question and the marks were very disappointing; only
half of the attempts gained over one mark out of twenty, and only a handful of
candidates gained over six marks.

Nonetheless, of the candidates who made a reasonable start, many were capable of
drawing a clear diagram of the position of the hoop after it had rolled, but few were able
to show how the position after it had rolled related to its initial position. This allowed
them to correctly determine the y-coordinate of P, but they became very unstuck when
attempting to determine the x-coordinate.

The next difficulty encountered was in calculating the components of the velocity of P,
as many candidates appeared unable to differentiate a function of θ with respect to t.

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.
For the determination of the kinetic energy, several candidates θ used the expected
method. It was also very encouraging to see a number of candidates correctly using the
formula total KE = linear KE + rotational KE, and then determining the rotational KE
1 .
using either moments of inertia or the explicit formula mr2 θ 2.
2

Finally, a small number of candidates correctly considered forces or energy and

deduced that the hoop rolls at constant speed.

Q10 This was the most popular mechanics question, and the question which gained
the best marks across the entire paper.

The sketch of the particle’s trajectories in the two different scenarios was generally well
done, with almost all candidates successfully completing the sketch. It was a little
disappointing, though, that very few attempted to justify their assumption that the
particle does, in fact, reach height h.

The next stage, using the “suvat” equations to deduce d, was generally either done
very well or very poorly. Of those who had difficulty, some were stuck trying to figure
out how to go about the question, others were unsure of which of the “suvat” equations
to use (despite all of the individual components of this question being very standard A-
1 2
level problems), while some derived a quadratic equation (having used s = ut + at )
2
but were incapable of then solving it.

Nonetheless, this question did require a sustained chain of logical steps, and it was
pleasing to see over a quarter of the candidates who attempted this question gaining
close to full marks on it.

Q11 This was the least popular of the mechanics questions, and of the candidates
who attempted it, only a handful made any progress beyond drawing a usually incorrect
sketch and writing down some equations.

The majority were aware that F = µR as the equilibrium is limiting. Unfortunately,

though, they often either missed forces from their diagram or drew at least one of the
frictional forces in the wrong direction. Another frequent problem was that they labelled
both normal reaction forces with the same variable R, thereby implicitly implying that
they are equal, whereas this is not the case. A small annoyance was the number of
candidates who used the notation Fr for friction; an unhealthy practice as it can so
easily be confused with F × r. Also, several failed to mark the angle α correctly on their
diagram.

After this, a small number of candidates correctly resolved in two directions and took
moments. Those who understood how to then manipulate the resulting equations to
eliminate most of the variables went on to produce essentially perfect solutions,
whereas everyone else became stuck at this point and found themselves unable to
progress any further.

No attempts using the theorem regarding three forces on a large body were seen,
which is a shame, as it made the problem significantly easier.

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Q12 This was by far the least popular question on the paper, as is often the case with
Probability and Statistics questions.

Of the candidates who attempted it, most successfully answered part (i), and a
significant number were also confident in the manipulation of sums required for part (ii).
Success was clearly dependent upon taking great care to ascertain the meaning of the
event X = r in terms of X1 and X2.

Part (iii) proved much more problematic, as almost no-one made use of both defining
inequalities for the median; one inequality on its own may appear to give the correct
answer, but is insufficient to gain the marks.

The handful of candidates who attempted part (iv) were generally successful in their
attempts.

Q13 This combinatorics question was attempted by close to half of all candidates, a
very encouraging statistic. About two-thirds of the attempts did not progress very far,
gaining five marks or fewer, but of those who did get further, the marks were fairly
evenly distributed.

For part (i), most attempts reached the stated answer, although a significant number
used very creative, if inaccurate or meaningless, ways of doing so. The majority of
candidates used counting methods, and many of these were successful to a greater or
lesser extent. The other method used by many candidates was to consider the
2
probability of the first wife sitting next to her husband ( ) and the conditional
5
probability of the spouse of the other person sitting next to the first husband sitting next
1
to them (this is ), and then multiplying these.
3

It is crucial at this point to reinforce that candidates must explain their reasoning in their
answers, especially when they are working towards a given answer. Simply writing
2 1 2
× = is woefully inadequate to gain all of the available marks; there must be a
5 3 15
justification of the reasoning behind it.

Parts (ii) and (iii) were found to be a lot more challenging. A number of candidates
attempted to construct probabilistic arguments, which are very challenging in this case.
The successful attempts all used pure counting arguments. The examiners often found
it challenging to decipher their thinking, though, as the explanations were often
somewhat incoherent. Those who used counting arguments usually made good
progress on both parts.

The favoured method for part (iii) was to use P(no pairs) = 1 − P(≥ 1 pair). It would have
certainly been worth checking the answer obtained using a direct method, as this would
have caught a number of errors.

The main errors encountered in good attempts at the later parts of the question were a
failure to consider all possible cases or a miscounting of the number of ways each
possible case could occur.

Overall, this question was answered well by a significant number of candidates.

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General Remarks
There were around 850 candidates for this paper – a slight increase on the 800 of the
past two years – and the scripts received covered the full range of marks (and beyond!).
The questions on this paper in recent years have been designed to be a little more
accessible to all top A-level students, and this has been reflected in the numbers of
candidates making good attempts at more than just a couple of questions, in the
numbers making decent stabs at the six questions required by the rubric, and in the
total scores achieved by candidates. Most candidates made attempts at five or more
questions, and most genuinely able mathematicians would have found the experience a
positive one in some measure at least. With this greater emphasis on accessibility, it is
more important than ever that candidates produce really strong, essentially-complete
efforts to at least four questions. Around half marks are required in order to be
competing for a grade 2, and around 70 for a grade 1.

The range of abilities on show was still quite wide. Just over 100 candidates failed to
score a total mark of at least 30, with a further 100 failing to reach a total of 40. At the
other end of the scale, more than 70 candidates scored a mark in excess of 100, and
there were several who produced completely (or nearly so) successful attempts at more
than six questions; if more than six questions had been permitted to contribute towards
their paper totals, they would have comfortably exceeded the maximum mark of 120.
While on the issue of the “best-six question-scores count” rubric, almost a third of
candidates produced efforts at more than six questions, and this is generally a policy
not to be encouraged. In most such cases, the seventh, eighth, or even ninth, question-
efforts were very low scoring and little more than a waste of time for the candidates
concerned. Having said that, it was clear that, in many of these cases, these partial
attempts represented an abandonment of a question after a brief start, with the
candidates presumably having decided that they were unlikely to make much successful
further progress on it, and this is a much better employment of resources.

As in recent years, most candidates’ contributing question-scores came exclusively from

attempts at the pure maths questions in Section A. Attempts at the mechanics and
statistics questions were very much more of a rarity, although more (and better)
attempts were seen at these than in other recent papers.

Comments on individual questions

Q1 The first question is invariably intended to be a gentle introduction to the paper,

and to allow all candidates to gain some marks without making great demands on either
memory or technical skills. As such, most candidates traditionally tend to begin with
question 1, and this proved to be the case here. Almost 700 candidates attempted this
question, making it (marginally) the second most popular question on the paper; and it
gained the highest mean score of about 14 marks.

There were still several places where marks were commonly lost. In (i), setting (x2, y2) =
(x1, y1) and eliminating y (for instance) leads to a quartic equation in x. There were two
straightforward linear factors easily found to the quartic expression, leaving a quadratic
factor which could yield no real roots. Many candidates failed to explain why, or show
that, this was so. In (ii), the algebra again leads to two solutions, gained by setting (x3,
y3) = (x1, y1). However, one of them corresponds to one of the solutions already found in
(i), where the sequence is constant, and most candidates omitted either to notice this or

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to discover it by checking. Another very common oversight – although far less important
in the sense that candidates could still gain all the marks by going the long way round –
was that the algebra in (ii) was exactly the same as that in (i), but with a = – x and b =
– y . For the very few who noticed this, the working for the second half of the question
was remarkably swift.

Q2 Noticeably less popular than Q1 – with only around 500 “hits” – and with a very
much poorer mean mark of about 8, it was rather obvious that many candidates were
very unsure as to what constituted the best partial fraction form for the given algebraic
fraction to begin with. Then, with very little direct guidance being given in the question,
candidates’ confidence seemed to ebb visibly as they proceeded, being required to turn
the resulting collection of single algebraic fractions into series, using the Binomial
Theorem, and then into a consideration of general terms. There was much fudging of
these general terms in order to get the given answers of either n + 1 or n + 2 for the
general term’s coefficients; even amongst those who did spot which one occurred when,
there was often little visible justification to support the conclusions. As a result of all the
hurdles to be cleared, those who managed to get to the numerical ending successfully
were very few in number.

Q3 This, the third most popular question on the paper, producing a mixed bag of
responses. It strikes me that, although the A-level specifications require candidates to
understand the process of proof by contradiction, this is never actually tested anywhere
by any of the exam. boards. Nonetheless, it was very pleasing to see that so many
candidates were able to grasp the basic idea of what to do, and many did so very
successfully. The impartial observer might well note that the situation in (i) is very much
tougher (in terms of degree) than that in (ii). However, candidates were very much more
closely guided in (i) and then left to make their own way in (ii).

Apart from the standard, expected response to (i) – see the SOLUTIONS document for
this – many other candidates produced a very pleasing alternative which they often
dressed up as proof by contradiction but which was, in fact, a direct proof. It was,
however, so mathematically sound and appealing an argument (and a legitimate
imitation of a p by c) that we gave it all but one of the marks available in this part of the
question. It ran like this:
Suppose w.l.o.g. that 0 < a ≤ b ≤ c < 1.
Then ab(1 – c) ≤ b2(1 – b) ≤ 274 by the previous result
(namely x2(1 – x) ≤ 4
27 for all x ≥ 0).
QED.
[Note that we could have used ab(1 – c) ≤ c2(1 – c) ≤ 274 also.]
It has to be said that most other inequality arguments were rather poorly constructed
and unconvincing, leaving the markers with little option but to put a line through (often)
several pages of circular arguments, faulty assumptions, dubious conclusions, and
occasionally correct statements with either no supporting reasoning or going nowhere
useful.

There was one remarkable alternative which was produced by just a couple of
candidates (that I know of) and is not included in the SOLUTIONS because it is such a
rarity. However, for those who know of the AM – GM Inequality, it is sufficiently
appealing to include it here for novelty value. It ran like this:

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4
Assume that bc(1 – a), ca(1 – b), ab(1 – c) > 27 .
Using the previous result, we have a (1 – a), b (1 – b), c (1 – c) ≤
2 2 2 4
27 .
Then, since all terms are positive, it follows that a ≤ bc, b ≤ ca, c ≤ ab so that
2 2 2

a2 + b2 + c2 ≤ bc + ca + ab. (*)
However, by the AM – GM Inequality (or directly by the Cauchy-Schwarz
Inequality),
a2 + b2 ≥ 2ab , b2 +c2 ≥ 2bc and c2 + a2 ≥ 2ca .
Adding and dividing by two then gives a2 + b2 + c2 ≥ ab + bc + ca, which
contradicts the conclusion (*), etc., etc.

Q4 Another very popular question, poorly done (600 attempts, mean score below 7).
Most efforts got little further than finding the gradient of the normal to the curve, and I
strongly suspect that this question was frequently to be found amongst candidates’ non-
contributing scorers. Using the tan (A – B) formula is a sufficiently common occurrence
on past papers that there is little excuse for well-prepared candidates not to recognise
when and how to apply it. Once that has been done, the question’s careful structuring
guided able candidates over the hurdles one at a time, each result relying on the
preceding result(s); yet most attempts had finished quite early on, and the majority of
candidates failed to benefit from the setters’ kindness.

Q5 This was the most popular question on the paper (by a small margin) and with
the second highest mean mark (12) of all the pure questions. Those who were able to
spot the two standard trig. substitutions s = sin x and c = cos x for the first two parts
generally made excellent progress, although the log. and surd work required to tidy up
the second integral’s answer left many with a correct answer that wasn’t easy to do
anything much useful with at the very end, when deciding which was numerically the
greater. The binomial expansion of (a + b)5 was handled very comfortably, as was much
of the following inequality work. However, the very final conclusion was very seldom
successfully handled as any little mistakes, unhelpful forms of answers, etc., prevented
candidates’ final thoughts from being sufficiently relevant.

Q6 This was the least popular of the pure maths questions. Although there were 300
starts to the question, most of these barely got into the very opening part before the
attempt was abandoned in favour of another question. Most attempts failed to show that
f(x) has a period of 4π. As mentioned, few proceeded further. Of those who did, efforts
were generally very poor indeed – as testified to by the very low mean mark of 4 – with
the necessary comfort in handling even the most basic of trig. identities being very
conspicuous by its absence. Part (iii) was my personal favourite amongst the pure
questions, as it contained a very uncommon – yet remarkably simple – idea in order to
get started on the road to a solution. The idea is simply this: f(x), being the sum of a
cosine term and sine term, is equal to 2 if and only if each of these separate terms is
simultaneously at its maximum of 1. That is, the question is actually two very easy trig.
equations disguised as one very complicated-looking one. Once realised, the whole
thing becomes very straightforward indeed, but only a few candidates had persevered
this far.

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Q7 In many ways, part (i) of the question was very routine, requiring little more than
technical competence to see the differential equation, using the given substitution,
through to a correct solution. Part (ii) then required candidates to spot a slightly different
substitution on the basis of having gained a feel for what had gone on previously. I had
thought that many more candidates would try something involving the square root of 1
+ x3 or the cube root of 1 + x2, rather than cube root of 1 + x3, but many solutions that I
saw went straight for the right thing. Once this had been successfully pushed through –
with the working mimicking that of (i) very closely indeed – it was not difficult to spot the
general answer required, unproven, in (iii). Overall, however, it seems that a lot of
candidates failed to spot the right thing for part (ii) and their solutions stopped at this
point. With almost 600 attempts, the mean score on this question was 10.

Q8 As with Q6, this was both an unpopular question and poorly done. Those
candidates who did do well generally did so after spotting that they could use the Angle
Bisector Theorem to polish off the first half of the question, expressing λ in terms of a
and b almost immediately. Predominantly, the whole thing relied almost exclusively
upon the use of the scalar product (or, alternatively, the Cosine Rule) and a bit of
manipulation. The fact that the mean mark on this question was below 7 is simply
indicative of the general lack of confidence amongst candidates where vectors are
concerned.

Q9 Of the applied maths questions, this was by far the most popular, with over 400
attempts. However, most of these were only partial efforts, with few candidates even
getting around to completing part (i) successfully, and the mean score ended up at
about 8. Most candidates were comfortable with the routine stuff to start with, quoting
and using the trajectory equation and using the identity sec2α = 1 + tan2α to get a
quadratic equation in tanα. For the remaining parts of the question, working was much
less certain, even given the helpful information about small-angle approximations, and
very few candidates were able to get a suitable approximation for tanα. Fewer still
could turn an angle in radians into one in degrees.

Q10 Though much less popular than Q9, the attempts at this question followed a
similar pattern, with most candidates coping pretty well with the routine opening
demands – the use of the two main principles governing collisions questions:
Conservation of Linear Momentum and Newton’s Experimental Law of Restitution – but
then falling down when a little more care and imagination were required in the parts that
followed. With some careful application of ideas relating to similar triangles and a bit of
inequalities work to follow, most candidates attempting these questions were just not up
to the task. Few got as far as working on the initial and final kinetic energies; of these
only a very small number noticed that there was a very quick way to go about it (see the
SOLUTIONS). I don’t recall seeing anyone successfully managing to get the right
answer after having taken the longer route.

Q11 This question attracted under 100 attempts and a mean mark of under 3. The
strong complaint I have made in the Report over recent years has consistently been that
candidates’ efforts on such questions have been seriously compromised by a disturbing
inability to draw a decent diagram at the outset. I’m afraid that this was a major
stumbling-block to successful progress with this question this year also. It was also a bit

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of a problem that candidates tended to confuse the acceleration of P relative to the

wedge with its absolute acceleration relative to the stationary surface on which the
wedge stood (say). As few decent attempts were made, it is difficult to be very specific
about what went on otherwise.

Q12 There were almost 200 attempts to Q12, and the mean score was the highest –
at 14 – of all the applied questions. This was partly due to the fact that the result of the
first part could be largely circumvented by anyone who knew a little bit about
expectation algebra, enabling them to write down E(X) straightaway. The simple
combinations of events, and their associated probabilities, in the final part of the
question were very confidently and competently handled by most candidates and many
polished the question off in its entirety relatively quickly.

Q13 Perhaps encouraged by the ease with which they had managed Q12, many of
these candidates went on to attempt this question also. Although the listing of relevant
cases was a fairly straightforward exercise, the handling of the binomial coefficients –
which certainly looked clumsy and unappealing – was coped with much less well, and
many mistakes were made in the ensuing algebra. In the very final part of the question,
( )
the idea that the calculus could lead to a nice, neat answer k = n(n − 1) that then
needed to be interpreted in terms of integer values, was just one step too far for most
takers. The eventual mean score of 8 on this question testifies to the difficulties found in
the algebra by most of the candidates who attempted it.

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General Remarks
Most candidates attempted five, six or seven questions, and scored the majority of their
total score on their best three or four. Those attempting seven or more tended not to do
well, pursuing no single solution far enough to earn substantial marks.

Comments on individual questions

Q1 This was the most popular question on the paper, and many earned good marks
on it. Nearly all the candidates followed the hint, and most then applied the same trick
with the third equation. Subsequent success depended on a candidate realising that
they had simultaneous equations in xy and x + y, although very rarely some managed to
solve directly in x and y.

Q2 About three fifths attempted this question, often obtaining the starred result and
the familiar S3(n) successfully, but with S4(n) tripping up many. Any that made progress
on part (ii) tended to be able to complete the whole question.

Q3 Just under half attempted this. Most were reluctant to use parametric
differentiation. Some found T’s coordinates successfully and got not further, but most
either made very little progress on the whole question, or got right through it.

Q4 Almost exactly the same number attempted this as question 3, but with much
less success. The initial inequality was frequently poorly justified, but some managed to
apply it correctly to obtain the starred result, and went on to do part (ii) respectably.
However, for most, it was a case of all or nothing.

Q5 In terms of attempts and success, this resembled question 2. Apart from some
that made no progress at all, the induction was accessible to many, as was the
expression for Tn(x). In both of these there were frequent gaps or inaccuracies even
though the solutions were understood in essence.

Q6 More than 80% attempted this, and with more success than any other question.
Having obtained the relation between x and p in each part, quite a few attempts then
treated these as differential equations rather than merely substituting back to find
expressions for y, and consequent inaccuracies lost marks.

Q7 Less than a fifth attempted this and frequently with little success except for
obtaining the initial result. The configuration for part (i) tripped up many, although some
skipped that to do part (ii) successfully.

Q8 Three fifths attempted this with most scoring about two thirds of the marks.
Apart from minor errors, the last part (expressing T in partial fractions etc.) was the
pitfall for most.

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Q9 Of the three Mechanics questions, this was the most popular with just under a
quarter of the candidates attempting it, but with least success. In spite of obtaining the
relation in the stem of the question, many failed to appreciate its consequence for the
acceleration-time graph in part (i) and as a consequence made little further progress. If
candidates managed part (i), then they tended to complete the question barring minor
errors, and the occasional assumption that the final simple case was simple harmonic
motion.

Q10 Just under a fifth attempted this, but many dealt successfully with the n short
strings case to earn about half the marks. Occasionally a candidate would obtain the
required length result for the heavy rope and fail to apply the same technique for the
elastic energy, but apart from minor errors, most that appreciated how to take the limit
had few difficulties.

Q11 Under a twelfth tried this. A number of different correct approaches were
successfully applied, and there were very few partially correct solutions.

Q12 Little more than a handful of candidates attempted this with three strong
attempts (near full marks) and the remainder making no headway at all.

Q13 About a ninth tried this. Apart from those who had no idea, there were three
categories of attempt. The first group obtained the first result but did not spot that
regardless of what happens in the first step, immediately after it there are 2n - 2 free
ends. The second group safely navigated the results for the general case but could not
see how to apply the approximation to obtain the result in the specific case, and the
final group had the satisfaction of finding the result. Most fell into the first category, with
fewer in the second, and a small number in the third.

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 STEP Mathematics (9465/94709475)
June 2008 Assessment Series

Grade boundaries

Paper Maximum S 1 2 3 U
Mark
Paper 1 120 81 65 43 29 0
(9465)
Paper 2 120 94 69 58 35 0
(9470)
Paper 3 120 82 63 52 34 0
(9475)

The cumulative percentage of candidates achieving each grade was as follows:

Paper S 1 2 3 U
Paper 1 6.5 17.1 45.1 69.7 100
(9465)
Paper 2 12.2 36.0 50.3 81.8 100
(9470)
Paper 3 13.0 38.0 56.9 82.1 100
(9475)

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STEP Solutions
2008

Mathematics
STEP 9465, 9470, 9475

239
 Contents

Step Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 3
STEP Mathematics II 37
STEP Mathematics III 49

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 STEP I, Solutions
June 2008

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 STEP I STEP Solutions June 2008

Question 1

What does it mean to say that a number x is irrational?

It means that we cannot write x = m/n where m and n are integers with n 6= 0.

Prove by contradiction statements A and B below, where p and q are real numbers.

A: If pq is irrational, then at least one of p and q is irrational.

B: If p + q is irrational, then at least one of p and q is irrational.

We first prove statement A.

Assume that pq is irrational, but neither p nor q is irrational, so that both p and q are
rational. But then pq is the product of two rational numbers, so is rational. This contradicts
that assumption that pq is irrational. So statement A is true.
Now for statement B we argue similarly.
Assume that p + q is irrational, but neither p nor q is irrational, so that both p and q are
rational. But then p + q is the sum of two rational numbers, so is rational. This contradicts
the assumption that p + q is irrational. So statement B is true.

Disprove by means of a counterexample statement C below, where p and q are real numbers.

C: If p and q are irrational, then p + q is irrational.

√ √
One example is p = 2, q = − 2.

If the numbers e, π, π 2 , e2 and eπ are irrational, prove that at most one of the numbers
π + e, π − e, π 2 − e2 , π 2 + e2 is rational.

We assume that the five given numbers are, indeed, irrational.

We have (π + e) + (π − e) = 2π, which is irrational (if p is irrational, then so is 2p). So by
statement B, at least one of π + e and π − e is irrational.
Similarly, (π 2 + e2 ) + (π 2 − e2 ) = 2π 2 , which is irrational. So by statement B again, at least
one of π 2 + e2 and π 2 − e2 is irrational.
Assume that both π + e and π 2 − e2 are rational. Then
π 2 − e2
π−e=
π+e
would also be rational. But we know that at least one of π + e and π − e is irrational, so
π + e and π 2 − e2 cannot both be rational. Similarly, we can’t have both π − e and π 2 − e2
rational.
Thus if two of the four numbers are rational, they must be π 2 + e2 and one of π ± e.

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Assume that π 2 + e2 and π + e are rational. Then (π + e)2 = (π 2 + e2 ) + 2eπ is the square
of a rational number, so is rational. But then 2eπ = (π + e)2 − (π 2 + e2 ) would be rational,
contradicting the irrationality of eπ. Thus we cannot have both π 2 + e2 and π + e rational.
Similarly, if π 2 + e2 and π − e are both rational, we would have 2eπ = (π 2 + e2 ) − (π − e)2
being rational, again a contradiction.
Thus at most one of these four numbers is rational.

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Question 2

√
The variables t and x are related by t = x + x2 + 2bx + c, where b and c are constants
and b2 < c. Show that
dx t−x
= ,
dt t+b
1
and hence integrate √ .
x2 + 2bx + c

We have, differentiating the given expression with respect to x:

dt 2x + 2b
=1+ √
dx 2 x2 + 2bx + c
2(x + b)
=1+
2(t − x)
(t − x) + (x + b)
=
t−x
t+b
= .
t−x
The required result follows on taking the reciprocal of both sides:
dx 1 t−x
= = .
dt dt/dx t+b

To find the integral, we use the given substitution for x, yielding:

Z Z
1 1 dx
√ dx = dt
2
x + 2bx + c t − x dt
1 t−x
Z
= dt
t−x t+b
Z
1
= dt
t+b
= ln |t + b| + k
 √ 
= lnx + b + x2 + 2bx + c + k
√

= ln x + b + x2 + 2bx + c + k
with the last line following as x2 + 2bx + c > x2 + 2bx + b2 = (x + b)2 , so the parenthesised
expression is positive.

Verify by direct integration that your result holds also in the case b2 = c if x + b > 0 but
that your result does not hold in the case b2 = c if x + b < 0.

With b2 = c, we have the integral

Z Z
1 1
√ dx = p dx
x2 + 2bx + b2 (x + b)2
Z
1
= dx
|x + b|

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We now consider the two cases discussed in the question. Firstly, if x + b > 0, then we have
Z Z
1 1
√ dx = dx
2
x + 2bx + b 2 |x + b|
Z
1
= dx
x+b
= ln(x + b) + k 0

(we don’t need absolute value signs as x + b is positive), whereas before we had
√
p

ln x + b + x2 + 2bx + c + k = ln x + b + (x + b)2 + k


= ln x + b + |x + b| + k


= ln x + b + (x + b) + k
= ln(2(x + b)) + k
= ln(x + b) + ln 2 + k

Thus our earlier formula works in the case that b2 = c and x + b > 0, where we take
k 0 = k + ln 2 (which we may do, as they are arbitrary constants).
Next, when x + b < 0, direct integration yields:
Z Z
1 1
√ dx = dx
x2 + 2bx + b2 |x + b|
Z
1
= − dx
x+b
= − ln |x + b| + k 0
= − ln(−(x + b)) + k 0

as x + b < 0. But now our earlier formula yields

√
p

ln x + b + x2 + 2bx + c + k = ln x + b + (x + b)2 + k


= ln x + b + |x + b| + k


= ln x + b − (x + b) + k
= ln 0 + k

which is not even defined. So the earlier result fails to give any answer in the case b2 = c
when x + b < 0.

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Question 3

Prove that, if c > a and d > b, then

ab + cd > bc + ad. (∗)

We have

ab + cd − bc − ad = (a − c)b + (c − a)d
= (c − a)(d − b)
>0 as c > a and d > b,

from which (∗) follows immediately.

(i) If x > y, use (∗) to show that x2 + y 2 > 2xy.

If, further, x > z and y > z, use (∗) to show that z 2 + xy > xz + yz and deduce that
x2 + y 2 + z 2 > xy + yz + zx.
Prove that the inequality x2 + y 2 + z 2 > xy + yz + zx holds for all x, y and z.

Letting a = b = y and c = d = x in (∗), which we can do as x > y, yields y 2 + x2 > yx + yx,

that is
x2 + y 2 > 2xy. (1)

Next, letting a = b = z, c = x and d = y in (∗) gives

z 2 + xy > zx + zy (2)

as we wanted.
Now adding the inequalities (1) and (2) gives us

x2 + y 2 + z 2 + xy > 2xy + yz + zx.

Subtracting xy from both sides yields our desired result:

x2 + y 2 + z 2 > xy + yz + zx. (3)

Finally, we have now proved inequality (3) when x > y > z, but we need to show that it is
true whatever the values of x, y and z. But the inequality is symmetric in x, y and z, meaning
that rearranging (permuting) the variables in any way does not change the statement. For
example, if we swap x and z, we get

z 2 + y 2 + x2 > zy + yx + xz,

which is exactly the same inequality.

So we can assume that the the values of x, y and z we are given satisfy x > y > z, without
changing the statement of the inequality, and we know that the inequality holds in this case.

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s t r
(ii) Show similarly that the inequality + + > 3 holds for all positive r, s and t.
t r s

We begin by assuming that r > s > t > 0, and set c = r, a = s, d = 1/s and b = 1/r. Then
by our assumption, we see that c > a and d > b, so (∗) gives
s r
+ > 1 + 1. (4)
r s

Now set c = s, a = t, d = 1/t and b = 1/r, and note that c > a and d > b, so that (∗) gives
t s s
+ >1+ . (5)
r t r
Adding the inequalities (4) and (5) gives
s r t s s
+ + + >1+1+1+ ,
r s r t r
so that
r t s
+ + > 1 + 1 + 1, (6)
s r t
as we want.
Now this inequality is true for r > s > t, and by exactly the same argument it will also hold
if s > t > r or t > r > s, by cycling the variables.
If r > t > s, then we have to start again, but the argument is almost identical.
We set c = r, a = t, d = 1/t and b = 1/r. Then by our assumption, we see that c > a and
d > b, so (∗) gives
t r
+ > 1 + 1. (7)
r t
We now set c = r, a = s, d = 1/s and b = 1/t, and note that c > a and d > b, so that (∗)
gives
t s s
+ >1+ . (8)
r t r
Once again, adding these two inequalities gives inequality (6), and by cycling the variables,
it is also true if t > s > r or s > r > t.
We have now shown (6) to be true for all six possible orderings of r, s and t, so it is true for
all possible (positive) values of r, s and t.

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Question 4

A function f(x) is said to be convex in the interval a < x < b if f 00 (x) > 0 for all x in this
interval.
2
(i) Sketch on the same axes the graphs of y = 3
cos2 x and y = sin x in the interval
0 6 x 6 2π.
The function f(x) is defined for 0 < x < 2π by
2
f(x) = e 3 sin x .

Determine the intervals in which f(x) is convex.

We note that cos2 x = 12 (cos 2x + 1) from the double angle formula, so that the graph of
2
3
cos2 x is a translated, stretched version of y = cos 2x.

1
2
3
y = 23 cos2 x

0 x1 π x2 π 3π 2π
2 2
y = sin x
−1

Now we have
2
f(x) = e 3 sin x
2
f 0 (x) = ( 23 cos x)e 3 sin x
2
f 00 (x) = ( 49 cos2 x − 23 sin x)e 3 sin x
2
= 23 ( 23 cos2 x − sin x)e 3 sin x .
2
As e 3 sin x > 0 for all x, we have

f 00 (x) > 0 if and only if 2

3
cos2 x − sin x > 0.

But we have just drawn a graph of the two functions y = 23 cos2 x and y = sin x, so we see
that f(x) > 0 when 0 6 x 6 x1 and when x2 6 x 6 2π, where x1 and x2 are the x-coordinates
of the points of intersection of the two graphs. So all we need to do is to solve the equation
2
3
cos2 x − sin x = 0

to determine the values of x1 and x2 .

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Using cos2 x = 1 − sin2 x and then factorising gives

2
3
cos2 x = sin x
⇐⇒ 2
3
(1 − sin2 x) = sin x
⇐⇒ 2 − 2 sin2 x = 3 sin x
⇐⇒ 2 sin2 x + 3 sin x − 2 = 0
⇐⇒ (2 sin x − 1)(sin x + 2) = 0
⇐⇒ sin x = 12
Therefore x1 = π6 and x2 = 5π
6
, and the function f(x) is convex in the intervals 0 < x < π
6
and 5π
6
< x < 2π.

(ii) The function g(x) is defined for 0 < x < 12 π by

g(x) = e−k tan x .

If k = sin 2α and 0 < α < π/4, show that g(x) is convex in the interval 0 < x < α,
and give one other interval in which g(x) is convex.

We have
g(x) = e−k tan x
g0 (x) = −k sec2 x.e−k tan x
g00 (x) = (k 2 sec4 x − 2k sec2 x tan x)e−k tan x
= k sec2 x(k sec2 x − 2 tan x)e−k tan x .
Therefore g 00 (x) > 0 when k sec2 x − 2 tan x > 0, that is, when k tan2 x − 2 tan x + k > 0. We
can solve the equality k tan2 x − 2 tan x + k = 0 using the quadratic formula:
√
2 ± 4 − 4k 2
tan x =
√2k
1 ± 1 − k2
=
pk
1 ± 1 − sin2 2α
= substituting k = sin 2α
sin 2α
1 ± cos 2α
= .
sin 2α
Thus we have two possibilities:
1 + cos 2α 2 cos2 α
tan x = = = cot α = tan( π2 − α)
sin 2α 2 sin α cos α
or
1 − cos 2α 2 sin2 α
tan x = = = tan α.
sin 2α 2 sin α cos α
π
Then, since 0 < α < 4
and 0 < x < π2 , we have x = α or x = π2 − α.
It follows, since k > 0 and α < π2 −α, that g(x) is convex, that is, g00 (x) > 0, when 0 < x < α
or when π2 − α < x < π2 . (In more detail, when x ≈ 0, g00 (x) ≈ k > 0, and when x ≈ π2 ,
g00 (x) ≈ k tan2 x > 0.)

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Question 5

The polynomial p(x) is given by

n−1
X
n
x + ar xr ,
r=0

where a0 , a1 , . . . , an−1 are fixed real numbers and n > 1. Let M be the greatest value of
|p(x)| for |x| 6 1. Then Chebyshev’s theorem states that M > 21−n .

(i) Prove Chebyshev’s theorem in the case n = 1 and verify that Chebyshev’s theorem
holds in the following cases:

(a) p(x) = x2 − 12 ;
(b) p(x) = x3 − x.

In the case n = 1, Chebyshev’s theorem states:

Let p(x) be the polynomial x + a0 , and let M be the greatest value of |p(x)| for
|x| 6 1. Then M > 1.

If a0 > 0, then when x = 1, p(1) = 1 + a0 > 1, so M > 1.

If a0 < 0, then when x = −1, p(−1) = −1 + a0 < −1, so |p(−1)| > 1 and M > 1.
Finally, if a0 = 0, then p(x) = x, so |p(x)| = |x|. It follows that |p(x)| = |x| 6 1 when
|x| 6 1 and p(1) = 1, so M = 1.
Thus in all cases M > 1.

Now to verify the theorem in the specified cases. The obvious approach is to find the
maximum absolute value of the function over the interval. It is important to verify that the
function’s maximum absolute value really is at least 21−n in both cases.

(a) p(x) = x2 − 12 is a quadratic whose minimum value is at x = 0. So we only need to

consider the value of p(0) and the values of p(x) at the endpoints of the interval: p(−1)
and p(1). We have p(0) = − 12 , p(−1) = p(1) = 12 , so − 12 6 p(x) 6 12 , and hence
|p(x)| 6 12 = 2−1 with the maximum value taken on by |p(x)| being 12 .
In this case, where n = 2, Chebyshev’s theorem states that M > 2−1 . Since we have
M = 2−1 in this case, Chebyshev’s theorem holds.

(b) Given p(x) = x3 − x we first look for stationary

√ points. We have p0 (x) = 3x2 − 1
so there are stationary points at x = ±1/ 3. We thus evaluate p(x)
√ at these√points
and √at the endpoints
√ x = ±1. We have√p(−1) = p(1) = 0, p(−1/
√ 3) = 2/3 3 and
p(1/ 3) = −2/3 3. Hence |p(x)| 6 2/3 3, so that M = 2/3 3.
As n = 3, we wish to show that M > 14 . But M 2 = 4
27
> 4
64
= 1
16
, so M > 1
4
as required.

A second approach, which is simpler and more direct, is to observe that all we need to do
is to find some value of x in the interval −1 6 x 6 1 for which |p(x)| > 21−n , for then we

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know that the maximum value of |p(x)| in this interval will be at least that. In (a), we have
|p(1)| = 12 > 2−1 and in (b), |p( 12 )| = | 18 − 12 | = 38 > 2−2 . So we are done.

(ii) Use Chebyshev’s theorem to show that the curve y = 64x5 +25x4 −66x3 −24x2 +3x+1
has at least one turning point in the interval −1 6 x 6 1.

1
Let p(x) = x5 + 64 (25x4 − 66x3 − 24x2 + 3x + 1) = y/64. The turning points of p(x) are
the same as the turning points of y. Then if we let M be the greatest value of |p(x)| for
|x| 6 1, we have M > 2−4 = 16 1
. Let x0 be the value of x in the interval −1 6 x 6 1 for
which |p(x0 )| = M , i.e., where |p(x)| takes its maximum value. (If there is more than one
such point, choose any of them to be x0 .)
1 3 1 1 1
Now p(−1) = 64 and p(1) = 64 , and so |p(1)| < 16
and |p(−1)| < 16
. But since |p(x0 )| > 16
,
we cannot have x0 = ±1, so −1 < x0 < 1.
1
Then x0 must be a local maximum or a local minimum: if p(x0 ) > 16 then it is at the the
greatest value of p(x) in the interval −1 6 x 6 1 (and is not at an endpoint); similarly, if
1
p(x0 ) 6 − 16 , then x0 is at the least value. Either way, x0 is a turning point as we wanted.

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Question 6

The function f is defined by

ex − 1
f(x) = , x > 0,
e−1
and the function g is the inverse function to f, so that g(f(x)) = x. Sketch f(x) and g(x)
on the same axes.

y y = f(x)

y = g(x)

0 x

Verify, by evaluating each integral, that

1
Z Z k
2 1
f(x) dx + g(x) dx = √ ,
0 0 2( e + 1)
1
where k = √ , and explain this result by means of a diagram.
e+1

We find g(x), the inverse of f(x), as follows, noting that y = f(x) if and only if x = g(y):
ex − 1
y = f(x) =
e−1
⇐⇒ (e − 1)y = ex − 1
⇐⇒ ex = (e − 1)y + 1


⇐⇒ x = ln (e − 1)y + 1


so that g(x) = ln (e − 1)x + 1 .

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Now we can evaluate the integrals. We have

1 1
ex − 1
Z Z
2 2
f(x) dx = dx
0 0 e−1
1 h x i 12
= e −x
e−1 0
1 1
(e 2 − 12 ) − (1 − 0)


=
e−1
1 1
e 2 − 32


=
e√−1
2 e−3
= .
2(e − 1)
R
For g(x), we can either use substitution and the standard result ln x dx = x ln x − x + c or
integration by parts, effectively deriving the result. We demonstrate both methods.
√
Using substitution, we set u = (e−1)x+1.
√ When x
√ = 0, u = 1, and when x = k = 1/( e+1),
we can easily calculate that u = ( e − 1) + 1 = e. Finally, du/dx = e − 1. Thus
Z k Z k

g(x) dx = ln (e − 1)x + 1 dx
0 0
√
Z e
dx
= ln u du
1 du
√
Z e
1
= ln u du
1 e−1
1 h i √e
= u ln u − u
e−1 1
1 √ √ √

= ( e ln e − e) − (ln 1 − 1)
e−1
1 √

= 1 − 12 e
e − 1√
2− e
= .
2(e − 1)

Alternatively we can use integration by parts. We first note that

 
e−1
ln((e − 1)k + 1) = ln √ +1
e+1
√

= ln ( e − 1) + 1
√
= ln( e)
= 12 .

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Then we can evaluate our integral as follows:

Z k Z k
g(x) dx = 1. ln((e − 1)x + 1) dx
0 0
Z k  
 k e−1
= x ln((e − 1)x + 1) 0 − x dx
0 (e − 1)x + 1
Z k
(e − 1)x + 1 − 1
= k ln((e − 1)k + 1) − dx
0 (e − 1)x + 1
Z k
1 1
= 2k − 1− dx
0 (e − 1)x + 1
 k
1 1
= 2k − x − ln((e − 1)x + 1)
e−1 0
 
1 1 
= 2k − k− ln((e − 1)k + 1) − (0 − ln 1)
e−1
 
1 1
= 2k − k −
2(e − 1)
1
= − 1k
2(e − 1) 2
1 1
= − √
2(e − 1) 2( e + 1)
√
1 e−1
= −
2(e − 1) 2(e − 1)
√
2− e
=
2(e − 1)

as we found using the substitution method.

We therefore have
1
k √ √
2 e−3 2− e
Z Z
2
f(x) dx + g(x) dx = +
0 0 2(e − 1) 2(e − 1)
√
e−1
=
2(e − 1)
1
= √
2( e + 1)

as we wanted.
Finally, to explain this result with the aid of a diagram, we want to fill in the two areas
indicated by the integrals on our sketch of the functions above. √ It would be useful to know
1 1/2
the value of f( 2 ) for this purpose: it is (e − 1)/(e − 1) = 1/( e + 1) = k, which is very
convenient. Since g(x) is the inverse of f(x), it follows likewise that g(k) = 12 . We can now
sketch the areas on our graph.

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y
y = f(x)
y = g(x)

1
2
k

0 k 1 x
2

R 1/2
In this sketch, the dark shaded area is the integral 0 f(x) dx and the striped area is the
Rk
integral 0 g(x) dx. We have reflected the dark shaded area in the line y = x to get the light
1
√ that the shaded and striped areas add to give a 2 × k
shaded area also shown. It is now clear
rectangle, so the area is k/2 = 1/2( e + 1), as we found.

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Question 7

The point P has coordinates (x, y) with respect to the origin O. By writing x = r cos θ and
y = r sin θ, or otherwise, show that,√if the line OP is rotated by 60◦ clockwise about O,
the new y-coordinate of P is 12 (y − 3 x). What is the new y-coordinate in the case of an
anti-clockwise rotation by 60◦ ?

The situation described is illustrated in the following diagram, where OP 0 is the image of OP
under the specified rotation, P 0 having coordinates (x0 , y 0 ).

P (x, y) P 0 (x0 , y 0 )

r
r θ θ − 60◦

Then we have
x = r cos θ
y = r sin θ
and
y 0 = r sin(θ − 60◦ )
We use the compound angle formula for sine to get
y 0 = r(sin θ cos 60◦ − cos θ sin 60◦ )
√
3
= 12 r sin θ − 2
r cos θ
√
3
= 12 y − 2
x
√
= 12 (y − 3 x).
Likewise, if the rotation is by 60◦ anticlockwise, we replace the θ − 60◦ by θ + 60◦ and repeat
the above to get √
y 0 = 12 (y + 3 x).

√
An equilateral triangle OBC has vertices at O, (1, 0) and ( 12 , 12 3), respectively. The
point P has coordinates (x, y). The perpendicular distance from P to the line through C
and O is h1 ; the perpendicular distance from P to the line through O and B is h2 ; and the
perpendicular distance from P to the line through B and C is h3 .
 √ 
Show that h1 = 1 y − 3 x and find expressions for h2 and h3 .
2

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h1
√ C
3
2
h3 P (x, y)

h2

O B
0 1

Clearly h2 = |y|. (We need to take the absolute value as P might lie under the x-axis.)
For h1 , consider rotating the entire shape by 60◦ clockwise about O. This will rotate OC
to the x-axis, and the perpendicular
√ from P to OC will become vertical.
 √ The
 transformed
1 1
y-coordinate of P is 2 (y − 3 x), as we deduced earlier, so h1 = 2 y − 3 x.
◦
Finally, for h3 , we start
√ by rotating anticlockwise by 60 around O. Then P√3ends up with
1
y-coordinate 2 (y + 3 x)  and the side BC ends up lying along the line y = 2 ; subtracting
1
√ √ 
these then gives h3 = 2 y + 3 x − 3 . 
[An alternative argument is to translate the whole diagram by one unit to the left first, so
that B moves to the origin and P moves
1
√ to (x − 1, y). Then rotating around the new B
gives P the new y-coordinate of 2 (y + 3(x − 1)), which is the same as before.]

1
√
Show that h1 + h2 + h3 = 2
3 if and only if P lies on or in the triangle OBC.

We have
1
   √   √ √ 

h1 + h2 + h3 = 2y  + y − 3 x + y + 3 x − 3
2
1
  √  √ √ 

= 2y  +  3 x − y  +  3 − 3 x − y  .
2

Using the triangle inequality, we then have

  √  √ √ 

h1 + h2 + h3 = 12 2y  +  3 x − y  +  3 − 3 x − y 
 √ √ √ 
> 12 (2y) + ( 3 x − y) + ( 3 − 3 x − y)
√
= 12 3,

with equality if and only if all of the bracketed terms are > 0 or all of the bracketed terms
are 6 0.
√
terms are negative or zero, then√2y 6√0, so y 6 0. And 3 x − y 6 0 implies
If all of the √
that x 6 y/ 3 6 0. But then we must have 3 − 3 x − y > 0, which is impossible. So we
cannot have all three terms negative or zero.
Therefore, if we have equality, we must have all three terms positive or zero. But 2y > 0
if
√ and only if P lies on or above the
√ x-axis, that is, on or above the line OB. Similarly,
3 x − y > 0 if and only if y 6 3 x, √ which is true √ if and√only if P lies on or below
the line OC (which has equation y = 3 x). Finally, 3 − 3 x − y > 0 if and only if

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√ √
y 6 3 − 3√x, which √ is true if and only if P lies on or below the line BC (which has
equation y = 3 − 3 x).
√
Putting these together shows that h1 + h2 + h3 = 12 3 if and only if P lies on or inside the
triangle OBC.

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Question 8

(i) The gradient y 0 of a curve at a point (x, y) satisfies

(y 0 )2 − xy 0 + y = 0. (∗)

By differentiating (∗) with respect to x, show that either y 00 = 0 or 2y 0 = x.

Hence show that the curve is either a straight line of the form y = mx + c, where
c = −m2 , or the parabola 4y = x2 .

Differentiating (∗) with respect to x, using the chain rule and product rule, gives

2y 0 y 00 − y 0 − xy 00 + y 0 = 0,

which, on cancelling terms and factorising, yields

(2y 0 − x)y 00 = 0,

so either y 00 = 0 or 2y 0 − x = 0.
We now solve these two differential equations. Firstly, by integrating y 00 = 0 twice with
respect to x, we get y 0 = a, so y = ax + b (where a and b are constants). Substituting this
back into (∗) gives
a2 − x.a + (ax + b) = 0,
so
a2 + b = 0,
which gives us the straight line y = mx + c with m = a and c = b = −a2 = −m2 .
In the other case, y 0 = 12 x, which on integrating gives y = 41 x2 + c. Substituting this back
into (∗) gives
( 12 x)2 − x( 12 x) + ( 14 x2 + c) = 0,
so c = 0 and y = 14 x2 , or 4y = x2 as required.

(ii) The gradient y 0 of a curve at a point (x, y) satisfies

(x2 − 1)(y 0 )2 − 2xyy 0 + y 2 − 1 = 0. (†)

Show that the curve is either a straight line, the form of which you should specify,
or a circle, the equation of which you should determine.

Differentiating (†) with respect to x, using the chain rule and product rule, gives:

2x(y 0 )2 + (x2 − 1).2y 0 y 00 − 2yy 0 − 2xy 0 y 0 − 2xyy 00 + 2yy 0 = 0,

which, on cancelling terms, yields

2(x2 − 1)y 0 y 00 − 2xyy 00 = 0.

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Finally, factorising brings us to our desired conclusion:

(x2 − 1)y 0 − xy y 00 = 0,

so either y 00 = 0 or y 0 (x2 − 1) − xy = 0.
We now solve these two equations. Again, integrating y 00 = 0 twice gives y 0 = a, so y = ax+b.
Substituting this back into (†) gives:
(x2 − 1).a2 − 2x(ax + b).a + (ax + b)2 − 1 = 0,
which is equivalent to
a2 x2 − a2 − 2a2 x2 − 2abx + a2 x2 + 2abx + b2 − 1 = 0,
so
−a2 + b2 − 1 = 0,
or b2 = a2 + 1. Thus the equation is satisfied by straight lines y = mx + c where c2 = m2 + 1.
In the other case, y 0 (x2 − 1) − xy = 0.
It looks as though we could solve this by separating the variables to give:
Z Z
1 x
dy = 2
dx,
y x −1
so ln y = 12 ln |x2 − 1| + c. Doubling and exponentiating then gives
y 2 = C|x2 − 1|.
To determine C and if its value depends upon whether |x| < 1 or |x| > 1, we ought to try
substituting back into (†). However, it is simpler to substitute in the result y 0 (x2 − 1) = xy
directly into (†).
Substituting y 0 = xy/(x2 − 1) back into (†) gives
 2  
2 xy xy
(x − 1) 2
− 2xy 2
+ y 2 − 1 = 0.
x −1 x −1
Expanding brackets then gives
x2 y 2 2x2 y 2
− + y2 − 1 = 0
x2 − 1 x2 − 1
so
x2 y 2
− 2
+ y 2 − 1 = 0.
x −1
Multiplying by x2 − 1 gives
−x2 y 2 + (y 2 − 1)(x2 − 1) = 0
so
−x2 y 2 + y 2 x2 − y 2 − x2 + 1 = 0
or
−x2 − y 2 + 1 = 0,
and we therefore deduce that the only other possible solution is the circle x2 + y 2 = 1.
We must finally check that the circle does, in fact, satisfy y 0 = xy/(x2 − 1). Differentiating
x2 + y 2 = 1 with respect to x gives 2x + 2yy 0 = 0, so y 0 = −x/y = −xy/y 2 = −xy/(1 − x2 ),
as required.

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Question 9

Two identical particles P and Q, each of mass m, are attached to the ends of a diameter
of a light thin circular hoop of radius a. The hoop rolls without slipping along a straight
line on a horizontal table with the plane of the hoop vertical. Initially, P is in contact with
the table. At time t, the hoop has rotated through an angle θ. Write down the position at
time t of P , relative to its starting point, in cartesian coordinates, and determine its speed
in terms of a, θ and θ̇. Show that the total kinetic energy of the two particles is 2ma2 θ̇2 .

Q
O a O
P
θ

x
P
time 0 time t

The diagram shows the hoop rolling to the right, indicating the positions of the hoop at
time 0 and time t.
Taking the origin to be at the initial position of P , the coordinates of the centre of the hoop
at time t are (aθ, a), since the hoop has rolled a distance aθ. Therefore P has coordinates
(a(θ − sin θ), a(1 − cos θ)) and position vector

rP = a(θ − sin θ)i + a(1 − cos θ)j.

The velocity vector of P is then

d
d

ṙP = a(θ − sin θ) i + a(1 − cos θ) j
dt dt

dθ d dθ d dθ
=a − (sin θ) i − a (cos θ) j
dt dθ dt dθ dt
= a(θ̇ − cos θ.θ̇)i + a sin θ.θ̇ j


= aθ̇ (1 − cos θ)i + sin θ j ,

and hence P has speed vP given by

vP2 = (aθ̇)2 (1 − cos θ)2 + (sin θ)2

= (aθ̇)2 (1 − 2 cos θ + cos2 θ + sin2 θ)

= (aθ̇)2 (2 − 2 cos θ)
= (aθ̇)2 .4 sin2 12 θ
= (2aθ̇ sin 12 θ)2 ,

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so that vP = 2a|θ̇ sin 12 θ|.

Similarly, the coordinates of Q are (a(θ + sin θ), a(1 + cos θ)), so Q has position vector

rQ = a(θ + sin θ)i + a(1 + cos θ)j.

Arguing as before, the velocity vector of Q is then

ṙQ = a(θ̇ + cos θ.θ̇)i − a sin θ.θ̇ j

= aθ̇ (1 + cos θ)i − sin θ j

so that Q has speed vQ given by

2
= (aθ̇)2 (1 + cos θ)2 + (sin θ)2


vQ
= (aθ̇)2 (1 + 2 cos θ + cos2 θ + sin2 θ)
= (aθ̇)2 (2 + 2 cos θ)
= (aθ̇)2 .4 cos2 12 θ.

Adding vP2 = (aθ̇)2 .4 sin2 12 θ to this gives the total kinetic energy as
1
2
mvP2 2
+ 12 mvQ = 12 m(vP2 + vQ
2
)
= 12 m (aθ̇)2 .4 sin2 12 θ + (aθ̇)2 .4 cos2 12 θ

= 12 m(aθ̇)2 .4
= 2ma2 θ̇2

as required.

Given that the only external forces on the system are gravity and the vertical reaction of
the table on the hoop, show that the hoop rolls with constant speed.

Consider the hoop as a single system. The only external forces on the hoop are gravity and
the normal reaction. Both of these are vertical, while the hoop only moves in a horizontal
direction. Therefore, no work is done on the hoop, so that GPE + KE is constant.
The gravitational potential energy of P and Q together, taking the centre of the hoop as
potential energy zero, gives mga(− cos θ) + mga(+ cos θ) = 0. So the GPE of the system is
constant, meaning that the kinetic energy is also constant.
Since the total kinetic energy is 2ma2 θ̇2 , it follows that θ̇ is constant, that is, the hoop rolls
with the constant speed aθ̇.

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Question 10

On the (flat) planet Zog, the acceleration due to gravity is g up to height h above the surface
and g 0 at greater heights. A particle is projected from the surface at speed V and at an
angle α to the surface, where V 2 sin2 α > 2gh. Sketch, on the same axes, the trajectories
in the cases g 0 = g and g 0 < g.

We know that the path of a projectile is parabolic when the gravity is constant. When
the gravity is less, the parabola will be “bigger”, as the projectile will travel higher before
returning to the ground, but the horizontal component of velocity will be unaffected.
So our sketch will consist of a parabola for the case g = g 0 and a pair of parabolas joined at
height h in the case g 0 < g. Note that the velocity does not change suddenly at height h, so
the curve will be “smooth” at this point. (Technically, it has a continuous first derivative.)
We need to check that the particle does reach height h before we draw a sketch. The vertical
component of velocity is initially V sin α. If the gravity were a constant g, then the maximum
height reached would be s, where the formula v 2 = u2 + 2as gives us 02 = V 2 sin2 α − 2gs,
so s = V 2 sin2 α/2g > h, so the particle does reach height greater than h.
So here, then, is the sketch:

g0 = g g0 < g

Show that the particle lands a distance d from the point of projection given by

V −V0 V0
 
d= + 0 V sin 2α,
g g
p
where V 0 = V 2 − 2gh cosec2 α.

We note that the horizontal speed is a constant V cos α, as there is no horizontal component of
acceleration. Therefore the distance travelled is this times the time travelled. By symmetry,
we find the time taken to reach the highest point on the trajectory, and then double it to
find the total time.
First part: below height h.
We use the “suvat” equations to determine the time taken and the vertical speed at height h.
Taking upwards as positive, we have s = h, u = V sin α, a = −g. Then v 2 = u2 + 2as gives

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v 2 = V 2 sin2 α − 2gh and v = u + at gives

p
V sin α − V 2 sin2 α − 2gh
t=
g
p !
V − V − 2gh cosec2 α
2
= sin α
g
V −V0
 
= sin α,
g
p
writing V 0 = V 2 − 2gh cosec2 α.
Second part: above height h.
p
This time, u = V 2 sin2 α − 2gh = V 0 sin α, v = 0 and a = −g 0 , so v = u + at gives

V 0 sin α
t= .
g0

Therefore, the total time taken to reach the highest point is

V −V0 V 0 sin α V −V0 V0

   
sin α + = + 0 sin α.
g g0 g g

Finally, we need to multiply this by 2 to get the total time taken and then by V cos α to get
the distance travelled, giving the distance

V −V0 V0
 
d=2 + 0 sin α.V cos α
g g
0 0
 
V −V V
= + 0 V sin 2α,
g g

using 2 sin α cos α = sin 2α.

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Question 11

A straight uniform rod has mass m. Its ends P1 and P2 are attached to small light rings
that are constrained to move on a rough circular wire with centre O fixed in a vertical
plane, and the angle P1 OP2 is a right angle. The rod rests with P1 lower than P2 , and with
both ends lower than O. The coefficient of friction between each of the rings and the wire
is µ. Given that the rod is in limiting equilibrium (i.e., on the point of slipping at both
ends), show that
1 − 2µ − µ2
tan α = ,
1 + 2µ − µ2
where α is the angle between P1 O and the vertical (0 < α < 45◦ ).
Let θ be the acute angle between the rod and the horizontal. Show that θ = 2λ, where λ is
defined by tan λ = µ and 0 < λ < 22.5◦ .

We present two methods for solving the problem. The first is a standard method using
resolution of forces; the second is to use standard results about three forces acting on √
a large
body. We specify that the length of the rod is 2`, so that the radius of the circle is ` 2.
Method 1: Resolving all the forces
We begin by drawing a clear sketch of the situation.

O
√
√ ` 2
` 2 F2
α N2
θ `
α
N1
θ P2
`
45◦ M
θ
P1 α
F1
mg

Note that as the rod is in limiting equilibrium, both of the frictional forces act to prevent it
from slipping towards the horizontal, and F1 = µN1 , F2 = µN2 . Also, we see that θ = 45◦ −α,
as the angle OP1 P2 is 45◦ (the triangle being isosceles), so 0 < θ < 45◦ .
We now resolve the forces in two directions. We could resolve in any two directions, but so
that we can exclude mg from at least one of the equations, we choose to resolve horizontally

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and vertically. Another sensible choice would have been to resolve along the directions of
OP1 and OP2 .

R(↑) N1 cos α − µN1 sin α + N2 sin α + µN2 cos α − mg = 0 (1)

R(→) N1 sin α + µN1 cos α − N2 cos α + µN2 sin α = 0 (2)

We also need to take moments. There are four obvious places about which we can take
moments: O, P1 , P2 and M . For completeness, we show what happens if we calculate
moments about all four points; clearly only one of these is necessary.
y √ √
M (O) mg.` sin(45◦ − α) − F1 .` 2 − F2 .` 2 = 0 (3)
y √ √
M (P1 ) mg.` cos(45◦ − α) − N2 .` 2 − F2 .` 2 = 0 (4)
y √ √
M (P2 ) N1 .` 2 − F1 .` 2 − mg.` cos(45◦ − α) = 0 (5)
y √ √ √ √
M (M ) N1 .`/ 2 − F1 .`/ 2 − N2 .`/ 2 − F2 .`/ 2 = 0 (6)

Our task is now to eliminate everything to find an expression for tan α in terms of µ. We
can use any one of the equations (3)–(6) to do this, but (6) appears to be the easiest to
work with. (With the others, we would have to use a compound angle formula such as
sin(45◦ − α) = sin 45◦ cos α − cos 45◦ sin α = √12 (cos α − sin α).)
Recalling that F1 = µN1 and F2 = µN2 , equation (6) then gives us

N1 − µN1 = N2 + µN2

so
N1 (1 − µ) = N2 (1 + µ), (7)
or equivalently
N1 1+µ
= . (8)
N2 1−µ
(9)

We can also solve equations (1) and (2) simultaneously to get the results

mg(cos α − µ sin α)
N1 =
1 + µ2
and
mg(sin α + µ cos α)
N2 = ,
1 + µ2
so
N1 cos α − µ sin α
= .
N2 sin α + µ cos α

Equating this expression with equation (8) yields

cos α − µ sin α 1+µ
= .
sin α + µ cos α 1−µ

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Dividing the numerator and denominator of the left hand side by cos α then gives
1 − µ tan α 1+µ
= .
tan α + µ 1−µ

A simple rearrangement of this then yields our desired result:

1 − 2µ − µ2
tan α = .
1 + 2µ − µ2

(Alternatively, one could use (7) to substitute for N2 in equation (2), after writing F1 = µN1
and F2 = µN2 . Factorising then gives

N1 (1 + µ)(sin α + µ cos α) − N1 (1 − µ)(cos α − µ sin α) = 0.

On dividing by N1 cos α and expanding the brackets, we end up with an expression in tan α
which we can again rearrange to reach our desired conclusion.)

Now if tan λ = µ with 0 < λ < 22.5◦ , we have (recalling that θ = 45◦ − α)

tan θ = tan(45◦ − α)
tan 45◦ − tan α
=
1 + tan 45◦ tan α
1 − tan α
= as tan 45◦ = 1
1 + tan α
2
1 − 1−2µ−µ
1+2µ−µ2
= 2
1 + 1−2µ−µ
1+2µ−µ2
(1 + 2µ − µ2 ) − (1 − 2µ − µ2 )
=
(1 + 2µ − µ2 ) + (1 − 2µ − µ2 )
4µ
=
2 − 2µ2
2µ
=
1 − µ2
2 tan λ
=
1 − tan2 λ
= tan 2λ.

Then since 0 < θ < 45◦ , it follows that θ = 2λ as required.

Method 2: Three forces on a large body theorem
We recall the theorem that if three forces act on a large body in equilibrium, and they are not
all parallel, then they are concurrent (i.e., they all pass through a single point). We combine
the normal reaction, N , and friction, F , at each end of the rod into a single reaction force, R.
This acts at an angle φ to the normal, where tan φ = F/N . In our case, F = µN as the rod
is on the point of slipping, so tan φ = µ. But the question defines λ to be the acute angle
such that tan λ = µ, from which it follows that φ = µ.
We now redraw the diagram showing only the total reaction forces, which we call R1 and R2
in this context, and we ensure that R1 , R2 and the weight mg pass through a single point X.

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R1
R2

O X

α λ
θ
θ 45◦
λ P2
` `
M
P1 45◦ − λ

mg

The rest of the question is now pure trigonometry. We apply the sine rule to the triangles
P1 M X and P2 M X, first noting that

∠M XP1 = 180◦ − (45◦ − λ) − (90◦ + λ) = 45◦ − θ + λ

and
∠M XP2 = 180◦ − (90◦ − θ) − (45◦ + λ) = 45◦ + θ − λ
to get
` MX
= (10)
sin(45◦ − θ + λ) sin(45◦ − λ)
` MX
◦
= . (11)
sin(45 + θ − λ) sin(45◦ + λ)
Then using the identity sin(90◦ − x) = cos x twice, once with x = 45◦ − θ + λ and then with
x = 45◦ + λ, we deduce from equation (11) that
` MX
= . (12)
cos(45◦ − θ + λ) cos(45◦ − λ)
We can now divide equation (12) by (10) to get

tan(45◦ − θ + λ) = tan(45◦ − λ). (13)

It follows immediately that 45◦ − θ + λ = 45◦ − λ as both angles are acute, so θ = 2λ, which
answers the second part of the question.
This result then leads us to conclude that
2 tan λ 2µ
tan θ = tan 2λ = 2
= .
1 − tan λ 1 − µ2

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Finally, as we know that α = 45◦ − θ, we can deduce that

tan α = tan(45◦ − θ)
tan 45◦ − tan θ
=
1 + tan 45◦ tan θ
1 − tan θ
= as tan 45◦ = 1
1 + tan θ
2µ
1 − 1−µ 2
= 2µ
1 + 1−µ2
(1 − µ2 ) − 2µ
=
(1 − µ2 ) + 2µ
1 − 2µ − µ2
= ,
1 + 2µ − µ2
and we are done.

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Question 12

In this question, you may use without proof the results:

n
X n
X
r= 1
2
n(n + 1) and r2 = 16 n(n + 1)(2n + 1).
r=1 r=1

The independent random variables X1 and X2 each take values 1, 2, . . . , N , each value
being equally likely. The random variable X is defined by
(
X1 if X1 > X2
X=
X2 if X2 > X1 .

2r − 1
(i) Show that P(X = r) = for r = 1, 2, . . . , N .
N2

We have X = r when either X1 = r and X2 < r, or X2 = r and X1 < r or X1 = X2 = r.

Therefore

P(X = r) = P(X1 = r ∩ X2 < r) + P(X2 = r ∩ X1 < r) + P(X1 = X2 = r)

1 r−1 1 r−1 1 1
= . + . + .
N N N N N N
2r − 1
= .
N2

Alternatively, one can argue as follows:

P(X = r) = P(X1 = r ∩ X2 6 r) + P(X2 = r ∩ X1 6 r) − P(X1 = X2 = r)

1 r 1 r 1 1
= . + . − .
N N N N N N
2r − 1
= .
N2

(ii) Find an expression for the expectation, µ, of X and show that µ = 67.165 in the case
N = 100.

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By the definition of expectation, we have

N
X
µ = E(X) = r.P(X = r)
r=1
N
X r(2r − 1)
=
r=1
N2
N
1 X 2
= 2 (2r − r)
N r=1
1 2 1


= N (N + 1)(2N + 1) − N (N + 1) using the given results
N2 6 2
1
N (N + 1)(4N + 2 − 3)
= 6
N2
(N + 1)(4N − 1)
= .
6N
In the case N = 100, this is 101 × 399/600 = 13 433/200 = 67.165 as required.

(iii) The median, m, of X is defined to be the integer such that P(X > m) > 12 and
P(X 6 m) > 12 . Find an expression for m in terms of N and give an explicit value
for m in the case N = 100.

We have
k
X 2r − 1
P(X 6 k) =
r=1
N2
1
2. 12 k(k + 1) − k


= 2
N
k2
= 2
N
so that
(k − 1)2
P(X > k) = 1 − P(X 6 k − 1) = 1 − .
N2
We are looking for the value of m which makes P(X > m) > 12 and P(X 6 m) > 12 . The
first condition gives

(m − 1)2
1− > 12
N2
(m − 1)2
⇐⇒ 6 12
N2
⇐⇒ (m − 1)2 6 12 N 2
N
⇐⇒ m−16 √
2
N
⇐⇒ m6 √
2
+ 1.

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The second condition, P(X 6 m) > 12 , yields

m2
> 12
N2
⇐⇒ m2 > 12 N 2
N
⇐⇒ m> √
2
.
√ √ √
So we have N/ 2 6 m 6 (N/ √2) + 1, thus m is the smallest integer greater than N/ 2
(which is not itself an integer as 2 is irrational). The smallest integer greater than or equal
√ x is called the ceiling of x, and is denoted by dxe, so we can state our result as
to a number
m = dN/ 2e.
√
In the case N = 100, m = d100/ 2e = d70.7 . . . e = 71.

(iv) Show that when N is very large,

√
µ 2 2
≈ .
m 3

We have formulæ for µ from part (ii) and for m from part (iii). Therefore we have, for
large N ,
√

µ (N + 1)(4N − 1)
= dN/ 2e
m 6N
(N + 1)(4N − 1)
≈ √
6N (N/ 2)
4N 2 + 3N − 1
= √
6N 2 / 2
4 + N3 − N12
= √
6/ 2
4
≈ √
6/ 2
√
2 2
= .
3

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Question 13

Three married couples sit down at a round table at which there are six chairs. All of the
possible seating arrangements of the six people are equally likely.
2
(i) Show that the probability that each husband sits next to his wife is 15
.

We call the couples H1 and W1 , H2 and W2 , H3 and W3 . We seat H1 arbitrarily, leaving

5! = 120 ways of seating the remaining five people. If each husband sits next to his wife,
then there are two seats in which W1 can sit, each of which leaves four consecutive seats for
the other two couples.
There are four choices for who to sit immediately next to W1 , and that forces the following
seat as well (being the spouse of that person).
Next, there are two choices for who to seat the other side of H1 , and the final person must
sit next to their spouse.
So there are 2 × 4 × 2 = 16 ways to have all of the husbands sitting next to their wives, with
16 2
a probability of 120 = 15 .

(ii) Find the probability that exactly two husbands sit next to their wives.

Assume to begin with that H1 and W1 are separated and the other two husbands are seated
next to their wives. Once H1 has been seated, there are five possible positions for W1 , as
shown in the following diagrams (there are two possibilities shown in each of the first two):

H1 H1 H1

W1 W1

W1 W1

W1

Clearly the first two possibilities do not work: in the first, H1 and W1 are next to each other.
In the second, whoever sits between H1 and W1 will be separated from their spouse. So
W1 must sit opposite H1 , one couple sits to the right of H1 and the other couple to his left.
There are two choices for which couple sits to the right of H1 , and two choices for whether
the husband or wife sits next to H1 ; similarly there are two choices for whether the husband
or wife of the third couple sits next to H1 . So in all, there are 2 × 2 × 2 = 8 ways to seat the
couples with H1 and W1 separated and the other couples together.
Similarly, there are 8 ways with couple 2 separated and 8 ways with couple 3 separated, so
there are 3 × 8 = 24 ways in total.
24 3
Thus the probability is 120
= 15
= 15 .

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(iii) Find the probability that no husband sits next to his wife.

Method 1: First find the probability of exactly one husband sitting next to his wife.
Let us assume that H3 and W3 are the only pair next to each other. Then in the above
diagrams, the left hand one fails as H1 and W1 are together. The right hand one also fails,
as if H3 and W3 are together, H2 and W2 must also be. So the only valid configuration is the
middle one, with either H2 or W2 between H1 and W1 and the other partner on the other
side of either H1 or W1 .
There are two choices for where W1 will sit, two choices for which of H2 or W2 will sit
between H1 and W1 , two choices for where the other partner will sit, and two choices for
which way round H3 and W3 will sit, giving 2 × 2 × 2 × 2 = 16 possibilities.
Finally, we need to multiply this by three, as any one of the three couples could be the
48
adjacent one, giving 3 × 16 = 48 possibilities, and hence a probability of 120 = 25 = 15
6
.
2 3 6 4
Thus the probability that no husband sits next to his wife is 1 − 15
− 15
− 15
= 15
.

Method 2: Find the probability directly.

If no husband sits next to his wife, there are two possible configurations as shown in these
diagrams (the arrows join husbands with their wives):

The number of ways of arranging the first case (fixing H1 at the top as usual) is 2×2×2 = 8,
as their are two ways of choosing which couple sits in which of the diagonal pairs of seats,
two ways of couple 2 sitting and two ways of couple 3 sitting.
For the second case, which is no longer totally symmetrical between the three couples, if H1
sits in the top seat, there are again 8 ways of seating the other two couples. As there are
three choices for which couple sits opposite each other, there are 3 × 8 = 24 ways in all.
32 4
Thus in total there are 8 + 24 = 32 ways, giving a total probability of 120
= 15
.

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June 2008

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STEP II STEP Solutions June 2008

STEP Mathematics II 2008: Solutions

1 (i) Given (xn+1 , yn + 1) = (xn2 – yn2 + 1, 2xn yn + 1), it is easier to remove the subscripts and

set x2 – y2 + 1 = x and 2xy + 1 = y. Then, identifying the y’s (or x’s) in each case, gives
1
y2 = x2 – x + 1 and y = . Eliminating the y’s leads to a polynomial equation in x;
1− 2x
namely, 4x4 – 8x3 + 9x2 – 5x = 0.

Noting the obvious factor of x, and then finding a second linear factor (e.g. by the factor
theorem) leads to x(x – 1)(4x2 – 4x + 5) = 0. Here, the quadratic factor has no real roots,
since the discriminant, Δ = 42 – 4.4.5 = – 64 < 0. [Alternatively, one could note that
4x2 – 4x + 5 ≡ (2x – 1)2 + 4 > 0 ∀x. ]

The two values of x, and the corresponding values of y, gained by substituting these x’s
1
into y = , are then (x , y) = (0 , 1) and (1 , – 1)
1− 2x

(ii) Now (x1 , y1) = (– 1 , 1) ⇒ (x2 , y2) = (a , b) and (x3 , y3) = (a2 – b2 + a , 2ab + b + 2).
Setting both a2 – b2 + a = – 1 and 2ab + b + 2 = 1, so that the third term is equal to the
−1
first, and identifying the b’s in each case, gives b2 = a2 + a + 1 and b = .
1 + 2a

One could go about this the long way, as before. However, it can be noted that the
algebra is the same as in (i), but with a = – x and b = – y. Either way, we obtain the
two possible solution-pairs: (a , b) = (0 , – 1) and (– 1 , 1).

However, upon checking, the solution (– 1 , 1) actually gives rise to a constant sequence
(and remember that the working only required the third term to be the same as the first,
which doesn’t preclude the possibility that it is also the same as the second term!),
so we find that there is in fact just the one solution: (a , b) = (0 , – 1).

2 The correct partial fraction form for the given algebraic fraction is
1+ x A B Cx + D
≡ + +
(1 − x ) 1 + x
2 2
( )
1 − x (1 − x) 2
1+ x2
,

although these can also be put together in other correct ways that don’t materially hinder
the progress of the solution. The standard procedure now is to multiply throughout by the
denominator of the LHS and compare coefficients or substitute in suitable values: which
leads to A = 12 , B = 1 , C = 12 and D = – 12 .

In order to apply the binomial theorem to these separate fractions, we now use index
notation to turn
1+ x
(
(1 − x) 2 1 + x 2
≡ A(1 − x) −1 + B (1 − x) −2 + Cx 1 + x 2
) ( )
−1
(
+ D 1+ x2 ) −1

into the infinite series

∞ ∞ ∞ ∞
1
2 ∑x
n =0
n
+ ∑ (n + 1) x
n =0
n
+ 1
2 ∑ (−1)
n =0
n
x 2 n+1 – 1
2 ∑ (−1)
n =0
n
x 2n .

It should be clear at this point that the last two of these series have odd/even powers only,
with alternating signs playing an extra part. The consequence of all this is that we need to

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STEP II STEP Solutions June 2008

examine cases for n modulo 4; i.e. depending upon whether n leaves a remainder of 0, 1, 2
or 3 when divided by 4.

For n ≡ 0 (mod 4), the coefft. of xn is 1

2 +n+1+0– 1
2 = n + 1;

A1 for n ≡ 1 (mod 4), coefft. of xn is 1

2 +n+1+ 1
2 – 0 = n + 2;

A1 for n ≡ 2 (mod 4), coefft. of xn is 1

2 +n+1+0+ 1
2 = n + 2;

A1 for n ≡ 3 (mod 4), coefft. of xn is 1

2 +n+1– 1
2 + 0 = n + 1.

11000 1 .1
For the very final part of the question, we note that = , is a cancelled form of
8181 0.9 × 1.01
2

our original expression, with x = 0.1 . (N.B. |x| < 1 assures the convergence of the infinite series
forms). Substituting this value of x into

1 + 3x + 4x2 + 4x3 + 5x4 + 7x5 + 8x6 + 8x7 + 9x8 + …

then gives 1.344 578 90 to 8dp.

3 (i) Setting
dy
= 81x2 – 54x = 0 for TPs gives (0 , 4) and ( 23 , 0) . You really ought to
dx
know the shape of such a (“positive”) cubic, and it is customary to find the crossing-
points on the axes: x = 0 gives y = 4, and y = 0 leads to x = – 1 and x = 23 (twice).
[If you have been paying attention, this latter zero for y should come as no surprise!]
The graph now shows that, for all x ≥ 0, y ≥ 0; which leads to the required result –
x2(1 – x) ≤ 274 – with just a little bit of re-arrangement.

In order to prove the result by contradiction (reduction ad absurdum), we first assume

that all three numbers exceed 274 . Then their product
bc(1 – a)ca(1 – b)ab(1 – c) > ( 274 )3.
However, this product can be re-written in the form
a2(1 – a). b2(1 – b.) c2(1 – c),
and the previous result guarantees that x2(1 – x) ≤ 274 for each of a , b , c, from which
it follows that
a2(1 – a). b2(1 – b.) c2(1 – c) ≤ ( 274 )3,
which is the required contradiction. Hence, at least one of the three numbers bc(1 – a),
ca(1 – b), ab(1 – c) is less than, or equal to, 274 .

(ii) Drawing the graph of y = x – x2 (there are, of course, other suitable choices, such as
y = (2x – 1)2 for example) and showing that it has a maximum at ( 12 , 14 ) gives
x(1 – x) ≤ 1
4 for all x.
The assumption that p(1 – q) , q(1 – p) > 1
4 ⇒ p(1 – p).q(1 – q) > ( 14 )2.
However, we know that x(1 – x) ≤ 1
4 for each of p and q, and this gives us that
p(1 – p).q(1 – q) ≤ ( 14 )2.
Hence, by contradiction, at least one of p(1 – q) , q(1 – p) ≤ 1
4 .

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⎛ dy dy ⎞
4 Differentiating implicitly gives 2⎜ x + y + ax + ay ⎟ = 0 , from which it follows that
⎝ dx dx ⎠

dy x + ay ax + y
=− and hence the gradient of the normal is .
dx ax + y x + ay
y ax + y
−
y x x + ay xy + ay 2 − ax 2 − xy
Using tan(A – B) on this and gives tan θ = = 2 .
x y ax + y x + axy + axy + y 2
1+ ×
x x + ay

However, we know that x2 + y2 + 2axy = 1 from the curve’s eqn., and so

tan θ = a y 2 − x 2 .

dθ ⎛ dy ⎞
(i) Differentiating this w.r.t. x then gives sec2θ = a⎜ 2 y − 2 x ⎟ . Equating this to
dx ⎝ dx ⎠
dy x + ay
zero and using =− from earlier then leads to a(x2 + y2) + 2xy = 0 .
dx ax + y

(ii) Adding x2 + y2 + 2axy = 1 and a(x2 + y2) + 2xy = 0 gives (1 + a)(x + y)2 = 1 .

(iii) However, subtracting these two eqns. instead gives (1 – a)(y – x)2 = 1 , and
multiplying these two last results together yields (1 – a2) (y2 – x2)2 = 1.

1
Finally, using tan θ = a y 2 − x 2 ⇒ (y2 – x2)2 = 2
tan2θ , and substituting this
a
a
into the last result of (iii) then gives the required result: tan θ = . All that
1− a2
remains is to justify taking the positive square root, since tan θ is | something |, which
is necessarily non-negative.

π/2 π/2
sin 2 x 2 sin x cos x
5 Using a well-known double-angle formula gives ∫0
1 + sin 2 x
dx = ∫0
1 + sin 2 x
dx , and

this should suggest an obvious substitution: letting s = sin x turns this into the integral
1
2s
∫ 1+ s
0
2
ds .

This is just a standard log. integral (the numerator being the derivative of the denominator),
leading to the answer ln 2 .

Alternatively, one could use the identity sin2x ≡ 1

2 − 12 cos 2 x to end up with
π/2
2 sin 2 x
∫
0
3 − cos 2 x
dx.

This, again, gives a log. integral, but without the substitution.

1
1
A suitable substitution for the second integral is c = cos x , which leads to ∫ 2−c
0
2
dc .

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STEP II STEP Solutions June 2008

Now you can either attack this using partial fractions, or you could look up what is a fairly
standard result in your formula booklet. In each case, you get (after a bit of careful log and
surd work)
1
2
(
ln 1 + 2 . )
Now (1 + 2 )5 = 1 + 5 2 + 20 + 20 2 + 20 + 4 2 = 41 + 29 2 (using the binomial
theorem, for instance), and
41 + 29 2 < 99 ⇔ 29 2 < 58 ⇔ 2 < 2 ,
which is obviously the case. Also, 1.96 < 2 ⇒ 1.4 < 2 . Thereafter, an argument such as
2 1.4
>1+ 2 ⇔ 2 > (1 + 2 ) ⇔ 128 > 41 + 29 2
7 5

⇔ 87 > 29 2 ⇔ 3 > 2
> 2 5 >1+ 2 .
2 7
from which it follows that 2

Taking logs in this result then gives 2 ln 2 > ln(1 + 2 ) ⇒ ln 2 >

1
2
( )
ln 1 + 2 ; and
π/2 π/2
sin 2 x sin x
∫
0
1 + sin 2 x
dx > ∫ 1 + sin
0
2
x
dx .

6 (i) Firstly, cos x has period 2π ⇒ cos (2x) has period π;

⎛ 3x ⎞
and sin x has period 2π ⇒ sin ⎜ ⎟ has period 43 π.
⎝ 2⎠
⎛ π⎞ ⎛ 3x π ⎞
Then f(x) = cos ⎜ 2 x + ⎟ + sin ⎜ − ⎟ has period 4π = lcm(π , 4
3 π).
⎝ 3⎠ ⎝ 2 4⎠

(ii) Any approach here is going to require the use of some trig. identity work. The most
⎛π ⎞
+ θ ⎟ = − sin θ so that f(x) = 0 reduces to
straightforward is to note that cos ⎜
⎝2 ⎠
⎛ π⎞ ⎛ 3x π ⎞ π ⎛ 3x π ⎞
cos ⎜ 2 x + ⎟ = cos ⎜ + ⎟ , from which it follows that 2 x + = 2nπ ± ⎜ + ⎟
⎝ 3⎠ ⎝ 2 4⎠ 3 ⎝ 2 4⎠
where n is an integer, using the symmetric and periodic properties of the cosine curve.
Taking suitable values of n, so that x is in the required interval, leads to the answers
31π π
x=– (from n = – 1, with the – sign), x = – (n = 0, with both + and – signs),
42 6
17π 41π
x= (n = 1, – sign) and x = (n = 2, – sign).
42 42
π
Since x = – is a repeated root (occurring twice in the above list), the curve of
6
y = f(x) touches the x-axis at this point.

For those who are aware of the results that appear in all the formula books, but which
seem to be on the edge of the various syllabuses, that I know by the title of the Sum-
⎛ A+ B⎞ ⎛ A− B⎞
and-Product Formulae, such as cos A + cos B ≡ 2 cos⎜ ⎟ cos⎜ ⎟ , there is a
⎝ 2 ⎠ ⎝ 2 ⎠
second straightforward approach available here. For example, noting that
⎛π ⎞ ⎛ π⎞ ⎛ 3π 3x ⎞
cos ⎜ − θ ⎟ = sin θ gives cos ⎜ 2x + ⎟ + cos ⎜ − ⎟ = 0 which (from the above
⎝2 ⎠ ⎝ 3⎠ ⎝ 4 2⎠

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STEP II STEP Solutions June 2008

⎛ x 13π ⎞ ⎛ 7 x 5π ⎞
identity) then gives 2 cos ⎜ + ⎟ cos ⎜ − ⎟ = 0, and setting each of these two
⎝ 4 24 ⎠ ⎝ 4 24 ⎠
cosine terms equal to zero, in turn, yields the same values of x as before, including the repeat.

⎛ π⎞
(iii) The key observation here is that y = 2 if and only if both cos ⎜ 2x + ⎟ = 1 and
⎝ 3⎠
⎛ 3x π ⎞
sin ⎜ − ⎟ = 1, simultaneously. So we must solve
⎝ 2 4⎠
⎛ π⎞ π 5π 11π
cos ⎜ 2x + ⎟ = 1 ⇒ 2 x + = 0 , 2π , 4π , … , giving x = , , … ; and
⎝ 3⎠ 3 6 6
⎛ 3x π ⎞ 3x π π 5π π 11π
sin ⎜ − ⎟ =1 ⇒ − = , , … , giving x = , ,….
⎝ 2 4⎠ 2 4 2 2 2 6
11π
Both equations are satisfied when x = , and this is the required answer.
6

dy x du
7 (i) Differentiating y = u 1 + x 2 gives = u. + 1 + x2 . ; so that
dx 1 + x2 dx

1 dy x 1 ⎧ ux du ⎫ x
= xy + becomes ⎨ + 1 + x 2 . ⎬ = xu 1 + x 2 + .
y dx 1+ x 2
u 1+ x ⎩ 1+ x
2 2 dx ⎭ 1 + x2
Simplifying and cancelling the common term on both sides leads to
1 du
. = xu 1 + x 2 .
u dx

This is a standard form for a first-order differential equation, involving the separation
of variables and integration:
1
∫ u 2 . du = ∫ x 1 + x dx ⇒ −
2 1 1
(
= 1 + x2
u 3
)3
2
(+ C).

3 1 + x2
Using x = 0 , y = 1 (u = 1) to find C leads to the final answer, y = .
(
4 − 1 + x2 )
3
2

(ii) The key here is to choose the appropriate function of x. If you have really got a feel for
what has happened in the previous bit of the question, then this isn’t too demanding. If
you haven’t really grasped fully what’s going on then you may well need to try one or
two possibilities first. The product that needs to be identified here is y = u 1 + x 3 ( )
1
3
.
Once you have found this, the process of (i) pretty much repeats itself.

= u. x 2 (1 + x 3 ) + (1 + x 3 )
dy −23 1 du 1 dy x2
3
means that = x2 y + becomes
dx dx y dx 1 + x3
= x 2u (1 + x 3 ) .
1 du 1
3
.
u dx

Separating variables and integrating:

∫ x (1 + x ) dx = − = (1 + x 3 ) 3 (+ C) ;
1 1 1
∫ u 2 . du =
1 4
2 3 3

u 4

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STEP II STEP Solutions June 2008

and x = 0 , y = 1 (u = 1) gives C and the answer y =

(
4 1 + x3 )
1
3

.
(
5 − 1 + x3 )
4
3

(iii) Note that the question didn’t actually require you to simplify the two answers in (i) and
(ii), but doing so certainly enables you to have a better idea as to how to generalise the
results:
(n + 1)(1 + x n ) n
1

y= .
(n + 2) − (1 + x n )
1+ 1n

8 It is never a bad idea to start this sort of question with a reasonably accurate diagram …
something along the lines of
A

P
θ Q
O θ

The first result is an example of what is known as the Ratio Theorem:

AP : PB = 1 – λ : λ ⇒ p = λa + (1 – λ)b .
Alternatively, it can be deduced from the standard approach to the vector equation of a straight
line, via r = a + λ(b – a).

Using the scalar product twice then gives

a • p = λa2 + (1 – λ)(a • b) and b • p = λ(a • b) + (1 – λ) b2 .
a•p b•p
Equating these two expressions for cos θ , = , re-arranging and collecting up
ap bp
like terms, then gives ab{λ(a + b) – b} = a • b {λ(a + b) – b}. There are two possible
consequences to this statement, and both of them should be considered. Either ab = a • b,
which gives cos 2θ = 1, θ = 0, A = B and violates the non-collinearity of O, A & B; or the
bracketed factor on each side is zero, which gives
b
λ= .
a+b

However, if you know the Angle Bisector Theorem, the working is short-circuited quite
dramatically:
AP OA (1 − λ )( AB) a b
= ⇒ = ⇒ b – b λ = aλ ⇒ λ = .
PB OB λ ( AB) b a+b

Next, AQ : QB = λ : 1 – λ ⇒ q = (1 – λ)a + λb.

Then
OQ2 = q • q = (1 – λ)2 a2 + λ2 b2 + 2λ(1 – λ) a • b

and OP2 = p • p = (1 – λ)2 b2 + λ2 a2 + 2λ(1 – λ) a • b .

[N.B. This working can also be done by the Cosine Rule.]
Subtracting:
OQ2 – OP2 = (b2 – a2) [λ2 – (1 – λ)2] = (b2 – a2) (2λ – 1)
and, substituting λ in terms of a and b into this expression, gives the required answer
b−a
= (b – a)(b + a) × = (b – a)2.
b+a

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9 (i) Using a modified version of the trajectory equation (which you are encouraged to have
gx 2
learnt), y = h + x tan α – 2
sec2α , and substituting in g = 10 and u = 40 gives
2u
gx 2 2
y = h + x tan α – sec α .
320

Setting x = 20 and y = 0 into this trajectory equation and using one of the well-
known Pythagorean trig. identities (sec2α = 1 + tan2α) leads to the quadratic equation
5t2 – 80t – (4h – 5) = 0
in t = tan α.
[Note that you could have substituted x = 20 and y = – h into the unmodified
trajectory equation and still got the same result here.]
Solving, using the quadratic formula, and simplifying then gives
tan α = 8 ± 63 + 54 h .

We reject tan α = 8 + 63 + 54 h , since this gives a very high angle of projection

and hence a much greater time for the ball to arrive at the stumps. Now, since α is
x 1 1
small, cos α ≈ 1, and the time of flight = = ≈ .
u cos α 2 cos α 2

(ii) h > 5
4 for tan α = 8 – 64 + ε ) < 0.

(iii) Now h = 2.5 ⇒ tan α = 8 – 64 + 1 = 8 – 8 (1 + 641 ) 2 . The Binomial Theorem then

1

allows us to expand the bracket, and it seems reasonable to take just the first term
past the 1: tan α = 8 – 8 (1 + 128
1
+ .....) , so that tan α ≈ – 161 . [We can ignore the
minus sign, since this just tells us that the projection is below the horizontal.]
Using tan α ≈ α for small-angles, and converting from radians into degrees using
the conversion factor 180/π ≈ 57.3 then gives α ≈ 3.6o .

10 On this sort of question, a good, clear diagram is almost essential, even when it is not asked-for.
A M D

α
a w a–y
γ
Y
a u
v y

β
B x X (b – x) N b C

(i) The two fundamental principles involved in collisions are the Conservation of Linear
Momentum (CLM) and Newton’s Experimental Law of Restitution (NEL or NLR).
For the collision at X, applying CLM || BC ⇒ mu sin α = mv cos β
.. NEL ⇒ eu cos α = v sin β

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Dividing these two gives tan β = e cot α or tan α tan β = e .

At Y, “similarly”, we have tan β tan γ = e . Hence α = γ (since all angles are acute).

(ii) A good approach here is to use similar Δs, and a bit of sensible labelling is in order
(see the diagram). Let BX = x (XN = b – x) and CY = y (DY = a – y). Then
x y b
tan α = , tan β = , tan γ = .
a 2b − x a− y
a ( x − b)
Using α = γ to find (e.g) y in terms of a, b, x ⇒ ax – xy= ab ⇒ y = .
x
Next, we use the result tan α tan β = e from earlier to get x in terms of a and b:
x a ( x − b) / x b(1 + 2e)
× = e ⇒ x – b = 2be – ex ⇒ x = ,
a 2b − x 1+ e
b(1 + 2e)
from which it follows that tan α = .
a(1 + e)

At this stage, some sort of inequality argument needs to be considered, and a couple of
obvious approaches might occur to you.
b(1 + 2e) b be b b(1 + 2e) 2b be 2b
I tan α = = + > and tan α = = − <
a (1 + e) a a(1 + e) a a(1 + e) a a(1 + e) a
b 2b
give < tan α < ; and the shot is possible, with the ball striking BC between N
a a
and C, whatever the value of e .
b 3b
II As e → 0, tan α → + and as e → 1, tan α → − , so that
a 2a
b 3b
< tan α < ; and the shot is possible, with the ball striking BC between N and
a 2a
the midpoint of NC, whatever the value of e.

(iii) There are two possible approaches to this final part. The first, much longer version,
involves squaring and adding the eqns. for the collision at X, and then again at Y, to
get
v2 = u2(sin2α + e2cos2α) and w2 = v2(sin2β + e2cos2β).
Now, noting that the initial KE = 12 mu 2 and the final KE = 12 mw 2 , the fraction of
1
mu 2 − 12 mw 2 w2
KE lost is 2
1 2
= 1 − 2
= 1 – (sin2α + e2 cos2α)(sin2β + e2cos2β)
2 mu u
tan 2 α + e 2 tan 2 β + e 2
=1– × .
sec 2 α sec 2 β
From here, we use tan α tan β = e and sec2α = 1 + tan2α to get
t 2 + e2 e2 / t 2 + e2 t 2 + e 2 e 2 (1 + t 2 ) / t 2
× × 2
(t + e 2 ) / t 2 = 1 – e , as required.
2
1– = 1 –
1+ t2 1 + e2 / t 2 1+ t2

However, it is very much quicker to note the following:

At X, the ↑-component of the ball’s velocity becomes e × initial ↑-component ,
and
at Y, the →-component of the ball’s velocity becomes e × initial →-component.
Hence its final velocity is eu and the fraction of the KE lost is then

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1
mu 2 − 12 me 2u 2
2
1 2
= 1 – e2.
2 mu

R
11 F

b F a
R

mg kmg

Once again, a good, clear diagram is an important starting-point, and the above diagram shows the
relevant forces – labelled using standard notations – along with the accelerations of P down the sloping
surface of the wedge (a) and the wedge itself along the plane (b).

(i) Noting the acceleration components of P are a cosθ – b (→) and a sin θ (↓), we
employ Newton’s Second Law as follows:
N2L → for P m(a cosθ – b) = R sinθ – F cosθ
N2L ↓ for P ma sin θ = mg – F sinθ – R cosθ
N2L ← for wedge kmb = R sinθ – F cosθ
a cos θ
From which it follows that a cosθ – b = kb ⇒ b = .
k +1

Alternatively, one could use N2L to note P’s → accln. component and also the
wedge’s accln. ←, but instead use
CLM ↔ km bt = m (a cosθ – b)t (where t = time from release)
and this again leads to the above result for b.

Now, for P to move at 45o to the horizontal, a cosθ – b = a sinθ . Then

a cos θ
b = a(cosθ – sinθ ) =
k +1
k
⇒ (k + 1)( cosθ – sinθ) = cosθ ⇒ k + 1 – (k + 1) tanθ = 1 and tanθ = .
k +1

When k = 3, tanθ = 34 , sinθ = 53 , cosθ = 54 and b = 15 a .

Substituting these into the first two equations of motion from (i), along with the use of
the Friction Law (in motion), which assumes that F = μR , gives
m( 54 a – b) = 53 R – 54 F or 3R – 4F = m(4a – 5b) = 3ma ⇒ R(3 – 4μ) = 3ma
and
5 ma = mg – 5 F – 5 R or 4R + 3F = m(5g – 3a) ⇒ R(4 + 3μ) = 5mg – 3ma.
3 3 4

Dividing, or equating for R :

4 + 3μ 5 g − 3a 5(3 − 4 μ ) g
= ⇒ (12 + 9μ)a = 5(3 – 4μ)g – (9 – 12μ)a ⇒ a = .
3 − 4μ 3a 3(7 − μ )

(ii) Finally, if tanθ ≤ μ , then both P and the wedge remain stationary. So, technically,
the answer is “nothing”.

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12 Clearly, X ∈ {0 , 1 , 2 , 3} and working out the corresponding probabilities is a good thing to do at

some point in this question (although it can, of course, be done numerically later when a value for p
has been found).
p(X = 0) = (1 – p)(1 – 13 p) (1 – p2)
p(X = 1) = p(1 – 1
3 p)(1 – p2) + (1 – p) 13 p(1 – p2) + (1 – p)(1 – 1
3 p)p2
= p(1 – p)( 43 + 5
3 p – p2)
2
p(X = 2) = p. 13 p(1 – p ) + p(1 – 1
3 p)p2 + (1 – p) 13 p.p2
= 1
3 p2(1 + 4p – 3p2)
p(X = 3) = 13 p4
[Of course, one of these could be deduced on a (1 – the sum of the rest) basis, but that can
always be left as useful check on the correctness of your working, if you so wish.]

Then E(X) = ∑ x. p( x) = 0 + p(1 – p)( 4

3 + 5
3 p – p2) + 2
3 p2(1 + 4p – 3p2) + p4
= 43 p + p2
Alternatively, if you have done a little bit of expectation algebra, it is clear that
E(X) = ∑
E ( X i ) = p + 13 p + p 2 = 43 p + p 2 .
Equating this to 4
3 ⇒ 0 = 3p2 + 4p – 4 ⇒ 0 = (3p – 2)(p + 2), and since 0 < p < 1 it follows that
2
p= 3 .

In the final part, you will need either (p0 and p1) or (p2 and p3):
35
p0 = 243 and p1 = 108
243 or
84
p2 = 243 16
and p3 = 243 .
Next, a careful statement of cases is important (with, I hope, obvious notation):
p(correct pronouncement) = p(G and ≥ 2 judges say G) + p(NG and ≤ 1 judges say G)
100 143 143 − 43t
=t. + (1 – t) . =
243 243 243
Equating this to 1
2 and solving for t ⇒ 243 = 286 – 86t ⇒ 86t = 43 ⇒ t = 1
2 .

Alternatively, let p(King pronounces guilty) = q .

Then “King correct” = “King pronounces guilty and defendant is guilty”
or “King pronounces not guilty and defendant is not guilty”
so that p(King correct) = qt + (1 – q)(1 – t).
Setting qt + (1 – q)(1 – t) = 12 ⇔ (2q – 1)(2t – 1) = 0 , and since q is not identically equal to
1 1
2 , t= 2 .

13 (i) p(B in bag P) = p(B not chosen draw 1) + p(B chosen draw 1 and draw 2)
⎛ k⎞ k k
= ⎜1 − ⎟+ ×
⎝ n⎠ n n+k
=
1
n(n + k )
((n − k )(n + k ) + k 2 )
n
=
n+k
This has its maximum value of 1 for k = 0, and for no other values of k. Since
k
p=1 − ≤ 1 and for k = 0, p = 1 but k > 0 for all p < 1).
n+k

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(ii) p(Bs in same bag) = p(B1 chosen on D1 and neither chosen on D2)
+ p(B1 chosen on D1 and both chosen on D2)
+ p(B1 not chosen on D1 and B2 chosen on D2)

n+ k −2 n+ k −2
k Ck k Ck −2 ⎛ k ⎞ k
= × n+ k
+ × n+ k
+ ⎜1 − ⎟ ×
n Ck n Ck ⎝ n⎠ n+k

Notice that, although the nCr terms look very clumsy, they are actually quite simple
once all the cancelling of common factors has been undertaken.

k n(n − 1) k k (k − 1) k (n − k )
= × + × +
n (n + k )(n + k − 1) n (n + k )(n + k − 1) n(n + k )

k ⎧ n 2 − n + k 2 − k + (n 2 + nk − n − nk − k 2 + k ) ⎫
= ⎨ ⎬
n⎩ (n + k )(n + k − 1) ⎭

2k (n − 1)
=
(n + k )(n + k − 1)

Differentiating this expression gives

dp (n 2 + 2nk + k 2 − n − k ) × 2(n − 1) − 2k (n − 1) × (2n + 2k − 1)
=
dk [(n + k )(n + k − 1)]2
= 0 when n2 + 2nk + k2 – n – k = 2nk + 2k2 – k since n > 2, n – 1 ≠ 0

⇒ k2 = n(n – 1) .
Now there is nothing that guarantees that k is going to be an integer (quite the contrary,
in fact), so we should look to the integers either side of the (positive) square root of
n(n – 1):
k= [ n(n − 1) ] and k = [ ]
n(n − 1) + 1 .
In fact, since n2 – n = (n – ½ )2 – ¼ , [n 2
]
− n) = n – 1 and we find that,

2(n − 1) 2 n −1
when k = n – 1 , p = =
(2n − 1)2(n − 1) 2n − 1

2n(n − 1) n −1
and when k = n , p = = also,
(2n)(2n − 1) 2n − 1

and k = n – 1 , n are the two values required.

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June 2008

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STEP Mathematics III 2008: Solutions

1. Following the hint yields

ax 2 + by 2 + ( a + b) xy = ( x + y )
1
3
which is + xy = ( x + y )
1 1
5 3

+ xy = ( x + y )
1 1 1
The same trick applied to the third equation gives .
7 3 5
The two equations can be solved simultaneously for xy and (x+y), giving
and ( x + y ) =
3 6
xy =
35 7

Thus x and y are the roots of the quadratic equation 35z 2 − 30z + 3 = 0
(x and y are interchangeable).
a and b are then found by substituting back into two of the original equations and the
full solution is
3 2 3 2 6
x= ± 30 = ±
7 35 7 7 5
3 2 3 2 6
y= m 30 = m
7 35 7 7 5
1 30 1 1 5
a= m = m
2 36 2 6 6
1 30 1 1 5
b= ± = ±
2 36 2 6 6

2. (i) On the one hand

n +1

∑ [(r + 1) ]
n n n n
− r k = ∑ (r + 1) − ∑ r k = ∑ r k − ∑ r k =(n + 1) whilst expanding
k k k

r =0 r =0 r =0 r =1 r =0
binomially yields

n
⎛ k⎞ n ⎛ k⎞ n ⎛ k ⎞ n k −1 n
k ∑ r k −1 + ⎜ ⎟ ∑ r k − 2 + ⎜ ⎟ ∑ r k − 3 +...+⎜ ⎟ ∑ r + ∑1
r =0 ⎝ 2⎠ r = 0 ⎝ 3⎠ r = 0 ⎝ k − 1⎠ r = 0 r =0

⎛ k⎞ ⎛ k⎞ ⎛ k ⎞
= kS k −1 ( n) + ⎜ ⎟ S k − 2 ( n) + ⎜ ⎟ S k − 3 ( n) +...+⎜ ⎟ S ( n) + ( n + 1)
⎝ 2⎠ ⎝ 3⎠ ⎝ k − 1⎠ 1
and hence the required result.

Applying this in the case k = 4 gives

⎛ 4⎞ ⎛ 4⎞
4 S 3 ( n) = ( n + 1) − ( n + 1) − ⎜ ⎟ S 2 ( n) − ⎜ ⎟ S1 ( n)
4

⎝ 2⎠ ⎝ 3⎠

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STEP III STEP Solutions June 2008

which, after substitution of the two given results and factorization, yields the familiar
1
S 3 ( n) = n 2 ( n + 1)
2

4
The identical process with k = 5 results in
n( n + 1)(6n 3 + 9n 2 + n − 1) = n( n + 1)( 2n + 1)(3n 2 + 3n − 1)
1 1
S 4 ( n) =
30 30

(ii) Applying induction, with the assumption that S t ( n) is a polynomial of degree

t + 1 in n for t < r for some r , and then considering (*),
( n + 1) r +1 − ( n + 1) is a polynomial of degree r + 1 in n ,
⎛ r + 1⎞
and each of the terms − ⎜ ⎟S ( n) is a polynomial of degree r + 2 − j in
⎝ j ⎠ r +1− j
n where j ≥ 2 , i.e. the degree is ≤ r . A sum of polynomials of degree ≤ r + 1 in n ,
is a polynomial of degree ≤ r + 1 in n , and there is a single non-zero term in n r +1
from just ( n + 1) so the degree of the polynomial is not reduced to < r + 1 , i.e. it is
r +1

r + 1 . (The initial case is true to complete the proof.)

k +1 n k +1 0
If, S k ( n) = ∑ ai n i = ∑ r k then S k ( 0) = a 0 + ∑ ai 0i = ∑ r k = 0 and so a 0 = 0
i =0 r =0 i =1 r =0

k +1 1 k +1
S k (1) = ∑ ai 1i = ∑ r k = 1 and so ∑a i = 1 as required.
i =0 r =0 i =0

dy b cosθ
3. =
dx − a sin θ
a sin θ
So the line ON is y = x
b cosθ
b sin θ
SP is y = ( x + ae)
a( cosθ + e)

Solving simultaneously by substituting for x to find the y coordinate of T,

b sin θ ⎛ b cosθ ⎞
y= ⎜ y + ae⎟
a( cosθ + e) a sin θ
⎝ ⎠
and using b 2 = a 2 (1 − e 2 ) to eliminate a 2 gives the required result.
b 2 cosθ
Then the x coordinate of T is .
a(1 + e cosθ )
b2 by
Eliminating θ using sec θ + e = and tan θ = ,
ax ax
2
⎛ b2 ⎞ ⎛ by ⎞
2

( x, y) satisfies ⎜⎝ ax − e⎟⎠ = 1 + ⎜⎝ ax ⎟⎠
and again using b 2 = a 2 (1 − e 2 ) , this time to eliminate b 2 , gives, following
simplifying algebra
( x + ae) 2 + y 2 = a 2 , as required.

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4. (i)

The graph of z = y has gradient 1 and passes through the origin.

⎛ y⎞ 1 ⎛ y⎞ 1
The graph of z = tanh⎜ ⎟ which has gradient sec h 2 ⎜ ⎟ ≤ for y ≥ 0 also passes
⎝ 2⎠ 2 ⎝ 2⎠ 2
though the origin and is asymptotic to z = 1 .
⎛ y⎞
Thus y ≥ tanh⎜ ⎟ for y ≥ 0 .
⎝ 2⎠
⎛ y⎞
2 sinh 2 ⎜ ⎟
x −1 cosh y − 1 ⎝ 2⎠ ⎛ y⎞
If x = cosh y , then = = = tanh⎜ ⎟
x +1 cosh y + 1 ⎛ y⎞ ⎝ 2⎠
2 cosh 2 ⎜ ⎟
⎝ 2⎠
⎛ y⎞ x −1
and as y ≥ tanh⎜ ⎟ for y ≥ 0 , ar cosh x ≥ for x ≥ 1 .
⎝ 2⎠ x +1
x −1 x −1 x −1 x −1
= = for x > 1, and (*) is obtained.
x +1 x +1 x −1 x2 − 1

(ii) By parts ∫ ar cosh xdx = xar cosh x − x 2 − 1 + c

x −1
and ∫ x2 − 1
dx = x 2 − 1 − ar cosh x + c ′

x −1
x x

Thus ∫ ar cosh xdx ≥ ∫

1 1 x2 − 1
dx for x > 1 gives

xar cosh x − x 2 − 1 ≥ x 2 − 1 − ar cosh x for x > 1, which rearranges to give result

(iii) Integrating (ii) similarly gives xar cosh x − x 2 − 1 ≥ 2 ( x 2 − 1 − ar cosh x )

for x > 1, which also can be rearranged as desired.

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5. There are a number of correct routes to proving the induction, though the
simplest is to consider ( (T
k +1
2
) (
( x )) − Tk ( x )Tk + 2 ( x ) − (Tk ( x )) − Tk −1 ( x )Tk +1 ( x )
2
)
For f ( x ) = 0 , (Tn ( x ) ) − Tn −1 ( x ) Tn +1 ( x ) = 0
2

Tn +1 ( x ) Tn ( x )
and so = provided that neither denominator is zero, leading to
Tn ( x ) Tn −1 ( x )
Tn ( x ) T ( x)
= 1 = r( x) ,
Tn−1 ( x ) T0 ( x )
T ( x ) Tn −1 ( x ) T ( x)
= (r ( x ) )
n
and so n × ×...× 1
Tn −1 ( x ) Tn − 2 ( x ) T0 ( x )
Tn ( x ) = (r ( x )) T0 ( x )
n
Thus

Substituting this result into (*) for n = 1,

((r( x)) 2
)
− 2 xr ( x ) + 1 T0 ( x ) = 0 , and as T0 ( x ) ≠ 0 , solving the quadratic gives
r( x) = x ± x 2 − 1

6. (i) Differentiating y = p 2 + 2 xp with respect to x gives

dp dp
p = 2 p + 2x + 2 p which can be rearranged suitably.
dx dx
dx 2
The differential equation + x = −2 has an integrating factor p 2
dp p
and integrating will give the required general solution.
3
Substituting x = 2, p = −3 , leads to A = 0 , i.e. p = − x which can be
2
3 2
substituted in the original equation and so y = − x .
4

dx 2 (ln p + 1)
(ii) The same approach as in part (i) generates + x=− ,
dp p p
which with the same integrating factor has general solution
1 1
x = − − ln p + Bp − 2
4 2
and particular solution
1 1
x = − ln p −
2 4

Again, substitution of ln p (and p) in the original equation leads to the solution

1 − 2 x − 21
which is y = − e
2

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7. The starting point c−a =

1
2
( )
1 + i 3 (b − a) leads to the given result.

Interchanging a and b gives 2c = ( a + b) + i 3( a − b) if A, B, C are described

clockwise.

(i) The clue to this is the phrase “can be chosen” and a sketch demonstrates that a
pair of the equilateral triangles need to be clockwise, and the other pair anti-clockwise

Applying the results in the stem of the question to this configuration,

2 p = ( d + e) + i 3 ( e − d )
2q = (e + f ) + i 3 (e − f )
2r = ( f + g ) + i 3( g − f )
2 s = ( g + d ) + i 3( g − d )

and so 2 PS = ( g − e) + i 3( g − e) = −2 RQ , PSQR is a parallelogram.

(The pairs could have been chosen with opposite parity leading to very similar
working.)

(ii) Supposing LMN is clockwise, U is the centroid of equilateral triangle LMH, V

of MNJ, and W of NLK, then
3u = l + m + h where 2h = ( l + m) + i 3( m − l ) with similar results for v and w.

[ ]
Both 6w , and 3 ( u + v ) + i 3( u − v ) can be shown to equal 3( n + l ) + i 3( l − n) and
so UVW is a clockwise equilateral triangle.

1
8. (i) p=−
2
⎛ 1 ⎞
(1 + px)S = 13 x
1 2x
with all other terms cancelling and so S = x ⎜ 1 − x⎟ =
3 ⎝ 2 ⎠ 3( 2 − x )

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Using the sum of a GP

1 ⎛ x n +1 ⎞
x ⎜ 1 − n +1 ⎟
1 1 1 1 3 ⎝ 2 ⎠
S n+1 = x + x 2 + x 3 +...+ x n +1 =
3 6 12 3 × 2n ⎛ x⎞
⎜1 − ⎟
⎝ 2⎠
1
Alternatively S n +1 = S − (a n + 2 x n + 2 +...) = S − x n +1 S
2 n +1
⎛ x n +1 ⎞ 2 x
= ⎜ 1 − n +1 ⎟
⎝ 2 ⎠ 3( 2 − x )

(ii) Using similar working to part (i)

18 + 8 p + 2q = 0
37 + 18 p + 8q = 0
5
so p = − , q = 1
2
and so (1 + px + qx 2 )T = 2 + 3x
⎛ 5 ⎞ 4 + 6x 4 + 6x
giving T = ( 2 + 3x ) ⎜ 1 − x + x 2 ⎟ = 2 =
⎝ 2 ⎠ 2 − 5x + 2 x ( 2 − x )(1 − 2 x )
−1

(1 − 2 x ) −1 − ⎛⎜ 1 −
14 8 x⎞
By partial fractions T = ⎟
3 3⎝ 2⎠
8⎛ ⎛ x⎞ ⎞
2 n
x ⎛ x⎞
and so Tn +1 =
14
3
( 2 n

3⎝
)
1 + 2 x + ( 2 x ) +...+( 2 x ) − ⎜ 1 + + ⎜ ⎟ +...+⎜ ⎟ ⎟
2 ⎝ 2⎠ ⎝ 2⎠ ⎠
⎛ ⎛ x ⎞ n +1 ⎞
⎜1 − ⎜ ⎟ ⎟
=
(
14 1 − (2 x )
n +1
) −
8 ⎝ ⎝ 2⎠ ⎠
3 1 − 2x 3 x
1−
2

Section B: Mechanics

9. When the particle starts to move, friction is limiting and so

mg sin πT0 − μmg = 0
i.e. μ = sin πT0

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STEP III STEP Solutions June 2008

(i)

When the particle comes to rest, the area under the acceleration-time graph is zero
1

i.e. ∫ g sin πt − μ
T0
o gdt = 0

Completing the manipulation and eliminating T0 using the relation at the start of the
question renders the required result.

(ii)

In the case μ = μ0 , the motion is periodic with period 2, the particle is

stationary in intervals (0,T0 ) , (11
, + T0 ) , (2,2 + T0 ) … , reversing its direction
of motion after times 1, 2, 3, … , and returning to its starting point at time 2
(and 4,6….)

In the case μ = 0 , the motion is simple harmonic motion (period 2)

superimposed on uniform motion , the particle instantaneously comes to rest at
time 2, 4, …but otherwise always moves in the positive x direction.
g
( x = 2 ( πt − sin πt ) )
π

56
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STEP III STEP Solutions June 2008

10. Considering the rth short string Tr = mg + Tr −1

λx
Also we have Tr = r , and T1 = mg
l
⎛ rmgl ⎞
∑ (l + xr ) = ∑ ⎜⎝ l + λ ⎟⎠
n n
Thus Tr = rmg and so the total length is given by
1 1

mgl n( n + 1)
= nl +
λ 2
λx r 2
2
⎛ lmg ⎞ r m2 g 2 l
n n 2
The elastic energy stored is ∑ = ∑ λ⎜ ⎟ = n(n + 1)(2n + 1)
1 2l 1
⎝ λ ⎠ 2l 12λ

For the uniform heavy rope, we let M = nm , L0 = nl , and consider the limit as n → ∞
⎛ M g L0 n( n + 1) ⎞ ⎛ Mg ⎞
L = lim⎜ L0 + ⎟ = L0 ⎜ 1 + ⎟
⎝ n λ n 2 ⎠ ⎝ 2λ ⎠
and the elastic energy stored is
⎛ m2 g 2 l ⎞ ⎛ M 2 g 2 L0 n(n + 1)( 2n + 1) ⎞ M 2 g 2 L0
lim⎜ n(n + 1)(2n + 1)⎟ = lim⎜ ⎟=
⎝ 12λ ⎠ ⎝ 12λ n3 ⎠ 6λ
2λ ( L − L0 )
2

and eliminating M using the result just found for L we obtain

3 L0

11. If the resistance couple (constant) is L, then using L = Iα for the second
Iω 0
phase of the motion, L = and rotational kinetic energy used up doing work
T
against the couple in the second phase gives
1
Iω 2 = L × n2 × 2π
2 0
Hence, eliminating L and simplifying gives the first result.

If the particle descends a distance h in the first phase of motion, then h = 2πrn1 . If
the particle has speed v at the end of the first phase, then v = rω 0
and using the work-energy principle,
1 1
mgh − L × n1 × 2π = Iω 0 2 + mv 2
2 2
Hence, eliminating h, v and ω 0 2 obtains the second result.

57
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STEP III STEP Solutions June 2008

Section C: Probability and Statistics

12.

∞ 0 ∞
1 1
M x (θ ) = ∫ e f ( x ) dx = ∫ e x ( 1+θ ) dx + ∫ e − x ( 1−θ ) dx
θx

−∞ −∞
2 0
2
1
[ ] 1
[ 1
] 1
∞
e x ( 1+θ ) −∞ − e − x ( 1−θ ) 0
0
= = + (requiring θ < 1)
2(1 + θ ) 2(1 − θ ) 2(1 + θ ) 2(1 − θ )

= (1 − θ 2 )
−1

Var ( X ) = M x ″ ( 0) − M x ′ ( 0) ( )
2

M x ′ (θ ) = 2θ (1 − θ )
2 −2
, M x ″ (θ ) = 2(1 − θ 2 ) + 8θ (1 − θ 2 )
−2 −3

and so M x ′ (0) = 0 , M x ″ (0) = 2 , Var ( X ) = 2

⎛ ⎞
Or alternatively, M x (θ ) = E ( e θX ) = E ⎜ 1 + θX + θ 2 X 2 +.....⎟
1
⎝ 2 ⎠
= (1 − θ 2 )
−1
= 1 + θ 2 + θ 4 +.....
⎛1 ⎞
and so E ( X ) = 0 , E ⎜ X 2 ⎟ = 1 , Var ( X ) = E ( X 2 ) − ( E ( X ) ) = 2
2

⎝2 ⎠

−n

( )
θ
n ⎛ ⎞ ⎛ θ2 ⎞
2n , then M T (θ ) = E ( e ) = E e ∑ ( i )
= ∏ E⎜e
θ Xi
θT
If T = Y
X 2n 2n
⎟ = ⎜1 − ⎟
i =1 ⎝ ⎠ ⎝ 2n ⎠

58
296
STEP III STEP Solutions June 2008

⎛ θ2 ⎞ ⎡ θ2 θ4 θ6 ⎤ θ2 θ4 θ6
log( M T (θ )) = − n log⎜ 1 − ⎟ = − n ⎢− − 2 − −.....⎥ = + + +.....
⎝ 2n ⎠ ⎣ 2n 8n 24n 3 ⎦ 2 8n 24n 2
θ2 ⎛θ 2 ⎞
Thus as n → ∞ , log( M T (θ )) → , and so M T (θ ) → exp⎜ ⎟
2 ⎝ 2⎠

P( Y ≥ 25) = 0 ⋅ 05 and P Y ( )
2n ≥ 1 ⋅ 96 = 0 ⋅ 05 and so
25 = 1 ⋅ 96 2n
252 625
2n = 2 ≈
1 ⋅ 96 4
625
n≈ ≈ 78
8

1
13. P( 1 ring created at first step) = ,
2n − 1
2n − 2
P( 0 rings created at first step) =
2n − 1
1 2n − 2 1
E( number of rings created at first step) = ×1+ ×0=
2n − 1 2n − 1 2n − 1

Regardless of what happens at first step, after the first step there 2n − 2 free ends.
Similarly after second step 2n − 4 free ends regardless, etc.
1 1 1 1 1
E( number of rings at end of process) = + + + +......+
2n − 1 2n − 3 2n − 5 2n − 7 1

Var( number of rings at end of process) =

1 ⎛ 1⎞
2 2 2
⎛ 1 ⎞
2 2
1 ⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1
−⎜ ⎟ + −⎜ ⎟ + −⎜ ⎟ + −⎜ ⎟ +......+ − ⎜ ⎟
2n − 1 ⎝ 2n − 1⎠ 2n − 3 ⎝ 2n − 3⎠ 2n − 5 ⎝ 2n − 5⎠ 2n − 7 ⎝ 2n − 7 ⎠ 1 ⎝ 1⎠
(as numbers of rings created at each step are independent)
2( n − 1) 2( n − 2) 2( n − 3) 2
= 2 + 2 + 2 +......+ 2
( 2n − 1) ( 2n − 3) ( 2n − 5) 3

1 1 1
For n = 40000 , E( number of rings created) = 1 +
+ +.....+
3 5 79999
1 1 1 1 1 ⎛1 1 1 ⎞
= 1 + + + + +.....+ _ ⎜ + +.....+ ⎟
2 3 4 5 80000 ⎝ 2 4 80000 ⎠
1
≈ ln 80000 − ln 40000
2
= 2 ln 20
≈6

59
297
The Cambridge Assessment Group is Europe's largest assessment agency
and plays a leading role in researching, developing and delivering
assessment across the globe. Our qualifications are delivered in over 150
countries through our three major exam boards.

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Cambridge Assessment is a not-for-profit organisation.

This mark scheme is published as an aid to teachers and students, to indicate

the requirements of the examination. It shows the basis on which marks were
awarded by the Examiners. It does not indicate the details of the discussions
which took place at an Examiners’ meeting before marking commenced.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the
relevant knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question

papers and the Report on the Examination.

Cambridge Assessment will not enter into any discussion or correspondence

in connection with this mark scheme.

© UCLES 2008

More information about STEP can be found at:

http://www.admissionstests.cambridgeassessment.org.uk/adt/

60
298
STEP Examiners’
Report
2009

Mathematics
STEP 9465, 9470, 9475

299
 2

Contents
Step Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 3
STEP Mathematics II 13
STEP Mathematics III 18
Explanation of Results 21

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 3

STEP 2009 Paper I Principal Examiner’s Report

There were significantly more candidates attempting this paper again this year (over 900 in
total), and the scores were pleasing: fewer than 5% of candidates failed to get at least 20
marks, and the median mark was 48.
The pure questions were the most popular as usual; about two-thirds of candidates attempted
each of the pure questions, with the exceptions of question 2 (attempted by about 90%) and
question 5 (attempted by about one third). The mechanics questions were only marginally
more popular than the probability and statistics questions this year; about one quarter of the
candidates attempted each of the mechanics questions, while the statistics questions were
attempted by about one fifth of the candidates.
A significant number of candidates ignored the advice on the front cover and attempted
more than six questions. In general, those candidates who submitted answers to eight or
more questions did fairly poorly; very few people who tackled nine or more questions gained
more than 60 marks overall (as only the best six questions are taken for the final mark).
This suggests that a skill lacking in many students attempting STEP is the ability to pick
questions effectively. This is not required for A-levels, so must become an important part of
STEP preparation.
Another “rubric”-type error was failing to follow the instructions in the question. In particu-
lar, when a question says “Hence”, the candidate must make (significant) use of the preceding
result(s) in their answer if they wish to gain any credit. In some questions (such as ques-
tion 2), many candidates gained no marks for the final part (which was worth 10 marks) as
they simply quoted an answer without using any of their earlier work.
There were a number of common errors which appeared across the whole paper. These
included a noticeable weakness in algebraic manipulations, sometimes indicating a serious
lack of understanding of the mathematics involved.p As examples, one candidate tried to
use the misremembered identity cos β = sin 1 − β 2 , while numerous candidates made
deductions of the form “if a2 + b2 = c2 , then a + b = c” at some point in their work. Fraction
manipulations are also notorious in the school classroom; the effects of this weakness were
felt here, too.
Another common problem was a lack of direction; writing a whole page of algebraic manip-
ulations with no sense of purpose was unlikely to either reach the requested answer or gain
the candidate any marks. It is a good idea when faced with a STEP question to ask oneself,
“What is the point of this (part of the) question?” or “Why has this (part of the) question
been asked?” Thinking about this can be a helpful guide.
One aspect of this is evidenced by pages of formulæ and equations with no explanation. It
is very good practice to explain why you are doing the calculation you are, and to write
sentences in English to achieve this. It also forces one to focus on the purpose of the
calculations, and may help avoid some dead ends.
Finally, there is a tendency among some candidates when short of time to write what they
would do at this point, rather than using the limited time to actually try doing it. Such
comments gain no credit; marks are only awarded for making progress in a question.
STEP questions do require a greater facility with mathematics and algebraic manipulation
than the A-level examinations, as well as a depth of understanding which goes beyond
that expected in a typical sixth-form classroom. It is wise to heed the sage advice on the

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 4

STEP Mathematics website, http://www.admissionstests.cambridgeassessment.org.

uk/adt/step:

From the point of view of admissions to a university mathematics course, STEP

has three purposes. . . . Thirdly, it tests motivation. It is important to prepare
for STEP (by working through old papers, for example), which can require con-
siderable dedication. Those who are not willing to make the effort are unlikely
to thrive on a difficult mathematics course.

Question 1
The start of this question was attempted successfully by the majority of candidates, though
a significant number failed to provide any justification of why 32 × 53 has only 10 factors.
Disturbingly, many candidates expanded 32 × 53 = 1125 and then attempted to factorise the
latter, rather than noticing that it is much easier to work from the already-factorised form.
Few of these candidates made any further progress in the question.
Only about half of the candidates were able to make any significant progress on the second
half of part (i); this required deducing a general formula for the number of factors. The
idea was that by working through the specific example given at the beginning, it would be
realised that every factor has the form 3a × 5b , and then a simple counting argument would
do the job. There were numerous incorrect formulæ used at this point; candidates did not
seem to understand that a formula cannot just be pulled from thin air, but requires some
justification.
Of those candidates who reached a correct formula, several became stuck attempting to solve
the equation (m + 1)(n + 1) = 12; expanding the left hand side is likely to lead nowhere
(especially in part (ii) of the question). The form mn + m + n = 11 used by some candidates
was particularly unhelpful, as it makes it very hard to justify that all solutions have been
found. Furthermore, to answer this part successfully, some systematic approach is required,
as we want the number of solutions to this equation. Several candidates slipped up at this
point by failing to realise that 0 is, in fact, an integer, giving m = 0, n = 11 and vice versa
as solutions.
Those candidates who understood part (i) generally made good progress on part (ii). How-
ever, a sizeable number of candidates failed to write down the most general form for N , often
writing things likes N = pm × q n or N = 3m × 5n . Also, many failed to consider all possible
factorisations of 428. Furthermore, they needed to give some justification for the choice of
primes and exponents corresponding to each factorisation of 428, and this was often lacking.
Nevertheless, there were some very good answers to this question, and those who understood
the principles involved generally made very good progress.

Question 2
This was the most popular question on the paper by far, being attempted by about 90% of
the candidates. It was also found to be one of the most straightforward questions; almost
60% of the attempts successfully completed the first two parts of the question, gaining 10 of
the available 20 marks in the process, and about 20% of the attempts gained 15 marks or

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above.
In the first part, most candidates understood the procedures required to find the tangent to a
curve given implicitly, although some went to the effort of taking the cube root of the original
equation. Disturbingly, a few of them could not do this correctly, and wrote y = x + a + b.
A much more common and equally concerning error was to differentiate the equation and
correctly deduce that dy/dx = x2 /y 2 , and then use this as the gradient without substituting
in the values of x and y at the point under consideration. The resulting purported equation
for the tangent,
x2
y − b = 2 (x + a),
y
is clearly not a linear equation. The candidates then rearranged it and figured out that some
substitutions were necessary to reach the given answer, and fudged their way to it. It is
always recommended that students go back through their answer to find the source of the
error, rather than trying to fudge, as the latter is almost certainly going to get no credit.
For the second part of the question, most candidates realised that they needed to solve the
two equations simultaneously (although there were some substitution errors encountered at
this point). The process of substituting one equation into another and rearranging to reach
the given cubic was done well, although there was a reasonable amount of fudging present
here, too. Also, some candidates attempted to perform the substitutions while keeping the
as and bs present in the equations; it is much simpler to substitute at the start, so that one
only needs to work with two variables (x and y).
A sizeable number of candidates gave up at this point. This is unfortunate, as with the
question saying “Hence”, the obvious thing to do is to solve the cubic. Even without any
idea of how to go beyond this, actually solving the cubic might have given a clue. Those who
attempted this step were reasonably successful, primarily once they realised that x = −1 is
a root of the cubic. Again, though, a number of candidates were let down by poor algebraic
manipulation at this point.
Finally, having reached the two solutions of the cubic, many candidates floundered, assuming
that x = 17 7
could not possibly give them a solution. Unfortunately, x = −1 does not
help, either. Those who did press ahead with x = 17 7
often struggled to determine the
corresponding value of y, frequently leading to some very long pieces of arithmetic as they
attempted to calculate x3 +a3 +b3 and then take the cube root, rather than using the tangent
equation 4y − x = 9. Often, the long arithmetical approaches failed to reach the correct
answer because of a simple slip, which was a shame. A significant number of candidates did,
however, manage to correctly reach the final conclusion.
As mentioned in the introduction, many candidates simply quoted a solution to this final
equation (namely 63 = 53 + 43 + 33 ), and this was given no credit, as the question had
specifically said “Hence”.

Question 3
This was found to be one of the hardest questions on the whole paper, with a median mark
of 2 and a mean mark of 3.9. A relatively small number of candidates appreciated the
subtleties involved and rest consequently produced fairly nonsensical answers.
For the first half of part (i), relatively few candidates successfully attempted to relate the two

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given equations. It was encouraging that the majority of candidates were able to correctly
quote the quadratic formula (especially as it is given in the formula book!), but there was
considerable difficulty in evaluating it in this case. It seemed that candidates were also
confused by the fact that the formula is given for the equation ax2 + bx + c = 0, whereas
here the constant term involves the variable a. Other candidates became confused about the
distinction between negative roots and non-real (complex) roots; negative numbers are real!
Of those who related the two equations, very few appreciated that the solutions of an equation
involving a square root may be different from the corresponding squared equation. This was
compounded by many students asserting that 1 + 4a < 0 if and only if a < 0 (where the “if
and only if” may have been implied). Few understood that proving that there was no real
solution when a < 0 was insufficient to show that there was one real solution when a > 0.
A small number of candidates proved this part of the question using an effective graphical
method, which was very impressive.
Interestingly, the second half of part (i) proved to be more straightforward for a number
of candidates, possibly because there were no square roots involved. However, some were
confused and thought that the conditions on a in the first half of part (i) would be inherited
in the second half.
Only a handful of students made any significant attempt at part (ii), and of those, some
used algebraic methods following the ideas of part (i) while others used graph-sketching
approaches, often quite successfully.

Question 4
This question was found to have a moderate level of difficulty. There were some very good
attempts as well as some very poor ones.
The initial part of the question confused some candidates; the sketches produced showed that
the statement: “The largest and smallest angles of the triangle are α and β, respectively”
caused a surprising amount of difficulty. The use of “respectively”, in particular, is standard
mathematical language and should be recognised—a number of candidates swapped the two
angles. (It might be that some genuinely thought that the angles were the other way round,
but most simply probably misunderstood the question.) There were also a fair number of
candidates who labelled the angle opposite p as either α or β, which prevented them from
making much further progress in the question.
Most candidates were able to correctly use the cosine rule in the context of this question,
which was very pleasing. Manipulating it to deduce the required equation was much harder,
though: as usual, the lack of algebraic fluency was the sticking point. There were many
careless errors, including sign errors, transcription errors and the like. Also, a number of
candidates did not simplify the algebraic fractions when they could, thereby making their
task even more challenging. Many demonstrated that they did not understand how to add
fractions, and the monstrous messes which resulted were quite painful to see.
The straightforward method for approaching this question was to substitute expressions for
cos α and cos β into the left and right hand sides of the required equation and to show that
they turned out to be equal. A significant number of candidates did not realise that this
would be useful, and instead produced pages of algebra leading nowhere.

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In the second part of the question, a number of candidates who had not succeeded in the
first part nonetheless realised that they could use the given result to solve the second part,
and some were successful at this.
There were some good attempts at the second part. The sticking points were once again
mostly algebraic: some did not know the double angle formulæ or they used incorrect ex-
pressions for them; of those who did know them, many then erred in their expansion of the
brackets. The next sticking point was solving the cubic; some did not attempt to do so,
while others made algebraic slips. Those who did succeed in deducing cos β usually went on
to correctly find the side lengths and hence the required ratio.

Question 5
This was the least popular of the pure questions, being attempted by barely one-third of the
candidates. The marks followed suit; one third of the attempts scored 0 while another third
scored 1: these were generally candidates who tried the question and then quickly gave up.
Of those who got past the initial stages of the question, the significant majority gained full
marks or very close to it.
The question essentially required only one idea, and was then a straightforward application
of “connected rates of change” or the chain rule. The idea needed was that r, h and ` are
related via Pythagoras, so that if A is fixed (i.e., it’s a constant), then h can be expressed
as a function of r.
Those students who realised that h is a function of r mostly got the whole question correct;
those who regarded h as a constant got nowhere.

Question 6
This was found to be another hard question. While the integral in part (i) looks deceptively
easy, many candidates had little idea how to approach it. Simplifying algebraic fractions
via division is a standard A-level technique, and students should recognise that this is a
useful approach when attempting to integrate rational functions. Others successfully used
the substitution u = x + 1 to simplify the integral.
One very common approach to part (i) was to use integration by parts. While this is often
a useful technique, here it proves fruitless, and many candidates spent much effort without
appearing to realise that they were not getting anywhere. Some thought they were making
progress, but had actually misapplied the parts formula.
Other candidates tried manipulating the integral in completely incorrect ways, for example
by trying to use the formula
R for the
R derivative of a quotient, or the derivative of a logarithm,
or stating things like f g dx = f g dx.
About one third of attempts scored zero on this question, and fewer than one third of the
attempts progressed beyond part (i). The main sticking point was part (ii), where few of the
candidates realised that the substitution was u = 1/x. There were two major hints in the
question itself: firstly that the limits were 1/m to m, and these remained unchanged, and
secondly that the 1/xn changed into un−1 , which is about the same as un . Many candidates
failed to spot this, and spent time trying other substitutions unsuccessfully. Even among

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those who did realise what they needed to do, many seemed unsure how to handle the limits
correctly when performing a substitution.
Some of part (iii) could be done even without successfully completing parts (i) and (ii).
Very few candidates realised this, though, and did not even attempt part (iii) if they had
got stuck earlier. Of those who did try part (iii), some used partial fractions instead of the
earlier results (which was perfectly acceptable). Some of these were very successful, whereas
others did not fully understand partial fractions. The most common error was writing
1 A B
= 3+ ,
x3 (1+ x) x 1+x

which cannot possibly get the correct answer.

Other candidates applied the results of parts (i) and (ii) with varying degrees of success. A
common error was being careless about the limits when trying to use the result of part (ii) in
the integral in (iii)(b); the limits are not of the form 1/m to m, and so more care is needed.

Question 7
Candidates found this question very approachable, and was one of the highest scoring ques-
tions on the paper. The standard integral at the start was generally well done, and about
three-quarters of candidates gained full marks for it. Common mistakes included incorrect
notation for integration by parts, forgetting to use the [· · · ]2π
0 notation when using parts
with a definite integral, and of course making sign errors.
The trigonometrical identity in part (i) was done perfectly by most candidates who tried it.
The following integral was usually performed successfully, and over half of all attempts got
this far. The errors which occurred were similar to above, along with the standard inability
to add and simplify fractions.
The integral in part (ii) caused significantly more problems. Most candidates wrote down
the product formula for sin A sin B, but then many got stuck: they could not see how to
proceed. It is very common in STEP questions for the later parts of a question to depend on
the ideas developed earlier parts. In this case, we end up after expansion with two integrals
like the one seen in part (i), but it seems as though this was not noticed.
1
The other common problem which occurred in this part was missing or losing the factor of 2
in this part; this is another example of algebraic carelessness costing marks.

Question 8
The first part of this question was generally done very well, but the second half stumped
most candidates. Most students confidently used the double angle formula for tan 2α, and
most realised that the line y = 43 x is tangent to the circle because it is at an angle of 2α to
the x-axis and passes through the origin. There were some candidates who misremembered
or misapplied the double angle formula, or were unable to explain why the circle touches the
line. In questions such as these, a good sketch is very helpful.
The second part of the question was done relatively poorly in comparison; fewer than a
quarter of the candidates gained more than one mark for it. Even after drawing a good sketch,

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many candidates showed little idea of how to proceed. They wrote pages of calculations
with no direction, and so gained no credit. It is not worth performing calculations without
a strategy; how will knowing the coordinates of the vertices, for example, help find the value
of t? (This was a very common calculation, perhaps because it is something the candidates
are very comfortable with.) About half of the attempts noted that the two given lines were
perpendicular, but then did not do anything useful with this fact.
Many students attempted to substitute the equation of the line 4y+3x = 15 into the equation
of the circle to eliminate y, but then went on to eliminate t by asserting that x = 2t. This
cannot succeed, as the point of tangency with 4y +3x = 15 clearly does not lie on the vertical
line through the centre of the circle.
Some candidates, though, successfully used the ideas of part (i) to find the equation of the
angle bisector of the x-axis and the side AB, and then found the value of t which had (2t, t)
lying on this bisector.
Another very common successful approach was to find the distance of the point (2t, t) from
the line 4y + 3x = 15 using the formula given in the formula book for the distance of a point
from a line.
It is encouraging to see a significant number of candidates offering creative (and correct)
solutions to a problem such as this.

Question 9
The start of this question was found to be fairly approachable. Most candidates were able
to draw a correct sketch of the situation. However, a significant number became unstuck at
this very early stage, getting the directions incorrect (for example, having Q at an angle of β
above the horizontal) or swapping P and Q or worse.
The next challenge was to correctly apply the “suvat” equations (motion with constant
acceleration) to this situation. This was done successfully for P by most candidates, although
some made sign errors; it is vital to always indicate which direction is positive in problems
such as these. The motion of Q, however, caused many problems: what is s when the
motion does not start from the ground? Some candidates determined the displacement from
the starting position and then tried adding or subtracting d, others introduced d later on,
while others seemed to ignore d entirely. The general principle is to be clear about the
meaning of one’s symbols; being explicit whether s is the vertical displacement upwards or
downwards from the ground or from D would have averted many of the problems.
Most candidates were able to show that cos α = k cos β, but the quadratic proved too much
for the majority of them. There were many attempts, but because of the earlier errors in
calculating the displacements of the particles or mistakes in algebraic manipulations, the
quadratic did not appear. Many did not realise that it was necessary to use the earlier result
about the cosines, while others asserted that sin α = k sin β.
About one-fifth of the attempts at the question reached the final part of the question. Of
those, most were able to determine a formula for T , but many became stuck at this point.
Once again, there were frequent attempts to rearrange formulæ, but few were successful.
Only a handful of candidates managed to progress beyond the requirement that sin2 α 6 1
to reach the desired conclusion.

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Question 10
This was (marginally) the most popular of the mechanics questions, attempted by close
to one third of the candidates. Those who attempted it generally gained reasonably good
marks, and there were a few very pleasing answers.
In the first part, however, there was a clear lack of understanding among the candidates
of the basic principles of mechanics. While they showed a good understanding of resolving
forces, the once commonplace mantra of “Apply Newton’s Second Law to each particle, then
combine as needed” seemed to be all but forgotten: too few candidates indicated the tension
in the string on their diagram. Many therefore treated the two particles as a single system
which could be regarded as one particle with various forces, but offered no justification for
this assertion. In this case, such an approach does work, but had the pulley not been smooth,
it would not have done. Furthermore, applying Newton’s Second Law at each particle is a
much more straightforward, less error-prone approach, and may have saved a number of
candidates from unfortunate sign errors.
Those who managed to write down a correct equation of motion were generally quite con-
fident to show the required inequality. However, most fudged the argument, saying things
like “the force on M is bigger than the force on m so M g sin α > mg cos α,” which does not
show an appreciation of the effect of tension.
The trigonometric manipulations required at the start of part (ii) were done fairly well,
and those who got this far were very comfortable differentiating the expression using the
quotient rule. However, having done so, very few were then able to perform the requisite
manipulations on the derivative to reach the result that tan3 α = 2, and hence deduce the
relationship between M and m.
It was also disappointing that of those good candidates who had reached this point, only
three realised that they needed to justify that this was a (local) maximum; it should be
standard for students to ascertain the nature of any stationary point they have found if they
require the maximum or minimum of a function.

Question 11
This was found to be another very hard question, with most attempts being fragmentary;
almost half of students gained 2 or less, though the handful who made it beyond half way
achieved either 19 or 20 marks.
The first part of the question required correctly stating the laws of conservation of momentum
and restitution, together with the formula for kinetic enery. Almost all candidates realised
this and attempted to write down the appropriate formulæ. It was particularly disappointing,
though, that so many candidates struggled to do even this correctly. Most were comfortable
with the conservation of momentum, but having learnt the law of restitution in the form:
speed of separation
e= ,
speed of approach
they were at a loss when all of the velocities were in the same direction, and consequently
introduced various sign errors. A number of candidates remembered the formula upside-down
(as speed of approach over speed of separation), and this was invariably fatal.

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The next stumbling block was the formula for loss of kinetic energy. A sizeable minority of
students wrote that the loss of KE in the first particle is 12 m(u−p)2 instead of 12 mu2 − 12 mp2 =
1
2
m(u2 − p2 ) and similarly for the second particle; this is a disturbing error, although it is
not clear whether it arose because of a belief that a2 − b2 = (a − b)2 or a lack of thought
about what loss of KE means.
A number of students compounded their problems by confusing M and m, often due to
careless handwriting. It is a cardinal principle of written mathematics that distinct symbols
should look obviously distinct!
Those students who made one of these errors generally failed to make any significant further
progress with the question.
The key step in part (i) was to manipulate the formula for loss of KE to eliminate M and q,
making use of the other two equations to achieve this. Most candidates who had reached this
point succeeded in doing this and reaching the required formula. Some made their lives more
difficult by expanding every bracket fully; it is a valuable skill to recognise when this would
be helpful and when not. For the final deduction in part (i), the instruction to “deduce”
the result u > p meant that no credit was given to those students who simply stated “it’s
obvious”; some justification on the basis of the formula for loss of KE was required.
The majority of candidates did not even attempt part (ii), which is a shame, as the ideas
were essentially the same as in part (i), albeit with a little more sophistication: the task was
to eliminate some of the variables. The first half required writing down the word statement
in symbols and eliminating either M or M , while the second half required elimination of p
and q followed by some rearrangement. Any attempt to perform these eliminations would
have led to the correct results.

Question 12
This was the more popular of the probability questions, and was attempted by almost a
quarter of the candidates. It gained the highest scores on the whole paper: the median mark
was 14.
The introductory part caused some problems. This is a standard result which should have
been recognised as such. Some candidates began with (x + y)2 and became stuck. Others
attempted to prove the result by induction, which will be challenging as x and y are any
real numbers.
Part (i) was generally done extremely well, with many candidates drawing an appropriate
tree diagram. However, some candidates failed to add the probabilities (more commonly
ab
when they had not drawn a tree diagram), asserting things such as P(same colour) = (a+b) 2
2ab
instead of (a+b)2 .
Part (ii) caused more problems. Many candidates decided that it would be useful to expand
(a + b + c)3 and wasted a lot of time doing so. Another common issue was not adequately
justifying the probability stated in the question; the arguments proposed for counting the
number of possibilities were often weak or confused. Students would have done well to either
draw a tree diagram (possibly abbreviated) or to explicitly list all of the possibilities. While
taking time, it would have ensured that they reached the correct answer.
Most of the candidates were able to calculate the probabilty of all three being the same

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colour correctly, although some left out the factor of 3.

The last part of the question caused the most difficulty. Some suggested that since there were
18 ways of having exactly two the same colour and 3 ways of having them all the same colour,
the former was 6 times more likely than the latter. This totally ignored the probabilities
of these events and the work done earlier, as well as the “deduce” in the question. Others
assumed a 6 b 6 c and then wrote a + x = b, a + y = c to try to relate it to the initial
inequality, but struggled to get further. Nevertheless, there were a good number of students
who correctly related the earlier inequality to the new context and succeeded in making the
required deduction.

Question 13
This was a very unpopular question and most attempts were fragmentary, while a small num-
ber were essentially perfect. Most attempts were very weak at conveying their combinatorial
ideas clearly, leading to confused and incorrect arguments. A few candidates completely
misunderstood what was being asked and tried to answer something different entirely.
For part (i), very few candidates realised that the best way to begin part (i) was to regard
the girls as a block dividing the boys into two sections. Even those who did spot this still
struggled to deduce the required probability. Most of those who counted the total number
of arrangements as (n + 3)! and the number of arrangements of the girls as 3! failed to see
that the girls could be positioned as a block in only (n + 1) positions.
For part (ii), most offerings were vain attempts to reach the given answer, but demonstrated
no understanding of the problem.

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STEP 2
General Remarks
Of the 1000+ entries for this paper, around 920 scripts actually arrived for marking, giving
another slight increase in the take-up for this paper. Of this number, five candidates scored a
maximum and seventy-five achieved a scoring total of 100 or more. At the other end of the scale,
almost two hundred candidates failed to reach the 40-mark mark. Otherwise, marks were spread
reasonably normally across the mark range, though there were two peaks at about 45 and 65 in
the distribution. It is comforting to find that the ‘post-match analysis’ bears out the view that I
gained, quite firmly, during the marking process that there were several quantum states of mark-
scoring ability amongst the candidature. Many (about one-fifth of the entry) struggled to find
anything very much with which they were comfortable, and marks for these candidates were
scored in 3s and 4s, with such folk often making eight or nine poor efforts at different questions
without ever getting to grips with the content of any one of them. The next “ability band” saw
those who either scored moderately well on a handful of questions or managed one really
successful question plus a few bits-‘n’-pieces in order to get up to a total in the mid-forties. To
go much beyond that score required a little bit of extra talent that could lead them towards the
next mark-hurdle in the mid-sixties. Thereafter, totals seemed to decline almost linearly on the
distribution.

Once again, it is clear that candidates need to give the questions at least a couple of minutes’
worth of thought before commencing answering. Making attempts at more than the six scoring
efforts permitted is a waste of valuable time, and the majority of those who do so are almost
invariably the weaker brethren in the game. Many such candidates begin their efforts to
individual questions promisingly, but get no more than half-a-dozen marks in before abandoning
that question in favour of another – often with the replacement faring no better than its
predecessor. In many such cases, the candidate’s best-scoring question mark would come from
their fifth, or sixth, or seventh, or …?, question attempted, and this suggests either that they do
not know where their strengths lie, or that they are just not taking the time to make any sensible
decision as to which questions to attempt and in what order to do them, adopting some sort of
hit-and-hope strategy. With the pleasing number of very high totals to be found, it is clear that
there are many places in which good marks were available to those with the ability to first
identify them and then to persevere long enough to be able to determine what was really going
on therein.

It is extremely difficult to set papers in which each question is pitched at an equivalent level of
difficulty. Apart from any other factors, candidates have widely differing strengths and
weaknesses; one student’s algebraic nuance can be the final nail in the coffin of many others, for
instance. Moreover, it has seemed enormously clear to me – more particularly so since the arrival
of modular A-levels – that there is absolutely no substitute for prolonged and determined practice
at questions of substance. One moment’s recognition of a technique at work can turn several
hours of struggle into just a few seconds of polishing off, and a lack of experience is always
painfully clear when marking work from candidates who are under-practised at either the art of
prolonged mathematics or the science of creative problem-solving. At the other, more successful,
end of the scale there were many candidates who managed to produce extraordinary amounts of
outstanding work, racking up full-, or nearly full-, marks on question after question. With the
marks distributed as they were, it seems that the paper was pitched appropriately at the intended
level, and that it successfully managed to distinguish between the different ability-levels to be
found among the candidates.

As in previous years, the pure maths questions provided the bulk of candidates’ work, with
relatively few efforts to be found at the applied ones. Moreover, many of these were clearly acts
of desperation.

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Comments on individual questions

Q1 The first question is usually intended to be a gentle introduction to the paper, and to allow
all candidates to gain some marks without making great demands on either memory or technical
skills. This year, however, and for the first time that I can recall in recent years, the obviously
algebraic nature of the question was enough to deter half the candidates from attempting it.
Indeed, apart from Q8, it was the least popular pure maths question. This was a great pity: the
helpful structure really did guide folks in the right direction, and any half-decent candidates who
did try it usually scored very highly on it. There were, nonetheless, a couple of stumbling-blocks
along the way for the less wary, and many candidates tripped over them. The point at which most
of the less successful students started to go astray was when asked to show that ABCD is a
rectangle. Lots of these folk elected to do so by working out distances … when the use of
gradients would have been much simpler. The really disappointing thing was that many simply
showed both pairs of opposite sides to be equal in length without realising that this only proved
the quadrilateral a parallelogram. The next major difficulty was to be found in the algebra, in
 
turning the area, 2  2   2 , into something to do with u and v. It was quite apparent that many
were unable to do so because they failed to appreciate that  and  were particular values of x
and y that satisfy the two original curve equations, so that  4   4  u and   v . Then,
squaring the area expression does the trick. Some got part of the way to grasping this idea, but
approached from the direction of solving x 4  y 4  u and xy  v as simultaneous equations;
the resulting surds-within-surds expressions for  and  were too indigestible for almost
anyone to cope with.

Q2 Another of the less popular pure maths questions. It is clear that many A-level students
are deeply suspicious of approximations and logarithms, and these plus the fact that y is a
“function of a function of a function” clearly signalled to many to pass by on the other side. Of
those who did take up the challenge here, almost all plumped automatically for differentiation in
(i), usually by taking logs first and then differentiating implicitly. Just a few knew how to
differentiate directly using the fact that ax = ex ln a. However, calculus was not actually required,
since the maxima and minima of y can be deduced immediately from knowledge of the sine
function. It then helped candidates enormously if they were able to work generally in deciding
what values of x gave these stationary points, not least because they would need some care in
figuring out which to use in (iv). It was a pleasant surprise to find that (ii) was generally handled
quite well, but sketches were poor – usually as a result of previous shortcomings – especially for
x < 0; many candidates did realise, almost independently of previous working it seemed, that the
right-hand ‘half’ of the curve oscillated increasingly tightly. In (iv), a lack of clarity regarding
the x-values, allied to an uncertainty over dealing with the logs, proved a great hindrance to the
majority. Also, it has to be said that, even amongst those with the right k’s to hand, a simple
diagram of what they were attempting to work with would undoubtedly have saved them a lot of
mark-spurning algebraic drivel.

Q3 Only marginally behind Q5 for popularity, this was a surprising hit amongst candidates.
It had been anticipated that a trig. question containing lots of surds would be a bit of a turn-off,
but this didn’t prove to be the case. Moreover, it turned out to be the highest scoring question on
the paper too. I had expressed reservations, during the setting process, that we had been a little
too helpful in flagging up what was needed at each stage of the process, and so it proved to be.
Most hiccups came at the outset, where proving even a simple identity such as this one was
beyond many, even those who continued very successfully. The only other trouble-spot came in
(ii) when lots of candidates (who should be applauded for trying to keep in the spirit of

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“showing” stuff) undertook this part by rationalising the denominator (twice!) to prove the given
result, while the sneakier types just multiplied across and verified it. Shame on them!
Q4 Considering the very poor marks gained on this question, it was surprisingly popular,
with almost 600 “hits”. Its essential difficulty lay in the fact that one can only go so far in this
question before requiring the ‘key of insight’ in order to progress further. And that was that, as
they say. Personally, this was my favourite question, as the key is such a simple one once it is
pointed out to you (clearly not an option in the exam., of course). Parts (i) and (ii) require
candidates to find p(1) = 1 and to show that p(x) has (x – 1)4 as a factor, and most did so
perfectly satisfactorily. The (strikingly similar) information then given in (iii) should then
suggest (surely?), to anyone with any sort of nous, that they are required to make similar further
deductions. Nope – apparently not. Even amongst the few who did then find p(–1) = –1 and
show that p(x) has (x + 1)4 as a factor, very few knew what to do with these facts. I think that
this is principally because most students work “on automatic” in examinations – a by-product of
the much (and rightly) criticised modular system – simply doing as they are told at each little
step of the way without ever having to stand back, even momentarily, and take stock of the
situation before planning their own way forwards. This is the principal shame with modular
assessment: the system prevents the very able from ever having to prove their ability whilst
simultaneously persuading the only modestly able that they are fantastic mathematicians when
they aren’t. A moment of thoughtful reflection on the nature of this strange creature that is p(x)
and what we now know about it reveals all. It is a polynomial of degree 9. Its derivative must
therefore be a polynomial of degree 8. And we know that p(x) has two completely distinct
factors of degree 4. Apart from the tendency to assume that a polynomial always commences
with a coefficient of 1, the rest (in principle) is just a matter of adding two 4s to get 8.

Q5 This was the most popular question on the paper, though only by a small margin, and the
second highest scoring. In fact, I can be very specific and state that almost all of the really
successful attempts scored 15 or 16 marks. The few marks lost were almost invariably in (ii),
where so very, very few picked up the hints as to the only, minor difficulty within the question.
Once again, this is almost certainly due to the mind-set of simply ploughing on regardless
without stopping to think about what is actually going on. Whilst understanding that nearly all
candidates will feel under considerable pressure to pick up as many marks as possible as quickly
as possible, NO-ONE who sits this paper should be of the view that they are not going to be
challenged to think. And, to be fair to the setting panel, we did put some fairly obvious signposts
up for those who might take the trouble to look for such things. For future STEP candidates, this
will make an excellent practice question for teachers to put their way. (If they are willing to learn
from their mistakes, and you think you can catch them out … this is a marvellous question to
use.) One pointer is in the change of limits, from (5, 10) to  54 , 10 ; the other is in the switch from
asking for integrals-to-be-evaluated to asking for areas. The crux of the matter is that most A-
level students believe that x 2  x rather than | x |. Once you realise that, the question is fiddly
but otherwise rather easy.

Q6 This was another very popular question, but the one with the lowest mean mark score of
all the pure questions, at about 7. I think that the initial enthusiasm of seeing something familiar
in the Fibonacci Numbers was more than countered by the inequalities work that formed the bulk
of the question. Nonetheless, I suspect that, if given the opportunity to talk it through after the
event, many candidates would admit that half of the marks on the question are actually
ludicrously easy to acquire and that they were really only put-off by appearances. For instance,
to show that S > any suitable lower-bound, one need only keep adding terms until the appropriate
figure is exceeded. For those reciprocals of integers that are not easily calculated, such as 131 , it is
perfectly reasonable to note something that they are greater than and us that in its place. Thus,
S  1  1  12  13  15  18  ...  1  1  12  14  15  101  ...  2  0.5  0.25  0.2  0.1  3.05  3

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works pretty easily (though it may not have scored full marks in (i) as a particular approach was
requested), and something similar could be made to work in showing that S > 3.2 in (ii). The
approach that the question was designed to direct candidates towards was that of stopping the
1 1 1 
direct calculation at some suitable stage, and using an inequality of the form  2   ,
Fi  i 2 
F
possibly alternating with the odd- and even-numbered terms, to make the remaining sum less
than a summable infinite GP. For further thoughts and possible developments of these ideas, I
would refer the reader to the Hints & Solutions document for this paper.

Q7 In many ways, this question is little more than an academic exercise, since I can see no
way in which these integrals would actually arise in any practical situation. I apologise for this.
However, it was a good test of candidates’ ability to stretch a general result in different
directions, probing them for increasing amounts of insight and perseverance. For future STEP-
takers, the opening result is a good one for which to find a generalisation, and it is a possibly
fruitful avenue to explore the “product rule of differentiation” for three terms (etc.), say y = pqr
in this case. Such an approach might have helped to prevent some of the ghastly mix-ups in
writing all the terms out that were to be found in the scripts. It was disappointing to see that so
few candidates seemed to think that they should tidy up the answer and demonstrate that the left-
over bits did indeed form a cubic polynomial, as required by the question. In the end, we gave
anyone the mark who simply observed that what was left was a cubic (if indeed that was the case
in their working). Thereafter, (i) is a straightforward application of the result, requiring
candidates only to identify the values of n, a and b. However, even here, it was rare to see folks
justifying the form of the cubic, which might have acted as a check for errors. In (ii), the
polynomial term is no longer cubic, so candidates were expected to try to see if an extra factor of
(x – 1) could be taken out to go with the other twenty-one (x – 1)s, which indeed it could.
Checking the cubic’s terms was rather more important here. The final integral, in (iii), was
difficult, and this was where candidates were ‘found out’ on this question. The obvious thing is
to try and extract some (x – 2) factor(s) from the quartic polynomial, but this doesn’t work.
Candidates may reflect that they shouldn’t have found this too much of a surprise, as that would
simply have been repeating the “trick” of (ii). Though only a small minority realised it, the next
most obvious possibility to try, having already found in (ii) that ‘the next case up’ gives a quartic
rather than a cubic polynomial, is to try some combination of the obvious answer and the next
one up, and this turns out to be exactly what is required.

Q8 The vectors question was the least popular pure maths question by a considerable way,
and only marginally more popular than most of the applied questions. In general, attempts were
short-lived and candidates usually gave up when the algebra got a bit “iffy”. Strangely, an awful
lot of attempts began very poorly indeed; the ranges of values of  and  had a geometric
significance relating to where P and Q lay on the lines AB and AC, and these were not well
grasped. Moreover, it was surprising to see diagrams in which the lines had not even been drawn,
often leaving the marker to guess whether the points were supposed to be on them or not. Many
responses to the next part were so bizarre that they were almost funny: a lot of candidates
thought that CQ etc. were vectors rather than lengths, and the “” was treated variously as a
scalar multiplication, the scalar product and the vector product. Oddly enough, they could all
lead to the required answer,  = 1, even legitimately (with a bit of care) though we were harsh
on statements that were actually nonsense. Very few made it to the later stages of the question.

Q9 Of the applied maths questions, this was (again) by far the most popular, drawing around
300 attempts. It also proved to be one of the highest-scoring of all the questions on the paper,
with a mean score of 12. Finding the position of the centre of mass was sensibly used by most

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candidates, and the first two parts yielded high marks. The last part attracted less confident
algebra, which is curious given that it involved much the same sort of work.

Q10 In general, one can be forgiven for approaching a collisions question in an automatic
way; applying the Conservation of Linear Momentum (CLM) and Newton’s Experimental Law of
Restitution (NEL or NLR) for both initial collisions. Most candidates did the routine stuff quite
well but then got bogged down in the ensuing algebra. The nice thing about this question is that
it can also be done without the need for the use of CLM at all. The NEL statements give a
relationship between the final velocities (v1 to v4, say) of the four particles, and then equating for
the times to the following collisions at O uses these velocities without ever requiring to have
them in terms of u. I was greatly surprised to see such a high proportion of the attempts using
some such suitably concise approach, and they were almost guaranteed full marks on the
question, and for very little time and trouble.

Q11 This question was the least popular question on the paper, and those trying it averaged
only 6 marks on it. The most surprising aspect of it is that so few could even write a decent N2L
statement to begin with, and they simply stood no chance thereafter. For those who made it to the
first-order, variables-separable differential equation, the work was much more promising, though
I suspect this is due to the fact that only the very able made it this far. The unpromising
integration of f(v) dv, where the f(v) turned out to be a linear-over-linear algebraic fraction, was
certainly unappealing to look at, but a simple substitution such as s = P – (n + 1)Rv reduces it to
a very simple piece of integration. As far as I recall it, most of the inequalities in (i) were fudged,
though it was very heart-warming indeed to see those excellent few who made it right to the end.
It is a pity that a last minute change to the question, prior to printing, which had been intended to
help candidates by giving them the final answer, then omitted the factor (n + 1) in its
denominator. Fortunately, we are talking about no more than twenty of the most able (and high-
scoring) candidates here; those who had explained it correctly, but then crossed-out the (n + 1)
since it didn’t appear on the question-paper, were given the final mark. As for those who were
slightly less honest and gave the proper explanation but (presumably deliberately) didn’t write
the missing factor in anywhere, in order to fudge it, we were mean and didn’t give them the final
mark.

Q12 There were only around 200 attempts to Q12, and the mean score was just 5, making it
the worst question on the paper for marks. Many of the attempts failed to get very far at all,
2
largely for finding that efforts to integrate e ax proved difficult. Personally, I had thought it was
giving too much away to ask for the sketch of the pdf at the beginning, in that it might just give
the game away that what was being handled here was just half of a normal distribution. I worried
in vain. Sadly, without this crucial observation, little or no progress was possible. Even with it, a
little care was still needed in handling the differences between this and N(0, 1), for which the
tables could be used.

Q13 This question was only marginally more popular than Q11, but those who did attempt it
were usually well rewarded with marks; candidates averaging almost 11 on it. The careful use of
binomial expansions, and remembering to use the result q = 1 – p throughout, made this an
eminently approachable question in principle. Those who stumbled did so over little arithmetical
slips, such as with the careless handling of minus signs. Once an error has been introduced,
although method marks are there to be had, the final satisfaction of getting to the answer is never
to be experienced without going back to correct it.

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STEP Mathematics III 2009: Report

The vast majority of candidates (in excess of 95%) attempted at least five questions,
and nearly a quarter attempted more than six questions, though very few doing so
achieved high scores (about 2%). Most attempting more than six questions were
submitting fragmentary answers, which, as the rubric informed candidates, earned
little credit.

Section A: Pure Mathematics

1. A popular question attempted by more than four fifths of the candidates, and
scoring as well as any question, and most successfully obtained expressions for p and
q. Quite a lot also obtained the quadratic equation and from it the sum and product of
roots for s and t. However, a common error at this stage was to overlook the
coefficient of the second degree term not being 1. For this reason, or otherwise,
because they didn’t know what to do many “fell at the last hurdle”, although a good
number completed the question successfully.

2. This was similar to question 1 in popularity and success. Virtually all got part
(i) correct, and many used the series correctly to obtain the value for a1. Quite a few
completed the question, although frequently candidates dropped 2 marks through not
looking at terms properly.

3. Though slightly more popular than the first 2 questions, the attempts scored
marginally less well. Candidates began well, though the limit of f '(t) was not well
done. The even function in (ii) was usually correctly justified. Part (iii) was
frequently not quite correctly justified, though some did so by sketching  
and . The sketch of f(t) often had the wrong gradient as it
approached the y axis, and asymptotes were frequently not identified.

4. About half the candidates attempted this, with similar levels of success to
question 3. Parts (i) and (iii) caused few problems though part (ii) did. There were
some errors in part (iv), but it was the last part using the four results that usually went
wrong.

5. This was the most popular question, with a few more attempts than question 3,
but with a level of success matching the first two questions. Many showed the first
two results correctly, and quite a few the third one. The last part tripped up many
candidates, though the most successful used the first approach in the mark scheme. A
number of candidates understood “independent of n” in the question to be given
information, and attempted to find a, b, and c by solving three simultaneous equations
for specific values of n. However, there were commonly errors in the values of the Sn
used. An efficient alternative solution is given in the mark scheme.

6. About a third of the candidates attempted this, though with less success than
any of its predecessors. Attempts were mostly “all or nothing”. Some candidates
thought that the cyclic quadrilateral property had to be that opposite angles are
supplementary, as the only property that they knew.

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7. Approximately two thirds of the candidates attempted this, earning roughly

half marks in doing so. Part (i) and finding the three expressions for P0, P1 & P2 from
part (ii) largely went well. The result involving Pn+1 saw most falling by the wayside,
especially those who attempted it by induction. Quite a few candidates did score all
but two marks in proving that Pn was a polynomial of degree n or less, but not
appreciating that there was still something to do regarding the leading term.

8. Roughly the same number attempted this as question 7, with slightly less
success. Usually, a candidate did not properly obtain the first three results, and so
would end up having apparently finished the whole question but in fact scoring only
two thirds marks. The problem was often that the limiting process was not fully
understood. In part (ii), there was often odd splitting going on to attempt the
integration by parts and this part often went wrong.

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Section B: Mechanics

9. The most popular of the three Mechanics questions, being attempted by a sixth
of the candidates, it was also the least successful, scoring only a quarter of the marks.
Quite a few candidates scored nothing at all, and quite a few got the result in part (i)
correctly, although by a variety of approaches, given that the uniform acceleration
equations can be combined in numerous ways. However, few made any headway
with the trajectory equation for part (ii).

10. The second least popular question on the paper attempted by a twelfth of the
entry, the success rate on it was comparable to questions 7 and 8. Mostly, candidates
successfully found and solved the differential equation, but finding T stumped them,
with a alternative approaches successfully used in a very small number of cases.

11. A little less popular than question 9, the success rate was between those of the
other questions in this section. Candidates tended to polarize scoring less than a third
of, or very nearly all if not all of the marks. Conservation of energy, uniform
acceleration and even constant speed were erroneous methods attempted. Even some
good candidates carried out the integration in part (i), and then failed to solve the
quadratic equation for x.

Section C: Probability and Statistics

12. About a tenth of the candidates attempted this, usually earning quarter marks.
Quite often the conditional bit in part (i) threw them, so they were 3 marks down
before they got into the question. 80% of the attempts did not obtain or use the pgf as
required. A small number of candidates really knew their stuff and did it very well.

13. A handful of candidates attempted this question with a couple making a good
stab at part (i), but otherwise it was the odd crumb, if even that, which was collected.

318
 21

Explanation of Results STEP 2009

All STEP questions are marked out of 20. The mark scheme for each question is designed to reward
candidates who make good progress towards a solution. A candidate reaching the correct answer will
receive full marks, regardless of the method used to answer the question.

All the questions that are attempted by a student will be marked. However, only the 6 best answers will be
used in the calculation of the final grade for the paper.

There are five grades for STEP Mathematics which are

S – Outstanding
1 – Very Good
2 – Good
3 – Satisfactory
U – Unclassified

Usually a candidate who answers 4 questions well (not perfectly) will be awarded a grade 1 for that paper.

The rest of this document presents, for each paper, the grade boundaries (minimum scores required to
achieve each grade), cumulative percentage of candidate achieving each grade, and a graph showing the
score distribution (percentage of candidates on each mark).

STEP Mathematics I (9465)

Grade boundaries
Maximum Mark S 1 2 3 U
120 95 72 58 35 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 5.0 17.4 32.4 78.1 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics I

319
 22

STEP Mathematics II (9470)

Grade boundaries
Maximum Mark S 1 2 3 U
120 98 71 61 39 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 9.1 29.6 47.3 80.9 100

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics II

STEP Mathematics III (9475)

Grade boundaries
Maximum Mark S 1 2 3 U
120 95 67 55 38 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 14.2 45.8 62.9 89.2 100

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics III

320
STEP Solutions
2009

Mathematics
STEP 9465, 9470, 9475

321
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the requirements of the examination. It shows the basis on which marks were
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in connection with this mark scheme.

© UCLES 2009

More information about STEP can be found at:

http://www.admissionstests.cambridgeassessment.org.uk/adt/

2
322
 Contents

Step Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 4
STEP Mathematics II 42
STEP Mathematics III 53

3
323
 STEP I, Solutions
2009

4
324
Question 1

A proper factor of an integer N is a positive integer, not 1 or N , that divides N .

(i) Show that 32 × 53 has exactly 10 proper factors.

It is not by accident that this question writes “ 32 × 53 ” and not “ 1125”: it is aiming to
suggest that it is much more straightforward to think about factors of a number if we are
given its prime factorisation to begin with. Also, note that the question does not ask us to
multiply out the factorisations at any point. In fact, there is no need to even give the factors
explicitly if you do not need to.

Determining the proper factors of 32 × 53 is straightforward: any factor must be of the form
3r × 5s with 0 6 r 6 2 and 0 6 s 6 3, giving the factors:
30 × 50 (= 1) (this is not a proper factor)
30 × 51 (= 5)
30 × 52 (= 25)
30 × 53 (= 125)
31 × 50 (= 3)
31 × 51 (= 15)
31 × 52 (= 75)
31 × 53 (= 375)
32 × 50 (= 9)
32 × 51 (= 45)
32 × 52 (= 225)
32 × 53 (= 1125) (this is not a proper factor)

Therefore there are 10 proper factors in total.

Alternatively, we could simply note that there are 3 possible values for the power of 3 (namely
0, 1 and 2) and 4 for the power of 5 (namely 0, 1, 2 and 3), making 3 × 4 = 12 factors.
Of these, two are not proper (1 and the number 32 × 53 itself), leaving 12 − 2 = 10 proper
factors.

(i) (cont.)
Determine how many other integers of the form 3m ×5n (where m and n are integers)
have exactly 10 proper factors.

Now that we have done this and understood how to count the factors of 32 × 53 , we can
answer the second part: the number of proper factors of 3a × 5b is (a + 1)(b + 1) − 2, as
the power of 3 in a factor can be 0, 1, . . . , a, and the power of 5 can be 0, 1, . . . , b. So we
require (a + 1)(b + 1) − 2 = 10, or (a + 1)(b + 1) = 12. Here are the possibilities:

5
325
 a+1 b+1 a b n = 3a × 5b
1 12 0 11 30 × 511
2 6 1 5 31 × 55
3 4 2 3 32 × 53
4 3 3 2 33 × 52
6 2 5 1 35 × 51
12 1 11 0 311 × 50

so there are 6 possibilities in total. This means that there are 5 other integers with the
required properties.
We use these same ideas in part (ii).

(ii) Let N be the smallest positive integer that has exactly 426 proper factors. Determine
N , giving your answer in terms of its prime factors.

Following the same ideas as in part (i), let n = 2a ×3b ×5c ×7d ×· · · be the prime factorisation
of the positive integer n. (Note that we should use a letter other than N to distinguish our
arbitrary integer from the special one that we seek.)
Then the number of factors of n is (a + 1)(b + 1)(c + 1)(d + 1) · · · , and we must subtract 2
to get the number of proper factors. Assuming now that n has 426 proper factors, we must
have
(a + 1)(b + 1)(c + 1)(d + 1) · · · − 2 = 426,
so
(a + 1)(b + 1)(c + 1)(d + 1) · · · = 428.
Now we can factorise 428 = 22 × 107, and 107 is prime. So the possible factorisations of 428
are 428 = 2 × 214 = 4 × 107 = 2 × 2 × 107, so there can be at most three prime factors
in n. We are seeking the smallest such n, so we choose the smallest possible primes, giving
the smaller ones higher powers and larger ones smaller powers. So the smallest values of n
for each possible factorisation of 428 are as follows:
With 428 = 428: n = 2427
With 428 = 2 × 214: n = 2213 × 3
With 428 = 4 × 107: n = 2106 × 33
With 428 = 2 × 2 × 107: n = 2106 × 3 × 5
Since we seek the smallest possible value, our answer is clearly 2106 × 3 × 5, as 2107 × 3 >
33 = 27 > 3 × 5 = 15.

6
326
Question 2

A curve has the equation

y 3 = x 3 + a3 + b3 ,
where a and b are positive constants. Show that the tangent to the curve at the point
(−a, b) is
b2 y − a2 x = a3 + b3 .

Differentiating the equation of the curve with respect to x gives

dy
3y 2 = 3x2 .
dx
dy dy
Substituting x = −a and y = b gives 3b2 dx
= 3a2 , so dx
= a2 /b2 . Then the standard
equation of a straight line gives
a2
y−b= (x + a),
b2
which easily rearranges into the form b2 y − a2 x = a3 + b3 , as required.

In the case a = 1 and b = 2, show that the x-coordinates of the points where the tangent
meets the curve satisfy
7x3 − 3x2 − 27x − 17 = 0 .

In the case a = 1, b = 2, the curve has equation y 3 = x3 + 9, and the tangent at (−1, 2) has
equation 4y − x = 9. We therefore substitute 4y = x + 9 into y 3 = x3 + 9 as follows (after
multiplying by 43 ):

64y 3 = 64x3 + 576

=⇒ (x + 9)3 = 64x3 + 576
=⇒ x3 + 27x2 + 243x + 729 = 64x3 + 576
=⇒ 63x3 − 27x2 − 243x − 153 = 0
=⇒ 7x3 − 3x2 − 27x − 17 = 0, on dividing by 9,

and this is the equation required.

Hence find positive integers p, q, r and s such that

p3 = q 3 + r 3 + s3 .

Now our equation looks hard to solve, but we know that there is a solution at x = −1, as
the curve and line are tangent at this point. In fact, since they are tangent, x = −1 must
be a double root. So we can take out a factor of (x + 1)2 to get

(x + 1)(7x2 − 10x − 17) = 0

=⇒ (x + 1)2 (7x − 17) = 0.

7
327
Thus either x = −1, which we aleady
know, or x = 17 7
. Since this point lies on the line
4y − x = 9, the y-coordinate is 17
7
+ 9 /4 = 20
7
. Thus, as this point also lies on the curve
3 3 3 3
y = x + a + b , we have
20 3 17 3
+ 13 + 23 .



7
= 7

Now multiplying both sides by 73 gives us our required result:

203 = 173 + 73 + 143 ,

so a solution is p = 20, q = 17, r = 7, s = 14.

8
328
Question 3

1
(i) By considering the equation x2 + x − a = 0 , show that the equation x = (a − x) 2
has one real solution when a > 0 and no real solutions when a < 0.

1
This looks somewhat confusing at first glance; why might x = (a − x) 2 , which can be
rearranged as the given quadratic, only have one solution whereas the quadratic can have
two? But we must remember that this equation involves a square root, and by convention,
this is the positive square root; therefore, real solutions must satisfy both x > 0 and a−x > 0,
even if the quadratic has other solutions in addition.

Consider the equation

1
x = (a − x) 2 . (∗)
2 2
If this is true, then squaring gives x = a − x, or x + x − a = 0. The solutions of this
quadratic are given by √
−1 ± 1 + 4a
x= ,
2
and these will be real solutions of (∗) if and only if x > 0 and a − x > 0, that is 0 6 x 6 a.
But as a − x = x2 , we will always have a − x > 0 for real solutions of the quadratic, so we
need only check that x > 0. For a solution to (∗), we therefore require the plus sign in the
quadratic formula, and we also need 1 + 4a > 1, so a > 0.
√

Thus, for a < 0, there are no real solutions to (∗), and for a > 0, x = −1 + 1 + 4a /2 is
the unique real solution.

An alternative approach is graphical. After we have shown that (∗) leads to the quadratic
x2 + x − a = 0, we see that the real solutions of (∗) correspond to those of the quadratic
where x > 0. We can therefore sketch the graph of y = x2 + x − a and observe where the
roots are.
The quadratic y = x2 + x has roots at x = 0 and x = −1 with a line of symmetry at x = − 12 .
The equation y = x2 + x − a is a simple translation by a vertically downwards, like this:

a = −1

a=0

a=1

It is therefore clear that, for a > 0, there is one root with x > 0, and for a < 0, there is no
such root.

9
329
 (i) (cont.)
Find the number of distinct real solutions of the equation

13
x = (1 + a)x − a

in the cases that arise according to the value of a.

Since cube-rooting is invertible, we have

13
x = (1 + a)x − a ⇐⇒ x3 = (1 + a)x − a.

We are thus trying to solve the cubic equation x3 − (1 + a)x + a = 0. Inspection reveals one
root, x = 1, so we can factorise the cubic as (x − 1)(x2 + x − a) = 0. Using the discriminant
of the quadratic factor, 1 + 4a, we find that x2 + x − a = 0 has 0, 1 or 2 real roots according
to whether 1 + 4a < 0, 1 + 4a = 0 or 1 + 4a > 0, respectively.
Hence the original equation has 1 real root if a < − 14 , 2 distinct real roots if a = − 14 (being
x = 1 and x = − 12 ), and 3 real roots if a > − 14 .
In the latter case, there is the possibility that they are not all distinct, though, if x = 1 is a
root of x2 + x − a = 0. This only happens when a = 2, and in this case, there are also only
2 distinct real roots.

(ii) Find the number of distinct real solutions of the equation

1
x = (b + x) 2

in the cases that arise according to the value of b.

This is very similar to part (i), with the only difference being that this time we have b + x
rather than a − x. The argument should therefore be fairly similar to part (i).

Starting with the equation

1
x = (b + x) 2 , (†)
we again square this to get x2 = b + x, or x2 − x − b = 0.
From the first form, we see that to have any solutions, we must have x > 0 and b + x > 0.
From the second, we see that the discriminant 1 + 4b > 0 and b + x = x2 > 0 as long as x is
real. So if b < − 14 , there are no solutions.
The solutions to the quadratic are
√
1± 1 + 4b
x= .
2
In the case b = − 14 , the repeated solution is x = 1
2
> 0, so there is one solution in this case.

10
330
In the case b > − 14 , we still require x > 0 for solutions. The smaller root is
√
1− 1 + 4b
x= ,
2
which is negative if b > 0 and non-negative if b 6 0.
Thus the equation (†) has no solutions if b < − 14 , one solution if b = − 14 or b > 0, and two
solutions if − 14 < b 6 0.

An alternative approach to this part of the question is again to draw a graph of the functions
involved. As in part (i), we draw the graph of y = x2 − x − b and determine how many roots
it has with x > 0. The graph y = x2 − x has roots at x = 0 and x = 1, and the graph of
y = x2 − x − b is a translation of this by −b in the y-direction, as shown in this sketch:

b = −2

b = b0

b=0
b=1

From these sketches it is clear that for b < b0 , there are no positive real solutions; at b = b0
there is one (repeated) positive real solution, for 0 > b > b0 there are two non-negative real
solutions, and for b > 0 there is one positive real solution. Finally, determining b0 is easy:
we want x2 − x − b0 to have a repeated root, and this means that x2 − x − b0 = (x − 12 )2 , so
that b0 = − 14 .

11
331
Question 4

The sides of a triangle have lengths p − q, p and p + q, where p > q > 0. The largest and
smallest angles of the triangle are α and β, respectively. Show by means of the cosine rule
that
4(1 − cos α)(1 − cos β) = cos α + cos β.

In situations like this, it’s often useful to draw a sketch to gain some clarity about what is
happening. Note that the largest angle is always opposite the longest side, and the smallest
angle is always opposite the shortest side.

C
α
p p−q

β
A p+q B

Using the cosine rule with the angles at A and C gives, respectively:

(p + q)2 = p2 + (p − q)2 − 2p(p − q) cos α (1)

(p − q)2 = p2 + (p + q)2 − 2p(p + q) cos β (2)

Then we need to manipulate these two equations in order to reach the desired result. There
are several ways to do this; we show two of them.

Approach 1: Determining the cosines and substituting

Equation (1) gives, upon rearranging:

p2 + (p − q)2 − (p + q)2
cos α =
2p(p − q)
p + (p − 2pq + q 2 ) − (p2 + 2pq + q 2 )
2 2
=
2p(p − q)
2
p − 4pq
=
2p(p − q)
p − 4q
= . (3)
2(p − q)

12
332
Likewise, equation (2) yields:

p2 + (p + q)2 − (p − q)2
cos β =
2p(p + q)
p + (p + 2pq + q 2 ) − (p2 − 2pq + q 2 )
2 2
=
2p(p + q)
2
p + 4pq
=
2p(p + q)
p + 4q
= . (4)
2(p + q)

Using equations (3) and (4), we now evaluate 4(1 − cos α)(1 − cos β) and cos α + cos β:
  
p − 4q p + 4q
4(1 − cos α)(1 − cos β) = 4 1 − 1−
2(p − q) 2(p + q)
p + 2q p − 2q
= 4. .
2(p − q) 2(p + q)
p2 − 4q 2
= 2
p − q2
and
p − 4q p + 4q
cos α + cos β = +
2(p − q) 2(p + q)
(p − 4q)(p + q) + (p + 4q)(p − q)
=
2(p − q)(p + q)
p − 3pq − 4q 2 + p2 + 3pq − 4q 2
2
=
2(p2 − q 2 )
2(p2 − 4q 2 )
=
2(p2 − q 2 )
p2 − 4q 2
= .
p2 − q 2
Therefore we have the required equality

4(1 − cos α)(1 − cos β) = cos α + cos β. (∗)

Approach 2: Determining q/p and equating

From equation (1) above, we can expand to get

p2 + 2pq + q 2 = p2 + p2 − 2pq + q 2 − 2p(p − q) cos α,

so that
p2 − 4pq = 2p(p − q) cos α.
We now divide by p2 to get

1 − 4q/p = 2(1 − q/p) cos α.

13
333
We can now rearrange this to find q/p:

(2 cos α − 4)q/p = 2 cos α − 1,

so
q 2 cos α − 1
= .
p 2 cos α − 4

Doing the same for equation (2) gives:

p2 − 2pq + q 2 = p2 + p2 + 2pq + q 2 − 2p(p + q) cos β,

so that
p2 + 4pq = 2p(p + q) cos β.
Again, dividing by p2 brings us to

1 + 4q/p = 2(1 + q/p) cos β,

yielding
(4 − 2 cos β)q/p = 2 cos β − 1,
so
q 2 cos β − 1
= .
p 4 − 2 cos β

Equating these two expressions for q/p now gives us

2 cos α − 1 2 cos β − 1
= ,
2 cos α − 4 4 − 2 cos β

so that (cross-multiplying and dividing by two):

(2 cos α − 1)(2 − cos β) = (cos α − 2)(2 cos β − 1).

Now we expand the brackets to get

4 cos α − 2 − 2 cos α cos β + cos β = 2 cos α cos β − 4 cos β − cos α + 2,

so that
4 − 4 cos α − 4 cos β + 4 cos α cos β = cos α + cos β,
and the left hand side factorises to give us our desired result:

4(1 − cos α)(1 − cos β) = cos α + cos β.

3
In the case α = 2β, show that cos β = 4
and hence find the ratio of the lengths of the
sides of the triangle.

Substituting α = 2β into (∗) gives:

4(1 − cos 2β)(1 − cos β) = cos 2β + cos β.

14
334
We use the double angle formula for cos 2β to write this expression in terms of cos β, giving:
4(2 − 2 cos2 β)(1 − cos β) = 2 cos2 β + cos β − 1,
so
8(1 + cos β)(1 − cos β)2 = (2 cos β − 1)(cos β + 1).
Since cos β 6= −1, we can divide by cos β + 1 to get
8(1 − cos β)2 = 2 cos β − 1,
so we can rearrange to get
8 cos2 β − 18 cos β + 9 = 0,
which factorises as
(4 cos β − 3)(2 cos β − 3) = 0.
3
Since cos β = 2
is impossible, we must have cos β = 34 , as required.
We now substitute this result into equation (2) to get
(p − q)2 = p2 + (p + q)2 − 2p(p + q). 34 .
Expanding this gives
p2 − 2pq + q 2 = p2 + p2 + 2pq + q 2 − 32 p2 − 32 pq,
so
1 2
2
p − 52 pq = 0,
which gives p = 5q. Hence the side lengths are p − q = 4q, p = 5q and p + q = 6q, which are
in the ratio 4 : 5 : 6.
An alternative way to do this last part is as √ follows.√We have α = 2β,√so cos α = 2 cos2 β−1 =
2 · ( 34 )2 − 1 = 18 . It follows that sin α = 18 63 = 38 7 and sin β = 14 7. We can now use the
sine rule to get
a c
= ,
sin A sin C
or a/c = sin A/ sin C. It follows that
p−q sin β
=
p+q sin α
1
√
7
= 43 √
8
7
2
= ,
3
giving 3(p − q) = 2(p + q), or p = 5q. The rest of the result follows as above.
A third way of doing this, and arguably the simplest, is to substitute into equation (4),
which gives:
3 p + 4q
= .
4 2(p + q)
Multiplying both sides by 4(p + q) to clear the fractions gives
3(p + q) = 2(p + 4q),
so that p = 5q. The rest of the argument again follows as above.

15
335
Question 5

A right circular cone has base radius r, height h and slant length `. Its volume V , and
the area A of its curved surface, are given by

V = 13 πr2 h, A = πr`.

(i) Given that A is fixed and√r is chosen so that V is at its stationary value, show that
A2 = 3π 2 r4 and that ` = 3 r.

Since A is fixed and r is allowed to vary, we rearrange A = πr` as ` = A/πr. Also, we can
draw a side view of the cone to determine the relationship between r, h and `:

`
h

So clearly, `2 = h2 + r2 .
Substituting this into ` = A/πr gives

A2
`2 = h2 + r2 = ,
π 2 r2
which we can rearrange to give h2 in terms of r and A:

A2
h2 = − r2 .
π 2 r2

Now V = 13 πr2 h, so we have

V 2 = 19 π 2 r4 h2
π 2 r4
 2 
A 2
= −r
9 π 2 r2
= 19 (A2 r2 − π 2 r6 ).

(Working with V 2 rather than just V allows us to avoid square roots.)

Differentiating with respect to r gives
dV
2V = 19 (2A2 r − 6π 2 r5 ).
dr
When V is at its stationary value, dV /dr = 0, so we require 2A2 r − 6π 2 r5 = 0. As r 6= 0,
we must have 6π 2 r4 = 2A2 , or A2 = 3π 2 r4 , as wanted.

16
336
Substituting this into our formula for `2 gives

A2
`2 =
π 2 r2
3π 2 r4
= 2 2
π r
= 3r2 ,
√
so ` = 3 r, as we wanted to show.

(ii) Given, instead, that V is fixed and r is chosen so that A is at its stationary value,
find h in terms of r.

We have `2 = h2 + r2 = A2 /π 2 r2 and V = 13 πr2 h. This time, V is fixed, so h = 3V /πr2 .

Thus

A2 = π 2 r2 (h2 + r2 )
 2 
2 2 9V 2
=π r +r
π 2 r4
9V 2
= 2 + π 2 r4
r
Differentiating as before gives

dA 18V 2
2A = − 3 + 4π 2 r3 ,
dr r
√
so dA/dr = 0 when 4π 2 r6 = 18V 2 , so 2πr3 = 3V 2. Finally, substituting this into our
formula h = 3V /πr2 gives
√
2πr3 / 2
h= 2
√ πr
= 2 r.

17
337
Question 6

(i) Show that, for m > 0,

m
x2 (m − 1)3 (m + 1)
Z
dx = + ln m .
1/m x+1 2m2

We note that the numerator of the fraction (x2 ) has a higher degree than the denominator
(x + 1), so we divide them first, getting x2 = (x + 1)(x − 1) + 1, so our integral becomes
Z m Z m
x2 1
dx = x−1+ dx
1/m x + 1 1/m x+1
h im
1 2
= 2 x − x + ln |x + 1|
1/m

= 2 m − m + ln |m + 1| − 12 m−2 − m−1 + ln |(1/m) + 1|

1 2

m4 − 2m3 − 1 + 2m m+1
= 2
+ ln
2m (1/m) + 1
3 2
(m + 1)(m − 3m + 3m − 1) m(m + 1)
= + ln
2m2 1+m
3
(m + 1)(m − 1)
= + ln m.
2m2
(An alternative approach is to use the substitution u = x + 1, which leads to exactly the
same result.)

(ii) Show by means of a substitution that

Z m Z m n−1
1 u
n
dx = du .
1/m x (x + 1) 1/m u + 1

Comparing the two integrals suggests that we should try the substitution u = 1/x. If we do
this, we get x = 1/u and dx/du = −1/u2 . Also, the limits x = 1/m and x = m become
u = m and u = 1/m respectively. So we have
Z m Z 1/m
1 1 dx
dx = du
1/m xn (x + 1) m (1/u)n (1/u + 1) du
1/m
un −1
Z
= du
m 1/u + 1 u2
1/m
un
Z
=− du
m u(1 + u)
Z m n−1
u
= du.
1/m u + 1

18
338
 (iii) Evaluate:
2
x5 + 3
Z
(a) dx .
1/2 x3 (x + 1)

This clearly relies on the earlier parts of the question, where m = 2. We can break the
integral into two parts and use the results of (i) and (ii) as follows:
Z 2 Z 2 Z 2
x5 + 3 x5 1
3
dx = 3
dx + 3 3
dx
1/2 x (x + 1) 1/2 x (x + 1) 1/2 x (x + 1)
Z 2 Z 2
x2 u2
= dx + 3 du
1/2 x + 1 1/2 u + 1
Z 2
x2
=4 dx
1/2 x + 1
(m + 1)(m − 1)3
 
=4 + ln m
2m2
= 4( 38 + ln 2)
= 32 + 4 ln 2.

An alternative way to approach this question is to ignore what has gone before and to use
partial fractions. We first divide to get

x5 + 3 x3 + 3
= x − 1 +
x3 (x + 1) x3 (x + 1)
and then express the final term using partial fractions:

x3 + 3 3 3 3 2
= − + − .
x3 (x + 1) x x2 x3 x + 1
Integrating then gives:
Z 2 Z 2
x5 + 3 3 3 3 2
3
dx = x − 1 + − + − dx
1/2 x (x + 1) 1/2 x x2 x3 x + 1
 2 2
x 3 3
= − x + 3 ln x + − 2 − 2 ln(x + 1)
2 x 2x 1/2

= (2 − 2 + 3 ln 2 + 32 − 38 − 2 ln 3) − ( 18 − 12 + 3 ln 12 + 6 − 6 − 2 ln 32 )
= 32 + 4 ln 2.

(iii) Evaluate:
2
x5 + x3 + 1
Z
(b) dx .
1 x3 (x + 1)

19
339
It is no longer so obvious how to proceed, as the limits are not of the form 1/m to m. So
we break up the integral as in (a) and repeat the substitution of part (ii) once again, noting
that the only change here is that the limits are different. We have:
Z 2 Z 2
x5 x2
3
dx = dx,
1 x (x + 1) 1 x+1

and there is not much we can do at this point, short of evaluating this integral as in part (i).
Next, we have
2 2
x3
Z Z
1
dx = dx
1 x3 (x + 1) 1 x+1
 2
= ln |x + 1| 1
= ln 3 − ln 2.

The third part gives us

2 1
u2
Z Z
1
3
dx = du,
1 x (x + 1) 1/2 u+1
by using the substitution of part (ii), but noting the the limits transform into 1/1 = 1 and
1/2, which are then reversed by the minus sign.
Adding all three terms and using part (i) with m = 2 now gives:
Z 2 5 Z 2 Z 1
x + x3 + 1 x2 u2
dx = dx + ln 3 − ln 2 + du
1 x3 (x + 1) 1 x+1 1/2 u + 1
Z 2
x2
= dx + ln 3 − ln 2
1/2 x + 1
3
= 8
+ ln 2 + ln 3 − ln 2
3
= 8
+ ln 3.

Again, this question can also be approached using partial fractions. Dividing gives
x5 + x 3 + 1 2x3 + 1
= x − 1 +
x3 (x + 1) x3 (x + 1)
and then partial fractions expansion gives us:
2x3 + 1 1 1 1 1
3
= − 2+ 3+ .
x (x + 1) x x x x+1
We now integrate to reach our final answer:
Z 2 Z 2
x5 + 3 1 1 1 1
3
dx = x−1+ − 2 + 3 + dx
1 x (x + 1) 1 x x x x+1
 2 2
x 1 1
= − x + ln x + − 2 + ln(x + 1)
2 x 2x 1
= (2 − 2 + ln 2 + 2 − 18 + ln 3) − ( 12 − 1 + 0 + 1 − 12 + ln 2)
1

= 38 + ln 3.

20
340
Question 7

Show that, for any integer m,

Z 2π
1
ex cos mx dx = 2π


e − 1 .
0 m2 + 1

This is a standard integral with standard techniques for solving it. The approach used here
is not the only possible one, but is fairly general.
R 2π
We write I = 0 ex cos mx dx and apply integration by parts twice, taking great care of the
signs, as follows:
Z 2π
I= ex cos mx dx
0
h 1 i2π Z 2π 1
x
= e · sin mx − ex · sin mx dx
m 0 0 m
Z 2π
1 2π 1 0 1
ex · sin mx dx



= m e sin 2πm − m e sin 0 −
0 m
Z 2π
1
=0− ex · sin mx dx
0 m
h 1 i2π Z 2π 1
x
= − e · 2 (− cos mx) + ex · 2 (− cos mx) dx
m 0 0 m
 1   1  Z 2π 1
= 2
e2π cos 2πm − 2
e0 cos 0 − ex · 2 cos mx dx
m m 0 m
1 2π
1
= 2 e − 1 − 2 I.
m m
Multiplying throughout by m2 gives

m2 I = e2π − 1 − I,

and now adding I to both sides and dividing by m2 + 1 gives the desired result.

We performed these integrations by writing ex cos mx dx in the form v du

R R
dx
dx, where
du x du x
dx
= cos mx and v = e . We could equally well have chosen dx = e and v = cos mx, and
would have ended up with the same conclusion.

(i) Expand cos(A + B) + cos(A − B). Hence show that

Z 2π
ex cos x cos 6x dx = 650
19
e2π − 1 .


0

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341
We have

cos(A + B) + cos(A − B) = cos A cos B − sin A sin B + cos A cos B + sin A sin B
= 2 cos A cos B.

Matching this to the integral we have been given, we set A = 6x and B = x (so that A − B
is positive, though this is not critical), giving

2 cos x cos 6x = cos 7x + cos 5x.

Thus our integral becomes

Z 2π
1 2π x
Z
x
e cos x cos 6x dx = e (cos 7x + cos 5x) dx
0 2 0
 
1 1 2π

1 2π


= e −1 + 2 e −1
2 72 + 1 5 +1
 
1 1 1
e2π − 1


= +
2 50 26
1 26 + 50 2π

= · e −1
2 1300

= 650 e2π − 1 ,
19

as required.

Z 2π
(ii) Evaluate ex sin 2x sin 4x cos x dx.
0

We are clearly asked to do the same type of trick again. Here is one way to proceed.

We are looking for the product of two sines, and this appears in the compound angle formula
for cosine. So we consider

cos(A + B) − cos(A − B) = cos A cos B − sin A sin B − cos A cos B − sin A sin B
= −2 sin A sin B.

We can therefore write 2 sin 2x sin 4x = cos 2x − cos 6x, using A = 4x and B = 2x. This
gives the integral as
Z 2π
1 2π x
Z
x
e sin 2x sin 4x cos x dx = e (cos 2x − cos 6x) cos x dx
0 2 0
1 2π x 1 2π x
Z Z
= e cos 2x cos x dx − e cos 6x cos x dx.
2 0 2 0

Now the second integral is exactly the one we evaluated in (i), and the first integral can be
approached in the same way, with A = 2x and B = x, so 2 cos 2x cos x = cos 3x + cos x.

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342
Therefore
2π
1 2π x
Z Z
x
e cos 2x cos x dx = e (cos 3x + cos x) dx
0 2 0
 
1 1 2π

1 2π


= e −1 + 2 e −1
2 32 + 1 1 +1
 
1 1 1
e2π − 1


= +
2 10 2
3
e2π − 1 .


= 10

Finally, subtracting the two integrals gives

Z 2π
1 2π x 1 2π x
Z Z
x
e sin 2x sin 4x cos x dx = e cos 2x cos x dx − e cos 6x cos x dx
0 2 0 2 0
= 12 · 10
3
e2π − 1 − 21 · 650
19
e2π − 1




2π
= 12 195 19


650
− 650
e − 1
2π
= 12 · 176


650
e − 1
44
e2π − 1 .


= 325

23
343
Question 8

(i) The equation of the circle C is

(x − 2t)2 + (y − t)2 = t2 ,

where t is a positive number. Show that C touches the line y = 0.

This is a circle with centre (2t, t) and radius t, therefore it touches the x-axis, which is
distance t from the centre.
Alternatively, we are looking to solve the simultaneous equations

(x − 2t)2 + (y − t)2 = t2
y = 0.

Substituting the second equation into the first gives (x − 2t)2 + t2 = t2 , or (x − 2t)2 = 0.
Since this only has one solution, x = 2t, the line must be tangent to the circle.

(i) (cont.)
Let α be the acute angle between the x-axis and the line joining the origin to the
centre of C. Show that tan 2α = 43 and deduce that C touches the line 3y = 4x.

We begin by drawing a sketch of the situation.

t
α
α
2t

Clearly, therefore, tan α = t/2t = 12 , so

2 tan α
tan 2α =
1 − tan2 α
1
=
1 − 14
= 43 .

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344
From the sketch, it is clear that by symmetry, the line through the origin which makes an
angle of 2α with the x-axis touches the circle C. Since the gradient of this line is tan 2α = 43
and it passes through the origin, it has equation y = 43 x, or 3y = 4x.

(ii) Find the equation of the incircle of the triangle formed by the lines y = 0, 3y = 4x
and 4y + 3x = 15.
Note: The incircle of a triangle is the circle, lying totally inside the triangle, that
touches all three sides.

This circle has the properties described in part (i), in that it touches both the x-axis (i.e.,
the line y = 0) and the line 3y = 4x, and it lies above the x-axis, so it must have the form
given. The only remaining condition is that it must touch the line 4y + 3x = 15. Two circles
with centre (2t, t) touch these three lines, but only one of them lies within the triangle, as
illustrated on this sketch, so we must take the one with the smaller value of t:

3y = 4x

4y + 3x = 15

O A

Approach 1: Algebraic substitution

Substituting 4y + 3x = 15 into the equation of C will give us the intersections of the line
and the circle. The line will be a tangent to C if and only if the discriminant of the resulting
quadratic equation is 0. Doing this, from the equation of C:

(x − 2t)2 + (y − t)2 = t2

we get:
(x − 2t)2 + ((15 − 3x)/4 − t)2 = t2 .
Multiplying both sides by 42 and expanding gives:

16x2 − 64tx + 64t2 + (15 − 3x)2 − 8(15 − 3x)t + 16t2 = 16t2 ,

so
16x2 − 64tx + 64t2 + 225 − 90x + 9x2 − 120t + 24tx + 16t2 = 16t2 .
Collecting terms in x gives:

25x2 − (90 + 40t)x + 225 − 120t + 64t2 = 0.

25
345
Since this has a repeated root, as the line is required to be tangent to the circle, the discrim-
inant must be zero, so

(90 + 40t)2 − 4 × 25(225 − 120t + 64t2 ) = 0,

so
(9 + 4t)2 − (225 − 120t + 64t2 ) = 0,
or
16t2 + 72t + 81 − (225 − 120t + 64t2 ) = 0,
giving
−48t2 + 192t − 144 = 0.
Dividing all the terms by −48 gives t2 − 4t + 3 = 0, which factorises as (t − 1)(t − 3) = 0.
Thus the two circles in the sketch are given by t = 1 and t = 3, and the incircle is clearly
the one with t = 1. So the incircle has equation

(x − 2)2 + (y − 1)2 = 1.

Alternative 2: Distance of a point from a line

In the formula book, we are given the result:

|n1 α + n2 β + n3 γ + d|
The perpendicular distance of (α, β, γ) from n1 x+n2 y+n3 z+d = 0 is p .
n21 + n22 + n23
Therefore the distance of the centre of the circle, (2t, t, 0), from the line 3x + 4y + 0z − 15 = 0
is given by
|3 × 2t + 4 × t + 0 − 15| |10t − 15|
√ = = |2t − 3|.
2
3 +4 +02 2 5
But the line is tangent to the circle, which has radius t, so we must have

|2t − 3| = t.

This equation has two solutions: 2t − 3 = t gives t = 3, and 2t − 3 = −t gives t = 1. As we

require the smaller solution, we must have t = 1, so that the incircle has equation

(x − 2)2 + (y − 1)2 = 1.

Alternative 3: Finding another angle bisector

As in part (i), using the notation in the diagram above, the line from A to I bisects the
angle OAB. So if we write OÂI = β, then OÂB = 2β.
Now tan 2β = 34 , so we solve
2 tan β 3
tan 2β = 2
=
1 − tan β 4
to find tan β. We get the quadratic equation

3 − 3 tan2 β = 8 tan β,

26
346
 1
which factorises as (3 tan β − 1)(tan β + 3) = 0, so tan β = 3
or tan β = −3. But β is acute,
so tan β = 31 . Thus the line AI has equation

y − 0 = − 13 (x − 5)

which can be written as 3y + x = 5.

The point I has coordinates (2t, t), and substituting this in gives 3t + 2t = 5, so t = 1 as
before.

Alternative 4: Using Euclidean geometry

We find that the intersection of the line 4y + 3x = 15 with the x-axis is at (5, 0). We
can also find the intersection of the two lines 4y + 3x = 15 and 3y = 4x by solving them
simultaneously:

12y + 9x = 45
12y − 16x = 0,

so 25x = 45, or x = 95 , so y = 12
5
. Thus the side of the triangle from the origin to the point
of intersection has length 3, the side from (5, 0) to the point of intersection has length 4
and the base has length 5 (using Pythagoras), so we have a 3-4-5 triangle. (The triangle is
right-angled as 4y + 3x = 15 and 3y = 4x are perpendicular.)
Consider now this figure, where we have drawn radii from the incentre to the three sides of
the triangle. Note that, since the triangle’s sides are tangents to the circle, the radii meet
at right angles, so ICBD is a square with all sides equal to the radius, t.

B
t t
C
D

t t 4−t
3−t I
t

O E A

Now AC = AE = 4 − t since these are both tangents from A to the circle; likewise, OE =
OD = 3 − t, so OA = OE + EA = (3 − t) + (4 − t) = 7 − 2t. But OA = 5, so 2t = 2 and
t = 1, giving the radius and hence the equation of the incircle.

Alternative 5: More Euclidean geometry

We begin in the same way by finding the side lengths of the triangle. We can then quote
the result that if r is the incircle radius and the side lengths of the triangle are a, b, c, then
these are related by
Area = 12 r(a + b + c).
1
In this case, the area is 2
× 3 × 4 = 6 and 12 (a + b + c) = 6, so r = 1 as required.

27
347
Question 9

Two particles P and Q are projected simultaneously from points O and D, respectively,
where D is a distance d directly above O. The initial speed of P is V and its angle of
projection above the horizontal is α. The initial speed of Q is kV , where k > 1, and
its angle of projection below the horizontal is β. The particles collide at time T after
projection.
Show that cos α = k cos β and that T satisfies the equation

(k 2 − 1)V 2 T 2 + 2dV T sin α − d2 = 0.

We begin by drawing a sketch showing the initial situation.

D
β
Q
kV
+ve
d

V
P
α
O

For components, we will write uP and uQ for the horizontal components of the velocities
of P and Q respectively, and vP and vQ for the vertical components (measured upwards).
For the displacements from O, we write xP and xQ for the horizontal components and yP
and yQ for the vertical components.
Resolving horizontally, using the “suvat” equations, we have

uP = V cos α,
uQ = kV cos β;
xP = V t cos α,
xQ = kV t cos β.

Likewise, vertically we have

vP = V sin α − gt,
vQ = −kV sin β − gt;
yP = V t sin α − 12 gt2 ,
yQ = d − kV t sin β − 12 gt2 .

At time T , the particles collide, so xP = xQ , giving

V T cos α = kV T cos β,

28
348
so that cos α = k cos β.
Also, yP = yQ , so
V T sin α − 12 gT 2 = d − kV T sin β − 12 gT 2 ,
which gives
V T sin α = d − kV T sin β.
Now k 2 sin2 β = k 2 − k 2 cos2 β = k 2 − cos2 α from the above, so we rearrange and then square
both sides of the previous equation to make use of this conclusion:

kV T sin β = d − V T sin α,

so
k 2 V 2 T 2 sin2 β = (d − V T sin α)2 ,
which then gives us

V 2 T 2 (k 2 − cos2 α) = d2 − 2dV T sin α + V 2 T 2 sin2 α.

Finally, subtracting the right hand side from the left gives

V 2 T 2 (k 2 − cos2 α − sin2 α) − d2 + 2dV T sin α = 0,

or
V 2 T 2 (k 2 − 1) + 2dV T sin α − d2 = 0, (∗)
as required.

Given that the particles collide when P reaches its maximum height, find an expression
for sin2 α in terms of g, d, k and V , and deduce that

gd 6 (1 + k)V 2 .

At the maximum height, we have vP = 0, so V sin α = gT . Substituting for T in (∗) gives

us  2  
2 V sin α 2 V sin α
V (k − 1) + 2dV sin α − d2 = 0.
g g
Multiplying through by g 2 and expanding brackets gives

(k 2 − 1)V 4 sin2 α + 2gdV 2 sin2 α − g 2 d2 = 0.

Thus
g 2 d2
sin2 α = .
(k 2 − 1)V 4 + 2gdV 2

For the inequality, the only thing we know for certain is that sin2 α 6 1, and this gives

g 2 d2
6 1,
(k 2 − 1)V 4 + 2gdV 2
so
g 2 d2 6 (k 2 − 1)V 4 + 2gdV 2 .

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349
It is not immediately clear how to continue, so we try completing the square for gd:

(gd − V 2 )2 − V 4 6 (k 2 − 1)V 4 ,

so that
(gd − V 2 )2 6 k 2 V 4 .
Now if a2 6 b2 , then a 6 |b|, so
gd − V 2 6 |kV 2 |.
But kV 2 > 0, so we are almost there:

gd − V 2 6 kV 2 ,

which finally gives us the required

gd 6 (1 + k)V 2 .

30
350
Question 10

A triangular wedge is fixed to a horizontal surface. The base angles of the wedge are α
and π2 − α. Two particles, of masses M and m, lie on different faces of the wedge, and are
connected by a light inextensible string which passes over a smooth pulley at the apex of
the wedge, as shown in the diagram. The contacts between the particles and the wedge
are smooth.

m
M

π
α 2
−α

m
(i) Show that if tan α > the particle of mass M will slide down the face of the wedge.
M

We start by drawing the forces on the picture, and labelling the masses as A (of mass M )
and B (of mass m). We let T be the tension in the string.

T
NA NB
T a
B
A Mg
a
mg π
−α
α 2

We now resolve along the faces of the wedge at A and B. (It turns out that resolving
normally to the face doesn’t help at all for this problem.)
RA (%) T − M g sin α = −M a (1)
RB (-) T − mg cos α = ma (2)
We are not interested in the tension in the string, so we subtract these equations (as (2)−(1))
to eliminate T , yielding
M g sin α − mg cos α = ma + M a.
Thus, dividing by M + m gives
M g sin α − mg cos α
a= . (3)
M +m
The mass M will slide down the slope if and only if a > 0, that is if and only if
M g sin α − mg cos α > 0.
Rearranging and dividing by g cos α gives M tan α > m, or
m
tan α > .
M

31
351
 2m
(ii) Given that tan α = , show that the magnitude of the acceleration of the particles
M
is
g sin α
tan α + 2
and that this is maximised at 4m = M 3 .
3

We simply substitute 2m/M = tan α into our formula (3) for a to find the magnitude of the
acceleration:
M g sin α − mg cos α
a=
M +m
2g sin α − (2m/M )g cos α
= multiplying by 2/M
2 + (2m/M )
2g sin α − g tan α cos α
=
2 + tan α
2g sin α − g sin α
=
2 + tan α
g sin α
=
2 + tan α

To maximise this with respect to α, we differentiate with respect to α (using the quotient
rule) and solve da/dα = 0:

da (2 + tan α)g cos α − sec2 α.g sin α

=
dα (2 + tan α)2
g(2 cos α + sin α − sec2 α sin α)
=
(2 + tan α)2
=0

so we require
2 cos α + sin α − sec2 α sin α = 0.
We can simplify this by dividing through by cos α, so that we are able to express everything
in terms of tan α:
2 + tan α − sec2 α tan α = 0,
so
2 + tan α − (1 + tan2 α) tan α = 0,
so that tan3 α = 2. Remembering that tan α = 2m/M , we finally get

8m3
= 2,
M3
which leads immediately to M 3 = 4m3 .
We finally need to ensure that this stationary point gives us a maximum; this is clear, since
as α → 0, a → 0 and as α → π2 , a → 0.

32
352
Question 11

Two particles move on a smooth horizontal table and collide. The masses of the particles
are m and M . Their velocities before the collision are ui and vi, respectively, where i is a
unit vector and u > v. Their velocities after the collision are pi and qi, respectively. The
coefficient of restitution between the two particles is e, where e < 1.

(i) Show that the loss of kinetic energy due to the collision is
1
2
m(u − p)(u − v)(1 − e),

and deduce that u > p.

Before the collision:

m M
ui vi

After the collision:

m M
pi qi

Conservation of momentum gives:

mu + M v = mp + M q. (1)

Newton’s Law of Restitution gives:

q − p = e(u − v). (2)

Now the loss of kinetic energy due to the collision is

E = initial KE − final KE
= ( 12 mu2 + 12 M v 2 ) − ( 12 mp2 + 12 M q 2 )
= 12 m(u2 − p2 ) + 12 M (v 2 − q 2 )
= 12 m(u − p)(u + p) + 12 M (v − q)(v + q).

Now from (1), we get M (v − q) = m(p − u), so we get

E = 12 m(u − p)(u + p) + 12 m(p − u)(v + q)

= 12 m(u − p) (u + p) − (v + q)

= 12 m(u − p) (u − v) − (q − p)

= 12 m(u − p) (u − v) − e(u − v)

= 12 m(u − p)(u − v)(1 − e).

Now, since the loss of energy cannot be negative, we have E > 0. But we are given that
e < 1 and u > v, so we must have u − p > 0, or u > p.

33
353
 (ii) Given that each particle loses the same (non-zero) amount of kinetic energy in the
collision, show that
u + v + p + q = 0,
and that, if m 6= M ,
(M + 3m)u + (3M + m)v
e= .
(M − m)(u − v)

The first particle loses an amount of kinetic energy equal to 12 m(u2 − p2 ); the second loses
1
2
M (v 2 − q 2 ), so we are given
1
2
m(u2 − p2 ) = 12 M (v 2 − q 2 ),

so
m(u2 − p2 ) − M (v 2 − q 2 ) = 0.
Again, we use M (v − q) = m(p − u), so that

m(u2 − p2 ) − M (v 2 − q 2 ) = m(u + p)(u − p) − M (v + q)(v − q)

= m(u + p)(u − p) − m(v + q)(p − u)
= m(p + q + u + v)(u − p)
= 0.

Since the amount of kinetic energy lost is non-zero, we have E > 0 (in the notation of
part (i)), so that u > p. Thus we must have p + q + u + v = 0.
We can also equate the loss of energy of each particle with 12 E, so:
1
2
m(u2 − p2 ) = 14 m(u − p)(u − v)(1 − e),

giving
u + p = 12 (u − v)(1 − e). (3)
We now need to eliminate p, and rearrange to get an expression for e. Equation (2) gives

M q − M p = M e(u − v),

and subtracting this from (1), mp + M q = mu + M v, gives

(m + M )p = (m − M e)u + (M + M e)v.

We multiply (3) by m + M to get:

(m + M )p + (m + M )u = 12 (M + m)(u − v)(1 − e),

and then substitute in our expression for (m + M )p to give us:

(m − M e)u + (M + M e)v + (m + M )u = 12 (M + m)(u − v)(1 − e).

We expand and rearrange to collect terms which are multiples of e:

2(mu − M ue + M v + M ve + mu + M u) = (M u − M v + mu − mv)(1 − e)

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354
so
(2M v − 2M u + M u − M v + mu − mv)e = M u − M v + mu − mv − 4mu − 2M v − 2M u,
which leads to
(M − m)(v − u)e = −M u − 3M v − 3mu − mv.

We can write the right hand side as −(M + 3m)u − (3M + m)v, so that assuming M 6= m,
we can divide by (M − m)(v − u) to get

(M + 3m)u + (3M + m)v

e= ,
(M − m)(v − u)

as we wanted.

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355
Question 12

Prove that, for any real numbers x and y, x2 + y 2 > 2xy.

We can rearrange the inequality to get

x2 − 2xy + y 2 > 0.

But the left hand side is just (x − y)2 , so this inequality becomes (x − y)2 > 0. This is clearly
true, as any real number squared is non-negative, and since this is just a rearrangement of
the desired inequality, that must also be true.

(i) Carol has two bags of sweets. The first bag contains a red sweets and b blue sweets,
whereas the second bag contains b red sweets and a blue sweets, where a and b are
positive integers. Carol shakes the bags and picks one sweet from each bag without
looking. Prove that the probability that the sweets are of the same colour cannot
exceed the probability that they are of different colours.

We can draw a tree diagram to represent this situation:

b
a+b R

a R a
a+b a+b
B
b b
a+b a+b R

B a
a+b
B

So
a b b a
P(same colour) = · + ·
a+b a+b a+b a+b
2ab
=
(a + b)2
a a b b
P(different colours) = · + ·
a+b a+b a+b a+b
a2 + b2
= .
(a + b)2

Since a2 + b2 > 2ab, it follows that the probability that the sweets are of the same colour
cannot exceed the probability that they are of different colours, as required.

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356
 (ii) Simon has three bags of sweets. The first bag contains a red sweets, b white sweets
and c yellow sweets, where a, b and c are positive integers. The second bag contains
b red sweets, c white sweets and a yellow sweets. The third bag contains c red sweets,
a white sweets and b yellow sweets. Simon shakes the bags and picks one sweet from
each bag without looking. Show that the probability that exactly two of the sweets
are of the same colour is
3(a2 b + b2 c + c2 a + ab2 + bc2 + ca2 )
,
(a + b + c)3

and find the probability that the sweets are all of the same colour. Deduce that the
probability that exactly two of the sweets are of the same colour is at least 6 times
the probability that the sweets are all of the same colour.

We argue in the same way:

P(exactly 2 same) = P(RRW) + P(RRY) + P(RWR)+

P(RYR) + P(WRR) + P(YRR)+
P(WWR) + P(WWY) + P(WRW)+
P(WYW) + P(RWW) + P(YWW)+
P(YYR) + P(YYW) + P(YRY)+
P(YWY) + P(RYY) + P(WYY)
aba abb acc
= 3
+ 3
+ +
(a + b + c) (a + b + c) (a + b + c)3
aac bbc cbc
3
+ 3
+ +
(a + b + c) (a + b + c) (a + b + c)3
bcc bcb bba
3
+ 3
+ +
(a + b + c) (a + b + c) (a + b + c)3
baa aca cca
3
+ 3
+ +
(a + b + c) (a + b + c) (a + b + c)3
cac caa cbb
3
+ 3
+ +
(a + b + c) (a + b + c) (a + b + c)3
ccb aab bab
3
+ 3
+
(a + b + c) (a + b + c) (a + b + c)3
3(a2 b + b2 c + c2 a + ab2 + bc2 + ca2 )
= .
(a + b + c)3

More simply, the probability that all three are the same colour is given by

P(all 3 same) = P(RRR) + P(WWW) + P(YYY)

abc bca cab
= 3
+ 3
+
(a + b + c) (a + b + c) (a + b + c)3
3abc
= .
(a + b + c)3

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357
Now to find the inequality we want, we apply the initial inequality again: we have a2 +
b2 > 2ab, so a2 c + b2 c > 2abc. Similarly, b2 + c2 > 2bc, so b2 a + c2 a > 2abc, and finally,
c2 b + a2 b > 2abc. Thus

3(a2 b + b2 c + c2 a + ab2 + bc2 + ca2 ) 3(a2 c + b2 c + b2 a + c2 a + c2 b + a2 b)

=
(a + b + c)3 (a + b + c)3
3(2abc + 2abc + 2abc)
>
(a + b + c)3
6(3abc)
= ,
(a + b + c)3

showing that the probability that exactly two of the sweets are of the same colour is at least
6 times the probability that the sweets are all of the same colour.

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358
Question 13

I seat n boys and 3 girls in a line at random, so that each order of the n + 3 children is as
likely to occur as any other. Let K be the maximum number of consecutive girls in the
line so, for example, K = 1 if there is at least one boy between each pair of girls.

(i) Find P(K = 3).

There are two equivalent ways to approach this question: either to regard the boys and
girls as all distinct, so that there are (n + 3)! possible orders, or to regard all boys as
n+3


indistinguishable and all girls as indistinguishable, so that there are 3 possible orders.
We use the latter approach here.

We note that, regarding the boys as indistinguishable and the girls as indistinguishable,
there are  
n+3 1
= (n + 3)(n + 2)(n + 1)
3 6
possible arrangements of the students.
If K = 3, this means that all three girls are adjacent. So the situation must be that there
are r boys, followed by 3 girls, followed by n − r boys, where r = 0, 1, . . . , n, so there are
n + 1 possibilities.
Thus
n+1
P(K = 3) = 1
6
(n
+ 3)(n + 2)(n + 1)
6
= .
(n + 2)(n + 3)

(ii) Show that

n(n − 1)
P(K = 1) = .
(n + 2)(n + 3)

Approach 1: Counting explicitly

To have K = 1, we must have each pair of girls separated by at least one boy, like this:

B. . . B} G |B.{z
| {z . . B} G |B.{z
. . B} G |B.{z
. . B}
r1 r2 r3 r4

where r1 > 0, r2 > 0, r3 > 0, r4 > 0 and r1 + r2 + r3 + r4 = n.

If r1 and r2 are fixed, then we must have r3 + r4 = n − (r1 + r2 ), so we can have r3 = 1, 2,
. . . , n − (r1 + r2 ), giving n − (r1 + r2 ) possibilities for r3 and r4 .

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359
Thus, if r1 is fixed, r2 could be 1, 2, . . . , n − r1 − 1 (but not n − r1 , as we must have r3 > 0).
Thus the number of possibilities for a fixed value of r1 is given by
n−r
X 1 −1 n−r
X 1 −1 n−r
X 1 −1

(n − r1 − r2 ) = (n − r1 ) − r2
r2 =1 r2 =1 r2 =1

= (n − r1 − 1)(n − r1 ) − 12 (n − r1 − 1)(n − r1 )
= 12 (n − r1 − 1)(n − r1 )
= 12 (n2 − 2nr1 + r12 − n + r1 ).

Now, r1 can take the values 0, 1, 2, . . . , n − 2 (as we need r1 > 0 and r2 > 0), giving the
total number of possibilities as
n−2
X
1
2
(n2 − 2nr1 + r12 − n + r1 )
r1 =0
n−2
X n−2
X n−2
X
= 1
2
(n2 − n) − 12 (2n − 1) r1 + 1
2
r12
r1 =0 r1 =0 r1 =0
1 2 1 1 1
= 2
(n − 1)(n − n) − 2 (2n − 1) · 2 (n − 2)(n − 1) + 12 (n − 2)(n − 1)(2n − 3)
1 2


= 12
(n − 1) 6(n − n) − 3(2n − 1)(n − 2) + (n − 2)(2n − 3)
= 1
12
(n − 1)(6n2 − 6n − 6n2 + 15n − 6 + 2n2 − 7n + 6)
= 1
12
(n − 1)(2n2 + 2n)
1
= 6
n(n − 1)(n + 1).

Thus we can finally deduce

1
− 1)(n + 1)
6
n(n
P(K = 1) = 1
(n
+ 3)(n + 2)(n + 1)
6
n(n − 1)
= .
(n + 2)(n + 3)

Approach 2: A combinatorial argument

We have to place each of the three girls either between two boys or at the end of the line,
and we cannot have two girls adjacent. We can think of the line as n boys with gaps between
them and at the ends, like this:
B B ... B B

Note that there are n + 1 gaps (one to the right of each
boy, and one at the left of the line).
Three of the gaps are to be filled with girls, giving n+1
3
= 16 (n + 1)n(n − 1) ways of choosing
them. Therefore
1
+ 1)n(n − 1)
6
(n
P(K = 1) = 1
(n
+ 3)(n + 2)(n + 1)
6
n(n − 1)
= .
(n + 2)(n + 3)

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360
Approach 3: Another combinatorial argument

We add one more boy at the right end of the line. In this way, we have a boy to the right of
each girl, as follows:
B. . . B} GB |B.{z
| {z . . B} GB |B.{z
. . B} GB |B.{z
. . B}
r1 r2 r3 r4
where this time, we have r1 > 0, r2 > 0, r3 > 0 and r4 > 0. Also, as there are now n + 1
boys, we have r1 + r2 + r3 + r4 = n + 1 − 3 = n − 2.
So if we think of GB as
one ‘person’, there are n − 2 boys and 3 GBs, giving n + 1 ‘people’
in total. There are n+1
3
= 16 (n + 1)n(n − 1) ways of arranging them, giving
1
6
+ 1)n(n − 1)
(n
P(K = 1) = 1
6
(n
+ 3)(n + 2)(n + 1)
n(n − 1)
= .
(n + 2)(n + 3)

(iii) Find E(K).

We could attempt to determine P(K = 2) directly, but it is far easier to note that K can
only take the values 1, 2 or 3. Thus

P(K = 2) = 1 − P(K = 1) − P(K = 3)

(n + 2)(n + 3) − n(n − 1) − 6
=
(n + 2)(n + 3)
n + 5n + 6 − n2 + n − 6
2
=
(n + 2)(n + 3)
6n
= .
(n + 2)(n + 3)
P
Since E(K) = k k.P(K = k), we can now calculate E(K):

E(K) = 1.P(K = 1) + 2.P(K = 2) + 3.P(K = 3)

n(n − 1) + 2 · 6n + 3 · 6
=
(n + 2)(n + 3)
2
n − n + 12n + 18
=
(n + 2)(n + 3)
2
n + 11n + 18
=
(n + 2)(n + 3)
(n + 2)(n + 9)
=
(n + 2)(n + 3)
n+9
= .
n+3

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361
 STEP II, Solutions
2009

42
362
1 Both graphs are symmetric in the lines y =  x , and x4 + y4 = u is also symmetric in the x- and
y-axes. These facts immediately enable us to write down the coordinates of B( , ), C(– , –)
and D(–  , – ). Remember to keep the cyclic order A, B, C, D correct, else this could lead to
silly calculational errors later on. The easiest way to show that ABCD is a rectangle is to work out
the gradients of the four sides (which turn out to be either 1 or –1) and then note that each pair of
adjacent sides is perpendicular using the “product of gradients = –1” result. Working with
distances is also a possible solution-approach but, on its own, only establishes that the
quadrilateral is a parallelogram. However, the next part requires you to calculate distances
anyhow, and we find that CB, DA have length ( + ) 2 while BA, DC are of length ( – ) 2 .
Multiplying these then give the area of ABCD as 2(2 – 2).

All of this is very straightforward, and the only tricky bit of work comes next. It is important to
think of  and  as particular values of x and y satisfying each of the two original equations. It is
then clear that (2 – 2)2 = 4 + 4 – 2(2 2) = u – 2v2, so that Area ABCD = 2 u  2v 2 .
Substituting u = 81, v = 4 into this formula then gives Area = 2 81  2  16 = 14, which is
intended principally as a means of checking that your answer is correct.

2(i) It is perfectly possible to differentiate a^(sin[ ex]) by using the Chain Rule (on a function of a
function of a function) but simplest to take logs. and use implicit differentiation. Then, setting
dy
= 0 and noting that  ex and ln a are non-zero, we are left solving the eqn. cos( ex) = 0 for
dx
the turning points. This gives ex = (2n + 1) 12   x = ln n  12  , y = a or 1a , depending upon
whether n is even or odd. Although not actually required at this point, it may be helpful to note
at this stage that the evens give maxima while the odds give minima. There is, however, a
much more obvious approach to finding the TPs that doesn’t require differentiation at all,
and that is to use what should be well-known properties of the sine function: namely, that
sin( . exp x )
ya has maxima when sin( e ) = 1, i.e.  e = (2n + 2 ) , and x = ln(2n + 2 ) for
x x 1 1

n = 0, 1, … , with ymax = a . Similarly, minima occur when sin( ex) = –1, i.e.  ex = (2n – 12 ) ,
and x = ln(2n – 12 ) for n = 1, 2, … , with ymin = 1a .

(ii) Using the addition formula for sin(A + B), and the approximations given, we have
sin( ex)  sin( +  x) = – sin( x)  –  x for small x,
leading to y  a  x  e  x ln a  1 –  x. ln a .

(iii) Firstly, we can note that, for x < 0, the curve has an asymptote y = 1 (as x  – , y  1+) .
Next, for x > 0, the curve oscillates between a and 1a , with the peaks and troughs getting ever 1
closer together . The work in (i) helps us identify the TPs: the first max. occurs when n = 0 at a
negative value of x [N.B. ln  12  < 0] at y = a; while the result in (ii) tells us that the curve is
approximately negative linear as it crosses the y-axis.

(iv) The final part provides the only really tricky part to the question , and a quick
diagram might be immensely useful here. Noting the relevant x-coordinates
x1 = ln 2k  32  , x2 = ln2k  12  , and x3 = ln2k  12  , a
1
a
the area is the sum of two trapezia (or rectangle – triangle) , and manipulating
 4k  1   4k  3  4   1  x1 x2 x3
ln   ln   ln1   leads to the final, given answer.
3 
 4k  3   4k  3   k4

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363
3 Using the “addition” formula for tan(A – B),
1  tan 2x cos 2x  sin 2x cos 2x  sin 2x cos 2x  sin 2x
LHS  tan 4  2x     
1  tan 2x cos 2x  sin 2x cos 2x  sin 2x cos 2x  sin 2x
1  2 sin 2x cos 2x 1  sin x
 (since c2 + s2 = 1)   sec x – tan x  RHS.
cos 2  sin 2
2 x 2 x
cos x
Alternatively, one could use the “t = tan( 12 -angle)” formulae to show that
1 t 2 2t (1  t ) 2 1  t 1  tan 2x
RHS       LHS.
1  t 2 1  t 2 (1  t )(1  t ) 1  t 1  tan 2x

(i) Setting x = 
4 in (*)  tan 8  2  1 . Then, using the addition formula for tan(A + B) with
3  2 1 3  2 1
 
, we have tan 1124  tan 3  8  =
A= 3 and B = 8

1  3 2 1  =
3  6 1
, as required.

(ii) Now, in the “spirit” of maths, one might reasonably expect that one should take the given
expression, rationalise the denominator (twice) and derive the given answer, along the lines …
3  2 1 3  2 1 1 3  6 1 2 2  3 3 1
    = 2 2  3 6.
1 3  6 1 3  6 1 3  6 3 1 3 1
However, with a given answer, it is perfectly legitimate merely to multiply across and verify that
 
3  6  1 2  2  3  6  3  2 1. 
11
(iii) Having got this far, the end is really very clearly signposted. Setting x = 24 in (*) gives

tan 48  sec 1124  tan 1124  1  t 2  t

 1  4  2  3  6  4 2  4 3  4 6  2 6  2 12  2 18 – 2  2  3  6   
= 15  10 2  8 3  6 6 – 2  2  3  6 
4(i) Writing p(x) – 1  q(x).(x – 1)5, where q(x) is a quartic polynomial, immediately gives p(1) = 1.

(ii) Diffg. using the product and chain rules leads to

p(x)  q(x).5(x – 1)4 + q(x).(x – 1)5  (x – 1)4.{5 q(x) + (x – 1) q(x)},
so that p(x) is divisible by (x – 1)4 .

(iii) Similarly, we have that p(x) is divisible by (x + 1)4 and p(– 1) = – 1 .

Thus p(x) is divisible by (x + 1) .(x – 1)  (x – 1) . However, p(x) is a polynomial of degree
4 4 2 4

eight, hence p(x)  k(x2 – 1)4 for some constant k . That is, p(x)  k x 8  4 x 6  6 x 4  4 x 2  1 .
 
Integrating term by term then gives p(x)  k 19 x 9  74 x 7  65 x 5  43 x 3  x + C, and use of both
315
p(1) = 1 and p(– 1) = – 1 help to find k and C; namely, k= and C = 0.
128

44
364
5 The very first bit is not just a giveaway mark, but rather a helpful indicator of the kind of result or
technique that may be used in this question:  
2
x  1  1  x  2 x  1 ; but pay attention to what
happens here. Most particularly, the fact that  
2
x  1  1  x  2 x  1 does NOT necessarily
mean that x  2 x  1  x  1  1 since positive numbers have two square-roots! Recall that
x 2  x and not just x. Notice that, during the course of this question, the range of values under
consideration switches from (5, 10) to  54 , 10 , and one doesn’t need to be particularly suspicious
to wonder why this is so. A modicum of investigation at the outset seems warranted here, as to
when things change sign.

(i) So … while x  2 x  1  x  1  1 seems a perfectly acceptable thing to write, since x  1 is

a necessary condition in order to be able to take square-roots at all here (for real numbers), simply
writing down that 
x  2 x 1   x 1 1  may cause a problem. A tiny amount of
exploration shows that x  1  1 changes from negative to positive around x = 2. Hence, in part
10

 2 dx = 2x
10
(i), we can ignore any negative considerations and plough ahead: I = 5
= 10.
5

(ii) Here in (ii), however, you should realise that the area requested is the sum of two portions, one of
which lies below the x-axis, and would thus contribute negatively to the total if you failed to take
this into account. Thus,
1 x 1 x 1 1
   
2 10 2 10
1
dx +  dx =  ( x  1)  1 dx +  1  ( x  1) 2 dx
 12
Area = 
1.25 x 1 2 x 1 1.25 2


= 2 x 1  x  2
1.25
 
 x  2 x  1 102 = 4¼ .

(iii) Now  
2
x  1  1  x  2  2 x  1  x  0 so we have no cause for concern here. Then
1 x 1  x 1 1
 
10 10
 12  12
I= 
x 1.25 x  1 x  1
dx =  (
x 1.25
x  1)  ( x  1) dx


= 2 x  1  2 x  1 110
.25
= 2 11  1   
6 If you don’t know about the Fibonacci Numbers by now, then … shame on you! Nevertheless, the
first couple of marks for writing down the next few terms must count as among the easiest on the
paper. (F1 = 1, F2 = 1), F3 = 2, F4 = 3, F5 = 5, F6 = 8, F7 = 13, F8 = 21, F9 = 34 and F10 = 55.

(i) If you’re careful, the next section isn’t particularly difficult either. Using the recurrence relation
1 1 1
gives   since Fi – 2 < Fi – 1 for i  4. Splitting off the first few terms then
Fi Fi  1  Fi 2 2 Fi 1
n
1 1 1  1 1  1 1 1  1 1 
leads to S =  >  1    ..... or   1    .....  , where the
i  1 Fi F1 F2  2 4  F1 F2 F3  2 4 
long bracket at the end is the sum-to-infinity of a GP. These give, respectively, S > 1 + 1 × 2 = 3
or 1 + 1 + 12  2 = 3. A simpler approach could involve nothing more complicated than adding
the terms until a sum greater than 3 is reached, which happens when you reach F5.

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365
 1 1 1 
A similar approach yields    for i  3 and splitting off the first few terms, this time
Fi 2  Fi 2 
separating the odd- and even-numbered terms, gives
n
1 1 1  1 1   1 1 
S=  =      .....      ..... 
i  1 Fi F1 F2  F3 F5   F4 F6 
1 1 1  1 1 1 
< 11 1    .....   1    .....
2 2 4  3 2 4 
1 1
= 1  1   2   2  3 23 .
2 3

(ii) To show that S > 3.2, we simply apply the same approaches as before, but taking more terms
initially before summing our GP (or stopping at F7 in the “simpler approach” mentioned
previously). Something like
1 1 1 1 1  1 1  1 1 1
S>     1    .....  = 1  1     2  3 307 > 3 306  3.2
F1 F2 F3 F4 F5  2 4  2 3 5
1
does the job pretty readily. Then, to show that S < 3 2 , a similar argument to those you have been
directed towards by the question, works well with little extra thought required:
1 1 1 1 1  1 1 1 
S < 1  1    1    .....  1    .....
2 3 5 2 4  8 2 4 
1 1 1 1
= 1  1     2   2  3 60 29
 3 12 .
2 3 5 8

1 1 1 
Returning to the initial argument, Fi < 2 Fi – 1 or    for i  4, we can extend this to
Fi 2  Fi  1 
3 1 2 1  5 1 3 1 
Fi > Fi – 1 or    for i  5, Fi < Fi – 1 or    for i  6, etc., simply
2 Fi 3  Fi  1  3 Fi 5  Fi  1 
by using the defining recurrence relation for the Fibonacci Numbers, leading to the general results
 F  1  F2 k  1  1
Fn >  2 k  Fn – 1 or   for n  2k + 1
 F2 k  1  Fn  F2 k  Fn  1
and
F  1  F2 k  1
Fn <  2 k  1  Fn – 1 or   for n  2k + 2.
 F2 k  Fn  F2 k  1  Fn  1

Fn 5 1
Since the terms   , the golden ratio, (being the positive root of the quadratic
Fn1 2
n
1 1 2
equation x2 = x + 1, we can deduce the approximation S  F
i 1

Fn  1
 since the geometric
i

1 1 1  
progression 1    ...    1   2 . Taking n = 9, (i.e. just using the first 10
  2
1    1 
1

Fibonacci Numbers which you were led to write down at the start),
9
1 1 2 614893 1 5 3
S    =    3.359 89,
i 1 Fi F10 185640 55 2
which is correct to 5 d.p. For further information on this number, try looking up the ‘Reciprocal
Fibonacci constant’ on Wikipedia, for instance.

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7 It is easy to saunter into this question’s opening without pausing momentarily to wonder if one is
going about it in the best way. Whilst many can cope with differentiating a “triple”-product with
ease, many others can’t. However, even for interests’ sake, one might stop to consider a general
approach to such matters. Differentiating y = pqr (all implicitly functions of x) as, initially, p(qr)
and applying the product-rule twice, one obtains y = pq r + p q r + p qr, and this can be used
here with p = (x – a)n, q = ebx and r = 1  x 2 without the need for a lot of the mess (and
subsequent mistakes) that was (were) made by so many candidates. Here, y = (x – a)n ebx 1  x 2
gives
dy x
= (x – a)n ebx + (x – a)n b ebx 1  x 2 + n(x – a)n – 1 ebx 1  x 2
dx 1 x 2

( x  a ) n  1 e bx
Factorising out the given terms  x( x  a)  b( x  a)(1  x 2
)  n(1  x 2 ) , and we
1 x 2

are only required to note that the term in the brackets is, indeed, a cubic; though it may prove
helpful later on to simplify it by multiplying out and collecting up terms, to get
q(x) = bx3 + (n +1 – ab)x2 + (b – a)x + (n – ab).

( x  4)14 e 4 x
(i) The first integral, I1 =  4 x 
 1 dx , might reasonably be expected to be a very
3

1 x 2

straightforward application of the general result, and so it proves to be. With n = 15, and taking
a = b = 4, so that q(x) = 4x3 – 1 (which really should be checked explicitly), we find
I1 = (x – 4)15 e4x 1  x 2 (+ C).

( x  1) 21 e12 x
(ii) This second integral, I2 =  12 x 4

 x 2  11 dx , is clearly not so straightforward,
1 x 2

since the bracketed term is now quartic. Of the many things one might try, however, surely the
simplest is to try to factor out a linear term, the obvious candidate being (x – 1).
Finding that 12x4 – x2 – 11  (x – 1)(12x3 + 12x2 + 11x + 11), we now try n = 23, a = 1, b = 12
to obtain q(x) = 12x3 + 12x2 + 11x + 11 and I2 = (x – 1)23 e12x 1  x 2 (+ C).

( x  2) 6 e 4 x
(iii) The final integral, I3 =  4 x 4
 x 3  2 dx, is clearly intended to be even less simple
1 x 2

than its predecessor. However, you might now suspect that “the next case up” is in there
somewhere. So, if you try n = 8, a = 2, b = 4, which gives
( x  2) 7 e 4 x ( x  2) 6 e 4 x
dy8
=  4 x 3  x 2  2 x =  4 x 4  7 x 3  4 x ,
dx 1 x 2
1 x 2

as well as the obvious target n = 7, a = 2, b = 4, which yields

( x  2) 6 e 4 x
dy7
=  4 x 3  2 x  1,
dx 1 x 2

It may now be clear that both are involved. Indeed,

 dy dy 
I3 =   8  2 7  dx = y8 + 2 y7 = x(x – 2)7 e4x 1  x 2 (+ C).
 dx dx 

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367
8 For the diagram, you are simply required to show P on AB, strictly between A and B; and Q on AC
on other side of A to C. The two given parameters indicate that CQ =  AC and BP =  AB .
Substituting these into the given expression, CQ  BP = AB  AC  AC . AB = AB . AC
1
   . [Notice that CQ, BP, etc., are scalar quantities, and hence the “” cannot be the

vector product!]

Writing the equation of line PQ in the form r = t p + (1 – t) q for some scalar parameter t and
substituting the given forms for p and q gives r = t a + t(1 – )b + (1 – t) a + (1 – t)(1– )c.
1  1 t   1
Eliminating    r =  t   a  t (1   )b  (1  t ) c . Comparing this to the
      
1 
given answer, we note that when t  from the b-component, 1  t  , etc., so that we
1   1
do indeed get r =  a  b  c , as required.

Since d – c = b – a , one pair of sides of opposite sides of ABDC are equal and parallel, so we can
conclude that ABDC is a parallelogram

9 (i) If you “break the lamina up” into a rectangle and a triangle (shapes whose geometric centres
should be well-known to you), with relative masses 2 and 1, and impose (mentally, at least) a
coordinate system onto the diagram, then the x-coordinate of the centre of mass is given by

x
 mi xi = 2  92  1 12 = 7.
 mi 3

(ii) A more detailed approach, but still along similar lines, might be constructed in the following,
tabular way:
Shape Mass Dist. c.o.m. from OZ
LH end 540 7 Note that each mass has been
RH end 540 7 calculated as
Front 41d 27
2 area  density ()
Back 40d 0
Base 9d 9
2

2  (540  )  7  41d  272  0  9dp  92

Then xE = , which (after much cancelling) simplifies to
1080   90d
2  60  7  66d 3(140  11d )
= = .
10(12  d ) 5(12  d )

A similar approach for the full tank gives

Object Mass Dist. c.o.m. from OZ
Tank 2880 27
4

Water 10800k 7
2880   274  10800kp  7 27  105k
and xF = = .
2880   10800k 4  15k

48
368
10 The standard approach in collision questions is to write down the equations gained when applying
the principles of Conservation of Linear Momentum (CLM) and Newton’s Experimental Law of
Restitution (NEL or NLR), and then what can be deduced from these.

For P1, 2 : CLM  m1 u = m1 v1 + m2 v2 and NEL  eu = v2 – v1.

Solving to determine the final speeds of P1 and P2 then yields
(m  em2 ) m (1  e)
v1 = 1 u and v2 = 1 u.
m1  m2 m1  m2
Similarly, for P4, 3 : CLM  m4 u = m4 v4 + m3 v3 and NEL  eu = v3 – v4, leading to
m (1  e) (m4  em3 )
v3 = 4 u and v4 = u.
m3  m4 m3  m 4

If we now write X = OP2 and Y = OP3 initially, and equate the times to the following collisions
at O, we have
m  m2 X  m3  m4 Y
(1st collision): 1
m1 (1  e)u m4 (1  e)u
and
(2nd collision): 1
m  m2 X  m3  m4 Y .
m1  em2 u m4  em3 u
Cancelling u’s and (1 + e)’s
 1
m  m2 X  m3  m4 Y and m1  m2 X  m3  m4 Y . (*)
m1 m4 m1  em2  m4  em3 
m1  em2 m4  em3
Dividing these two (or equating for X / Y)   , which simplifies to
m1 m4
m2 m3
 . Finally substituting back into one of the equations (*) then gives
m1 m4
 m   m 
X 1  2   Y 1  3   X = Y .
 m1   m4 

Rather surprisingly, however, the momentum equations turn out to be totally unnecessary here.
Consider …
Collision P1, 2 : NEL  eu = v2 – v1
Collision P4, 3 : NEL  eu = v3 – v4 so that v2 – v1 = v3 – v4 (*).
X Y X Y
Next, the two equated sets of times are  and   X v3 = Y v2 and X v4 = Y v1 .
v 2 v3 v1 v 4
Subtracting: X v3  v 4   Y v 2  v1   X = Y from (*).

49
369
11 N2L  FT – (n + 1)R = (n + 1)Ma , where FT is the tractive, or driving, force of the engine.
P
 (n  1) R
P  (n  1) Rv
Using P = FT . v then gives a = v or . Note here that, for a > 0
M (n  1) M (n  1)v
we require P > (n + 1)Rv.

dv dv P  (n  1) Rv
Writing a = gives = which is a “variables separable” first-order
dt dt M (n  1)v
M (n  1)v V
M (n  1)v
T
differential equation: dv = dt   dv =  1.dt ( = T ).
P  (n  1) Rv 0
P  (n  1) Rv 0

Some care is needed to integrate the LHS here, and the simplest approach is to use a substitution
such as s = P – (n + 1)Rv, ds = – R(n + 1) dv to get
M Ps M P  M
T=  
ds
= 2 
 1 ds = P lns   s
R s  R(n  1) (n  1) R  s  (n  1) R 2
M
= P lnP  (n  1) Rv   P  (n  1) Rv V0
(n  1) R 2

 MP
= lnP  (n  1) Rv   P  (n  1) Rv  P ln P  P  0
(n  1) R 2
 MP  P  (n  1) Rv  MV
= ln 
(n  1) R 2
 P  R
More careful algebra is still required to manipulate this into a form in which the given
approximation can be used:
 MP  ( n  1) Rv  MV
T= ln1  
( n  1) R 2
 P  R
 MP (n  1) Rv 1  (n  1) Rv  2  MV
     ..... 

(n  1) R 2 P 2  P  R
 
MV (n  1) MV 2 MV
=  ..... 
R 2P R
so that PT  12 (n  1) MV 2 , and this is just the statement of the Work-Energy Principle, namely
“Work Done = Change in (Kinetic) Energy”, in the case when R = 0.

When R  0, WD against R = WD by engine – Gain in KE  (n + 1)RX = PT – 12 (n  1) MV 2 .

[Unfortunately, a last-minute change to the wording of the question led to the omission of one of
the (n + 1)s.]

50
370
12 (i) This whole question is something of a “one-trick” game, I’m afraid, and relies heavily on being
able to spot that X is just half of a normal distribution. The Standard Normal Distribution N(0, 1)
x
1  12 t 2
is given by P(X  x) = 
2 
e dt. Once the connection has been spotted, the accompanying

pure maths work is fairly simple, including the sketch of the graph. This is particularly important
2
since the function e kx cannot be integrated analytically.

1
(ii) Substituting t = 2x , dt = 2 dx and equating to (being just the positive half of a normal), gives
2
 1 2  
1 2 t 1 2 x 2 1 2
   e  2 x dx 
2
e dt = 2e dx =  .
2 0 2 0 2 0 4
1 2 4
Since total probability = 1, we have  and k = .
k 4 2


2 x 2  1 2 x 2   1 1
(iii) Thereafter, E(X) = k 0 xe d x = k  4 e  = 4 k 
0 2
.

  1 2
 
1 
Also, E(X 2) = k  x  x e  2 x dx = k  xe 2 x    e 2 x dx  using integration by parts
2 2

0  4 0 04 
 1 2  1
= k 0    = .
 4 4  4
1 1  2
Then Var(X) = E(X 2) – E2(X) =  or .
4 2 4

m
1 4
 e 2 x dx , and this
2
(iv) For the median, we want to find the value m of x for which =
2 2 0
requires to undo some of the above work in order to be able to use N(0, 1) and the statistics tables
provided in the formula book.
m 2m
 12 t 2
dt = 22m   12  or   12 m  
1 2 2 x 2 1 3
2

2 0
 2e d x = 2 
2 0
 e
4
Use of the N(0, 1) tables then gives 2m  0.6745 (0.675-ish) and m = 0.337 or 0.338 .

51
371
13 For A: p(launch fails) = p(>1 fail) = 1 – p0 – p1 = 1 – q4 – 4q3p
so that E(repair) = x p(x) = 0.q4 + K.4q3p + 4K(1 – q4 – 4q3p)

= 4K q 3 p  (1  q)(1  q  q 2  q 3 )  4q 3 p 

= 4Kp 1  q  q 2  2q 3 
For B: p(launch fails) = p(>2 fail) = 1 – p0 – p1 – p2 = 1 – q6 – 6q5p – 15q4p2
so that E(repair) = x p(x)
= 0.q6 + K.6q5p + 2K.15q4p2 + 6K(1 – q6 – 6q5p – 15q4p2)

= 6K q 5 p  5q 4 p 2  (1  q )(1  q  q 2  q 3  q 4  q 5 )  6q 5 p  15q 4 p 2 
Extracting the p and obtaining the remaining in terms of q only,

= 6Kp q 5  5q 4 (1  q )  1  q  q 2  q 3  q 4  q 5  6q 5  15q 4 (1  q ) 
= 6Kp 1  q  q 2
 q 3  9q 4  6q 5 
Setting Rep(A) = 2
3
  
Rep(B)  12Kp 1  q  q 2  2q 3 = 2Kp 1  q  q 2  q 3  9q 4  6q 5 
Clearly, p = 0 is one solution and the rest simplifies to
0 = 3q (1 – 3q + 2q ) = 3q3(1 – q)(1 – 2q).
3 2

1
We thus have p = 1, 0, 2 , with the 0 and 1 being rather trivial solutions.

52
372
 STEP III, Solutions
2009

53
373
STEP Mathematics III 2009: Solutions

Section A: Pure Mathematics

1. The result for p can be found via calculating the equation of the line SV
ms − nv
( y − ms = ( x − s) ) or similar triangles. The result for q follows from that for p
s−v
(given in the question) by suitable interchange of letters to give
( m − n) tu
q=
mt − nu

As S and T lie on the circle, s and t are solutions of the equation

λ2 + ( mλ − c) = r 2 i.e. (1 + m 2 )λ2 − 2mcλ + (c 2 − r 2 ) = 0
2

c2 − r 2 2mc
and so from considering sum and product of roots, st = 2 , and s + t =
1+ m 1 + m2
c2 − r 2 2nc
Similarly uv = 2 , and u + v = can be deduced by interchanging letters.
1+ n 1 + n2

( m − n) sv ( m − n) tu
Substituting from the earlier results p+q = + which can
ms − nv mt − nu
( m − n)
be simplified to
( ms − nv )( mt − nu)
( stm( u + v ) − nuv( s + t ) )
and then substituting the sum and product results yields the required result.

2 (i) The five required results are straightforward to write down, merely observing
that initial terms in the summations are zero.

(ii) Substituting the series from (i) in the differential equation yields that
− a1 + 3a 3 x 2 + (8a 4 + 4a 0 ) x 3 +..... = 0 , after having collected like terms.
Thus, comparing constants and x 2 coefficients a1 = 0 and a 3 = 0
Comparing coefficients of x n−1 , for n ≥ 4 , n( n − 1) a n − na n + 4a n−4 = 0 which gives
the required result upon rearrangement.

−1
With a 0 = 1 , a 2 = 0 , and as a1 = 0 , and a 3 = 0 , we find a 4 = , a5 = 0 , a 6 = 0 ,
2!
1
a 7 = 0 , a8 = , etc.
4!
Thus y = 1 − ( x 2 ) + ( x 2 ) − ( x 2 ) +..... = cos( x 2 )
1 2 1 4 1 6

2! 4! 6!

With a 0 = 0 , a 2 = 1 , y = ( x 2 ) − ( x ) + 5! ( x ) − 7! ( x ) +..... = sin( x 2 )

1 2 3 1 2 5 1 2 7
3!

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374
3. (i) Substituting the power series and tidying up the algebra yields
1
f (t) = and so lim f ( t ) = 1 .
⎛ t ⎞ t →0
⎜ 1 + +.....⎟
⎝ 2! ⎠
1 ⎛ 1 1⎞
−
− t ⎜ − ⎟ −.....
( e − 1) − te
t
2 ⎝ 2 ! 3!⎠
t
−1
Similarly, f ′( t ) = = and so lim f ′( t ) =
(e − 1)
t 2
⎛ t ⎞
⎜ 1 + +.....⎟
2
t →0 2
⎝ 2! ⎠
(Alternatively, this can be obtained by de l’Hopital.)

1
(ii) If we let g ( t ) = f ( t ) + t , then simplifying the algebra gives
2
t (e t + 1)
g( t ) =
2(e t − 1)
after which it is can be shown by substituting –t for t
that g ( −t ) is the same expression.

(iii) If we let 1 , and find its stationary point, sketching the

graph gives

Hence 1 1 and so 1 1 0. (Alternatively, a sketch with and

will yield the result.)
Thus 0, with equality only possible for 0, but we know
−1
lim f ′(t ) = and so, in fact, f(t) is always decreasing i.e. has no turning points.
t →0 2

55
375
Considering the graph of . It passes through (0,1), is symmetrical
and approaches as ∞ and thus is

Therefore the graph of also passes through (0,1), and has

asymptotes 0 and and thus is

4. (i) Substituting into the definition yields the Laplace transform as

∞ ∞

∫ e − st e −bt f ( t ) dt = ∫ e −t s+b f ( t )dt = F ( s + b)

( )

0 0

(ii) Similarly, a change of variable in the integral using u at yields the result.

(iii) Integrating by parts yields this answer.

(iv) A repeated integration by parts obtains

F ( s ) = 1 − s 2 F ( s)
which leads to the stated result.

56
376
Using the results obtained in the question, the transform of cosqt is
⎛ sq ⎞ s ( s + p)
q −1 ⎜ 2 2 ⎟= 2 2 ,and so the transform of e
− pt
cos qt is
⎝ s q + 1⎠ s + q ( s + p) 2 + q 2
5. The first result may be obtained by considering
( x + y + z) − ( x 2 2
+ y 2 + z 2 ) = 2( yz + zx + xy ) ,
the second by
( x 2 + y 2 + z 2 )( x + y + z) = x 3 + y 3 + z 3 + ( x 2 y + x 2 z + y 2 z + y 2 z + z 2 x + z 2 y)
and the third by
( x + y + z) 3
= ( x 3 + y 3 + z 3 ) + 3( x 2 y + x 2 z + y 2 z + y 2 z + z 2 x + z 2 y ) + 6 xyz

Considering sums and products of roots, we can deduce that x satisfies the cubic
equation 0 , as do y and z by symmetry. Multiplying by ,
, with similar results for y and z. Summing these yields

Alternatively,

1.
to give the result.

6. Using Euler, e iβ − e iα = ( cos β − cosα ) + i( sin β − sin α )

and so
e iβ − e iα = ( cos β − cos α ) + ( sin β − sin α )
2 2 2

which can be expanded, and then using Pythagoras, compound and half angle
formulae this becomes
4 sin 2 ( β − α )
1
2
e iβ − e iα = 2 sin ( β − α ) as both expressions are positive.
1
2
Alternative methods employ the factor formulae.

e iα − e iβ e iγ − e iδ + e iβ − e iγ e iα − e iδ
⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞
= 2 sin⎜ (α − β )⎟ 2 sin⎜ (γ − δ )⎟ + 2 sin⎜ ( β − γ )⎟ 2 sin⎜ (α − δ )⎟
⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠
which by use of the factor formulae and cancelling terms may be written
⎛ ⎛1 ⎞ ⎛1 ⎞⎞
2⎜ cos⎜ (α − β − γ + δ )⎟ − cos⎜ ( β − γ + α − δ )⎟ ⎟
⎝ ⎝2 ⎠ ⎝2 ⎠⎠
and then again by factor formulae,

57
377
 ⎛1 ⎞ ⎛1 ⎞
2 sin⎜ (α − γ )⎟ 2 sin⎜ ( β − δ )⎟
⎝2 ⎠ ⎝2 ⎠
which is

e iα − e iγ e iβ − e iδ as required.

Thus, the product of the diagonals of a cyclic quadrilateral is equal to the sum of the
products of the opposite pairs of sides ( Ptolemy’s Theorem).

7. (i) This result is simply obtained using the principle of mathematical induction.
The n = 1 case can be established merely by obtaining f1 and f2 from the definition,
and then substituting these along with f0.

(ii)
P0 ( x ) = (1 + x 2 )
1
=1
1+ x2
− 2x
P1 ( x ) = (1 + x 2 )
2
= −2 x
(1 + x 2 ) 2
6x 2 − 2
P2 ( x ) = (1 + x 2 )
3
= 6x 2 − 2
(1 + x ) 2 3

dPn ( x )
Pn +1 ( x ) − (1 + x 2 ) + 2(n + 1) xPn ( x )
dx
which differentiating Pn by the product rule and substituting
= (1 + x 2 )
n+2
(
f n +1 ( x ) − (1 + x 2 ) (1 + x 2 )
n +1 n
)
f n +1 ( x ) + ( n + 1) 2 x(1 + x 2 ) f n ( x ) + 2( n + 1) x(1 + x 2 )
n +1
f n ( x)
which is zero.

Again using the principle of mathematical induction and the result just obtained, it can
be found that Pk +1 ( x ) is a polynomial of degree not greater than k + 1 .

Further, assuming that Pk ( x ) has term of highest degree, ( − 1 ) ( k + 1 ) ! x , as

k k

dP ( x )
Pn +1 ( x ) − (1 + x 2 ) n + 2(n + 1) xPn ( x ) = 0 , the term of highest degree of Pk +1 ( x ) is
dx
( − 1) k ( k + 1) ! kx k −1 x 2 − 2( k + 1) x( − 1) k ( k + 1) ! x k
= ( − 1) ( k + 2) ! x k +1 as required.
k +1

(The form of the term need not be determined, but it must be shown to be non-zero.)

8. (i) Letting x = e − t ,
lim ln lim 1 lim 0

and so letting 1 , lim ln 0.

Thus, lim lim 1

58
378
(ii) Integrating by parts,
ln 1 ln
ln
1 1

1 1
0 0 ln
1 1
!
So …

x 2 (ln x )
1 1 1 2

(iii) ∫ x x dx = ∫ e x ln x dx = ∫ 1 + x ln x +
0 0 0
2!
+..... dx

2 3 4
1 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞
= 1 + I 1 + I 2 +..... = 1 − ⎜ ⎟ + ⎜ ⎟ − ⎜ ⎟ +..... as required.
2! ⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠

Section B: Mechanics

9. (i) With V as the speed of projection from P, x and y the horizontal and
vertical displacements from P at a time t after projection, and T the time of flight from
P to Q, then
cos , sin , cos , and

So tan tan , and tan tan

Thus tan tan 2 tan 2 tan

(ii) Using the trajectory equation written as a quadratic equation in tan ,

tan tan 0 , giving tan tan , and

tan tan 1 1 tan .

Applying the compound angle formula and substituting, tan cot

So, , and as 0 ,0 ,0 ,
.
Reversing the motion we have, , and therefore,

0 ,0 , , , and ,
so , or as required

59
379
10. Supposing that the particle P has mass m, the spring has natural length l, and
modulus of elasticity λ,
If the speed of P when it hits the top of the spring is v, then 2
By Newton’s second law, the second-order differential equation is thus
and so with initial conditions that 0,
2 , when 0.
has complementary function cos sin

where , and particular integral , where .

The initial conditions yield, and √2
So cos √2 sin .

cos √2 sin may be expressed in the form cos where

√
2 , and tan

So cos

0 next when , that is when 2

So 2 2 2 2 tan and 2 2 tan .

11. (i) Conserving momentum yields 1 and so

1 Written as 1 , separating variables and integrating
, but as 0 , when 0, 0
√
So 0 , and so , except 0, and thus
√

(ii) 1

So, 1 and as 0, 0 , and we have .

Thus as required.

Separating variables and integrating

and as 0 , when 0, 0
So 0 , and so , except 0, and thus

60
380
If 1 2 is a perfect square, then x will be linear in t and will be
constant, i.e. if 4 4 0 , that is
√
(in which case , and as expected.)

Otherwise, , and as ∞, ,

a constant, as required.

Section C: Probability and Statistics

12. (i) , | , and so

∑

∑ ∑ using the sum of an infinite GP.

∑
(ii) ∏

0 , 1 ,…,

and so … (infinite GP)

Thus ∏

…
, , and
and so 1 , 1 1 1 2
…
and for 0, 1, 2, …
! !

(Alternatively, is coefficient of in which can be expanded

binomially to yield the same result.)

13.
(i) cos cos 2 cos
Therefore,
So , for 1 1
√

So
If , √1 equiprobably, so 0, 0 and thus
, 0 , and hence , 0.

61
381
X and Y are not independent for if , √1 only, whereas without the
restriction, Y can take all values in 1,1 .
(ii) ∑ ∑ 0 , and 0 similarly.
∑ ∑ ∑ as , are independent and
each have expectation zero.
∑ 0 from part (i), and so 0. Thus , 0 , and
hence , 0 as required.

For large n , ~ 0, approximately, by Central Limit Theorem.

Thus,

| | | | | | 2 | | 1 · 960 0 · 95
√

62
382
STEP Examiners’ Report
2010

Mathematics
STEP 9465/9470/9475

October 2010

383
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 3
STEP Mathematics II 15
STEP Mathematics III 19
Explanation of Results 22

384
 STEP 2010 Paper I: Principal Examiner’s Report

Introductory comments
There were significantly more candidates attempting this paper than last year (just over
1000), and the scores were much higher than last year (presumably due to the easier first
question): fewer than 2% of candidates scored less than 20 marks overall, and the median
mark was 61.
The pure questions were the most popular as usual, though there was much more varia-
tion than in some previous years: questions 1, 3, 4 and 6 were the most popular, while
question 7 (on vectors) was intensely unpopular. About half of all candidates attempted
at least one mechanics question, and 15% attempted at least one probability question.
The marks were unsurprising: the pure questions generally gained the better marks, while
the mechanics and probability questions generally had poorer marks.
A sizeable number of candidates ignored the advice on the front cover and attempted
more than six questions, with a fifth of candidates trying eight or more questions. A good
number of those extra attempts were little more than failed starts, but suggest that some
candidates are not very effective at question-picking. This is an important skill to develop
during STEP preparation. Nevertheless, the good marks and the paucity of candidates
who attempted the questions in numerical order does suggest that the majority are being
wise in their choices. Because of the abortive starts, I have often restricted my attention
below to those attempts which counted as one of the six highest-scoring answers, and
referred to these as “significant attempts”.
The majority of candidates did begin with question 1 (presumably as it appeared to be
the easiest), but some spent far longer on it than was wise. Some attempts ran to over
eight pages in length, especially when they had made an algebraic slip early on, and used
time which could have been far better spent tackling another question. It is important to
balance the desire to finish the question with an appreciation of when to stop and move
on.
Many candidates realised that for some questions, it was possible to attempt a later
part without a complete (or any) solution to an earlier part. An awareness of this could
have helped some of the weaker students to gain vital marks when they were stuck; it is
generally better to do more of one question than to start another question, in particular
if one has already attempted six questions. It is also fine to write “continued later” at
the end of a partial attempt and then to continue the answer later in the answer booklet.
As usual, though, some candidates ignored explicit instructions to use the previous work,
such as “Hence”, or “Deduce”. They will get no credit if they do not do what they are
asked to! (Of course, “Hence, or otherwise, show . . . ” gives them the freedom to use any
method of their choosing; often the “hence” will be the easiest, but in Question 5 this
year, the “otherwise” approach was very popular.)
On some questions, some candidates tried to work forwards from the given question and
backwards from the answer, hoping that they would meet somewhere in the middle. While
this worked on occasion, it often required fudging, and did bring to mind a recent web
comic: http://xkcd.com/759.

385
It is wise to remember that STEP questions do require a greater facility with mathe-
matics and algebraic manipulation than the A-level examinations, as well as a depth of
understanding which goes beyond that expected in a typical sixth-form classroom. STEP
candidates are therefore recommended to heed the sage advice on the STEP Mathematics
website, http://www.admissionstests.cambridgeassessment.org.uk/adt/step:

From the point of view of admissions to a university mathematics course,

STEP has three purposes. . . . Thirdly, it tests motivation. It is important to
prepare for STEP (by working through old papers, for example), which can
require considerable dedication. Those who are not willing to make the effort
are unlikely to thrive on a difficult mathematics course.

Common issues
There were a number of common errors and issues which appeared across the whole paper.
The first was a lack of fluency in algebraic manipulations. STEP questions often use
more variables than A-level questions (which are more numerical), and therefore require
candidates to be comfortable engaging in extended sequences of algebraic manipulations
with determination and, crucially, accuracy. This is a skill which requires plenty of practice
to master.
Along with this comes the need for explanations in English: a sequence of formulæ or
equations with no explicit connections between them can leave the reader (and writer)
confused as to the meaning: Does one statement follow from the other? Are they equiv-
alent statements? Or are they perhaps simultaneous equations? For example, writing
x = 2 followed by x2 = 4 is not the same as writing x = 2 followed by 2x = 4, and both
are different from writing x = 2 followed by y = 3. In some cases, this cost marks, in
particular when a candidate was required to show that “A if and only if B”. Brief con-
nectives or explanations (“thus”, “so”, “∴” or “⇒”) would help, and sometimes longer
sentences are necessary. The solutions booklet is more verbose than candidates’ solutions
need to be, but give an idea of how English can be used.
Another related issue is legibility. Many candidates at some point in the paper lost marks
through misreading their own writing. Common confusions this year included muddling
their symbols, the most common being: M and m; V and v; u and n; u and N ; x and n
(primarily among Oriental candidates); α and 2; a and 9; s, S and 5. In other years, z
and 2 have also caused confusion. It is sad that, at this stage, candidates are still wasting
marks because of bad writing habits.
A couple of basic (primary school) arithmetical processes caused some problems this
year, namely long multiplication and fractions. While there is rarely a need for long
multiplication in STEP examinations, some of the candidates attempted to use it at
various points, and it was shocking to see that they were not competent with a compact
hand-written method, some even still using a chunking or grid-type method. For fractions,
the situation was similar, the most noticeable thing being the lack of cross-cancellation
3
or any simplification when multiplying fractions. (For example, when calculating 16 × 10 27
,
1 5 5 30
it is far simpler to first cancel to 8 × 9 giving 72 than to multiply directly to get 432 .)
Quadratics caused a few raised eyebrows when candidates tried to solve equations of the

386
form (blah)2 = 9 by expanding brackets and solving the resulting quadratic instead of
simply saying blah = ±3.
Graph sketches were again weak. Students need much more practice with sketching graphs
of interesting functions (beyond the standard A-level fare of quadratics, cubics, reciprocals
of linears and x2 , and the basic trigonometrical functions). Sketching functions should
involve consideration of all of their main features: their axis-crossings, their stationary
points, their asymptotic behaviour and even such basics as whether they are positive or
negative in various regions. Reciprocals of quadratics and cubics are good for learning
this, as well as more sophisticated functions. Students preparing for STEP would be
advised to make up such functions, attempt to sketch them, and then check their answers
on a graphical calculator or using software such as Geogebra (which is free, and can be
found at http://www.geogebra.org/).
Finally, a strong reminder that it is vital to draw appropriate, clear, accurate diagrams
when attempting mechanics questions: it was shocking how many candidates attempted
to solve the collision question without a diagram or a moments question with a tiny, rough
sketch!

Question 1
This was by far the most popular question, attempted by almost all candidates. It was
very pleasing to see so many perfect solutions, showing that many candidates had a good
command of basic algebraic manipulation.
The first part of the question was answered very well. Some candidates failed to take
enough care with their algebra, the most common errors being to either lose the 4a term
entirely or to replace it by 4.
There were many different approaches used to tackle the second part, the majority of them
being effective. The most common conceptual error was assuming that the solutions had
to be integers. As the question did not say this, such attempts gained relatively few marks
unless supported by a full algebraic solution.
There were a few recurring technical errors and other ineffective approaches.
The first was to implicitly assume that the second equation could be rewritten in the
same way as the first without checking the consistency of all six equations (most notably,
checking that a + bc2 = 6).
The second was to fail to square root an equation correctly: from (y − 2x)2 = 4, for
example, a significant number of students gave only the single solution y − 2x = 2.
A significant number of students made arithmetical errors when solving the equations
which left them bogged down in messy calculations, costing them only a few accuracy
marks but lots of time.
Also, several took long-winded methods which involved multiplying out expressions such
as (y − 2x)2 = 4 to reach quadratic equations; as mentioned above, candidates should
certainly be aware that they ought to simply take square roots.
Most candidates who reached a pair of simultaneous equations for (y−2x)2 and (x−y+2)2

387
correctly solved for one of the terms, but a surprising number then substituted their
answer(s) back into one of the original quadratic equations in x and y to eliminate a
variable, giving themselves far more algebraic slog than actually necessary.
Overall, the mean mark for this question was in excess of 14/20, making it the most
successfully answered question on the paper.

Question 2
This was a fairly popular question, attempted by around two-thirds of candidates. It
was pleasing to see how many of them were correctly able to differentiate the expression
given at the start of the question. Again, the earlier comments on algebraic accuracy bear
repeating at this point: a number of candidates became unstuck here through algebraic
or sign errors, and this was a repeating theme throughout this question.
The inequalities proved challenging: setting the derivative equal to zero was an obvious
step, and then the resulting quadratic is crying out for consideration of the discriminant.
Even many of those who got this far failed to adequately explain their solution of the
quadratic inequality, jumping straight to the given answer.
Moving on to the graph sketches, it became obvious that different candidates have differ-
ent understandings of what is expected. At the very least, a graph sketch should indicate
(where possible) the coordinates of stationary points and axis crossings, as well as asymp-
totic behaviour. Many candidates, pleased with their success at finding the x-coordinates
of the stationary points, then stopped and did not calculate the y-coordinates. The lattar
would have made the decision about the nature of the stationary points essentially trivial
and would have helped them draw more accurate sketches, especially in part (ii).
On a positive note, most of the candidates who reached this part of the question deter-
mined the turning points and vertical asymptotes correctly, though some thought that
part (ii) had no turning points (despite having shown that it does earlier on).
The mean score on this question was noticeably lower than on several of the other pure
mathematics questions, suggesting that graph-sketching is an area which requires more
attention from candidates during their preparation.

Question 3
This was a very popular question, and the marks were generally very encouraging. A
significant majority of attempts simply rattled through the first part of the question,
showing confidence and competence with their trigonometric identities. A few candidates’
solutions for this part lasted several pages, but the majority were very swift and efficient.
For the main part of the question, most candidates started off very well by attempting to
equate the lines’ gradients, though a few thought that the gradient is given by the formula
(x2 − x1 )/(y2 − y1 ). A number of candidates tried to use vector methods by showing that
−→ −→
P Q.SR
= 1,
|P Q|.|SR|

388
 −→
which involves significant (and messy) algebraic manipulation. (And if they used RS
−→
instead of SR, they were unlikely to try to equate to −1 instead of 1.) A few tried to
−→ −→
show that P Q = SR, which is more restrictive than what is required. In general, the
scalar (dot) product method is good for identifying perpendicular vectors, but far less
useful for parallel vectors.
Most candidates who followed the gradient method then reached the intermediate conclu-
sion that tan 12 (q + p) = tan 12 (s + r), but did not know how to continue correctly. They
either simply used the given answer or concluded that 12 (q + p) + kπ = 12 (s + r) where
k = 0, 1 or 2. Very few said that k could be any integer or gave any justification for their
restriction to the given possibilities.
Finally, almost no candidates appreciated that the question’s requirement to show that
the lines are parallel if and only if the given condition was met meant that they had
to either show that their argument was reversible (using ⇐⇒ connectives or some other
indication that it was reversible), or explain why r + s − p − q = 2π implies that the lines
are parallel.

Question 4
This was another very popular question and, in spite of the challenges it posed, was
answered well by many of the candidates.
The majority of those who attempted the question knew how to perform an integration by
substitution, and many were able to do so correctly in this challenging example. Almost
every candidate was correctly able to determine dx/dt and substitute for x correctly in the
integrand. The next step, simplifying the resulting expression, proved more challenging,
and several candidates slipped up at this point (for example by forgetting a square root
sign).
R
Those who reached an integral of the form −2/(t2 − 1) dt generally used the given result
to perform the integration, though some ignored the note and proceeded to use partial
fractions; this did not gain them any extra credit.
The majority either ignored the absolute value signs given in the note, quietly dropped
them in the next step, or integrated their partial fraction expansion without the use of
absolute value signs. While it is true, in this case, that they can (and should) be dropped,
this did require some minimal justification.
p
The simplification of the logarithm expression with t substituted with (x + 1)/x proved
challenging. A good number of candidates wisely expanded the given right-hand side and
showed
p that it gave the same as the left-hand side, while others floundered with the expres-
sion 1 + 1/x. This question called for a deep understanding of simplification of fractions
together with an equally sophisticated understanding of rationalising denominators, and
this proved too hard for many candidates.
Most candidates attempted the second part of the question, even if they had not completed
the first part.
R A2 disturbing number of candidates had misremembered the formula for
volume as 2πy dx, and were therefore out by a factor of 2 throughout the rest of the
question. They were only lightly penalised for this error.

389
There was some noticeable difficulty experienced in expanding y 2 , with many also making
their lives somewhat harder by writing y as a single fraction before squaring. Even among
the correct expansions, fewer attempts than expected noted that the integrand could be
written as x1 + x+1
1
−2×(∗), where (∗) was the integrand from the first part of the question.
Therefore many candidates ended up becoming stuck at this point.
Of those candidates who reached the correct integrand, most were able to correctly sub-
stitute in the limits and get the signs correct (though this did require care, and not all
managed it). The next step, simplifying the logarithms, left many struggling. While
most were fairly comfortable with the rules for logs, far fewer were happy with simpli-
fying as they went or cancelling common factors (“cross-cancelling”) when multiplying
fractions, leading some to work with fractions with large numerators and denominators.
This approach frequently ended with arithmetical errors. Very few candidates applied the
9
logarithm rules to logs of fractions, such as writing ln 16 = ln 9 − ln 16 = 2 ln 3 − 4 ln 2.
This would have made the arithmetic far simpler and would have given them far more
chance of reaching the correct answer.

Question 5
This was a fairly unpopular question, attempted by only about 40% of the candidates
(and being one of the best six attempts of fewer than one-third of candidates). The
marks were also poor, with the median mark of the significant attempts being 3/20.
About one-sixth of all attempts failed to gain a single mark, even though writing out the
binomial expansion of (1 + x)n using binomial coefficients would have been sufficient for
this. About one-quarter of candidates failed to make any further progress beyond this
point, though another quarter substituted x = 1 and reached the result of part (i) before
giving up.
Those who persevered generally did reasonably well, with many correct answers to parts
(ii) and (iii), and a good number succeeding on part (iv) as well. It was surprising how
many successfully answered parts (ii) and (iii) using binomial coefficient manipulations
without any reference back to part (i).
In part (ii), a number of candidates attempted to differentiate 2n to get n.2n−1 . Others,
who used the binomial manipulation approach, were careless about the first and last terms
of the sum or made no attempt to justify why all of the corresponding middle terms were
equal.
On this note, some candidates expanded both sides of the identity they were trying to
prove and equated the corresponding terms without any understanding or indication of
why they were the same; this gained relatively few marks, especially in part (iv) where
the terms do not even correspond in this way.
In part (iii), those candidates who used a calculus-based approach frequently failed to
consider the constant of integration, and could not, therefore, justify the need for the −1
term.
Finally, whereas a good number of the calculus-based approaches succeeded on part (iv),
no candidate who used a binomial-manipulation method managed to extend their tech-
niques to this case.

390
Question 6
This was another popular and well-answered question. The median mark was 13, and
around one-fifth of attempts gained full marks.
Almost all candidates were fine with the first step and showing that (∗) holds. A significant
majority were also comfortable with deducing (∗∗), though there were some who had
difficulties in applying the product rule twice or appreciating that u was a function of x
rather than a constant.
The final part of the question, however, caused numerous difficulties for the majority of
candidates. First of all, some simply did not understand what was being asked of them,
and thought of v as a constant (even though they had appreciated that u itself is a function
of x). Then a significant number thought that d2 u/dx2 = v 2 rather than dv/dx, belying
a lack of understanding of the meaning of a second derivative. Those who overcame these
hurdles reached a correct first order ODE with no constant term, but many struggled to
solve it; even those who correctly separated the variables could not figure out how to
integrate x−2
x−1
, even though this is a standard A-level integration question. (Some tried
integration by parts, with a predicatable lack of success.)
Those who managed to integrate to determine ln v then went on to exponentiate, mostly
successfully, though a number forgot about the arbitrary constant or ended up with an
expression of the form v = f (x) + c instead of v = cf (x).
Those who reached this point generally appreciated that they now needed to write v =
du/dx and integrate once more, and most realised that now was the time to integrate
byRparts. Unfortunately, many forgot to multiply their original arbitrary constant by the
dv
− u dx dx term of the parts formula.
For the last step, a number of candidates thought that it was sufficient to simply plug
y = Ax + Bex into the differential equation; this, of course, gained no credit, as the
question had explicitly said “Hence show that . . . ”.

Question 7
This was a very unpopular question, attempted by only 20% of candidates and being
one of the six best questions of only 10%. It was also the worst-scoring: of the signifi-
cant attempts, the median mark was 1/20. However, those who managed to get beyond
the start of the question generally did quite well, resulting in an upper quartile mark
(among significant attempts) of 11/20. It generally indicated a very poor understanding
of vectors among those students who attempted this question. Only a handful of students
successfully completed the question.
Of those who attempted the question, many drew a decent diagram, which is very helpful
in understanding what is being asked, though few realised that C lies strictly inside
triangle OAB. Many also realised that they could simply write down the formula for q
directly by symmetry.
Most, however, appeared to be incapable of writing down the equation of a straight line
in vector form and were therefore unable to proceed any further.

391
Those who did often made their lives more difficult by writing the equation of OA as
r = a+λa rather than choosing the origin as their fixed point, giving the simpler equation
r = λa. While the former is clearly correct, it is messier to work with and therefore more
likely to lead to errors later.
A very common confusion was to write down the equations of the lines but to call them
−→ −−→
OA and BC instead of OA and BC; this sometimes led to candidates trying to equate the
−→ −−→
vectors as OA = BC, which was fairly nonsensical (and other candidates tried equating
the vectors without writing down the equations of the lines).
Another frequent piece of nonsense was an attempt to divide one vector by another or to
add a vector to a scalar.
Few candidates used notation carefully, and many suffered for it. It is strongly recom-
mended that students are taught to always distinguish their vector variables from their
scalar variables by underlining, under-squiggling or using arrows above them, and that
this is insisted upon. It makes it far less likely, then, that a student would write something
*
like a + λ = µ or a b (both of which were frequently seen).
˜˜
Some candidates attempted to write the formula for p in words and hoped that, somehow,
this would be sufficient justification. This was joined by several attempts to work forwards
from what was known and backwards from the desired result, which together with some
glue in between was meant to provide a convincing argument for the result.
Those who successfully completed the first part were generally successful with finding the
position vector of r as well.
The final part of the question was attempted by relatively few candidates. There was a
mixed level of success. Most realised that they needed to find the position vector of s,
and this was done fairly well; the algebra was the trickiest part here, as the ideas were
the same as earlier. The last step, proving the equality of the ratios, was a little trickier,
and there were a few attempts to divide vectors or to ignore problems with signs.

Question 8
This was a moderately popular question and candidates obtained a broad spread of marks
on it.
For the first part, almost all candidates successfully argued that a3 is divisible by 3, but
a significant number could not give a reasonably convincing explanation for why a itself
is divisible by 3. We did not require a perfect argument, but there had to be a mention
of 3 being prime. It was common to see this either
√ asserted
√ √ (“since 3 | a3 , then 3 | a”) or,
3 3 3
less commonly, something creative like: “a = a3 = 3 3c3 − b3 , and as a is an integer,
a must be divisible by 3”.
A good number of candidates went on to correctly explain why b and c were divisible
by 3, but quite a few talked about b3 = 3c3 − 13 a3 and c3 = 19 a3 + 13 b3 being divisible by 3
without any justification for these assertions; the idea of writing a multiple of 3 as 3k for
some k was appreciated by some candidates but overlooked by others. A few candidates
also made basic algebraic errors when rearranging the equation a3 + 3b3 = 9c3 ; more care
is needed!

392
It was nice to see that a fair number of candidates knew about modular arithmetic and
could use it to construct effective arguments both here and in part (ii).
A number of candidates showed some serious misconceptions about√divisibility,
√ with
√ a
3 3
common error being 3 | (r + s) ⇒ 3 | r and 3 | s. A few expanded a + b as a + b,
3

and there were even occurrences of r + s = t ⇒ r = t/s.

Other somewhat common failings were attempting to prove general statements by using
particular examples (such as “take a = 42”), making unsubstantiated assertions (for
example “since a3 /3 is an integer, so is a3 /9”), and misuse of “similarly”: just because
27 | a3 , it does not follow “similarly” that 27 | b3 and 27 | c3 unless it has already been
shown that 3 | b and 3 | c.
The final step of the argument, the infinite descent argument, was poorly understood and
served to differentiate the stongest candidates from the rest. Again, we did not expect a
perfect argument, but merely some explanation that a non-zero solution would give rise
to another, smaller, non-zero solution, and so on, which is impossible.
For part (ii), most candidates who attempted it realised that there were very few possi-
bilities for the final digits of fourth powers, though a few went to the effort of calculating
84 and 94 explicitly rather than just considering final digits. Some thought that 5r4 could
only end in 5, forgetting the possibility that it might end in 0. Nevertheless, many un-
derstood what they were meant to do and tried to argue that p and q must both be
multiples of 5. Some succeeded, but others made errors in their logic and did not consider
all possible cases, for example, many candidates considered several the case 5 - p but not
5 | p. Others effectively said “if 5 | p, then we must have 5 | q,” but did not show that
we must have 5 | p. Finally, as in part (i), there were some who argued that 5 | r4 but
did not explain how to deduce that 5 | r, and many who were stuck on the final proof by
infinite descent.
It was perhaps unsurprising that so many candidates appealed to Fermat’s Last Theorem
in their attempts to prove these results, even though it was not at all relevant to the
question. (It was equally unsurprising that there was almost no use of Fermat’s Little
Theorem to help with the calculations required in part (ii) by stating that a4 ≡ 1 (mod 5)
whenever 5 - a.)

Question 9
This was the most popular of the mechanics questions, being attempted by about one-third
of candidates (though this question counted towards the final mark of only two-thirds of
these). The mean mark was about 6/20.
A number of the attempts struggled to calculate the distance of the centre of mass from
the wall, though most were able to do so using a quick sketch.
Unfortunately, many candidates gave up at this point, unsure of what to do next. It
should be second-nature that for a large-body question, the “right” thing to do is “resolve
twice, moments once”. Others tried this but failed to do so correctly: first of all, many
drew poor or confusing diagrams; it is vital that candidates draw diagrams which are clear
enough to understand what is going on at every point. It was sadly common to see friction
labelled as F r, where this could easily be confused with F × r in some contexts: students

393
should always be taught to use single-letter variable names. Further, many students were
inconsistent with their force labelling: some labelled the reaction as RA at A but the
friction as RB at B (where A and B are the points of contact) or similar gaffes—this, of
course, led to confusion and errors later. Others used the same variable for two different
reaction forces or two different friction forces. Yet others left out one of the forces.
Even with the hurdle of an accurate diagram overcome, many only resolved once rather
than twice, and a few tried taking moments around two or three different points but never
resolving. (The latter can be made to work, but is usually far more effort than necessary.)
Taking moments was also found to be challenging: most attempts failed by forgetting
a force or by not understanding the meaning of “perpendicular distance”. (For exam-
ple, referring to the diagram in the sample solutions, when taking moments about B,
candidates would have R1 contributing R1 × 2a or occasionally R1 × 2a cos α instead of
R1 × 2a sin α.) Some candidates also got their signs wrong. I personally encourage my
students to always indicate the orientation in which they are taking moments (clockwise
y
or anticlockwise, indicated with a small curved arrow as in M (A)) and to place all of the
moments on the same side of the equation with the appropriate sign; this will also help if
they ever come to learn about moments of inertiaP and angular acceleration, as they will
then be more confident working with the formula F d = I θ̈.
A surprising number switched the a and b at some point in the question.
Most of those who correctly reached this point were able to make good progress towards
the required conclusions (though a few, sadly, did not attempt the final part of the
question, even though it was fairly straightforward). They showed a good command
of the trigonometric identities required and were confident in manipulating the equations
to eliminate the forces.
Unsurprisingly, there were almost no candidates who used the Three Forces Theorem
approach.

Question 10
This question was attempted by about 20% of the candidates, though many became stuck
fairly early on in the question.
For the first part of the question, the majority of candidates showed a good understanding
of how to differentiate a vector, though close to half got no further than finding the velocity
and acceleration (often correctly, often with errors). Most of the rest then went on to
evaluate scalar products to find the angle between two vectors, usually successfully. A
small number took the elegant geometric-trigonometric approach very successfully.
Those who reached this point sketched the path of the particle, but with little thought
for what they had just done: almost no candidates had a diagram in which the direction
of motion was at 45◦ to the position of the particle even at the start point. Some drew a
vague swirl, but most at least indicated one or two coordinates (usually correctly). Only
two candidates were awarded full marks for their sketch, which required just a couple
of coordinates and a clear (explicit or implicit) indication of the direction of motion
somewhere along the path.

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The final part of the question was very poorly done. Few even attempted it, and of those

who did, many thought that a time delay of T meant that rQ = et+T cos(t + T )i + · · ·
instead of using t − T . Those who got this far often concluded that P Q is proportional
to et simply because they reached an equation of the form
−→ p
|P Q| = et × some expression involving t and T ,
without simplifying the square root to eliminate t.

Question 11
This was the least popular of the mechanics questions, yet the best-answered.
Nonetheless, the very standard start of the question proved to be a major stumbling-block
for many candidates, with over a third unable to write down a pair of correct equations
for the collision. The reason for this was very simple: about half of candidates failed to
draw a diagram; this led to them trying to keep the directions in their heads, with the
predictable consequence that most had inconsistent signs in their equations of momentum
and restitution. It was almost impossible to make any significant further progress in such
cases. (It also made the examiners’ lives significantly harder, but they were not penalised
for this!)
The majority of those who did write down two correct equations were generally able to
solve them and reach correct expressions for the velocities of the two particles after the
collision. (Although the question had asked for the speeds rather than the velocities, full
marks were awarded for just determining the velocities.)
Many candidates who reached this point only found the conditions for one of the two
particles to change direction (deducing the 2em > M − m required), but did not give an
adequate (or any) explanation for why the lighter particle also changed direction.
A common problem in this first half was that candidates again misread their own writing,
confusing M and m. In the second half, some candidates similarly confused V and v.
There were few candidates who attempted the second half of the question. Almost no
candidates gave a convincing explanation for the given equation (making little or no
reference to the circular track or the change of direction at every collision). Those who
understood what was going on (whether or not they explained it well) and went on to try
to determine v and V were frequently careless in their counting of collisions, ending up
with expressions involving e2n rather that e2n+1 .

Question 12
The two probability questions were each attempted by about 10% of candidates. This
was the better-answered of them.
The majority of candidates were able to write down the definition of E(X) in this context,
but very few figured out how to expand the definition to produce the required result.
In the second part of the question, many candidates were capable of determining P(X > 4)
by a variety of arguments, often involving writing P(X > 4) = 1 − P(X 6 3) and then

395
considering cases. Unfortunately, most of these arguments did not generalise easily, so
that the required result of P(X > n) = pn−1 + q n−1 was not deduced, and this left most
students stuck at this point. Also, quite a few were unable to determine P(X = 2) or
P(X = 3) correctly, which did not help either.
Most candidates, even those who had become stuck earlier, attempted the final part of
the question. There seemed to be a widespread understanding that pq is maximised at
p = q = 12 , or that pq
1
> 4, but very few actually proved this (and justification was
required).

Question 13
This question was the most poorly answered on the paper, with over half of attempts
scoring no marks.
Nonetheless, most candidates were capable of writing down the pdf of a Poisson distribu-
tion, but only a minority understood that they needed to consider two different Poisson
distributions to make any progress.
Worse still, it was very common to see candidates writing things like: “Let X be the
number of texts received. Then P(1 < X < 2) = · · · .” This shows a total lack of
understanding of what the Poisson distribution is doing: there is no time period given
in the definition of X, and how could the number of texts lie strictly between 1 and 2?
Were the candidates to have let X be the waiting time until the first text, this would have
made sense, but at this level, most candidates have not yet met this concept.
Even those who progressed beyond this point and actually managed to reach the required
quadratic in eλ generally became stuck when trying to show that there are two positive
values of λ: they showed (or tried to show) that eλ > 0 rather than the necessary eλ > 1.
Those who tried the second part generally did not appreciate that eλ1 and eλ2 are the
two roots of the equation pe2λ − eλ + 1 = 0, and often used the two possible roots for
each of eλ1 and eλ2 , leading to some nonsensical answers; very few reached the required
expression for λ1 + λ2 . Also, none of the candidates who reached this point seemed to
know (or use) the result that the product of roots of ax2 + bx + c = 0 is c/a; this is a very
useful tool for students to have.
Finally, in the last part of the question, very few of the candidates were capable of finding
an event involving the two phones equivalent to “the first text arrives between 1 and 2
hours”, leaving them unable to make any meaningful progress. Drawing a Venn diagram or
listing possibilities would have been of help, but there was little evidence that candidates
used techniques such as these.

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9470 STEP II 2010 Report
General Remarks
There were just under 1000 entries for paper II this year, almost exactly the same number as last
year. Of this number, more than 60 scored over 90% while, at the other end of the scale, almost
200 failed to score more than 40 marks. In hindsight, many of the pure maths questions were a
little too accessible and lacked a sufficiently tough ‘difficulty gradient’, so that scores were
slightly higher than anticipated. This was reflected in the grade boundaries for the “1” and the
“2” (around ten marks higher than is generally planned) in particular. Next year’s questions may
be expected to be a little bit more demanding, but only in the sense that the final 5 or 6 marks on
each question should have rather more bite to them: it should certainly not be the case that all
questions are tougher to get into at the outset.

Most candidates attempted the requisite number of questions (six), although many of the weaker
brethren made seven or eight attempts, most of which were feeble at best and they generally only
picked up a maximum of 5 or 6 marks per question. It is a truth universally acknowledged that
practice maketh if not perfect then at least a whole lot better prepared, and choosing to waste
time on a couple of extra questions is not a good strategy on the STEPs. The major down-side of
the present modular examination system is that students are not naturally prepared to approach
the subject holistically; ally this to the current practice of setting highly-structured, fully-guided
questions requiring no imagination, insight, depth or planning from A-level candidates in a
system that fails almost nobody and rewards even the most modestly able with high grades in a
manner reminiscent of a dentist giving lollipops to kids who have done little more than been
brave and seen the course through, it is even more important to ensure a full and thorough
preparation for these papers. The 20% of the entry who seem to be either unprepared for the
rigours of a STEP, or unwittingly possessed of only a smattering of basic advanced-level skills,
seems to be remarkably steady year-on-year, even in a year when their more suitably prepared
compatriots found the paper appreciably easier than usual.

As in previous years, the pure maths questions provided the bulk of candidates’ work, with
relatively few efforts to be found at the applied ones.

Comments on individual questions

Q1 This question was attempted by almost two-thirds of the candidature, with a mean mark
of around 11 12 . Whilst most attempts were very successful, a lot of marks were lost by poorly
structured working, where the candidate got themselves confused in some way or another. The
only two common conceptual difficulties were the oversight of the equal gradients at the point of
contact and the lack of a suitable circle equation to start working with. Apart from these, most
candidates’ work went smoothly and successfully, although sign errors often cost them at least
one of the final three answer marks.

Q2 This was the most popular question on the paper, drawing an attempt from almst every
candidate. There were several proofs of the initial trigonometric identities using de Moivre’s
Theorem but most settled for the more standard cosine and sine of (2x + x). Personally, I was
against the inclusion of the given answer of cos– 1  16  in (ii) as it led to what struck me as an
unwelcome dichotomy of approaches. Most candidates opted to verify that the two polynomials
in “c” that arose gave the same numerical answer, and this working was not entirely
straightforward – in the event, lots of candidates failed to show the markers that they had done
the working correctly for both expressions – whereas my original intention had been that they
should collect terms up into a single polynomial equation and factorise it by first spotting the
(repeated) factor (c – 1) hinted at in (i).
There was one important mathematical oversight that many candidates made during this
question, and it was due to not reading the question sufficiently carefully. The wording of the

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9470 STEP II 2010 Report
question in (ii) clearly states that Eustace’s misunderstanding of the integration of powers of the
sine function was for n = 1, 2, 3, … . Unfortunately, rather a lot of candidates thought that he
would then have integrated sin x (i.e. the case n = 1) correctly as –cos x. We concocted a mark-
scheme for this eventuality which allowed candidates ‘follow-through’ for 6 out of the 10 marks
allocated here, but the self-imposed penalty of four marks could not be avoided as it was just no
longer possible to get, for instance, the given answer.
Finally, there is a bit of an apology to make: at some final stage of the printing process,
the bit of the question that identified  as lying in the range 0 to  got removed; this left
candidates having to think about general solutions rather than just the two decently small ones
that had been looked-for when the question was first written. Nevertheless, not only was this the
most popular question for number of attempts, it was also the most successful for candidates with
a man score of almost 15.

Q3 One doesn’t need to be too devoted a mathematician to recognise the Fibonacci numbers
in this question, and many candidates clearly recognised this sequence. However, they were still
required to answer the question in the way specified by the wording on the paper and a lot of
attempts foundered at part (ii). This was the second most frequently attempted question, yet drew
the second worst marks, averaging just over 8. Most attempts got little further than (i), and many
foundered even here due to a lack of appreciation of the difference of two cubes factorisation.
Things clearly got much worse in (ii) when far too many folks seemed incapable of attempting a
 6
binomial expansion of 1 5 ; many who did manage a decent stab at this then repeated the
 
6
work for 1 5 . Very few sorted this out correctly and, as a result, there were relatively few
stabs at part (iii).

Q4 This question received about the same number of “hits” as Q1 and came out with an
average mark only fractionally lower. For the majority, the introductory work was successfully
completed along with the rest of (i), although a lot of candidates’ working was very unclear in
the first integral, involving logarithms. One or two marks were commonly lost as the correct
answer of 12 could easily have been guessed from the initial result, and the working produced by
the candidates failed to convince markers that it had been obtained legitimately otherwise. The
fault was often little more than failing either to identify the relevant “f(x)” or to show it
implicitly by careful presentation of the working of the log. function.
The excellent part (ii) required candidates to mimic the method used to find the opening
result rather than repeat its use in a new case, and this was only accessible to those with that
extra bit of insight or determination.

Q5 This was the least popular question on the paper and attracted the lowest average score of
about 7. This is partly explained by the way that, like Q3 and Q6 particularly, it drew a lot of
attempts from desperate weaker students who started, only to give up before too long (in order,
presumably, to try yet another question in some hit-and-miss approach, scrambling for odd
marks here and there). Of those who persevered, there were plenty of marks to be had. Little
more was required than the use of the scalar product, a careful application of algebra, and a
modest grasp of the geometrical implications of what the working represented.

Q6 Of the pure maths questions on the paper, only Q5 and this one attracted attempts from
under half the candidature; this despite the fact that it is obviously (to the trained eye, at least)
the easiest question on the paper. Parts (i) and (ii) require nothing more than GCSE
trigonometry, and (iii) can be done in one line if one knows a little bit about geometric centres of
3-d shapes. Clearly 3-dimensional objects, and the associated trig., are sufficiently daunting to
have put most folks off either completely or early on in the proceedings, and the average mark
scored here was under 10.

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9470 STEP II 2010 Report
Q7 This proved to be the second most popular question on the paper, both by choice and by
success. I imagine that its helpful structure probably contributed significantly to both. Part of the
problem is that there are ways to do this using methods not on single maths specifications, so it
was necessary to be quite specific. Nonetheless, there were still areas where marks were
commonly lost; in (i), candidates were required to show that both TPs lie below the x-axis and,
while one of the y-coordinates was obviously negative (being the sum of three negative terms),
the other one was only obviously so by completing the square. The problems found by
candidates, even in the first case, just highlights the widespread difficulty found by students
when dealing with inequalities.

Q8 This was the sixth most popular of the pure maths questions, with an average score of just
under 11. Many of the early marks were easily gained, although the sketches were often unclear
enough to warrant a loss of marks – in particular, the fact that the required function oscillates
between e – x and –e – x was seldom made obvious; indeed, a clear indication that the function’s
zeroes occurred at (regular) intervals of  units on the x-axis was also poorly indicated. Although
most candidates were happy to attempt integration by parts successfully, and then subtract areas,
the limits of integration, xn and xn + 1, were only occasionally correctly identified. This meant that
a lot of the following work, whilst correct in method, was seldom likely to get to the correct,
given answer. The final piece of work, even though it could be found using this given answer,
was poorly attempted.

Q9 Despite the fact that this question required two pieces of identical working in order to
obtain the given results (the second following almost immediately from the first, if reasoned
appropriately) and that each could be obtained by considering either distances or times, this was
a very unpopular question, eliciting only 153 attempts scoring an average of under 7 marks (the
poorest average mark of all questions). The key observation was that the two particles are always
at the same height, hence share a common vertical component of speed.

Q10 Eliciting exactly the same number of responses as Q9, this question was found a little
easier, but only because the first part was very standard A-level “collisions” work. Applying this
first result repeatedly required only clear thinking and clear presentation, and those who
persevered generally scored quite highly and opened up the prospect of some straightforward
log. work in the final part of (ii). The biggest hurdle to a completely successful solution usually
arose in poor numerical justification of the final answer.

Q11 This was the most popular of the applied maths questions, shortly ahead of Q12 for the
number of “hits” received but still well behind the popularity of any of the pure questions.
Having initially expected that candidates would recognise a ‘3-force problem’ and use Lami’s
Theorem or a triangle of forces, no-one did. Instead, attempts merely went for the “resolve twice
and take moments” strategy, which worked very well in principle, but were often hampered by
lack of care over the angles involved. Uses of the sin/cos(A  B) formulae were good, although a
lot of candidates got a bit confused before arriving at the given result. The final piece of work
was just a bit of pure maths. Interestingly enough, just one or two candidates appealed to a result
I had never heard of before: the “Cot Rule”, which (upon investigation) turned out to be perfectly
legitimate.

Q12 Only marginally less popular than Q12, and scoring marginally better on average (10.6
against 10.4), at least the first half of the question was straightforward work on a continuous
probability distribution. Those who kept their nerve in (ii) when dealing with the median found
many of the later marks were easily acquired also, the biggest hurdle to complete success (again)
being the poor skills on display when justifying results involving inequalities.

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9470 STEP II 2010 Report
Q13 This question worked almost exactly like Q12, in the sense that the start required some
straightforward (probability) work, followed by an extension that needed only careful handling,
before finishing with some poorly-handled inequalities work. The biggest problem for candidates
lay in their lack of care to show that their chosen values of p and q actually satisfied any claimed
conditions.

T F Cross
Principal Examiner

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STEP Mathematics III 2010: Report

About 80% of candidates attempted at least five questions, and well less than 20% made genuine
attempts at more than six. Those attempting more than six questions fell into three camps which
were those weak candidates who made very little progress on any question, those with four or
five fair solutions casting about for a sixth, and those strong candidates that either attempted 7th
or even 8th questions as an “insurance policy” against a solution that seemed strong but wasn’t,
or else for entertainment!

Section A: Pure Mathematics

1. This was a very popular question, and the first two parts usually scored full marks. The
expression of D in part (iii) caused some problems with inaccurate algebra which then made the
last two results unobtainable. Those that simplified D most neatly were in a stronger position to
finish the question, though “if and only if” was frequently ignored, or only lip-service was paid
to it. Consequently, scores were well-spread.

2. The most popular question, the scoring rate was very similar to the first. Quite a few
candidates did not take the hint provided in part (i) to express cosh a in terms of exponentials in
order to perform the integration. However, apart from those that did not correctly substantiate
the given result, many handled the partial fractions and exponentials well, and quite a number
dealt with the infinite limit impressively. Problems arose later in the question with manipulating
logarithms and the instruction to express answers in terms of hyperbolic functions was either
overlooked or beyond their capacity.

3. Just over half the candidates attempted this question with most scores being quarter, half
or three quarters in equal shares. Most candidates understood the idea of the question, the
definition of a primitive root, and many wrote the roots of unity in (modulus) -argument form or
exponential form. Failure to present a logical argument in parts (ii) and (iv) was a common
problem and C6(x) tripped up quite a few.

4. This was a popular question, though it was not generally well scored upon, with very few
candidates earning full marks. Most began strongly, and finished by finding the values of b
correctly. However, basic sign errors did prevent some from achieving the numerical pay-off.
Part (ii) was, as expected, found trickier than part (i). Overall, the non-triviality of “if and only
if” was rarely addressed as an issue in either part.

5. This question resembled question 3 in popularity and success. Most were able to derive
line equations reliably, and address the intersection problem. (Those that used an equally valid
vector formulism had a low success rate for no apparent reason.) Very few addressed whether or
not factors that were being divided by were non-zero. Mistaking m for n and vice versa, careless
algebraic errors, and overlooking which equation represented which line caused problems in
trying to find T. The idea of explaining the construction verbally in the last part exposed that
many candidates are not used to expressing a formal argument in words. The nicety of this
question is that whilst all candidates will have encountered geometrical constructions involving
straight edge and compass, few will have previously met one that only requires a straight edge.

401
6. About a tenth of the candidates attempted this, with less success than nearly all other
questions on the paper. Part (i) caused few problems, but at some point in part (ii), errors were
frequently made or lack of attention to which of the two angles in parts (i) and (ii) was being
employed in which rotation, and so even those few that knew how to attempt part (iii) were
thwarted.

7. Just over 60% attempted this question, achieving moderate success. The opening result
was well done, but the two similar equations foundered frequently on incorrect differentiation. If
these two were correctly obtained, then the conjecture and induction were usually correct.
Appreciating that the final expression was actually a polynomial, and what this entails, passed
most by.

8. Three quarters of the candidates had a go at this, with moderate success. Most
understood the method intended for part (i) and were aware of the method of using an integrating
factor. Algebraic slips led to incorrect simultaneous equations in part (i), and few dealt with the
non-uniqueness of R(x) satisfactorily. Having found the integrating factor for part (ii), most did
not proceed further. Some candidates introduced a sign error into part (ii) which trivialized the
left hand side to a differential of a product. A small number of candidates produced elegant
solutions to part (ii) using the tan half angle substitution.

Section B: Mechanics

9. Less than a fifth of the candidates attempted this, though it was the most popular of the
non-Pure questions. Candidates were largely fairly successful or struggled to get started. Some
of those failing to get anywhere equated the normal reaction on P to the component of P’s
weight, completely ignoring the radial acceleration, and others got the sign of the force wrong.
Nearly every candidate failed to justify imposing the non-negative condition on the normal
reaction when .

10. Only 5% of the candidates attempted this and it was another case of nearly all or nothing.
Even the mostly successful candidates rarely handled the small oscillation algebraic
manipulation correctly, often overlooking using the small angle result throughout the expression,
so very few obtained the correct period though the principle was understood.

11. Slightly fewer attempted this than question 9, and this question was least well scored
upon of any on the paper. Generally, candidates got through part (i) successfully and then either
gave up or got right through the question. Common errors were the misapplication of
conservation of momentum, failure to distinguish directions which led to negative signs which
were then mis-handled to obtain the quoted answer in (i), and even strong candidates failing to
appreciate that acceleration was constant making the later parts all susceptible to constant
acceleration formulae and thus not requiring less direct approaches. The brief description at the
end was usually restricted to only trivially considering the block, and few gave any thought to
the bullet.

402
Section C: Probability and Statistics

12. Although this was marginally less popular than question 11, the success achieved was
similar to that on the first two questions. A small number of candidates didn’t get started but
most found the first parts straightforward and dealt with the manipulation and summation of the
geometric series correctly. Many found an incorrect “shortcut” on the last part, despite having a
good idea how to attempt it correctly having completed the earlier parts.

13. This was the least popular question with little more than a couple of handfuls of attempts.
In view of the small number of attempts, there were no detectable trends though oddly, the very
few candidates who mastered this question conspired to avoid full marks by making minor
algebraic inaccuracies having dealt with all the trickier aspects.

403
Explanation of Results STEP 2010

All STEP questions are marked out of 20. The mark scheme for each question is designed to reward
candidates who make good progress towards a solution. A candidate reaching the correct answer will
receive full marks, regardless of the method used to answer the question.

All the questions that are attempted by a student are marked. However, only the 6 best answers are used in
the calculation of the final grade for the paper.

There are five grades for STEP Mathematics which are:

S – Outstanding
1 – Very Good
2 – Good
3 – Satisfactory
U – Unclassified

The rest of this document presents, for each paper, the grade boundaries (minimum scores required to
achieve each grade), cumulative percentage of candidate achieving each grade, and a graph showing the
score distribution (percentage of candidates on each mark).

STEP Mathematics I (9465)

Grade boundaries
Maximum Mark S 1 2 3 U
120 103 84 70 47 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 4.8 16.9 35.4 77.7 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics I

www.admissionstests.cambridgeassessment.org.uk
404
STEP Mathematics II (9470)

Grade boundaries
Maximum Mark S 1 2 3 U
120 105 79 64 40 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 9.6 30.2 48.1 82.4 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics II

STEP Mathematics III (9475)

Grade boundaries
Maximum Mark S 1 2 3 U
120 78 56 46 29 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 15.3 41.7 60.1 88.7 100.0

Distribution of scores

3.5

3.0

2.5
Percent

2.0

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics III

www.admissionstests.cambridgeassessment.org.uk
405
406
STEP Solutions
2010

Mathematics
STEP 9465/9470/9475

October 2010

407
The Cambridge Assessment Group is Europe's largest assessment agency
and plays a leading role in researching, developing and delivering
assessment across the globe. Our qualifications are delivered in over 150
countries through our three major exam boards.

Cambridge Assessment is the brand name of the University of Cambridge

Local Examinations Syndicate, a department of the University of Cambridge.
Cambridge Assessment is a not-for-profit organisation.

This mark scheme is published as an aid to teachers and students, to indicate

the requirements of the examination. It shows the basis on which marks were
awarded by the Examiners. It does not indicate the details of the discussions
which took place at an Examiners’ meeting before marking commenced.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the
relevant knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question

papers and the Report on the Examination.

Cambridge Assessment will not enter into any discussion or correspondence

in connection with this mark scheme.

© UCLES 2010

More information about STEP can be found at:

http://www.atsts.org.uk

408
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 4
STEP Mathematics II 44
STEP Mathematics III 54

409
Question 1

Given that

5x2 + 2y 2 − 6xy + 4x − 4y ≡ a(x − y + 2)2 + b(cx + y)2 + d,

find the values of the constants a, b, c and d.

We expand the right hand side, and then equate coefficients:

5x2 + 2y 2 −6xy + 4x − 4y
≡ a(x − y + 2)2 + b(cx + y)2 + d
≡ a(x2 − 2xy + y 2 + 4x − 4y + 4) + b(c2 x2 + 2cxy + y 2 ) + d
≡ (a + bc2 )x2 + (2bc − 2a)xy + (a + b)y 2 + 4ax − 4ay + 4a + d,

so we require

a + bc2 = 5
2bc − 2a = −6
a+b=2
4a = 4
−4a = −4
4a + d = 0.

The fourth and fifth equations both give a = 1 immediately, giving b = 1 from the third
equation. Then the second equation gives c = −2 and the final equation gives d = −4.
We must also check that this solution is consistent with the first equation. We have
a + bc2 = 1 + 1 × (−2)2 = 5, as required. (Why is this necessary? Well, if the second
equation had begun with 7x2 + · · · , then our method would still have given us a = 1, etc.,
but the coefficients for the x2 term would not have matched, so we would not have been
able to write the second equation in the same way as the first.)
We thus deduce that

5x2 + 2y 2 − 6xy + 4x − 4y ≡ (x − y + 2)2 + (−2x + y)2 − 4.

Solve the simultaneous equations

5x2 + 2y 2 − 6xy + 4x − 4y = 9, (1)

6x2 + 3y 2 − 8xy + 8x − 8y = 14. (2)

Spurred on by our success in the first part, we will rewrite the first equation in the
suggested form:
(x − y + 2)2 + (y − 2x)2 − 4 = 9. (3)

410
We are led to wonder whether the same trick will work for the second equation, so let’s
try writing:

6x2 + 3y 2 − 8xy + 8x − 8y ≡ a(x − y + 2)2 + b(cx + y)2 + d.

As before, we get equations:

a + bc2 = 6
2bc − 2a = −8
a+b=3
4a = 8
−4a = −8
4a + d = 0.

(We can write these down as the right hand side is the same as before.)
This time, a = 2 from both the fourth and fifth equations, so we get b = 1 from the
third equation. The second equation gives us c = −2. Finally, the sixth equation gives
us d = −8.
We must now check that our solution is consistent with the first equation, which we have
not yet used. The left hand side is a + bc2 = 2 + 1 × (−2)2 = 6, which works, so we can
write the second equation as

2(x − y + 2)2 + (y − 2x)2 − 8 = 14.

(If we had not checked for consistency, we might have wrongly concluded that 183x2 +
3y 2 − 8xy + 8x − 8y can also be written in the same way.)
These two equations now look remarkably similar! In fact, let’s move the constants to the
right hand side and write them together:

(x − y + 2)2 + (y − 2x)2 = 13
2(x − y + 2)2 + (y − 2x)2 = 22.

We now have two simultaneous equations which look almost linear. In fact, if we write
u = (x − y + 2)2 and v = (y − 2x)2 , we get

u + v = 13
2u + v = 22

which we can easily solve to get u = 9 and v = 4.

Therefore, we now have to solve the two equations

(x − y + 2)2 = 9 (4)
(y − 2x)2 = 4. (5)

We can take square roots, so that (4) gives x − y + 2 = ±3 and (5) gives y − 2x = ±2.

411
Thus we now have four possibilities (two from equation (4), and for each of these, two from
equation (5)), and we solve each one, checking our results back in the original equations.

2x − y x − y + 2 x y LHS of (1) LHS of (2)

2 3 1 0 9 14
2 −3 7 12 9 14
−2 3 −3 −4 9 14
−2 −3 3 8 9 14

Therefore we see that the four solutions are (x, y) = (1, 0), (7, 12), (−3, −4) and (3, 8).

An alternative is to observe that equation (2) looks almost double equation (1), so we
consider 2 × (1) − (2):
4x2 + y 2 − 4xy = 4.
But the left hand side is simply (2x − y)2 , so we get 2x − y = ±2.
Substituting this into equation (3) gives us

(x − y + 2)2 + 4 − 4 = 9,

so that x − y + 2 = ±3.
Thus we have the four possibilities we found in the first approach, and we continue as
above.

Yet another alternative approach is to subtract (2) − (1) to get

x2 + y 2 − 2xy + 4x − 4y = 5,

so that
(x − y)2 + 4(x − y) = 5.
Writing z = x − y, we get the quadratic z 2 + 4z − 5 = 0, which we can then factorise to
give (z + 5)(z − 1) = 0, so either z = 1 or z = −5, which gives x − y = 1 or x − y = −5.
Substituting x − y = 1 into (3) now gives

(1 + 2)2 + (y − 2x)2 − 4 = 9,

so that (y − 2x)2 = 4; substituting x − y = −5, on the other hand, would lead us to

(−5 + 2)2 + (y − 2x)2 − 4 = 9,

and again we deduce (y − 2x)2 = 4.

We have again reached the same deductions as in the first approach, so we continue from
there.

412
Question 2

x − a
The curve y = ex , where a and b are constants, has two stationary points.
x−b
Show that
a−b<0 or a − b > 4.

We begin by differentiating using first the product rule and then the quotient rule:
d x − a x (x − a) x
 
dy
= e + e
dx dx x − b (x − b)
(x − b).1 − (x − a).1 x (x − a) x
= e + e
(x − b)2 (x − b)
(a − b) x (x − a)(x − b) x
= e + e
(x − b)2 (x − b)2
x2 − (a + b)x + (ab + a − b) x
= e .
(x − b)2
dy
Now solving dx = 0 gives x2 −(a+b)x+(ab+a−b) = 0. Since the curve has two stationary
points, this quadratic must have two distinct real roots. Therefore the discriminant must
be positive, that is
(a + b)2 − 4(ab + a − b) > 0,
and expanding gives a2 − 2ab + b2 − 4a + 4b > 0, so (a − b)2 − 4(a − b) > 0. Factorising
this last expression gives
(a − b)(a − b − 4) > 0,
so (sketching a graph to help, possibly also replacing a − b with a variable like x), we see
that we must either have a − b < 0 or a − b > 4.

(i) Show that, in the case a = 0 and b = 12 , there is one stationary point on either
side of the curve’s vertical asymptote, and sketch the curve.

 x 
x
We are studying the curve y = 1 e .
x− 2
We have a − b = − 12 < 0, so the curve has two stationary points by the first part of the
question. The x-coordinates of the stationary points are found by solving the quadratic

x2 − (a + b)x + (ab + a − b) = 0,

as above.
Substituting in our values for a and b, we get x2 − 12 x − 12 = 0, so 2x2 − x − 1 = 0,
which factorises to (x − 1)(2x + 1) = 0. Thus there are stationary points at (1, 2e) and
(− 12 , 12 e−1/2 ).
The vertical asymptote is at x = b, that is at x = 12 .

413
Therefore, since the two stationary points are at x = 1 and x = − 12 , there is one stationary
point on either side of the curve’s vertical asymptote.
We note that the only time the curve crosses the x-axis is when x = a, so this is when
x = 0, and this is also the y-intercept in this case.
As x → ±∞, y ∼ ex (meaning y is approximately equal to ex ; formally, we say that y is
asymptotically equal to ex ), as the fraction (x − a)/(x − b) tends to 1.
We can also note where the curve is positive and negative: since ex is always positive,
y > 0 whenever both x − a > 0 and x − b > 0, or when both x − a < 0 and x − b < 0, so
y < 0 when x lies between a and b and is positive or zero otherwise.
Using all of this, we can now sketch the graph of the function. The nature of the stationary
points will become clear from the graphs. In the graph, the dotted lines are the asymptotes
(x = 12 and y = ex ) and the red line is the graph we want, with the stationary points
indicated.
y

(1, 2e)

(− 12 , 12 e−1/2 )
0 1 x
2

9
(ii) Sketch the curve in the case a = 2
and b = 0.

x − 9 
2
This time, we are studying the curve y = ex .
x
Proceeding as in (i), we have a − b = 92 > 4, so again, the curve has two stationary points.
The x-coordinates of the stationary points are given by solving the quadratic

x2 − (a + b)x + (ab + a − b) = 0,

as above.
Substituting our values, we get x2 − 92 x + 92 , so 2x2 − 9x + 9 = 0. Again, this factorises
nicely to (x − 3)(2x − 3) = 0, giving stationary points at ( 32 , −2e3/2 ) and (3, − 12 e3 ).
The vertical asymptote is at x = b, that is at x = 0. This time, therefore, the stationary
points are both to the right of the vertical asymptote.
The x-intercept is at x = a, that is, at ( 92 , 0). There is no y-intercept as x = 0 is an
asymptote.
Again, as x → ±∞, y ∼ ex .

414
As in (i), y < 0 when x lies between a and b and is positive or zero otherwise.
Using all of this, we can now sketch the graph of this function. Note that the asymptote
y = ex is much greater than y until x is greater than 20 or so, as even then (x−a)/(x−b) ≈
15/20, and only slowly approaches 1. We don’t even attempt to sketch the function for
such large values of x!

0 9 x
2

( 32 , −2e3/2 )

(3, − 12 e3 )

415
Question 3

Show that
sin(x + y) − sin(x − y) = 2 cos x sin y
and deduce that
sin A − sin B = 2 cos 12 (A + B) sin 12 (A − B).

We use the compound angle formulæ (also called the addition formulæ) to expand the
left hand side, getting:

sin(x + y) − sin(x − y) = (sin x cos y + cos x sin y) − (sin x cos y − cos x sin y)
= 2 cos x sin y,

as required.
For the deduction, we want A = x + y and B = x − y, so x = 12 (A + B) and y = 12 (A − B),
solving these two equations simultaneously to find x and y. Then we simply substitute
these values of x and y into our previous identity, and we reach the desired conclusion:

sin A − sin B = 2 cos 12 (A + B) sin 12 (A − B).

(This identity is known as one of the factor formulæ.)

Show also that

cos A − cos B = −2 sin 12 (A + B) sin 12 (A − B).

Likewise, we have

cos(x + y) − cos(x − y) = (cos x cos y − sin x sin y) − (cos x cos y + sin x sin y)
= −2 sin x sin y,

so again substituting x = 12 (A + B) and y = 12 (A − B) gives

cos A − cos B = −2 sin 12 (A + B) sin 12 (A − B).

416
 The points P , Q, R and S have coordinates (a cos p, b sin p), (a cos q, b sin q),
(a cos r, b sin r) and (a cos s, b sin s) respectively, where 0 6 p < q < r < s < 2π,
and a and b are positive.
Given that neither of the lines P Q and SR is vertical, show that these lines are parallel
if and only if
r + s − p − q = 2π.

Remark: The points P , Q, R and S all lie on an ellipse, which can be thought of as a
stretched circle, as their coordinates all have xa = cos θ and yb = sin θ, so they satisfy the

2
2
equation xa + yb = 1.

The lines P Q and SR are parallel if and only if their gradients are equal (and neither are
vertical, so their gradients are well-defined), thus
b sin q − b sin p b sin s − b sin r
P Q k RS ⇐⇒ =
a cos q − a cos p a cos s − a cos r
sin q − sin p sin s − sin r
⇐⇒ =
cos q − cos p cos s − cos r
2 cos 2 (q + p) sin 12 (q − p)
1
2 cos 12 (s + r) sin 12 (s − r)
⇐⇒ =
−2 sin 12 (q + p) sin 12 (q − p) −2 sin 12 (s + r) sin 12 (s − r)
cos 12 (q + p) cos 12 (s + r)
⇐⇒ =
− sin 12 (q + p) − sin 12 (s + r)
⇐⇒ cot 12 (q + p) = cot 12 (s + r)
⇐⇒ 12 (q + p) = 12 (s + r) + kπ for some k ∈ Z
⇐⇒ q + p = s + r + 2kπ for some k ∈ Z
⇐⇒ r + s − p − q = 2nπ for some n ∈ Z.
The last four lines could have also been replaced by the following:
P Q k RS ⇐⇒ · · ·
cos 12 (q + p) cos 12 (s + r)
⇐⇒ =
− sin 12 (q + p) − sin 12 (s + r)
⇐⇒ cos 12 (q + p) sin 12 (s + r) = cos 12 (s + r) sin 12 (q + p)
⇐⇒ sin 12 (s + r) cos 12 (q + p) − cos 12 (s + r) sin 12 (q + p) = 0
⇐⇒ sin 12 ((q + p) − (s + r)) = 0
1
⇐⇒ 2
(q + p − s − r) = π for some k ∈ Z
⇐⇒ r + s − p − q = 2nπ for some n ∈ Z.

We are almost there; we now only need to show n = 1 in the final line. We know that
0 6 p < q < r < s < 2π, so r +s < 4π and 0 < p+q < r +s, so that 0 < r +s−p−q < 4π,
which means that n must equal 1 if P Q and RS are parallel.
Thus P Q and RS are parallel if and only if r + s − p − q = 2π.

417
Question 4

1
Use the substitution x = , where t > 1, to show that, for x > 0,
t2 −1
Z
1 √ √

p dx = 2 ln x + x + 1 + c.
x(x + 1)
 
1 1 t − a
Z
[Note: You may use without proof the result dt = ln   + constant. ]
t2 − a2 2a  t + a 

Using the given substitution, we first use the chain rule to calculate
dx 2t
= −(t2 − 1)−2 · 2t = − 2 .
dt (t − 1)2

(We could alternatively have used the quotient rule to reach the same conclusion.)
We can now perform the requested substitution, simplifying the algebra as we go:
1 1 dx
Z Z
p dx = q · dt
x(x + 1) 1
· t2 dt
t2 −1 t2 −1
1 −2t
Z
= t

· 2 2
dt
2 (t − 1)
Z t −1
−2
= 2
dt
t −1
 
1  t − 1 
= −2 × ln  +c using the given result
2 t + 1
 
t + 1
= ln 
  + c.
t − 1

At this point, we wish to substitute t for x, so we rearrange the original substitution to

get r r
1 x+1
t= +1= .
x x
This now yields: p 
 
t + 1  x+1 + 1 
ln   + c = ln  p x

 + c.

t−1   x+1
x
− 1
We note immediately that we can drop the absolute value signs, since both the numerator
and
p denominator of the fraction are positive (the denominator is positive as t > 1 or
(x + 1)/x√> 1). So we get, on multiplying the numerator and denominator of the
fraction by x to clear the fractions,

418
   √ √ !
t + 1 x+1+ x
ln   + c = ln √ √ +c
t − 1 x+1− x
√ √
√ √
!
x+1+ x x+1+ x
= ln √ √
√ √
+c
x+1− x x+1+ x
√ √
2 !
x+1+ x
= ln +c
(x + 1) − x
√ √
2
= ln x + 1 + x + c
√ √

= 2 ln x + 1 + x + c,

which is what we were after.

The section of the curve

1 1
y=√ −√
x x+1
between x = 18 and x = 16 9
is rotated through 360◦ about the x-axis. Show that the
volume enclosed is 2π ln 54 .

R 9/16
To find the volume of revolution, we need to calculate the definite integral 1/8
πy 2 dx:
Z 9/16
πy 2 dx
1/8
9/16  2
1 1
Z
=π √ −√ dx
1/8 x x+1
9/16
1 1 1
Z
=π − 2p + dx
1/8 x x(x + 1) x + 1
h √ √
i9/16
= π ln x − 4 ln x + x + 1 + ln(x + 1) using the above result
1/8
 p p
  p p

9 9 25 25 1 1 9 9
= π ln 16 − 4 ln 16
+ 16
+ ln 16
− π ln 8
− 4 ln 8
+ 8
+ ln 8
   
3 3 5 5 1 1 3 3


= π 2 ln 4 − 4 ln( 4 + 4 ) + 2 ln 4 − π 2 ln 2 2 − 4 ln 2 2 + 2 2 + 2 ln 2 2
√ √ √ √


= π (2 ln 3 − 2 ln 4) − 4 ln 2 + (2 ln 5 − 2 ln 4) −
√ √ √

π −2 ln(2 2) − 4 ln 2 + (2 ln 3 − 2 ln(2 2)
= π(2 ln 3 − 4 ln 2 − 4 ln 2 + 2 ln 5 − 4 ln 2)−
π(−3 ln 2 − 2 ln 2 + 2 ln 3 − 3 ln 2)
= π(−4 ln 2 + 2 ln 5)
= 2π(−2 ln 2 + ln 5)
= 2π ln 45 .

419
Question 5

By considering the expansion of (1 + x)n where n is a positive integer, or otherwise,

show that:
       
n n n n
(i) + + + ··· + = 2n ;
0 1 2 n

We take the advice and begin by writing out the expansion of (1 + x)n :
       
n n 0 n 1 n 2 n n
(1 + x) = x + x + x + ··· + x , (∗)
0 1 2 n

where we have pedantically written in x0 and x1 in the first two terms, as this may well
help us to understand what we are looking at.
Now comparing this expansion to the expression we are interested in, we see that the only
difference is the presence of the xs. If we substitute x = 1, we will get exactly what we
want:        
n n n n n
(1 + 1) = + + + ··· + ,
0 1 2 n
as all powers of 1 are just 1.

       
n n n n
(ii) +2 +3 + ··· + n = n2n−1 ;
1 2 3 n

For the rest of the question, there are two very distinct approaches, one via calculus and
one via properties of binomial coefficients.

Approach 1: Use calculus

This
one looks a little more challenging, and we must observe carefully nthat

r there is no
n
0
term.

Comparing to the binomial expansion, we see that the term r x has turned
n
into r .r. Now, setting x = 1 will again remove the x, but where are we to get the r
from? Calculus gives us the answer: if we differentiate with respect to x, then xr becomes
rxr−1 , and then setting x = 1 will complete the job. Now differentiating (∗) gives
       
n−1 n 0 n 1 n 2 n
n(1 + x) = .1x + .2x + .3x + · · · + .nxn−1 ,
1 2 3 n
so by setting x = 1, we get the desired result.

Approach 2: Use properties of binomial coefficients

We know that  
n n!
=
r (n − r)!r!

420
so we can manipulate this formula to pull out an r, using r! = r.(r − 1)! and similar
expressions. We get
 
n n!
=
r (n − r)!r!
1 n!
= .
r (n − r)!(r − 1)!
n (n − 1)!
= .
r (n − r)!(r − 1)!
n n−1
 
=
r r−1
so that r nr = n n−1



r−1
. This is true as long as r > 1 and n > 1, so we get
n−1 n−1 n−1
           
n n n
+2 + ··· + n =n +n + ··· + n
1 2 n 0 1 n−1
= n.2n−1
where we have used the result from part (i) with n − 1 in place of n to do the last step.

       
n 1 n 1 n 1 n 1
2n+1 − 1 ;


(iii) + + + ··· + =
0 2 1 3 2 n+1 n n+1

Approach 1: Use calculus

Spurred on by our previous success, we see that now the n2 x2 term gives us n2 . 13 , so we

think of integration instead. Integrating (∗) gives

       
1 n+1 n 1 n 1 2 n 1 3 n 1
(1 + x) = .x + . x + . x + ··· + . xn+1 + c .
n+1 0 1 2 2 3 n n+1
We do need to determine the constant of integration, so we put x = 0 to do this; this
gives      
1 n+1 n n 1 n 1
1 = .0 + . .0 + + · · · + . .0 + c,
n+1 0 1 2 n n+1
1
so c = n+1
.
Now substituting x = 1 gives
       
1 n+1 n 1 n 1 n 1 n 1
(1 + 1) = + ++ + ··· + + .
n+1 0 2 1 3 2 n+1 n n+1
1
Finally, subtracting n+1
from both sides gives us our required result.
(An alternative way to think about this is to integrate both sides from x = 0 to x = 1.)

Approach 2: Use properties of binomial coefficients

We can try rewriting our identity so that the 1r stays with the r − 1 term; this gives us
1 n−1
   
1 n
= .
n r r r−1

421
 n n−1



Unfortunately, though, our expressions involve r−1 terms rather than r−1
terms, but
we can fix this by replacing n by n + 1 to get
   
1 n+1 1 n
= .
n+1 r r r−1

We substitute this in to get

     
n 1 n 1 n
+ +··· +
0 2 1 n+1 n
     
1 n+1 1 n+1 1 n+1
= + + ··· +
n+1 1 n+1 2 n+1 n+1
1
= (2n+1 − 1),
n+1
where we have again used the result of part (i), this time with n + 1 replacing n.

       
n 2 n 2 n 2 n
(iv) +2 +3 + ··· + n = n(n + 1)2n−2 .
1 2 3 n

Approach 1: Use calculus

This looks similar to (ii), in that we have increasing multiples. So we try differentiating (∗)
twice, giving us:
       
n−2 n n 0 n 1 n
n(n − 1)(1 + x) = .1.0 + .2.1x + .3.2x + · · · + .n(n − 1)xn−2 .
1 2 3 n

Unfortunately, though, the coefficient of nr is r(r − 1) rather than the r2 we actually

want. But no matter: we can just add r and we will be done, as r2 = r(r − 1) + r, and
we know from (ii) what terms like n2 .2 sum to give us. So we have, putting x = 1 in our

above expression:
       
n−2 n n n n
n(n − 1)(1 + 1) = .1.0 + .2.1 + .3.2 + · · · + .n(n − 1).
1 2 3 n

Now adding the result of (ii) gives

     
n−2 n−1 n n n
n(n − 1)2 + n2 = .(1.0 + 1) + .(2.1 + 2) + .(3.2 + 3) + · · · +
1 2 3
 
n
.(n(n − 1) + n)
n
so        
n−2 n n 2 n 2 n
(n(n − 1) + 2n)2 = + .2 + .3 + · · · + .n2 .
1 2 3 n

The left side simplifies to n(n + 1)2n−2 , and thus we are done.

422
An alternative (calculus-based) method is as follows. The first derivative of (∗), as we
have seen, is
       
n−1 n 0 n 1 n 2 n
n(1 + x) = .1x + .2x + .3x + · · · + .nxn−1 .
1 2 3 n

Now were we to differentiate again, we would end up with terms like n(n − 1)xn−2 , rather
than the desired n2 xk (for some k). We can remedy this problem by multiplying the whole
identity by x before we differentiate, so that we are differentiating
       
n−1 n 1 n 2 n 3 n
nx(1 + x) = .1x + .2x + .3x + · · · + .nxn .
1 2 3 n

Differentiating this now gives (using the product rule for the left hand side):
       
n−1 n−2 n 2 0 n 2 1 n 2 2 n
n(1+x) +n(n−1)x(1+x) = .1 x + .2 x + .3 x +· · ·+ .n2 xn−1 .
1 2 3 n

Substituting x = 1 into this gives our desired conclusion (after a small amount of algebra
on the left hand side).

Approach 2: Use properties of binomial coefficients

As this looks

similar to

the result of part (ii), we can start with what we worked out there,
namely r nr = n n−1
r−1
, giving us

n−1 n−1 n−1 n−1

             
n 2 n 2 n
+2 +· · ·+n =n +n.2 +n.3 +· · ·+n.n
1 2 n 0 1 2 n−1

Taking out the factor of n leaves us having to work out

n−1 n−1 n−1 n−1

       
+2 +3 + ··· + n
0 1 2 n−1

This looks very
similar to the problem of part (ii) with n replaced
by n − 1, but now the
multiplier of nr is r + 1 rather than r, and there is also an n
0
term. We can get over
n n n




the first problem by splitting up (r + 1) r as r r + r , so this expression becomes

n−1 n−1 n−1

     
+2 + · · · + (n − 1) +
1 2 n−1
n−1 n−1 n−1 n−1
       
+ + + ··· +
0 1 2 n−1

The first line is just part (ii) with n replaced by n − 1, so that it sums to (n − 1).2n−2 , and
the second line is just 2n−1 = 2.2n−2 by part (i). So the answer to the original question
(remembering the factor of n we took out earlier) is

n (n − 1).2n−2 + 2.2n−2 = n(n − 1 + 2).2n−2 = n(n + 1).2n−2 .

423
Question 6

Show that, if y = ex , then

d2 y dy
(x − 1) 2 − x + y = 0. (∗)
dx dx

dy d2 y
If y = ex , then = ex and = ex . Substituting these into the left hand side of (∗)
dx dx2
gives
d2 y dy
(x − 1) 2 − x + y = (x − 1)ex − xex + ex = 0,
dx dx
so y = ex satisfies (∗).

In order to find other solutions of this differential equation, now let y = uex , where u
is a function of x. By substituting this into (∗), show that

d2 u du
(x − 1) 2
+ (x − 2) = 0. (∗∗)
dx dx

We have y = uex , so we apply the product rule to get:

dy du x
= e + uex
dx dx
 du 
= + u ex
dx
d2 y  d2 u x du x   du x x

= e + e + e + ue
dx2 dx2 dx dx
2
du x du x
= 2 e + 2 e + uex
dx dx
 d2 u du 
= + 2 + u ex .
dx2 dx
(If you know Leibniz’s Theorem, then you could write down d2 u/dx2 directly.)
We now substitute these into (∗) to get
 d2 u du   du 
x
(x − 1) + 2 + u e − x + u ex + uex = 0.
dx2 dx dx

Dividing by ex 6= 0 and collecting the derivatives of u then gives

d2 u du
(x − 1) 2
+ (2x − 2 − x) + (x − 1 − x + 1)u = 0,
dx dx
which gives (∗∗) on simplifying the brackets.

424
 du
By setting = v in (∗∗) and solving the resulting first order differential equation
dx
for v, find u in terms of x. Hence show that y = Ax + Bex satisfies (∗), where A and B
are any constants.

du d2 u dv
As instructed, we set = v, so that 2
= , which gives us
dx dx dx
dv
(x − 1) + (x − 2)v = 0.
dx
This is a standard separable first-order linear differential equation, so we separate the
variables to get
1 dv x−2
=−
v dx x−1
and then integrate with respect to x to get
1 x−2
Z Z
dv = − dx.
v x−1
Performing the integrations now gives us
1 
Z 
ln |v| = − 1 − dx
x−1
= −x + ln |x − 1| + c,
which we exponentiate to get
|v| = |k|e−x |x − 1|,

where k is some constant, so we finally arrive at

v = ke−x (x − 1).

We now recall that v = du/dx, so we need to integrate this last expression once more to
find u. We use integration by parts to do this, integrating the e−x part and differentiating
(x − 1), to give us
Z
u = ke−x (x − 1) dx
Z
= k(−e )(x − 1) − k(−e−x ).1 dx
−x

= k(−e−x )(x − 1) − ke−x + c

= −kxe−x + c,

which is the solution to (∗∗).

Now recalling that y = uex gives us y = −kx + c ex as our general solution to (∗). In
particular, letting k = −A and c = B, where A and B are any constants, shows that
y = Ax + Bex satisfies (∗), as required.

425
Question 7

Relative to a fixed origin O, the points A and B have position vectors a and b, respec-
tively. (The points O, A and B are not collinear.) The point C has position vector c
given by
c = αa + βb,
where α and β are positive constants with α + β < 1. The lines OA and BC meet
at the point P with position vector p, and the lines OB and AC meet at the point Q
with position vector q. Show that
αa
p= ,
1−β
and write down q in terms of α, β and b.

The condition c = αa + βb with α + β < 1 and α and β both positive constants means
that C lies strictly inside the triangle OAB. Can you see why?

We start by sketching the setup so that we have something visual to help us with our
thinking.

Q
R
C
A
O P

The line OA has points with position vectors given by r1 = λa, and the line BC has
points with position vectors given by
−−→ −−→
r2 = OB + µBC = b + µ(c − b) = (1 − µ)b + µc.

The point P is where these two lines meet, so we must have

p = λa = (1 − µ)b + µc
= (1 − µ)b + µ(αa + βb)
= (1 − µ + βµ)b + αµa.

426
Since a and b are not parallel, we must have 1 − µ + βµ = 0 and αµ = λ. The first
equation gives (1 − β)µ = 1, so µ = 1/(1 − β). This gives λ = α/(1 − β), so that
αa
p= .
1−β

Now swapping the roles of a and b (and hence also of α and β) will give us the position
vector of Q:
βb
q= .
1−α

Show further that the point R with position vector r given by

αa + βb
r=
α+β
lies on the lines OC and AB.

We could approach this question in two ways, either by finding the point of intersection
of OC and AB or by showing that the given point lies on both given lines. We give both
approaches.

Approach 1: Finding the point of intersection

We require r to lie on OC, so r = λc, and r to lie on AB, so r = (1 − µ)a + µb, as before.
Substituting for c and equating coefficients gives

αλa + βλb = (1 − µ)a + µb,

so that

αλ = 1 − µ
βλ = µ.

Adding the two equations gives (α + β)λ = 1, so λ = 1/(α + β) and hence

c αa + βb
r= = .
α+β α+β

Approach 2: Showing that the given point lies on both lines

The equation of line OC is r1 = λc, and

αa + βb 1
r= = c
α+β α+β
is of the required form, so R lies on OC.

427
 −→
The equation of the line AB can be written as r2 −a = µAB, so we want r−a = µ(b−a).
Now we have
αa + βb α + β
r−a= − a
α+β α+β
−βa + βb
=
α+β
β
= (b − a),
α+β

which is of the form µ(b − a), so R lies on both lines.

OQ OS
The lines OB and P R intersect at the point S. Prove that = .
BQ BS

S lies on both OB and P R, so we need to find its position vector, s. Once again, we
require s = λb = (1 − µ)p + µr, so we substitute for p and r and compare coefficients:

s = λb = (1 − µ)p + µr
αa αa + βb
= (1 − µ) +µ
1−β α+β
(1 − µ)α(α + β)a + µ(1 − β)(αa + βb)
=
(1 − β)(α + β)
((1 − µ)α(α + β) + µα(1 − β))a + µ(1 − β)βb
=
(1 − β)(α + β)
α(α + β − µα − 2µβ + µ)a + µ(1 − β)βb
=
(1 − β)(α + β)

Since the coefficient of a in this expression must be zero, we deduce that

α+β
µ= ,
α + 2β − 1
so that
µ(1 − β)β
s= b
(1 − β)(α + β)
α+β β
= b
α + 2β − 1 (α + β)
β
= b
α + 2β − 1

−→ −−→
Now, since OQ and BQ are both multiples of the vector q, we can compare the lengths OQ
and BQ in terms of their multiples of q. This might come out to be negative, depending
on the relative directions, but at the end, we can just consider the magnitudes.

428
 βb
We thus have, since q = ,
1−α
OQ  β  .  β 
= −1
BQ 1−α 1−α
 β  . β − 1 + α
=
1−α 1−α
β
= ,
β−1+α
while
OS  β . β 
= −1
BS α + 2β − 1 α + 2β − 1
 β  .  −α − β + 1 
=
α + 2β − 1 α + 2β − 1
β
= .
−α − β + 1
Thus these two ratios of lengths are equal, as the magnitude of both of these is
β
.
1 − (α + β)

429
Question 8

(i) Suppose that a, b and c are integers that satisfy the equation

a3 + 3b3 = 9c3 .

Explain why a must be divisible by 3, and show further that both b and c must
also be divisible by 3. Hence show that the only integer solution is a = b = c = 0.

We have a3 = 9c3 − 3b3 = 3(3c3 − b3 ), so a3 is a multiple of 3. But as 3 is prime, a itself

must be divisible by 3. (Why is this? If 3 divides a product rs, then 3 must divide either
r or s, as 3 is prime. Therefore since a3 = a.a.a, and 3 divides a3 , it follows that 3 must
divide one of the factors, that is, 3 must divide a.)
Now we can write a = 3d, where d is an integer. Therefore we have

(3d)3 + 3b3 = 9c3 ,

which, on dividing by 3, gives

9d3 + b3 = 3c3 .
By the same argument, as b3 = 3(c3 − 3d3 ), it follows that b3 , and hence also b, is divisible
by 3.
We repeat the same trick, writing b = 3e, where e is an integer, so that

9d3 + (3e)3 = 3c3 .

We again divide by 3 to get

3d3 + 9e3 = c3 ,
so that c3 , and hence also c, is divisible by 3.
We then write c = 3f , where f is an integer, giving

3d3 + 9e3 = (3f )3 .

Finally, we divide this equation by 3 to get

d3 + 3e3 = 9f 3 .

Note that this is the same equation that we started with, so if a, b, c are integers which
satisfy the equation, then so are d = a/3, e = b/3 and f = c/3. We can repeat this process
indefinitely, so that a/3n , b/3n and c/3n are also integers which satisfy the equation. But
if a/3n is an integer for all n > 0, we must have a = 0, and similarly for b and c.
Therefore the only integer solution is a = b = c = 0.
[In fact, we can say even more. If a, b and c are all rational, say a = d/r, b = e/s, c = f /t
(where d, e, f are integers and r, s, t are non-zero integers), then we have
 d 3  e 3  f 3
+3 =9 .
r s t

430
Now multiplying both sides by (rst)3 gives

(dst)3 + 3(ert)3 = 9(f rs)3 ,

with dst, ert and f rs all integers, and so they must all be zero, and hence d = e = f = 0.
Therefore, the only rational solution is also a = b = c = 0.]

(ii) Suppose that p, q and r are integers that satisfy the equation

p4 + 2q 4 = 5r4 .

By considering the possible final digit of each term, or otherwise, show that p and q
are divisible by 5. Hence show that the only integer solution is p = q = r = 0.

We consider the final digit of fourth powers:

a a4 2a4
0 0 0
1 1 2
2 6 2
3 1 2
4 6 2
5 5 0
6 6 2
7 1 2
8 6 2
9 1 2

So the last digits of fourth powers are all either 0, 5, 1 or 6, and of twice fourth powers
are all either 0 or 2.
Also, 5r4 is a multiple of 5, so it must end in a 0 or a 5.
Therefore if 2q 4 ends in 0 (that is, when q is a multiple of 5), the possibilities for the final
digit of p4 + 2q 4 are

(0 or 1 or 5 or 6) + 0 = 0 or 1 or 5 or 6,

so it can equal 5r4 (which ends in 0 or 5) only if p4 ends in 0 or 5, which is exactly when
p is a multiple of 5.
Similarly, if 2q 4 ends in 2 (so q is not a multiple of 5), the possibilities for the final digit
of p4 + 2q 4 are
(0 or 1 or 5 or 6) + 2 = 2 or 3 or 7 or 8,
so it can not be equal to 5r4 (which ends in 0 or 5).
Therefore, if p4 + 2q 4 = 5r4 , we must have p and q both being multiples of 5.
Now as in part (i), we write p = 5a and q = 5b, where a and b are both integers, to get

(5a)4 + 2(5b)4 = 5r4 .

431
Dividing both sides by 5 gives
53 a4 + 2.53 b4 = r4 ,
where we are using dot to mean multiplication, so as before, r4 must be a multiple of 5
as the left hand side is 5(52 a4 + 2.52 b4 ). Thus, since 5 is prime, r itself must be divisible
by 5. Then writing r = 5c gives

53 a4 + 2.53 b4 = (5c)4 ,

which yields
a4 + 2b4 = 5c4
on dividing by 53 .
So once again, if p, q, r give an integer solution to the equation, so do a = p/5, b = q/5
and c = r/5. Repeating this, so are p/5n , q/5n , r/5n , and as before, this shows that the
only integer solution is p = q = r = 0.
[Again, the same argument as before shows that this is also the only rational solution.]

This is an example of the use of Fermat’s Method of Descent, which he used to prove one
special case of his famous Last Theorem: he showed that x4 + y 4 = z 4 has no positive
integer solutions. In fact, he proved an even stronger result, namely that x4 + y 4 = z 2
has no positive integer solutions.
Another approach to solving the first step of part (ii) of this problem is to use modular
arithmetic, where we only consider remainders when dividing by a certain fixed number.
In this case, we would consider arithmetic modulo 5, so the only numbers to consider are
0, 1, 2, 3 and 4, and we want to solve p4 + 2q 4 ≡ 0 (mod 5), where ≡ means “leaves the
same remainder”. Now a quick calculation shows that p4 ≡ 1 unless p ≡ 0, while 2q 4 ≡ 2
unless q ≡ 0, so that

p4 + 2q 4 ≡ (0 or 1) + (0 or 2) ≡ 0, 1, 2 or 3 (mod 5)

with p4 + 2q 4 ≡ 0 if and only if p ≡ q ≡ 0.

Incidentally, Fermat has another theorem relevant to this problem, which turns out to
be relatively easy to prove (Fermat himself claimed to have done so), and is known as
Fermat’s Little Theorem. This states that, if p is prime, then ap−1 ≡ 1 (mod p) unless
a ≡ 0 (mod p). In our case, p = 5 gives a4 ≡ 1 (mod 5) unless a ≡ 0 (mod 5), as we
wanted.

432
Question 9

2a 2b
α

The diagram shows a uniform rectangular lamina with sides of lengths 2a and 2b leaning
against a rough vertical wall, with one corner resting on a rough horizontal plane. The
plane of the lamina is vertical and perpendicular to the wall, and one edge makes an
angle of α with the horizontal plane. Show that the centre of mass of the lamina is a
distance a cos α + b sin α from the wall.

We start by redrawing the sketch, labelling the corners and indicating the centre of mass
as G, as well as showing various useful lengths.

A G
a C
b
2a 2b
α
B

It is now clear that the distance of G from the wall is a cos α (horizontal distance from
wall to midpoint of AB) plus b sin α (horizontal distance from midpoint of AB to G), so
a total of a cos α + b sin α.
Also, in case it is useful later, we note that the vertical distance above the horizontal
plane is, by a similar argument from the same sketch, a sin α + b cos α.

The coefficients of friction at the two points of contact are each µ and the friction is
limiting at both contacts. Show that

a cos(2λ + α) = b sin α,

where tan λ = µ.

There are two approaches to this. One is to indicate the reaction and friction forces
separately, while the other is to use the Three Forces Theorem. We show both of these.

433
Approach 1: All forces separately

We start by sketching the lamina again, this time showing the forces on the lamina,
separating the normal reactions from the frictional forces.

F1

A R1 G
C
R2
W
α
F2 B

We now resolve and take moments:

R(↑) F1 + R2 − W = 0
R(→) R1 − F2 = 0
y
M (A) W (a cos α + b sin α) − R2 .2a cos α + F2 .2a sin α = 0

Since friction is limiting at both points of contact, we have F1 = µR1 and F2 = µR2 .
Substituting these gives:
R(↑) µR1 + R2 − W = 0 (1)
R(→) R1 − µR2 = 0 (2)
y
M (A) W (a cos α + b sin α) − 2aR2 cos α + 2aµR2 sin α = 0 (3)

Equation (2) gives R1 = µR2 , so we can substitute this into (1) to get W = (1 + µ2 )R2 .
Substituting this into (3) now leads to

(1 + µ2 )R2 (a cos α + b sin α) = 2aR2 (cos α − µ sin α).

We can clearly divide both sides by R2 , and we are given that tan λ = µ, so we substitute
this in as well, to get

(1 + tan2 λ)(a cos α + b sin α) = 2a(cos α − tan λ sin α).

We spot 1 + tan2 λ = sec2 λ, and so multiply the whole equation through by cos2 λ, as the
form we are looking for does not involve sec λ:

a cos α + b sin α = 2a(cos2 λ cos α − sin λ cos λ sin α).

Since the form we are going for is b sin α = a cos(2λ + α), we make use of double angle
formulæ, after rearranging:

b sin α = a(2 cos2 λ cos α − 2 sin λ cos λ sin α) − a cos α

= a((2 cos2 λ − 1) cos α − 2 sin λ cos λ sin α)
= a(cos 2λ cos α − sin 2λ sin α)
= a cos(2λ + α),

434
and we are done with this part.
Approach 2: Three Forces Theorem

The ‘Three Forces Theorem’ states that if three (non-zero) forces act on a large body in
equilibrium, and they are not all parallel, then they must pass through a single point.
(Why is this true? Let’s say two of the forces pass through point X. Taking moments
about X, the total moment must be zero, so the moment of the third force about X must
be zero. Therefore, the force itself is either zero or it passes through X. Since the forces
are non-zero, the third force must pass through X.)
In our case, we have a normal reaction and a friction force at each point of contact. We
can combine these into a single reaction force as shown in the sketch. Here we have
written N for the normal force, F for the friction and R for the resultant, which is at an
angle of θ to the normal.

R
F

θ
N

We see from this sketch that tan θ = F/N = µN/N = µ. In our case, since tan λ = µ we
must have θ = λ.
We can now redraw our original diagram with the three (combined) forces shown:

R2
R1
D
X

λ
A G C
λ
2a
α
O P B
W

We can now use the Three Forces Theorem is as follows. Looking at the diagram, we
know that the distance OB = 2a cos α = OP + P B. Now we know OP = a cos α + b sin α,
so we need only calculate P B.
But P B = P X tan λ (using the triangle P BX), and

P X = OA + height of X above A
= 2a sin α + (a cos α + b sin α) tan λ.

435
Putting these together gives

OB = 2a cos α = OP + P B
= a cos α + b sin α + (2a sin α + (a cos α + b sin α) tan λ) tan λ
= a cos α(1 + tan2 λ) + b sin α(1 + tan2 λ) + 2a sin α tan λ
= a cos α sec2 λ + b sin α sec2 λ + 2a sin α tan λ.

We can now rearrange to get

b sin α sec2 λ = 2a cos α − a cos α sec2 λ − 2a sin α tan λ.

Since we want an expression for b sin α, we now multiply by cos2 λ to get

b sin α = 2a cos α cos2 λ − a cos α − 2a sin α sin λ cos λ

= a((2 cos2 λ − 1) cos α − (2 sin λ cos λ) sin α)
= a(cos 2λ cos α − sin 2λ sin α)
= a cos(2λ + α).

An alternative argument using the Three Forces Theorem proceeds by considering the
distance XP . Using the left half of the diagram, we have

XP = OA + OP tan λ
= 2a sin α + (a cos α + b sin α) tan λ.

From the right half of the diagram, we also have XP = P B/ tan λ, and P B = 2a cos α −
OP = a cos α − b sin α, so that

2a sin α + (a cos α + b sin α) tan λ = (a cos α − b sin α)/ tan λ.

Now multiplying by tan λ and collecting like terms gives

b sin α(1 + tan2 λ) = a cos α(1 − tan2 λ) − 2a sin α tan λ.

Then using 1 + tan2 λ = sec2 λ and then multiplying by cos2 λ gives us

b sin α = a cos α(cos2 λ − sin2 λ) − 2a sin α sin λ cos λ

= a cos α cos 2λ − a sin α sin 2λ
= a cos(α + 2λ),

as we wanted.

π
Show also that if the lamina is square, then λ = 4
− α.

We have a = b as the lamina is square, so that our previous equation becomes

a sin α = a cos(2λ + α).

436
Dividing by a gives
sin α = cos(2λ + α).
Now we can use the identity sin α = cos( π2 − α), so that
π
− α = 2λ + α.
2
(Being very careful, we should check that we can take inverse cosines of both sides to
deduce this equality. This will be the case if both 0 < π2 − α < π2 and 0 < 2λ + α < π2 .
But 0 < α < π2 so the first inequality is clearly true. For the second inequality, we have
0 < λ < π2 so that 0 < 2λ + α < 3π2
. But since the cosine of this is positive (being sin α),
it must lie in the range 0 < 2λ + α < π2 as required.)
Subtracting α and dividing by 2 now gives our desired result:
π
− α = λ.
4

437
Question 10

A particle P moves so that, at time t, its displacement r from a fixed origin is given by

r = et cos t i + et sin t j.

Show that the velocity of the particle always makes an angle of π4 with the particle’s
displacement, and that the acceleration of the particle is always perpendicular to its
displacement.

To find the velocity, v, and acceleration, a, we differentiate with respect to t (using the
product rule).
We have

r = et cos t i + et sin t j

v = dr/dt = et cos t − et sin t i + et sin t + et cos t j

a = dv/dt = (et cos t − et sin t − (et sin t + et cos t))i+

(et sin t + et cos t) + (et cos t − et sin t j

= (−2et sin t)i + (2et cos t)j.

We can rewrite these, if we wish, by taking out the common factors:

r = et (cos t)i + (sin t)j

v = et (cos t − sin t)i + (sin t + cos t)j

a = 2et (− sin t)i + (cos t)j) .

From these, we can easily find the magnitudes of the displacement, velocity and acceler-
ation:
p
|r| = et (cos t)2 + (sin t)2 = et
p
|v| = et (cos t − sin t)2 + (sin t + cos t)2
p
t
= e 2 cos2 t + 2 sin2 t
√
= et 2
p
|a| = 2et (− sin t)2 + (cos t)2 = 2et .

We can now find the angles between these using a.b = 2|a||b| cos θ; firstly, for displace-
ment and velocity we have

r.v = e2t cos t(cos t − sin t) + sin t(sin t + cos t)

= e2t (cos2 t + sin2 t)

= e2t , while
t t
√
r.v = e .e 2 cos θ,

438
 √ π
so that cos θ = 1/ 2, so that θ = 4
as required.
Next, for displacement and acceleration we have
r.a = 2e2t cos t(− sin t) + sin t cos t = 0,

so they are perpendicular.

Geometric-trigonometric approach

There is another way to find the angles involved which does not use the scalar (dot)
product.
Recall that the velocity is v = et (cos t − sin t)i + (sin t + cos t)j . We can use the

“R cos(θ + α)” technique, thinking of cos t − sin t as 1 cos t − 1 sin t, so that

√
cos t − sin t = 2 √12 cos t − √12 sin t


√
= 2(cos t cos π4 − sin t sin π4 )
√
= 2 cos(t + π4 )
√
sin t + cos t = 2 √12 sin t + √12 cos t


√
= 2(sin t cos π4 + cos t sin π4 )
√
= 2 sin(t + π4 )
√
Thus v = 2et cos(t + π4 )i + sin(t + π4 )j , so v is at an angle of π4 with r.

π π
Likewise, a makes an angle of 4
with v, and so an angle of 2
with r.

Sketch the path of the particle for 0 6 t 6 π.

One way of thinking about the path of the particle is that its displacement at time t is
t
et from the origin and at an


given by r = e (cos t)i + (sin t)j , so that it is at distance


angle of t (in radians) to the x-axis (as (cos t)i+(sin t)j is a unit vector in this direction).
Thus its distance at time t = 0 is e0 = 1, and when it has gone a half circle, its distance
is eπ , which is approximately e3 ≈ 20. So the particle moves away from the origin very
quickly!
Another thing to bear in mind is that its velocity is always at an angle of π4 to its
displacement. Since it is moving away from the origin, its velocity is directed away from
the origin, so initially it is moving at an angle of π4 above the positive x-axis.
As we sketch the path, we also indicate the directions of the velocities at the times t = 0,
t = π2 and t = π.

eπ/2

−eπ 1 x

439
 A second particle Q moves on the same path, passing through each point on the path
a fixed time T after P does. Show that the distance between P and Q is proportional
to et .

We write rP = r for the position vector of P and rQ for the position vector of Q. We
therefore have

rP = et cos t i + et sin t j

rQ = et−T cos(t − T ) i + et−T sin(t − T ) j,

and so we can calculate |rP − rQ |2 :

2
2
|rP − rQ |2 = et cos t − et−T cos(t − T ) + et sin t − et−T sin(t − T )
= e2t cos2 t − 2et et−T cos t cos(t − T ) + 2e2(t−T ) cos2 (t − T ) +

e2t sin2 t − 2et et−T sin t sin(t − T ) + e2(t−T ) sin2 (t − T )

= e2t − 2e2t−T (cos t cos(t − T ) + sin t sin(t − T ) + e2(t−T )

= e2t − 2e2t−T cos(t − (t − T )) + e2t−2T

= e2t 1 − 2e−T cos T + e−2T ,

so that p
|rP − rQ | = et 1 − 2e−T cos T + e−2T ,
which is clearly proportional to et , as required, since T is a constant.

440
Question 11

Two particles of masses m and M , with M > m, lie in a smooth circular groove
on a horizontal plane. The coefficient of restitution between the particles is e. The
particles are initially projected round the groove with the same speed u but in opposite
directions. Find the speeds of the particles after they collide for the first time and show
that they will both change direction if 2em > M − m.

This begins as a standard collision of particles question. ALWAYS draw a diagram

for collisions questions; you will do yourself (and the examiner) no favours if you try to
keep all of the directions in your head, and you are very likely to make a mistake. My
recommendation is to always have all of the velocity arrows pointing in the same direction.
In this way, there is no possibility of messing up the Law of Restitution; it always reads
v1 − v2
v1 − v2 = e(u2 − u1 ) or = e, and you only have to be careful with the signs of
u2 − u1
the given velocities; the algebra will then keep track of the directions of the unknown
velocities for you.

u1 = u u2 = −u
Before M m

v1 v2
After M m

Then Conservation of Momentum gives

M u1 + mu2 = M v1 + mv2

and Newton’s Law of Restitution gives

v2 − v1 = e(u1 − u2 ).

Substituting u1 = u and u2 = −u gives

M v1 + mv2 = (M − m)u (1)

v2 − v1 = 2eu. (2)

Then solving these equations (by (1) − m × (2) and (1) + M × (2)) gives
(M − m − 2em)u
v1 = (3)
M +m
(M − m + 2eM )u
v2 = . (4)
M +m
The speeds are then (technically) the absolute values of these, but we will stick with these
formulæ as they are what are needed later.

441
Now, the particles both change directions if v1 and v2 have the opposite signs from u1
and u2 , respectively, so v1 < 0 and v2 > 0. Thus we need

M − m − 2em < 0 and (5)

M − m + 2eM > 0. (6)

But (6) is always true, as M > m, so we only need M − m < 2em from (5).

After a further 2n collisions, the speed of the particle of mass m is v and the speed of
the particle of mass M is V . Given that at each collision both particles change their
directions of motion, explain why

mv − M V = u(M − m),

and find v and V in terms of m, M , e, u and n.

The fact that the particles both change their directions of motion at each collision means
that if they have velocities v1 and v2 after some collision, they will have velocities −v1 and
−v2 before the next collision. This is because they are moving around a circular track,
and therefore next meet on the opposite site, and hence are each moving in the opposite
direction from the one they were moving in. (We do not concern ourselves with precisely
where on the track they meet, and we are thinking of our velocities as one-dimensional
directed speeds.)
Therefore, mvm + M vM is constant in value after each collision, where vm is the velocity
of the particle of mass m, and vM that of the particle of mass M , but it reverses in
sign before the next collision. So after the first collision, it it M u − mu to the right (in
our above sketch), and hence after an even number of further collisions, it will still be
M vM + mvm = M u − mu to the right. But after an even number of further collisions,
the particle of mass M is moving to the left, so vM = −V , vm = v. Thus

mv − M V = (M − m)u.

Also, since there are a total of 2n+1 collisions, we have, by 2n+1 applications of Newton’s
Law of Restitution,
V + v = e2n+1 (u + u).
Solving these two equations simultaneously as before then yields

(2me2n+1 − M + m)u
V =
M +m
2n+1
(2M e + M − m)u
v= .
M +m

442
Question 12

A discrete random variable X takes only positive integer values. Define E(X) for this
case, and show that
X∞
E(X) = P (X > n).
n=1

For the definition of E(X), we simply plug the allowable values of X into the definition
of E(X) for discrete random variables, to get
∞
X
E(X) = n P(X = n).
n=1

Now, we can think of n, P(X = n) as the sum of n copies of P(X = n), so that we get
∞
X
E(X) = n P(X = n)
n=1
= 1.P(X = 1) + 2.P(X = 2) + 3.P(X = 3) + 4.P(X = 4) + · · ·
= P(X = 1) +
P(X = 2) + P(X = 2) +
P(X = 3) + P(X = 3) + P(X = 3) +
P(X = 4) + P(X = 4) + P(X = 4) + P(X = 4) + · · ·

Adding each column now gives us something interesting: the first column is P(X =
1) + P(X = 2) + P(X = 3) + · · · = P(X > 1), the second column is P(X = 2) + P(X =
3) + · · · = P(X > 2), the third column is P(X = 3) + P(X = 4) + · · · = P(X > 3), and
so on. So we get

E(X) = P(X > 1) + P(X > 2) + P(X > 3) + P(X > 4) + · · ·

X∞
= P(X > n),
n=1

as we wanted.
An alternative, more formal, way of writing this proof is as follows, using what is some-
times called “summation algebra”:

443
 ∞
X
E(X) = nP (X = n)
n=1
∞ X
X n
= P(X = n) summing n copies of a constant
n=1 m=1
X
= P(X = n) writing it as one big sum
16m6n<∞
X∞ X∞
= P(X = n) see below
m=1 n=m
X∞
= P(X > m),
m=1

which is the sum we wanted. For the penultimate step, note the we are originally summing
all pairs of values (m, n) where n is any positive integer and m lies between 1 and n, so
we have 1 6 m 6 n < ∞, as written on the third line. This can also be thought of as
summing over all pairs of values (m, n) where m is any positive integer (i.e., 1 6 m < ∞),
and n is chosen so that m 6 n < ∞, that is, we are summing on n from m to ∞.
[One final technical note: we are allowed to reorder the terms of this infinite sum because
all of the summands (the things we are adding) are non-negative. If some were positive
and others were negative, we might get all sorts of weird things happening if we reordered
the terms. An undergraduate course in Analysis will usually explore such questions.]

I am collecting toy penguins from cereal boxes. Each box contains either one daddy
penguin or one mummy penguin. The probability that a given box contains a daddy
penguin is p and the probability that a given box contains a mummy penguin is q,
where p 6= 0, q 6= 0 and p + q = 1.
Let X be the number of boxes that I need to open to get at least one of each kind of
penguin. Show that P(X > 4) = p3 + q 3 , and that
1
E(X) = − 1.
pq

We ask ourselves: what needs to happen to have X > 4? This means that we need to
open at least 4 boxes to get both a daddy and a mummy penguin. In other words, we
can’t have had both a daddy and a mummy among the first three boxes, so they must
have all had daddies or all had mummies. Therefore P(X > 4) = p3 + q 3 .
This immediately generalises to give P(X > n) = pn−1 + q n−1 , at least for n > 3. For
n = 1, P(X > 1) = 1, and for n = 2, P(X > 2) = 1 = p1 + q 1 , as we argued above.

444
Therefore, we have
∞
X
E(X) = P(X > n)
n=1
= 1 + (p1 + q 1 ) + (p2 + q 2 ) + (p3 + q 3 ) + · · ·
= (1 + p + p2 + p3 + · · · ) + (1 + q + q 2 + q 3 + · · · ) − 1
1 1
= + −1 adding the geometric series
1−p 1−q
1 1
= + −1 as p + q = 1
q p
p+q
= −1
qp
1
= −1 again using p + q = 1.
pq

Hence show that E(X) > 3.

1
To show that E(X) > 3, we simply need to show that pq
> 4. But this is the same as
showing that pq 6 14 , by taking reciprocals.
Now, recall that q = 1 − p, so we need to show that p(1 − p) 6 14 . To do this, we rewrite
the quadratic in p by completing the square:

p(1 − p) = p − p2 = 1
4
− (p − 12 )2 .

Since (p − 12 )2 > 0 for all p (even outside the range 0 < p < 1), we have p(1 − p) 6 41 , as
required, with equality only when p = q = 12 .
This can also be proved using calculus, or using the AM–GM inequality, or by writing
1/pq = (p + q)2 /pq and then rearranging to get (p − q)2 > 0.

445
Question 13

The number of texts that George receives on his mobile phone can be modelled by
a Poisson random variable with mean λ texts per hour. Given that the probability
George waits between 1 and 2 hours in the morning before he receives his first text is p,
show that
pe2λ − eλ + 1 = 0.
Given that 4p < 1, show that there are two positive values of λ that satisfy this
equation.

Let X be the number of texts George receives in the first hour of the morning and Y be
the number he receives in the second hour.
Then X ∼ Po(λ) and Y ∼ Po(λ), with X and Y independent random variables.
We thus have
P(George waits between 1 and 2 hours for first text) = P(X = 0 and Y > 0)
= P(X = 0).P(Y > 0)
= e−λ .(1 − e−λ )
= p,
so that e−λ − e−2λ = p.
Multiplying this last equation by e2λ gives eλ − 1 = pe2λ ; a straightforward rearrangement
yields our desired equation.
(This equation can also be deduced by considering the waiting time until the first text;
this is generally not studied until university, though.)
The solutions of the quadratic equation in eλ are given by
√
λ 1 ± 1 − 4p
e = .
2p
But we are given that 4p < 1, so that 1 − 4p > 0 and there are real solutions. We need
to show, though, that the two values of eλ that we get are both greater than 1, so that
the resulting values of λ itself are both greater than 0.
We have
√
1± 1 − 4p p
> 1 ⇐⇒ 1 ± 1 − 4p > 2p
2p
p
⇐⇒ ± 1 − 4p > 2p − 1.
Now, since 4p < 1, we have 2p√< 12 , so 2p − 1 < 0, from which it follows that for the
positive√sign in the inequality, 1 − 4p > 0 > 2p − 1. It therefore only remains to show
that − 1 − 4p > 2p − 1. But
p p
− 1 − 4p > 2p − 1 ⇐⇒ 1 − 4p < −(2p − 1)
⇐⇒ 1 − 4p < (2p − 1)2
⇐⇒ 1 − 4p < 4p2 − 4p + 1,

446
which is clearly true as 4p2 > 0. (We were allowed to square between the first and second
lines as both sides are positive.)
Thus the two solutions to our quadratic in eλ are both greater than 1, so there are two
positive values of λ which satisfy the equation.

The number of texts that Mildred receives on each of her two mobile phones can be
modelled by independent Poisson random variables but with different means λ1 and λ2
texts per hour. Given that, for each phone, the probability that Mildred waits between
1 and 2 hours in the morning before she receives her first text is also p, find an expression
for λ1 + λ2 in terms of p.

Each phone behaves in the same way as George’s phone above, so the two possible values
of λ are those found above. That is, the values of eλ1 and eλ2 are the two roots of
pe2λ − eλ + 1 = 0.
We know that the product of the roots of the equation ax2 + bx + c = 0 is c/a,1 so in our
case, eλ1 eλ2 = 1/p, so that eλ1 +λ2 = 1/p, giving

λ1 + λ2 = ln(1/p) = − ln p.

Find the probability, in terms of p, that she waits between 1 and 2 hours in the morning
to receive her first text.

Let X1 be the number of texts she receives on the first phone during the first hour and
Y1 be the number of texts that she receives on the first phone during the second hour.
Then X1 and Y1 are both distributed as Po(λ1 ), so

P(X1 = 0) = e−λ1
P(Y1 = 0) = e−λ1 .

Now let X2 and Y2 be the corresponding random variables for the second phone, so we
have

P(X2 = 0) = e−λ2
P(Y2 = 0) = e−λ2 .

We must now consider the possible situations in which she receives her first text between
1 and 2 hours in the morning. She must receive no texts on either phone in the first hour,
and at least one text on one of the phones in the second hour. We use the above result
that λ1 + λ2 = − ln p, so that e−λ1 −λ2 = p.

1
Why is this? If the roots of ax2 + bx + c = 0 are α and β, then we can write the quadratic as
a(x − α)(x − β) = a(x2 − (α + β)x + αβ), so that c = aαβ, or αβ = c/a. Likewise, b = −a(α + β) so that
α + β = −b/a.

447
Thus

P(first text between 1 and 2 hours)

= P(X1 = 0 and X2 = 0 and Y1 > 0 or Y2 > 0 or both)


= P(X1 = 0).P(X2 = 0). 1 − P(Y1 = 0 and Y2 = 0)


= P(X1 = 0).P(X2 = 0). 1 − P(Y1 = 0).P(Y2 = 0)
= e−λ1 .e−λ2 . 1 − e−λ1 .e−λ2

= e−λ1 −λ2 .(1 − e−λ1 −λ2 )

= p(1 − p),

and we are done.

Alternative approach

This approach uses a result which you may not have come across yet: the sum X + Y of
two independent Poisson random variables X ∼ Po(λ) and Y ∼ Po(µ) is itself a Poisson
variable with X + Y ∼ Po(λ + µ).
Since the number of texts received on the two phones together is the sum of the number
of texts received on each one, the total can be modelled by a Poisson random variable
with mean Λ = λ1 + λ2 texts per hour.
Then the probability of waiting between 1 and 2 hours in the morning for the first text is
given by q, where
qe2Λ − eΛ + 1 = 0,
using the result from the very beginning of the question, replacing p with q and λ with Λ.
Since Λ = λ1 + λ2 = ln(1/p) from above, eΛ = 1/p.
Therefore
eΛ − 1
q=
e2Λ
1/p − 1
=
(1/p)2
= p2 (1/p − 1)
= p(1 − p).

448
 Hints & Solutions for STEP II 2010

1 When two curves meet they share common coordinates; when they “touch” they also share a
common gradient. In the case of the osculating circle, they also have a common curvature at the
dy d2 y
point of contact. Since curvature (a further maths topic) is a function of both and , the
dx dx 2
question merely states that C and its osculating circle at P have equal rates of change of gradient.
It makes sense then to differentiate twice both the equation for C and that for a circle, with
equation of the form (x – a)2 + (y – b)2 = r2, and then equate them when x = 14  . The three
resulting equations in the three unknowns a, b and r then simply need to be solved simultaneously.

dy d2 y
For y = 1 – x + tan x , = – 1 + sec2x and 2
= 2 sec2x tan x .
dx dx
2
dy d2 y  dy 
2 2 2
For (x – a) + (y – b) = r , 2(x – a) + 2(y – b) = 0 and 2 + 2(y – b) + 2  = 0 .
 dx 
2
dx dx
When x = 14  , y = 2  14  and so  14   a   2  14   b   r 2 ;
2 2

dy ( x  a)
=   1 then gives a relationship between a and b;
dx ( y  b)

d2 y 4
and =4=  gives the value of b.
dx 2
2( y  b)
Working back then gives a and r.

Answers: The osculating circle to C at P has centre  14   12 , 5

2  14   and radius 1
2
.

2 The single-maths approach to the very first part is to use the standard trig. “Addition” formulae for
sine and cosine, and then to use these results, twice, in (i); firstly, to rewrite sin3x in terms of
sin3x so that direct integration can be undertaken; then to express cos3x in terms of cos3x in
order to get the required “polynomial” in cosx. Using the given “misunderstanding” in (ii) then
leads to a second such polynomial which, when equated to the first, gives an equation for which a
couple of roots have already been flagged. Unfortunately, the several versions of the question that
were tried, in order to help candidates, ultimately led to the inadvertent disappearance of the
interval 0 to  in which answers had originally been intended. This meant that there was a little bit
more work to be done at the end than was initially planned.

cos3x = cos(2x + x) = cos2x cos x – sin2x sin x = (2c2 – 1)c – 2sc.s = (2c2 – 1)c – 2c(1 – c2)
= 4c3 – 3c .
sin3x = sin(2x + x) = sin2x cos x + cos2x sin x = 2sc.c + (1 – 2s2)s = 2s(1 – s2) + s(1 – 2s2)
= 3s – 4s3

 
(i) I () =  7 sin x  8 sin x  dx =  sin x  2 sin 3x  dx =  cos x  cos 3x 
3 2
3 0
0 0

= – cos  – 23 (4cos3 – 3 cos) + 1 + 2

3 = – 83 c3 + c + 5
3

and I () = 0 when c = 1 ( = 0)

449
 (ii) J () =  sin
7
2
2
x  84 sin 4 x 0 = 7
2 (1 – cos2 ) – 2(1 – cos2)2 = – 2c4 + 1
2 c2 + 3
2

I () = J ()  0 = 12c4 – 16c3 – 3c2 + 6c + 1 = (c – 1)2 (2c + 1)(6c + 1)

Thus cos  = 1,  = 0; cos  = – 12 ,  = 23  ; and cos  = – 16 ,  =  – cos – 1 ( 16 ).

Answers:   2n , 2n  23  , (2n  1)  cos 1 1

6

3 You don’t have to have too wide an experience of mathematics to be able to recognise the
Fibonacci Numbers in a modest disguise here. (However, this is of little help here, as you should
be looking to follow the guidance of the question.) In (i), you are clearly intended to begin by
substituting n = 0, 1, 2 and 3, in turn, into the given formula for F n , using the four given terms of
the sequence. You now have four equations in four unknowns, and the given result in (i) is
intended to help you make progress; with (ii) having you check the formula in a further case. In
the final part, you should split the summation into two parts, each of which is an infinite geometric
progression.

(i) F 0 = 0  0 = a + b or b = – a . Then F 1 = 1  1 = a( – ).

[F 2 = 1  1 = a(2 – 2)   +  = 1 is needed later]
and F 3 = 2  2 = a(3 – 3) = a( – ) (2 +  + 2) by the difference of two cubes
= 1.(2 +  + 2)  2 +  + 2 = 2
1
Then, using any two suitable eqns., e.g. any two of  = – 1,  –  =
and  +  = 1, and
a
solving simultaneously gives a =
1
5
, b=–
1
5
1
2
1
,  = 1 5 ,  = 1  5 .
2
   

(ii) Using the formula F n = a n + b  n =

2
1
n
5
(1  5 ) n  (1  5 ) n  with n = 6 ; the Binomial
Theorem gives 1  5  6
 1  6 5  15.5  20.5 5  15.5 2  6.5 2 5  5 3 = 576 + 256 5 .

Similarly, 1  5 
6
 576  256 5 so that F 6 = 6
1
2 5
512 5 = 8.  

a   a   1    
 n n
Fn 1  1  1
(iii)  n  1 =        =  using the
n0 2 2 n0  2  2 n0  2  2 5  1  14 1  5  2 5  1  14  
1  5  
S formula for the two GPs;
1  4  1  4 
=      .
2 5 3 5  2 5 3 5 
2 3 5  2 3 5  2 2 5
Rationalising denominators then yields     =   = 1.
5  9  5  5  9  5  5  4 

450
4 Hopefully, the obvious choice is y = a – x for the initial substitution and, as with any given
result, you should make every effort to be clear in your working to establish it. Thereafter, the two
integrals that follow in (i) use this result with differing functions and for different choices of the
upper limit a. Since this may be thought an obvious way to proceed, it is (again) important that
your working is clear in identifying the roles of f(x) and f(a – x) in each case. In part (ii), however,
it is not the first result that is to be used, but rather the process that yielded it. The required
substitution should, again, be obvious, and then you should be trying to mimic the first process in
this second situation.

(i) Using the substn. y = a – x , dy = – dx and (0, a)  (a, 0) so that

f (a  y ) f (a  y )
a 0 a
f ( x)
0 f ( x)  f (a  x) dx = a f (a  y)  f ( y) .– dy = 0 f (a  y)  f ( y) dy
f (a  x)
a
=  f ( x)  f (a  x) dx , since the x/y interchange here is nothing more than a re-labelling.
0

f ( x)  f (a  x)
a a
Then 2 I =  dx =  1. dx = x a = a  I = 1
a.
f ( x)  f (a  x) 0 2
0 0

For f(x) = ln(1 + x) , ln(2 + x – x2) = ln[(1 + x)(2 – x)] = ln(1 + x) + ln(2 – x)
1
f ( x)
and ln(2 – x) = ln(1 + [1 – x]) = f(a – x) with a = 1 so that  f ( x)  f (1  x) dx =
0
1
2 .

 /2  /2  /2
sin x sin x sin x
0
sin x  14  
dx = 
0
sin x. 12  cos x. 1
dx = 2 
0
sin x  sin  2   x 
1
dx = 14  2 .
2

(ii) For u =
1 1
, du =  2 dx and  12 , 2  2, 12  .
x x
2
1 sin x
2
1 x sin x
0.5
.sin  u1 
1
Then  . 0.5 x 2 . sin x  sin  1x  dx = 2 sin  u1   sin u  .– du
u
dx =
0.5
x sin x  sin  1x 
2
1 sin  u1  2
1 sin  1x 
=  . du or  x sin x  sin  1x  dx
.
0.5
u sin u  sin  u1  0.5
2

 x dx = ln x0.5 = 2 ln 2
1
Adding then gives 2 I = 2
 I = ln 2 .
0.5

5 The opener here is a standard bit of A-level maths using the scalar product, and the following
parts use this method, but with a bit of additional imagination needed. In 3-dimensions, there are
infinitely lines inclined at a given angle to another, specified line, and this is the key idea of the
final part of the question. Leading up to that, in (i), you need only realise that a line equally
inclined to two specified (non-skew) lines must lie in the plane that bisects them (and is
perpendicular to the plane that contains, in this case, the points O, A and B). One might argue that
the vector treatment of “planes” is further maths work, but these ideas are simple geometric ones.

451
 (1, 1, 1)  (5,  1,  1) 1
cos2 = 
3. 27 3

(m, n, p)  (1, 1, 1) ( m, n, p)  (5,  1,  1)

(i) l 1 equally inclined to OA and OB iff 
m2  n2  p 2 . 3 m 2  n 2  p 2 . 27
i.e. 3(m + n + p) = 5m – n – p or m = 2(n + p).

mn p
For l 1 to be the angle bisector, we also require (e.g.) = cos  , where
m2  n2  p2 . 3
cos2 = 2 cos2 – 1 = 1
3  cos  = 2
3
, so that m + n + p = m2  n2  p 2 . 2 .

Squaring both sides: m2 + n2 + p2 + 2mn + 2np + 2pm = 2(m2 + n2 + p2)

 2mn + 2np + 2pm = m2 + n2 + p2
Setting m = 2n + 2p (or equivalent) then gives 2np + (2n + 2p)2 = (2n + 2p)2 + n2 + p2
which gives (n – p)2 = 0  p = n , m = 4n .
 m  4
   
Thus  n    1  , or any non-zero multiple will suffice.
 p  1
   

(ii) If you used the above method then you already have this relationship; namely,
2uv + 2vw + 2wu = u2 + v2 + w2 .
Thus, 2xy + 2yz + 2zx = x2 + y2 + z2 gives all lines inclined at an angle cos – 1 2
3
to OA and
hence describes the surface which is a double-cone, vertex at O, having central axis OA .

6 Although it seems that 3-dimensional problems are not popular, this is actually a very, very easy
question indeed and requires little more than identifying an appropriate right-angled triangle and
using some basic trig. and/or Pythagoras. There are thus so many ways in which one can approach
the three parts to this question that it is difficult to put forward just the one.
D
(i) Taking the midpoint of AB as the origin, O,
with the x-axis along AB and the y-axis along
OC, we have a cartesian coordinate system to
help us organise our thoughts. C

Then A =  12 , 0, 0 , B =  12 , 0, 0 ,

C = 0, 2
3

, 0 by trig. or Pythagoras, and A
P

P = 0, 6
3
, 0 . The standard distance formula
O
3 6 2
then gives PA (or PB) = 3 and PD = 3 or 3 .
B
1 3
(ii) The angle between adjacent faces is (e.g.) DOC = cos – 1  6  in right-angled triangle
 1 3
2 
–11
DOP, which gives the required answer, cos 3 .

452
 (iii) D The centre, S, of the inscribed sphere must, by symmetry,
2
3
 6
3
lie on PD, equidistant from each vertex.
3
6
r By Pythagoras, x2 = 1
12 +  6
9 2 3
6
x  x2   x= 4
6
.
r
Then r = x sin(90o – (ii)) = 1
3 x= 6
12 .
3
S 6
Alternatively, if you know that the sphere’s centre is at
r x The centre of mass of the tetrahedron, the point (S)
with position vector 14 (a  b  c  d) , then the answer
3
P 6 A 1 6
is just 4 DP = 12 .

7 The first two parts of the question begin, helpfully, by saying exactly what to consider in order to
proceed, and the material should certainly appear to be routine enough to make these parts very
accessible. Where things are going in (iii) may not immediately be obvious but, presumably, there
is a purpose to (i) and (ii) which should become clear in (iii).

dy
(i) y = x3 – 3qx – q(1 + q)  = 3(x2 – q) = 0 for x   q .
dx
 
When x =  q , y =  q q  1 < 0 since q > 0
2

q , y =  q  q  1 < 0
2
When x =  since q > 0 and q  1
Since both TPs below x-axis, the curve crosses the x-axis once only (possibly with sketch)

q q2 q3
(ii) x = u +  x3 = u 3  3uq  3  3
u u u
q2 q3 q2
0 = x3 – 3qx – q(1+ q) = u 3  3uq  3  3 – 3qu – 3  q  q 2
u u u
   
3
q 2
 u3 + 3  q(1  q) = 0 or u 3  q(1  q) u 3  q 3  0
u

u =
q(1  q)  q 2 (1  q) 2  4q 3
3
2
=
q
2
1  q  1  2 q  q 2  4q  
=
q
2
 2 q
2

1  q  1  q  = 1  q  (1  q ) = q or q2
1 2 1 2
giving u = q 3 or q 3 and x = q 3 + q 3

(iii)  + = p ,  = q   3   3       3 (   ) = p3 – 3qp .

3

One root is the square of the other   = 2 or  = 2  0   2      2 .   

  
Then 0 =  2      2   3   3    ( ) 2  p 3  3qp  q(1  q)
1 2
 p= q + q . 3 3

453
8 When asked to draw sketches of graphs, it is important to note the key features. The first curve is a
standard “exponential decay” curve; the second has the extra factor of sinx. Now sinx oscillates
between –1 and 1, and introduces zeroes at intervals of . Thus, C 2 oscillates between C 1 and –C 1 ,
with zeroes every  units along the x-axis. This sketch of the two curves should then make it clear
that the x i that are then introduced are the x-coordinates of C 2 ’s maxima, when sinx = 1. [It is
important to be clear in your description of x n and x n+1 in terms of n as these are going to be
substituted as limits into the area integrals that follow.] The integration required to find one
representative area will involve the use of “parts”, and the final summation looks like it must be
that of an infinite GP.

y = e–x

O  2 3 4
y = – e–x

–1

The curves meet each time sin x = 1 when x = 2n + 

2 ( n = 0, 1, 2, …).
(4n  3) (4n  1)
Thus x n = and x n + 1 = .
2 2

 e  
sin x dx attempted by parts =  e  x . cos x   e  x . cos x dx or  e  x . sin x   e  x . sin x dx   
x

 
(depending on your choice of ‘1st’ and ‘2nd’ part) =  e  x . cos x  e  x . sin x   e  x . sin x dx  .  
 
x 
Then I =  e (cos x  sin x) – I (by “looping”) =  12 e (cos x  sin x)
x

xn 1 xn 1

 e   cos x  sin x  x e cos x  sin x  2 x

xn 1
x x x x x
An =  e sin x dx =  e  12 e n
or 1
2 n
xn

= 12 e
 12  ( 4 n  1)
0  1  2  12 e   ( 4 n  3) 0  1  2 =
1
2 1
2 e
 12  ( 4 n  1)
 1  e  2

Note that A 1 = 12 e
 52 
e 2

1 and A n + 1 = e 2 A n so that


A
n 1
n 
= A1 1  e 2   e 2   ...
2

e 2
= 12 e
 52 
e 2
1   1
1 e  2
= 1  2
2 e
5
e 2
 1 2
e 1
 1
(using the S of a GP formula) = 12 e 2

454
9 Once you have written down all relevant possible equations of motion, this question is really quite
simple; the two results you are asked to prove arise from considering either times or distances to
the point of collision. There is, however, one crucial realisation to make in the process, without
which further progress is almost impossible; once noted, it seems terribly obvious, yet it probably
doesn’t usually fall within the remit of standard A-level examination questions.

   
For P 1 , x1  0 , x1  u cos  , x1  ut cos  , y1   g , y1  u sin   gt , y1  ut sin   12 gt 2
   
For P 2 , x 2  0 , x 2  v cos  , x 2  vt cos  , y 2   g , y 2  v sin   gt , y 2  vt sin   12 gt 2

u sin  u 2 sin 2 
Now P 1 is at its greatest height when y 2  0  t =  y1 = h = and it follows
g 2g
that u sin = 2 gh

Note that if the two particles are at the same height at any two distinct times (one of which is t = 0
here), then their vertical speeds are the same throughout their motions. Thus u sin = v sin .

2v sin 
y 2 = 0, t  0  t = . This is the time when P 2 would land. Also, the collision occurs
g
b
when x 2 = b  t = is the time of the collision.
v cos 
Then t P2 1
2  range < t(collision) < t P2 range  (or by distances)
v sin  b 2v sin  v 2 sin  cos  2v 2 sin  cos 
 < <  <b<
g v cos  g g g
(v sin  ) 2 2(v sin  ) 2
 cot  < b < cot  . Using u sin = v sin = 2 gh then gives
g g
2 gh 4 gh
cot  < b < cot   2h cot  < b < 4h cot  .
g g

One could repeat all this work for P 1 , but this is not necessary. Since the particles are at their
maximum heights simultaneously (see the above reasoning) and would achieve their “ranges”
simultaneously also, we have 2h cot  < a < 4h cot  .

10 I always feel that collisions questions are very simple, since (as a rule) there are only the two main
principles – Conservation of Linear Momentum and Newton’s Experimental Law of Restitution –
to be applied. Such is the case here. Part (ii) is only rendered more difficult by the introduction of
a number of repetitions, and then the question concludes with some pure mathematical work using
logarithms.

(i) u Using CLM: bmu = bmv B + mv A .

B A Using NEL: u = vA – vB .
bm m 2bu (b  1)u
vB vA Solving simultaneously: v A = and v B = .
b 1 b 1
 2 
Then v A =  1 u  2u – as b  , and v A < 2u always.
1 b 

455
 (ii)
u = u1 u2 = v2 un = vn
B 1 n m B 2 n – 1 m … … … Bn  m A m
v1 v2 vn v

 2   2   2 
2

Using the results of (i), v 2 = u 2 =  u ; u 3 =  u 2    u ; … etc. …

  1    1   1
n 1
 2   2   2   2 
n

all the way down to u n =  u n  1    u and v =  u n    u.

  1   1   1   1
2
Since u n =  1 , as  > 1, it follows that v can be made as large as possible.
 1

In the case when  = 4, v =  85 n u > 20u requires n log  85  > log 20  n 

log 20
.
log 85 
Now log 2 = 0.30103  log 8 = 3log 2 = 0.90309
and log 5 = log 10 – log 2 = 1 – 0.30103 = 0.69897
so that log  85  = log 8 – log 5 = 0.20412.
1.30103
Also log 20 = log 10 + log 2 = 1 + 0.30103 = 1.30103, so we have n  .
0.20412
Since 6  0.20412 = 1.22472 and 7  0.20412 = 1.42884, n min = 7.

11 A few years ago, a standard “three-force” problem such as this would have elicited responses
using Lami’s Theorem; since this tidy little result seems to have lapsed from the collective A-level
consciousness, I shall run with the more popular, alternative Statics-question approach of
resolving twice and taking moments. In order to get started, however, it is important to have a
good, clear diagram suitably marked with correct angles. The later parts of the question consist
mostly of trignometric work.
R T
Res. T sin(  ) + R sin( + ) = W
B
l  Res. T cos(  ) = R cos( + )
C 
l A W.2l cos = T.3l sin

2l 

A
W
T cos(    ) 3T sin 
Substituting to eliminate T ’s (e.g.)  T sin(  ) + sin( + ) =
cos(   ) 2 cos 
 2 cos cos  . cos   sin  . sin  sin  . cos   cos  . sin  
+ 2 cos cos  . cos   sin  . sin  sin  . cos   cos  . sin  
= 3 sin cos  . cos   sin  . sin  

456
 Dividing by cos cos cos
 2(cos   tan  . sin  )(tan  . cos   sin  )  2(cos  tan  . sin  )(tan  . cos   sin  )
= 3 tan  (1  tan  . tan  )
Multiplying out, cancelling and collecting up terms, and then dividing by tan tan then gives
the required answer 2 cot + 3 tan = cot .

1
 = 30o,  = 45o  cot = 2.1  3. = 2 3 ,
3
o
and tan15 = tan(60 – 45 ) =
3 1
o
 o  
3 1
2

 2 3 
1
.
1 3 3 1 2 3

12 In some ways, the pdf f(x) couldn’t be much simpler, consisting of just two horizontal straight-
line segments (in the non-zero part). Part (i) is then relatively routine – use “total prob. = 1” to
find the value of k, before proceeding to find E(X); and the trickiest aspect of (ii) is in the
inequalities work. You also need to realise that the median could fall in either of the two non-zero
regions. For (iii), it is necessary only to follow through each possible value of M relative to E, the
expectation.

Since the pdf is only non-zero between 0 & 1, and the area under its graph = 1, if a, b are both <
(>) 1 then the total area will be < (>) 1. Since we are given that a > b, it must be the case that
a > 1 and b < 1.

1 b
1 k 1
(i) 1 =  f ( x) dx =  a dx +  b dx = ax  + bx  = ak + b – bk  k =
k 1
.
0 0 k 0 k
a b
1 k 1
 ax 2  k  bx 2  1 ak 2 b bk 2
E(X) = 0 xf ( x ) dx = 0 ax dx + k bx dx =  2 

+ 
0  2 k
 =
2
 
2 2

b ( a  b)  1  b  ba  b 2  1  2b  b 2 1  2b  ab
2

=    = = .
2 2 a b 2( a  b ) 2(a  b)

a  ab
(ii) If ak  1
(i.e. M  (0, k)) then  1
 2a – 2ab  a – b  a + b  2ab
ab
2 2

1
1
and aM = 2 or M = .
2a
If ak  (i.e. M  (k, 1)), and noting that this is equivalent to a + b  2ab ,
1
2

1
then ak + (M – k)b = 12 or (1 – M)b = 12  M = 1 –
2b

1  2b  ab 1 a  2ab  a 2 b  a  b b(1  a ) 2
(iii) If a + b  2ab , then  – M =  = = >0
2(a  b) 2a 2 a ( a  b) 2a ( a  b)
and the required result follows.

1  2b  ab 1 b  2b 2  ab 2  2ab  2b 2  a  b
If a + b  2ab , then  – M = 1  =
2(a  b) 2b 2b(a  b)
a (1  b) 2
= > 0 as required.
2b(a  b)

457
13 This question is really little more than examining the various cases that arise for each outcome and
then doing a little bit of work algebraically. The result of part (i) is somewhat counter-intuitive, in
that Rosalind should choose to play the more difficult opponent twice, while one intutively feels
she should be playing the easier opponent. The real issue, however, is that she needs to beat both
opponents (and not just win one game): examining the probabilities algebraically makes this very
obvious. Part (ii) is a nice adaptation, where there is a cut-off point separating the cases when one
strategy is always best from another situation when either strategy 1 or 2 can be best. Here, it is
most important to demonstrate that the various conditions hold, and not simply state a couple of
probabilities and hope they do the job. [It is perfectly possible to do (iii) by “trial-and-error”, but I
have attempted to reproduce below an approach which incorporates a method for deciding the
matter.]

(i) P(W PPQ ) = P(W P W Q –) + P(L P W Q W P ) = p . q. 1 + (1 – p)qp = pq(2 – p).

Similarly, P(W PQQ ) = pq(2 – q) and P(W PPQ ) – P(W PQQ ) = pq(q – p) > 0 since q > p. Thus,
P(W PPQ ) > P(W PQQ ) for all p, q and “Ros plays Pardeep twice” is always her best strategy.

(ii) SI: P(W 1 ) = P(W Q W P – –) + P(W Q L P W P –) + P(W Q L P L P W P )

= pq + pq(1 – p) + pq(1 – p)2 or pq(3 – 3p + p2)
SIII: P(W 3 ) = pq(3 – 3q + q2) similarly.
SII: P(W 2 ) = P(W P W Q – –) + P(L P W P W Q –) + P(W P L Q W Q –) + P(L P W P L Q W Q )
= pq + pq(1 – p) + pq(1 – q) + pq(1 – p)(1 – q)
= pq(4 – 2p – 2q + pq) or pq(2 – p)(2 – q) .

P(W 1 ) – P(W 3 ) = pq(q – p) 3  [ p  q ] > 0 since q > p and p + q < 2 < 3 so that SI is
always better than S3

 
P(W 1 ) – P(W 2 ) = pq p 2  p  1  pq  2q = pq (2  p )(q  p )  (1  p ) 
1 p 1
> 0 whenever q – p >  1 .
2 p 2 p
1
Now p + 12 < q < 1  0 < p < 12  13 < 1  < 12 , so that SI always better than SII
2 p
when q – p > 12 .

1 p
P(W 1 ) – P(W 2 ) > < 0  q – p > < .
2 p
1 p 3
Take p = 1
,q= 1
 q–p= 1
< 1
and  7 > 14 so SII is better than SI.
4 2 4 2
2 p
1 p 3
Take p = 14 , q = 34    q – p = 12   < 12 and  so choosing
2 p 7
1 p 3
 < 73  12  141 (say 161 ) will give p = 14 , q = 16
11
and q – p = 167 >  so that
2 p 7
SI is better than SII.

[I believe that q – p > k has k = 12 as the least positive k which always gives SI better than SII,
but it is a long time ago that the problem was originally devised and I may be wrong.]

458
STEP Mathematics III 2010: Solutions

Section A: Pure Mathematics

1. The first two parts are obtained by separating off the final term of the summation and
expanding the brackets respectively giving , and
1

(the latter given in the question) .

By comparison with the expression for B,
1
1
which by substituting for
1

from the expression for B gives

1
1
.
Substituting for C from the initial result, the required expression can be obtained which can most
neatly be written  
1
1

Thus 1 yielding the first inequality.

Also, and this quadratic expression is only negative if and

only if .
Rearranging the inequality to make xn+1 the subject yields the required result.

2. The expression of cosh a in exponentials enables the integral to be written as

1
1
which can in turn can be expressed as
1

and so employing partial fractions this is

1

459
The evaluation of this with simplification of logarithms yields
1 1
2 sinh 1
giving the required result.
In part (ii), the same technique can be employed for both integrals giving, in the first case
1

1
coth
2 cosh 2
and in the second
1

1 1 1
tan tan

1
2 sinh
2 sinh 2 2
or alternatively

4 cosh
2

3. The two primitive 4th roots of unity are so 1

1, 1 1 1 so 1,
1 1 1 so 1
1 1 1 so 1
1 1 1 1 1 1 so 1

In part (ii), 0⇒ 1⇒ 1 so n is a multiple of 8, and as there are 4 primitive

8th roots of unity, n must be 8.

1⇒ 1 0⇒ 1 ⋯ 1
1 is the only non-primitive root as no power of any other root less than the pth equals unity,
because p is prime, so ⋯ 1

No root of 0 is a root of 0 for any . (For if , by the definition of

, there is no integer t such that 1 when 1. Similarly, if .)
Thus if ≡ , and if 0, then 0 or 0 , so
or .

460
If , then ≡ , and so ≡ 1 which is not possible for positive s, and
likewise in the alternative case.

4. (i) As satisfies both equations, 0 and 0, so subtracting

these the desired result is simply found.
If 0, then we may divide by , and find that
satisfies 0. But also,

and
so satisfies 0.
On the other hand if there is a common root, then it is found at the start of the question and as it
satisfies 0, the required result is found.
If 0 and , then and so the two
equations are one and trivially have a common root. Alternatively, if there is a common root and
, then the initial subtraction yields , and so the result is trivially true.

(ii) If 0, then 0 and

0 have a common root from (i), and so then do 0 and
0 which is the required result.
On the other hand, if the two equations have a common root  , then 0
and 1 0 , and thus so does
1 0 which is a quadratic equation and we can use
the result from (i) again.
Using , , , in the given condition, we obtain a cubic equation in b,
0, which has a solution 1, meaning the other two can be simply
√
obtained as .

5. The line CP can be shown to have equation 1 and so R is 0,

So, similarly, S must be ,0 .
Thus RS has equation 1 1 and PQ has equation .
As the coordinates of T satisfy both equations, they satisfy their difference which is
0. As RS and PQ intersect, which yields
0 and hence 0 implying that T’s coordinates
satisfy 0 giving the desired result. (Alternatively, 0⇔ 0 ,
which is a contradiction.)
The construction can be achieved more than one way, but one is to label the given square ABCD
anti-clockwise, choose points on AB and AD different distances from A, label them P and Q,
construct CP and CQ, and find their intersections with AD and AB, R and S, respectively, and
find the intersection of PQ and RS, label it T, then TA is perpendicular to AC. Rotating the
labelling through a right angle and repeating three more times achieves the desired square.

461
6. is cos , sin , 0 , is cos cos , sin cos , sin , is sin , cos , 0 , is
sin , cos , 0 , is 0,0,1 and is cos sin , sin sin , cos .
The scalar product ∙ gives the quoted result immediately. The direction of the axis can
1 cos cos
be found from the vector product 0 sin cos giving the direction of the axis as
0 sin
0
sin .
sin cos

7. The initial result can be obtained by differentiating y directly twice obtaining

sin sin
√
cos sin sin sin and substituting into the LHS.
(Slightly more elegant is to rearrange as cos sin , differentiate and then square to
obtain 1 1 and then differentiate a second time.)
The two similar results are 1 3 1 0 and
1 5 4 0, which lead to the conjecture
1 2 1 0 which is proved simply by induction.

Using 0 , we find that 1, 0, , 0, 4

and so the Maclaurin series commences 1 ⋯
! !

Now replacing x by sin ,

cos 1 ⋯ 1 sin sin ⋯
! ! ! !
All the odd differentials are zero, and the even ones are 1 2 … 2 ,
so if m is even all the terms are zero from a certain point (when 2 ) and thus the series
terminates and is a polynomial in sin , of degree m.

8. Substituting for , the desired integral is seen to be the reverse of the quotient rule, i.e.

To choose a suitable function in part (i), substitution of and

1 2 3 in the given expression yields a quadratic equation, and equating the
coefficients of the powers of x gives 5 3 2 , 2 3 , 3 2 .
These three equations are linearly dependent and so their solution is not unique.
Choosing, for example 0, 3, 2 and then 1, 1, 1 gives solutions
which are related by 1 i.e. the same bar the

arbitrary constant.

462
(ii) Rearranging the equation to be solved as , the

integrating factor is
As a result, the RHS we require to integrate is
Repeating similar working to part (i), except with 1 cos 2 sin and
sin cos , gives three linearly dependent equations,
5 2 , 3 2 ,4
Choosing e.g. 4, 5, 0, the solution is 4 5 sin 1 cos 2 sin

Section B: Mechanics

9. Resolving radially inwards for the mass P, sin   ,

where R is the normal reaction of the block on P, and v is the (common) speed of the masses
when OP makes an angle with the table.

Conserving energy, sin 0 , and making the subject of

this formula to substitute in the first equation re-arranged for R,
sin   is found.
Remaining in contact requires this expression to be non-negative for all , 0 .
Considering the graphs of asin and for 0 ,
asin 0, ∀ , 0 if and only if asin 0 for
so 0 for all , 0 if and only if 3 sin 2 0 which gives the
required result.

10. Resolving perpendicularly to OB, cos , where the tension in the

elastic string is . The sine rule
Putting these three results together gives the required expression.
Also from the sine rule, , so for and small, yielding the desired
result.
From this result,  may be made the subject of the formula, so that the result
1 sin , which for small angles becomes
1 can be written

and hence the period is 2 .

463
11. If the acceleration of the block is , and the acceleration of the bullet is , then
and ,
so the relative acceleration

The initial velocity of the bullet relative to the block is – and the final velocity of the bullet
relative to the block is 0. If the time between the bullet entering the block and stopping moving
through the block is T, then using” “, 0
For the block, the initial velocity is 0 , the final velocity is , and again using ,

and so

as required.

If the distance moved by the block whilst the bullet is moving through the block is ,
using” 2 “, 2 andso

Once the bullet stops moving through the block, the next initial velocity of block/bullet is , the
final velocity is 0, the acceleration is – , so the distance moved using
“ 2 ” is given by 0 2 i.e.
Thus the total distance moved is  

If , then the block does not move, and the bullet penetrates to a depth .

Section C: Probability and Statistics

12. 1 ⋯ ⋯ which is 1 plus an infinite GP. Summing that GP

and making S the subject produces the displayed result.

1 2 1 3 1 ⋯ 1 ⋯ so making use of the first

result with 1, 1 ,

464
 1 1 or alternatively, ⋯ which
both lead to the required result.
1 or alternatively, ⋯

The expected number of shots, S, is given by

1 2 3 4 5 ⋯
1 3 5 ⋯ 2 1 2 ⋯
which using the initial result of the question 2
and can be shown to simplify to the required expression.

13. , 0
1 1 0
1 1
1 1

As 0 and 1,
,
, ,

1 1

0 is given.
1 implies 1
, implies  as seen before.
, implies 1
and hence , , 1

for 1,2,3 as ,
, and
, , as a linear transformation does not affect correlation.

465
466
STEP Examiners’ Report
2011

Mathematics
STEP 9465/9470/9475

September 2011

467
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 3
STEP Mathematics II 15
STEP Mathematics III 19
Explanation of Results 22

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 STEP 2011 Paper I: Principal Examiner’s Report

Introductory comments
There were again significantly more candidates attempting this paper than last year (just
over 1100), but the scores were significantly lower than last year: fewer than 2% of
candidates scored above 100 marks, and the median mark was only 44, compared to 61
last year. It is not clear why this was the case. One possibility is that Questions 2 and 3,
which superficially looked straightforward, turned out to be both popular and far harder
than candidates anticipated. The only popular and well-answered questions were 1 and 4.
The pure questions were the most popular as usual, though there was noticeable variation:
questions 1–4 were the most popular, while question 7 (on differential equations) was fairly
unpopular. Just over half of all candidates attempted at least one mechanics question,
which one-third attempted at least one probability question, an increase on last year. The
marks were surprising, though: the two best-answered questions were the pure questions
1 and 4, but the next best were statistics question 12 and mechanics question 9. The
remainder of the questions were fairly similar in their marks.
A number of candidates ignored the advice on the front cover and attempted more than
six questions, with a fifth of candidates trying eight or more questions. A good number of
those extra attempts were little more than failed starts, but still suggest that some can-
didates are not very effective at question-picking. This is an important skill to develop
during STEP preparation. Nevertheless, the good marks and the paucity of candidates
who attempted the questions in numerical order does suggest that the majority are being
wise in their choices. Because of the abortive starts, I have generally restricted my at-
tention to those attempts which counted as one of the six highest-scoring answers in the
detailed comments.
On occasions, candidates spent far longer on some questions than was wise. Often, this
was due to an algebraic slip early on, and they then used time which could have been far
better spent tackling another question. It is important to balance the desire to finish a
question with an appreciation of when it is better to stop and move on.
Many candidates realised that for some questions, it was possible to attempt a later
part without a complete (or any) solution to an earlier part. An awareness of this could
have helped some of the weaker students to gain vital marks when they were stuck; it is
generally better to do more of one question than to start another question, in particular
if one has already attempted six questions. It is also fine to write “continued later” at
the end of a partial attempt and then to continue the answer later in the answer booklet.
As usual, though, some candidates ignored explicit instructions to use the previous work,
such as “Hence”, or “Deduce”. They will get no credit if they do not do what they
are asked to! (Of course, a question which has the phrase “or otherwise” gives them
the freedom to use any method of their choosing; often the “hence” will be the easiest,
though.)
It is wise to remember that STEP questions do require a greater facility with mathe-
matics and algebraic manipulation than the A-level examinations, as well as a depth of
understanding which goes beyond that expected in a typical sixth-form classroom. STEP

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STEP I 2011 Principal Examiner’s Report

candidates are therefore recommended to heed the sage advice on the STEP Mathematics
website, http://www.admissionstests.cambridgeassessment.org.uk/adt/step:

From the point of view of admissions to a university mathematics course,

STEP has three purposes. . . . Thirdly, it tests motivation. It is important to
prepare for STEP (by working through old papers, for example), which can
require considerable dedication. Those who are not willing to make the effort
are unlikely to thrive on a difficult mathematics course.

Students will also benefit from reading the detailed STEP I solutions which I have written
over the previous few years after attempting the papers; these are available from the “Test
Preparation” section of the above website.

Common issues
There were a number of common errors and issues which appeared across the whole paper.
The first was a lack of fluency in algebraic manipulations. STEP questions often use more
variables than A-level questions (which tend to be more numerical), and therefore require
candidates to be comfortable engaging in extended sequences of algebraic manipulations
with determination and, crucially, accuracy. This is a skill which requires plenty of practice
to master.
Another area of weakness is logic. The first section of the STEP Specification specifically
lists: “Mathematical vocabulary and notation: including: ‘equivalent to’; ‘necessary and
sufficient’; ‘if and only if’; ‘⇒’; ‘ ⇐⇒ ’; ‘≡’”. A lack of confidence in this area showed up
several times. In particular, a candidate cannot possibly gain full marks on a question
which reads “Show that X if and only if Y” unless they provide an argument which shows
that Y follows from X and vice versa.
Along with this comes the need for explanations in English: a sequence of formulæ or
equations with no explicit connections between them can leave the reader (and writer)
confused as to the meaning: Does one statement follow from the other? Are they equiv-
alent statements? Or are they perhaps simultaneous equations? For example, writing
x = 2 followed by x2 = 4 is not the same as writing x = 2 followed by 2x = 4, and both
are different from writing x = 2 followed by y = 3. Brief connectives or explanations
(“thus”, “so”, “∴”, “⇒” or “ ⇐⇒ ”) would help, and sometimes longer sentences are
necessary. The solutions booklet is more verbose than candidates’ solutions need to be,
but gives an idea of how English can be used.
In some cases, the need for explanations is even greater. Where a question instructs the
candidate to prove a statement, writing down an equation without justification is likely to
gain the candidate few (if any) marks. For example, it may well suffice to write “Taking
moments around A” or even just “M (A)” to indicate the source of the following equation.
In a pure mathematics question, something like “Substituting (1) into (2):” may well be
adequate.
Another related issue continues to be legibility. Many candidates at some point in the
paper lost marks through misreading their own writing. Common confusions include
muddling their symbols, the most common being: M and m; V and v; u and n; u and N ;

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STEP I 2011 Principal Examiner’s Report

x and n; α and 2; a and 9; s, S and 5; and occasionally z and 2. It is sad that, at this
stage, candidates are still wasting marks because of bad writing habits. A particularly
striking example is shown in this candidate’s work:

This apparently reads

u(M − 3m) + 2u(M + m)
M +m
One frequent error was dividing by zero. On several occasions, an equation of the form
xy = xz appeared, and candidates blithely divided by x to reach the conclusion y = z.
This may or may not be true, depending upon whether or not x could be zero. A better
approach in general is to rearrange to get x(y − z) = 0 and to deduce that either x = 0
or y = z. Alternatively, if it is known with certainty that x 6= 0, it is fine to divide by x,
but one must explicitly indicate that x 6= 0.
Again, I give a strong reminder that it is vital to draw appropriate, clear, accurate dia-
grams when attempting some questions, mechanics questions in particular: it was shock-
ing how many candidates attempted to solve a collision question without a diagram or a
moments question with a tiny, rough sketch!

Question 1
This was an easy and very popular first question, attempted by almost all of the candi-
dates. It was also the most successfully answered, with a median mark of 14.

(i) Most candidates differentiated the equation implicitly, reaching the specified gradi-
ent with ease. Some decided to rearrange the equation into the form y = · · · before
differentiating; this was frequently unsuccessful as the final manipulations required
were relatively tricky.
The next step, showing that p = ±q, was also generally answered well, even by
candidates who had become stuck on the first part of the question. It was surprising
to see many solutions using implicit differentiation to find the gradient of the straight
line ax + by = 1, rather than rearranging the equation to get y = mx + c and then
reading off the answer.
The final step was also generally answered well, though a number of candidates
could not see how to use the previous result (p = ±q) to progress. Furthermore,
there were a few candidates who managed to deduce one of the possibilities but not
the other; this was a little strange, as the argument was essentially identical.

(ii) While the majority of candidates used the result that the products of normal gra-
dients is −1, relatively few paid enough attention to notice that the equation of the
curve was different to that of part (i). This led them to conclude that a2 q 2 = −b2 p2 ,

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STEP I 2011 Principal Examiner’s Report

from which they were not able to make any progress. (Note here that the left hand
side is always positive and the right hand side is always negative.)
Of those who overcame this hurdle and attempted the question posed, some suc-
ceeded in reaching the specified conclusion, while most deduced that a2 q 2 = b2 p2
and then became stuck, unable to see how to progress. This is presumably because
this equations were more complex than in the first part of the question. Also, some
candidates who gave otherwise good answers did not consider both possible cases
aq = +bp and aq = −bp.

Question 2
This was a very popular question which was answered spectacularly poorly: almost half
of candidates scored 0. Of the other half, the median score was 8, with them succeeding
on the first part but getting no further.
R
For the first part, most candidates (presumably familiar with examples such as xex dx)
attempted to integrate by parts, even though that was not the most appropriate method
in this case. Those who did use parts frequently tried to integrate x/(1 + x) and tended
to get nowhere. Very few understood that a second integration by parts was necessary
to finish the job. Nevertheless, most of those who did attempt to use parts showed that
they did have a reasonable working knowledge of the technique.
There were a number of shocking R errors which
R recurredR frequently. The first was to use
the erroneous “rule” thatR f(x)g(x) dxR= f(x) dx. g(x) dx, so there were a number
ex 1
of candidates who wrote 1+x dx = ex 1+x dx or similar. Another common error was
that E, a definite integral, would appear inside integrals
R or evaluations, giving statements
1
or expressions such as [xE]0 = E(1) − E(0) or E dx. There were also a number of
candidates who applied parts and ended up with integrals inside integrals or integrals
inside evaluations ([· · · ]10 ).
A number of candidates also failed to distinguish between indefinite and definite integrals.
Parts (i) and (ii)
There were relatively few attempts at these later parts, though some who had been
stumped by the first part of the question did succeed with this part.
Some candidates tried to apply the techniques of the first part to this integral, but many
realised that a substitution was necessary and successfully executed it.
Most candidates who were successful with part (i) went on to complete the whole question.

Question 3
This was another very popular but poorly answered question. While most candidates were
able to gain some marks, few proceeded beyond the initial part of the question, giving a
median mark of 5 and an upper quartile of 7.
Candidates generally did well at the first part of this question, with the majority using
the compound angle formulæ to expand the sines on the left hand side of the identity.

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STEP I 2011 Principal Examiner’s Report

A number got stuck at this point, either because they did not know sin 13 π or cos 13 π
or because they made algebraic errors. About a quarter of candidates were unable to
complete the proof as they did not show that sin 3θ is identical to 3 sin θ − 4 sin3 θ or
equivalent; some simply stated the result with no justification. On the other hand, a
number of candidates offered very nice arguments using de Moivre’s theorem, which was
very nice to see.
Some candidates used the factor formulæ, but these were in the minority.

(i) Many candidates who reached this point were able to differentiate the identity cor-
rectly, which was pleasing. A significant number failed at this hurdle, though. Of
those who did differentiate correctly, though, the majority failed to realise that they
could divide their new identity involving cosines by the original identity to reach
the stated result. Since the rest of the question depended upon this idea, barely a
quarter of candidates made any further progress. There was a strong hint though:
the presence of cot θ should suggest thinking about cos θ/ sin θ.

(ii) Of those who had become stuck earlier, very few attempted this part, even though
the first half was totally independent of part (i). Of those who did, a significant
number became stuck after performing the given substitution, not realising that
they could then use the identity sin( π2 − x) = cos x. There were also a number
of candidates who successfully derived the result from scratch using the compound
angle formula for tan.
Few candidates made it as far as the cosec equation. Of those who did, and realised
that they needed to perform another division or equivalent, few were comfortable
enough with the trigonometric manipulations involved to reach an expression in-
volving cosec 2θ and thence to reach the stated result.

Question 4
This was a popular question, attempted by two-thirds of candidates. It was also one of
the most successfully answered, with a median mark of 11.
Candidates were very good at differentiating to find the
√ coordinates of T , though there
were some issues. Those who rearranged to find y = 4ax generally did not handle the
possibility that y could be negative. There were also a number of candidates who are still
confused when trying to find the equation of a tangent: they used the general expression
for dy/dx rather than substituting in the values of x and y at the point of tangency. This
gave them a “straight line” with equation y −2ap = 2ay
(x−ap2 ) which was then sometimes
rearranged to give a quadratic.
The vast majority were fine with this step, though, and went on to successfully find the
coordinates of T . Some used the symmetry of the situation to simply write down the
equation of the second tangent, while others determined it from scratch.
There was one sticking point, though: at this level of work, candidates are expected to
take care when dividing to ensure that they are not dividing by zero. A mark was therefore
awarded for stating that p − q 6= 0 or p 6= q when dividing by it, but very few candidates
did so.

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STEP I 2011 Principal Examiner’s Report

When it came to deducing the given formula for cos φ, most candidates made a good
start, with the dot-product approach more popular than the cosine rule. However, there
was a need for some fluent algebraic manipulations, in particular the ability to factorise.
This should have been made a little easier by knowing the desired final result, but most
candidates became bogged down at this point and were unable to deduce the given ex-
pression. The dot-product approach, with its slightly simpler algebra, was generally more
successful.
The final part of the question, requiring candidates to deduce that the line F T bisects the
angle P F Q, produced many spurious attempts. Few candidates appreciated the symmetry
of the situation, and so went on to calculate cos(∠T F Q) from scratch. Others attempted
to find cos(∠P F Q), presumably hoping to use a double angle formula or similar. These
approaches were sometimes successful.
There were also candidates who attempted to answer this part by using right-angled
trigonometry in one of the triangles, or by identifying similar or congruent triangles, even
though none of these approaches made sense in this situation.

Question 5
This was a moderately popular question, attempted by half of the candidates. Most made
a reasonably good start, but became stuck after deducing the value of I; only half of the
candidates gained more than 7 marks.
The first part was answered fairly poorly. Most were able to correctly sketch the graph
of y = sin x (though there were a few who could not) and y = kx, but very few made any
attempt to justify from their sketch the required result. The most common mark for this
part was 1 out of 4.
A handful of candidates decided to rewrite the equation as sin x/x = k and went on to
draw a graph of y = sin x/x, a far more challenging task which was successfully performed
by some of the candidates.
The integration part (deducing the formula for I) was generally answered very well. There
were some who did not understand how to split up the integral or the significance of x = α
to this part, but the majority correctly handled the absolute value and the necessity to
change the signs of the integrand in the integral from α to π.
Quite a number used geometrical arguments, considering a triangle from x = 0 to x = α,
a trapezium from x = α to x = π and two regions under the curve; most of these also
reached the given result.
Determining the stationary points of the function was found to be very challenging. While
the differentiation was generally performed correctly, many struggled to factorise the
resulting expression. Even among those who did, it was very common to divide by one of
the factors, ignoring the possibility that it might be zero. Those who were lucky enough
to divide by sin α − α cos α deduced the correct value for α at the minimum. Though they
gained no immediate credit for this, they were able to continue attempting the rest of the
question. Others, though, were less fortunate and asserted that the minimum occurred
where sin α = α cos α; they were unable to make any further progress.

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STEP I 2011 Principal Examiner’s Report

It was very unusual to see any sort of decent justification that sin α = α cos α has no
solutions in the required range.
Most of the students who reached this point correctly evaluated I at the stationary point
to reach the given value.
Finally, students were required to show that the stationary point is a minimum, which
few attempted. A significant amount of care was required for each of the approaches, and
a small number did so successfully. A few only evaluated I or dI/dα on one side of the
stationary point rather than on both. Students would do well to remember that there
are at least three different general approaches to determining the nature of a stationary
point, and that different methods might be more or less successful in different situations.

Question 6
This was another fairly popular question, but there were many very weak attempts; the
median mark was 5.
The first part was answered very poorly. A significant number of candidates only worked
out the first few terms and showed that they satisfied the given formula, without making
any attempt to justify the formula in general. The majority attempted to write down the
general term, with varying degrees of success: many forgot to take account of the minus
sign, and so did not include (−x)r . Another common error was to write expressions
involving (−n)!. Few candidates gave any justification for removal of the minus signs,
and solutions which correctly dealt with the case r < 2 were rare indeed.

(i) This part was answered very well by the majority of candidates. A number at-
tempted to factorise the numerator as 1 − x + 2x2 = (1 + x)(1 − 2x) or other
incorrect ways. Very few answers were careful about the boundary cases, namely
where r = 0 and r = 1, and so most candidates only achieved 4/5 on this part.
(ii) Only a minority of candidates made any progress on this part: most candidates
were unable to correctly identify the rule r2 /2r−1 for the terms of the sequence. Of
those who did, the most common approach was to relate the sequence to that of
part (i) as in the first approach described above. Some candidates were able to do
the manipulations correctly, but there were a significant number who made slips
along the way (for example, using r2 /2r or leaving out the initial term). Some used
the second approach or a variant of it.
It was very pleasing to see some students use the third approach described, many of
whom were successful.

Question 7
This was by far the least popular Pure Mathematics question, attempted by only one-third
of candidates. The marks achieved were poor; the median mark was 3.

(i) For this first part of the question, very few candidates were able to justify the form of
differential equation given. Many candidates talked about the water level decreasing

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STEP I 2011 Principal Examiner’s Report

until it reached α2 H, which is not what actually happens (it continually increases
until it reaches that point). This lack of understanding had a knock-on effect later
on, too, in that they were unable to generate the required differential equation in
part (ii).
Nevertheless, from the given DE, a good number were able to separate the variables
and at least make progress towards a solution, though only a minority were able to
complete the task.
A number of candidates muddled H and h; this sloppiness prevented them from
getting the right answer. It was also unfortunate that the symbols for proportionality
and the Greek letter alpha are similar; students had to take extra care to keep those
distinct as well.
Many candidates used the given result and approximation for ln(1 + x) to deduce
the stated approximation.

(ii) Few candidates were able to generate the√correct differential equation for this part,
with the offering of dh/dt = c(α2 H − h) being far more common. (We were
generous and only deducted a few marks if they made progress with this incorrect
equation.) Of those who attempted to solve either the correct or this variant differ-
ential equation, very few could work out how√to integrate the resulting expression.
Even when they used the substitution u = h, they were usually unable to inte-
grate the fractional linear expression resulting from
√ it. Some
√ were more successful,
especially when they used the substitution v = α H − h.
A number of students wisely picked up a couple of marks by demonstrating the
validity of the final approximation on the basis of the solution given in the question.

Question 8
This was a moderately popular pure mathematics question, attempted by about half of
the candidates. There were a number of good solutions, though the median mark was
only 6.

(i) (a) Most realised that n2 + 1 > 0 and deduced that m3 > n3 , though few gave
any sort of explanation of how m > n follows from this. A number of students
split into four cases depending on the signs of m and n; this was frequently a
laborious but correct approach.
The next part, requiring the expansion of (n+1)3 , was well answered, though few
understood the significance of “if and only if”: most only provided an argument
for one of the two directions.
Most were able to combine the two conditions, and were also comfortable with
solving quadratic inequalities.
(b) Few candidates understood the significance of the condition n < m < n + 1
where both n and m are supposed to be integers, leaving most unable to do this
part. This also led them to struggle with part (ii).

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STEP I 2011 Principal Examiner’s Report

(ii) Relatively few candidates understood what they were trying to do in this part. Many
tried to replicate the approach given in part (i) with varying degrees of success.
There were a few concerningqerrors which occurred in a number of scripts. One
1
was the arithmetic faux pas 2
= 14 , the other was the belief that f(x) = f(y) is
equivalent to or implied by f 0 (x) = f 0 (y).
Nonetheless, there were also a variety of other correct approaches, for example, some
showed that q − 2 6 p 6 q + 2 for all q, and then checked the five possible cases
p = q − 2, etc., to ascertain all possible solutions.

Question 9
This was the most popular Mechanics question, attempted by about 40% of the candi-
dates. It was well-answered overall; though the median mark was only 7, over a quarter
of candidates achieved 14 or more.
In attempting to find tan θ, most candidates confidently drew a sketch of the situation
and correctly wrote down the equations of motion. Some did not clearly indicate the
meanings of their symbols, and this sometimes led to confusion later; some used x for
time, which was bizarre.
The greatest stumbling block for the majority of candidates was the algebraic manipula-
tions. Once they had reached the equation gd21 /2v 2 cos2 θ = d1 tan θ − d2 , many seemed
unsure how to proceed. And of those who did, a significant number were unfamiliar with
the factorisation of a3 − b3 , leaving them unable to complete this part despite being given
the answer.
As a general rule, when using the “suvat” equations, it is worth indicating which equa-
tion is being used, and specifying the direction (horizontal or vertical) which is being
considered.
A significant number of candidates did not even attempt the final part of the question
(finding the range of the particle); it is unclear why this was the case.
Of those who did, many were successfully able to use their earlier work to determine the
range. A number became stuck because of algebraic errors, but about 10% of attempts
scored full marks.
It was also delightful to see the quadratic equation approach being successfully used
at least once; there are often significantly different ways of approaching a problem in
Mathematics, and this was a wonderful example.

Question 10
This Mechanics question was attempted by about one-quarter of candidates, but it was
found to be fairly difficult with almost half of candidates scoring 5 or fewer marks.
The first part of this question led to many wordy solutions which did not reach the nub
of the problem. Perhaps the wording “Explain why” rather than “Show that” or “Prove
that” was part of the cause of this.

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STEP I 2011 Principal Examiner’s Report

Many candidates realised that no energy is lost in the bounce, but few went on to make
correct deductions from this. An easy and frequent mistake was to say that both A and
B have the same kinetic energy when they collide; this is only true if they have the same
mass.
The next part begins as a standard collision question, and many candidates were very
comfortable with it. Many, however, had sign errors in their conservation of momentum
or equation of restitution equation, preventing any significant further progress. This
was particularly frequent among those candidates who did not draw a diagram with the
velocities (and their directions) before and after the collision clearly marked. Of those
who did, many were able to show that at least one of the particles moved upwards after
the collision. Surprisingly few were able to show that both moved upward, in spite of this
being a standard type of STEP question.
As mentioned in the overview, another frequent cause of difficulty in this question was a
confusion between ‘M ’ and ‘m’; at this level, it is crucial that students have developed a
mathematical writing style which is clear and avoids such errors.
A number of candidates made good attempts to find the maximum height of B. Many,
however, simply wrote down a series of calculations with no indication of what they were
attempting to calculate. As the answer was given, full marks could only be awarded if
the argument was clearly correct. It also turned out to be possible to reach the given
answer through a straightforward, yet incorrect, calculation; such solutions received very
few marks.

Question 11
This question was attempted by about one-third of candidates, but was fairly poorly
answered. About one-third of attempts scored 0 or 1, and the median mark was therefore
only 4. Nevertheless, there were many good solutions, reaching about two-thirds of the
way through, including a number of perfect scores.
An overall comment for this question is again that candidates need to explain their work.
Particularly in questions where the answer is given, little credit will be given for sim-
ply writing down an equation which leads to the required answer in one step unless a
justification can be seen.
For the first part of the question, most candidates drew decent diagrams, allowing them
to proceed, but a fair number drew something which was either inaccurate (placing A
below B or drawing the rod horizontally, for example), severely incomplete (no forces) or
too small and scribbly to be useful. These led to subsequent difficulties when attempting
to resolve forces or take moments.
About 20% of candidates stopped after drawing the diagram, gaining them either 0 or 1
(the modal score).
A number of candidates did not appreciate that the tensions in the two parts of the string
were equal, and were therefore unable to proceed.
A common error seen when taking moments was something like 3d(mg) = 7d(T sin β); this
could be made to give the ‘correct’ answer, but received little credit. Another common

478
STEP I 2011 Principal Examiner’s Report

error was to forget to include any forces in the moments equation.

Of those candidates who proceeded further than the diagram, most made good progress
towards finding the length of the string.
A common assumption was that ∠AP B was a right angle, without giving any justification
of this. A similar, though far less common, assertion was that AP/P B = AG/GB. Both
of these happen to be true, but candidates are required to prove them to gain any credit
for their argument.
Many different approaches were seen; the more common ones are described in the solutions
above, and there were many variants of these.
When it came to finding the angle of inclination, a small number of candidates success-
fully did so. There were many attempts which fudged their working to reach the stated
conclusion. A number stated that the angle AP B is bisected by the line P G without
giving any justification; such attempts gained few marks.

Question 12
Generally, Statistics questions are generally the least popular on STEP papers, and this
year was no exception; this question was answered by fewer than one-quarter of candidates.
In spite of this (or perhaps because of it), it was one of the most successfully answered
on the paper, with a median mark of 11 and an upper quartile of 16.

(i) A few candidates failed to get started and scored no marks at all, or misread the
question and gave the probability of failure instead, but the majority gained full
marks on this part.

(ii) Most candidates were able to enumerate the possible cases of success or the cases of
failure. However, a significant proportion did not provide any justification that they
had considered all possible cases, and so lost at least 2 marks. In part (iii), this
lack led a significant number to overlook one or more cases, and so get the wrong
answer.
Several candidates failed to realise that the probability of the second person having
a £2 coin depends upon the first person’s coin.

(iii) Those who succeeded on part (ii) generally made good progress on this part as well;
see the comments above.

Question 13
This question was attempted by around 20% of candidates, but was only counted as one
of the best six questions for about three-quarters of them. Even among those, it was
the most poorly answered question on the whole paper, with one-third of the attempts
gaining only 0 or 1 mark, a median of 4 marks and an upper quartile of 8 marks.

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STEP I 2011 Principal Examiner’s Report

(i) In this first part, many candidates did not even attempt to find k, and algebraic
errors during the integration were common. A number of attempts to find k were
obviously wrong, as they gave a negative area, but candidates did not notice this.
√
Most were unable to integrate x 4 − x2 , with failed attempts to integrate by parts
far outnumbering correct integrations of this expression.
A number of candidates attempted to calculate the median rather than the mean.
Other bizarre interpretations of the term “mean” were also seen.

(ii) Few candidates made it this far. Of those who did, there were some very good
solutions. One of the hardest parts was getting the logic correct: the phrasing of
the question as “Show that X if Y” was often misinterpreted to mean “Show that if
X then Y”, though the majority of the marks were awarded in such cases.

(iii) The handful of students who made it this far were generally successful at this part,
too.

Dr Julian Gilbey
Principal Examiner (Marking)
August 2011

480
STEP 2011 Paper 2 Report

General Remarks
There were just under 1000 entries for paper II this year, almost exactly the same number as last
year. After the relatively easy time candidates experienced on last year’s paper, this year’s
questions had been toughened up significantly, with particular attention made to ensure that
candidates had to be prepared to invest more thought at the start of each question – last year saw
far too many attempts from the weaker brethren at little more than the first part of up to ten
questions, when the idea is that they should devote 25-40 minutes on four to six complete
questions in order to present work of a substantial nature. It was also the intention to toughen up
the final “quarter” of questions, so that a complete, or nearly-complete, conclusion to any
question represented a significant (and, hopefully, satisfying) mathematical achievement.
Although such matters are always best assessed with the benefit of hindsight, our efforts in these
areas seem to have proved entirely successful, with the vast majority of candidates concentrating
their efforts on four to six questions, as planned. Moreover, marks really did have to be earned:
only around 20 candidates managed to gain or exceed a score of 100, and only a third of the
entry managed to hit the half-way mark of 60.

As in previous years, the pure maths questions provided the bulk of candidates’ work, with
relatively few efforts to be found at the applied ones. Questions 1 and 2 were attempted by
almost all candidates; 3 and 4 by around three-quarters of them; 6, 7 and 9 by around half; the
remaining questions were less popular, and some received almost no “hits”. Overall, the highest
scoring questions (averaging over half-marks) were 1, 2 and 9, along with 13 (very few attempts,
but those who braved it scored very well). This at least is indicative that candidates are being
careful in exercising some degree of thought when choosing (at least the first four) ‘good’
questions for themselves, although finding six successful questions then turned out to be a key
discriminating factor of candidates’ abilities from the examining team’s perspective. Each of
questions 4-8, 11 & 12 were rather poorly scored on, with average scores of only 5.5 to 6.6.

Comments on individual questions

Q1 The first question is invariably set with the intention that everyone should be able to
attempt it, giving all candidates something to get their teeth into and thereby easing them into the
paper with some measure of success. As mentioned above, this was both a very popular question
and a high-scoring one. Even so, there were some general weaknesses revealed in the curve-
sketching department, as many candidates failed to consider (explicitly or not) things such as the
gradient of the curve at its endpoints and, in particular, the shape of the curve at its peak (often
more of a vertex than a maximum). It was also strange that surprisingly many who had the
correct domain for the curve and had decided that the single point of intersection of line and
curve was at x = 1 still managed to draw the line y = x + 1 not through the endpoint at x = 1.
Most other features – domain, symmetry, coordinates of key points, etc. – were well done in (i).
Unfortunately, those who simply resort to plotting points are really sending quite the wrong
message about their capabilities to the examiners.
Following on in (ii), the majority of candidates employed the expected methods and were
also quite happy to plough into the algebra of squaring-up and rearranging; however, there were
frequently many (unnecessarily) careless errors involved. The only other very common error was
in the sketch of the half-parabola y  2 1  x , due to a misunderstanding of the significance of
the radix sign.

481
Q2 Personally, this was my favourite question, even though it was ultimately (marginally)
deflected from its original purpose of expressing integers as sums of two rational cubes. Given
that the question explicitly involves inequalities (which are, as a rule, never popular) and cubics
rather than quadratics, it was slightly surprising to find that it was the most popular question on
the paper. However, although the average score on the question was almost exactly 10, these two
issues then turned out to be the biggest stumbling-blocks to a completely successful attempt as
candidates progressed through the question, both in establishing the given inequalities and then
in the use of them. In particular, it was noted that many candidates “proved” the given results by
showing that they implied something else that was true, rather than by deducing them from
something else known to be true; such logical flaws received little credit in terms of marks. The
purpose of this preliminary work was to enable the candidates to whittle down the possibilities to
a small, finite list and then provide them with some means of testing each possibility’s validity.
This help was often ignored in favour of starting again. In general, though, part (i) was done
reasonably well; as was (ii) by those who used (i)’s methodology as a template.
Only a very few candidates were bold enough to attempt (ii) successfully without any
reference to (i)’s methods; indeed, this arithmetic approach was how the question was originally
posed (as part (i), of course) before proceeding onto the algebra. Noting that the wording of the
question does not demand any particular approach in order to find the required two solutions to
the equation x3 + y3 = 19z3, a reasonably confident arithmetician might easily note that
19  23 = 152 = 33 + 53 and 19  33 = 513 = 13 + 83
and it isn’t even necessary to look very far for two solutions. For 10 marks, this is what our
transatlantic cousins would call “ a steal”.

Q3 Again, despite the obvious presence of inequalities in the question, this was another very
popular question, and was generally well-handled very capably in part (i), where the structure of
the question provided the necessary support for successful progress to be made here. Part (ii) was
less popular and less well-handled, even though the only significant difference between this and
(i)(c) was (effectively) that the direction of the inequality was reversed. Although the intervals
under consideration were clearly flagged, many candidates omitted to consider that, having
shown the function increasing on this interval, they still needed to show something simple such
as f(0) = 0 in order to show that f(x)  0 on this interval. A few also thought that f (x) increasing
implied that f(x) was also increasing.

Q4 This question was the first of the really popular ones to attract relatively low scores
overall. In the opening part, it had been expected that candidates would employ that most basic
of trig. identities, sinA = cos(90o – A), in order to find the required values of , but the vast
majority went straight into double-angles and quadratics in terms of sin instead, which had
been expected to follow the initial work; this meant that many candidates were unable to explain
convincingly why the given value of sin18o was as claimed.
Despite the relatively straightforward trig. methods that were required in this question,
with part (ii) broadly approachable in the same way as the second part of (i), the lack of a clear-
minded strategy proved to be a big problem for most attempters, and the connection between
parts (ii) and (iii) was seldom spotted – namely, to divide through by 4 and realise that sin5
must be  12 . Many spotted the solution  = 6o, but few got further than this because they were
stuck exclusively on sin30o =  12 .

482
Q5 This vectors question was neither popular nor successful overall. For the most part this
seemed to be due to the fact that candidates, although they are happy to work with scalar
parameters – as involved in the vector equation of a line, for instance – they are far less happy to
interpret them geometrically. Many other students clearly dislike non-numerical vector
questions. Having said that, attempts generally fell into one of the two extreme camps of ‘very
good’ or ‘very poor’. More confident candidates managed the first result and realised that a
“similarity” approach killed off the second part also, although efforts to tidy up answers were
frequently littered with needless errors that came back to penalise the candidates when they
attempted to use them later on. Many candidates noted that D was between A and B, but failed to
realise it was actually the midpoint of AB. In the very final part, it was often the case that
candidates overlooked the negative sign of cos, even when the remainder of their working was
broadly correct.

Q6 This was another very popular question attracting many poor scores. There were several
very serious errors on display, including the beliefs that
f ( x) n1 f ( x) n1

n
f ( x) dx = or .
(n  1) (n  1)f ( x)
 
The understanding that the original integral needed to be split as  f ( x)  f ( x)f ( x) n dx before
attempting to integrate by parts was largely absent, with many substituting immediately for
f ( x)f ( x) in terms of f ( x) , which really wasn’t helpful at all. Those who got over this initial
hurdle generally coped very favourably with the rest of the question.
In (i), it was quite common for candidates to omit verifying the result for tan x.

Q7 The initial hurdle in this question involved little more than splitting the series into
separate sums of powers of  and , leading to easy sums of GPs. Many missed this and spent a
lot of wasted time playing around algebraically without getting anywhere useful. In (ii), many
candidates applied (i) once, for the inner summation, but then failed to do so again for the second
time, and this was rather puzzling. Equally puzzling was the lack of recognition, amongst those
who had completed most of the first two parts of the question successfully, that the sum of the
odd terms in (iii) was still a geometric series. Almost exactly half of all candidates made an
attempt at this question, but the average score was only just over 5/20.

Q8 This was the least popular of the pure maths questions, probably with good reason, as it
included a lengthy introduction and a diagram. In the first part, despite showing candidates that
the point where the string leaves the circle is in the second quadrant, the necessary coordinate
geometry work provided a considerable challenge. The second part, finding the maximum of x
by standard differentiation techniques, proved to be relatively straightforward and a lot of
candidates managed to get full marks for this work. The third part presented the core challenge
of this question, in the sense that not many candidates seemed to have understood how to set the
limits of the parametric integral, and ‘benefit of the doubt’ had to be fairly generously applied to
those who switched signs when it suited them. The next part of the question involved applying
integration by parts in order to evaluate the integrals but surprisingly few candidates managed to
do so entirely successfully. Some of the common issues were the signs, that now needed to be
fully consistent, and the application of parts twice after using double-angle formulae. The notion
of the “total area swept out by the string” was also not so well understood, with only a very few
realising that they needed to integrate from t = 0 to t = 12  as well. Most remembered to subtract
the area of the semi-circle though.

Q9 Almost half of all candidates attempted this question, and scores averaged over 12/20. In
the majority of cases, the first two parts of the question proved relatively straightforward
conceptually, although there was the usual collection of errors introduced because of a lack of

483
care with signs/directions. It was only the final part of the question that proved to be of any great
difficulty: most candidates realised that they had to show that the given expression for B’s
velocity was always positive, but a lot of their efforts foundered on the lack of appreciation that
the term 1  4e 2 could be positive or negative.

Q10 This was the second most popular of the mechanics questions. The first couple of parts to
the question were fairly routine in nature, but then the algebra proved too demanding in many
cases, principally when it came to dealing with a quadratic equation in t which had non-
numerical coefficients. Candidates also found it a struggle to know when to use g and H instead
of u and  in the working that followed. A good number of candidates understood the nature of
the problem as the two particles rose and fell together, although it transpired (unexpectedly) that
there was another difficult obstacle to grasp in working with two distances. Even amongst
essentially fully correct solutions, very few indeed arrived at the correct final answer for tan .

Q11 This was the least popular of all the questions on the paper, receiving under 40 “hits”.
The fact that it clearly involved both vectors and 3-dimensions was almost certainly responsible
for the reluctance of candidates to give it a go. Those who managed to get past the initial stage of
sorting out directions and components usually did very well, but most efforts foundered in the
early stages. It was not helpful that some of these efforts confused angles to the vertical with
those to the horizontal. Almost no-one verified that the given vector in (i) was indeed a unit
vector.

Q12 Around a quarter of all candidates made an attempt at this question, though the average
score was very low. Parts (ii) and (iii) were managed quite comfortably, on the whole, but it was
(i) that proved to be difficult for most of those who attempted the question. The real difficulty lay
in establishing the given result for w, as the event to which it corresponded was defined
recurrently. As it happens, most wayward solutions left the straight-and-narrow by misreading
the rules of the match to begin with.

Q13 This question was almost as unpopular as question 11, receiving under 70 attempts, very
few of which ventured an opening opinion as to what skewness might measure. Those who could
handle expectations lived up to them and scored well; the rest just found the question a little too
overwhelming in its demands.

T F Cross
Principal Examiner

484
STEP Mathematics III 2011: Report

The percentages attempting larger numbers of questions were higher this year than formerly.
More than 90% attempted at least five questions and there were 30% that didn’t attempt at
least six questions. About 25% made substantive attempts at more than six questions, of
which a very small number indeed were high scoring candidates that had perhaps done extra
questions (well) for fun, but mostly these were cases of candidates not being able to complete
six good solutions.

Section A: Pure Mathematics

1. As might be expected, this was a very popular question, in fact the most popular being
attempted by very nearly all the candidates. Fortunately, it was also generally well-attempted,
with scores well above those for other questions. Apart from frequent algebraic errors and
overlooking terms, especially when using results from a previous part that required
adaptation, the main difficulties were in showing that (*) in part (ii) did indeed lead to a first
order differential equation in , and the consequent solution of that equation. Part (iii) was
generally well done. At the other end of the scale, some candidates did leave their answers to
part (i) in the form ln .

2. This was quite a popular question, being attempted by 70% of candidates. Scores
were polarized, though overall the mean score was below half marks, much the same as half
of the questions on the paper. Most candidates successfully dealt with the stem. Attempts at
part (i) were in equal proportions, applying the stem or a variant of the standard proof of the
irrationality of the square root of 2, though some of the latter overlooked the fact that it was
the nth root being discussed. Parts (ii) and (iii) saw three methods employed. One method
was to consider the location of the real roots then apply the stem, the second being to re-
arrange the expression to equal the integer and consider factors (again applying the stem). In
both these cases, failure to consider all cases lost marks, and there were frequent lacks of
rigour. However, considering x being odd or even, when used, was particularly slick and
successful.

3. The second most popular question, attempted by 80% of the cohort, with a similar
level of success to question 2. The significance of the condition 4 was ignored by
many candidates, and the fact that it does not apply in the last part of the question was often
similarly overlooked. Whilst a and b were generally found correctly, the rest of the first part
was often missing. Though there were frequent numerical errors, many candidates correctly
found the given solution of the equation, though the other two eluded most, with a common
error being to assume that the other two were xw and xw2.

4. About two thirds of the candidates tried this, with very slightly greater success than
questions 2 and 3. They found part (i) tricky, especially understanding the integral of the
inverse function. Also, commonly, they thought the condition was that . However,
part (ii) was done better, most errors being due to taking the inverse incorrectly, and of
course, the verification frequently went wrong due to the false condition. Most realised the
function to use in part (iii) but there was plenty of inaccuracy in working this part, though the
final deduction caused few worries.

5. Less than a third of the candidates attempted this. There were quite a few perfect
scores, however the vast majority scored less than a quarter of the marks, which was the

485
mean mark. The general result at the start of the question was the key to success. Those that
stumbled with handling four variables in terms of the fifth one, and the consequent calculus,
did not attempt to make further progress into the rest of the question.

6. This was quite popular, with attempts from three quarters of the candidates, and
slightly more success than questions like 2 and 3. Needing to prove three equalities, many
got close to doing two well and, with the others splitting half and half between getting close
to all three or just one. A small number of candidates made several attempts without always
having any sense of direction and often proved a particular pair equal both ways round. The
other weaknesses were in dealing with the limits when changing variable and evaluating the
definite term (which was zero!) when employing integration by parts.

7. The popularity and success rate of this was very similar to question 6. Quite a few
failed to realise the importance of 1 1 as part of the induction, and even
if they did tripped up on that part of the working. Part (ii) generally went well and the result
in Cn and Dn was found more easily. Very few had a problem with part (iii) but a small
number failed totally to see what it was about.

8. The response rate of this was similar to question 4, but with success rate similar to
question 2. Most students did reasonably well getting half to three quarters of the marks by
finding u and v and doing part (i), and then getting hold of (ii) and (iii) or not. Part (iv)
rightly discriminated the strong candidates from the generality. A few alternative methods
were tried but mostly they had their limitations. Details like omitted points from loci and the
negative sign that arises when using the cosine double angle formula frequently lost marks.

Section B: Mechanics

9. About a sixth of candidates tried this, and on average with slightly less success than
question 2. Of the attempts, about a third were close to completely correct, and nearly all the
others were barely doing more than grasping at crumbs, reflecting the fact that candidates
either did or did not know what they were doing. There was negligible middle ground.

10. Just under a quarter of candidates offered something on this, with relatively little
success and less than 20 candidates earning good marks. As with question 9, it tended to be a
case of “all or nothing”. Of the good solutions, half based their working on the motion of and
relative to the centre of mass of the system, and the other half on setting up simultaneous
differential equations for the displacements of the particles. Of the poor attempts, most
usually drew some kind of diagram, but then didn’t use it to identify a sensible coordinate
system, or positive direction, and there were common confusions over displacements x and
extensions x. Energy approaches usually got nowhere.

11. The least popular question on the paper, attempted by about 4%, but with similar
mean score to question 2 (and several others). Mostly, they did pretty well in finding the
couple, and using the initial trigonometric relation and its consequences to do so. At that
point they tended not to know how to proceed to the last part, though there were some very
good and simple solutions from considering energy.

486
Section C: Probability and Statistics

12. This question ran a close second to number 11 for unpopularity, but reflected the
same level of success. Most attempts followed the method of the question, and if they got off
on the right foot, often got most of the way through. Some struggled with the algebra for the
variance result, and a few tripped up on the standard pgf for the number of tosses to the first
head. Strangely, having found the pgf for Y successfully, and used it or the results of the
question for expectation and variance, the final probabilities were often wrong, and not
merely from overlooking the initial case.

13. This too was fairly unpopular, being attempted by about 10% of the candidates. Of
these no more than a dozen got it largely correct, but there was only one totally correct
solution as the detail for the non unique case frequently tripped even the better candidates.
The mean score was only about a third of the marks available as most candidates got part (i)
largely correct, barring some simplifying errors and not obtaining the non unique solution.
Fewer candidates had the correct probabilities for part (ii) and so were unable to proceed,
though a few were wrong merely by a constant which cancelled to give the correct ratio.

487
Explanation of Results STEP 2011

All STEP questions are marked out of 20. The mark scheme for each question is designed to reward
candidates who make good progress towards a solution. A candidate reaching the correct answer will
receive full marks, regardless of the method used to answer the question.

All the questions that are attempted by a student are marked. However, only the 6 best answers are used in
the calculation of the final grade for the paper.

There are five grades for STEP Mathematics which are:

S – Outstanding
1 – Very Good
2 – Good
3 – Satisfactory
U – Unclassified

The rest of this document presents, for each paper, the grade boundaries (minimum scores required to
achieve each grade), cumulative percentage of candidate achieving each grade, and a graph showing the
score distribution (percentage of candidates on each mark).

STEP Mathematics I (9465)

Grade boundaries
Maximum Mark S 1 2 3 U
120 86 66 47 28 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 5.6 18.5 46.9 81.6 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics I

www.admissionstests.cambridgeassessment.org.uk
488
STEP Mathematics II (9470)

Grade boundaries
Maximum Mark S 1 2 3 U
120 83 62 49 29 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 10.0 30.9 52.8 85.3 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics II

STEP Mathematics III (9475)

Grade boundaries
Maximum Mark S 1 2 3 U
120 91 65 52 30 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 14.9 39.7 57.8 85.6 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics III

www.admissionstests.cambridgeassessment.org.uk
489
490
STEP Solutions
2011

Mathematics
STEP 9465/9470/9475

October 2011

491
The Cambridge Assessment Group is Europe's largest assessment agency
and plays a leading role in researching, developing and delivering
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countries through our three major exam boards.

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Cambridge Assessment is a not-for-profit organisation.

This mark scheme is published as an aid to teachers and students, to indicate

the requirements of the examination. It shows the basis on which marks were
awarded by the Examiners. It does not indicate the details of the discussions
which took place at an Examiners’ meeting before marking commenced.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the
relevant knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question

papers and the Report on the Examination.

Cambridge Assessment will not enter into any discussion or correspondence

in connection with this mark scheme.

© UCLES 2010

More information about STEP can be found at:

http://www.atsts.org.uk

492
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 4
STEP Mathematics II 46
STEP Mathematics III 56

493
 STEP I 2011 Solutions

Dr Julian Gilbey, Principal Examiner (Marking)

Question 1

a b ay 2
(i) Show that the gradient of the curve + = 1, where b 6= 0, is − 2 .
x y bx

We begin by differentiating the equation of the curve (ax−1 + by −1 = 1) implicitly with respect
to x, to get
dy
−ax−2 − by −2 = 0,
dx
so that
b dy a
− 2 = 2,
y dx x
giving our desired result
dy ay 2
= − 2.
dx bx

An alternative, but more complicated method, is to rearrange the equation first to get y in terms
of x before differentiating. We have, on multiplying by xy,

ay + bx = xy, (1)

so that (x − a)y = bx, which gives

bx
y= .
x−a
We can now differentiate this using the quotient rule to get

dy b(x − a) − bx.1 −ab

= 2
= .
dx (x − a) (x − a)2

The challenge is now to rewrite this in the form required. We can rearrange equation (1) to get
(x − a)y = bx, so that (x − a) = bx/y. Substituting this into our expression for the derivative
then gives
dy ab aby 2 ay 2
=− = − = −
dx (bx/y)2 b2 x2 bx2
as required.

494
STEP I 2011 Question 1 continued

a b
The point (p, q) lies on both the straight line ax + by = 1 and the curve + = 1,
x y
where ab 6= 0. Given that, at this point, the line and the curve have the same gradient,
show that p = ±q .

Rearranging the equation of the straight line ax + by = 1 as y = −( ab )x + 1

b shows that its
gradient is −a/b.
Then using the above result for the gradient of the curve, we require that

aq 2 a
− 2
=− ,
bp b

so q 2 /p2 = 1, that is p2 = q 2 or p = ±q.

Show further that either (a − b)2 = 1 or (a + b)2 = 1 .

Since (p, q) lies on both the straight line and the curve, it must satisfy both equations, so

a b
ap + bq = 1 and + = 1.
p q

Now if p = q, then the first equation gives (a + b)p = 1 and the second gives (a + b)/p = 1, and
multiplying these gives (a + b)2 = 1.
Alternatively, if p = −q, then the first equation gives (a − b)p = 1 and the second equation gives
(a − b)/p = 1, and multiplying these now gives (a − b)2 = 1.

(ii) Show that if the straight line ax + by = 1, where ab 6= 0, is a normal to the curve
a b
− = 1, then a2 − b2 = 12 .
x y

We can find the derivative of this curve as above. A slick alternative is to notice that this is
identical to the above curve, but with b replaced by −b, so that

dy ay 2
= 2.
dx bx

The gradient of the straight line is −a/b as before, so as this line is normal to the curve at the
point (p, q), say, we have
aq 2  a 
− = −1
bp2 b
as perpendicular gradients multiply to −1; thus a2 q 2 /b2 p2 = 1, or a2 q 2 = b2 p2 .
a
We therefore deduce that aq = ±bp, which we can divide by pq 6= 0 to get p = ± qb .
Now since (p, q) lies on both the straight line and the curve, we have, as before,

a b
ap + bq = 1 and − = 1.
p q

495
STEP I 2011 Question 1 continued

Now if ap = qb , the second equation would become 0 = 1, which is impossible. So we must have
a b
p = − q , giving
a a
+ = 1,
p p
a
so that p = − qb = 12 , giving p = 2a and q = −2b.
Substituting this into the equation of the straight line yields

a.2a + b.(−2b) = 1,
1
so that a2 − b2 = 2 as required.

496
Question 2

1
ex
Z
The number E is defined by E = dx .
0 1+x
Show that
1
xex
Z
dx = e − 1 − E,
0 1+x
1
x 2 ex
Z
and evaluate dx in terms of e and E.
0 1+x

Approach 1: Using polynomial division

Using polynomial division or similar, we find that we can write

x 1
=1− .
1+x 1+x
Therefore our first integral becomes
Z 1 Z 1
xex 1  x
dx = 1− e dx
0 1+x 0 1+x
Z 1 Z 1 x
x e
= e dx − dx
0 0 1+x
 1
= ex 0 − E
= e − 1 − E,

as required.
We can play the same trick with the second integral, as

x2 1
=x−1+ ,
1+x 1+x
so that
1 Z 1
x 2 ex 1  x
Z
dx = x−1+ e dx
0 1+x 0 1+x
Z 1 Z 1 1
ex
Z
= xex dx − ex dx + dx.
0 0 0 1+x

Now we can use integration by parts for the first integral to get
Z 1 Z 1
 1
xex dx = xex 0 − ex dx
0 0
= e − (e − 1)
= 1.

Therefore
1
x2 e x
Z
dx = 1 − (e − 1) + E = 2 − e + E.
0 1+x

497
STEP I 2011 Question 2 continued

Approach 2: Substitution

We can substitute u = 1 + x to simplify the denominator in the integral. This gives us

Z 1 Z 2
xex (u − 1)eu−1
dx = du
0 1+x 1 u
Z 2
eu−1
= eu−1 − du.
1 u
The first part of this integral can be easily dealt with. The second part needs the reverse
substitution to be applied, replacing u by 1 + x, giving
2 ex
eu−1

1
− dx = e − 1 − E.
1+x

This is essentially identical to the first approach. The second integral follows in the same way.

Approach 3: Integration by parts

Integration by parts is trickier for this integral, as it is not obvious

R dvhow to breakR upduour integral.
We use the parts formula as written in the formula booklet: u dx dx = uv − v dx dx.
There are several ways which work (and many which do not). Here is a relatively straightforward
approach. For the first integral, we take
x dv
u= and = ex
1+x dx
so that
du 1
= and v = ex .
dx (1 + x)2
We then get
1
xex 1
Z 1
xex ex
Z  
dx = − 2
dx
0 1+x 1+x 0 0 (1 + x)
Z 1
1 ex
= 2e − 2
dx.
0 (1 + x)

The difficulty is now integrating the remaining integral. We again use parts, this time taking
dv 1
u = ex and =
dx (1 + x)2
so that
du 1
= ex and v=− .
dx 1+x
This gives
1
ex 1
Z 1
ex ex
Z  
dx = − − − dx
0 (1 + x)2 1+x 0 0 1+x
= − 12 e + 1 + E.

Combining this result with the first result then gives

Z 1
xex
dx = 12 e − (− 12 e + 1 + E) = e − 1 − E.
0 1+x

498
STEP I 2011 Question 2 continued

For the second integral, we use a similar procedure, this time taking

x2 dv
u= and = ex
1+x dx
so that
du 2x + x2
= and v = ex .
dx (1 + x)2
We then get
1  2 x 1 Z 1
x2 e x x e (2x + x2 )ex
Z
dx = − dx
0 1+x 1+x 0 0 (1 + x)2
Z 1
(2x + x2 )ex
= 12 e − dx.
0 (1 + x)2

The integral in the last step can be handled in several ways; the easiest is to write

2x + x2 x2 + 2x + 1 − 1 1
2
= =1−
(1 + x) (1 + x)2 (1 + x)2

and then use the earlier calculation of ex /(1 + x)2 dx to get

R

1 1
x2 e x ex
Z Z 
x
dx == 12 e − e − dx
0 1+x 0 (1 + x)2
= 12 e − [ex ]10 + (− 12 e + 1 + E)
= −e + 1 + 1 + E
= 2 − e + E.

499
STEP I 2011 Question 2 continued

Evaluate also, in terms of E and e as appropriate:

1−x
1
e 1+x
Z
(i) dx ;
0 1+x

This integral looks to be of a vaguely similar form, but with a more complicated exponential
part. We therefore try the substitution u = 1−x
1+x and see what we get.
1−x
If u = , then
1+x
du −(1 + x) − (1 − x) −2
= = ,
dx (1 + x)2 (1 + x)2
dx
so that du = − 12 (1 + x)2 . Also, when x = 0, u = 1, and when x = 1, u = 0.
1−x
We can also rearrange u = to get
1+x

(1 + x)u = 1 − x
so ux + x = 1 − u
1−u
giving x= .
1+u

Thus
1−x
1 0
e 1+x eu
Z Z
− 12 (1 + x)2 du


dx =
0 1+x 1 1+x
Z 1
1 u
= 2 e (1 + x) du reversing the limits
0
1
1−u
Z  
1 u
= 2e 1+ du
0 1+u
1  
2
Z
1 u
= 2e du
0 1+u
1
eu
Z
= du
0 1+u
= E.

500
STEP I 2011 Question 2 continued

Evaluate also in terms of E and e as appropriate:

√
2 2
ex
Z
(ii) dx
1 x

√
Again we have a different exponent, so we try substituting
√ u = x2 , so that x = u, while
du dx 1
dx = 2x and so du = 2x . Also the limits x = 1 and x = 2 become u = 1 and u = 2, giving us
√
2 2 2
ex eu 1
Z Z
dx = du
1 x 1 x 2x
2
eu
Z
= du.
1 2u

This is very similar to what we are looking for, except that it has the wrong limits and a
denominator of 2u rather than u + 1 or perhaps 2(u + 1). So we make a further substitution:
u = v + 1, so that v = u − 1 and du/dv = 1, giving us
√
2 2 2
ex eu
Z Z
dx = du
1 x 1 2u
Z 1
ev+1
= dv
0 2(v + 1)
e 1 ev
Z
= dv
2 0 v+1
eE
= ,
2
where on the penultimate line we have written ev+1 = e.ev and so taken out a factor of e/2.
It is also possible to evaluate this integral more directly by substituting u = x2 − 1, so that
x2 = u + 1. The details are left to the reader.

501
Question 3

Prove the identity

4 sin θ sin( 13 π − θ) sin( 13 π + θ) = sin 3θ . (∗)

We make use of two of the factor formulæ:

2 sin A sin B = cos(A − B) − cos(A + B)

2 sin A cos B = sin(A + B) + sin(A − B)

(These can be derived by expanding the right hand sides using the addition formulæ, and then
collecting like terms.)
Then initially taking A = 13 π − θ and B = 13 π + θ and using the first of the factor formulæ gives

4 sin θ sin( 13 π − θ) sin( 13 π + θ) = 2 sin θ cos(−2θ) − cos( 23 π)

= 2 sin θ(cos 2θ + 12 )
= 2 sin θ cos 2θ + sin θ.

We now use the second factor formula with A = θ and B = 2θ to simplify this last expression to


sin 3θ + sin(−θ) + sin θ = sin 3θ,

as required.

An alternative approach is to expand the second and third terms on the left hand side using the
addition formulæ, giving:

4 sin θ sin( 13 π − θ) sin( 13 π + θ)

= 4 sin θ(sin 13 π cos θ − cos 13 π sin θ)(sin 13 π cos θ + cos 13 π sin θ)
√
√
= 4 sin θ 23 cos θ − 12 sin θ 23 cos θ + 12 sin θ

= 4 sin θ( 34 cos2 θ − 14 sin2 θ)

= 3 sin θ cos2 θ − sin3 θ,

while

sin 3θ = sin(2θ + θ)
= sin 2θ cos θ + cos 2θ sin θ
= 2 sin θ cos2 θ + (cos2 θ − sin2 θ) sin θ
= 3 sin θ cos2 θ − sin3 θ.

Thus the required identity holds.

502
STEP I 2011 Question 3 continued

(i) By differentiating (∗), or otherwise, show that

√
cot 19 π − cot 29 π + cot 49 π = 3.

We can differentiate a product of several terms using the product rule repeatedly. In general,
we have
d du dv dw
(uvwt . . .) = vwt . . . + u wt . . . + uv t... + ···
dx dx dx dx
In our case, we are differentiating a product of three terms, and we get

4 cos θ sin( 13 π − θ) sin( 13 π + θ) − 4 sin θ cos( 13 π − θ) sin( 13 π + θ) +

4 sin θ sin( 13 π − θ) cos( 13 π + θ) = 3 cos 3θ.

Now we are aiming to get an expressing involving cot, so we divide this result by (∗) to get

cot θ − cot( 13 π − θ) + cot( 13 π + θ) = 3 cot 3θ.

We now let θ = 19 π to get

√ √
cot 19 π − cot 29 π + cot 49 π = 3 cot 13 π = 3/ 3 = 3,

and we are done.

(ii) By setting θ = 16 π −φ in (∗), or otherwise, obtain a similar identity for cos 3θ and deduce
that
cot θ cot( 13 π − θ) cot( 13 π + θ) = cot 3θ .

Setting θ = 16 π − φ in (∗) as instructed gives

4 sin( 16 π − φ) sin( 16 π + φ) sin( 12 π − φ) = sin 3( 16 π − φ).

To get cosines from this expression, we will need to use the identity sin( 12 π − x) = cos x. So we
rewrite this as

4 sin 12 π − ( 13 π + φ) sin 12 π − ( 13 π − φ) sin( 12 π − φ) = sin( 12 π − 3φ)

which allows us to apply our identity to get

4 cos( 13 π + φ) cos( 13 π − φ) cos φ = cos 3φ,

which is a similar identity for cos 3φ. Replacing φ by θ and reordering the terms in the product
gives
4 cos θ cos( 13 π − θ) cos( 13 π + θ) = cos 3θ.
Now dividing this identity by (∗) gives our desired identity for cot:

cot θ cot( 13 π − θ) cot( 13 π + θ) = cot 3θ. (†)

(Note that there is no factor of 4 in this expression.)

503
STEP I 2011 Question 3 continued

Show that √
cosec 19 π − cosec 59 π + cosec 79 π = 2 3 .

As before, we differentiate the expression (†) which we have just derived to get

− cosec2 θ cot( 13 π − θ) cot( 13 π + θ) +

cot θ cosec2 ( 13 π − θ) cot( 13 π + θ) −
cot θ cot( 13 π − θ) cosec2 ( 13 π + θ) = −3 cosec2 3θ.

When we negate this identity and then divide it by (†), we will have lots of cancellation and we
will be left with terms of the form cosec2 x/ cot x. Now

cosec2 x 1 sin x 1 2
= 2 . = = = 2 cosec 2x,
cot x sin x cos x sin x cos x sin 2x
so that the division gives us

2 cosec 2θ − 2 cosec 2( 13 π − θ) + 2 cosec 2( 13 π + θ) = 6 cosec 6θ.

To get the requested equality, we halve this identity and set 2θ = 19 π so that
√
cosec 19 π − cosec 59 π + cosec 79 π = 3 cosec 13 π = 3. √23 = 2 3

as required.

504
Question 4

The distinct points P and Q, with coordinates (ap2 , 2ap) and (aq 2 , 2aq) respectively, lie on
the curve y 2 = 4ax. The tangents

to the curve at P and Q meet at the point T . Show that
T has coordinates apq, a(p + q) . You may assume that p 6= 0 and q 6= 0.

We begin by sketching the graph (though this may be helpful, it is not required):

y
P

T φ
F x

The equation of the curve is y 2 = 4ax, so we can find the gradient of the curve by implicit
differentiation:
dy
2y = 4a,
dx
and thus
dy 2a
= ,
dx y
as long as y 6= 0. (Alternatively, we could write x = y 2 /4a and then work out dx/dy = 2y/4a;
taking reciprocals then gives us the same result.)
Therefore the tangent at the point P with coordinates (ap2 , 2ap) has equation
2a
y − 2ap = (x − ap2 ),
2ap
which can easily be rearranged to give

x − py + ap2 = 0.

Since y = 0 would require p = 0, we can ignore this case, as we are assuming that p 6= 0. [In
y
fact, if y = p = 0, we can look at the reciprocal of the gradient, dx
dy = 2a , and this is zero, so the
line is vertical. In this case, our equation gives x = 0, which is, indeed, a vertical line, so our
equation works even when p = 0.]
Thus the tangent through P has equation x − py + ap2 = 0 and the tangent through Q has
equation x − qy + aq 2 = 0 likewise.
We solve these equations simultaneously to find the coordinates of T . Subtracting them gives

(p − q)y − a(p2 − q 2 ) = 0.

505
STEP I 2011 Question 4 continued

Since p 6= q, we can divide by p − q to get

y − a(p + q) = 0,

so y = a(p + q) and therefore x = py − ap2 = ap(p + q) − ap2 = apq.

Thus T has coordinates apq, a(p + q) , as wanted.

The point F has coordinates (a, 0) and φ is the angle T F P . Show that
pq + 1
cos φ = p
(p2 + 1)(q 2 + 1)

and deduce that the line F T bisects the angle P F Q.

In the triangle T F P , we can use the cosine rule to find cos φ:

T P 2 = T F 2 + P F 2 − 2.T F.P F. cos φ,

so that
TF2 + PF2 − TP2
cos φ = .
2.T F.P F
Now using Pythagoras to find the distance between two points given their coordinates, we obtain

2
2
T F 2 = a(pq − 1) + a(p + q)
= a2 (p2 q 2 − 2pq + 1 + p2 + 2pq + q 2 )
= a2 (p2 q 2 + p2 + q 2 + 1)
= a2 (p2 + 1)(q 2 + 1)

2
F P 2 = a(p2 − 1) + (2ap)2
= a2 (p4 − 2p2 + 1 + 4p2 )
= a2 (p4 + 2p2 + 1)
= a2 (p2 + 1)2

2
2
T P 2 = a(pq − p2 ) + a(p + q − 2p)

2
2
= ap(q − p) + a(q − p)
= a2 (p2 + 1)(q − p)2

Thus

T F 2 + F P 2 − T P 2 = a2 (p2 + 1)(q 2 + 1 + p2 + 1 − q 2 + 2pq − p2 )

= 2a2 (1 + p2 )(1 + pq)

so that
2a2 (1 + p2 )(1 + pq)
cos φ = p
2a2 (p2 + 1)(q 2 + 1)(p2 + 1)2
1 + pq
=p
(p + 1)(q 2 + 1)
2

as we wanted.

506
STEP I 2011 Question 4 continued

An alternative approach is to use vectors and dot products to find cos φ. We have
−−→ −→
F P .F T = F P.F T. cos φ

(where the dot on the left hand side is the dot product, but on the right is ordinary multiplica-
tion), so we need only find the lengths F P , F T as above and the dot product. The dot product
is
 2
−−→ −→
  
ap − a apq − a
F P .F T = . = (ap2 − a)(apq − a) + 2ap.a(p + q)
2ap − 0 a(p + q) − 0
= a2 (p2 − 1)(pq − 1) + 2a2 (p2 + pq)
= a2 (p3 q − p2 − pq + 1 + 2p2 + 2pq)
= a2 (p3 q + p2 + pq + 1)
= a2 (p2 (pq + 1) + pq + 1)
= a2 (p2 + 1)(pq + 1)

Therefore we deduce
−−→ −→
F P .F T
cos φ =
F P.F T
a2 (p2 + 1)(pq + 1)
= p
a(p2 + 1).a (p2 + 1)(q 2 + 1)
pq + 1
=p
(p + 1)(q 2 + 1)
2

as required.
Now to show that the line F T bisects the angle P F Q, it suffices to show that φ is equal to the
angle T F Q (see the sketch above).
Now we can find cos(∠T F Q) by using the above formula and swapping every p and q in it, as
this will swap the roles of P and Q.
But swapping every p and q does not change the formula, so cos(∠T F Q) = cos(∠T F P ), and
so ∠T F Q = ∠T F P as both angles are strictly less than 180◦ and cosine is one-to-one in this
domain.
Thus the line F T bisects the angle P F Q, as required.

507
Question 5

Given that 0 < k < 1, show with the help of a sketch that the equation

sin x = kx (∗)

has a unique solution in the range 0 < x < π.

We sketch the graph of y = sin x in the range 0 6 x 6 π along with the line y = kx.
d
Now since dx (sin x) = cos x, the gradient of y = sin x at x = 0 is 1, so the tangent at x = 0 is
y = x. We therefore also sketch the line y = x.

y y=x
y = kx

0 α π x

It clear that there is at most one intersection of y = kx with y = sin x in the interval 0 < x < π,
and since 0 < k < 1, there is exactly one, as the gradient is positive and less that that of
y = sin x at the origin. (If k 6 0, there would be no intersections in this range as kx would be
negative or zero; if k > 1, the only intersection would be at x = 0.)

Let Z π 
I= sin x − kx dx .
0

Show that
π 2 sin α
I= − 2 cos α − α sin α ,
2α
where α is the unique solution of (∗).

It is a pain to work with absolute values (the “modulus function”), so we split the integral
into two integrals: in the interval 0 6 x 6 α, sin x − kx > 0, and in the interval α 6 x 6 π,
sin x − kx 6 0. So
Z π
 
I= sin x − kx dx
Z0 α Z π
   
=  sin x − kx dx +
 sin x − kx dx
Z0 α Z πα
= sin x − kx dx + − sin x + kx dx
0 α
α  π
= − cos x − 12 kx2 0 + cos x + 12 kx2 α


= (− cos α − 12 kα2 ) − (− cos 0 − 0) + (cos π + 12 kπ 2 ) − (cos α + 12 kα2 )

= −2 cos α − kα2 + 12 kπ 2

508
STEP I 2011 Question 5 continued

π 2 sin α
= −2 cos α − α sin α +
2α
where the last line follows using kα = sin α so that k = (sin α)/α, and we have reached the
desired result.

Show that I, regarded as a function of α, has a unique stationary value and that this stationary
value is a minimum. Deduce that the smallest value of I is
π
−2 cos √ .
2

We differentiate I to find its stationary points. We have

π 2 α cos α − sin α
 
dI
= + 2 sin α − sin α − α cos α
dα 2 α2
π2
= (sin α − α cos α) − 2 (sin α − α cos α)
2α
π2
 
= 1 − 2 (sin α − α cos α)
2α
dI
√
so dα = 0 if and only if 2α2 = π 2 or sin α = α cos α. The former condition gives α = ±π/ 2,
while the latter condition gives tan α = α.
A quick sketch of the tan graph (see below) shows that tan α = α has no solutions in the range
d
0 < α < π (though α = 0 is a solution); the sketch uses the result that dx (tan x) = sec2 x, so
the tangent to y = tan x at x = 0 is y = x.

y y=x

π
2 π
x

√ √
Thus the only solution in the required range is α = π/ 2 (and note that π/ 2 < π).
To ascertain whether it is a maximum, a minimum or a point of inflection, we could either
look at the values of I or dI/dα at this point and either side or we could consider the second
derivative.
√
Either way, we will eventually have to work out the value of I when α = π/ 2, so we will do so
now:

509
STEP I 2011 Question 5 continued

√
π 2 sin(π/ 2) π π π
I= √ − 2 cos √ − √ sin √
2π/ 2 2 2 2
π π π π π
= √ sin √ − 2 cos √ − √ sin √
2 2 2 2 2
π
= −2 cos √ .
2

Approach 1: Using values either side

sin α
The function I is not well-defined when α = 0, but if we know that α → 1 as α → 0, we can
2
deduce that as α → 0, I → −2 + π2 > 52 > 2 (using π > 3).
When α = π, we have I = 2.
√ √
Since at α = π/ 2, we have I = −2 cos(π/ 2) < 2, this must be the minimum value of I.
Alternatively,
√if we wish to consider the value of dI/dα, we need to know the sign of sin α−α cos α
near α = π/ 2. Now since π2 < α < π, cos α < 0 and so this expression is positive. Therefore
√ √ √
for α slightly less that π/ 2, dI/dα < 0 and for α > π/ 2, dI/dα > 0, so that α = π/ 2 is a
(local) minimum.

Approach 2: Using the second derivative

We have
d2 I π2 π2
 
= 3 (sin α − α cos α) + 1 − 2 (cos α − cos α + α sin α)
dα2 α 2α
2 π2
 
π
= 3 (sin α − α cos α) + 1 − 2 α sin α
α 2α
√
Now when α = π/ 2, so that π 2 /2α2 = 1, we have

d2 I π2
= (sin α − α cos α).
dα2 α3
√ d2 I
Since α = π2 2 > π2 , we have sin α > 0 and cos α < 0, so dα2
> 0 and I has a local minimum at
this value of α.

510
Question 6

Use the binomial expansion to show that the coefficient of xr in the expansion of (1 − x)−3 is
1
2 (r + 1)(r + 2) .

Using the formula in the formula book for the binomial expansion, we find that the xr term is
−3 (−3)(−4)(−5) . . . (−3 − r + 1)
 
(−x)r = (−1)r xr
r r!
3.4.5. · · · .(r + 2) r
= x
r!
3.4.5. · · · .(r + 2) r
= x
1.2.3.4.5. · · · .r
(r + 1)(r + 2) r
= x
1.2
so the coefficient of xr is 12 (r + 1)(r + 2). But the argument as we’ve written it assumes that
r > 2 (as we’ve left ourselves with “1.2” in the denominator),
so we need to check that this
this also holds for r = 0 and r = 1. But this is easy, as −3 0 1
0 (−1) = 1 = 2 × 1 × 2 and
−3 1 1


1 (−1) = 3 = 2 × 2 × 3.
Alternatively, we could have argued
3.4.5. · · · .(r + 2) r 1.2.3.4.5. · · · .(r + 2) r (r + 1)(r + 2) r
x = x = x
r! 1.2.r! 1.2
and this would have dealt with the cases r = 0 and r = 1 automatically, as we are not implicitly
assuming that r > 2.

(i) Show that the coefficient of xr in the expansion of

1 − x + 2x2
(1 − x)3

is r2 + 1 and hence find the sum of the series

2 5 10 17 26 37 50
1+ + + + + + + + ··· .
2 4 8 16 32 64 128

We have
1 − x + 2x2
= (1 − x + 2x2 )(a0 + a1 x + a2 x2 + · · · + ar xr + . . . )
(1 − x)3
where ar = 12 (r + 1)(r + 2). Thus

1 − x + 2x2
= a0 + a1 x + a2 x2 + · · · + ar xr + · · ·
(1 − x)3
− a0 x − a1 x2 − · · · − ar−1 xr − · · ·
+ 2a0 x2 + · · · + 2ar−2 xr + · · ·
= a0 + (a1 − a0 )x + (a2 − a1 + 2a0 )x2 + · · ·
+ (ar − ar−1 + 2ar−2 )xr + · · ·

511
STEP I 2011 Question 6 continued

Thus the coefficient of xr for r > 2 is

ar − ar−1 + 2ar−2 = 12 (r + 1)(r + 2) − 12 r(r + 1) + (r − 1)r

= 12 (r2 + 3r + 2 − r2 − r + 2r2 − 2r)
= 12 (2r2 + 2)
= r2 + 1

as required. Also, the coefficient of x0 is a0 = 1 = 02 + 1 and the coefficient of x1 is a1 − a0 =

3 − 1 = 2 = 12 + 1, so the formula r2 + 1 holds for these two cases as well. Therefore, the
coefficient of xr is r2 + 1 for all r > 0.

Now we can sum our series: it is

2 5 10 17 02 + 1 12 + 1 22 + 1 r2 + 1
1+ + + + + ··· = + + + · · · + + ···
2 4 8 16 20 21 22 2r
= (02 + 1) + (12 + 1)( 12 ) + · · · + (r2 + 1)( 12 )r + · · ·
1 − 12 + 2( 12 )2
=
(1 − 12 )3
1
= 1
(8)
= 8.

(ii) Find the sum of the series

9 25 9 49
1+2+ +2+ + + + ··· .
4 16 8 64

The denominators look like powers of 2, so we will rewrite the terms using powers of 2:
9 25 9 1 4 9 16 25 36 49
1+2+ +2+ + + ··· = + + + + + + + ···
4 16 8 1 2 4 8 16 32 64
and it is clear that the general term is r2 /2r−1 , starting with the term where r = 1.
We can rewrite this in terms of the series found in part (i) by writing

r2 r2 r2 + 1 1
r−1
= 2 · r
= 2 · r
−2· r,
2 2 2 2
so our series becomes
   
2 5 10 17 1 1 1 1
2 + + + + ··· − 2 + + + + ···
2 4 8 16 2 4 8 16
   
2 5 10 17 1 1 1 1
=2 1+ + + + + ··· − 2 1 + + + + + ···
2 4 8 16 2 4 8 16
=2·8−2·2
= 12,

where on the second line, we have introduced the term corresponding to r = 0, and on the
penultimate line, we have used the result from (i) and the sum of the infinite geometric series
1 + 12 + 14 + 18 + · · · = 1/(1 − 12 ) = 2.

512
STEP I 2011 Question 6 continued

An alternative approach is to begin with the result of part (i) and to argue as follows.
We have
∞
1 − x + 2x2 X 2
= (r + 1)xr
(1 − x)3
r=0
∞
X ∞
X
= r 2 xr + xr
r=0 r=0
∞
X ∞
X
=x r2 xr−1 + xr .
r=0 r=0
P∞ 2 ( 1 )r−1 , 1
But our required sum is r=0 r 2 so we put x = 2 into this result and get
∞ ∞
1 − 12 + 2( 12 )2 X X
= 1
2 r2 ( 12 )r−1 + ( 12 )r .
(1 − 12 )3 r=0 r=0

The last term on the right hand side is our geometric series, summing to 2. The left hand side
evaluates to 8, and so we get
∞
X
8= 2 1
r2 ( 12 )r−1 + 2.
r=0

Thus our series sums to 12, as before.

A third approach is to observe that the series can be written as ∞

P 2 r
P∞ 2
r=0 (r + 1) x = r=0 (r +
r 1
2r + 1)x with x = 2 , then to look for a polynomial p(x) of degree at most 2 such that the
coefficient of xr in the expansion of p(x)/(1 − x)3 is exactly r2 + 2r + 1, using methods like
those in part (i). (The polynomial needs to be of degree at most 2 so that the terms are also
correct for r = 0 and r = 1 in addition to the general term being correct.) This turns out to
give p(x) = x + 1, so that the sum is ( 12 + 1)/(1 − 12 )3 = 12.

513
Question 7

In this question, you may assume that ln(1 + x) ≈ x − 12 x2 when |x| is small.
The height of the water in a tank at time t is h. The initial height of the water is H and
water flows into the tank at a constant rate. The cross-sectional area of the tank is constant.

(i) Suppose that water leaks out at a rate proportional to the height of the water in the
tank, and that when the height reaches α2 H, where α is a constant greater than 1, the
height remains constant. Show that
dh
= k(α2 H − h),
dt
for some positive constant k. Deduce that the time T taken for the water to reach height
αH is given by  
1
kT = ln 1 +
α
and that kT ≈ α−1 for large values of α.

Since the tank has constant cross-sectional area, the volume of water within the tank is propor-
tional to the height of the water.
Therefore we have the height increasing at a rate a − bh, where a is the rate of water flowing in
divided by the cross-sectional area, and b is a constant of proportionality representing the rate
of water leaking out. In other words, we have
dh
= a − bh.
dt
dh
Now, when h = α2 H, dt = 0, so a − bα2 H = 0, or a = bα2 H, giving

dh
= bα2 H − bh = b(α2 H − h).
dt
Hence if we write k = b, we have our desired equation.
We can now solve this by separating variables to get
1
Z Z
dh = k dt
α2 H − h
so that
− ln(α2 H − h) = kt + c.
At t = 0, h = H, so
− ln(α2 H − H) = c,
which finally gives us
kt = ln(α2 H − H) − ln(α2 H − h).

514
STEP I 2011 Question 7 continued

Now at time T , h = αH, so that

kT = ln(α2 H − H) − ln(α2 H − αH)
 2
α H −H

= ln
α2 H − αH
 2
α −1

= ln
α2 − α
 
α+1
= ln
α
 
1
= ln 1 +
α
as required.
1
When α is large, so that α is small, this is
 
1
kT = ln 1 +
α
1 1
≈ − 2
α 2α
1
≈ .
α

√
(ii) Suppose that the rate at which water leaks out of the tank is proportional to h (instead
of h), and that when the height reaches α2 H, where α is a constant greater than 1, the
height remains constant. Show that the time T 0 taken for the water to reach height αH
is given by
√ √
  
0 1
cT = 2 H 1 − α + α ln 1 + √
α
√
for some positive constant c and that cT 0 ≈ H for large values of α.

We proceed just as in part (i).

This time we have
dh √
= a − b h,
dt
√
where a and b are some constants. Now, when h = α2 H, dh = 0, so a − b α2 H = 0, which
√ dt
yields a = bα H. We thus have
dh √ √ √ √
= bα H − b h = b(α H − h).
dt
So if this time we write c = b, we have our desired differential equation.
We again solve this by separating variables to get
1
Z Z
√ √ dh = c dt.
α H− h
√ dh
To integrate the left hand side, we use the substitution u = h, so that h = u2 and du = 2u.
This gives us
1
Z
√ · 2u du = ct.
α H −u

515
STEP I 2011 Question 7 continued

We divide the numerator by the denominator to get

√ √
−2(α H − u) + 2α H
Z
ct = √ du
α H −u
√
2α H
Z
= −2 + √ du
α H −u
√ √
= −2u − 2α H ln α H − u + c0


√ √ √ √

= −2 h − 2α H ln α H − h + c0

where c0 is a constant.
√ √
An√alternative way of doing
√ this step is to use the√substitution v = α H − h, so that h =
(α H − v)2 = α2 H − 2αv H + v 2 and dh/dv = − 2α H + 2v. This gives us

Z
1 √
ct = (−2α H + 2v) = ct
v
√
−2α H
Z
= + 2 dv
v
√
= −2α H ln v + 2v + c0
√ √ √
√ √

= −2α H ln α H − h + 2 α H − h + c0

where c0 is again a constant.

At t = 0, h = H, so √ √ √ √

c0 = 2 H + 2α H ln α H − H .

Now at time T 0 , h = αH, so that

√ √ √ √ √ √ √ √

cT 0 = −2 αH − 2α H ln α H − αH + 2 H + 2α H ln α H − H


√ √ !
√ √ √ α H− H
= 2 H(1 − α) + 2α H ln √ √
α H − αH
 √ √
√ √ ( α + 1)( α − 1)
 
= 2 H 1 − α + α ln √ √
α( α − 1)
√
√ √
 
α+1
= 2 H 1 − α + α ln √
α
√ √
  
1
= 2 H 1 − α + α ln 1 + √
α
as required.
√
When α is large, 1/ α is small, so this gives
√ √
  
0 1
cT = 2 H 1 − α + α ln 1 + √
α
√ √
  
1 1
≈2 H 1− α+α √ −
α 2α
√ √ √
≈ 2 H 1 − α + α − 12


√
≈ H.

516
Question 8

(i) The numbers m and n satisfy

m 3 = n 3 + n2 + 1 . (∗)

(a) Show that m > n. Show also that m < n + 1 if and only if 2n2 + 3n > 0 . Deduce
that n < m < n + 1 unless − 32 6 n 6 0 .

As n2 > 0, we have
m3 = n3 + n2 + 1
> n3 + 1
> n3
so m > n as the function f(x) = x3 is strictly increasing.
Now
m < n + 1 ⇐⇒ m3 < (n + 1)3
⇐⇒ n3 + n2 + 1 < n3 + 3n2 + 3n + 1
⇐⇒ 0 < 2n2 + 3n
so m < n + 1 if and only if 2n2 + 3n > 0.
Combining these two conditions, n < m always, and m < n + 1 if and only if 2n2 + 3n > 0, so
n < m < n + 1 unless 2n2 + 3n 6 0.
Now 2n2 + 3n = 2n(n + 32 ) 6 0 if and only if − 32 6 n 6 0, so n < m < n + 1 unless − 32 6 n 6 0.

(b) Hence show that the only solutions of (∗) for which both m and n are integers are
(m, n) = (1, 0) and (m, n) = (1, −1).

If solution to (∗) has both m and n integer, we cannot have n < m < n + 1, as there is no
integer strictly between two consecutive integers. We therefore require − 32 6 n 6 0, so n = −1
or n = 0.
If n = −1, then m3 = 1, so m = 1.
If n = 0, then m3 = 1, so m = 1.
Thus the only integer solutions are (m, n) = (1, 0) and (m, n) = (1, −1).

(ii) Find all integer solutions of the equation

p3 = q 3 + 2q 2 − 1 .

We try a similar argument here. We start by determining whether p > q:

p > q ⇐⇒ p3 > q 3
⇐⇒ q 3 + 2q 2 − 1 > q 3
⇐⇒ 2q 2 > 1
⇐⇒ q 2 > 1
2

517
STEP I 2011 Question 8 continued

so that p > q unless q 2 6 12 , and q 2 6 1

2 if and only if − √12 6 q 6 √1 .
2
We now determine the conditions under which p < q + 1:

p < q + 1 ⇐⇒ p3 < (q + 1)3

⇐⇒ q 3 + 2q 2 − 1 < q 3 + 3q 2 + 3q + 1
⇐⇒ 0 < q 2 + 3q + 2

so p < q + 1 unless q 2 + 3q + 2 6 0. This condition becomes (q + 1)(q + 2) 6 0, so −2 6 q 6 −1.

Thus q < p < q + 1 unless − √12 6 q 6 √1
2
or −2 6 q 6 −1.
If p and q are both integers, this then limits us to three cases: q = 0, q = −1 and q = −2.
If q = 0, then p3 = −1, so p = −1.
If q = −1, then p3 = 0, so p = 0.
If q = −2, then p3 = −1, so p = −1.
Hence there are three integer solutions: (p, q) = (−1, 0), (p, q) = (−1, −2) and (p, q) = (0, −1).

518
Question 9

A particle is projected at an angle θ above the horizontal from a point on a horizontal plane.
The particle just passes over two walls that are at horizontal distances d1 and d2 from the
point of projection and are of heights d2 and d1 , respectively. Show that

d21 + d1 d2 + d22
tan θ = .
d1 d2

We draw a sketch of the situation:

B
d2
v d1
θ
d1 d2 x

We let the speed of projection be v and the time from launch be t. We resolve the components
of velocity to find the position (x, y) at time t:
R(→) x = (v cos θ)t (1)
R(↑) y = (v sin θ)t − 12 gt2 (2)

At A (distance d1 from the point of projection), we find

(v cos θ)t = d1
(v sin θ)t − 12 gt2 = d2

so that
d1
t=
v cos θ
giving
1
v sin θ gd2
d1 − 22 12 = d2 ,
v cos θ v cos θ
so that
gd21
d2 = d1 tan θ − .
2v cos2 θ
2

This can be rearranged to get

gd21
= d1 tan θ − d2 . (3)
2v 2 cos2 θ
(An alternative is to first eliminate t from equations (1) and (2) first to get

gx2
y = x tan θ − (4)
2v 2 cos2 θ

519
STEP I 2011 Question 9 continued

and then substitute x = d1 and y = d2 into this formula.)

Likewise, at B we get (on swapping d1 and d2 ):

gd22
= d2 tan θ − d1 . (5)
2v 2 cos2 θ
Multiplying (3) by d22 gives the same left hand side as when we multiply (5) by d21 , so that

(d1 tan θ − d2 )d22 = (d2 tan θ − d1 )d21 .

Expanding this gives

d1 d22 tan θ − d32 = d21 d2 tan θ − d31 .
Collecting terms gives:
(d1 d22 − d21 d2 ) tan θ = d32 − d31 ,
and we can factorise this (recalling that a3 − b3 = (a − b)(a2 + ab + b2 )) to get

d1 d2 (d2 − d1 ) tan θ = (d2 − d1 )(d22 + d2 d1 + d21 ).

Dividing by d1 d2 (d2 − d1 ) 6= 0 gives us our desired result:

d21 + d1 d2 + d22
tan θ = .
d1 d2

Find (and simplify) an expression in terms of d1 and d2 only for the range of the particle.

The range can be found by determining where y = 0, so (v sin θ)t − 12 gt2 = 0. This has solutions
t = 0 (the point of projection) and t = (2v/g) sin θ. At this point,

2v 2 sin θ cos θ
x = (v cos θ)t =
g
2v cos2 θ
2
= tan θ.
g

We have written sin θ = cos θ tan θ because equation (3) gives us a formula for the fraction part
of this expression: we get
d21
x= tan θ.
d1 tan θ − d2

Alternatively, using equation (4), we can solve for y = 0 to get x = 0 or

gx
tan θ = .
2v 2 cos2 θ
Since x = 0 at the start, the other solution gives the range. Using equation (3) to write

g d1 tan θ − d2
= ,
2v 2 cos2 θ d21

we deduce that
d21 tan θ
x=
d1 tan θ − d2

520
STEP I 2011 Question 9 continued

as before.

We can now simply substitute in our formula for tan θ, simplify a little, and we will be done:
 2
d1 + d1 d2 + d22

2
d1
d1 d2
x=  2
d1 + d1 d2 + d22

d1 − d2
d1 d2
d1 (d21 + d1 d2 + d22 )
=
(d21 + d1 d2 + d22 ) − d22
d1 (d21 + d1 d2 + d22 )
=
d21 + d1 d2
d21 + d1 d2 + d22
=
d1 + d2

Alternative approach

An entirely different approach to the whole question is as follows. We know that the path of
the projectile is a parabola. Taking axes as in the above sketch, the path passes through the
three points (0, 0), (d1 , d2 ) and (d2 , d1 ). If the equation of the curve is y = ax2 + bx + c, then
this gives three simultaneous equations:

0 = 0a + 0b + c
d2 = d21 a + d1 b + c
d1 = d22 a + d2 b + c.

The first gives c = 0, and we can then solve the other two equations to get a and b. This gives

d22 − d21 d1 + d2
a= 2 =−
d1 d2 − d22 d1 d1 d2
d31 − d23 d1 + d1 d2 + d22
2
b= =
d21 d2 − d22 d1 d1 d2

Then the gradient is given by dy/dx = 2ax + b, so at x = 0, the gradient dy/dx = b, which gives
us tan θ (as the gradient is the tangent of angle made with the x-axis). The range is given by
solving y = 0, so x(ax + b) = 0, giving x = −b/a = (d21 + d1 d2 + d22 )/(d1 + d2 ) as before.

521
Question 10

A particle, A, is dropped from a point P which is at a height h above a horizontal plane. A

second particle, B, is dropped from P and first collides with A after A has bounced on the
plane and before A reaches P again. The bounce and the collision are both perfectly elastic.
Explain why the speeds of A and B immediately before the first collision are the same.

Assume they collide at height H < h. The perfectly elastic bounce means that there was no loss
of energy, so A has the same total energy at height H on its upwards journey as it did when
travelling downwards. We can work out the speeds at the point of collision, calling them vA and
vB for A and B respectively. We write M for the mass of A and m for the mass of B (as in the
next part of the question). We have, by conservation of energy
2
M gH + 12 M vA = M gh
2
mgH + 12 mvB = mgh
2 = 2(gh − gH) and v 2 = 2(gh − gH), so |v | = |v | and the speeds of A and B are
so that vA B A B
the same.

The masses of A and B are M and m, respectively, where M > 3m, and the speed of the
particles immediately before the first collision is u. Show that both particles move upwards
after their first collision and that the maximum height of B above the plane after the first
collision and before the second collision is
4M (M − m)u2
h+ .
(M + m)2 g

This begins as a standard collision of particles question, and so I will repeat the advice from
the 2010 mark scheme: ALWAYS draw a diagram for collisions questions; you will do yourself
(and the marker) no favours if you try to keep all of the directions in your head, and you are
very likely to make a mistake. My recommendation is to always have all of the velocity arrows
pointing in the same direction. In this way, there is no possibility of getting the signs wrong
v1 − v2
in the Law of Restitution: it always reads v1 − v2 = e(u2 − u1 ) or = e, and you only
u2 − u1
have to be careful with the signs of the given velocities. The algebra will then keep track of the
directions of the unknown velocities for you.

A diagram showing the first collision is as follows.

uB = −u vB

B m B m

Before uA = u After vA

A M A M

522
STEP I 2011 Question 10 continued

Then Conservation of Momentum gives

M uA + muB = M vA + mvB
so
M u − mu = M vA + mvB ,

and Newton’s Law of Restitution gives

vB − vA = 1(uA − uB )

(using e = 1 as the collision is perfectly elastic). Substituting uA = u and uB = −u gives

M vA + mvB = (M − m)u (1)

vB − vA = 2u. (2)

Then solving these equations (by (1) − m × (2) and (1) + M × (2)) gives
(M − 3m)u
vA = (3)
M +m
(3M − m)u
vB = . (4)
M +m

To show that both particles move upwards after their first collision, we need to show that vA > 0
and vB > 0. From equation (3) and M > 3m (given in the question), we see that vA > 0; from
equation (4) and 3M − m > 9m − m > 0 (as M > 3m), we see that vB > 0. Thus both particles
move upwards after their first collision.
To find the maximum height of B between the two collisions, we begin by finding the maximum
height that would be achieved by B following the first collision assuming that there is no second
collision. We then explain why the second collision occurs during B’s subsequent downward
motion and deduce that it reaches that maximum height between the collisions.
The kinetic energy (KE) of B before the first collision is 21 mu2 and after the first collision is

3M − m 2 2
 
1 2 1
2 mvB = 2 m M +m
u ,

so that B has a gain in KE of 12 mvB 2 − 1 mu2 . When B is again at height h above the plane,
2
which is where it was dropped from, it now has this gain as its KE. (This is because the KE
just before the first collision has come from the loss of GPE; when the particle is once again at
height h, this original KE ( 12 mu2 ) has been converted back into GPE.)
The particle B can therefore rise by a further height of H, where
2 !
−

2 3M m
mgH = 12 mvB − 12 mu2 = 12 mu2 −1 ,
M +m
so
u2 9M 2 − 6M m + m2 M 2 + 2M m + m2
 
H= −
2g (M + m)2 (M + m)2
u2 8M 2 − 8M m
 
=
2g (M + m)2
4u2 M (M − m)
 
= .
g (M + m)2

523
STEP I 2011 Question 10 continued

Thus the maximum height reached by B after the first collision, assuming that the second
collision occurs after B has started falling is

4M (M − m)u2
h+H =h+ .
(M + m)2 g

Finally, we have to explain why A does not catch up with B before B begins to fall. But this
is easy: B initially has a greater upward velocity than A (as vB − vA = 2u > 0), so the height
of B is always greater than the height of A. Therefore they can only collide again after A has
bounced on the ground and is in its ascent while B is in its descent.

Alternative approach: using constant acceleration and “suvat”

An alternative approach is to use the formulæ for constant acceleration (“suvat”), as follows.
Just before collision, B has speed u, so the height H of B at this point is given by the “suvat”
equation v 2 = u2 + 2as, taking positive to be downwards:

u2 = 02 + 2g(h − H),

giving H = h − u2 /2g.
Immediately after the collision, B has velocity upwards given by equation (4) above. At the
maximum height, hmax , the speed of B is zero, so we can determine the maximum height using
v 2 = u2 + 2as again; this time, we take positive to be upwards, so a = −g:
2
(3M − m)u

2
0 = − 2g(hmax − H).
M +m

(Note that hmax − H > 0.)

Rearranging this gives

(3M − m)u 2
 
1
hmax = H +
2g M +m
u2 u2 (3M − m) 2
 
=h− +
2g 2g M +m
2 (3M − m)2 − (M + m)2
 
u
=h+
2g (M + m)2
u2 8M 2 − 8M m
 
=h+
2g (M + m)2
4M (M − m)u2
=h+
(M + m)2 g

as required.

524
Question 11

A thin non-uniform bar AB of length 7d has centre of mass at a point G, where AG = 3d.
A light inextensible string has one end attached to A and the other end attached to B. The
string is hung over a smooth peg P and the bar hangs freely in equilibrium with B lower
than A. Show that
3 sin α = 4 sin β ,
where α and β are the angles P AB and P BA, respectively.

We begin by drawing a diagram of the situation, showing the forces involved (the tension in the
string, which is the same at A and B since the peg is smooth, and the weight of the bar acting
through G). Clearly BG = 4d, which we have shown as well.
We have indicated the angles α, β and φ as defined in the question, and have also introduced
the angle θ as angle AGP . The point M is the foot of the perpendicular from P to AB, which
is used in some of the methods of solution.

P π P
π−θ−α 2 − (α − φ)
θ−β π
2 −φ−β
T T

T T
A α A α
M φ
θ π−θ
3d 3d
G G φ
4d β 4d β
W B W B
Diagram 1: Using θ for angle Diagram 2: Using φ for angle
between rod and vertical between rod and horizontal

Note that we have drawn the sketch with the weight passing through P . This must be the case:
both tensions pass through P and the system is in equilibrium. So taking moments around P
shows that W times the distance of the line of force of W from P must be zero, so that W acts
through P .
The simplest way of showing that 3 sin α = 4 sin β is to take moments about G:
y
M (G) T.3d sin α − T.4d sin β = 0

so that 3 sin α = 4 sin β.

An alternative approach is to apply the sine rule to the triangles P AG and P BG and resolve
horizontally; this, though, is a somewhat longer-winded method.

525
STEP I 2011 Question 11 continued

Given that cos β = 45 and that α is acute, find in terms of d the length of the string and show
that the angle of inclination of the bar to the horizontal is arctan 17 .

From cos β = 45 , we deduce sin β = 35 , and hence sin α = 4

3 sin β = 45 . Thus cos α = ± 35 , and
since α is acute, cos α = 35 .
There are numerous ways of finding the length of the string, l, in terms of d. We present a few
approaches here.

π
Approach 1: Show that ∠AP B = 2

To find the length of the string, we first find the angle AP B.

One method is to note that ∠AP B = π − α − β, so

sin(π − α − β) = sin(α + β)
= sin α cos β + cos α sin β
4 4 3 3
= 5 × 5 + 5 × 5
= 1,
π
so ∠AP B = 2 and the triangle AP B is right-angled at P .
Alternatively, as sin α = cos β, we must have α + β = π2 , so ∠AP B = π2 .
Thus AP = AB cos α = 7d. 35 = 21
5 d and BP = AB sin α = 7d. 45 = 28
5 d, so the string has length
( 21 28 49
5 + 5 )d = 5 d.

Approach 2: Trigonometry with the perpendicular from P

In the triangle AP B, we draw a perpendicular from P to AB, meeting AB at M . Then

P M = AP sin α = BP sin β. (This can also be shown directly by applying the sine rule:
AP/ sin β = BP/ sin α.)
We also have AB = AM + BM = AP cos α + BP cos β = 7d.
Now using our known values of sin α, etc., these equations become 45 AP = 35 BP so that BP =
4 3 4
3 AP , and 5 AP + 5 BP = 7d.
Combining these gives 35 AP + 16
15 AP = 7d, so AP =
21
5 d and hence BP = 28
5 d and l = AP +BP =
49
5 d.

Approach 3: Cosine rule

We apply the cosine rule to the triangle AP B, and we write x = AP and y = BP for simplicity.
This gives us

x2 = AB 2 + y 2 − 2AB.y cos β or
y 2 = AB 2 + x2 − 2AB.x cos α.

There are different ways of continuing from here. The most straightforward is probably to begin
by showing that BP = 43 AP or y = 43 x as in Approach 2. This then simplifies the two equations

526
STEP I 2011 Question 11 continued

to give

( 34 y)2 = (7d)2 + y 2 − 2(7d).y. 45 or

( 43 x)2 2
= (7d) + x − 2
2(7d).x. 35 .

We can then expand and rearrange these quadratics to get

7 2
16 y − 56
5 dy + 49d2 = 0 or
7 2 42 2
9x + 5 dx − 49d = 0.

Dividing by 7 and clearing fractions (multiplying by 80 and 45 respectively) gives

5y 2 − 128dy + 560d2 = 0 or
5x2 + 54dx − 315d2 = 0.

These quadratics turn out to factorise as

(y − 20d)(5y − 28d) = 0 or
(x + 15d)(5x − 21d) = 0.
28
The first equation gives two possibilities: y = 20d or y = 5 d, whereas the second only gives
one: x = 21
5 d.
For the first equation, y = 20d would imply x = 34 y = 15d, but then we would have

AB 2 + x2 − y 2 49d2 + 225d2 − 400d2

cos α = = < 0,
2AB.x 14d.15d
which is not possible as α is acute.
21 28 49
So we must have x = 5 d and y = 5 d, and hence l = x + y = 5 d.

Approach 4: Sine rule

Using the sine rule on the triangle AP B, we have

AB AP BP
= = .
sin(π − α − β) sin β sin α

We use sin(π − φ) = sin φ and the addition (compound angle) formula to write

sin(π − α − β) = sin(α + β)
= sin α cos β + cos α sin β
4 4 3 3
= 5 × 5 + 5 × 5
= 1,

so that the sine rule becomes

7d AP BP
= = ,
1 3/5 4/5
21 28 49
giving AP = 5 d, BP = 5 d and hence l = AP + BP = 5 d.
We now find φ, the angle of inclination of the bar to the horizontal. Referring to the above
diagrams, we have φ = π2 − θ, so tan φ = cot θ.
Here again are several approaches to this problem.

527
STEP I 2011 Question 11 continued

Approach 1: Resolving forces horizontally and vertically

We resolve horizontally to get

R(→) T sin(π − θ − α) − T sin(θ − β) = 0.

Therefore we get
sin(θ + α) = sin(θ − β).

We now use the addition formula for sine to expand these, and then substitute in our values for
sin α, etc., giving:
sin θ cos α + cos θ sin α = sin θ cos β − cos θ sin β
so
3
5 sin θ + 45 cos θ = 4
5 sin θ − 35 cos θ
giving
7 1
5 cos θ = 5 sin θ.
Dividing by cos θ now gives cot θ = 17 , hence the angle made with the horizontal is given by
tan φ = 17 , yielding φ = arctan 17 as required.

Alternatively, using φ instead of θ in the original equations, we get

R(→) T sin( π2 + φ − α) − T sin( π2 − φ − β) = 0,

which simplifies (on dividing by T 6= 0 and using sin( π2 − x) = cos x) to

cos(α − φ) − cos(φ + β) = 0.

The rest of the argument follows as before.

Approach 2: Resolving forces parallel and perpendicular to the rod

We resolve parallel to the rod to get

R(&) T cos α + W cos θ − T cos β = 0

and perpendicular to the rod to get

R(%) T sin α − W sin θ + T sin β = 0.

Rearranging these gives:

W cos θ = −T cos α + T cos β

W sin θ = T sin α + T sin β.

Dividing these equations gives

− cos α + cos β
cot θ =
sin α + sin β
−3 + 4
= 45 35
5 + 5
= 17 .

528
STEP I 2011 Question 11 continued

Since tan φ = cot θ, as we noted above, we have tan φ = 17 , so φ = arctan 17 as required.

Approach 3: Dropping a perpendicular

In the triangle AP B, we draw a perpendicular from P to AB, meeting AB at M . Then

P M = AP sin α = 21 4 84 21 3 63
5 d. 5 = 25 d and AM = AP cos α = 5 d. 5 = 25 d. As AG = 3d, it follows
12
that M G = AG − AM = 25 d.
Then (see the diagram above) we have tan θ = P M/M G = 84
 12
25 d 25 d = 7 so that cot θ = tan φ =
1 1
7 , giving φ = arctan 7 as required.

Approach 4: P G bisects ∠AP B

As in approach 1 above, we resolve horizontally (using diagram 2) to get

T sin( π2 − (α − φ) − T sin( π2 − φ − β) = 0.

Since both of the angles involved here are acute (as the triangle BGP has an obtuse angle at G),
they must be equal, giving α − φ = φ + β, so that 2φ = α − β.
Hence we have

cos 2φ = cos α cos β + sin α sin β

= 35 . 45 + 45 . 35
24
= 25 .

We therefore deduce using cos 2φ = 2 cos2 φ − 1 that cos2 φ = 49 2 1

50 and sin φ = 50 . It follows that
1
tan2 φ = sin2 φ/ cos2 φ = 49 , giving tan φ = 17 (the positive root as φ is acute) or φ = arctan 17
as required.

529
Question 12

I am selling raffle tickets for £1 per ticket. In the queue for tickets, there are m people each
with a single £1 coin and n people each with a single £2 coin. Each person in the queue
wants to buy a single raffle ticket and each arrangement of people in the queue is equally
likely to occur. Initially, I have no coins and a large supply of tickets. I stop selling tickets if
I cannot give the required change.

(i) In the case n = 1 and m > 1, find the probability that I am able to sell one ticket to
each person in the queue.

I can sell one ticket to each person as long as I have a £1 coin when the single person with a £2
coin arrives, which will be the case as long as they are not the first person in the queue. Thus
the probability is
1 m
1− = .
m+1 m+1

(ii) By considering the first three people in the queue, show that the probability that I am
m−1
able to sell one ticket to each person in the queue in the case n = 2 and m > 2 is .
m+1

This time, I can sell to all the people as long as I have one £1 coin when the first £2 coin is
given to me and I have received at least two £1 coins (in total) by the time the second £2 coin
is offered.
So we consider the first three people in the queue and the coin they bring; in the table below,
“any” means that either coin could be offered at this point. (This called also be represented as
a tree diagram, of course.) The probabilities in black are those of success, the ones in red are
for the cases of failure. Only one or the other of these needs to be calculated.

1st 2nd 3rd Success? Probability

m−1
  
m m m+2
£1 £1 any yes × =
m+2 m+1 2 2
m−1 m−1
  
m 2 m+2
£1 £2 £1 yes × × =
m+2 m+1 m 1 2
 
m 2 1 m+2
£1 £2 £2 no × × =1
m+2 m+1 m 2
  
2 m+1 m+2
£2 any any no =
m+2 1 2

To determine the probabilities in the table, there are two approaches. The first is to find the
probability that the kth person brings the specified coin given the previous coins which have
been brought; this is the most obvious method when this is drawn as a tree diagram. The
second approach is to count the number of possible ways of arranging the remaining coins
and to

divide it by the total number of possible arrangements of the m + 2 coins, which is
m+2 1
2 = 2 (m + 2)(m + 1).

530
STEP I 2011 Question 12 continued

Therefore the probability of success is

m(m − 1) 2(m − 1) (m + 2)(m − 1) m−1

+ = = .
(m + 2)(m + 1) (m + 2)(m + 1) (m + 2)(m + 1) m+1

Alternatively, we could calculate the probability of failure (adding up the probabilities in red)
and subtract from 1 to get

2 2 2 1 + (m + 1) 2 m−1
1− − =1− . =1− = .
(m + 2)(m + 1) m + 2 m+2 m+2 m+2 m+2

(iii) Show that the probability that I am able to sell one ticket to each person in the queue
m−2
in the case n = 3 and m > 3 is .
m+1

This time, it turns out that we need to consider the first five people in the queue to distinguish
the two cases which begin with £1, £1, £2, £2; the rest of the method is essentially the same
as in part (ii).

1st 2nd 3rd 4th 5th Success? Probability

m−1 m−2
  
m m m+3
£1 £1 £1 any any yes × × =
m+3 m+2 m+1 3 3

m m−1 3 m−2
£1 £1 £2 £1 any yes × × ×
m+3 m+2 m+1 m
m−1
  
m+3
=
2 3

m m−1 3 2 m−2
£1 £1 £2 £2 £1 yes × × × ×
m+3 m+2 m+1 m m−1
m−2
  
m+3
=
1 3

m m−1 3 2 1
£1 £1 £2 £2 £2 no × × × ×
m+3 m+2 m+1 m m−1
 
m+3
=1
3

m 3 m−1 m−2
£1 £2 £1 £1 any yes × × ×
m+3 m+2 m+1 m
m−1
  
m+3
=
2 3

m 3 m−1 2 m−2
£1 £2 £1 £2 £1 yes × × × ×
m+3 m+2 m+1 m m−1
m−2
  
m+3
=
1 3

m 3 m−1 2 1
£1 £2 £1 £2 £2 no × × × ×
m+3 m+2 m+1 m m−1

531
STEP I 2011 Question 12 continued

 
m+3
=1
3
  
m 3 2 m m+3
£1 £2 £2 any any no × × =
m+3 m+2 m+1 1 3
  
3 m+2 m+3
£2 any any any any no =
m+3 2 3

(Alternatively, the four cases beginning £1, £2 can be regarded as £1, £2 followed by 2 people
with £2 coins and m − 1 people with £1 coins, bringing us back into the case of part (ii). So the
m 3
probability of success in these cases is m+3 × m+2 × m−2
m , where the final fraction comes from
the result of part (ii).)
Therefore the probability of success is
1
m(m − 1)(m − 2) + 3(m − 1)(m − 2)+
(m + 3)(m + 2)(m + 1)


6(m − 2) + 3(m − 1)(m − 2) + 6(m − 2)
m−2

= m(m − 1) + 6(m − 1) + 12
(m + 3)(m + 2)(m + 1)
m−2
= (m2 + 5m + 6)
(m + 3)(m + 2)(m + 1)
m−2
= .
m+1

Similarly, the probability of success can be calculated by considering the probability of failure:
the probability of success is therefore
6 6 6m 3
1− − − −
(m + 3)(m + 2)(m + 1) (m + 3)(m + 2)(m + 1) (m + 3)(m + 2)(m + 1) m + 3
12 + 6m + 3(m + 1)(m + 2)
=1−
(m + 3)(m + 2)(m + 1)
3m2 + 15m + 18
=1−
(m + 3)(m + 2)(m + 1)
3(m + 2)(m + 3)
=1−
(m + 3)(m + 2)(m + 1)
3
=1−
m+1
m−2
= .
m+1

There seems to be a pattern in these results, and one might conjecture that the probability of
m+1−n
being able to sell one ticket to each person in the general case m > n is . This turns
m+1
out to be correct, though the proof uses significantly different ideas from those used above.

532
Question 13

In this question, you may use without proof the following result:
Z p p
4 − x2 dx = 2 arcsin( 12 x) + 12 x 4 − x2 + c .

A random variable X has probability density function f given by


2k√
 −a 6 x < 0
f(x) = k 4 − x2 06x62

0 otherwise,


where k and a are positive constants.

(i) Find, in terms of a, the mean of X.

R∞
We know that −∞ f(x) dx = 1, so we begin by performing this integration to determine k.
We have
Z 0 Z 2 p 0 p 2
4 − x2 dx = 2kx −a + k 2 arcsin x2 + 12 x 4 − x2 0
 
2k dx + k
−a 0


= 2ak + k (2 arcsin 1 + 0) − (2 arcsin 0 + 0)
= 2ak + kπ
= k(2a + π)
= 1,

so k = 1/(2a + π).
We can now work out the mean of X; we work √ in terms of k until the very end to avoid ugly
1
calculations. We can integrate the expression x 4 − x2 = kx(4 − x2 ) 2 either using inspection
(as we do in the following) or the substitution u = 4 − x2 , giving du/dx = −2x, so that the
R0 1 3 0
integral becomes k 4 − 12 u 2 du = k − 13 u 2 4 = 83 k.


Z ∞
E(X) = xf(x) dx
−∞
Z 0 Z 2 p
= 2kx dx + kx 4 − x2 dx
−a 0
0 3 2
kx2 −a + − k3 (4 − x2 ) 2
 
= 0
3
2
= (0 − ka ) + (0 − (− k3 × 4 ))
2

= −ka2 + 83 k
8
− a2
3
=
2a + π
8 − 3a2
= .
3(2a + π)

533
STEP I 2011 Question 13 continued

1
(ii) Let d be the value of X such that P(X > d) = 10 . Show that d < 0 if 2a > 9π and find
an expression for d in terms of a in this case.

1
We have d < 0 if and only if P(X > 0) < P(X > d) = 10 , so we consider P(X > 0). Using the
above integration (or noting that X is uniform for x < 0), we have

P(X > 0) = 1 − P(X < 0)

= 1 − 2ak
2a
=1−
2a + π
π
= .
2a + π
1
Therefore P(X > 0) < 10 if and only if

π 1
< ;
2a + π 10
that is 10π < 2a + π, or 2a > 9π. Putting these together gives d < 0 if and only if 2a > 9π.
In this case, as d < 0, we have

P(X > d) = 1 − P(X < d) = 1 − 2k(d − (−a)),

1 9
so 1 − 2k(d + a) = 10 , so d + a = 10 /2k, giving

9
d= −a
20k
9(2a + π)
= −a
20
9π − 2a
= .
20
Note that, since 2a > 9π, this gives us d < 0 as we expect.

An alternative approach is to calculate the cumulative distribution function first. We have


0

 x < −a

2k(x + a) −a 6 x < 0
F(x) = 1 1
√

 k 2a + 2 arcsin 2 x + 2 x 4 − x 2 06x62



1 x>2

(though only the part with −a 6 x 6 0 is actually needed).

9 9
Then we solve F(d) = 10 . If it turns out that d < 0, then we have F(d) = 2k(d + a) = 10 , which
rearranges to give d = (9π − 2a)/20 as above. Now if 2a > 9π, then (9π − 2a)/20 < 0 so that
9
F ((9π − 2a)/20) = 10 and d < 0, as required.

534
STEP I 2011 Question 13 continued

√
(iii) Given that d = 2, find a.

We note that now d > 0, so we have to integrate to find a explicitly. We get

√ Z 2 p p 2
P(X > 2) = √ k 4 − x2 dx = 2 arcsin x2 + 12 x 4 − x2 √2

2
√
2
√
2
√

= k (2 arcsin 1 + 0) − (2 arcsin 2 + 2 4 − 2)
π
= k(π − 2 − 1)
= k( π2 − 1)
π
2 −1
=
2a + π
1
= 10 .

Thus 10( π2 − 1) = 2a + π, so that 2a = 4π − 10, giving our desired result: a = 2π − 5.

√
Alternatively, one could calculate P(X < 2) in the same manner and find a such that this
9
equals 10 .
As a check, it is clear that 2a = 4π − 10 < 9π, so d > 0 from part (ii).

535
9470 STEP II 2011

Q1 (i) There are several routine features of a graph that one should look to consider on any curve-
sketching question: key points, such as where the curve meets or cuts either of the coordinate axes,
symmetries (and periodicities for trig. functions), asymptotes, and turning-points are the usual
suspects. In this case, the given function involves square-roots as well, so the question of the
domain of the function also comes into question. Considering all such things for
y  1  x  3  x should help you realise the following:
* the RHS is only defined for –3  x  1 (so the endpoints are at (–3, 2) and (1, 2));
 
* the graph is symmetric in the line x = –1, with its maximum at  1, 2 2 ; NB it must be a
maximum since 2 2  2 so there is no need to resort to calculus to establish this;
* the curve is thus  –shaped, and the gradient at the endpoints is infinite. This last point wasn’t
of great concern for the purposes of the question, so its mention was neither rewarded nor its
lack penalised: however, this is easily determined by realising that each term in the RHS is of
1
1
the form X 2 , so their derivatives will be of the form X 2 which, when evaluated at an endpoint
1
will give one of them of the form symptomatic of an asymptote.
0
A quick sketch of y = x + 1 shows that there is only the one solution at x = 1.

(ii) Each side of this second equation represents an easily sketchable curve. Indeed, the RHS is
essentially the same curve as in (i), but defined on the interval [–3, 3]. The LHS is merely a
“horizontal” parabola, though only its upper half since the radix  
sign denotes the positive
square-root. These curves again intersect only the once, when x < 0. Resorting to algebra …
squaring, rearranging suitably and squaring again then yields a quadratic equation in x having one
positive and one negative root.

Q2 The required list of perfect cubes is 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, though there were
no marks for noting them.

(i) In this question, it is clearly important to be able to factorise the sum of two cubes. So, in this
first instance x + y = k, (x + y)(x2 – xy + y2) = kz3  x 2  (k  x) x  (k  x) 2  z 3  0 , which
gives the required result upon rearrangement. One could either treat this as a quadratic in x and
4z 3  k 2
deal with its discriminant or go ahead directly to show that  ( y  x) 2  0 which
3
4z  k
3 2
immediately gives that is a perfect square and also that z 3  14 k 2 ; and the other half of
3
the required inequality comes either from z 3  k 2  3xy  k 2 (since x, y > 0) or from noting that
the smaller root of the quadratic in x is positive. Substituting k = 20 into the given inequality then
4z 3  k 2
yields 100  z 3  400  z = 5, 6, 7 ; and the only value of z in this list for which is a
3
perfect square is z = 7, which then yields the solution (x, y, z ) = (1, 19, 7). Although not a part of
the question, we can now express 20 as a sum of two rational cubes in the following way:
3 3
 1   19 
20 =      .
7  7 

(ii) Although this second part of the question can be done in other ways, the intention is clearly
that a similar methodology to (i)’s can be employed. Starting from
x + y = z2 , (x + y)(x2 – xy + y2) = kz.z2  x 2  z 2  x x  z 2  x   kz  0
2

536
 4kz  z 4
we find that is a perfect square, and also that k  z 3  4k . With k = 19, 19  z 3  76
3
 z = 3 or 4. This time, each of these values of z gives

z 76  z 3 
a perfect square, yielding the
3
two solutions (x, y, z ) = (1, 8, 3) and (6, 10, 4). Thus we have two ways to represent 19 as a sum
3 3 3 3
1 8 3 5
of two rational cubes: 19 =      and      .
 3  3 2 2

Purely as an aside, interested students may like to explore other possibilities for x 3  y 3  kz 3 .
One that never made it into the question was
 
x + y = kz, ( x  y ) x 2  xy  y 2  kz.z 2  x 2  (kz  x) x  (kz  x) 2  z 2  0

 3 x  3kzx  z (k  1)  0 . Then x 
2 2 2 3kz  9k 2 z 2  12 z 2 (k 2  1) 1
6

 6 z 3k  12  3k 2 , 
requiring 12 – 3k2  0 i.e. k2  4  k = 1 or 2.
When k = 1: x 3  y 3  z 3 has NO solutions by Fermat’s Last Theorem;

and when k = 2: x 3  y 3  2 z 3 has (trivially) infinitely many solutions x = y = z .

Q3 This question is all about increasing functions and what can be deduced from them. It involves
inequalities, which are never popular creatures even amongst STEP candidates. Fortunately, you
are led fairly gently by the hand into what to do, at least to begin with.

(i) (a) f (x) = cos x – {x.–sin x + cos x} = x sin x  0 for x  [0, 1

2
 ] , and since f(0) = 0 it
follows that f(x) = sin x – x cos x  0 for 0  x   .
1
2

(i) (b) A key observation here is that the “1” is simply a disguise for
d
x  , so you are actually
dx
being given that
d
arcsin x   d x  in the given interval; in other words, that f(x) = arcsin x – x
dx dx
is an increasing function. Since f(0) = 0 and f increasing, f(x) = arcsin x – x  0 for 0  x < 1, and
the required result follows.

x sin x  x cos x
(i) (c) Writing g(x) =  g (x) = 2
> 0 for 0 < x < 12  using (a)’s result.
sin x sin x
Now, it may help to write u = arcsin x, just so that it looks simpler to deal with here. Then u  x by
(b)’s result  g(u)  g(x) since g(x)  0 and the required result again follows.

(ii) There is a bit more work to be done here, but essentially the idea is the same as that in part (i),
only the direction of the inequality seems to be reversed, so care must be taken. An added
difficulty also arises in that we find that we must show that f   0 by showing that it is increasing
tan x x sec 2 x  tan x 2 x  sin 2 x
from zero. So g(x) = , g (x) =  . Examining f(x) = 2x – sin2x
x x2 2 x 2 cos 2 x
(since the denominator is clearly positive in the required interval): f(0) = 0 and f (x) = 2 – 2cos2x
 0 for 0 < x < 12   f  0  g (x)  0  g increasing. Mimicking the conclusion of (i) (c),
the reader should now be able to complete the solution.

537
Q4 (i) Using sin A = cos(90o – A) gives  = 360n  (90o – 4 ) – Note that you certainly should be
aware of the periodicities of the basic trig. functions  5 = 360n + 90o or 3 = 360n + 90o .
These give either  = 72n + 18o   = 18o , 90o , 162o or  = 120n + 30o   = 30o , 150o .

Now using the double-angle formulae for sine (twice) and cosine, we have c = 2.2sc.(1 – 2s2). We
can discount c = 0 for  = 18o, so that 1 = 4s(1 – 2s2) which gives the cubic equation in s = sin,
8s3 – 4s + 1 = 0  (2s – 1)(4s2 +2s – 1) = 0. Again, we can discount c = 12 for  = 18o) which
leaves us with sin18o the positive root (as 18o is acute) from the two possible solutions of this
5 1
quadratic; namely, sin18o = .
4

(ii) Using the double-angle formula for sine, we have 4s2 + 1 = 16s2(1 – s2)  0 = 16s4 – 12s2 + 1
12  80 3  5
 s2 =  . At first, this may look like a problem, but bear in mind that we want
32 8
2
6  2 5  5  1
2
it to be a perfect square. Proceeding with this in mind, s =    so that we have

16  4 
 5  1
the four answers, sin x =   .

 4 

(iii) To make the connection between this part and the previous one requires nothing more than
division by 4 to get sin 2 x  14  sin 2 2 x , and the solution x = 3 = 18o, 5 = 30o   = 6o
immediately presents itself from part (ii). However, in order to deduce a second solution (noting
that  = 45o is easily seen to satisfy the given equation), it is important to be prepared to be a bit
flexible and use your imagination. The other possible angles that are “related” to 18o and might
satisfy (ii)’s equation, can be looked-for, provided that sin5 =  12 (and there are many
possibilities here also). A little searching and/or thought reveals
 5  1
sin x =     3 = 180o + 18o = 198o also works, since 5 = 330o has sin5 =  12 ,

 4 
and the second acute answer is  = 66o.

Q5 The simplest way to do this is to realise that OA is the bisector of BOC, so that A is on the
diagonal OA of parallelogram OBAC (in fact, since OB = OC, it is a rhombus)  b + c = a
for some  (giving the first part of the result). Also, as BC is perpendicular to OA, (b – c) . a = 0
ab
 (2b – a) . a = 0   = 2 .
 aa 

 bc 
Similarly (replacing a by b and b by c in the above), we have d = kb – c where k = 2 
bb
 b  a  b  b   ab    ab  
= 2   2    2  d =  2    2 b  a  b  = b  a where
 bb  bb  bb 
 ab   a  b2 
  2   
bb
  1 or 4 a  ab  b    1 .
 

Now A, B and D are collinear if and only if AD  b  (  1)a is a multiple of AB  b  a

 t(b – a) = b  (  1)a for some t ( 0).

538
 Comparing coefficients of a and b then gives (t =)  =  + 1.

In the case when  = – 12 ,  = 1

2
and D is the midpoint of AB.

 a  b 2   ab
2

Finally,  = 1
 1
 4   1  4   1 , and using the scalar product formula
 a  ab  b 
2 2   ab 
ab 3
cos   gives cos    . [Note that a . b has the same sign as  .]
ab 8

To begin with, it is essential to realise that the integrand of I    f ( x)  f ( x) dx must have its
2 n
Q6
two components split up suitably so that integration by parts can be employed. Thus
 n

I =   f ( x)  f ( x) f ( x) dx = f ( x) 
1
n 1
 f ( x)n  1     f ( x) 1  f ( x)n  1  dx.
n 1
 
Now (and not earlier) is the opportune moment to use the given relationship f ( x)  kf ( x) f ( x) ,

so that I = f ( x) 
1
 f ( x)n  1    kf ( x)  1  f ( x)n  2  dx, which is now directly integrable
n 1  n 1 
as f ( x) 
1
 f ( x)n  1  1
 k  f ( x)
n3
(+ C).
n 1 (n  1)(n  3)

(i) For f(x) = tan x, f (x) = sec2x and f (x) = 2 sec2x tan x = kf ( x) f ( x) with k = 2.
sec 2 x tan n  1 x 2 tan n  3 x
Also, differentiating I =  gives
n 1 (n  1)(n  3)
dI

dx n  1
1
sec 2 x.(n  1) tan n x. sec 2 x  2 sec x. sec x tan x. tan n  1 x 

–
1
(n  1)(n  3)
2(n  3) tan n  2 x. sec 2 x  = sec4x tannx =  f ( x)2   f ( x)n as required,
although this could be verified in reverse using integration. Using this result directly in the first
given integral is now relatively straightforward:
sin 4 x sec 2 x tan 5 x 2 tan 7 x
 cos 8 x   C .
4 4
d x = sec x tan x dx =
5 35

(ii) Hopefully, all this differentiating of sec and tan functions may have helped you identify the
right sort of area to be searching for ideas with the second of the given integrals.
If f(x) = sec x + tan x, f (x) = sec x tan x + sec2x = sec x(sec x + tan x)
and f (x) = sec2x(sec x + tan x) + sec x tan x(sec x + tan x)
= sec x (sec x + tan x)2 = kf ( x) f ( x) with k = 1.

 sec xsec x  tan x  dx =  sec xsec x  tan x   sec x  tan x 

6 2 4
2
Then dx
sec x(sec x  tan x) 6 (sec x  tan x) 7
=  C
5 35

539
Q7 (i) Once you have split each series into sums of powers of  and  separately, it becomes clear
= 1     2  ...   n   1     2  ...   n 
n
that you are merely dealing with GPs. Thus b
r 0
r

n 1  1  n 1  1 1 n  1
=
 1

 1
=   1   n  1  1 , since  – 1 = 2 and  – 1 =  2
2
1 n
n 1  1  n 1  1 1
= a n  1  2 and, similarly,  a r =  = bn  1 .
2 r 0 2 2 2

(ii) There is no need to be frightened by the appearance of the nested sums here as the ‘inner sum’
has already been computed: all that is left is to work with the remaining ‘outer sum’ and deal
2n
 m  2n  1  1 2n  1
carefully with the limits:    a r     bm  1    bm (since b0 = 0)
m0r 0  m0 2  2 m0

=
1  1

2 2

a 2n  2  2  = 2 n  2   2 n  2  2 =

1
2
1 n 1 2
2
  
 2    n  1
n 1 2
since  = –1  
and n + 1 is even when n is odd = bn  1  when n is odd. However, when n is even, n + 1 is odd
1 2

2
2n
 m
 1
and    a r  = bn  1   2 or a n  1  .
2 1 2

m0r 0  2 2

2
 n 
(iii) We already have the result   a r  = bn  1  , so the only new thing is
1 2

r 0  2

=   3  5  ...  2 n  1      3   5  ...   2 n  1  , which is still the sum of two

n
a
r 0
2r  1

 2 n  2  1   2 n  2  1
GPs, merely with different common ratios, having sum  .
2  1  2 1
Now 
2  1  3  2 2  1  2 1  2  2  and 
 2  1  3  2 2  1  2 1  2  2 , 
   2 n  2  2 = bn  1  when n is odd, and bn  1   2 when n is even.
n
1 2n  2 1 1
so  a 2 r  1 =  2 2

r 0 2 2 2
2
 n  n
Thus   a r  – a 2r  1 = 0 when n is odd / = 2 when n is even.
r 0  r 0

Q8 The string leaves the circle at C(–cos , sin ).

Since the radius of the circle is 1, Arc AC =  – t =  (so cos = – cos t and sin = sin t).
Then B = (–cos + t sin , sin + t cos ) = (cos t + t sin t , sin t – t cos t).

dx
  sin t  t cos t  sin t  t cos t by the Product Rule; = 0 when t = 0, (x, y) = (1, 0) or
dt
t = 12  , (x, y) =  12  , 1 . This is xmax so t0 = 12  .

The required area under the curve and above the x-axis is
1
2
 1
2
  

 sin t  t cos t t cos t dt =   t 2 1  cos 2t  dt

dx
A =  y dt = 1
2 t sin 2t dt +  1
2
 dt  1
 1

2 2

using the double-angle formulae for sine and cosine. As the integration here may get very messy,
it is almost certainly best to evaluate this area as the sum of three separate integrals:

540
  
1 1   1 1 1   3
 2 t sin 2t dt = 4

t cos 2t 1 
 
1 4 cos 2t dt = 4

t cos 2t 
8
sin 2t 
1
=
8
;
1
2
 2  2
2

1 1   7 3

 2t
2
dt =  t 3  =
1
 6  1  48
2 2
 
1   1 3 3
 t cos 2t dt =  t 2 sin 2t    2 t sin 2t dt = 0 – 
1 2
and 2
= using a previous answer.
1
 4 1 1 8 8
2 2 2

7 3 3
Thus A =  .
48 4

For the total area swept out by the string during this process (called Involution), we still need to
3 
add in the area swept out between t = 0 and t = 12  , which is  
(there is, of course, no
48 4
need to repeat the integration process), and then subtract the area inside the semi-circle. Thus the
7 3 3   3    3
total area swept out by the string is  +      (area inside semi-circle) = .
48 4  48 4  2 6

Q9 Collisions questions are always popular, as there are only two or three principles which are to be
applied. It is, nonetheless, good practice to say what you are attempting to do. Also, a diagram,
though not an essential requirement, is almost always a good idea, if only since it enables you to
specify a direction which you are going to take to be the positive one, especially since velocity
and momentum are vector quantities. Once these preliminaries have been set up, the rest is fairly
easy. By CLM, 3mu = 2mVA + mVB and NEL/NLR gives e.3u = VB – VA . Solving these
simultaneously for VA and VB yields VA = u(1 – e) and VB = u(1 + 2e).

Next, after its collision with the wall, B has speed | f VB | away from the wall.

For the second collision of A and B, by CLM (away from wall), fmVB – 2mVA = 2mWA – mWB,
and NEL/NLR gives WA + WB = e(VA + f VB). Substg. for VA & VB from before in both of these
equations  2WA – WB = u f (1  2e)  2(1  e) and WA + WB = eu(1  e)  f (1  2e). Solving
these simultaneously for WA (not wanted) and WB then gives WB = 13 u 2 1  e 2  f 1  4e 2 , as    
required.

Noting that 1  4e 2 can be negative, zero, or positive, it may be best (though not essential) to
consider the possible cases separately:
if e = 12 , WB = 13 u2 34   f 0   12 u  0 ;
if 1
2    
< e < 1, WB = 13 u 2 1  e 2  f 4e 2  1 > 0 for all e, f since each term in the bracket is > 0;
if 0 < e < 1
2 , 1 – e2 > 3
4 and WB > 13 u32  f 1  4e 2   13 u32  1  1  0 .

541
 u sin 
Q10 The maximum height of a projectile is when y  u sin   gt  0  t  . Substituting this
g
u 2 sin 2 
into y  ut sin   gt  H = 1
2
2
(although some people learn it to quote).
2g

u 2 sin 2 
When the string goes taut, its length l is given by l = 1
2 H . But l is also given from the
4g
y-component of P’s displacement as l = ut sin   12 gt 2 , which gives the quadratic equation
2u sin   4u 2 sin 2   4 gH
gt 2  (2u sin  )t  H  0 in t. Solving by the quadratic formula, t =
2g

=
2 2 gH  8 gH  4 gH
2g
=
1
g

2 gH  gH =
H
g
  
2  1 , where we take the smaller of the

two roots since we want the first time when an unimpeded P is at this height.

At this time, P’s vertical velocity is v = y  u sin   g

H
g
 
2 1 = 2 gH  gH  2 1 
u sin 
 gH or . Thus, the common speed of P/R after the string goes taut, by CLM, is 1
2 gH
2
u sin 
or .
2 2

When the string goes slack, we must consider the projectile motion of R, which has initial velocity
u sin 
components u cos  and  . [Note that both P and R move in this way, so P no longer
2 2
u sin  g
interferes with R’s motion.] R’s vertical displacement is zero when y R  t  t 2 = 0 (t  0)
2 2 2
u sin 
 t (and this is the extra time after the string has gone slack). The total distance
g 2

travelled by R is thus D = x1 + x2, where x1 = u cos 

u sin 
g 2

2  1 and x2 = u cos  u sin 
g 2
u 2 sin  cos 
= .
g

Finally, setting D = H  tan = 2.

542
Q11 (i) The saying goes that “a picture paints a thousand words” and tis is especially true in mechanics
questions, if for no better reason than it gives the solver a clear indication of angles/directions for
– in this case – the forces involved. The relevant diagrams are as follows:

O A O B O C A

k k k Tsin30o
T U V j

30o 60o
O iO
P P P  
Vsin60o
U cos

B C
2
It might also be wise to note the sines and cosines of the given angles: tan = 2  sin  
3
1 2 1 2 2
and cos   , and tan =  sin   and cos   . Having noted these carefully, it
3 4 3 3
is now reasonably straightforward to state that the vector in the direction of PB is

 U cos  . cos  i  U cos  . sin  j  U sin  k =  i 

1 2 2
j k.
3 3 3
Note that the question requires you to verify that this vector has magnitude 1.

(ii) The forces involved are now readily written down …

 1 2 2 
TB =   i  j k U follows from (i)’s answer. Also,
 3 3 3 
1
TA = T sin 30 j  T cos 30 k = T j  3 k ,
2
 
1 2 2 1 
TC = V sin 60 cos  i  V sin 60 sin  j  V cos 60 k = V  i j  k 
2  3 3 
and W = – W k .

(iii) Having set the system up in vector form, the fundamental Statics principle involved is that
TA + TB + TC + W = 0 .
Comparing components in this vector equation gives
1 6
( i ) 0 U  V  0  U V 6
3 3
1 2 3 5 3
(j) T U V  0  (using U  V 6 ) T  V
2 3 6 3
3 6 1 5 3
(k) T U  V  W  (using U  V 6 and T  V)
2 3 2 3
W 3 W 6 W
T , U , V  .
3 5 5

543
Q12 It is important in these sorts of contrived games to read the rules properly: in this case, you must
ensure that you are clear what is meant by ‘match’, ‘game’ and ‘point’. Then, a careful listing of
cases is all that is required.

(i) P(re-match) = P(XYX) + P(YXY) = p(1 – p)2 + (1 – p)3 = (1 – p)2.

P(Y wins directly) = P(YY) + P(XYY) = (1 – p)p + p(1 – p)p = p(1 – p)(1 + p) or p 1  p 2  .
 
Thus, P(Y wins) = w = p 1  p 2  w(1  p ) 2 , and you will note the way starting the match again
leads to a recurrent way of describing Y’s chances of winning. Re-arranging this then gives
p1  p 2  p 1  p 2  p 1  p 2  1  p 2
w = = = for p  0.
1  (1  p) 2 1  (1  p) 1  (1  p)  p(2  p) 2  p

Next, w   1  
2 1  p 2  (2  p ) p (1  2 p )
 , and since 2 – p > 0, w  12 has the same sign as
2
2( 2  p ) 2( 2  p )
1 – 2p and hence as 12  p . Hence, w > 12 if p < 12 and w < 12 if p > 12 .

To be fair at this point, the final demand of part (i) ended up being rather less demanding than was
originally intended, as the answer is either “Yes” or “No” … though you would of course, be
expected to support your decision; no marks are given for being a lucky guesser! The following
calculus approach is thus slightly unnecessary, as one can simply provide an example to show that
w can decrease with p. The following, more detailed analysis had been intended.
dw (2  p)(2 p)  1  p 2 (1)
 =
1

p2  4 p 1 
1

[2  p]2  3 .  
dp (2  p ) 2
(2  p) 2
(2  p ) 2

dw dw
Then > 0 for 0 < p < 2  3 and < 0 for 2  3 < p  1.
dp dp

For a fair game, Y’s expectation should be 0. Thus, using E(gain) = g i  P( g i ) , where gi is the
“gain function” for Y, with w = 5
12 when p = 2
3 , we have 0 = (k )  125  (1)  127  k = 1.4 .

When p = 0, the results run YXY ... re-match ... YXY ... re-match ... and the match never ends.

Q13 Firstly, skewness is a measure of a distribution’s lack of symmetry.

(i) For the next part, you should understand how the “expectation” function behaves.
       
E  X    = E X 3  3X 2  3 2 X   3 = E X 3  3E X 2  3 2 E X    3
3

= E X   3 
3 2
  2   3 2 .   3 using E[X] =  and E[X2] = 2 + 2
= E X   3
3 2
  3 , as required.

For a given distribution, this next bit of work is very routine indeed.
2  1
  1  1
1 1
E X    2 x 2 dx =  x 3  = 23   ; E X 2   2 x 3 dx =  x 4  = 1
2   2  181 ; and
0 3  0 0 2  0
 3. 23 . 181  278
  2  1
1 2
2 2
E X 3   2 x 4 dx =  x 5  = 2
5 ; all of which then lead to   5
= when
0 5  0 18 18
1
5
substituted into the given (previously deduced) formula.

544
 x
(ii) Here, F(x) =  2 x dx = x 2 (0  x  1)  F – 1(x) = x (0  x  1)
0

F 1 109   2 F 1  12   F 1 101    3  2 5 1 4  2 5
3 2 1

 D  10 2 10
=   2 5.
F 1
   F 101 
9
10
1 3
10
 1
10
3 1 2

M
M is given by  2 x dx = 1
2  M2 = 1
2  M= 1
2
(OR by M = F – 1( 12 ) = 1
2
).

 
0

3 23  1

And P = 1
2
 6 2 9.
3 2

In order to establish the given inequality “chain”, we must show that D > P and P >  (there is no
point in proving that D > ). One could reason this through by considering approximants to 2
and 5 , but care must be taken not to introduce fallacious “roundings” which don’t support the
direction of the inequality under consideration. The alternative is to establish a set of equivalent
numerical statements; for example, to show that D > P …
2  5  6 2  9  11  5  6 2
 121  5  22 5  72 (after squaring, since both sides are positive)
 54  22 5 or 27  11 5  729  605 (again, squaring positive terms)
and this final result clearly IS true, so the desired inequality is established.

545
STEP Mathematics III 2011: Solutions

Section A: Pure Mathematics

1. (i) The differential equation can be solved either by separating variables or using
an integrating factor. In either case, , or the negative of it is required, and this can
be found either by re-writing as 1 or using the substitution, 1.
Thus the solution is 1 .

(ii) The substitution yields ′ , and

′′ 2 ′ .

Substituting these expressions in the differential equation and simplifying gives

1 2 0 which is effectively the first order differential equation

from part (i) with ′.

So ′ 1 , which is an exact differential (or integration by parts could be used),

and so as required.

(iii) Part (ii)’s substitution gives 1 which using the

integrating factor from part (i) gives 1 , and thus

1 . Alternatively, the solution to part (ii) is the complementary

function and a quadratic particular integral should be conjectured, which in view of the cf
need only be , yielding the same result.

2. As 0, 0 , which, when evaluated, gives every term but one to

be an integer, and so, that term, , must be an integer, and as p and q are integers with no
common factor greater than 1, this can only happen if 1 , giving the required deduction.

(i) To show that the nth root of 2 is irrational, consider 2, and evaluate
1 and 2 , then apply the stem of the question.

(ii) Considering the turning points of 1 , there can only be one real
root. Evaluating 2 and 1 and applying the stem gives the result.

(iii) Considering the graphs of and 5 7 , for 3 , that these cannot

intersect for 0 can be observed by noting their signs for 0 1 ∙ 4 , and their
gradients for 1 ∙ 4 . For 0 , and n even, it is sufficient to consider signs, whereas
for n odd, it is enough to evaluate 5 7 for 2 , and 1 3,
depending on the case, and then applying the stem. The case 2, can be demonstrated by
completing the square and showing that there are no real roots.

546
Part (i) could be demonstrated by a minor variant to the usual proof for the irrationality of the
square root of 2. Parts (ii) and (iii) could be shown by applying the stem and then
considering the left hand side of each equation for the cases n even and n odd.

3. Considering the quadratic equation 0 , the condition 4

shows, by considering the discriminant, that the roots are unequal. Supposing that
3 can be written as , and equating coefficients
generates the four equations
1
3 3 0
3 3 3

The first pair may be solved simultaneously to give and .

Substitution yields and , or alternatively,
and and so α and β satisfy 0 i.e. 0.
For 8, 48 , 4 2 0.
Hence α and β are the roots of 8 48 64 0 , i.e. 6 8 0 and
wlog 2 , 4, 2, 1.
So 24 48 0 can be re-arranged as 2
√ √ √
As 1 , √2 , √2 , √2 and so , ,
√ √ √

If 2 and , 4 so the first part cannot be used.

However, 3 2 0 can be readily factorised as 2 0 and so
(repeated) or 2

4. (i) is the area between the curve , the x axis, and the line
is the area between the curve , the y axis, and the line .

The sum of these areas is greater than or equal to the area of the rectangle, with equality
holding if .

(ii) With , the sum of the two integrals is

547
 But as 1, , and so the required result follows by applying the result
of part (i).
If , simple algebra shows , so and
equality is verified.

(iii) sin satisfies the conditions of part (i)

So 1 cos , and, by parts, sin √1 1
which together give the required result.

Choosing 0, and , part (i) gives 0 sin √1 1 which
can be re-arranged to give the required result.

5.
tan

and hence integrating gives the result.

A is cos , sin and B is cos , sin

cos cos sin sin using (*) which

leads directly to .
Replacing –a by b gives
As , these expressions can be equated to give .
The area between curves C and D is which by substitution gives
as required.

6. Using the substitution tanh , then it can be shown that , by making

use of 2 sinh cosh sinh to obtain the integrand, and tanh ln to
obtain the limits.
If instead, integration by parts is used differentiating tanh and integrating ,and

employing tanh ln to demonstrate that tanh ln 0, .

The substitution can be used to demonstrate that .

(Alternatively, starting from U, the substitution 2 tanh obtains , the

substitution ln obtains , and the substitution 2 followed by integration
by parts yields ; starting from V, by parts it can be shown that , using the
substitution that , and the substitution tanh that ; or starting
from X, the substitution ln gives , integration by parts gives , and
the substitution tanh gives .)

7. (i) The induction requires 1 and

1 1 .
1 √ 1 √ 1

548
 2 1 2 1 and so 2 1 and 2 , and
1 2 1 1 2 1 the result is true for 2.
Evaluating using 1 then 2 1 2 1
and 2 2 1 , and so substituting and simplifying,
1 1 1 by the induction.

(ii) √ 1 √ √ 1 √ 1

√ 1 1 √ which is of required form because

and 1 are integers and

1 1 1

1 1 as required.

Trivially the case 1 is true.

(iii) In the case that n is even,

1 1 1 1 1
as required,
and in the case that n is odd, √ 1 √ 1

1 as required.

8. using the complex conjugate, so

and

(i) If tan , 0 , then sin , and cos θ, using the general

result and so 1 but the point i.e. 0,1 is not included.

(ii) If 1 1 , and 0 , then it is the same locus as (i) except ,

and so it is the semi-circle that is the part of 1 below the u axis.

(iii) 0 , then 0 and , and as 1 1 which is that part of the v

axis below the u axis, i.e. ∞ 0.

(iv) Let 2 tan and 1 , so as ∞ ∞, , then

1
sin and 2
1 cos θ , so the locus is the circle
excluding the point θ , which is 0,1 .

549
Section B: Mechanics

9. For the initial equilibrium position, suppose , considering potential energy,

with potential energy zero level at O, 4 cos 3 sin , for
equilibrium, 0 , giving tan .
Then conserving energy,
4 cos 3 sin 7 4 cos 3 sin which having
substituted for gives 7 8 cos 6 sin 10

(i) Resolving radially in general for Q, if R is the contact force,

4 cos 4 , so when , 0 , and thus 4 cos 4
and so substituting for and in the energy result gives 15 cos 6 sin 10 .

(ii) Resolving tangentially for Q, 4 sin 4 and for P,

3 cos 3 so eliminating between them and re-arranging,
sin cos as required.

10. Suppose Q is displaced x and P is displaced y, and let ,

then and .
Adding and integrating leads to .
Subtracting gives and so sin from solving the
differential equation and employing the initial conditions that when 0, 0, 0,
and .
Thus, sin and sin . When the string next
returns to length a , sin 0, and so as required.
So at this time, , and 0.
The total time between the impulse and the subsequent collision is .

11. On the one hand the distance between the point on the disc vertically below , 0,0
and P is sin as the string length b makes an angle with the vertical. On the other, it is
2 sin , the third side of an isosceles triangle with two radii a at an angle , and hence the
required result.
The horizontal component of the tension in each string is sin and it acts at a
perpendicular distance cos from the axis of symmetry. Thus the couple is
sin cos . Resolving vertically, cos . Substituting for T in the
expression for the couple and then using bsin 2 sin to eliminate , gives the
required result.
The initial potential energy relative to the position where the strings are vertical is
1 cos . This is converted into kinetic energy . Equating these
expressions and once again using bsin 2 sin to eliminate , gives the required
result.

550
Section C: Probability and Statistics

12. As , ′ ′ ′ , and as 1 1, ′ 1
, ′ 1 , and ′ 1 , the first result follows.
Similarly, ′′ ′′ ′ ′ ′′ , and
1 1 1
′′ 1 ′ 1 ′ 1 ′′ 1

as required.
A fair coin tossed until a head appears is distributed so . The PGF for
the number of heads when a fair coin is tossed once is . Thus .

Using the results 2 1 , and 2 1.

, being the coefficient of in , is for 1 , and for 0 .

13. (i) , 1
1
and so . The most probable value of X is the minimum value of r such
that 1 , because this expression decreases as r increases. All the factors are
positive so it is simple to rearrange the algebra to obtain 1 so .
The answer is not unique when there is a value of r such that 1 , in which case,
, which will only happen if divides 1 .

(ii) Using the same strategy as for part (i) , ,

1 , and so .
Again, the most probable value of X is the minimum value of r such that
1, giving , and this is not unique if 2 divides
1 1 .

551
552
STEP Examiners’ Report
2012

Mathematics
STEP 9465/9470/9475

November 2012

553
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 3
STEP Mathematics II 7
STEP Mathematics III 10
Explanation of Results 13

554
Comments on individual questions 

Q1   

This was a popular question and most candidates managed to find expressions for   and   easily. 

Many did not make any progress beyond this point however. Of those that continued with the 
question many managed to find the minimum value of   easily, but the minimum value of 
 caused more trouble. The majority of candidates did not realise that they could simply 
differentiate   to find the minimum value and so attempted a more complicated 
differentiation. In many cases they were successful, but then failed to find the value of   to 
complete the question. 

Q2 

This was again a popular question with candidates achieving varying levels of success. The first part 
of the question was generally well attempted, although a number of candidates who identified that 
the equation could be rewritten as   assumed that this would result in a sharp change 
of gradient at the points of intersection with the  ‐axis (as would be seen in the graph of 
). 

Some candidates attempted to regard the equation as a quadratic in   to identify the number of 
roots for different cases, but this did not sufficiently distinguish between the different cases in the 
solutions provided. A better method is to consider the intersections of two graphs as the value of   
is varied. Another common mistake was to give just part of the solution set in the case   or to 
assume that it is not possible for there to be three solutions. 

In the second part of the question most candidates correctly found the two possible values of  , but 
often just chose one of the values to work with for the next part. Graphically, the difference 
between the two cases is a reflection in the  ‐axis and, since the intersections with a horizontal line 
are being found, the sets of values of   will be the same in both cases. This was not explained in 
choices that decided to explore just one of the two cases. 

Q3 

The first part of this question was generally well answered by those candidates who recognised the 
concept of placing the required area between two triangles. There were a number of answers where 
the graph was sketched, but no further progress was made however. 

Of those who solved the first part of the question, many were able to do some work on the second 
part of the question, but took the upper limit of the integral as   (consistent with the method 
for the first part of the question, but not required here as   is the upper limit to choose) and 
therefore arrived at far more complicated expressions than were needed. Often this led to a 
doomed attempt to handle the algebra that was left. 

555
For those who correctly chose the limits of the integration the part of the inequality whose solution 
involved   still proved a challenge for some. 

Q4 

This question was generally well attempted, with many candidates able to obtain the equations of 
the tangents and normals. The point of intersection of the tangents did not cause too much 
difficulty, but the intersection of the normal was problematic for many candidates who struggled to 
simplify the expressions and therefore did not reach a point where it was obvious that the point lay 
on the line  . 

The final part of the question is much easier if the expressions are simplified as they are 
encountered and for this the factorisation of the difference of two cubes is a useful thing to know. It 
is also useful to realise that the information that   means that the expression   
can be factorised as  . 

Q5 

This was another popular question, with attempts by a large number of candidates. Integration by 
parts was required for both of the main methods for the first integral, either preceded or followed 
by the application of some trigonometric identities. A number of candidates managed to obtain the 
correct answers for each of the first two integrals, but then struggled to relate them to the final part 
of the question, in some cases ignoring the different limits and in others incorrectly manipulating the 
logarithms. 

Q6 

This question was not attempted by many candidates, possibly due to the apparent three‐
dimensional nature of the problem (although it reduces to a two‐dimensional problem immediately). 
Many candidates solved the first part of the problem through applications of Pythagoras theorem in 
the various right‐angled triangles that can be identified rather than using the similarity that is 
present. A number of candidates got the expressions for the tangent and cotangent confused. 

Where candidates attempted the second part they were generally more successful, although care 
needed to be taken over the individual steps. Relating the identity to the first part of the question 
involved an understanding of the trigonometric graphs and this was done successfully by a number 
of candidates. 

Q7 

The first two parts of this question were generally well answered, although a number of candidates 
were confused about the order in which to substitute the variables into the equation and thus got 
answers with   and   confused. The third part proved more complex, with the factorisation of the 
final expression causing problems for some candidates. 

Q8 

556
This was a very popular question with attempts by most of the candidates. In many cases the first 
substitution was correctly carried out and the resulting differential equation solved, but then no 
progress was made on the second part. 

The candidates who realised that the same substitution would work for the second part managed to 
get to the reduced differential equation. Although the integration for this differential equation was 
more complicated, many of the candidates who reached this stage managed to evaluate the integral 
(although sometimes by longer methods than were needed). 

Q9 

This question was quite popular. The first part of the question involves the application of the 
formulae for motion under uniform acceleration and this was generally well carried out, although a 
number of candidates did not justify the choice of the positive square root. A number of candidates 
also took a longer approach to the calculation, calculating the time to reach the highest point and 
then the time for the downward journey. Many candidates realised that differentiation of the 
expression for the range was required, although some decided to differentiate with respect to  , 
rather than  , making the task more difficult. The differentiation requires a degree of care to make 
sure that the signs are correctly managed and many candidates did manage to complete this 
successfully. 

Q10 

For many candidates the first part of the question was solved correctly. The manipulation of the 
expressions involving sums and differences of square roots was more complicated for a number of 
candidates and the derivation of the formula required in the second part was less successfully 
carried out. The final part of the question involved the substitution of the values given into the 
formula, and providing that this was done with care the correct answer was generally found 
successfully. 

Q11 

This question began with some quite familiar calculations involving inclined planes and many 
candidates who attempted the question were able to reach the solution. In some cases the required 
value of   was assumed rather than solving the simultaneous equations. 

The second part of the question was less familiar and some candidates did not realise that the fact 
that the pulley can move means that there will be different accelerations at different points in the 
system. They therefore attempted to calculate one value for the acceleration that worked for all of 
their equations. 

Q12 

This question was not attempted by many candidates. In some cases it was not understood that the 
function had to be integrated to find the probabilities. The identification of the probabilities required 
for the conditional probability calculation was also problematic. Where it was there were some good 
answers, although the algebraic manipulation proved a little complicated for some candidates. 

557
Q13 

This question was not attempted by many candidates. There were some good answers showing a 
clear thought process to reach the required value, but many of the other solutions offered suffered 
from a lack of explanation of the method meaning that the ideas being applied were difficult to 
follow. 

558
9470 STEP II 2012 Report
General Remarks
There were just over 1000 entries for paper II this year, almost exactly the same number as last
year. Overall, the paper was found marginally easier than its predecessor, which means that it
was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50
candidates scored 100 marks or more, with more than 400 gaining at least half marks on the
paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40
marks. It was pleasing to note that the advice of recent years, encouraging students not to make
attempts at lots of early parts to questions but rather to spend their time getting to grips with the
six that can count towards their paper total, was more obviously being heeded in 2012 than I can
recall being the case previously.

As in previous years, the pure maths questions provided the bulk of candidates’ work, with
relatively few efforts to be found at the applied ones. Questions 1 and 2 were the most popular
questions, although each drew only around 800 “hits” – fewer than usual. Questions 3 – 5 & 8
were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and
Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most
successfully attempted question on the paper – with very little interest shown in the rest of
Sections B or C.

Comments on individual questions

Q1 The first question is set with the intention that everyone should be able to attempt it, but
20% of candidates were clearly put off by the algebraic nature of this year’s opener. It was ,
nonetheless, the most popular question on the paper, possibly due to the lack of any advanced
techniques. The specifically numerical parts of the question were generally more confidently,
and hence successfully, handled than the general ones. So, for instance, formulae for the various
coefficients in (i) were often not correct, even when high marks were scored on the question.
When numerical answers went astray, it was usually due to incorrect signs in the early stages,
with candidates failing to realise that all terms were positive. The very final demand (for the
coefficient of x66) was the real test, not only of candidates’ resilience and nerve but also of their
grasp of where the various contributions were coming from. From a marking point of view, very
few candidates gave particularly clear methods, and it was usually difficult for the markers to
decipher the underlying processes from what appeared to be merely a whole load of numbers
written down and added up.

Q2 This turned out to be the second most popular question and the highest scoring of the
pure questions. Explanations apart, most candidates held their nerve remarkably well to produce
careful algebra leading to correct answers. The added “trap” in part (i) – in that each answer
contained an arbitrary constant – caught many out.

Q3 This was another popular question, scoring just over half of the marks on average.
Although most efforts to establish the given initial result were eventually successful, many made
1
hard work of it, failing to notice the obvious result that if t = x 2 1  x then  x 2  1  x .
t
1
The first integral could then be found by realising that f(t) = 2 was the relevant function here,
t
or by repeating the substitution already used.
Quite a few candidates thought that the second integral followed from the first, which
was unfortunate, as it didn’t. However, most efforts at this second integral were unsuccessful
anyhow, with candidates usually getting 3 of the 10 marks for setting up the substitution and then
often going round in circles. The main problem lay in using sin and cos instead of tan and sec, or

-1-
559
9470 STEP II 2012 Report

in continuing with x 2 1  x without identifying a suitable function f(t). It was helpful to find
this, but not essential.

Q4 This was quite a popular question, as candidates seemed to like using the log series, and
appreciated the helpful structuring of the question. However, inequalities are seldom entirely
confidently handled, and explanations (wherever they are required) are generally rather feeble.
Thus, several marks were often not picked up, sometimes because the candidates did not think
they had to consider addressing issues such as whether the series was valid in this case. Part (i)
was usually fairly well done, with (ii) providing more of a challenge. Part (iii) required only
informal arguments, but many scored only 1 of the 2 marks allocated here due to being a bit too
vague about what was going on.

Q5 Another popular question, but scoring a relatively low average mark overall; however,
this was partly due to the high number of partial attempts, and good efforts usually scored around
14 marks. Surprisingly, the greatest difficulty was found in the differentiation in part (ii). There
were lots of marks for the curve-sketching, and several easy features to work with: principally
the symmetry and the asymptotes (and the behaviour of the functions on either side of these). For
many who struggled, the biggest problem lay in where to put the y-axis, which was largely
immaterial. As mentioned already, differentiation attempts were rather poor on the whole, with
muddling of the Chain, Product and Quotient Rules. Even for those who differentiated correctly,
extracting the factor (2x – a – b) proved too tough, despite the fact it might have been obvious
with a bit of thought.

Q6 This was the second least popular of the pure maths questions, partly (it seems) because
many candidates did not know what was meant by the term cyclic quadrilateral. The other
immediate hurdle was that candidates needed to know that “opposite angles of a cyclic quad. are
supplementary”. We know this because of the large number of “attempts” that got no further than
an initial diagram and a bit of working. Thus, the question was even less popular than the raw
figures show. In reality, the question involved little more than some GCSE-level trigonometry,
the difference of two squares factorisation and the result sin2 + cos2 = 1. Those who overcame
the initial hurdles scored highly.

Q7 This question was the least popular of the pure maths questions by a considerable margin,
and attempts at it were usually fairly poor. In fact, very few candidates got beyond the opening
(given) result. The barrier to further progress was almost invariably the failure to realise that all
points on a circle centre O and radius 1 have position vectors that satisfy xx = 1.

Q8 Well over a half of all candidates attempted this question but, on average, it proved to be
the least well scoring. The initial inequality was usually well handled, but most of the remaining
parts of the question were poorly handled in very circuitous ways, with few candidates being
very clear in either what they were trying to prove or how. The first, given, result simply follows
from equating for q2 in two successive cases of the given recurrence definition. This result was
then supposed to help with the following result, but almost no-one seemed to realise this, and
attempts at inductive proofs were quite common at this stage (usually unsuccessfully).
Candidates’ confidence had clearly ebbed away well before the final part of the question, and so
there were very few attempts at the two cases of the final paragraph.

Q9 Despite its obviously algebraic nature, and incorporating inequalities, this was a
remarkably popular question with candidates, more than 400 of whom chose to do it. Moreover,
it also proved to be the most successful question on the paper, with the average mark exceeding a
score of 12. Marks that were lost generally arose from a lack of care with signs (directions) or a
failure to justify the direction of the inequality from the physical nature of the situation.

-2-
560
9470 STEP II 2012 Report

Q10 This was the least popular question on the paper, attracting the poorest efforts and having
the weakest mean score (under 4 marks). Around half of attempts foundered at the very outset by
failing to have “the vertical plane containing the rod ... perpendicular to the axis of the cylinder”.
Those candidates who resolved horizontally and vertically, instead of parallel and perpendicular
to the rod, invariably ended up with a terrible mess that they simply couldn’t sort; a few forget to
take moments at all and were thus unable to make much progress towards the answers required.

Q11 There were many very capable attempts at this question, taken by around a quarter of all
candidates. At some stage, a general approach was required to the use of the principle of
conservation of linear momentum; some resorted to “pattern-spotting” (which lost them a couple
of marks) and others to an inductive approach, collision by collision, which worked well though
was generally a lengthier bit of work. Some candidates mixed up n with N in the following part,
while others incorrectly considered > rather than  and ended up missing the answer by 1. A
little bit of care was needed with the summation in the final part, and there was a bit of fiddling
going on in order to get the given answer. Nevertheless, it was pleasing to see the principles
understood well, even if the details were less carefully attended to.

Q12 This was not a popular question, and most attempts petered out after part (i), which was
usually handled very well, even when the situation was split into more cases than was strictly
necessary. Indeed, few made much of a serious attempt at (ii), mainly because they were either
finding the range of p such that (i)’s given answer was equal to 73 , or solving 2.5 < E(X) < 3.5,
where X was the number of days that the light was on.

Q13 This question drew very little interest from candidates. Most attempts gained the first
couple of answers and then differentiated to find the pdf of Y. In the attempts to find E(Y) and
E(Y 2), most candidates rightly attempted integration by parts, although some coefficients went
astray when either making a substitution or comparing the integrals with the corresponding ones
of the standard normal distribution. Slightly surprisingly, it was relatively common to find E(Y 2)
correct but the variance incorrect, as candidates failed to make this modest extra step without
error.

T F Cross
Principal Examiner

-3-
561
STEP 3 2012 Examiners’ report 

The number of candidates attempting more than six questions was, as last year, about 25%, though 
most of these extra attempts achieved little credit. 

1.  In spite of the printing error at the start of the question, two thirds of the candidates 
attempted this question.  Most candidates earned a quarter of the marks by obtaining z in terms of y 
in part (i), and then went no further.  Some candidates realised the significance of the first line of the 
question that the expression given was an exact differential, and those that did frequently then 
scored highly.  Some candidates found their own way through having obtained z in terms of y in part 
(i), then making y the subject substituted back to find a second order differential equation for z, 
which they then solved and hence completed the solution to each part. 

2.  Three quarter s of the candidates attempted this question, making it the second most 
popular, and one of the two most successful.  Generally, part (i) was successful attempted, though at 
times marks were dropped through insufficient explanation and quite a few struggled to deal with 
the “remainder term”.  Some candidates expanding the brackets worked with the second and third, 
the fourth brackets etc., only including the first bracket last.  About half the candidates considered 
the product of all of the denominators in part (ii) and replicated the method for the first part, whilst 
others used the results from part (i), replacing    by    and employing the factorisations of sums 
and differences of cubes.  Full marks were not uncommon on this question, nor were half marks. 

3.  Two thirds of candidates attempted this question, but generally, with only moderate success 
earning just less than half marks.  The vast majority of candidates (more than 85%) did not observe 
that, regardless of the case, the two parabolas “touch exactly once”, dropping 4 or 5 marks 
immediately.  However, most managed to obtain the three results in part (ii), though a few seemed 
to forget to derive that for k.  Unaccountably, many threw away the final marks, only considering the 
case   1 .  

4.  Just over 70% of the candidates attempted this, with marginally less success than question 3.  
Lots of attempts relied on manipulating series for e, and would have struggled had the first two 
results not been given, and even so, there were varying levels of success and conviction.  This 
approach fell apart in the this part with the cubic term.  Some candidates used a generating function 
method successfully with an  ∑  .  However, whilst this worked well for part (i), it got 
!
very nasty for part (ii).  There were lots of sign errors with the log series in part (ii), having begun 
well with partial fractions. 

5.  This was only very slightly more popular than question 4, though with the same level of 
success.  A lot of candidates scored just the first 5 marks, getting as far as completing the 
simplification in part (i) (b) , but then, being unable to apply it for the final result, and then making 
no progress with part (ii).  The biggest problem was that candidates ignored the definitions given at 
the start of the question, most notably that “a and b are rational numbers”.   The other common 
problem was that candidates chose a simple value for    such as     or     rather than for  cos  such 
as    .  In part (ii), quite frequently, candidates substituted   √2  , and   √2   and 
some then successfully found solutions.  For part (ii) (c), a method using  cosh  and  sinh  was not 
unexpected, although the comparable one with  sec  and  tan  was quite commonly used too. 

562
6.  Two thirds attempted this, with less success than its three predecessors.  Very few indeed 
scored full marks, for even those that mastered the question rarely sketched the last locus correctly, 
putting in a non‐existent cusp.  Most candidates managed the first part, good ones the second part 
too, and only the very best the third part.  Quite a few assumed the roots were complex and then 
used complex conjugates, with varying success.  Many candidates lost marks through careless 
arithmetic and algebraic errors.  Given that most could do the first part, it was possible for 
candidates to score reasonably if they took care and took real parts and imaginary parts correctly. 

7.  Two thirds attempted this too, with marginally greater success than question 2.  Most did 
very well with the stem, though a few were unable to obtain a proper second order equation.  Those 
that attempted part (i) were usually successful.  The non‐trivial exponential calculations in part (ii) 
caused problems for some making computation mistakes whilst others were totally on top of this.  
Part (iii) tested the candidates on two levels, interpreting the sigma notation correctly, and 
recognising and using the geometric series.  Some managed this excellently. 

8.  This was the most popular question attempted by over 83% of candidates, and the third 
most successful with, on average, half marks being scored.  Part (i) caused no problems, though 
some chose to obtain the result algebraically.  Part (ii) was not well attempted, with a number 
stating the two values the expression can take but failing to do anything else or failing with the 
algebra.  Part (iii) was generally fairly well done although frequently the details were not quite tied 
up fully. 

9.  The second least popular question attempted by only a couple of dozen candidates with very 
little success, less than any other question.  The problem was none of the candidates appreciated 
how to handle the algebra to obtain the first result, even if they had obtained the equations of 
motion.  Unfortunately, they rarely had the full set of equations of motion.  As a consequence, they 
made no progress on the second part. 

10.  This was the most popular of the non‐Pure questions, being attempted by about a sixth of 
candidates, and with more success than all but three questions on the paper.  Many attempts failed 
to include a clear, legible, accurate diagram, and so an unclear mess of variables invariably failed to 
lead to satisfactory conclusions.  On the other hand, the general standard of mechanics was above 
average, and the initial energy equation was usually correct.  Many candidates came to grief with 
the general energy equation, confusing signs.  A good number of strong candidates ploughed 
straight through correctly, and all who did so, then gained credit at the end for using the 
discriminant to demonstrate that R is non‐zero. 

11.  This was slightly less popular than question 10, and slightly less success was achieved.  Most 
candidates correctly evaluated the kinetic and potential energies of the particle, and the kinetic 
energy of the rope.  However they had more difficulty finding the potential energy of the rope, and 
put themselves at an unnecessary disadvantage by not explaining their logic.  There were different 
ways of splitting up the rope, which one they used they frequently failed to make clear, and likewise 
those calculating potential energy relative to a reference point failed to make the choice of that 
point clear.  The second part of the question was done very well using the result given for the first 
part.  The last part was fairly easy, but quite a few candidates did not justify the logic fully. 

563
12.  Under  9% of candidates attempted this, though the level of success was comparable with 
that achieved in questions 3 to 6, 10 and 11.  The derivation of the pdf was, in many cases, the 
stumbling point whether being found directly, or via the cpf, lacking clear explanation.  The 
expectation caused few problems.  The second part reflected the first in each respect. 

13.  Even fewer attempted this than question 9.  It was the second least successfully attempted 
question.  Generally, part (i) was reasonably attempted although a number of attempts were very 
unconvincing as candidates failed to approach this as conditional probability.  Hardly any got 
properly to grips with the second part, though some cashed in with the final variance result. 

564
Explanation of Results STEP 2012

All STEP questions are marked out of 20. The mark scheme for each question is designed to reward
candidates who make good progress towards a solution. A candidate reaching the correct answer will
receive full marks, regardless of the method used to answer the question.

All the questions that are attempted by a student are marked. However, only the 6 best answers are used in
the calculation of the final grade for the paper.

There are five grades for STEP Mathematics which are:

S – Outstanding
1 – Very Good
2 – Good
3 – Satisfactory
U – Unclassified

The rest of this document presents, for each paper, the grade boundaries (minimum scores required to
achieve each grade), cumulative percentage of candidates achieving each grade, and a graph showing the
score distribution (percentage of candidates on each mark).

STEP Mathematics I (9465)

Grade boundaries
Maximum Mark S 1 2 3 U
120 93 77 54 35 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 5.9 18.9 53.7 82.4 100.0

Distribution of scores
3.0%

2.5%

2.0%
Percent

1.5%

1.0%

0.5%

0.0%
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics I

www.admissionstests.cambridgeassessment.org.uk
565
STEP Mathematics II (9470)

Grade boundaries
Maximum Mark S 1 2 3 U
120 91 72 60 31 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 9.0 25.7 41.9 85.2 100.0

Distribution of scores
2.5%

2.0%
Percent

1.5%

1.0%

0.5%

0.0%
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics II

STEP Mathematics III (9475)

Grade boundaries
Maximum Mark S 1 2 3 U
120 84 65 53 32 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 12.7 39.2 59.6 85.2 100.0

Distribution of scores
5.0%

4.0%
Percent

3.0%

2.0%

1.0%

0.0%
0 10 20 30 40 50 60 70 80 90 100 110 120
Score on STEP Mathematics III

www.admissionstests.cambridgeassessment.org.uk
566
We very much regret a printing error in the STEP Mathematics Paper III taken on 27 June, which resulted in
question 1 (an optional question) being affected.

We have taken steps to identify the candidates who may have been affected and to ensure that this does
not affect their applications for university places.

The mark scheme for question 1 was adjusted to reflect the change to the question, and markers were
trained accordingly. After marking was complete, the scripts of all candidates who attempted question 1,
together with any representations made on behalf of candidates, were considered individually by a team led
by the Chief Examiner. Where a candidate was found to have been placed at a disadvantage, appropriate
adjustments were made following the grading meeting.

We are undertaking a full review of our processes to make sure that errors of this kind are not repeated.

www.admissionstests.cambridgeassessment.org.uk
567
STEP Solutions
2012

Mathematics
STEP 9465/9470/9475

November 2012

568
The Cambridge Assessment Group is Europe's largest assessment agency
and plays a leading role in researching, developing and delivering
assessment across the globe. Our qualifications are delivered in over 150
countries through our three major exam boards.

Cambridge Assessment is the brand name of the University of Cambridge

Local Examinations Syndicate, a department of the University of Cambridge.
Cambridge Assessment is a not-for-profit organisation.

This mark scheme is published as an aid to teachers and students, to indicate

the requirements of the examination. It shows the basis on which marks were
awarded by the Examiners. It does not indicate the details of the discussions
which took place at an Examiners’ meeting before marking commenced.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the
relevant knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question

papers and the Report on the Examination.

Cambridge Assessment will not enter into any discussion or correspondence

in connection with this mark scheme.

© UCLES 2010

More information about STEP can be found at:

http://www.atsts.org.uk

569
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 4
STEP Mathematics II 8
STEP Mathematics III 18

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STEP I ‐ Hints and Solutions 

Question 1 

The relationships for    and   can be obtained by substituting the coordinates of the three known 

points into the equation of the line, or by using the formula for calculating the gradient from a pair 
of points. 

The formula for the sum of the distances is then easy to find and differentiation with respect to   
will allow the minimum value to be found. 

Similarly, the distance for the second part can be written as   and again differentiation can 
be used to find the value of   for which the minimum occurs. Since the minimum value of 
 occurs at the same value of   as the minimum value of  , the differentiation 
can be simplified by just differentiating   with respect to  . It is then a simple matter to 
write this answer in a form similar to that of the first part. 

Question 2 

To sketch the graph it is important to know where the stationary points are. Either by considering 
the graph of   or by differentiating it can be seen that there are two minima and one 
maximum. 

The equation in the second part can be rearranged to show that the solutions correspond to the 
intersections of the graph of   and a straight line. The case requiring care is the two 
solution case as this must include the straight line which touches the two minima. 

For the next part of the question, differentiation of the equation twice leads quickly to the two 
possible values of  . Both cases then need to be considered, but it should be clear that one graph is 
a reflection of the other in the  ‐axis, so the sets of values for   will be the same for both cases. 

The final graph should clearly have a minimum for some negative value of  .   will still be 0 at 
, so there will be two points of inflexion. 

Question 3 

The sketch of the graph, including a chord and tangent should not cause much difficulty. Adding the 
line   should show that the area under the graph lies between two triangles, both with a base 
of length   and with heights   and  . Integrating the function between the two limits and then 
rearranging will give the correct relationship. 

For the second part of the question a different diagram is needed, this time showing the area under 
a curve contained within a trapezium and with a trapezium contained within it. The differentiation 

571
and integration of   will produce the   expressions required in the final answer and so the 
vertical lines   and   can be used to define the regions. 

Question 4 

The equations of the tangents at   and   should be easy to find and then the solution of 

simultaneous equations will give the required coordinates for  . Similarly, the equations for the 
normal should be easy to find, but it is more difficult to find simplified expressions for the solution to 
the simultaneous equations (which are useful for the final part of the question). The factorisation of 
the difference between two cubes,   will be useful for avoiding a lot 
of algebraic manipulation. If this expression is simplified correctly then the final result becomes 
straightforward. 

Question 5 

The most obvious approaches to the first integration are to integrate by parts or to use the identity 
 and then make a substitution. The second integration can also be evaluated by 
integrating by parts, but the identity for   is not as useful. 

For the third integration it is necessary to rewrite the integral in a form from which the previous 
results can be applied. The first point to note is that the expression within the logarithm is not a 
simple cosine function and so the first step to making the expression similar to those used previously 

is to rewrite it in the form  . Once this is done, the substitution of  , with 

some knowledge of the relationship between sine and cosine graphs should reduce the integral to a 
combination of the two previous ones. 

Question 6 

By writing down expressions for the height of the pole using the tangent of each of the angles of 
elevation, the problem is quickly reduced to a two dimensional problem about lines within a circle. 
The simplest way to tackle this is to observe that the triangles in the diagram are similar to each 
other, but approaches working with various right angled triangles also lead to the correct solution. 

The proof of the identity should be straight forward for those familiar with the commonly used 
trigonometric relationships and the inequality is then easily found by considering the consequence 
of the constraint placed on p and q. Given that cot is a decreasing function in the required range of 
values, the final result follows easily. 

Question 7 

By substituting   for all of the terms in the recurrence relation the result for the first part should 
follow easily. For the second part, the two values   and   can be substituted in both orders into 
the relation, giving two equations in   and  . Solving these two equations then leads to the correct 
set of possibilities. 

572
Following the same principle, the substitution of  ,   and   into the relation can be done in three 
different ways, leading to three simultaneous equations. Solving these equations gives an equation 
in   and  , which can be solved to give the two different cases. The case where   is easy 
to check, and for the case where   it should be noticed that this 
can only occur if  . 

Question 8 

The new differential equation follows quite easily once the substitution given has been followed. It is 
an example of a differential equation that can be solved by separating the variables and so by 
evaluating the two integrals that are reached the required form can be obtained. 

The same substitution will also reduce the second differential equation to a simpler form and it is 
again an example that can be solved by separating the variables. It should be clear that partial 
fractions are an appropriate method for evaluating the integral that is required. 

Question 9 

The first part of the question is a straightforward calculation involving the equations for motion 
under uniform acceleration. It is important to explain the reason for choosing the positive square 
root however. The final result follows from correct use of an identity for  . 

It is also easy to find an expression for the range once the time has been calculated and further 
application of an identity for   will give an expression for this in terms of  . Differentiation 

with respect to   will then give the result that the maximum value occurs when  . The final 

part of the question can be solved by substituting the appropriate values of   into the formula for 
the range. 

Question 10 

Although the equation looks complicated, calculations for the time that it takes the stone to drop 
and the time for the sound to return allow the first relationship to be deduced quite easily. The 
second relationship can be shown by simplifying the expression for   and showing that it is equal to 

. This can then be rearranged to give  . The final part of the question is then a 
substitution of the values given into the formula to obtain the estimate. 

Question 11 

While the diagram may look a little more complicated than the standard questions on this topic, the 
first section of this question requires the usual steps to establish a pair of simultaneous equations. 
The difference in the second part of the question is that the acceleration cannot be assumed to be 
constant (as the pulley in the middle of the diagram is able to move) and the important extra 
relationship that is needed is the relationship between the accelerations at the three points (the two 
particles and the pulley). 

573
Question 12 

The probabilities for a failure in each year long period need to be calculated by evaluating the 
integral and from these it is possible to construct a tree diagram from which the probability can be 
calculated. The final part of the question is simply the calculation of a conditional probability. As 
always with conditional probability the important step is to deduce which two probabilities need to 
be calculated. 

Question 13 

There are a number of ways to approach this problem. The most obvious is to work out how many 
possibilities there are for each number of digits. A clear method for categorising these is needed to 
work out the number of possibilities in each case. For example, if there are 4 different digits then 
there are five choices for the digits to be used and four choices for the digit to be repeated.  There 
are then ten choices for the positions of the repeated digits and 3! choices for the order of the 
remaining digits. This gives 1200 altogether. 

574
 Hints & Solutions for Paper 9470 (STEP II) June 2012

Question 1
To be honest, the binomial expansions of (1  x)n, in the cases n = 1, 2, are used so frequently within AS-
and A-levels that they should be familiar to all candidates taking STEPs. Replacing x by xk is no great
further leap.

 
2
The general term in 1  x 6 is easily seen to be (n + 1) x6n and the x24 term in 1  x 6 
1  x 3 comes  
2
1

from 1.x24 + 2x6.x18 + 3x12.x12 + 4x18.x6 + 5x24.1, so that the coefficient of x24 is 1 + 2 + 3 + 4 + 5 = 15,
arising from a sum of triangular numbers. Thus, the coefficient of xn is
 0 if n  6k  {1, 2, 4, 5}
1
 2 (k  1)(k  2) if n  6k  3
 1 (k  1)(k  2) if n  6k
2
which is most easily described without using n directly, as here.

   
In (ii), f(x) = 1  2 x 6  3x 12  4 x 18  5 x 24  ... 1  x 3  x 6  x 9  ... 1  x  x 2  x 3  ... and the x24 term
comes from
1.1.5x24 + 1.x6.4x18 + 1.x12.3x12 + 1.x18.2x6 + 1.x24.1
+ x3.x3.4x18 + x3.x9.3x12 + x3.x15.2x6 + x3.x21.1
+ x6.1.4x18 + x6.x6.3x12 + x6.x12.2x6 + x6.x18.1
+ x9.x3.3x12 + x9.x9.2x6 + x9.x15.1
+ x12.1.3x12 + x12.x6.2x6 + x12.x12.1
+ x15.x3.2x6 + x15.x9.1
+ x18.1.2x6 + x18.x6.1
+ x21.x3.1
+ x24.1.1
giving the coefficient of x24 as 15 + 2  (10 + 6 + 3 + 1) = 55.

However, there are lots of ways to go about doing this. For instance ...
Note that, because every non-multiple-of-3 power in bracket 3 is redundant, the x24 term comes from
considering f(x) = 1  x 6   1  x 
2 3 2
  
= 1  2 x 6  3 x 12  4 x 18  5 x 24  ... 1  2 x 3  3 x 6  4 x 9  ... .
Again, every non-multiple-of-6 power in this 2nd bracket is also redundant, so one might consider only
 
f(x) = 1  3x 6  5 x 12  7 x 18  9 x 24  ... 1  2 x 6  3x 12  4 x 18  5 x 24  ... 
from which the coefficient of x24 is simply calculated as 1  5 + 3  4 + 5  3 + 7  2 + 9  1 = 55. This
result, in some form or another, gives the way of working out the coefficient of x6n for any non-negative
n
integer n. It is immediately obvious that it is  (n  1  r )(2r  1) which turns out to be the same as
r 0
n 1

r
r 1
2
 16 (n  1)(n  2)(2n  3) .The proof of this result could be by induction or direct manipulation of the

standard results for r and r2.

The coefft. of x25 is 55, the same as for x24, since the extra x only arises from replacing 1 by x, x3 by x4,
etc., in the first bracket’s term (at each step of the working) and the coefficients are equal in each case.

In the case when n = 11, the coefficient of x66 is 12  1 + 11  3 + 10  5 + … + 2  21 + 1  23 = 650.

575
Question 2
Firstly, pq ( x)  has degree mn.

(i) Deg[p(x)] = n  Deg[p(p(x))] = n2 & Deg[p(p(p(x)))] = n3.

Deg[LHS]  max n 3 , n  while RHS is of degree 1. Therefore the LHS is not constant so n = 1 and p(x)
is linear. Setting p(x) = ax + b  p(p(x)) = a(ax + b) + b = a2x + (a + 1)b and
p(p(p(x))) = a[a2x + (a + 1)b] + b = a3x + (a2 + a + 1)b.
Then a x + (a + a + 1)b – 3ax – 3b + 2x  0  (a3 – 3a + 2)x + (a2 + a – 2)b  0
3 2

 (a – 1)(a2 + a – 2)x + (a2 + a – 2)b  0

 (a2 + a – 2)[(a – 1)x + b]  0
 (a + 2)(a – 1)[(a – 1)x + b]  0
We have, then, that a = –2 or 1. In either case, b takes any (arbitrary) value and the solutions are thus
p1(x) = –2x + b and p2(x) = x + b.

(ii) Deg[RHS] = 4 while Deg[LHS]  max n 2 , 2n, n  , so it follows that n = 2 and p(x) is quadratic.
Setting p(x) = ax2 + bx + c, we have
2p(p(x)) = 2a(ax2 + bx + c)2 + 2b(ax2 + bx + c) + 2c
= 2a a 2 x 4  2abx 3  2acx 2  b 2 x 2  2bcx  c 2  + 2b(ax2 + bx + c) + 2c
 
3 p( x)  = 3 a 2 x 4  2abx 3  2ac  b 2 x 2  2bcx  c 2 and –4p(x) = – 4ax2 – 4bx – 4c.
2

Thus, LHS = 2a 3  3a 2 x 4  4a 2 b  6ab x 3  2ab 2  4a 2 c  2ab  3b 2  6ac  4a x 2

+ 4abc  2b 2  6bc  4b x  2ac 2  2bc  2c  3c 2  4c  ,
while the RHS = x4.
Equating terms gives
x4 ) 2a3 +3a2 – 1 = 0  (a  1) 2 (2a  1)  a = –1 or 12
x3 ) 2ab(2a + 3) = 0  b = 0
2
x ) 2a(2ac + 3c – 2) = 0  c = 2 when a = –1; i.e. p1(x) = – x2 + 2
OR c = 12 when a = 12 ; i.e. p2(x) = 12 x 2  1 .
Note that there are two sets of conditions yet to be used, so the results obtained need to be checked
(visibly) for consistency:
x1 ) 2b(2ac + b + 3c – 2) = 0 checks and x0 ) c(2ac + 3c – 2) = 0 checks also.

Question 3
1
It helps greatly to begin with, to note that if t  x 2  1  x , then  x 2  1  x . These then give the
t
dx 1 1  2
result x  12 t  12 t 1 , from which we find   t and (changing the limits) x : (0, )  t : (1, ),
dt 2 2

f 
  
1 1  1 
so that x  1  x dx =  f t   2 1  2  dt = 2  f x 1  2  dx , as required.
2 1

0 1  t  1  x 


1 1

x 
For the first integral, I1 = 2
dx , we are using f(x) = in the result established initially.
0
2
1  x x2
 
1 
 x 
 1  1  1
Then I1 = 1
2 1 1  x 2 . x 2 dx = 1
2
2
 x  4 dx = 1
2  
 x 3x 3  =
1
2
0  1  13   23 .
1  1

576
In the case of the second integral, the substitution x = tan  dx = sec2 d . Also 1  x 2  sec and
the required change of limits yields 0, 12    0,   . We then have
1
 1

sec
3
 
2 2
1
I2 =  d = 0  sec  tan   d [Note the importance of changing to sec and tan]
0 1  sin  
3

1
 
2
sec x2 1
0 sec  tan  3 .sec  d = 
x 
2
= 3
dx .
0
2
1  x
 1
1
t  
 t   t  1 , so that
2 2
We now note, matching this up with the initial result, that we are using f(t) =
t3 2t 4
 
 t 2 1  t 2 1 1 
 t   1 2
I2 = 12   2 . 4  dt = 1
4
2
 2t  4  t 6 dt = 1
4   
 t 3t 3 5t 5  =
1
4
0  1  23  15   157 .
1
t   2t  1 1

Question 4
(i) This first result is easily established: For n, k > 1, nk + 1 > nk and k + 1 > k so (k  1)  n k 1  k  n k
1 1
 k 1
 k (since all terms are positive).
(k  1)n kn
 1 1 1 1 1 1 1
Then ln1     2  3  4  5  ... (a result which is valid since 0 <  1 )
 n  n 2n 3n 4n 5n n
1  1 1   1 1  1
   2  3    4  5   ...  since each bracketed term is positive, using
n  2n 3n   4n 5n  n
A1
1 n
1  1
the previous result. Exponentiating then gives 1   e n  1    e .
n  n

(ii) A bit of preliminary log. work enables us to use the ln(1 + x) result on
 2 y 1  1   1   1 1 1 1 1 
ln  = ln1    ln1   =       ... 
 2y 1
2 3 4 5
 2y   2 y   2 y 2( 2 y ) 3(2 y ) 4( 2 y ) 5(2 y ) 
 1 1 1 1 1 
–    2
 3
 4
 5
 ... 
 2 y 2( 2 y ) 3(2 y ) 4( 2 y ) 5(2 y ) 
 1 1 1  1
= 2  3
 5
 ...  (since all terms after the first are positive).
 2 y 3(2 y ) 5(2 y )  y
1
Again, note that we should justify that the series is valid for 0 <  1 i.e. y > 1
2 in order to justify the
2y
y
 2 y  1
use of the given series. It then follows that ln  > 1, and setting y = n + 1
(the crucial final step)
 2y 1
2

n 1 n 1
 2n  2  2  1 2
gives ln   1  1    e.
 2n   n

577
(iii) This final part only required a fairly informal argument, but the details still required a little bit of care
in order to avoid being too vague.
n 1 n 1 n n
 1 2  1  1 2  1  1
As n  , 1    1    1    1    1   1   from above and e is squeezed
 n  n  n  n  n
into the same limit from both above and below.

Question 5
With any curve-sketching question of this kind, it is important to grasp those features that are important
and ignore those that aren’t. For instance, throughout this question, the position of the y-axis is entirely
immaterial: it could be drawn through any branch of the curves in question or, indeed, appear as an
 1 
asymptote. So the usually key detail of the y-intercept, at  0 , 2  in part (i), does not help decide
 a 1 
what the function is up to. The asymptotes, turning points (clearly important in part (ii) since they are
specifically requested), and any symmetries are important. The other key features to decide upon are the
“short-term” (when x is small) and the “long-term” (as x   ) behaviours.

In (i), there are vertical asymptotes at x = a – 1 and x = a + 1; while the x-axis is a horizontal asymptote.
There is symmetry in the line x = a (a consequence of which is the maximum TP in the “middle” branch)
1
and the “long-term” behaviour of the curve is that it ultimately resembles the graph of y = 2 .
x

(ii) Differentiating the function in (ii) gives

g ( x) 
2
    
( x  b) ( x  a ) 2  1  ( x  a ) ( x  b) 2  1
 2

( x  a )  1 ( x  b)  1
2 2 2

and setting the numerator = 0  ( x  a)( x  b)x  a  x  b  x  a  x  b  0 . Factorising yields
a  b  (a  b) 2  4ab  4
 
(2x – a – b) x  (a  b) x  (ab  1)  0 , so that x  (a  b) or
2 1
2
2
.

In the first case, where b > a + 2 (i.e. a + 1 < b – 1), there are five branches of the curve, with 4 vertical
asymptotes: x = a  1 and x = b  1. As the function changes sign as it “crosses” each asymptote, and the
1
“long-term” behaviour is still to resemble y = 2 , these branches alternate above and below the x-axis,
x
with symmetry in x  2 (a  b) .
1

In the second case, where b = a + 2 (i.e. a + 1 = b – 1), the very middle section has collapsed, leaving
only the four branches, but the curve is otherwise essentially unchanged from the previous case.

Question 6
A A quick diagram helps here, leading to the important observation, from the
 GCSE geometry result “opposite angles of a cyclic quad. add to 180o”, that
b BCD = 180o –  . Then, using the Cosine Rule twice (and noting that
a cos(180o – ) = – cos ):
D in BAD: BD2 = a2 + d 2 – 2ad cos
B in BCD: BD2 = b2 + c2 + 2bc cos
a2  b2  c2  d 2
b c Equating for BD2 and re-arranging gives cos =
2(ad  bc)
C

578
Next, the well-known formula for triangle area,   12 ab sin C , twice, gives Q = 12 ad sin   12 bc sin  ,
2Q 4Q
since sin( – ) = sin . Rearranging then gives sin = or .
ad  bc 2(ad  bc)

Use of sin  + cos  = 1 

2 16Q 2
2

a2  b2  c2  d 2 
2

= 1 and this then gives the printed

4(ad  bc) 2 4(ad  bc) 2

result, 16Q 2 = 4ad  bc   a 2  b 2  c 2  d 2 .
2

2

  
Then, 16Q 2 = 2ad  2bc  a 2  b 2  c 2  d 2 2ad  2bc  a 2  b 2  c 2  d 2 by the difference-of-two-
squares factorisation
 
= [b  c] 2  [a  d ] 2 [a  d ] 2  [b  c] 2 
= [b  c]  [a  d ][b  c]  [a  d ][a  d ]  [b  c][a  d ]  [b  c]
using the difference-of-two-squares factorisation in each large bracket
= b  c  d  a a  b  c  d a  c  d  b a  b  d  c  .
Splitting the 16 into four 2’s (one per bracket) and using 2s = a + b + c + d

 Q2 =
2s  2a  2s  2b  2s  2c  2s  2d   ( s  a)(s  b)(s  c)(s  d ) .
2 2 2 2
Finally, for a triangle (guaranteed cyclic), letting d  0 (Or s – d  s Or let D  A), we get the result
known as Heron’s Formula:  = s ( s  a )( s  b)( s  c) .

Question 7
Many of you will know that this point G, used here, is the centroid of the triangle, and has position vector
g = 13 x1  x 2  x 3  .
Then GX 1  x1  g 2x1  x 2  x 3  and so GY1   13 1 2x1  x 2  x 3  , where 1 > 0.
1
3

Also OY1  OG  GY1  13 x1  x 2  x 3   13 1 2x1  x 2  x 3   13 [1  2 ]x1  [1  1 ](x 2  x 3 )  , the first
printed result.

The really critical observation here is that the circle centre O, radius 1 has equation | x |2 = 1 or x . x = 1,
where x can be the p.v. of any point on the circle.
Thus, since OY1  OY1  1 , we have

1 1
9
(1  2 ) 1
2
 2(1  1 ) 2  2(1  21 )(1  1 )x1 (x 2  x 3 )  2(1  1 ) 2 x 2  x 3 
 9  1  41  41  2  41  21  2(1  21 )(1  1 )x1 (x 2  x 3 )  2(1  1 ) 2 x 2  x 3
2 2

 0  3(1  1 )(1  1 )  (1  21 )(1  1 )x1 (x 2  x 3 )  (1  1 ) 2 x 2  x 3

As 1 > 0, 0  3(1  1 )  (1  21 )x1 (x 2  x 3 )  (1  1 )x 2  x 3
 0  3  31  x1 x 2  x 2  x 3  x 3  x1   1 (x 2  x 3 )  21 x1 x 2  x1 x 3 

3       
 1  , using   x 2  x 3 ,   x 3  x1 and   x1  x 2 .
3    2  2

3        3       
Similarly,  2  and 3  .
3    2  2 3    2  2

579
 GX i 1 GX 1 GX 2 GX 3 1 1 1 9  (     )  4(     )
Using  (i = 1, 2, 3),      
GYi i GY1 GY2 GY3 1 2 3 3       
9  3(     )
= = 3.
3       
n
GX i
[Interestingly, this result generalises to n points on a circle:   n .]
i  1 GYi

Question 8
 2   2  q2
    q (> 0)   2  2   2  q 2   2   2  q 2  2   2  the opening

 2   2  q2
result, 20.


un  q 2
2

un 1  etc.  u n  u n  1u n  1  q 2  u n  1  u n  2 u n (since the result is true at all stages) and

2 2

un 1
equating for q2  u n u n  u n  2   u n  1 u n  1  u n  1  .

un  un  2 un 1  un  1 u  un 1
Now this gives  which  n  1 is constant (independent of n). Calling
un 1 un un
this constant p gives u n  1  pu n  u n  1  0 , as required. In order to determine p, we only need to use the
un 1  un 1
fact that p = for all n, so we choose the first few terms to work with.
un
 2  q2

 2  q2 u  u2   2   2  q2
u2 = and p  0 =  .
 u1  
 2  q2  2   2  q2
Alternatively, u2 =  = = p –   p 
 
2
  2  q2 
    2  q 2
 q   q  
= 
2 2 2 2 2
and u3 =  p    p 
    2  q2 
  
  
 2
 q2      q  =
2 2
 2  q2  2
2 2

   q 
= 2 2

since  2  q 2  0 as u2 non-zero (given). Since p is consistent for any chosen , , the proof follows
inductively on any two consecutive terms of the sequence.

Finally, on to the given cases.

  2   2  q2 
If  >  + q, u n  1  u n  ( p  1)u n  u n  1    1u n  u n  1
  
> (2  1)u n  u n 1 by the initial result
> u n  u n 1
Hence, if u n  u n  1 > 0 then so is u n  1  u n . Since  > , u 2  u1  0 and proof follows inductively.

580
 If  =  + q then p = 2 and u n  1  u n  u n  u n  1 so that the sequence is an AP.
Also, u0 =  , u1 =  + q, u2 =  + 2q, …  the common difference is q (and we still have a strictly
increasing sequence, since q > 0 given).

Question 9
In the standard way, we use the constant-acceleration formulae to get
x = ut cos and y = 2h – ut sin – 12 gt2 .
a ga 2
When x = a, t  . Substituting this into the equation for y  y  2h  a tan   2 sec 2  .
u cos  2u
As y > h at this point (the ball, assuming it to be “a particle”, is above the net), we get
ga 2 1 2(h  a tan  )
h  a tan   2 sec 2   2  , as required.
2u u ga 2 sec 2 

For the next part, we set y = 0 in y = 2h – ut sin – 1

2 gt2 and solve as a quadratic in t to get
 2u sin   4u 2 sin 2   16 gh
t ... (the positive root is required).
2g
 u 2 sin 2   4 gh  u sin  
Setting x = (u cos )t and noting that x < b , u cos   b
 g 
 
bg
 u 2 sin 2   4 gh   u sin  .
u cos 

There are several ways to proceed from here, but this is (perhaps) the most straightforward.
b 2 g 2 sec 2 
Squaring  u 2 sin 2   4 gh  2
 2bg tan   u 2 sin 2 
u
b 2 g sec 2 
Cancelling u 2 sin 2  both sides & dividing by g  4h   2b tan 
u2
1 2(2h  b tan  ) 1
Re-arranging for   2
u 2
b 2 g sec 2  u
1 2(h  a tan  ) 2(2h  b tan  ) 2(h  a tan  )
Using the first result, 2  , in here  
u a g sec 
2 2
b 2 g sec 2  a 2 g sec 2 
Re-arranging for tan  ab(b  a) tan   hb 2  2a 2  , which leads to the required final answer
hb 2  2a 2 
tan   . However, it is necessary (since we might otherwise be dividing by a quantity that
ab(b  a )
could be negative) to explain that b > a (we are now on the other side of the net to the projection point)
else the direction of the inequality would reverse.

581
Question 10
As with many statics problems, a good diagram is essential to successful progress. Then there are
relatively few mechanical principles to be applied ... resolving (twice), taking moments, and the standard
“Friction Law”. It is, of course, also important to get the angles right.

Taking moments about M :

R1 F2 R1 a sin = R2 a sin + F1 a cos + F2 a cos

Using the Friction Law : F1 =  R1 and F2 =  R2

R2 O Dividing by cos and re-arranging

r R1 tan = R2 tan +  R1 +  R2
B  (R1 – R2) tan =  (R1 + R2)
r a

a M
 
A

F1 W (or mg)

For the second part, it seems likely that we will have to resolve twice (not having yet used this particular
set of tools), though we could take moments about some other point in place of one resolution. There is
also the question of which directions to resolve in – here, it should be clear very quickly that “horizontally
and vertically” will only yield some very messy results.

Moments about O :  (R1 – R2) r = W r sin sin

Resolving // AB : (R1 – R2) cos +  (R1 + R2) sin = W sin
(Give one A1 here if all correct apart from a – sign)
Resolving r AB : (R1 + R2) sin –  (R1 – R2) cos = W cos
Note that only two of these are actually required, but it may be easier to write them all down first and then
decide which two are best used.
( R  R2 ) cos    ( R1  R2 ) sin 
Dividing these last two eqns.  tan   1
( R1  R2 ) sin    ( R1  R2 ) cos 
( R1  R2 ) cos   ( R1  R2 ) tan  sin 
Using first result,  (R1 + R2) = (R1 – R2) tan  tan  
tan 
( R1  R2 ) sin    ( R1  R2 ) cos 

cos   tan  sin 
 tan   . (There is no need to note that R1 ≠ R2 for then the rod would hve to be
tan 
sin    cos 

positioned symmetrically in the cylinder.)
 cos 2   sin 2   
Multiplying throughout by  cos  tan   = and, using
sin    cos 
2 2 2
1  cos    2 cos 2 
2

a   r2
cos   gives tan    2
r  a 2 1   2 
.
a2 2 a 
2
r
1  2    2 
r r 
 R  R2 
Finally, tan      1  tan  , from the first result, < tan   <  .
 R1  R2 

582
Question 11
Again, a diagram is really useful for helping put ones thoughts in order; also, we are going to have to
consider what is going on generally (and not just “pattern-spot” our way up the line).

Before i th collision After i th collision

u Vi – 1 Vi


i

Pi Bi – 1 Bi

Mass im k  12 i(i  1) m Mass k  12 i(i  1) m

Using the principle of Conservation of Linear Momentum,
CLM  m u + M Vi – 1 = (M + im) Vi (NB V0 = 0) leads to
u 2u 3u nu 2nu
V1  , V2  , V3  , ... , Vn   .
k 1 k 1 2 k 1 2  3 k  2 n (n  1) 2k  n (n  1)
1

u u u

Alternatively, CLM  for all particles gives mu  2m   3m   ......  nm   k  12 n(n  1) mV ,
2 3 n
2nu
and rearranging for V = Vn yields Vn  .
2k  n (n  1)

u 2nu u
The last collision occurs when Vn  , i.e. 
n 1 N ( N  1)  n (n  1) n  1
 2n (n  1)  N ( N  1)  n (n  1)  n (n  1)  N ( N  1)  there are N collisions.

2
N
u N
1
Now, the total KE of all the Pi ’s is 
i 1
1
2 (i m)   =
i
1
2 mu 2  .
i 1 i
2
 u  2 N 
The final KE of the block is N ( N  1)mV N = N ( N  1)m   12 mu  .
1 2 1

 N 1  N 1
2 2

N
1  N 
Therefore, the loss in KE is the difference: 12 mu 2  – 12 mu 2  .
i 1 i  N 1 
N 1
N 1  1 1 1 1  1 1
 1 , the loss in KE is 12 mu 2 1    ...   1   = 2 mu    .
2
Since
N 1 N 1  2 3 N N 1 i2  i 

583
Question 12
This can be broken down into more (four) separate cases, but there is no need to:
P(light on) = p  34  12 + (1 – p)  14  12 = 18 (1  2 p ) , and then the conditional probability
1
(1  p) (1  p )
P(Hall | on) = 18 = .
8 (1  2 p ) (1  2 p )

To make progress with this next part of the question, it is important to recognise the underlying binomial
distribution, and that each day represents one such (Bernouilli) trial. We are thus dealing with B(7, p1),
(1  p )
where p1 = is the previously given answer.
(1  2 p)

For the modal value to be 3, we must have P(2) < P(3) < P(4); that is,
7 7 7 7
 ( p1 ) 2 (1  p1 ) 5   ( p1 ) 3 (1  p1 ) 4 and  ( p1 ) 4 (1  p1 ) 3   ( p1 ) 3 (1  p1 ) 4 .
 2  3  4  3
(1  p )
Using p1 = gives
(1  2 p )
2 5 3 4
 1 p   3p   1 p   3p 
21     35     33 p   51  p   p  145
1  2 p  1 2 p  1 2 p  1 2 p 
and
4 3 3 4
 1 p   3p   1 p   3p 
35     35     1  p   3 p   p  14 .
1 2 p  1 2 p  1  2 p  1 2 p 

Question 13
Working with the distribution Po( = ky2), P(no supermarkets) = e  ky and P(Y < y) = 1 – e  ky .
2 2

Differentiating w.r.t. y to find the pdf of Y  f(y) = 2kπ y e  ky , as given. Then
2


2kπ y 2 e  ky dy . Using Integration by Parts and writing 2kπ y 2 e  ky as y  2kπ y e  ky 
2 2 2
E(Y) =   
0
  
gives E(Y) =  y  e  ky   e  ky dy = 0 + e  ky dy. It is useful (but not essential) to use the
2 2 2
   0  
0 0

1  12 x 2 1  1
simplifying substitution x = y 2k at this stage to get  e dx =  (by the
2
2k 0 2 k 2 k
given result, relating to the standard normal distribution’s pdf, at the very beginning of the question).

2kπ y 3 e  ky dy , and using Integration by Parts and, in a similar way to earlier,
2
Next, E(Y 2) = 
0
 
writing 2kπ y 3 e  ky as y 2  2kπ y e  ky  , E(Y 2) =
2 2  y 2  e  ky 2   2 y e  ky 2 dy
   0 0

 

 
1  ky 2  1   ky 2  1
=0+  2k ye =  e  (using a previous result, or by substitution) =
k 0 k  0 k

1 1 4 
 Var(Y) =   , the given answer, as required.
k 4k 4k

584
STEP 3 2012 Hints and Solutions 

1.  The stem integrates to give    and for part (i) using   1 , the stem gives 

  which can solved for z using the initial conditions, and the integrated stem is a first 
order differential equation for y, which when solved, again with the initial conditions, produces the 
required result.  Part (ii) follows the same pattern, with   2 instead, which has solution  
  . 

2.  The simplification in the opening is   1   , obtained by repeated use of the 

difference of two squares.  A simple algebraic rearrangement, followed by taking a limit, the 
logarithm of both expressions, and differentiation produces the other three results in part (i).  Part 
(ii) can be obtained by replacing    by    in ln 1 ∑ ln 1   from part (i), 
factorising the difference and the sums of cubes and subtracting that part (i) result before 
differentiating.  An alternative is to replicate part (i) using instead the product 

  1 1 1 1 … 1   , but then a little extra 
care is required with the rearrangement, and consideration of the limit. 

3.  The two parabolas, with vertices oriented  in the direction of the positive axes, touch in the 
third quadrant in case (a), and in the first quadrant in the other three cases.  In case (b), there are 
intersections in the second and third quadrants, in (c) in the third and fourth, and in (d), the trickiest 
case, they are in the third quadrant and in the first, between the touching point and the vertex of 
the parabola on the x axis.  The first result of part (ii)is obtained by eliminating y between the two 
equations, the second by differentiating and equating gradients, and the third, by eliminating   , 
(  , from the first result using the second. 

√
The cases that arise are   1 ,  21 ,  2  (d),    ,   √13   ,  2  (d), 
√
and   ,   √13   ,  0  (a). 

4.  Writing     as     and then cancelling the first fraction, give exponential series.  

! ! !
Similarly,  1   can be written as   1 3 1, and   2 1   as  8 1 2
12 1 2 1 , the latter giving the result  21 1 .  Using partial fractions,    can 

be written as  1  , the first term giving a GP, and the other two, log series.  The result for 
part (ii) is thus  12 16 ln 2  . 

5.  For non‐integer rational points, it makes sense to use values of  cos   and  sin   based on 

Pythagorean triples such as 3, 4, 5  or 5, 12, 13,  The technique for (i) (b) can be used for (ii) (a), 
merely by changing the value of m, whereas for (ii) (b), a slightly more involved expression is needed 
such as  cos √ sin  ,  sin √ cos  .  For (ii) (c), there are two alternatives 
that work sensibly,  cosh √ sinh  ,  sinh √ cosh    
or   sec √ tan  ,   tan √ sec .   
 
 

585
A completely different approach for the last part of the question is to write 
   7  and to choose  √2 with nearly any choices of rational 
a and b possible.  Then, as    , and numbers of the form  √2 are a field over the 
operations    and   ,     has to be of the correct form, and then solving for x and y, they 
likewise have to be of the required form. 
Some possible solutions are  
(i)(a)  1,0  and   ,  , (b)  1,1  , 1  , and   ,  (using  1 , cos  ) 
(ii)(a)  1, √2  and  √2, √2  (using 2 , cos  ) , (b)  √2, √2   (using 
3 , 1, cos  ), (c)  √2 , √2   (using  2 ,  3 and cosh  ,  
sinh   or  sec  , tan  ) 
 
6.  Substituting     for z in the quadratic equation and equating real and imaginary parts 
yields the first two results, the imaginary gives two situations, one as required and the other 
substituted into the real gives the second.  The first of these two results substituted into the real 
gives a circle radius 1, centre the origin, whilst the second gives the real axis without the origin.  The 
second quadratic equation succumbs to the same approach giving the real axis without the origin 
(again), and a circle centre  1,0  radius 1 also omitting the origin.  The same approach in the third 
√
case yields the real axis with     and considering the discriminant,  0 and   2.  

On the other hand,  2  produces    

   
 
7.  The second order differential equation   7 6 0  may be obtained by 
differentiating the first equation, then substituting for    using the other equation, and then doing 
likewise for   using the first again.  The solutions for   and   follow in the usual manner.  Parts (i) 

and (ii) yield   2 2  and     respectively.  Part (iii) 
merely requires the sum to be expanded and two geometric progressions emerge so that  
  . 
 
8.  1 , 2 , 3 ,
5  so both expressions in part (i) equal ‐1.  Considering either 
  or    , 
and applying the recurrence relation, both can be found to be zero.  \so the given expression is 
shown to be 1 if n is odd and ‐1 if n is even.  The tan compound angle formula enables the proof in 
part (iii) to be completed once the initial recurrence relation and the result from part (ii) have been 
applied to the expression obtained following algebraic simplification.  Rearranging the result and 
substituting into the required sum gives, by the method of differences,   ∑ tan  

586
9.  Eliminating   ,   ,   , and    between the five equations    ,                  
 ,   ,   , and     yields the required 
result for    .  The only difference in the second part is that the second equation becomes 

 , and so the same elimination yields    , 
which can be seen to be smaller than that in part (i).  The first four equations in this second part give 
 , and so as  0 , the required result follows. 
 
10.  After motion commences, the next at rest position has the string at     to the vertical.  
Conserving energy between the two at rest positions gives   3  .  Conserving energy for the 
general position and resolving radially, bearing in mind that the angle of the radius to the vertical is 
twice the angle of the string to the vertical, and using a double angle formula gives the required 
result.  The discriminant being negative or completing the square demonstrates that the reaction 
force is always positive. 
 
11.  Various approaches can be used to find the energy terms.  If potential energy zero level is 
taken to be at P, then the initial potential energy is   2   .  When the particle has fallen a 
distance x, the kinetic energy of the particle is    , the potential energy of the particle is   , 
the potential energy of the part of the stationary piece of string of length x is   2  , the 

potential energy of the remaining piece of (doubled up) string is   1 2  , 

and the kinetic energy of the shorter moving piece is   2 . This yields the first result and 
differentiation of it yields the second.  As   0 2  ,     , as  0  and the 

denominator is twice a square    0  , the final result follows. 

12.  The sketched region contained by AB, and the two line segments connecting A and B to the 
centre of the triangle.  A simple approach for the pdf is via   ∝ , finding the 
constant of proportionality and then differentiating to give the required result.    . 

For the second part,  192 , and so   . 

13.  That   |   yields the part (i) result.  Similarly, that 

| 0 | 0  yields the first result of part (ii), and that 

  | | | 0 0 | 0 0  yields the second.   

Using    , and so 

  | | | | | |  

587


STEP Examiner’s Report 2013

Mathematics

STEP 9465/9470/9475

October 2013

588
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 3
STEP Mathematics II 8
STEP Mathematics III 11
Explanation of Results 14

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STEP 1 2013 Examiners’ Report 

General Comments
Around 1500 candidates sat this paper, a significant increase on last year. Overall, responses were
good with candidates finding much to occupy them profitably during the three hours of the
examination. In hindsight, two or three of the questions lacked sufficient ‘punch’ in their later parts,
but at least most candidates showed sufficient skill to identify them and work on them as part of
their chosen selection of questions. On the whole, nearly all candidates managed to attempt 4-6
questions – although there is always a significant minority who attempt 7, 8, 9, … bit and pieces of
questions – and most scored well on at least two. Indeed, there were many scripts with 6 question-
attempts, most or all of which were fantastically accomplished mathematically, and such excellence
is very heart-warming.

Comments on individual questions

[Examiner’s note: in order to extract the maximum amount of profit from this report, I would firmly
recommend that the reader studies this report alongside the Hints and Solutions supplied
separately.]

Q1 This question is all about using substitutions to simplify the working required to solve
various increasingly complicated looking equations. It was the most popular question on the paper,
essentially attempted by every candidate (as is the intention). The obvious pitfall of not realising
that the square-root sign indicates the “non-negative square-root” of a quantity was clearly flagged
at the outset. Thus, the only remaining hurdle to fully complete success lay in the need to check the
validity of solutions once found. The mean score on this question was 14/20, and this question thus
represented a successful entré to the paper for almost everyone.

The use of the quadratic formula and the method of completing the square appeared in almost equal
measure throughout the question, although a significant minority of candidates opted to rearrange
and square in both (ii) and (iii). This was not a major obstacle to success in (ii) but led to a quartic
equation in (iii) with which few candidates knew how to make successful progress. The final hurdle
for most candidates lay in a final justification that any roots found (up to four of them, depending
upon the method chosen) were genuinely valid. It is very easy to explain, without the use of direct
verification, that the two roots found via the substitution method are good, but very few candidates
made any attempt to justify their results.

Q2 This was another very popular question, attempted by more than a 1000 of the candidates.
The initial difficulties arose in the interpretation of the integer-part (or floor) function. Candidates’
graphs revealed the difficulties and uncertainties associated with the use of such a function. In
particular, the lack of “jumps” at the endpoints of each unit interval was very prevalent, and many
candidates effectively assumed that the function is an even function. There was also considerable
uncertainty in how to represent whether endpoints were “in” or “out” – the usual convention being
closed dots for “included” and open dots for “not-included”. Also, many candidates failed to show
in their sketch that the function was zero in the interval 0  x < 1, and others drew straight line
segments instead of portions of a reciprocal curve in each unit segment. Pleasingly, however, (at

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least from the candidates’ point of view), it was possible to get quite a few of these bits wrong and
still go on to answer correctly many of the following parts of the question. Thus it was that the
mean mark on the question, at 9/20, was still a respectable one.

In parts (ii) and (iii), it was only necessary for candidates to realise with which portion of the
function they had to deal in order to be undertaking the correct algebra, and the ten marks allocated
to these parts of the question were generally those from which the majority of candidates were
scoring the bulk of their marks. Only the very last part of the question required much thought, and
candidates were not helped by an unwillingness to set down in writing any of their underlying
thoughts, merely opting for statements that seemed to come from nowhere obvious. It was
unfortunate that some considered the function to be defined only on the interval –3  x  3, which
was simply that required for the sketch.

Q3 This vector question was actually very straightforward, though its unfamiliar appearance clearly
put most candidates off, with only around 350 of them making an attempt at it. There were nine
marks available for the first two parts, which were technically undemanding, and it is no
coincidence that the mean score on the question was around 9½/20. I suspect that, for the most part,
this was considered by candidates to be one of those questions that are done towards the end of the
examination in order to bump up their paper total by getting the easier marks at the beginning of the
question, with no real intention of making a complete attempt. Candidates usually gave up part-way
through (iii) where a stab at the “corresponding result for X * (Y * Z)” was required of them, which
was actually just (X * Y) * (X * Z). I imagine this highlights the lack of students’ familiarity with
such properties as distributivity when considering binary operations.

Q4 This was another very popular and high-scoring question (attracting over 1200 attempts and
with a mean mark of more than 10/20). The first part to this question involved two integrals which
can be integrated immediately by “recognition”, although many students took a lot of time and
trouble to establish the given results by substitution and surprising amounts of working. Those
candidates who had found these easy introductory parts especially troubling usually did not proceed
far, if at all, into part (ii). Those who did venture further usually picked up quite a lot of marks.

One of the great advantages to continued progress in the question is that the two integrals in part (ii)
can be approached in so many different ways – the examiners worked out more than 25 slightly
different approaches, depending upon how, and when, one used the identity sec2x = 1 + tan2x; how
one split the “parts” in the process of “integrating by parts”; and even whether one approached the
various secondary integrals that arose as a function of sec x or tan x. This meant that, with care,
most of the marks were accessible, although many candidates clearly got into a considerable tangle
at some stage of proceedings. The most common “howler” was the mix-up between the definite
integrals (i.e. numerical values) given in (i) and their associated unevaluated indefinite integrals
(i.e. functions) which formed part of a subsequent integral.

Q5 This question was usually found to be amongst candidates’ chosen six, attracting almost a
thousand attempts, though on the whole it produced the lowest mean score of the popular pure
maths questions, weighing in at under 7½/20. The initial attraction of the question was undoubtedly
the obvious “circle” nature of the given quadratic form when k = 0, meaning that part (i) was very

591
familiar territory. Unfortunately, there were very few marks allocated to this bit. Part (ii) drew a lot
of unsuccessful work, especially as candidates seemed ill-inclined to extend the requested
factorisation from that of 3x2 + 3y2 + 10xy into that of the full quadratic expression. Even amongst
those who did make that extra step, there were relatively few that grasped the geometric
consequences of the result that AB = 0  A = 0 or B = 0 meant that the solutions amounted to a
line-pair. The question’s demand for a sketch of the solutions meant that most of the marks were
only awarded for candidates who had made this geometric interpretation.

Part (iii) was the genuinely tough part of the question, but substantial help was offered to enable
candidates to make a start on it, which most duly employed. However, working forwards and
backwards through the given substitutions did not make for easy reading and it was clear that many
candidates did not realise the given locus of Q is that of a standard parabola. Several marks were
gained by most candidates, but few made a thorough fist of it.

Q6 Although this question attracted a few more attempts than Q3, it was the lowest-scoring of the
pure maths questions. Confident use of the sigma-notation is clearly in short supply and this was,
perhaps, that feature of the question that deterred most candidates from attempting it. Also, many
attempts were simply from those candidates cherry-picking the opening three marks for proving the
standard “Pascal’s Triangle” result, mostly by proving it directly from the definition of the binomial
coefficient in terms of factorials (which we had decided to allow when setting the paper). This
almost invariably accounted for 3 of the 6.7 marks gained on average for the question as a whole.

Those who proceeded further than this opening result generally fell into a couple of very wide traps:
a careless handling of the terms at the ends of the series (which, being 1, could be replaced by other
binomial coefficients that were also 1) and a failure to consider odd and even cases separately. A
final obstacle, were one needed, lay in the oversight of establishing the validity of the relationship
between the Bn’s and the Fn’s for their starting terms – surprisingly, many candidates failed to
evaluate B0 and B1 correctly.

Q7 Around 1300 candidates attempted this question, making it the second most popular question on
the paper. It was also the second highest-scoring question on average which, if nothing else, pays
tribute to the candidates’ ability to spot the right questions to attempt. In hindsight, this was
possibly a little too straightforward; this was undoubtedly partly due to the appearance of similar
questions (on what are known as homogeneous differential equations) in recent years’ STEPs, but
also to the fact that part (ii) could be solved by the use of the given approach for part (i). It was part
(iii) that required of candidates a stretch of the imagination – the use of y = ux2 – but even this
helped make the question more approachable, as this substitution could also be used to solve part
(ii) if it turned out that candidates got imaginative a bit earlier than anticipated.

For those making essentially correct attempts at parts (ii) and (iii), the only final hurdle to complete
success lay in the hoped-for statement of a domain for the functions which had been found as
solutions. We allowed as obvious the taking of non-negative square-roots (since the given “initial”
values of y are positive – though, in general, candidates should be encouraged to state that they
recognise they are doing this) but expected candidates to indicate a suitable interval for the x values
in each case: the hint lay in the given answer to part (i).

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Q8 Almost 1200 candidates made an attempt at this question, making it fourth favourite, and the
mean score on it was 9.3/20 which, if nothing else, suggests that it wasn’t quite as easy as folks
considered it to be. To begin with, there is a lot to do for the relatively few marks available, and
minor slips over domains and ranges subsequently proved quite costly. Apart from the obvious
errors from those candidates who thought the order of composition occurred the other way around,
and the few who took “ab” to mean the product of the functions a and b, the usual slip-ups were:
thinking that x 2  x , when it is actually | x |, and not realising that the domain of the composite
function fg is just the domain of g. In (ii), although the functions fg and gf look the same (both are
| x |), their domains and ranges are different: fg has domain ℝ and range y  0, while gf has domain |
x |  1 and range y  1.

A lack of a clear grasp of the domains and ranges of h and k in part (iii) was partly responsible for
the poor sketches, although the ability to recognise the asymptote y = 2x was also widespread.
There were even occasions, when sketching the curve for k, that a correctly drawn asymptote was
subsequently labelled as y = –2x simply because of its appearance in the quadrant in which x and y
are both negative.

Q9 This question was the most popular of the applied questions, drawing well over 500
responses, and the most successful of the mechanics questions, with an average score of 8.8/20. It
proved to be a surprisingly good discriminator, giving a good range of marks. The use of constant-
acceleration formulae for the projectile motion provided a routine and straightforward start to the
question, but this was followed by the momentum equation for the collision, which proved trickier,
with quite a few candidates getting to mucos – Mvcos = MwB – mwA but no further. A lot of
candidates resorted to writing down the result mucos = Mvcos without any attempt to justify it.
The second result then found many candidates going round in algebraic circles, and very few indeed
managed to find the answer (not given) to the very final part of the question.

Q10 This question proved to be the least popular question on the paper, eliciting a mere 150
responses. The mean score of 7.7 on it was almost entirely drawn from the first six marks allocated
for obtaining the given result, and then for setting vn = 0 in the following part. This does raise the
thorny issue – during the setting process – of the extent to which (intermediate) answers should be
given in the question, as candidates clearly find great comfort in having something to work towards,
but are otherwise surprisingly weak. Here, for instance, almost any tiny slip-up in working, signs,
etc., inevitably had disastrous consequences for a candidate’s prospect of successful continuation
with the work and very few indeed progressed much beyond the first result.

Q11 A combination of some obviously tricky trigonometry and inequalities meant that this
mechanics question was both unpopular and low-scoring, despite the given answer in (i). Only 300
candidates attempted it, and they averaged a score of 5/20, with most of the marks being scored at
the beginning with correct statements regarding the resolution of forces vertically and horizontally.
In (ii), it was important for candidates to realise (a fact clearly indicated by the question’s wording)
that the condition W > Tsin( + ) would no longer hold; those that recognised the change in the
kinematics did not have too much trouble in working the problem through to its end. However,

593
there were too few who had made it to the end of (i) intact, and these candidates had given up
already without proceeding into part (ii).

Q12 This probability question drew more than 350 responses, scoring just over half-marks on
average. There are many ways to go about part (i), of varying degrees of sophistication: those
opting for elaborate tree diagrams tended to be the least successful. The final part of (i) was really
intended as a test of whether candidates realised that this is the same situation viewed “in reverse”,
so the answer is the same. Very few candidates spotted the symmetry argument or got it correct by
longer methods. Those who had obtained the given result of (i) by one of the more sophisticated
methods had little difficulty in employing a similar argument in (ii), although some did mix up the
roles of the n and the k. A few did the general method and then substituted particular values. Those
who did use a general approach here then fared very well in part (iii) and they usually went on to
apply Stirling’s approximation correctly.

Q13 After Q10, this was the least popular question on the paper, and supplied the poorest average
score on the whole paper of only 2½/20. I have little doubt that the principal reason for both these
factors is the lack of any helpful structure or given answers within the question. Essentially, this
problem is that of the set-up for a game of Solitaire, but stripped of its context. In this game, when a
standard pack of playing cards, suitably shuffled, is laid out at the start, there are seven piles of
cards, and each pile has its final card face up. This particular question is looking for cards of the
same colour (red or black) and denomination (number or J, Q, K and Ace), giving the 26 pairs. This
was, of course, entirely by-the-by as far as candidates were concerned.

Unfortunately, most attempts at this question were abandoned very early on as candidates realised
they didn’t really know what to do. Surprisingly, very few even took the trouble to note that the
defined discrete random variable X could only take the values 0, 1, 2 or 3. Following attempts to
work out the probability for any these outcomes almost invariably consisted of a jumble of fractions
and factorials but without any obvious plan to them, and certainly without any explanatory
indicators as to what might actually be intended. Only P(X = 0), being the easiest of the four cases
to evaluate, was calculated with any degree of success by any of the candidates who attempted the
question.

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STEP 2 2013 Examiners’ report

All questions were attempted by a significant number of candidates, with questions 1 to 3 and 7 the
most popular. The Pure questions were more popular than both the Mechanics and the Probability
and Statistics questions, with only question 8 receiving a particularly low number of attempts within
the Pure questions and only question 11 receiving a particularly high number of attempts.

1. This was the most popular of all of the questions. Overall part (i) of this question was well
answered, although there were a number of candidates who were not able to find the tangent and
intercept even in this first case. Very few attempts at part (ii) of this question involved the use of
sketches. While many attempts at part (iii) recognised the link in the final part with part (ii) of the
question, many of the explanations in this section were not well enough explained to gain full marks.
In the final part it was pleasing to note that many candidates realised that the conditions implied
that the intersection with the y-axis was at a negative value.

2. This was the second most popular question on the paper and the average score was half of
the marks. Despite the instruction in the first part of the question to use a substitution a significant
number of candidates chose to use integration by parts to establish the result. There were some sign
errors in the integrations, but most candidates managed to reach the final result in the first part of
the question. The second part of the question was found to be the hardest, with induction the most
popular method, although the process was often not fully explained. The final part of the question
did not appear to be too problematic for those that reached it. However, algebraic mistakes, such as
factors disappearing, resulted in some marks being lost. Similarly, mistakes in the arithmetic in the
final part of the question were not uncommon.

3. This question was again popular and had an average score of about half of the marks. In the
first part almost all candidates were able to sketch the correct shape of graph, but some did not
provide suitable explanations to accompany these or included additional cases that were not asked
for. A number of candidates attempting the second part of the question reached one of the results
by squaring an inequality without considering the signs and many assumed that the result of part (i)
implied that c must be negative. Only about half of the candidates attempted part (iii), and many of
those who did did not use sketches in their solutions. Solutions to part (iv) generally involved
guessing of the values of a, b and c followed by a check that the conditions were met.

4. This question received a relatively small number of attempts compared to the other Pure
Mathematics questions. On average candidates who attempted this question only received a quarter
of the marks available. Some candidates did not manage to write down the correct equation of the
line or did not appreciate that the phrase “unit radius” means that the radius is 1. Many candidates
produced loci for the second part of the question without any indication of a method. In the final
part of the question the significance of the restrictions on the value of b were not appreciated by
many of the candidates.

5. This was one of the more successfully attempted questions on the paper and the Pure
Mathematics question with the highest average mark. While some candidates struggled with the
application of the chain rule throughout this question, many were able to complete the first part of
the question without much difficulty. Showing that f satisfied the required conditions in part (i) was
generally well done, but the sketching of the graph was found to be more difficult, with a number of

595
candidates not identifying the asymptotes and some thinking that part of the graph would drop
below the x-axis. Most of the candidates who attempted part (iii) found the roots of the equation
successfully, but a large number forgot to exclude the roots when solving the inequality. In the final
part, many identified x=3 as a solution, but those who split the fraction into two equations (one for
the numerator equalling 343 and one for the denominator equalling 36) did not check that the
solution worked for both parts. Those who used the symmetries established in part (i) were then
able to identify the other roots easily, while those who attempted algebraic solutions for the other
roots were generally not successful.

6. The algebra required for the first part of the question proved to be quite challenging for a
number of candidates, but most were able to reach the required answer. The proof by induction in
the second part of the question was generally well done, although a number of candidates did not
write up the process clearly. In the final part of the question it was clear that many candidates had
identified the relationship between the sequences and Fibonacci numbers and some candidates
therefore stated that the limit would be the golden ratio, but without any supporting calculations. In
the final part there were few responses which clearly explained that the new sequence would still
satisfy the conditions required if it were started at a later term.

7. This question was attempted by a large number of candidates, only slightly fewer than
question 2, and was one of the more successful ones with an average score above half of the marks.
While some candidates proved the converse of the required result, part (i) of the question was
generally done well, although a surprising number of candidates did not write down the numerical
solutions when asked. Those students who realised the way to write x and y in terms of m and n
reached the result of part (ii) easily, while others sometimes spent a lot of effort on this making little
or no progress. In part (iii) many candidates spotted the difference of two squares, but some did not
realise that there would be two ways to factorise . Only very few students were able to solve the
final part of the question.

8. Candidates attempting question 8 generally received either a very low or a very high score.
Many attempts did not progress further than an attempt to sketch the graph and identify the
rectangle to be used. There were also some attempts that confused the line with a
transformation of the curve . In the second part of the question there were some
difficulties with the differentiation of , but those candidates who successfully completed this
section did not in general have any difficulties with the remainder of the question.

9. The average score on this question was below a quarter of the marks as a large number of
attempts did not make progress beyond the first few steps of the solution, achieving just the marks
for the resolution of forces required in the first part of the question. Many candidates forgot some
of the forces involved and very few decided to take moments. Some of the more clever solutions
took moments about one of the contact points, which removes the need for one of the steps
resolving forces.

10. This was the least popular of the Mechanics questions. The first part of the question was
generally well answered and many candidates were able to apply the result of part (i) to the
particular case identified in part(ii). Part (iii) was found to be more challenging, but some candidates
did manage to provide a convincing argument for their answer.

596
11. This was the most popular of the Mechanics questions and also the most successfully
answered question on the paper with candidates scoring on average three quarters of the marks.
Candidates appeared to be very comfortable with the concepts of conservation of momentum and
the law of restitution and were able to progress through the series of calculations required without
too much difficulty. There were some errors in the algebra, but the majority of candidates were able
to work through accurately to the end of the question.

12. This was the least popular of all the questions. Many of those who did attempt the question
succeeded in calculating the expressions for the expectations, but the simplification of the
calculation for the variance proved more tricky. A good number of the candidates managed to reach
the final part of the question, but few were able to provide a valid argument for the final result.

13. Many candidates were able to complete the parts of the question that related to the early
cases, but some struggled to generalise the expressions for the probabilities in the cases required in
part (iii) of the question. Of those that reached the correct expressions many struggled to establish
the required relationships between them.

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STEP 3 2013 Examiners’ report 

With the number of candidates submitting scripts up by some 8% from last year, and whilst 
inevitably some questions were more popular than others, namely the first two, 7 then 4 and 5 to a 
lesser extent , all questions on the paper were attempted by a significant number of candidates.   
About a sixth of candidates gave in answers to more than six questions, but the extra questions were 
invariably scoring negligible marks.  Two fifths of the candidates gave in answers to six questions. 

1.  Most candidates attempted this question, making it the most popular and it was also the 
most successful with a mean score of about two thirds marks.  The first two standard results caused 
few problems, nor did the integration, but some struggled to simplify to the single inverse tan form.  
In the final part, common errors were failure to reduce to the  0  case, confusion with the index 
e.g.   2 sin  instead of the correct result, or for those that were more successful, 
algebraic inaccuracies let them down.  Some attempted a recursive formula to evaluate  
sin   with varying success.  Most attempting the last part saw the connection between   
and the main result of the question. 

2.  This was the second most popular question, attempted by six out of every seven candidates, 
with only marginally less success than its predecessor.  The first differential equation was proved 
correctly and many successfully completed the general result by induction, although there were 
some problems with the initial case.  Some had difficulty finding the correct coefficients for the odd 
powers of   in the Maclaurin series but the last part produced a variety of errors and few correct 
answers.  Such errors included  sin  , forgetting to divide    by     , and attempting to 
evaluate the series using   1 . 

3.   A seventh of the candidates attempted this, making this the second least popular Pure 
question, though with on average, half marks being scored, it was the third most successful of the 
Pure questions.  Some candidates found the scalar product of    with itself to 
obtain the stem correctly, whilst some found its product with     or    , in which case they did not 
always appreciate the importance of symmetry.  Part (i) caused few problems.  Part (ii) saw a few 
errors with consideration of     signs, though some candidates used geometric considerations and 
then rotations correctly to obtain the results.  The last part separated the sheep from the goats. 

4.  Just over two thirds of candidates attempted this with moderate success, approximately one 
third marks.  Most succeeded with the opening result but even so, some lacked full explanation.  
Whilst most wrote down the correct form for the roots, few correctly expressed all the roots in the 
given range.  Surprisingly, there was very limited understanding of the connection between the roots 
and the factors of  1   so the general result was not well answered.  Conversely, part (i) was 
well‐answered with the exception of those who did not deal with the powers of   satisfactorily.  Part 
(ii) was beyond most candidates mainly because they failed to cancel the factor  1  .  However, 
those that managed to deal with this aspect generally answered the whole question very well. 

5.  Nearly as many attempted this as question 4, but only achieving a quarter of the marks 
making it the least successfully answered question.  Almost all missed the point of the question 
given in the first sentence, and made other assumptions, which frequently only applied to primes 
rather than integers in general.  As a consequence, most did not satisfactorily justify their results.  

598
They generally fared better tackling the second part of (i), though some tried to prove the statement 
in the wrong direction.  They approached (ii) better though few gave a valid argument why    . 

6.  About half attempted this with marginally more success than question 4.  Many candidates 
tried to write     or similar and likewise for     and then tried to expand which involved a 
lot more work than dealing with conjugates directly.  Some tried to use the cosine rule rather than 
the triangle inequality from the diagram.  In general, the first result and parts (i) and (ii) were well 
done but only the strongest candidates did better than pick up the odd mark here and there in trying 
to obtain the inequality.  A lot of mistakes were made mishandling inequalities, but even those who 
could do this correctly overlooked the necessity of substantiating that th4 square roots are positive 
and that the denominator is non‐zero. 

7.  Three quarters attempted this with more success than question 6 but less than question 3.  
Sadly, it was not uncommon for candidates to fail to differentiate   correctly.  Many established 
that   0  but then   1 , when   1 ,   0 , and   0  giving a maximum which was 
not sufficient and missed the point of the squared     term in    ,  with consequences for the 
rest of the question.  Many followed the stationary points line of logic correctly by considering the 
maximum and minimum values in part (i).  Having established the constant value of    , some 
candidates attempted to solve the differential equation, usually by incorrect methods.  The errors of 
part (i) were largely replicated in part (ii).  There were fewer attempts at part (iii), and a number fell 
at the first hurdle through not obtaining the correct    .  Further, numerous candidates assumed 
rather than proved that  5 cosh 4 sinh 3 0 . 

8.  A seventh answered this question, making it the second least attempted question scoring a 
third of the marks possible.  The first result evaded many candidates who did not identify and 
calculate the geometric progression, although a few did employ the fact that the sum of the roots of 
unity is zero.  The result for     caused few problems and was for many candidates the only success 
in the question.  Those that attempted the length of the chord were comfortable with the algebra of 
trigonometry namely  cos cos  , and  2 cos 1 cos 2  .  There was mixed success 
with completing the final result. 

9.  About a fifth attempted this, with the same success as question 7.  Common errors were 
false attempts for the volume at the beginning using hemisphere and cones, and in the last part 
approximating     small rather than     small.  Many candidates were successful as far as the 
equilibrium but couldn’t deal with the small oscillations successfully. 

10.  The number of candidates attempting this was almost identical to that attempting question 
3 with marginally more success making it the third best attempted question.  Most obtained the 
moment of inertia correctly, and many found the angular velocity correctly.  Provided that they had 
correctly applied conservation of angular momentum, and Newton’s law of elasticity, they almost all 
worked out the required result.  Some attempted to use conservation of linear momentum whilst 
others did not use conservation of angular momentum correctly.  Most then knew how to 
differentiate, but many made computation errors.  Even if they got the correct quadratic equation at 
the end, many solve d it wrongly.  Very few showed that the feasible solution did indeed generate a 
maximum. 

599
11.  A fifth of the candidates attempted this question, with marginally less success than question 
3.  Most that attempted this question managed to achieve the first two results successfully, unless 
they got the diagram wrong.  However, the final result was found trickier as some forgot to include 
the gravitational potential energy, some failed to evaluate the correct elastic potential energy and 
there were many mistakes made handling the surds. 

12.  This was the least popular question, attempted by a ninth of the candidates, with slightly 
less success than question 8.  The immediate problem was many made no mention of probabilities in 
order to calculate expectations.  Throughout, there was very poor justification, which included 
treating the random variables as though they were independent and compensating errors which led 
to given results.  Most progressed no further than part (a) of (ii) at best and many had    . 

13.  The number attempting this was very similar to that attempting question 3 with the same 
level of success as question 11.  In general, candidates attempted both parts of (a) correctly, and 
then likewise part (i) of (b) then stopped.  However, part (b) (ii) tripped up many.  Some successfully 
dealt with part (iii) without having managed (ii). 

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Explanation of Results STEP 2013

All STEP questions are marked out of 20. The mark scheme for each question is designed to reward
candidates who make good progress towards a solution. A candidate reaching the correct answer will
receive full marks, regardless of the method used to answer the question.

All the questions that are attempted by a student are marked. However, only the 6 best answers are
used in the calculation of the final grade for the paper.

There are five grades for STEP Mathematics which are:

S – Outstanding
1 – Very Good
2 – Good
3 – Satisfactory
U – Unclassified

The rest of this document presents, for each paper, the grade boundaries (minimum scores required
to achieve each grade), cumulative percentage of candidates achieving each grade, and a graph
showing the score distribution (percentage of candidates on each mark).

STEP Mathematics I (9465)

Grade boundaries
Maximum Mark S 1 2 3 U
120 100 82 64 40 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 4.8 18.6 45.0 81.6 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics I

www.admissionstestingservice.org

601
STEP Mathematics II (9470)

Grade boundaries
Maximum Mark S 1 2 3 U
120 100 79 67 32 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 8.0 24.8 38.3 85.4 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics II

STEP Mathematics III (9475)

Grade boundaries
Maximum Mark S 1 2 3 U
120 85 63 48 27 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 12.3 36.6 56.7 85.4 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics III

www.admissionstestingservice.org
602


603


STEP Solutions 2013

Mathematics

STEP 9465/9470/9475

October 2013

604
The Admissions Testing Service is a department of Cambridge English
Language Assessment, which is part of Cambridge Assessment, a not-for-
profit department of the University of Cambridge.

Cambridge English Language Assessment offers the world’s leading

qualifications for learners and teachers of the English language. Over 4 million
people from 130 countries take Cambridge English exams every year.

This mark scheme is published as an aid to teachers and students, to indicate

the requirements of the examination. It shows the basis on which marks were
awarded by the Examiners. It does not indicate the details of the discussions
which took place at an Examiners’ meeting before marking commenced.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the
relevant knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question

papers and the Report on the Examination.

The Admissions Testing Service will not enter into any discussion or
correspondence in connection with this mark scheme.

© UCLES 2013

More information about STEP can be found at:

www.stepmathematics.org.uk

605
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 4
STEP Mathematics II 13
STEP Mathematics III 19

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STEP 1 2013 Hints & Solutions 

Q1 This question is all about using substitutions to simplify the working required to solve
various increasingly complicated looking equations. To begin with, you are led gently by the hand
in (i), where the initial substitution has been given directly to you. You are also reminded that y
must be non-negative, since x denotes the positive square-root of x (positive unless x is 0, of
course). The result is obviously a quadratic equation, y 2  3 y  12  0 , and is solvable by (for
 3  11
instance) use of the quadratic formula. However, only one of the apparent solutions, y  ,
2
2
 11  3 
is positive, so the other is rejected and we proceed to find that x    = 5  32 11 .

 2 
In (ii) (a), the approach used in (i) should lead you to consider the substitution y = x2,
which gives the quadratic equation y 2  10 y  24  0 . This, in turn, yields y = x  2 = –12 or 2.
Again, this must be non-negative, so we find that x  2  2 and x = 2.
In (ii) (b), it should not now be too great a leap of faith to set y  2 x 2  8 x  3 which, with
a bit of modest tinkering, yields up y 2  2 y  15  0  y = 2 x 2  8 x  3 = –5, 3. Again, y  0, so
2 x 2  8 x  3 = 3  (with cancelling) x 2  4 x  6  0  x  2  10  x 2  14  4 10 . The
final step here is to check that both apparent solutions work in the original equation, since the
squaring process usually creates invalid solutions. [Note that it is actually quite easy to see that both
solutions are indeed valid, but this still needs to be shown, or otherwise explained. Longer methods
involving much squaring usually generated four solutions, two of which were not valid.]

Q2 The “leading actor” throughout this question is the integer-part function, x  , often referred
to as the floor-function. Purely as an aside, future STEP candidates may find it beneficial to play
around with such strange, possibly artificial, kinds of functions as part of their preparation because,
although they clearly go beyond the scope of standard syllabuses at this level, they are within reach
and require little more than a willingness to be challenged. [Note that some care needs to be taken
when exploring such things. In the case of this floor-function, at least a couple of function plotting
software programs that recognise the “INT” function do so incorrectly for x < 0: for instance,
interpreting INT(–2.7) as –2 rather than –3.]
The key elements of the sketch in (i) are as follows. The jump in the value of x  whenever
x hits an integer value means that the graph is composed of lots of “unit” segments, the LH end of
which is included but not the RH end. The usual convention for signalling these properties is that
the LH endpoint has a filled-in dot while the RH endpoint has an open dot. Then, in-between
n
integer values of x, each segment of the curve is of the form and thus appears to be a portion of a
x
reciprocal curve.
The purpose of parts (ii) and (iii) is to see if you can use your graph to decide how to solve
some otherwise fairly simple equations: the key is to have a clear idea as to where the various
portions of the graph exist. The analysis looks complicated, but candidates were actually only

607
required to pick the appropriate n’s and write down the relevant answers (so the working only
needed to reflect what was going on “inside one’s head”). Note that, for n  x  n  1 , x   n so
n n
f(x) = . Also,  f ( x )  1 for x > 0, and f(x)  1 for x < 0, so f(x) = 127 only in [1, 2), yielding
x n 1
1 7
the equation  in (ii)  x = 127 .
x 12
n 17
Similarly,   24n > 17n + 17  n > 2 73 , i.e. n  3; so f(x) = 17 24 only in [1, 2)
n  1 24
1 17 2 17
and [2, 3). In [1, 2), f ( x)    x = 17 24
and in [2, 3), f ( x)    x = 17 48
. Next, for
x 24 x 24
n n 4
x < 0, 1  f ( x)  , and   –4n – 4 > –3n  n < –4; so f(x) = 43 only in [–4, –3),
n 1  n 1 3
[–3, –2), [–2, –1) and [–1, 0). However, since f(–3) = 1 there is no solution in [–4, –3). Otherwise,
3 4 2 4
in [–3, –2), f ( x)    x =  94 ; in [–2, –1), f ( x )    x =  32 ; and in [–1, 0),
x 3 x 3
1 4
f ( x)    x =  34 .
x 3
n 9 8 9
For (iii),  for n > 9 so f(xmax) = 109 in [8, 9) and f ( x)    x = 809 .
n  1 10 x 10
Only the very last part required any great depth of insight, and the ability to hold one’s
nerve. The equation f(x) = c has exactly n roots when the horizontal line y = c cuts the curve that
number of times. That is …
n n 1 n 1 n
… when x > 0 : c ; … when x < 0: c , n  2; … and c  2 for n = 1.
n 1 n2 n n 1

Q3 This vector question is tied up with the geometric understanding that, for distinct points with
position vectors x and y, the point with p.v. x + (1 – )y cuts XY in the ratio (1 – ): (though it is
important to realise that this point is only between X and Y if 0 <  < 1). Part (i) tests (algebraically)
the property of commutativity (whether the composition yields different results if the order of the
application of the operation is changed):
XY = YX  x + (1 – )y = y + (1 – )x  (2 – 1)(x – y) = 0  (since x  y)  = 12 .
Part (ii) then explores the property of associativity (whether the outcome is changed when
the order of the elements involved in two successive operations remains the same but the pairings
within those successive operations is different). Here we have
(XY)Z =  (x + (1 –  )y) + (1 – )z =  2x + (1 – )y + (1 – )z
and X(YZ) = x + (1 – )[y + (1 – )z] = x + (1 – )y + (1 – )2z
Thus, (XY)Z – X(YZ) = (1 – )(x – z) and the two are distinct provided   0, 1 or X  Z.
Part (iii) now explores a version of the property of distributivity (although usually referring
to two distinct operations): (XY)Z = 2 x   (1   )y  (1   )z , and
(XZ)(YZ) = [x + (1 –  )z][y + (1 –  )z] = 2 x   (1   )z +  (1   )y  (1   ) 2 z
= 2 x   (1   )y  (1   )z , and the two are always equal.
Next, X(YZ) = x   (1   )y  (1   ) 2 z , and

608
 (XY)(XZ) = [x + (1 –  )y][x + (1 –  )z]
= 2 x   (1   )y +  (1   )x  (1   ) 2 z
= 2 x   (1   )y  (1   )z .
Hence X(YZ) = (XY)(XZ) also.
In (iv), you will notice that the condition 0 <  < 1 comes into play, so that P1 cuts XY
internally in the ratio (1 – ):. Following this process through a couple more steps shows us that
Pn cuts XY in the ratio (1 – n):n , which is easily established inductively.

Q4 The first part to this question involved two integrals which can readily be integrated by
“recognition”, upon spotting that
d
f (tan x)   f (tan x)  sec 2 x and d f (sec x)  f (sec x)  sec x tan x .
dx dx
(They can, of course be integrated using suitable substitutions, etc.) Thus, we have
 1 n 1  1
 tan x. sec x dx =  n  1 tan x  = n  1
n 2

n n 1 1 
and  sec x. tan x dx =  sec x. sec x tan x dx =  sec n x  =
2 1
.
  n

n  n
The two integrals in part (ii) can be approached in many different ways – the examiners
worked out more than 25 slightly different approaches, depending upon how, and when, one used
the identity sec2x = 1 + tan2x, how one split the “parts” in the process of “integrating by parts”, and
even whether one approached the various secondary integrals that arose as a function of sec x or
tan x. Only one of these approaches appears below for each of these two integrals..
 /4
 sec 4 x   / 4  / 4 sec 4 x  1
 /4

0 –   J , where J =  sec
4 4
x sec x tan x dx = x .  dx (by parts) = x dx.
 4  0 0
4 4 4 0

 /4  /4
 1
Then, J =  sec
2
x dx +  sec
2
 
x tan 2 x dx = tan x  13 tan 3 x =
4
3
, and our integral is  .
4 3
0 0

 
Next,  x 2 sec 2 x. tan x dx = x 2 . 12 tan 2 x –  2 x . 12 tan 2 x dx (by parts) =
2
32

  x sec 2 x  1 
 /4
2
 K   x dx, where K =  x sec
2
= x dx.
32 0


Then, K = x . tan x –  tan x dx = x tan x – ln(sec x) =
1
 ln 2 , so that this last integral is
4 2
2  1  
2
2  1
–   ln 2  + or   ln 2 .
32  4 2  32 16 4 2

Q5 This question simply explores the different possibilities that arise when considering curves
of a particular quadratic form. In (i), with a zero product term, we have equal amounts of x2 and y2,

609
and this is symptomatic of a circle’s equation: x 2  3 x  y 2  y  0  x  32    y  12  
2 2
1
2

2
10 ,
which is a circle with centre  32 ,  12  and radius 12 10 . This circle passes through the points (0, 0),
(0, –1) & (–3, 0).
In (ii), with k = 10
3 , we have 3 x  y x  3 y  3  0 . This factorisation tells us that we have
the line-pair y = –3x & x + 3y = –3; the first line passing through the origin with negative gradient,
while the second cuts the coordinate axes at (0, –1) and (–3, 0).
Part (iii) is the genuinely tough part of the question, but help is given to point you in the
right direction. When k = 2, we have  x  y   3 x  y  0 , and using  = 45o in the given
2

X Y X Y
substitution gives x  y  X 2 and y  x  Y 2  x  and y  . Then
2 2
3 X  3Y X Y
x  y 2  3 x  y  0 becomes 2X 2    0  2 X 2  2 2 X  Y 2 or  
2 2
 2
2 X  1  1  Y 2 . This is now in what should be a familiar form for a parabola, with axis of
1 x y 1
symmetry X   (substituting back) 
i.e. x  y   1 . For the sketch, we must
2 2 2
rotate the standard parabola through 45o anticlockwise about O in order to get the original parabola
x 2  2 xy  y 2  3 x  y  0 .
For those who have encountered such things, all three curves here are examples of conic
sections.

Q6 It should be pretty clear that this question is all about binomial coefficients. The opening
result – the well-known Pascal Triangle formula for generating one row’s entries from those of the
previous row – is reasonably standard and can be established in any one of several ways. The one
 n  1
intended here was as follows: the coefficient of xr in (1 + x)n + 1 is   , and this is obtained
 r 
from (1 + x)(1 + x)n, where the coefficient of xr comes from
 n  r 1  n  r 
1  x  .....    x    x  ..... 

  r  1 r 
and the required result follows.
In the next stage, for n even, write n = 2m so that
 2m   2m  1  2m  2   m  1  m 
B2m + B2m + 1 =          .....      
 0   1   2   m  1  m 
 2m  1  2m   2m  1  2m  2   m  1
+             .....    .
 0   1   2   3   m 
and, pairing these terms suitably, this is
 2m  1  2m   2m    2m  1  2m  1  m  1  m  1  m
                .....          .
 0    0   1    1   2    m  1   m    m
 2m  1  2m  2   m   m  1
Now, using the opening result, and the fact that             1 , we have
 0   0   m   m  1

610
  2m  2   2m  1   2m   m  2   m  1
          .....       ,
 0   1    2   m   m  1
m  1  2( m  1)  j 
and this is just    = B2m + 2, as required.
j0  j 
In the case n odd, write n = 2m + 1, so that
 2m  1  2m   2m  1  m  2   m  1
B2m + 1 + B2m + 2 =          .....      
 0   1   2   m 1  m 
 2m  2   2m  1  2m   2m  1  m  2   m  1
+             .....      
 0   1   2   3   m   m  1
which gives, upon pairing terms suitably,  
 2m  2   2m  1  2m  1  2m   2m   m  2   m  2   m  1  m  1
               .....            
 0   0   1   1   2   m  1   m     m   m  1 
 2m  3   2m  2    2m  1  m  3   m  2 
=           .....      
 0   1    2   m   m  1 
 2m  2   2m  3 
using the opening result and the fact that       1
 0   0 
m  1  2( m  1)  1  j 
=    = B2m + 3.
j0  j 
To complete an inductive proof, we must also check that the starting terms match up properly, but
this is fairly straightforward. We conclude that, since B0 = F1 , B1 = F2 and Bn & Fn satisfy the
same recurrence relation, we must have Bn = Fn + 1 for all n.

Q7 In (i), there is a generous tip given to help you on your way with this question. Starting from
dy du du 1
y = ux we have ux , so that the given differential equation becomes u  x   u or
dx dx dx u
1
 u du   x dx upon separation of variables. You are now in much more familiar territory and may
y2
proceed in the standard way: 12 u 2  2  ln x  C  y 2  x 2 2 ln x  2C  . Using the given
2x
conditions x = 1, y = 2 to determine C = 2 then gives the required answer y  x 2 ln x  4 .
However, there is one small detail still required, namely to justify the taking of the positive square-
root, which follows from the fact that y > 0 when x = 1. (Note that you were given x > e – 2 , for the
validity of the square-rooting to stand, so it is not necessary to justify this. However, it should serve
as a hint that a similar justification may be required in the later parts of the question.)
In (ii), either of the substitutions y = ux or y = ux2 could be used to solve this second
differential equation. In each case, the method then follows that of part (i)’s solution very closely
indeed; separating variables, integrating, eliminating u and substituting in the condition x = 1, y = 2
to evaluate the arbitrary constant. The final steps require a justifying of the taking of the positive
square-root and a statement of the appropriate condition on x in order to render the square-rooting a
valid thing to do. The answer is y  x 5 x 2  1 for x > 1
5
.

611
 In (iii), only the substitution y = ux2 can be used to get a variable-separable differential
1 1
equation, which boils down to  u du   2 dx  12 u 2   D . Using x = 1, y = 2 (u = 2) to
x x
evaluate the constant D leads to the answer y  x 6 x 2  2 x for x > 1
3 .

Q8 This question is all about composition of functions and their associated domains and ranges.
Whilst being essentially a very simple question, there is a lot of scope for minor oversights. One
particular pitfall is to think that x 2  x , when it is actually | x |. Also, when considering the
composite function fg, it is essential that the domain of g (the function that “acts” first on x) is
chosen so that the output values from it are suitable input values for f. You should check that this is
so for the four composites required in part (i).
In (ii), the functions fg and gf look the same (both are | x |) but their domains and ranges are
different: fg has domain ℝ and range y  0, while the second has domain | x |  1 and range y  1.

In (iii), the essentials of the graph of h are: it starts from (1, 1) and increases. Since x 2  1
is just (x – a tiny bit) after a while, the graph of h approaches y = 2x from below. Using similar
reasoning, the graph of k for x ≥ 1, also starts at (1, 1) but decreases asymptotically to zero. (It is
well worth noting that x  x 2  1 is the reciprocal of x  x 2  1 , since their product is 1.)
However, this second graph has a second branch for x  –1, which is easily seen to be a rotation
about O (through 180o) of the single branch of h, this time approaching y = 2x from above. Finally,
note that the domain of kh is x  1, and since the range of h is y  1, the range of kh is 0 < y  1.

Q9 This question incorporates the topics of collisions and projectiles, each of which consists of
several well-known and oft quoted results. However, much of the algebraic processing can be
shortcut by a few insightful observations. To begin with, if the two particles start together at ground
level and also meet at their highest points, then they must have the same vertical components of
velocity. Thus u sin   v sin  . Then, if they both return to their respective points of projection, the
collision must have ensured that they both left the collision with the same horizontal velocity as
when they arrived; giving, by Conservation of Linear Momentum, that mucos = Mvcos. Dividing
these two results gives the required answer, m cot = M cot.
u sin  v sin 
The collision occurs when t   (from the standard constant-acceleration
g g
u 2 sin  cos
formulae) and at the point when A has travelled a distance b = and B has travelled a
g
v 2 sin  cos  1
distance  (v sin  )(v cos  ) , the sum of these two distances being denoted d.
g g
Md
Substituting for the brackets using the two initial results then gives b  , as required.
mM
u 2 sin 2 
Moreover, the height of the two particles at the collision is given by y = , so that
2g

612
 1 u 2 sin  cos  sin  1
h=   = b tan  .
2 g cos  2

Q10 This question makes more obvious use of the standard results for collisions, but also ties
them up with Newton’s 2nd Law (N2L) of motion and, implicitly, the Friction Law and resolution
of forces (in the simplest possible form). Thus, if R is the normal contact reaction force of floor on
puck, F the frictional resistance between floor and puck, we have (in very quick order) the results
R = mg, F = R = mg and, by N2L, mg = –ma, where a is the puck’s acceleration. The constant-
acceleration formula v2 = u2 – 2as then gives wi  1  vi  2gd (where vi is the speed of the puck
2 2

when leaving the i-th barrier, for i = 0, 1, 2, …, and wi is its speed when arriving at the i-th barrier,
for i = 1, 2, 3, …). Also, Newton’s (Experimental) Law of Restitution (NEL or NLR) gives
vi  1  r.wi  1 , from which it follows that vi  1  r 2 vi  2r 2 gd , as required.
2 2

Iterating with this result, starting with v1  r 2 v 2  2r 2 gd , then leads to the general result
2

 
v n  r 2 n v 2  2r 2 gd 1  r 2  r 4  ...  r 2 n  2 . The large bracket is the sum-to-n-terms of a GP,
2

1  r 2n
namely , and you simply need to set vn = 0 and tidy up in order to obtain the result
1 r 2
v2
r 2n 
1  r 2n r 2 .
2gd 1 r2
v2  r2 
by k and re-writing the expression for r2n   2  , we simply have to
Replacing
2 gd 
 r  k 1 r
2
 
 r2 
 
take logs and solve for n to get n = 
ln 2

r  k 1 r2   . Setting r = e – 1 in this result, and tidying
2 ln r
up, then gives n = 12 ln 1  k e  1 .
2

v2
When r = 1, the distance travelled is just nd, and we have v2 = 2 gnd and n  k.
2gd

Q11 Since we are told that  +  < 1

2
 , the two tensions T sin and T cos are effectively
components of a notional force (T), inclined slightly to the right of the normal contact reaction force
R, say. Hence, if there is motion, it will take place to the right. Then, resolving vertically and
horizontally for the block, calling the frictional force (acting to the left) F, we have
R + T sin cos + T cos sin = W and F + T sin sin = T cos cos
or, using trig. identities, W = R + T sin( + ) and F = T cos( + ). Since W > T sin( + ), it
follows that R > 0 so the block does not rise. Otherwise, F ≤  R and using  = tan for equilibrium,
we have T cos( + ) ≤ tan W  T sin(   ) i.e. W tan ≥ T tan sin( + ) + T cos( + )
 W sin ≥ T sin sin( + ) + T cos( + ) cos = T cos( +  –  ).

In the next part, W = T tan 12 (   )  R = T sin 12 (   ) – T sin( + ) < 0 so R = 0

(and F = 0) as the block lifts from ground. Taking unit vectors i and j in the directions  and 

613
   T sin  sin     T cos  cos    0 
respectively: TA =   , TB =   , W =   , and the
 T sin  cos    T cos  sin     T tan 2 (   ) 
1

 T cos(   ) 
resultant force on the block is TA + TB +W =   , which is in the
 T sin(   )  T tan 2 (   ) 
1

 sin(   )  tan 12 (   ) 
direction tan 1   (relative to i). Since  +  = 2 , this is the direction
 cos(   ) 
 sin 12 (   ) 
 2 sin 12 (   ) cos 12 (   )  
1 sin 2 (   )2 cos 2 (   )  1
1  cos 12 (   )   1 2 1

tan  tan  
 cos(   )  cos(   ) . cos 2 (   )
1 
   
 
 tan tan 2 (   )  = 2 (   )   , noting that 2 cos 2 12 (   )  1  cos(   ) .
1 1 1

Q12 As with all such probability questions, there are many ways to approach the problem. The
one shown here for part (i) is one that generalises well to later parts of the question. Suppose the
3 3 3 3
container has 3R, 3B, 3G tablets. Then the probability is    for one specified order (e.g.
9 8 7 56
9
RBG). We then multiply by 3! = 6 for the number of permutations of the 3 colours to get . The
28
final part of (i) is really a test of whether you realiuse that this is the same situation viewed “in
reverse”, so the answer is the same.
Using the method above, with a suitable notation, in part (ii) we have
n n n 2n 2 n3
P3(n) =    3! = or .
3n 3n  1 3n  2 3n  13n  2  3n 
 
3
Then in (iii), P(correct tablet on each of the n days) = P2(n)P2(n – 1)P2(n – 2) … P2(2)P2(1)
 n 2 . 2 !   (n  1) 2 . 2 !   (n  2) 2 . 2 !   2 2 . 2 ! 12 . 2 !
=  
  
    ...    
 2n(2n  1)   (2n  2)(2n  3)   (2n  4)(2n  5)   4 . 3   2 .1 
n 2n
( n !) 2 2 n 2n n  2n 
= or .Then, using n!  2n   and (2n)!  4 n   , we have
( 2n) !  2n  e  e 
 
n
n 2n
2 n   2n
e 2n 2n  2 n n
Prob. = =  n .
2 n2n 2n
2 n  2 2n
2
4 n  2n
e
Q13 First, note that x  {0, 1, 2, 3}, so there are four probabilities to work out (actually, three,
and the fourth follows by subtraction from 1). The easier ones to calculate are X = 0 and X = 3, so
let us find these first.
For P(X = 0): the 7 pairs from which a singleton can be chosen can be done in 26C7 ways;
7
26
C7  27
then, we can choose one from each pair in 2 ways; so that P(X = 0) = 52
.
C7

614
 26.25.24.23.22.21.20 52 52.51.50.49.48.47.46 3520
Now 26C7 = & C7 =  P(X = 0) = .
7 .6 .5 .4 .3 .2 .1 7 .6 .5 .4 .3 .2 .1 5593
P(X = 3): the 3 pairs from 26 can be chosen in 26C3 ways; then the one singleton from the
remaining 23 pairs can be chosen in 23C1 = 23 ways, and the one from that pair in 21 ways ; so that
26
C 3  23 C1  21 5
P(X = 3) = 52
= (similarly for calculation).
C7 5593
For P(X = 1): the 1 pair can be chosen in 26C1 = 26 ways; the 5 pairs from which a singleton
is chosen can be done in 25C5 ways; and the singletons from those pairs in 25 ways; so that
26  25 C 5  2 5 1848
P(X = 1) = 52
= .
C7 5593
For P(X = 2): the 2 pairs can be chosen in 26C2 ways; then the 3 pairs from which a
singleton is chosen can be done in 24C3 ways; and the singletons from those pairs can be chosen in
26
C 2  24 C 3  2 3 220
23 ways; so that P(X = 2) = 52
= .
C7 5593

Then E(X) =  x . p( x) =
1
0  3520  1  1848  2  220  3  5
5593
2303 7  7  47 7
=   .
5593 7  17  47 17

615
STEP 2 2013 Hints and Solutions

Question 1.

The gradient of a line from a general point on the curve to the origin can be calculated easily and the
gradient of the curve at a general point can be found by differentiation. Setting these two things to
be equal will then lead to the correct value of . A similar consideration of gradients to the origin
will establish the second result and if the line intersects the curve twice then a sketch will illustrate
that there must be one intersection on each side of the point of contact found in the first case. A
similar process will establish the result for part (ii).

For part (iii) the gradient of the line must be smaller than the gradient of the line through the origin
which touches the curve, so the intersection with the y-axis must be at a positive value. This means
that the conditions of part (ii) are met, which allows for the comparison between and to be
made.

The condition given in part (iv) is equivalent to stating that the line is parallel to the one found at the
very beginning of the question. This implies that the intersection with the y-axis is at a negative
value and so an adjustment to the steps taken in part (ii) will establish the required result.

Question 2.

The obvious substitution in the first part leads easily to the required result. It should then be easy to
establish the second result by making the integral into the sum of two integrals and noting that
taking out a common factor leaves ( ) to be simplified. Integration by parts will lead to the
next result after which taking out one of the factors of ( ) will allow the integral to be split into
a difference of two integrals.

The result in part (ii) is most easily proved by induction. It is necessary to fill in the gap in the
factorial on the denominator by multiplying both the numerator and denominator by the missing
even number. In alternative approaches, it needs to be remembered that the product of the even
numbers up to and including can be written as

The final part is a straightforward substitution, although care needs to be taken with the signs. The
final result can be obtained using the relationship established in part (i) as none of the reasoning
requires to be an integer.

616
Question 3.

For it to be possible for the cubic to have three real roots it must have two stationary points. Since
the coefficient of is positive it must have a specific shape. A sketch will show that only the two
cases given will result in an intercept with the y-axis at a negative value.

In order for the cubic in part (ii) to have three positive roots, both of the turning points must be at
positive values of . Differentiation will allow most of the results to be established. The condition
that is needed to ensure that the leftmost root is also positive.

The condition implies that there must be a turning point at a positive value of . The shape
of the graph is as in part (i), but this time the intersection with the y-axis is at a positive value. This is
sufficient to deduce the signs of the roots.

For part (iv) it is easiest to note that changing the value of does not (as long as remains negative)
change whether or not the conditions of (*) are met. As this represents a vertical translation of the
graph any example of a case satisfying (*) can be used to create an answer for this part by
translating the graph sufficiently far downwards.

Question 4.

The equations of the line and circle are easily found and so the second point of intersection (and so
the coordinates of M) can be easily found. The two parts of this question then involve regarding the
coordinates of M as parametric equations.

In part (i) is the parameter and is restricted so that the point that the line passes through is inside
the circle. This gives a straight line between the points which result from the cases and
. The length of this line can be determined easily from the coordinates of its endpoints.

In part (ii) it is again quite easy to eliminate the parameter from the pair of equations and the shapes
of the loci should be easily recognised. In part (b) however, the restriction on the values of need to
be considered as the locus is not the whole shape that would be identified from the equation.

617
Question5.

Simple applications of the chain rule lead to relationships that will allow the three cases of zero
gradients to be identified in part (i).

In part (ii) the relationships follow easily from substitution and therefore the three stationary points
identified in part (i) must all exist. By considering the denominator there are clearly two vertical
asymptotes and the numerator is clearly always positive. Additionally, the numerator is much larger
than the denominator for large values of . Given this information there is only one possible shape
for the graph.

In part (iii) the solutions of the first equation will already have been discovered when the
coordinates of the stationary points in part (ii) were calculated. The range of values satisfying the
first inequality should therefore be straightforward. One of the solutions of the second equation
should be easy to spot, and consideration of the graph shows that there must be a total of six roots.
Applying the two relationships about the values of will allow these other roots to be found. The
solution set for the inequality then follows easily from consideration of the graph.

Question 6.

The definition of the sequence can be used to find a relationship between and and
therefore also a relationship between and . Taking the difference of these then leads to the
required result.

It is clear from the definition of the sequence that, if one term is between 1 and 2, then the next
term will also be between 1 and 2. This is then easy to present in the form of a proof by induction for
part (ii).

The result of part (i) shows that the sequence in part (iii) is increasing and the result proved in part
(ii) shows that it is bounded above. The theorem provided at the start of the question therefore
shows that the sequence converges. Similarly the second sequence is bounded below and
decreasing (and therefore if the terms are all multiplied by -1 a sequence will be generated which is
bounded above and increasing). Therefore the second sequence also converges to a limit.

The relationship between and established in part (i) can then be used to find the value of
this limit and, as it is the same for both the odd terms and the even terms, the sequence must tend
to the same limit as well.

Finally, starting the sequence at 3 will still lead to the same conclusion as the next term will be
between 1 and 2 and all further terms will also be within that range, so all of the arguments will still
hold for this new sequence.

618
Question 7.

A solution of the equation should be easy to spot and a simple substitution will establish the new
solution that can be generated from an existing one. This therefore allows two further solutions to
be found easily by repeated application of this result.

In part (ii) write and and then substitute into (*). With some simplification the
required relationship will be established.

Since is a prime number there is only two ways in which it can be split into a product of two
numbers ( and ). The right hand side of the equation is clearly a difference of two
squares and therefore a pair of simultaneous equations can be solved to give expressions for and
. Finally, the expression for is similar to the relationship established in part (ii), so solutions to
the original equation can be used to generate values of , and which satisfy this equation.

Question 8.

Begin by calculating the largest area of a rectangle with a given width and then maximize this
function as the width of the rectangle is varied. The definition of can be reached by setting the
derivative of the area function to 0.

The definition of involves the differentiation of an integral of which uses the variable as the
upper limit. The derivative of ( ) is therefore ( ). The next statement relates the area bounded
by the curve and the line ( ) with the area of the largest rectangle with edges parallel to the
axes that can fit into that space, so the first area must be greater and since that integral is equal to
( ) ( ) the result that follows is easily deduced.

The final part of the question involves finding expressions for ( ) and ( ) and then simplifying
the relationship established at the end of part (ii).

Question 9.

Resolving the forces vertically will establish the first result. For the second part of the question it can
be established that all of the frictional forces are equal in magnitude by taking moments about the
centre of one of the discs. Resolving forces vertically and horizontally for the discs individually will
then lead to simultaneous equations that can be solved for the magnitudes of the reaction and
frictional forces.

Since the discs cannot overlap there is a minimum value that can take and the value of is
increasing as increases. This allows the smallest possible value of the frictional force between the
discs to be calculated and therefore it can be deduced that no equilibrium is possible if the
coefficient of friction is below this minimum value.

619
Question 10.

Following the usual methods of considering horizontal and vertical parts of the motion will lead to
the first result (some additional variables will need to be used, but they will cancel out to reach the
final result.

If and are the same point then the result in part (i) can be applied for this point which will give
an equation which is easily solved to give once the double angle formula has been applied.

For the final part it is possible to find the times at which the particle reaches each of the two points.
The two equations reached can then be used to find an expression for the difference between the
time at which the particle reaches each of the two points and then it can easily be deduced whether
this is positive or negative, which will show which point is reached first.

Question 11.

The standard methods of conservation of momentum and the law of restitution will allow the
speeds after the second collision to be deduced. A third collision would have to be between the first
and second particles and this will only happen if the velocity of the first particle is greater than that
of the second one.

Providing a good notation is chosen to avoid too much confusion, it is possible to find the velocities
after the third collision and then consider the velocities of the second and third particles to
determine whether or not there is a fourth collision.

Question 12.

The formula for the expectation of a random variable should be well known and both of the
expectations can easily be written in terms of and .

Similarly, the formula for variance should be well known and so it is a matter of rearranging the
sums in such a way as to reach the forms given in the question. Note that the definitions of and
are such that .

Since the ( ) ( ) the equation in the final part of the question can be rewritten in
terms of the variables defined at the start of the question. It can then be shown that this is not
possible for any non-zero value of .

620
Question 13.

An alternating run of length 1 must be two results showing the same side of the coin. It is then easy
to show that the probability is as given. Similarly a straight run of length 1 must be two different
results (in either order) and so the probability can again be calculated easily. The terms involved are
those in the expansion of ( ) and so starting with the statement that ( ) then
relationship between the two probabilities can be established.

An alternating run of length 2 must be one result followed by the other one twice, while a straight
run of length 2 must be two identical results followed by the other one. They will therefore be
calculated by the same sums (with the products in a different order each time) so the probabilities
must be equal. By considering the ways in which runs of length 3 can be obtained it is clear that
these two probabilities must also be equal.

An alternating run of length must be of each of the two possibilities followed by a repeat of
whichever came last. A straight run of length must be of one of the possibilities followed by 1
of the other. Taking the difference between these two probabilities gives an expression which can be
seen to always have the same sign, which will determine which probability is greater. A similar
method will also work for the final case.

621
STEP 3 2013 Hints and Solutions 

1.  The first two results, whilst not necessarily included in current A2 specifications, are 
standard work.  Applying them,   2   , which can then be 

evaluated using a change of variable to give   tan tan  .  To simplify this to 

√ √ √

obtain the required result,  tan tan tan  must be simplified using the relevant 

√ √
compound angle formula. 

It is fairly straightforward to show that    2 sin  , so applying this for   2, 1, 0  

√
and applying the main result of the question to evaluate   , gives   2 

2.  It is elegant to multiply by the denominator, then differentiate implicitly, and finally multiply 
by the same factor again to achieve the desired first result.  The general result can be proved by then 
using induction, or by Leibnitz, if known.  The general result can be used alongside the expression for  
, and the first derived result with the substitution  0  to show that the general term of the 
!
Maclaurin series for even powers of x is zero, and for odd powers of x is    .  Thus, as  
!
√
⋯ the required infinite sum is     with     , that is   . 
! !

3.  The scalar product of     with  ∑  , which is of course zero, can be expanded giving  

. 1  and three products   .  which are equal by symmetry, giving the required result.  
Expanding the expression suggested in (i), gives  ∑ . 2 . .  , which, bearing in mind 
that   . 1 ,   . 1 , and that   . ∑ 0 , gives the correct result.  Considering that  
.  ,   . 1 , and that  a is positive, enables the given values to be found.  Similarly  
√ √
.  ,   .  ,  and   . 1  yields   , , ,  .  In (iii), using the 
√

logic of (i),  . 4 1 .  , as required.  Expanding this, and using 
the coordinates of   and those of     that have been found, 

√ √ √ √ √
  ∑ 16 4  
√ √

16 4   which is sufficient. 

4.   The initial result is obtained by expanding the brackets and expressing the exponentials in 
trigonometric form.   The (2n)th roots of ‐1 are    ,   1 , which lead to the 
factors of   1  and these paired using the initial result give the required result.  Part (i) follows 
directly from substituting     in the previous result, and as n is even,   1 2 .  Using the 
given factorisation in part (ii), the general result can be simplified by the factor 

   2 cos 1 1.  Again substituting     , and that  cos cos  gives 

the evaluation required. 

622
5.  Writing     as     , and employing the permitted assumption, as    and    are 
coprime,    divides    .  Repetitions of this argument imply finally that     divides     .  Letting  
 ,    .  Continuing this argument similarly gives the result    . As a 
consequence,   1, and thus     and    must both be  1 .  Thus if   √   where    and    are 
coprime, it is rational and can be written in lowest terms,  then     and so   1 and thus   
√   is an integer.  Otherwise,  √   cannot be written as   , that is, it is irrational. 
  For (ii), using the same logic as in part (i), as    divides    ,     divides   , so  
, for some    . Likewise,   ′ , for some integer   ′ , and thus  ′ 1 ,  
so   1 , and   .  If      is a prime factor of    , then     divides    , and so     too.  
Writing    , using the logic of the very first part of the question, if     does not divide    ,    
divides   , and repetition of this argument leads to a contradiction.  So    is a prime factor of    . 
  and    is the highest power of     that divides     .  So    , and    .  So     
divides    , but as   and    are coprime,     divides     and thus    .  By the given result, this 
means   1 , and as     is only divisible by  1 ,   1 .  If     is a positive rational    , such that  
  is rational, then     so   1 and      is a positive integer. 
 
6.    The opening result is the triangle inequality applied to OW, OZ, and WZ where OW and OZ 
are represented by the complex numbers w and z.   
| ∗ ∗ ∗ ∗
Part (i) relies on using  |  ,   , | | | | | | , and 
∗ ∗
substituting   2| |  .  Having obtained the desired equation , the reality of E is 
apparent from the reality of the other terms and its non‐negativity is obtained from the opening 
result of the question.  Part (ii) relies on the same principles as part (i). 
The inequality can be most easily obtained by squaring it, and substituting for both numerator and 
denominator on the left hand side using parts (i) and (ii), and algebraic rearrangement leads to  
1 | | 1 | | 0 which is certainly true.  The argument is fully reversible as | | 1 , and  
| | 1 , | ∗ | 1, and so 1 ∗
0 so the division is permissible, and the square rooting of 
the inequality causes no problem as the quantities are positive.  The working follows identically if 
| | 1 , and  | | 1 . 
 
7.  As   2   is zero for all    ,   is constant, and     using the 

initial conditions.  The deduction follows from the non‐negativity of    .  In part (ii), it can be 

shown that   2 0  for   0,  and as initially    , the deduction for  

cosh  follows in the same way as that in part (i).  In part (iii), the choice of    relies on  
2 cosh 2 sinh   so   2 sinh cosh  .  Then  

  2 5 cosh 4 sinh 3 2 3  ,  and initially    .  
The final result can be deduced as in the previous parts,  with the additional consideration that  
sinh 0 . 
 
8.  The sum is evaluated by recognising that it is a geometric progression with common ratio 
⁄ ⁄
  which may be summed using the standard formula and as  1 0 , the denominator 

623
is non‐zero so the sum is zero.  By simple trigonometry,   cos  .  As    ,    .  
Thus      where   1 ⁄  .  Simplifying,    .  The 
summation of the reciprocals of this expression is simply found using a double angle formula and 
then by expressing  the trigonometric terms as the real part of the sum at the start of the question. 
 
9.  The volume is obtained as a volume of revolution     which gives the 
result.  Similarly, Newton’s 2nd law gives     which simplifies to the 
required result.  Substituting    when   0  gives    .  Substituting     
yields    , so for small    this approximates to SHM with period  

   . 
 
10.  The initial result can be obtained in a number of different ways, but probably use of the 
parallel axes rule is the simplest.  Conserving angular momentum about P,   

3    where     is the velocity of the particle after impact, 
and     is the angular velocity of the beam after the impact, and by Newton’s experimental law of 
impact   .  Eliminating    between these two equations gives the quoted 
expression for    .  Substituting   2  , for maximum    ,  0 .  This gives a quadratic 
equation, with solutions    and    .  The latter is not feasible and the former can be 
shown to generate a maximum which equates to the given result. 

 
11.  As the distance from the vertex to the centre of the equilateral triangle is    , the extended 

length of each spring is     giving the tension in each as     which simplifies to the 

given result.  Resolving vertically  3 sin 3  , and using the result for    , substituting    , 
and rationalising the denominator gives the required value for    .  Conserving energy, when    , 

gravitational potential energy is   3 tan  , elastic potential energy is  

1 , whereas when    , gravitational potential energy is   3 tan  , elastic 

potential energy is   1 , and kinetic energy is     hence giving    . 

!
12.  1  , so    .  There are     arrangements of the As and Bs, and the 
! !
!
number of arrangements with a B in the  1   th place and an A in the   th place is   , 
! !

so   1   for  2  , and     if   1 .  These combine to give  

  correctly.   

624
 1  only if the first letter is an A, the  1  th letter is a B, and the   th letter is an A.  This 
! !
has probability     giving    correctly.  
! ! ! !

  1  only if the  1  th letter is a B, and the   th letter is an A, the  1  th letter is a B, and 

! !
the   th letter is an A which has probability     so    , and 
! ! ! !

thus  ∑ 1   and so  

   ∑ ∑ ∑ 1   which yields the required result. 

∑ ∑ ∑ 2   so      which can be used to 
obtain     correctly. 

13.  integrating  0  between limits of 0 and    gives the result of (a) (i), and 

integrating the left hand side by parts yields part (ii).  As    is a probability density function, 
1 , which can be evaluated using the result of (a) (ii) with  2  and so  
2 .    2 2 2   and using (a) (ii) ,  
2 1  , as 
 , integration by parts gives  
 .  Part (iii) is derived from part (ii) by rearranging  

1 2   and making     the subject, then translating   1  to    . 

625


626


STEP Examiner’s Report 2014

Mathematics

STEP 9465/9470/9475

October 2014

627
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 3
STEP Mathematics II 8
STEP Mathematics III 12
Explanation of Results 15

628
 SI 2014 Report
General Comments
More than 1800 candidates sat this paper, which represents another increase in uptake for
this STEP paper. The impression given, however, is that many of these extra candidates are just not
sufficiently well prepared for questions which are not structured in the same way as are the A-level
questions that they are, perhaps, more accustomed to seeing. Although STEP questions try to give
all able candidates “a bit of an intro.” into each question, they are not intended to be easy, and (at
some point) imagination and real flair (as well as determination) are required if one is to score well
on them. In general, it is simply not possible to get very far into a question without making some
attempt to think about what is actually going on in the situation presented therein; and those
students who expect to be told exactly what to do at each stage of a process are in for a shock. Too
many candidates only attempt the first parts of many questions, restricting themselves to 3-6 marks
on each, rather than trying to get to grips with substantial portions of work – the readiness to give
up and try to find something else that is “easy pickings” seldom allows such candidates to acquire
more than 40 marks (as was the case with almost half of this year’s candidature, in fact).
Poor preparation was strongly in evidence – curve-sketching skills were weak, inequalities
very poorly handled, algebraic capabilities (especially in non-standard settings) were often pretty
poor, and the ability to get to grips with extended bits of working lacking in the extreme; also, an
unwillingness to be imaginative and creative, allied with a lack of thoroughness and attention to
detail, made this a disappointing (and, possibly, very uncomfortable) experience for many of those
students who took the paper.
On the other side of the coin, there was a very pleasing number of candidates who produced
exceptional pieces of work on 5 or 6 questions, and thus scored very highly indeed on the paper
overall. Around 100 of them scored 90+ marks of the 120 available, and they should be very proud
of their performance – it is a significant and noteworthy achievement.

Comments on individual questions

[Examiner’s note: in order to extract the maximum amount of profit from this report, I
would firmly recommend that the reader studies this report alongside the Hints and Solutions
supplied separately.]

Q1 Traditionally, question 1 is intended to be the most generous and/or helpful question on the
paper, in order to permit as many candidates as possible to get started in a reasonably friendly
situation, and thereby pick up at least 10 marks on the paper; giving them a positive start to the
examination. This year, however, despite the high rate of popularity (over 80% of the candidature
attempted Q1), there were several surprises in store for the examiners.
Firstly, it was not nearly so popular as it appears from the proportion of attempts, as it
turned out that many of these attempts were either weak or inconsequential, petering out the
moment the work became algebraic rather than numerical. The other surprise was how poorly the
very simple ideas were handled. Many candidates clearly did not know what constituted a proof in
these settings, when little more than (say) a statement such as 2k  1  (k  1) 2  k 2 in (ii) was
perfectly sufficient. Quite a few went on to attempt what was clearly intended to be an inductive
proof, having already written the correct (and wholly adequate) result, sometimes in each of parts
(ii), (iii) and (iv).

629
 Furthermore these sorts of mistakes were often preceded by incorrect numerical work in part
(i), including offerings that ignored the initial prompting regarding the use of non-negative integers,
such as “12 = 7 2   37  . In other parts of the question, candidates would resort to providing
2

counterexamples to results that had not been suggested; such as, in part (iv) producing a
counterexample (such as “6”) to refute the notion that “every number of the form 4k + 2 can be
written as the difference of two squares” when the question actually required them to show that no
number of this form has the proposed property, so offering one example represented a considerable
misunderstanding of mathematical ideas and terminology. Part (iv) also suffered from the common
misconception that factorising 4k + 2 as 2(2k + 1) immediately meant that 2k + 1 had to be prime.
Candidates who had seen and used modular arithmetic had a bit of an advantage in (iv) but,
in fact, there was very little evidence of such. Partly balancing the widespread lack of (pretty basic)
number theoretic appreciation were the few candidates who found this all very straightforward, as
had been intended to be the case. Overall, however, this question provided a very disappointing
range of responses, and the mean score of under 5/20 underlines this fact.

Q2 This was another very popular question, attempted by around 80% of candidates, and
producing much greater success, although it has to be noted that this usually extended to work up to
the end of part (iii), after which the integration efforts had become so prolonged and involved that
many candidates simply moved on elsewhere rather than plough on into (iv). This all led to a mean
score on the question of almost exactly 10/20.
There were two perfectly acceptable approaches to part (i); integrating the LHS by parts ,
after writing ln(2 – x) as the product ln(2 – x).1 or, instead, differentiating the RHS in order to
verify the result. The first method is probably that which is most “in the spirit of the game”, though
the second is, perhaps, the shrewder tactic. For those adopting the first approach, some difficulties
were encountered when the extra “2 – x” failed to appear naturally, indicating a failure to appreciate
the nature of arbitrary constants (namely, that “ – x + C” could equally well be written have been
written as “ – x + 2 + C”).
The curve-sketching was handled in the usual mixed way, with many candidates clearly
well-prepared and dealing admirably with crossing-points on the axes and asymptotes, while others
at the other end of the ability range seemed capable only of plotting points and “joining the dots”.
There were also many who thought that, towards asymptotes, the graph of a function should only
approach positive infinity, and this led to a -shaped central portion. However, there were very few
sketch-attempts that failed to pick up at least 3 or 4 point-scoring features of the graph at hand.
Following this, attempts at part (iii) again generally picked up quite a few of the available
 
marks by using the correct limits, separating ln 4  x 2 as ln(2 – x) + ln(2 + x), and then adapting
the given result of (i) to the second term (although this frequently took far longer than should have
been the case). However, when it came to part (iv), most candidates decided they’d had enough and
went elsewhere. Many who started (iv) were clearly thrown by the extra set of modulus-brackets
(even though they were intended to make life easy by rendering every part of the first curve
positive). The extra bits of area then came simply from the use of a new pair of limits and the given
result for evaluating the integrand in the vicinity of the asymptote.

630
Q3 This was, by far, the most popular question on the paper, with around 90% of candidates
making an attempt at it. The mean score on it was just under 9/20. There was much on this question
that made it straightforward, although there were several points at which candidates either
overlooked something or were not sufficiently careful in their explanation. Almost all candidates
managed part (i) successfully and managed to obtain the quartic equation in (ii). Many of these,
however, assumed they had made a mistake since the question gives a cubic instead, and some of
them tried, often repeatedly, the same working again rather than try to remove the “obvious” factor
of (b – 1). Continuing with the given result again allowed candidates to employ some fairly routine
skills, and many did so successfully (though often without important bits of explanation).
Part (iii) proved to be rather less successful for most candidates, as they simply substituted
straightaway for p and q and then found themselves in difficulties with the ab term that arose.
Finally, it was (once again) clear that the majority of candidates are really not very comfortable
with inequalities and very few of them managed to establish both “halves” of the required result.

Q4 This question was very unpopular indeed, almost certainly due to the lack of given structure.
It attracted the attention of less than half of all candidates, and many of these attempts got no further
than the first two lines of working involving the use of the Cosine Rule. This led to a mean score of
4/20 on this question and made it the second lowest scoring question on the paper.
Those candidates who did then differentiate generally overlooked the fact that they should
d
have been differentiating with respect to time; fortunately, with being constant, there was very
dt
little penalty in terms of both marks lost and in straying from a profitable path of progress through
the working.
As mentioned already, attempts that went significantly beyond this point were few and far
between, with relatively few of such candidates realising that they needed to turn the resulting
equation into a quadratic in cos , and even fewer managing to factorise it appropriately. For the
very final part of the question, quite a few managed to obtain the correct (given) answer
legitimately, even though they had been unsuccessful with the bulk of the question’s working until
then. This, at least, demonstrates a shrewd grasp of examination technique and generally earned
them 3 or 4 marks at the end.

Q5 As already indicated in the comments for Q3, candidates generally do not like working with
inequalities and this question is riddled with them. Q5 thus turned out to be the second least popular
of the pure questions, and scored poorly with a mean score of under 5/20, again largely due to a
lack of progress beyond the first part of the question.
Even in (i), there was a tendency to dive straight in to the sketch without having worked out
any useful points on the curve, including missing the obvious point (a, 0). Using this would have
led easily towards a factorisation of f(x) into three linear factors, though most preferred to find the
turning points instead (which approach works equally well). Though not crucial to the following
working, the special case a = 0 remained almost universally unaddressed.
Each of parts (ii) and (iii) can be approached via the AM-GM Inequality though very few did
so as we had instructed candidates to use the result of part (i). It was disappointing to see so few
serious attempts at part (ii) since all that is required is to set a = y and then the “x + 2y” is
practically waving at you. Part (iii) required a bit more thought (setting p = x and then q + r = 2a)
but both imagination and determination seemed in short supply by this stage of the question.

631
Q6 Around half of the candidates attempted this question, though successful working was (yet
again) almost entirely limited to the opening part. Even here, there were far too many candidates
who were unable to turn 4sin2 cos2 into sin22. Many of those who did spot this simplification
had difficulties trying to find u2 in a similar form – thereby completely overlooking the fact that
starting with sin2(A) at any stage of the process clearly had to yield sin2(2A) at the next. This also
applies to the inductive step, which thus requires almost no further working – a position which was
carefully avoided by all those who ploughed on into the standard format of a proof by induction
without thinking about what they had just established.
Part (ii), was not often attempted. Some candidates substituted for vn but not vn+1; others
thought that un and un2 were the same thing and collected their coefficients up together; and,
generally, the algebra was clearly found very unappealling. The very last part of the question
required a “hence” approach, so those candidates who simply set about the sequence numerically
scored only one mark. Even those who played the game according to the “hence” overlooked the
need to check that the given condition was satisfied.

Q7 Of the pure maths questions, this was by far the least popular, with attempts from only
around a quarter of the candidature. Very few attempts proceeded beyond the opening stages.
Surprisingly – despite the fact that the whole question can be done with little more than the
knowledge of how to split a line segment in a given ratio, vector equations of lines joining two
given points, and the finding of a point of intersection of two lines – most attempts began with
incorrect statements of the position vectors d and e in terms of b and a (respectively). Most
candidates wisely gave up at this point, though some persevered, but with little success. The mean
score of under 3/20 on this question reflects the paucity and brevity of efforts.
Geometers amongst the readership may notice that this result follows from a combination of
Menelaus’ Theorem and Ceva’s Theorem. A handful of candidates noticed it too, and scored most
of the marks for relatively little effort.

Q8 In hindsight, this question was a little too straightforward, and could well have been placed
earlier on in the paper. Nonetheless, around two-thirds of all candidates attempted it, and marks
were generally very high, making it the second highest-scoring question on the paper (and only
marginally behind Q2) at just under 10/20. Finding equations of lines and intersections in the
coordinate geometry setting was clearly much more in candidates’ comfort zone than the vector
setting of Q7, although there were problems caused by the surfeit of minus signs, and many
repeated their working for La when finding Lb rather than simply changing the a’s into b’s.
Parts (ii) and (iii) were also handled well, though slightly less confidently than (i). Part of
this was due to the lack of clear explanations given by candidates as to what had been done or
found, or a failure to realise that there was a need to justify that “ c ” satisfied the same conditions
as the a from earlier on. Sadly, some failed to give the x a new label (c here), and persisted to
substitute x’s as part of a gradient into what then became a non- linear formula. Overall, however,
this was a good question for candidates and most managed to make substantial progress most of the
way through it.

632
Qs.9-11 The mechanics questions in section B always seem to prove more popular than those
in the probability & statistics section C, and this was again the case this year. Qs. 9 & 11 each drew
around 700 attempts – I suspect because they look the more standard settings – while Q10 attracted
the attention of almost 550. However, those who persevered with Q10 generally scored more marks
(Q9’s means score was just under 6½/20; Q11’s a mark less; while Q10’s was around 9½/20),
largely because of the numerical “pay-off” at the end of the question.
The key point in Q9 was to realise that the given expression for T meant that T was the time
when the particle “turned round” horizontally. Those candidates who spotted this then had the
opportunity to score lots of marks, as the sketches related to the point at which the particle turned
back relative to its highest point and its landing-point. Those who approached the problem from a
purely algebraic direction usually struggled with the significances of the three (four) cases.
Those candidates who did well on Q10 were generally those who set their working out in a
more structured manner. For instance, a diagram to illustrate the assigned (symbols for the) speeds
and directions of the objects involved in the collision generally helps prevent mistakes involving
signs when applying the principles of Conservation of Linear Momentum and Newton’s
(Experimental) Law of Restitution. Indeed, without such a clear indication, it is often very difficult
indeed for the markers to follow working involving symbols that just appear from nowhere!
Unfortunately, Q11 suffered particularly from precisely this issue also, and most marks
gained on it came from successful attempts at the single-pulley scenario given in part (i) – for which
we allocated six marks. Efforts at part (ii) often saw candidates failing to produce diagrams
indicating directions for the various accelerations (etc.) and simply resorting to writing down
several vague statements based on vague interpretations of “resolving” ideas and/or N2L. The
failure to grasp that there needed to be some notion of relative accelerations involved for the two
masses attached to pulley P1 meant that most efforts were fatally flawed anyhow. Strangely, yet
illustrating again the shrewdness of exam. technique amongst some of the candidates, it was
possible to attempt the very final part just by taking the two given answers and running with them
(although many failed to appreciate that an ‘if and only if’ proof needed two directions of
reasoning).

Qs12 & 13 These questions elicited the least amount of interest from candidates (250 and 320
attempts respectively), though marks were generally in line with those for Qs. 9 & 11 (also
respectively) at around 6½/20 and 5½/20.
The big hurdle in Q12 was the modulus function, which many candidates simply ignored.
Those who were happy to use it properly generally gained the required result and used it to find that
k = 7 (though some failed to discount k = 1 along the way). A further obstacle arose as most
candidates who continued into the second part of the question failed to account for all six outcomes
when calculating P(X > 25).
In Q13, apart from the usual sign errors, most candidates correctly found g(x) and h(x). A
popular approach for the mean was to throw the word “centroid” at the problem and hope that this
sufficed. For the median, around half of attempts failed to consider the easy (symmetric) case when
m = c = ½(a + b). The case c > ½(a + b), corresponding to when m lies under the first line-
segment, was handled very well; the other case very poorly, since most candidates tried to work
from the LHS up rather than from the RHS down. A very few realised that working down from the
top actually made this just a “write down” by switching a for b suitably.

633
 S2 2014 Report
General Comments 

There were good solutions presented to all of the questions, although there was generally less 
success in those questions that required explanations of results or the use of diagrams and graphs to 
reach the solution. Algebraic manipulation was generally well done by many of the candidates 
although a range of common errors such as confusing differentiation and integration and simple 
arithmetic slips were evident. Candidates should also be advised to use the methods that are asked 
for in questions unless it is clear that other methods will be accepted (such as by the use of the 
phrase “or otherwise”). 

Comments on individual questions. 

Question 1 

While the first part of the question was successfully completed by many of the candidates, there 
were quite a few diagrams drawn showing the point P further from the line AB than Q. Those who 
established the expression for  cos  were usually able to find an expression for  sin  and good 
justifications of the quadratic equation were given. The case where P and Q lie on the lines AC 
produced and BC produced caused a lot of difficulty for many of the candidates, many of whom tried 
unsuccessfully to create an argument based on similar triangles. 

The condition for (*) to be linear in   did not cause much difficulty, although a number of candidates 
did not give the value of cos . Many candidates realised that the justification that the roots 
were distinct would involve the discriminant, although some solutions included the case where the  
discriminant could be equal to 0 were produced. However, very few solutions were able to give a 
clear justification that the discriminant must be greater than 0. 

In the final part some candidates sketched the graph of the quadratic rather than sketching the 
triangle in the two cases given. In the second case many candidates did not realise that Q was at the 
same point as C. 

Question 2 

This was one of the more popular questions of the paper. Most candidates successfully showed that 
the first inequality was satisfied, but when producing counterexamples, some failed to show that 
either  0 or  0 for their chosen functions. In the second part many candidates did not 
attempt to choose values of  ,   and  , but substituted the general form of the quadratic function 
into the inequality instead. In the case where the function involved trigonometric functions, many of 
those who attempted it were able to deduce that  , but several candidates made 
mistakes in the required integration. Those who established two inequalities were able to decide 
which gives the better estimate for  . 

   

634
Question 3 

Many candidates produced a correct solution to the first part of the question. There were a number 
of popular methods, such as the use of similar triangles, but an algebraic approach finding the 
intersection between the line and a perpendicular line through the origin was the most popular. 
Some candidates however, simply stated a formula for the shortest distance from a point to a line. 
Establishing the differential equation in the second part of the question was generally done well, but 
many candidates struggled with the solution of the differential equation. A common error was to 
ignore the case  0 and simply find the circle solution. 

The final part of the question was attempted by only a few of the candidates, many of whom did not 
produce an example that satisfied all of the conditions stated in the question, in particular the 
condition that the tangents should not be vertical at any point was often missed. 

Question 4 

Many candidates were able to perform the given substitution correctly and then correctly explain 
how this demonstrates that the integral is equal to 1. The second part caused more difficulty, 
particularly with candidates not able to state the relationship between arctan  and arctan . 
Attempts to integrate with the substitution  arctan  often resulted in an incorrect application 
of the chain rule when finding  . 

In the final part of the question many candidates attempted to use integration by parts to reach the 
given answer. 

Question 5 

This was the most popular question on the paper and the question which had the highest average 
score. Most candidates correctly solved the differential equation in the first part of the question, but 
many then calculated the constant term incorrectly. In the second part of the question most 
candidates were able to find the appropriate values of a and b, but then did not see how to apply 
the result from part (i) and so did the integration again or just copied the answer from the first part. 
Some candidates again struggled to obtain the correct constant for the integration and others did 
not substitute the correct values for the point on the curve (taking  ,  as  1,1  rather than 
, ). 

Question 6 

This was one of the less popular of the pure maths questions, but the average mark achieved on this 
paper was one of the highest for the paper. The first section did not present too much difficulty for 
the majority of candidates, with a variety of methods being used to show the first result such as 
proof by induction or use of  cos sin . In the second part of the question many of the 
candidates struggled to explain the reasoning clearly to show the required result. Most candidates 
who reached the final part of the question realised that the previous part provides the basis for a 
proof by induction. 

635
Question 7 

This was another of the less popular pure maths questions. The nature of this question meant that 
many solutions involved a series of sketches of graphs with very little written explanation. Most 
candidates were able to identify that the sloping edges of   would have the same gradient 
as the sloping edges of  , but many did not have both sloping edges overlapping for the two 
graphs. In some cases only one sloping edge of   was drawn. A large number of candidates 
who correctly sketched the graphs identified the quadrilateral as a rectangle, rather than a square. 
In the second part of the question, sketches of the case with one solution often did not have the 
graph of  | | meeting the  ‐axis at one corner of the square identified in part (i), although 
many candidates were able to identify the different cases that could occur. Unfortunately in the final 
part of the question very few candidates used the result from the first part of the question and so 
considered a number of possibilities that do not exist for any values of a, b, c and d. 

Question 8 

This was the least popular of the pure maths questions and also the one with the lowest average 
score. Many of the candidates were able to show the required result at the start of the question, 
although very few candidates explained that   could be either of the two integers when the range 
included two integers. Parts (i) and (ii) were then quite straightforward for most candidates, 
although many calculated the range of values but did not justify their choice in the case where there 
were two possibilities. In the final two parts of the question some candidates mistakenly chose the 
value 0 when asked for a positive integer. 

Question 9 

This question was not attempted by a very large number of candidates and the average score 
achieved was the lowest on the paper. While there were a number of attempts that did not proceed 
beyond drawing a diagram to represent the situation, the first part of the question was done well by 
a large number of candidates. Many were also able to adjust the result for the case when the 
frictional force acts downwards. Unfortunately, in the final part of the question many candidates 
continued to use  , not realising that this only applies in the critical case and so there were 
very few correct solutions to this part of the question. 

Question 10 

This was the most popular of the mechanics questions and also the one that had the best average 
score, although candidates did struggle to get very high marks on the question particularly on the 
final parts. The first part of the question asks for a derivation of the equation for the trajectory which 
was familiar to many candidates, although in some cases the result was obtained by stating that it is 
a parabola and knowledge of the maximum value and the range. Many candidates who successfully 
obtained the Cartesian equation then struggled with the differentiation with respect to  , instead 
finding the maximum height for a constant value of  . Unfortunately, this made the remainder of the 
question insoluble. Some candidates decided to differentiate with respect to   instead, which did 
not cause any serious problems, although it did require more work. A few candidates used the 
discriminant rather than differentiation, but did not provide any justification of this method. 

636
Candidates were able to draw the graph, but many did not label the area that was asked for in the 
question. Those who reached the final part of the question and considered the distance function for 
the position during the flight used differentiation to work out the greatest distance. However, many 
did not realise that the maximum value of a function can be achieved at an end‐point of the domain 
even with a derivative that is non‐zero. 

Question 11 

Many candidates who attempted this question struggled, particularly due to a difficulty in drawing a 
diagram to represent the situation. From these incorrect diagrams candidates often reached results 
where one of the signs did not match that given in the question. The calculation of the acceleration 
was found to be difficult by many of the candidates, although those who understood that 
differentiation of the coordinates of P would give the acceleration were then able to complete the 
rest of the question correctly. Those candidates that attempted the final part of the question were 
able to solve it correctly. 

Question 12 

This was the least popular question on the paper. A large number of candidates who attempted this 
question seemed unable to work out where to start on the first part of the question. Much of the 
rest of the question requires working with the hazard function defined at the start of the question 
and so many candidates who attempted these parts were able to do the necessary integration to 
solve the differential equations that arose. A common error among those who attempted part (iv) 
was to ignore the “if and only if” statement in the question and only show the result one way round. 

Question 13 

This was the more popular of the two questions on Probability and Statistics, but as in previous years 
it still only attracted answers from a very small number of candidates. The average mark for this 
question was also quite low, often due to a difficulty in explaining the reasoning behind some of the 
parts of the question. Many candidates were able to find the expression for  4  and most 
were then able to obtain the general formula required in part (i) of the question, although a number 
of candidates did not include the correct number of factors in the answer. Parts (ii) and (iii) did not 
cause too much difficulty, but the final part required a clear explanation to gain full marks. 

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STEP 3 2014 Examiners’ report 

A 10% increase in the number of candidates and the popularity of all questions ensured that all 
questions had a good number of attempts, though the first two questions were very much the most 
popular.  Every question received at least one absolutely correct solution.  In most cases when 
candidates submitted more than six solutions, the extra ones were rarely substantial attempts. Five 
sixths gave in at least six attempts. 

1.  This was the most popular question on the paper, being attempted by approximately 14 out 
of every 15 candidates.  It was the second most successfully attempted with a mean score of half 
marks.  The stem of the question caused no problems, but a common mistake in part (i) was to 
attempt derivatives to obtain the desired result.  Most candidates came unstuck in part (ii), making it 
much more difficult for themselves by attempting to work with expressions in a, b, and c rather than 
using the log series working with q and r, and as a result making sign errors, putting part (iii) beyond 
reach, and although they could find counterexamples for the claim in part (iv), they did so without 
the clear direction that working with the expressions in q and r would have made obvious. 

2.  This was only marginally less popular than question 1, but was the most successfully 
attempted with a mean of two thirds marks.  Most that attempted the question were able to do the 
first two parts easily, but could not find a suitable substitution to do the last part.  In about a tenth 
of the attempts, a helpful substitution was made in part (iii) which then usually resulted in successful 
completion of the question.  Modulus signs were often ignored, or could not be distinguished from 
usual parentheses, and the arbitrary constant, even though it appeared in the result for part (i), was 
frequently overlooked.  A few did not use the correct formulae for  cosh 2  , instead resorting to the 
trigonometric versions.  A handful of candidates attempted partial fractions in the last part having 
correctly factorised the quartic, but this did not use the previous parts as instructed. 

3.  About half of the candidates attempted this, but it was the second least successfully 
attempted with a mean score just below a quarter marks.  Most managed the first result, with those 
not doing so falling foul of various basic algebraic errors.  The second result of part (i) was often 
answered with no justification.  The second part was poorly done with a variety of approaches 
attempted such as obtaining a distance function, and then using completing the square or 
differentiation, or investigating the intersections of the circle and parabola.  Few considered the 
geometry of the parabola and its normal which would have yielded the results fairly simply. 

4.  Two thirds of the candidature attempted this but with only moderate success earning just a 
third of the marks.  The very first result was frequently obtained although some fell at the first 
hurdle through not appreciating that they needed to use  sec 1 tan   , or else that there 
was then an exact differential.  The second result in part (i) was ‘only if’ whereas many read it, or 
answered it, as ‘if’.  In part (ii), most spotted    .  There were many inappropriate functions 
suggested for the last part of the question, many which ignored the requirement that   0 , 1. 

5.  This was a moderately popular question attempted by half the candidates, with some 
success, scoring a little below half marks.  There were some basic problems exposed in this question 
such as the differences between a vector and its length, the negative of a vector and the vector, and 
the meaning of ‘if and only if’ resulting in things being shown in one direction only throughout the 
question.  Part (i) was generally well done, but in part (ii), it was commonly forgotten that there 

638
were two conditions for XYZT to be a square.  Approaches using real and imaginary parts (breaking 
into components) were not very successful. 

6.  The third most popular question, as well as the third most successful being only marginally 
behind question one in marks, having been attempted by about 70%.  Many did not use the required 
starting point, instead resorting to monotonicity or drawing pictures (graphs) which were not proofs.  
In parts (i) especially and (ii) as well, candidates failed to use the result that   0  , cavalierly 
using   0 , or even   1  without justification.  Many made complicated choices of 
functions for (i ) and (ii), and then got lost in their differentiations, and finally there was frequent 
lack of care to ensure that quantities dividing inequalities were positive. 

7.  Roughly two fifths of the candidates attempted this with a mean score of just over three 
marks making it the least well attempted question on the paper.  Most could do part (i), which is 
GCSE material, but frustratingly quite a few stated that the triangles were similar with no 
justification.  Part (ii) was by far the most poorly attempted part with a lot of hand‐waving 
arguments.  Part (iii) was done well by virtue of only the best candidates making it past part (i) with 
75% of solutions containing good proofs by contradiction for the first result and the last two parts 
were pretty well done. 

8.  Just fewer than half the candidates attempted this scoring just over a third of the marks.  
Many managed all but part (iii) easily but few managed that last part, and most did not try it.  In part 
(i), having correctly used the result from the stem, there was frequently not enough care taken in 
extending this to the full sum.   A not infrequent error of logic was that   ∑ 1 1  and  
lim 1 ∞  somehow implies that  ∑ 1   does not converge. 
→

9.  A fifth attempted this, scoring at the same level as question 8.  The first differentiation and 
the verification of the initial conditions were managed but very few bothered to check that the 
equation of motion was satisfied.  Most obtained the first displayed result but few realised that     
was the angle of depression rather than elevation and this generated plenty of sign errors.  A few did 
achieve the very final result. 

10.  This was attempted by a quarter of the candidates with scores just below those achieved in 
question 5.  Good candidates could obtain full marks in less than a page of working whilst weak ones 
spent a lot of effort trying to solve differential equations for    and     because they hadn’t spotted 
the change of variables to     and    .  The vast majority could obtain the first equation, 
often using the given result as a guide.  However, there was frequent confusion between extension 
and total length of the springs.  In addition, sign errors in     prevented the next part from working 
out.  Lots did not realise to work with     and    , but those that saw SHM in one of these, 
saw it in the other.  Likewise, with initial conditions, quite a few overlooked   0 , 0 , which 
prevented them solving for the constants, and also the sign was often overlooked in the condition 
  .  In attempting the last result, some used the factor formula, which worked but was 
unnecessary.  Quite often, they stumbled over the final step of logic ending up with apparent 
contradictions such as  √3 1 , which is of course false, but did not demonstrate full understanding. 

639
11.  Just marginally more popular than question 10, it was attempted with the same level of 
success.  Provided that a correct figure (and it didn’t matter whether P was above the level of B or 
not) was drawn, and that resolving was correctly conducted, then candidates could obtain the two 
tensions in general, in which case the inequality frequently followed.  However, the geometric result 
stumped many; the few completing it did so via the cosine rule and completing the square.  At this 
point, the final results usually followed for candidates still on track. 

12.  Less than 8% tried this, scoring just over a quarter of the marks.   Very few got the question 
totally correct, but a number got it mostly right.  Nearly all managed the median of Y, but the 
probability density function of Y caused some to stumble.  However the mode result, apart from 
some poor differentiation, was mostly alright.  The explanation in part (iii) eluded some candidates 
who were otherwise strong.  Applying the mode result in part (iv) to obtain     surprisingly tripped 
up some merely through inaccurate differentiation.  As one would hope, anyone that got as far as 
part (iv) spotted that the median of X was  . 

13.  Just a handful of candidates more attempted this than question 12, but scoring marginally 
less with one quarter marks.  A small number did just part (i), but otherwise candidates tended to 
either score zero or nearly all of the marks.  There was some very shaky logic in finding the first 
result of part (iii) and then failing to deduce, as required, the probability result.  Quite often, the 
working for     in part (iv), whilst usually correct, was extremely convoluted. 

640
Explanation of Results STEP 2014

All STEP questions are marked out of 20. The mark scheme for each question is designed to reward
candidates who make good progress towards a solution. A candidate reaching the correct answer will
receive full marks, regardless of the method used to answer the question.

All the questions that are attempted by a student are marked. However, only the 6 best answers are
used in the calculation of the final grade for the paper.

There are five grades for STEP Mathematics which are:

S – Outstanding
1 – Very Good
2 – Good
3 – Satisfactory
U – Unclassified

The rest of this document presents, for each paper, the grade boundaries (minimum scores required
to achieve each grade), cumulative percentage of candidates achieving each grade, and a graph
showing the score distribution (percentage of candidates on each mark).

STEP Mathematics I (9465)

Grade boundaries
Maximum Mark S 1 2 3 U
120 90 63 43 28 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 3.8 18.9 48.8 79.6 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics I

www.admissionstestingservice.org

641
STEP Mathematics II (9470)

Grade boundaries
Maximum Mark S 1 2 3 U
120 95 74 64 30 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 6.7 22.9 33.6 82.2 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics II

STEP Mathematics III (9475)

Grade boundaries
Maximum Mark S 1 2 3 U
120 81 59 48 28 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 11.0 35.4 52.2 84.5 100.0

Distribution of scores

4.0
3.5
3.0
2.5
Percent

2.0
1.5
1.0
0.5
0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics III

www.admissionstestingservice.org
642


643


STEP Solutions 2014

Mathematics

STEP 9465/9470/9475

October 2014

644
The Admissions Testing Service is a department of Cambridge English
Language Assessment, which is part of Cambridge Assessment, a not-for-
profit department of the University of Cambridge.

Cambridge English Language Assessment offers the world’s leading

qualifications for learners and teachers of the English language. Over 4 million
people from 130 countries take Cambridge English exams every year.

This mark scheme is published as an aid to teachers and students, to indicate

the requirements of the examination. It shows the basis on which marks were
awarded by the Examiners. It does not indicate the details of the discussions
which took place at an Examiners’ meeting before marking commenced.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the
relevant knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question

papers and the Report on the Examination.

The Admissions Testing Service will not enter into any discussion or
correspondence in connection with this mark scheme.

© UCLES 2014

More information about STEP can be found at:

www.stepmathematics.org.uk

645
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 4
STEP Mathematics II 11
STEP Mathematics III 18

646
STEP 1 2014 Hints and Solutions

Q1 This question is the traditional starter, involving ideas that should certainly be familiar to all
candidates. That doesn’t necessarily mean it is easy, just that everyone should be able to make a
start with it, and make some good progress thereafter. In fact, parts (i), (ii) and (iii) are each
especially amenable to all; and (i) is intended to help get you started along this particular road by
having you first write out some numerical examples of differences of two squares, before asking
you to move on to the algebra.
To begin with, (ii) requires only the observation that each odd number is the difference of
consecutive squares, 2k – 1  k2 – (k – 1)2; and then (iii) relies on noting that multiples of 4 arise
from the difference of squares of numbers that are two apart, namely 4k  (k + 1)2 – (k – 1)2. Part
(iv) develops these ideas further, although it is important now to deconstruct the problem into its
building blocks; in this case, that means examining the four possible cases for a2 – b2 when a and b
are either odd or even, none of which cases yield an even answer that is not automatically a multiple
of 4. Alternatively, this is very easily addressed by examining squares modulo 4.
Part (v) draws on ideas of factorisation of (positive) integers into factor-pairs. Here, you are
given the product pq where both p and q are odd. Using the difference of two squares factorisation,
if pq = a2 – b2 = (a – b)(a + b), then either a – b = 1 and a + b = pq OR a – b = p and a + b = q
(taking p < q w.l.o.g.), giving the (exactly) two required factorisations. However, if p = 2, then pq
is a multiple of 2, but not 4, and we have already shown this case to be impossible.
The final part of the question pulls all these ideas together in a numerical example, and we
need only prime-factorise 675 = 3352 and note that this yields (3 + 1)(2 + 1) = 12 factors and hence
six factor-pairs.

Q2 The main ideas behind this question are relatively straightforward, but there are many
difficulties in the execution of them. In (i), an integral of the form  ln(x) dx would usually have to
be approached by writing it as the product  ln( x)  1 dx and integrating by parts. On this occasion,
however, with a given result, it is perfectly possible to verify by differentiation. Curve-sketching
(rather than plotting) is a key skill and one that needs practice. The things to look for are crossing-
points on the two coordinate axes (which arise here when ln(1) appears), asymptotes (from when
we get ln(0)), symmetries and – if further detail is required – turning points. It should be obvious
here, for instance, that the curve is an even function, so there is a turning point when x = 0. In (iii),
you are given the area to be found so that you can check you are doing everything correctly before
you go on to answer the rather tougher part (iv). You should first realise (from your sketch-graph)
that it is required to integrate ln(4 – x2) between  3 and 3 , though you should always be on the
lookout for opportunities to make the working easier – here, you could integrate between 0 and 3
and then double the answer. Next, there is a very strong hint supplied by part (i) to split the log.
term up as ln(2 – x) + ln(2 + x), the first term of which has already been done and the second can be
deduced from it with a bit of care.
Part (iv) actually asks nothing new. Curve B is just curve A with all portions drawn above
the x-axis and it is only required that you deal with the extra bit(s) of area, and the awkwardness of
finding an area up to the asymptote is covered for you by the footnote that follows (iv).

647
Q3 This question was actually devised to address what happens when students misunderstand or
mis-apply a “rule” of mathematics and it turns out to give the right answer. Part (i) starts you off
gently: integrating both terms, squaring the RHS and solving very quickly gives b = 43 . Part (ii)
develops in much the same way, but with a non-zero lower limit to the integrals, and we
immediately see that the algebra gets much more involved. Importantly, it should be very clear that
whatever expression materialises must have (b – 1) as a factor (since setting b = a would definitely
give a zero area, thus trivially satisfying the given integral statement). This leads to the required
cubic equation.
The final part of (ii) requires a mixture of different ideas (and can be done in a number of
different ways). The most basic approach to demonstrating that a cubic curve has only one zero is to
illustrate that both of its TPs lie on the same side of the x-axis (or to show there are no TPs). The
popular Change-of-Sign Rule for continuous functions can be used to identify the position of this
zero.
Having got you started with some simple lower limits, part (iii) develops matters more
generally, and derives the (perhaps) surprising result that the exploration of this initial “stupid idea”
requires b and a to be “not too far apart” to an extent that is easily identifiable.

Q4 This is quite a sweet little question, and deals with the movements of the hands of a clock.
dx ab sin 
Using the Cosine Rule and differentiating gives  . However, it is
d b 2  a 2  2ab cos 
dx d2x dx d2x d
important to note that we can only work with and , rather than and , since is
d d 2
dt dt 2
dt
constant (so that extra terms in the use of the Chain Rule simply cancel). Then, differentiating again
with respect to  , equating to zero and solving leads to the quadratic equation in C = cos :
 
abC2 – b 2  a 2 C + ab = 0
which can be factorised (or solved otherwise). Only one of these solutions gives | C | < 1, and
a
substituting cos = gives the required result. [In the normal course of events, one should justify
b
that this is a maximum value rather than a minimum – but it is rather clear that it must be so in this
d 11
case.] Finally,  6  is the constant rate of change of , the angle between the two hands, and
dt
solving cos116    12 gives a time of 112 hours  11 minutes, as required.

648
Q5 This is another question that asks you to deploy your graph-sketching skills – in association
with the supporting algebra, of course – and finding the coordinates of the TPs of the family of
cubics given here, at (a, 0) and (–5a, 108a3) makes it clear that the requested result holds.
However, the case a = 0 (despite its relative triviality) still needs to be addressed separately.
Using (i)’s result with a = y then immediately gives 27xy2  (x + 2y)3  33 so that xy2  1.
Equality holds (from part (i)) when x = a = y and from when x + 2y = 3  x = y = 1.
Part (iii) requires a little more care and perseverance, but setting x = p and 2a = q + r gets
you off to a good start. There is now a little bit of “working to one side” needed if you are to
qr
2

convince yourself that    qr . Those students who have previously encountered the
 2 
Arithmetic Mean – Geometric Mean Inequality will have spotted this straightaway; otherwise, this
result becomes obvious upon rearrangement, as it is true  (q – r)2 ≥ 0, which is clearly true.
Having to write it out in this way does have the advantage of highlighting that equality holds if and
only if q = r here (leading to p = q = r for the main result).

Q6 Applying the given recurrence relation leads to

u1 = 4 sin2 (1 – sin2) = 4 sin2 cos2 = sin2(2) ,
though the question becomes very difficult indeed without the use of the two very basic
trigonometric identities involved to this point. A moment’s thought will help you avoid unnecessary
further working at this stage, as the determination of u2 from u1 = sin2(some angle) should
obviously give us the result u2 = sin2(twice that angle) in exactly the same way. Thus u2 = sin2(4)
and, in general, un = sin2(2n). Establishing the result inductively requires no more working than
that which was given two sentences ago, with the n = 1 “baseline” case having been given to you.
In part (ii), we approach a more awkward looking recurrence relation by transforming it
using the given substitution, and then comparing it with the required form. This gives the following
relationships:
p = 4, q – 2p = 4 and (q – 1)  + r = p 2
from which the required result follows. Upon checking, the final r.r. satisfies the appropriate
conditions, and since u0 = sin2  14   , we have un = sin2 2 n . 14   and so vn = 4sin2 2 n . 14   – 1.
However, it should be noted that sin 2 n . 14   = 0 for all n ≥ 2, so that {vn} = {1, 3, –1, –1, –1 …}.

649
Q7 When it boils down to it, this vectors question actually involves little more than the use of a
standard ratio result, vector equations of lines, and finding intersections of lines. Note first that, if a
p q
point T divides a line segment XY in the ratio p : q = : , then the respective position
pq pq
 q   p 
vectors of these points are related thus: t =  x   y . Indeed, it is precisely this result
 pq  pq
that leads to the vector equation of a line in the form r = (1 – )a + d (for the line AD here).
Writing BE’s equation similarly and substituting for d and e in terms of r, s, a and b then gives the
point of intersection of AD and BE by equating the two and comparing terms in a and b, deriving
the given form for g.
The equation of the line OG is then r = g, which meets AB at F, which is known to cut AB
in the ratio t : 1, which gives us two forms for the position vector of F. Comparing terms again
yields the required answer.
Incidentally, those students and teachers more familiar with standard results in Euclidean
Geometry will spot that this problem is actually an application of, first, Menelaus’ Theorem and
then Ceva’s Theorem.

 1
Q8 Part (i) directs you to finding the equation of La, which is y  1  ( x  a ) , and it follows
 a
 1
immediately that Lb has equation y  1  ( x  b) similarly. Solving simultaneously yields their
 b

point of intersection at ab, (1  a)(1  b)  . As b  a, this point of intersection  a 2 , (1  a ) 2 , and 
it is clear that a = x , so that 0 < 
x < 1 and 0 < x < 1, and that y  1  x . 2

Differentiating this gives

dy
dx
 1
1
which, at the point C  
c , (1  c ) 2 , gives a tangent
x

with equation y  1  c 2 
 1 
1 
  
 x  c . This rearranges into the form y  1 
1 
 x  c , 
 c  c
which is simply L c
where 0 < c < 1, as required.

650
Q9 This question provides rather a nice twist to the standard sort of projectiles question. Firstly,
there is a non-zero horizontal component of the acceleration. However, since this doesn’t affect the
U sin  2U sin  U cos 
vertical motion, it is still the case that TH = and TL = . Next, the time T =
g g kg
is introduced, which is quickly identified as the time when the horizontal component of the velocity
U sin  1
is zero. Writing it in the form T =  at an early stage is really helpful, partly because
g k tan 
U sin 
it pulls out the common factor of , but mostly as it identifies the factor of k tan which is
g
the major determining feature of the question. Having done this, you should realise that the three
cases do little more than ask you to sketch what happens depending upon when the wind (for surely
that is what is supplying this opposing force on the projectile) causes the particle to “trun round”
horizontally, relative to when the maximum height and greatest range are achieved. In the first case,
we get a “shortened” parabola compared to the usual shape of a non-wind-resisted projectile (so that
T isn’t reached before landing); in the second, the particle turns round horizontally before landing;
and in the third, the wind is so strong that the particle begins to be blown backwards before
reaching its greatest height. In the case of the “afterthought”, the particle is constrained to move up
and down in a straight line, returning to the point of projection.

Q10 In this question, we examine the dropping of two balls together, smaller on top of larger,
which leads to the surprising outcome where the smaller ball bounces so much higher than one
would expect it to. To begin with, we examine the bounce of the larger ball when it falls on its own.
Since it falls a distance H, it hits the ground with speed u given by u2 = 2gH, either by energy
considerations or by using the (“suvat”) constant acceleration formulae. It then leaves the ground
with speed v = eu according to Newton’s (Experimental) Law of Restitution, attaining a height H1
given by v2 = 2g(H1 – R). Substituting for v and then u gives the printed answer.
In the extended situation, we again use u  2 gH (for both balls) and v = eu (for the larger
ball after it bounces on the ground), though these do not need to be substituted immediately into the
“collision” statements that are gained when using the principle of Conservation of Linear
Momentum and N(E)LR for the subsequent collision between the balls. Solving simultaneously, the

upwards speed of the smaller ball is found to be x 

Me
 2 Me  m 
2
u . Using x2 = 2gd, where d is
M m
the distance travelled by the small ball’s centre of mass, and u2 = 2gH, we deduce the result that
2
 Me 2  2 Me  m 
d    H ; and then h = 2R + r + d.
 M m 
M
In the final part of the question, substituting in the given numbers yields the answer  9.
m
[Of course, this part of the question has really been asked back-to-front, as the real issue is to
explore the speed of the smaller ball after the collision; however, the numerical working would have
been much less favourable that way round.]

651
Q11 Part (i) of this question is a very straightforward single-pulley scenario, with the tension in
the string equal throughout its length, which is constant, and the accelerations of the particles equal
(relative to the string). Using Newton’s Second Law for each particle gives the acceleration (as
4 Mmg
shown) and the tension in the string given by T  .
M m
In the second part of the question, we apply exactly the same assumptions and principles in
exactly the same way, but there is a complication imposed now by the relative accelerations of the
two particles on the P1 pulley system. If we assume that the P1 system has the “same” acceleration
as the particle on the LHS of P, then (assuming that the m1 particle accelerates downwards within
this sub-system) the two particles on the RHS have accelerations b – a2 and b + a2 respectively.
Also, since P1 is taken to have zero mass, the tension in the main string is twice the tension in the
sub-system. Without this set-up, the following working is most unlikely to be meaningful in any
mark-scoring capacity. N2L applied several times, for the different particles, then leads to a set of
equations that gives the second printed answer.
With two given answers, the very final part of the question can be done as a “stand alone”
piece of work. Notice, however, that this is an if and only if proof and thus requires either two
separate arguments or one in which every step is reversible. In point of fact, it transpires that a1 = a2
if and only if (m1 – m2)2 = 0, which is equivalent to m1 = m2.

Q12 As with all such questions, one can make this much easier to deal with by being systematic,
and by presenting one’s working in a clear and coherent manner. To begin with, for instance, set out
a table of the six possible outcomes, along with their associated probabilities. Of course, the reason
why you are then asked for E(X 2) rather than E(X) is clearly because squaring negates any concerns
about the sign of whatever is inside a pair of “modulus” lines, so that they can simply be replaced
by ordinary brackets. You are then told that the (given) answer is a positive integer, which restricts
(k – 1) to being a multiple of 6 … namely, 1 or 7. Checking each of these in the possible E(X)’s that
arise enables you to eliminate k = 1 and confirm that k = 7.
Now that this information is known, it is possible to draw up the probability distribution for
21 4
X (again, a table works very well), work out the probabilities P(X > 25) = 144 and P(X = 25) = 144 ,
and then evaluate the expected payout
E(W) =  w  P( w)  = w  144
21
  1  1444   0
for the gambler (where W represents “winnings”). For this to be in the casino’s favour, this
expression must be negative, giving w < 7 and so the largest integer value of w is 6.

652
Q13 Although you are not asked for a sketch, a quick diagram might well help prevent stupid
mistakes. Since the area under the triangle’s two sloping sides is equal to 1, it follows that the
2
height of the triangle is h  , and so the gradient of the first line is this divided by (c – a) and
ba
the equation – that is, g(x) – follows immediately. Without further working, h(x) can be written
down immediately by substituting b for a appropriately (i.e. not in the bit of the expression for h;
and remembering that c – b is negative).
In (i), it is definitely not enough to think “Aha! I recognise that expression” and simply
throw the word “centroid” at the problem and hope to get all the points. You won’t. The question
makes it clear that it is intended for you to do the work to show that this expression is the mean.
(For a start, the values a, b and c lie on a line and aren’t “masses” placed at the vertices of the
triangle.) The value of E(X) must be found by integrating over the two regions, and then sorted out
algebraically.
For the final part of the question, it is important first to spend a moment making sure that
you can identify the different possible cases that arise. These depend upon whether c is < , = , or >
than 12 ( a  b) . The equality case is the easiest, and the easiest to overlook, since then m = c (by
symmetry). In the first case, the median lies under the first of the sloping line segments, and will be
gained by setting the area of a triangle equal to 12 . In a similar way to earlier, the third case can be
dealt with by “working down from b” instead of “working up from a). These give, respectively,
ma 1
2 (b  a )( c  a ) and m  b  1
2 (b  a )(b  c ) .

653
STEP II 2014

Question 1:

Drawing a diagram and considering the horizontal and vertical distances will establish the
relationships for 𝑥 cos 𝜃 and 𝑥 sin 𝜃 easily. The quadratic equation will then follow from use of the
identity cos2 𝜃 + sin2 𝜃 ≡ 1. The same reasoning applied to a diagram showing the case where P
and Q lie on AC produced and BC produced will show that the same equation is satisfied.
1
(*) will be linear if the coefficient of 𝑥 2 is 0, so therefore cos(𝛼 + 𝛽) will need to equal − , which
2
gives a relationship between 𝛼 and 𝛽. For (*) to have distinct roots the discriminant must be
positive. Using some trigonometric identities it can be shown that the discriminant is equal to
4(1 − (sin 𝛼 − sin 𝛽)2 ) and it should be easy to explain why this must be greater than 0.

The first case in part (iii) leads to 𝑥 = √2 ± 1 and so there are two diagrams to be drawn. In each
case the line joining P to Q will be horizontal.

√3
The second case in part (iii) is an example where (*) is linear. This leads to 𝑥 = 3
. Therefore Q is at
the same point as C and so the point P is the midpoint of AC.

Question 2:

𝜋 𝜋 𝜋 𝑛2 𝜋
By rewriting in terms of cos 2𝑛𝑥 it can be shown that ∫0 sin2 𝑛𝑥 𝑑𝑥 = 2 and ∫0 𝑛2 cos2 𝑛𝑥 𝑑𝑥 = 2
.
Therefore (*) must be satisfied as 𝑛 is a positive integer. The function 𝑓(𝑥) = 𝑥 does not satisfy (*)
and 𝑓(0) = 0 but 𝑓(𝜋) ≠ 0. The function 𝑔(𝑥) = 𝑓(𝜋 − 𝑥) will therefore provide a
counterexample where g(𝜋) = 0, but 𝑔(0) ≠ 0.

In part (ii), 𝑓(𝑥) = 𝑥 2 − 𝜋𝑥 will need to be selected to be able to use the assumption that (*) is
satisfied. The two sides of (*) can then be evaluated:
𝜋
𝜋5
∫ 𝑥 4 − 2𝜋𝑥 3 + 𝜋 2 𝑥 2 𝑑𝑥 =
0 30
𝜋
𝜋3
∫ 4𝑥 2 − 4𝜋𝑥 + 𝜋 2 𝑑𝑥 =
0 3

Substitution into (*) then leads to the inequality 𝜋 2 ≤ 10.

To satisfy the conditions on 𝑓(𝑥) for the second type of function, the values of 𝑝, 𝑞 and 𝑟 must
satisfy 𝑞 + 𝑟 = 0 and 𝑝 + 𝑟 = 0. Evaluating the integrals then leads to 𝜋 ≤ 22
7
.

2
Since (22
7
) < 10, 𝜋 ≤ 22
7
leads to a better estimate for 𝜋 2 .

654
Question 3:

By drawing a diagram and marking the shortest distance a pair of similar triangles can be used to
𝑐⁄ 𝑑
show that 𝑚
= 𝑐 , which simplifies to 𝑑 = 𝑐⁄ 2 .
𝑐√𝑚2 +1⁄ √𝑚 + 1
𝑚

For the second part, the tangent to the curve at the general point (𝑥, 𝑦) will have a gradient of 𝑦′
and so the 𝑦-intercept will be at the point (0, 𝑦 − 𝑥𝑦′). Therefore the result from part (i) can be
(𝑦 − 𝑥𝑦′)
applied using 𝑚 = 𝑦′ and 𝑐 = 𝑦 − 𝑥𝑦′ to give 𝑎 = ⁄ , which rearranges to give
√(𝑦′)2 + 1
the required result.

Differentiating the equation then gives 𝑦 ′′ (𝑎2 𝑦 ′ + 𝑥(𝑦 − 𝑥𝑦 ′ )) = 0 and so either 𝑦 ′′ = 0 or

𝑎2 𝑦 ′ + 𝑥(𝑦 − 𝑥𝑦 ′ ) = 0.

If 𝑦 ′′ = 0 then the equation will be of a straight line and the 𝑦-intercept can be deduced in terms of
𝑚.

If 𝑎2 𝑦 ′ + 𝑥(𝑦 − 𝑥𝑦 ′ ) = 0, then the differential equation can be solved to give the equation of a
circle.

Part (iii) then requires combining the two possible cases from part (ii) to construct a curve which
satisfies the conditions given. This must be an arc of a circle with no vertical tangents, with straight
lines at either end of the arc in the direction of the tangents to the circle at that point.

Question 4:

In part (i), if the required integral is called 𝐼 then the given substitution leads to an integral which can
be shown to be equal to −𝐼. This means that 2𝐼 = 0 and so 𝐼 = 0.
1
𝑏 arctan 𝑢
In part (ii), once the substitution has been completed, the integral will simplify to ∫1/𝑏 𝑑𝑢.
𝑢
1 𝜋 1 𝑏 𝜋
Since arctan 𝑥 + arctan (𝑥) = 2 the integral can be shown to be equal to 2 ∫1/𝑏 2𝑥 𝑑𝑥, which then
simplifies to the required result.

In part (iii), making with the substitution in terms of 𝑘 and simplifying will show that the integral is
equivalent to
∞
𝑘𝑢2
∫ 2 2 2 2
𝑑𝑢
0 (𝑎 𝑢 + 𝑘 )

Therefore choosing 𝑘 = 𝑎2 , the integral can be simplified further to

1 ∞ 𝑢2 1 ∞ 1 1 ∞ 𝑎2
∫ 𝑑𝑢 = ∫ 𝑑𝑢 − ∫ 𝑑𝑢
𝑎2 0 (𝑎2 + 𝑢2 )2 𝑎2 0 𝑎2 + 𝑢2 𝑎2 0 (𝑎2 + 𝑢2 )2
∞ 1
The result then follows by using the given value for ∫0 𝑎 2 +𝑥 2
𝑑𝑥.

655
Question 5:

Using the substitution 𝑦 = 𝑥𝑢, the differential equation can be simplified to

𝑑𝑢 1 + 4𝑢 − 𝑢2
𝑥 =
𝑑𝑥 𝑢−2
𝑦
This can be solved by separating the variables after which making the substitution 𝑢 = 𝑥 and
substituting the point on the curve gives the required quadratic in 𝑥 and 𝑦.
𝑑𝑌 𝑑𝑦
In part (ii), 𝑑𝑋 can be shown to be equal to 𝑑𝑥. The values of 𝑎 and 𝑏 need to be chosen so that the
right hand side of the differential equation has no constant terms in the numerator or denominator.
This leads to the simultaneous equations:

𝑎 − 2𝑏 − 4 = 0

2𝑎 + 𝑏 − 3 = 0
𝑑𝑌 𝑋−2𝑌
Solving these and substituting the values into the differential equation gives 𝑑𝑋 = 2𝑋+𝑌, and so

𝑑𝑋 2𝑋 + 𝑌
=
𝑑𝑌 𝑋 − 2𝑌

This is the same differential equation as in part (i), with 𝑥 = 𝑌 and 𝑦 = 𝑋. Most of the solution in
part (i) can therefore be applied, but the point on the curve is different, so the constant in the final
solution will need to be calculated for this case.

Question 6:

One of the standard trigonometric formulas can be used to show that

1 1
sin (𝑟 + 2) 𝑥 − sin (𝑟 − 2) 𝑥 = 2 cos 𝑟𝑥 sin 12𝑥.

Summing these from 𝑟 = 1 to 𝑟 = 𝑛 will then give the required result.

In part (i), the definition can be rewritten as 𝑆2 (𝑥) = sin 𝑥 + 12 sin 2𝑥. The stationary points can then
be evaluated by differentiating the function. The sketch is then easy to complete.

For part (ii), differentiating the function gives 𝑆𝑛′ (𝑥) = cos 𝑥 + cos 2𝑥 + ⋯ + cos 𝑛𝑥. Applying the
result from the start of the question, this can be written as
sin(𝑛+12𝑥) − sin 12𝑥
𝑆𝑛′ (𝑥) =
2 sin 12𝑥

Since sin 12𝑥 ≠ 0 in the given range, the stationary points are where sin (𝑛+12)𝑥 − sin 12𝑥 = 0. This can
1
then be simplified to the required form by splitting sin (𝑛+12)𝑥 into functions of 𝑛𝑥 and 2 𝑥 and noting
that sin 12𝑥 ≠ 0 and cos 12𝑥 ≠ 0 in the given range, so both can be divided by. By noting that the
1
difference between 𝑆𝑛−1 (𝑥) and 𝑆𝑛 (𝑥) is 𝑛 sin 𝑛𝑥 the result just shown can be used to show the
final result of part (ii). Part (iii) then follows by induction.

656
Question 7:

By considering the regions 𝑥 ≤ 𝑎, 𝑎 < 𝑥 < 𝑏 and 𝑥 ≥ 𝑏, 𝑓(𝑥) can be written as

𝑎 + 𝑏 − 2𝑥 𝑥≤𝑎
𝑓(𝑥) = 𝑏−𝑎 𝑎<𝑥<𝑏
2𝑥 − 𝑎 − 𝑏 𝑥≥𝑏

Therefore the graph of 𝑦 = 𝑓(𝑥) will be made up of two sloping sections (with gradients 2 and -2
and a horizontal section). The graph of 𝑦 = 𝑔(𝑥) will have the same definition in the regions 𝑥 ≤ 𝑎
and 𝑥 ≥ 𝑏, with the sloping edges extending to a point of intersection on the 𝑥-axis. The
quadrilateral with therefore have sides of equal length and right angles at each vertex, so it is a
square.

In part (ii), sketches of the cases where 𝑐 = 𝑎 and 𝑐 = 𝑏 show that these cases give just one
solution. If 𝑎 < 𝑐 < 𝑏 there will be no solutions and in the other regions there will be two solutions.

In part (iii) the graphs for the two sides of the equation can be related to graphs of the form of 𝑔(𝑥)
(apart from the section which is replaced by a horizontal line) in the first part of the question. Since
𝑑 − 𝑐 < 𝑏 − 𝑎, the horizontal sections of the two graphs must be at different heights so the number
of solutions can be seen to be the same as the number of intersections of the graphs of the form of
𝑔(𝑥).

Question 8:

The coefficients from the binomial expansion should be easily written down. It can then be shown
that
𝑐𝑟+1 𝑏(𝑛 − 𝑟)
=
𝑐𝑟 𝑎(𝑟 + 1)

This will be greater than 1 (indicating that the value of 𝑐𝑟 is increasing) while 𝑏(𝑛 − 𝑟) > 𝑎(𝑟 + 1),
𝑛𝑏−𝑎 𝑐𝑟+1 𝑛𝑏−𝑎 𝑐 𝑛𝑏−𝑎
which simplifies to 𝑟 < . Similarly, = 1 if 𝑟 = and 𝑟+1 < 1 if 𝑟 > . Therefore the
𝑎+𝑏 𝑐𝑟 𝑎+𝑏 𝑐𝑟 𝑎+𝑏
𝑛𝑏−𝑎
maximum value of 𝑐𝑟 will be the first integer after 𝑎+𝑏 (and there will be two maximum values for
𝑛𝑏−𝑎
𝑐𝑟 if 𝑎+𝑏 is an integer. The required inequality summarises this information.

In parts (i) and (ii) the values need to be substituted into the inequality. Where there are two
possible values, it needs to be checked that they are equal before taking the higher if this has not
been justified in the first case.

In part (iii) the greatest value will be achieved when the denominator takes the smallest possible
value, therefore 𝑎 = 1, and then in part (iv) the greatest value will be achieved by maximising the
numerator. Since the maximum possible value of 𝐺(𝑛, 𝑎, 𝑏) is 𝑛, 𝑏 ≥ 𝑛 will achieve this maximum.

657
Question 9:

Once a diagram has been drawn the usual steps will lead to the required result:

Resolving vertically:
𝐹 + 𝑇 cos 𝜃 = 𝑚𝑔
Resolving horizontally:
𝑇 sin 𝜃 = 𝑅
Taking moments about A:
𝑚𝑔(𝑎 cos 𝜑 + 𝑏 sin 𝜑) = 𝑇𝑑 sin(𝜃 + 𝜑)
Limiting equilibrium, so 𝐹 = 𝜇𝑅:
𝜇𝑇 sin 𝜃 + 𝑇 cos 𝜃 = 𝑚𝑔
Therefore:
𝑇𝑑 sin(𝜃 + 𝜑) = 𝑇(𝜇 sin 𝜃 + cos 𝜃)(𝑎 cos 𝜑 + 𝑏 sin 𝜑)
And so:
𝑑 sin(𝜃 + 𝜑) = (𝜇 sin 𝜃 + cos 𝜃)(𝑎 cos 𝜑 + 𝑏 sin 𝜑)

If the frictional force were acting in the opposite direction, then the only change to the original
equations would be the sign of 𝐹 in the first equation. Therefore the final relationship will change to
𝑑 sin(𝜃 + 𝜑) = (−𝜇 sin 𝜃 + cos 𝜃)(𝑎 cos 𝜑 + 𝑏 sin 𝜑)

For the final part, the first and third of the equations above can be used to show that
𝑇𝑑 sin(𝜃 + 𝜑)
𝐹= − 𝑇 cos 𝜃
𝑎 cos 𝜑 + 𝑏 sin 𝜑
𝑎+𝑏 tan 𝜑
Since 𝐹 > 0 if the frictional force is upwards, this then leads to the condition 𝑑 > tan 𝜃+tan 𝜑. Since
the string must be attached to the side 𝐴𝐵, 𝑑 cannot be bigger than 2𝑏, which leads to the final
result of the question.

Question 10:

Consideration of the motion horizontally and vertically and eliminating the time variable leads to a
Cartesian equation for the trajectory:

𝑔𝑥 2
𝑦 = 𝜆𝑥 − (1 + 𝜆2 )
2𝑢2

The maximum value can be found either by differentiation or by completing the square. Completing
the square gives:

𝑔𝑥 2 𝑢2 𝑢2 𝑔𝑥 2
𝑦=− (𝜆 − ) + −
2𝑢2 𝑔𝑥 2𝑔 2𝑢2

𝑢2 𝑔𝑥 2
Which shows that 𝑌 = 2𝑔 − 2𝑢2 . If this graph is sketched then the region bounded by the graph and
the axes will represent all the points that can be reached.

The maximum achievable distance must lie on the curve and the distance, 𝑑, of a point on the curve
2
𝑢2 𝑔𝑥 2
can be shown to satisty 𝑑2 = (2𝑔 + 2𝑢2 ) , which must be maximised when 𝑥 takes the maximum
value possible.

658
Question 11:

A diagram shows that the coordinates of 𝑃 are (𝑥 + (𝐿 − 𝑥) sin 𝛼 , −(𝐿 − 𝑥) cos 𝛼)

Therefore, by differentiating the 𝑦-coordinate of 𝑃 shows that the vertical acceleration of 𝑃 is

𝑥̈ cos 𝛼 and applying Newton’s Second Law gives

𝑇 cos 𝛼 − 𝑘𝑚𝑔 = 𝑘𝑚𝑥̈ cos 𝛼

A similar method for the horizontal motion of 𝑃 and 𝑅 gives the two equations

𝑇 sin 𝛼 = −𝑘𝑚(1 − sin 𝛼)𝑥̈

𝑇 − 𝑇 sin 𝛼 = −𝑚𝑥̈

For part (ii), eliminating 𝑇 from the last two equations gives the required relationship. A sketch of
𝑥
the graph of 𝑦 = (1−𝑥)2 will then show that for any value of 𝑘 there is a possible value between 0
and 1 for sin 𝛼.

In part (iii), elimination of T from the two equations formed by considering the motion of 𝑃 gives the
required result.

Question 12:

The required probability in the first part is given by

𝑃(𝑡 < 𝑇 < 𝑡 + 𝛿𝑡) 𝐹(𝑡 + 𝛿𝑡) − 𝐹(𝑡)

=
𝑃(𝑇 > 𝑡) 1 − 𝐹(𝑡)

In the case of small values of 𝛿𝑡, 𝐹(𝑡 + 𝛿𝑡) − 𝐹(𝑡) ≈ 𝑓(𝑡)𝛿𝑡, which leads to the correct probability.
1
In part (ii), differentiation gives 𝑓(𝑡) = , and substituting into the definition of the hazard function
𝑎
1
gives ℎ(𝑡) = 𝑎−𝑡. Both graphs are simple to sketch.

𝐹′(𝑡) 1
In part (iii), using the definition of the hazard function gives 1−𝐹(𝑡) = 𝑡 . Integrating gives
− ln|1 − 𝐹(𝑡)| = ln|𝑘𝑡|, and so the probability density function can be found by rearranging to find
𝐹(𝑡) and then differentiating.

A similar method in part (iv) shows that if ℎ(𝑡) is of the form stated then 𝑓(𝑡) will be of the given
form. Similarly, if 𝑓(𝑡) has the given form then ℎ(𝑡) can be shown to have the form stated.

In part (v), a differential equation can again be written using the definition of the hazard function
and this can again be solved by integrating both sides with respect to 𝑡.

659
Question 13:

Considering the sequence of events for 𝑋 = 4, the 1st, 2nd and 3rd numbers must all be different and
then the 4th must be the same as one of the first three. The probability is therefore

1 2 3
𝑃(𝑋 = 4) = (1 − ) (1 − )
𝑛 𝑛 𝑛

The same reasoning applied to 𝑋 = 𝑟 gives

1 2 𝑟−2 𝑟−1
𝑃(𝑋 = 𝑟) = (1 − ) (1 − ) ⋯ (1 − )
𝑛 𝑛 𝑛 𝑛

The result of part (i) is then found by observing that the probabilities of all possible outcomes add up
to 1.

Substituting the probabilities into the formula for 𝐸(𝑋) gives

2 1 2 1 2 3 1 2 𝑛−1
𝐸(𝑋) = + 3 (1 − ) + 4 (1 − ) (1 − ) + ⋯ + (𝑛 + 1) (1 − ) (1 − ) ⋯ (1 − )
𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛

For part (iii) observe that any case where 𝑋 ≥ 𝑘 will have the first 𝑘 − 1 numbers all different from
each other. Therefore

1 2 𝑘−2
𝑃(𝑋 ≥ 𝑘) = (1 − ) (1 − ) ⋯ (1 − )
𝑛 𝑛 𝑛

The first formula in part (iv) can be shown by considering 𝑘𝑃(𝑌 = 𝑘) to be equal to the sum of 𝑘
copies of 𝑃(𝑌 = 𝑘) and then regrouping the sum for 𝐸(𝑌). Finally this gives two different
expressions for 𝐸(𝑌), which must be equal to each other:

2 1 2 1 2 3 1 2 𝑛−1
+ 3 (1 − ) + 4 (1 − ) (1 − ) + ⋯ + (𝑛 + 1) (1 − ) (1 − ) ⋯ (1 − )
𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛 𝑛
1 1 2 1 2 𝑛−1
= 1 + 1 + (1 − ) + (1 − ) (1 − ) + (1 − ) (1 − ) ⋯ (1 − )
𝑛 𝑛 𝑛 𝑛 𝑛 𝑛

Rearranging and using the result from part (i) then gives the required result.

660
STEP 3 2014 Hints and Solutions 

1.  The stem results are obtained through algebraic expansion and equating coefficients.  Using 
the expression   1 1 1  for  1  , manipulating the logarithm 
of the product, and the series expansions for expressions like  ln 1   yields the 
displayed result.  In parts (ii), (iii), and (iv), it is simplest to find    ,   ,  , 
  , and     by expanding the series for  ln 1  , and 
choosing a counter‐example, selecting  a, b and c so that   0 . 

2.  The first part is solved using the given method, the formula  cosh 2 2 cosh 1 , and 

then employing partial fractions or the standard form quoted in the formula book.  The 
second part requires the substitution,   sinh  , the formula  cosh 2 1 2 sinh  , 
√
and a standard form to give   tan √2   .  The third part can be approached by 
making the substitution     and division of the resulting fraction in the numerator and 
denominator by    to give half the difference of the integrals in the first two parts.  
Alternatively, a similar style of working with the substitution     results in a sum 
instead of a difference. 

3.  (i) Given that the shortest distance between the line and the parabola will be zero if they 
meet, investigating the solution of the equations simultaneously , and the discriminant of 
the resulting quadratic equation, the first result of the question is the case that they do not 
meet.  The closest approach is the perpendicular distance of the point on the parabola 
where the tangent is parallel to the line, so using the standard parametric form, it is the 
perpendicular distance of   ,   from   , giving the required result with care 
being taken over the sign of the numerator bearing in mind the inequalities. 

(ii) The shortest distance of a point on the axis from the parabola, is either the distance from 
the vertex to the point, or the distance along one of the normals (which are symmetrically 
situated) which is not the axis.  If the normal at   , 2  passes through   , 0  , then 
2  .  From this it can be simply shown that shortest distance is     if   2 , and is 
2   if   2 . 

Then for the circle, the results follow simply, that the shortest distance will be    if  
 , and 0 otherwise if   2 , and 2   if  4   or 0 otherwise if 
2 . 

4.  Expanding the bracket in the integral    , and employing  sec 1 tan   yields    plus 

the integral of a perfect differential which can be evaluated simply.  For   0 , 
  tan 0  over the interval which can be solved using an integrating factor and then 
the condition   0 , 1 enables the arbitrary constant to be evaluated giving the 
required result.  In part (ii), similar working can be undertaken with the integral which is to 
be considered, given    .  The argument requires no discontinuity in the interval so  
 .  The function   cos   can be shown to meet the requirement. 

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5.  ABCD is a parallelogram if and only if    which yields the required result.  To be a 
square as well, angle   90  , and  | | | | , so    .  Treating the 
two results as simultaneous equations to be solved for     and     in terms of     and  , the 
second result of the stem can be shown with reversible logic.  For part (i) the same logic can 
be used for    as just used for  .  From the stem, XYZT is a square if and only if  
 , and 

  and given the working for X in part (i), these can be shown to be true 
treating Y, Z, and T similarly 

6.  Starting from   ′′ 0  for  0    ⟹ ′′ 0  where  0  , with 

the given conditions yields   0 , and then repeating the argument with   ′   
instead gives   0 .  Choosing   1 cos cosh   and applying the applying the 
stem of the question for  0  , gives the required inequality for   0 /2  in 
particular.  For part (ii), choosing   sin sinh   (in which case   ′′ 2  , 
where    was the suggested choice for part (i)) and   sin cosh   provide the 
desired results once care is taken with positivity of functions over the required interval when 
dividing inequalities. 

7.  Part (i), the intersecting chords theorem, is basic bookwork relying on angle properties in 
circles to establish similar triangles and hence the result.  Part (ii) can be obtained by 
considering that     lies on     and so     , that    also lies on    
producing a similar result and then equating these two expressions, finally rearranging to 
give (*).  Assuming that   0  and using (*) leads to    
which, in view of the distinctness of the four points    and the intersection of     and   

  at    , leads to the contradiction   0 .  Re‐writing    as  

  and similarly, using (*), as     and re‐writing, the expression can be 
shown to be the position vector of    .The final result comes from applying (i) using the 
information just gained and calculating both expressions by taking scalar products of the 
vectors whose magnitudes are quoted in (i). 

8.  The initial result is obtained by extending the given inequality so that each term of the sum 
is compared with     and    .  Part (i) is obtained using the stem, the given 
function,  2 , and summing the sums.  The deduction relies on considering the lower limit 
of the sum.  The same approach applies to part (ii), with the new function given and 
considering the upper limit which is obtained as a geometric progression.  Counting the 
number of elements of   1000  gives the method for obtaining    using the same 
function as part (i) except  0  if     has one or more 2 s in its decimal representation 
and with   10 , again with the sum of a geometric progression.  The final result is 
particularly attractive, demonstrating how few terms need  to be removed from the non‐
convergent harmonic progression (of part (i)) in order to produce a convergent sequence. 

9.  e    and a further differentiation yields m m mk  .  Using  r. j 0 

.
  obtains the first displayed result after re‐arrangement, as does  tan  the second. 
.

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   tan tan   can be shown to be   sinh kT kT   which leads to the two 
final inequalities. 

10.  The first result is obtained by considering Newton’s second law applied to the mass X under 
the tension in PX and the thrust of XY.      is similarly obtained considering 
Y.  Subtracting the two equations gives a SHM second order differential equation for   , 
and adding them gives similar for   .  Solving these using initial conditions give  
cos   and    cos √3   .  The final result is particularly elegant, 
and possibly a little surprising that a conservative oscillating system does not return to its 
starting position.  Treating the previous two results as simultaneous equations for    and   , 
and solving    , yields 1 cos √3   and  1 cos  , so that √3 2   and  
2   for non‐zero integers     and    , yielding the contradiction  √3  . 

11.  Resolving vertically and horizontally, and solving the resulting simultaneous equations and 
then tidying up the trigonometric expressions yields     and  

 .  Trivially, the former is positive, but the same condition 
applied to the latter, given that its denominator and the common factor of the numerator 
can be shown to be positive, yields the first required inequality.  The geometric inequality 
could be proved, as candidates tended to, by use of the cosine rule and then completing the 
square to obtain   cos cos  .  However, use of the sine rule and 
the maximum of the sine function, or the shortest distance of B from AP gives   sin  , 
which along with  cos sin 1 , give the required inequality.  In the particular case, 
it can be shown that    and the knowledge of the unattainable maximum value of 
∝
the cosine function along with the geometric inequality previously obtained leads to the 
final inequalities.  The geometry is that the strings are perpendicular, which can be 
appreciated by considering the equality case of    sin  . 

12.  The first result,    , is obtained merely by considering probabilities, and the given 

pdf of Y can be obtained by standard techniques or by consideration of changing the variable 
in the integral of the pdf of X.  The mode result relies on differentiation of the pdf of Y 
equated to zero to give a stationary value.  The explanation in part (iii) is simply that the 
required integral is merely that of the pdf of a Normal variable with mean   .  The 
expectation of Y is obtained in the standard manner, using an integral and the pdf of Y, and 
then a change of variable, in which exponential terms can be combined so as to use the 
explained result having completed the square in the exponent.  Using the three previous 
parts gives    ,    because X is symmetric, and, as stated,  

   , hence satisfying part (iv). 

13.  The first result is a trivial application of the definition of a probability generating function, 
and the second similarly.  In order to obtain the first printed result in part (iii), it is necessary 
to obtain a similar result to those in parts (i) and (ii)  giving    as the score is one higher 
and then applying the conditionality of the probabilities of these three results which is done 

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 by considering  the probability of a score     in the three cases to give the coefficient of   .  
Re‐arranging the formula for    , either differentiation or the binomial theorem can be 
used to find the required probability formula.  Finding   1 ⁄   and the 
knowledge that   1  enables the result of part (iii) to be rearranged to that of 
part (iv). 

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STEP Examiner’s Report 2015

Mathematics

STEP 9465/9470/9475

October 2015

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Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 3
STEP Mathematics II 8
STEP Mathematics III 11
Explanation of Results 14

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 SI 2015Report
General Comments
The aim of this report is to account for how well, or how poorly, the candidates performed
this year while at the same time attempting to indicate their corresponding areas of strength, or
weakness. If I should concentrate marginally more in the direction of the candidates’ weaknesses,
the reader should understand that it is with the hope that both future STEP candidates, and the
teachers preparing them for this examination in years to come, will have the opportunity to focus on
those areas of common weakness in an effort to ensure a better preparedness for what is
unquestionably the most demanding of examinations in the UK for students of pre-university years.
For the record, the scripts are marked by a team of postgraduate mathematical students –
many working towards a doctorate in this, or another, closely-related subject – who spend days
poring over the scripts, working in small teams under the supervision of the Principal Examiner and
carefully appointed “question captains”. Their powers of concentration are truly phenomenal and
they not only appreciate the need for mathematical rigour but (due to having once been in exactly
this position themselves) are also deeply sympathetic towards the candidates; making every effort
first to understand what has been presented to them by the candidates and then to reward genuinely
good mathematics when it appears, no matter how hastily and/or messily it has been set down onto
paper. Thus it is that the comments produced by the Principal Examiner within this Report are
merely summaries of what these markers have passed on to him (or her) at the end of the marking
period. Moreover, since the candidates’ backgrounds are entirely unknown to the markers, any
comments – critical or otherwise – cannot possibly be taken to have been directed towards
specifically chosen targets.
More than 2000 candidates sat SI this year, which represents another increase of around
10% over last year’s entry numbers. Once again, however, it is sadly the case that many of these
candidates have simply not prepared sufficiently well to be in a position to emerge from the
experience with any amount of positive feelings of success at the results of their efforts. In the first
instance, many candidates (this year, more than half of the entry) attempt more than the
recommended six questions. This automatically penalises them for the time that they have spent on
extra questions whose marks will not count towards their final total – remember that only the
highest-scoring six questions count towards a candidate’s final total; note also that a grade 1 can
usually be obtained from four questions which have been completed reasonably successfully, or
from two questions done completely correctly plus four “halves”, or from any in-between
combination of question-scores. Thus, it is strongly advised that candidates spend a few minutes at
some stage of the examination reading the questions carefully with a view to deciding which of
them they would best attempt.
Overall, this year’s paper worked out in very much the same way as had the 2014 paper,
with a mean score of around 43-44% and with approximately 45% of candidates failing to exceed a
total of 40 marks; though totals in excess of 100 marks were slightly down on 2014. Part of the
reason for this is that the five applied maths questions were very much non-standard this year, and
this prevented a lot of very easy marks (for routine beginnings) being picked up by candidates
attempting these questions. The mean score for Qs.9-13 thus fell from 6½ in 2014 to 2½ in 2015.
Another trend of recent years is the widespread dislike for the vectors questions, which have
become both unpopular and very low-scoring for candidates.

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 Points of general application regarding candidates’ attempts this year are little different to
usual – far too many candidates produce only fragmentary attempts (often, as mentioned above, to
almost every question they attempt) at solutions, with little apparent intent to persevere beyond the
first obstacle. Presentation was also particularly poor this year, with most candidates making life
hard both for themselves and for the markers who genuinely wish to find credit-worthy mathematics
in order to award the marks available. In long questions such as these, with the barest minimum of
structure provided, candidates need a lot of prior practice at past STEP questions in learning how to
supply their own.
Curve-sketching skills continue to be a weakness, as candidates tend to veer away from
justifying what they have drawn; algebra and calculus skills are very mixed, and it is was especially
clear this year how little candidates like being required to formulate their own solution-strategies –
no doubt being the result of an over-reliance on being told exactly what to do, as is customarily the
case in AS- and A-level papers.
On the other side of the coin, there was a very pleasing number of candidates who produced
exceptional pieces of work on 5 or 6 questions (or more), and thus scored very highly indeed on the
paper overall. Around 80 of them scored 90+ marks of the 120 available, and they should be very
proud of their performance – it is a significant and noteworthy achievement.

Comments on individual questions

[Examiner’s note: in order to extract the maximum amount of profit from this report, I
would firmly recommend that the reader studies this report alongside the Hints and Solutions or
Marking Scheme supplied separately.]

Q1 Traditionally, question 1is intended to be the most generous and/or helpful question on the
paper and, most years, it is attempted by almost everyone. This year – despite the fact that it is
obviously the question most similar to one that might appear on an A-level paper – the lack of given
structure, and the requirement for sketching, clearly put many candidates off, and only three-
quarters of candidates even started it. Once started, however, it proved to be the highest-scoring
question for candidates (as was the intention), eliciting a mean score of over 13½ out of 20. Most
serious attempts were thus highly successful, and it was generally only in part (ii) that marks were
commonly lost in bulk. Sadly, many who did move into (ii) with some measure of success,
overlooked the fact that it was a fairly straightforward follow-on from all the information gained or
used in part (i), and many attempts were thus successful but unnecessarily long-winded.

Q2 This was the most popular question of all, attempted by around 85% of candidates, and
producing a mean success rate of just over 10 out of 20. Part (i) was almost always concluded
successfully, and by the anticipated method of the use of the addition-formulae for sine and cosine;
a method then used sensibly, along with the well-known double-angle formulae, in (ii) to establish
that cosα was indeed a root of the given cubic equation. For many candidates, though, the other
roots of this cubic were often “found” by guesswork, and many candidates thought it appropriate to
“cancel” x with cosα in some strange way, rather than resort to the use of the quadratic formula.
Part (iii) was frequently not attempted at all, and many who did boldly venture forth therein did so
by not using the results of parts (i) and (ii) as instructed. A few rather presumptuously assumed that
cos 15o was a solution, when a little care would have revealed that it is, in fact, 2cos 15o that fits
the bill; the key being that 2cos 45o is the 2 at the end of the given equation.

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Q3 This question was attempted by just over half of the candidature, but produced a mean score
of only 4 marks ... largely due (I suspect) to the twin difficulty of a lack of supplied structure and
the poor ability of candidates to do their own modelling. In many instances, the two cases b < 3a
and b> 3a, given in the question, proved to be unhelpful as many candidates chose specific values
of b in each of these ranges as “exemplar” values of b and then supposed that this sufficed in
establishing the boundary cases; when, in fact, the given information was intended to guide where
they were to end up rather than from where they should begin. It was also especially disappointing
to see that so many candidates struggled to explain what they were doing, thinking that some poorly
labelled diagram would ‘do the trick’. The poor thinking behind the diagrams usually meant, for
instance, that one of the two possible scenarios for when the guard stood at the midpoint of a side
was completely unconsidered. Rather strangely – and disastrously in terms of scoring any marks at
all – it was very common indeed for candidates to have considered the area of the courtyard that
was visible to the guard, despite the very clear reference to the “length of the perimeter” in the
question.

Q4 Amongst the pure maths questions, this one was least popular, with less than a third of
candidates making an attempt at it, and producing a mean score of 4½ marks. Unlike Q3, there was
a lot of helpful information given in the question, and key intermediate results also. Those
candidates who realised that the gradient of the rod was tan answered the first part quite
acceptably, although it was relatively common to see the gradient of the curve, 12 x , being used as
the “m” in the formula y = mx + c for a straight line.
It is, unfortunately, often the case that when an answer is given with the view that it will
prove helpful to candidates, that they then miraculously manage to obtain it through any means
possible, and there were many wayward attempts to justify the given final answer without any clear
supporting evidence: the biggest difficulty arose from the need to use Ax = 0 in order to eliminate b.

Q5 More than 60% of candidates attempted this question, and scores were relatively healthy,
with a mean of almost 7½. Those candidates who realised that x was being treated as a constant
within the integrals generally found it fairly straightforward to make good progress; those who
didn’t were doomed to failure from the outset. The sketches were generally completed fairly
successfully, though few candidates managed to be entirely convincing, especially in (ii), where the
modulus function needed to be employed (although some candidates thought the matter through
sufficiently carefully without it by considering the various intervals of the domain of g).

Q6 This vectors question proved both unpopular and low-scoring, eliciting a mean of only 2.3
out of 20. In many cases, this was because candidates started their “solution” with a diagram before
abandoning it altogether and moving on elsewhere. Most of the remaining attempts assumed that
the quadrilateral was a square, rectangle or parallelogram to begin with – whether through a
misreading of the question or through an inability to deal with a general scenario it is hard to say.
Moreover, the usual convention of underlining vector quantities (thereby distinguishing them from
scalar ones) was almost universally avoided, and this made it extremely difficult to give serious
consideration to much of what was written, as candidates moved from scalar to vector and back
again.

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Q7 This was another popular question, since most candidates were able to make some progress
with the ideas involved, though few seemed to have a particularly thorough grasp. The change in
the variable being considered – from x to f(x) and then to a – was evidently the source of much of
the confusion, though I am sure that candidates would have made better progress with a more
carefully laid out plan for working through the different possibilities. In the end, it all boiled down
to the fact that a (continuous) function takes its maximum value on a finite interval at either an
endpoint or at a maximum turning-point. Thereafter, it was important to do some sensible
comparisons using inequalities. Thus, there were easy enough marks to be had and the mean score
of 7.7 out of 20 was the third highest on any question, after questions 1 & 2.

Q8 This was another very popular question, with three-quarters of the candidature making a
start at it. However, the mean score of 5.4 almost certainly arose from the acquisition of the 3 marks
allocated to the bit of introductory bookwork plus 2 or 3 marks gained by considering a suitable
pairing of terms for considering S in (ii). Inductive proofs were unnecessary in (i), and almost
invariably went wrong in (ii); this was a shame when the appearance of the term ( N  m) k in the
given expression really indicated for the use of the binomial theorem. It should have been relatively
straightforward to apply the given initial result of (ii) in the two cases that followed, but each
required the addition of an extra 0k term … perversely, to make the odd number of terms into an
even one, and then v.v. (since it is now the isolated middle term that is crucial), and this extra leap
of intuition was clearly where the difficulty lay. The final arguments relied on a sound grasp of
what had gone before, and most candidates had given up by this point.

Qs.9-11 These mechanics questions were – as mentioned earlier – very non-standard, and
hence found very difficult. Of the relatively few attempts appearing from candidates, almost none of
them got further than a hesitant start. There were two or three easy marks to be had in Q9 in finding
the general time for any one bullet to land, but very few candidates were able to cope with replacing
a specific launch angle with the variable angle given
In Q10, the real key to successful progress was to avoid worrying about any constants of
proportionality (such as that introduced by the unstated width of the bus), so very few candidates
managed to produce the early given result in a satisfactorily justifiable way. Introducing an extra
proportionality relationship for the journey time was then a leap too far, even for those who had
started well. Moreover, there were some candidates who only seem to be able to maximise or
minimise a function by using calculus, and this provided an extra layer of unnecessary clutter here.
Fewer than 200 candidates started this question, and most of these attempts had little more than a
sketchy diagram for the markers to consider.
Q11 attracted double the number of attempts of Q10, but had a marginally lower mean
score, and this was slightly surprising. A few years ago, statics questions such as this would have
been gobbled up with glee by many candidates, happy to collect some very easy mechanics marks.
The great hurdle for the weaker candidates remains the widespread inability to draw a good diagram
with all relevant forces marked on it in appropriate directions. Sadly, here, almost all diagrams
failed to include all of the relevant forces, and decent progress beyond that point was, therefore,
essentially impossible. Remarkably few candidates managed even to explain satisfactorily that the
two frictional forces were equal (by taking moments about the central axis of each of the two
cylinders).

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Qs12 & 13 These questions were also less routine than has usually been the case in recent years,
although they were nowhere near as demanding as the 2-3 mark mean score might suggest. In Q12,
despite the reference to the Poisson Distribution in the introduction, part (i) required a simple
statement of a Binomial term. Part (ii) then proved difficult as it became clear that few candidates
could manipulate a summation of terms in order to establish a result they might have anticipated
being allowed to quote in an ordinary A-level examination. A few candidates managed part (iii)
perfectly adequately without having gone very far with (ii), and this represented a shrewd use of
“examination technique” on their part.
Most attempts at Q13 got little further on in the question than writing the simple, general
term for P(A); namely  65 n  1  16  . Even though the Geometric Distribution is not expected here, the
probability of the run of independent events “n – 1 failures followed by a success” should be within
the scope of any STEP candidate (who has studied any small amount of probability and/or
statistics). Many candidates decided that (ii) and (iii) also required a general expression for each of
P(B) and P(B  C), whereas these were clearly intended to be numerical, and a brief ‘symmetry’
argument quickly reveals their respective probabilities to be 12 and 13 . Parts (iv) and (v) were
tougher, but required the candidates to see that each was the sum of an infinite number of terms, and
the helpfully given series expansion at the end of the question helped wrap these up. One third of all
candidates made an attempt at this question, but almost none of them got around to either of these
final two parts.

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STEP 2 2015 Report 

As in previous years the Pure questions were the most popular of the paper with questions 1, 2 and 
6 the most popular. The least popular questions on the paper were questions 8, 11 and 13 with 
fewer than 250 attempts for each of them. There were many examples of solutions in this paper that 
were insufficiently well explained given that the answer to be reached had been provided in the 
question. 

Question 1 

This was a popular question, but a number of common errors resulted in a relatively low average 
score for the attempts made. A number of candidates did not appreciate that it is necessary in the 
first part to show both that the gradient is positive for all relevant values of x and to check the value 
when x=0. Additionally, many candidates failed to note that the next part of this question instructed 
them to use the result shown in the second section of part (i) and instead used a graphical method. 
Other common errors included an incorrect use of the chain rule in the second part leading to a sign 
error and incorrect statements of formulae for the sums. 

Question 2 

The average mark for question 2 was the highest on the paper with a large number of good solutions 
produced using a wide variety of different methods. However, many solutions did not explain clearly 
the method being used – it is advisable to make every step of the solution clear, especially in the 
case of questions where the answer that is to be reached has been given. In many cases the 
diagrams drawn were not sufficiently large to allow candidates to work easily on the question and 
on a number of occasions the sizes of two angles were reversed in the diagram leading to other 
points being in the wrong position on the diagram. 

Question 3 

On the whole attempts at this question were good with a significant number of candidates obtaining 
full marks. In the first part of the question a number of candidates did not interpret the difference 
between successive terms of the sequence as triangles which included a particular length edge and 
chose to enumerate all possible cases – if this was carried out correctly it was still possible to achieve 
full marks on this section. Even when unsuccessful in this part of the question many candidates were 
able to write down correct expressions for the general cases. The proof by induction was generally 
well done, although a number of candidates failed to justify the first case fully (which can easily be 
done by enumerating all of the cases). The final part of the question (the corresponding result for an 
odd number of rods) was not attempted by all candidates. Of those that did, those who attempted 
to use induction rather than applying the result from earlier in the question struggled to reach the 
correct answer. 

Question 4 

This was a generally well attempted question, although it was a common error to draw graphs that 
were not continuous, even in some cases with statements that they were continuous. Marks were 
also lost through mislabelling of points on the graphs or through incorrect attempts to use 
arguments based on graphical transformations to deduce the shape of the graph. A number of 
candidates when trying to find the stationary points stated that they were going to differentiate a 
function, but then integrated it. 

   

673
Question 5 

This question had one of the lower average marks for the Pure maths questions on the paper. Most 
candidates were able to produce a proof by induction for the first part, but the vast majority failed 
to realise that there was more that needed to be done to prove the result stated in terms of arctan. 
As is the case for a number of other questions, candidates need to give a clear explanation of each 
step of the solution. Where candidates identified the relationship between the two parts of the 
question the second part was generally well attempted. 

Question 6 

This was the most popular question on the paper with over 1000 attempts made. The first section 
did not present significant difficulties to candidates and the integration was generally well 
completed, although occasionally with an error in the factor. The second part proved difficult for a 
number of candidates who failed to change the variable in the integral correctly, or in some cases 
did not change the variable in every position that it occurred. Other candidates did not apply a 
correct result for dealing with the trigonometric functions involved or did not clearly show how the 
required result was reached as the solution jumped through several steps to a statement of the 
result asked for in the question. There were very few successful attempts at the final part of the 
question, but they did include a variety of methods for evaluating the integral once the substitution 
had been made. 

Question 7 

The first part of this question was generally well answered, although a significant number of answers 
did not give the equation of the new circle. The case in part (i) where the two circles have the same 
radius was often not considered and the explanations for there not being such a circle in some cases 
were often not sufficiently clear. A significant number of candidates made the incorrect assumption 
in the second part that the centres of the three circles must lie on a straight line or attempted this 
part of the question with incorrect methods, such as equating the equations of the two given circles. 
In the final part of the question not all candidates realised that y2 must be positive and were unable 
to obtain the required inequality by any other means. 

Question 8 

This was one of the least attempted questions on the paper and the average score for the question 
was quite low. However, there were a number of very good answers to the question. Part (i) was 
answered correctly by the majority of candidates, but part (ii) was approached in a much more 
complicated manner than necessary by many candidates, attempting to work out the equation of 
the line rather than comparing vectors in its direction. Where the vectors were considered, solutions 
could have been made clearer by better grouping of the terms. A number of solutions referred to 
division of vectors rather than comparing coefficients. In the final part some candidates did not 
identify the simplest relationship between the vectors to ensure that Q lies halfway between P and 
R. Generally, more complicated relationships did not lead to correct solutions to this part of the 
question. 

   

674
Question 9 

This was the most popular of the Mechanics questions, but many candidates struggled to achieve 
good marks. In the first section many candidates had difficulties in finding the correct angles to work 
with – a clear diagram is very helpful in tackling this problem. Candidates often introduced new 
notation to help with the steps toward the solution, but this was sometimes poorly chosen and 
made solution of the problem more difficult. Explanations of the methods being used were also 
often poor – in particular the triangles being used at different stages were not clearly identified. 
There were also a number of errors when taking moments or when recalling exact values of the sine 
and cosine functions. There were a number of good attempts at the second part of the question, but 
a large number of candidates calculated the kinetic energy incorrectly. 

Question 10 

This question received generally very poor attempts, including a large number of partial attempts. 
The majority of attempts failed to get the correct expression of the velocity in the first part and this 
limited the number of marks that could be awarded for the remainder of the question. A very small 
number of attempts were awarded full marks and there were a considerable number of attempts in 
which correct methods were attempted following an incorrect solution to the first part of the 
question. 

Question 11 

This was the least popular question on the paper. Many answers to the first part did not give good 
explanations of the method for obtaining the velocity of A. Similarly, in the second part there were a 
number of statements such as “conservation of velocity” or “conservation of the modulus of 
momentum” used to support the answer without sufficient explanation to show that a valid method 
was being applied. Those candidates who attempted to use the equations of motion under uniform 
acceleration were unable to reach the solution. Part (iii) was very poorly answered with almost no 
correct solutions offered. In the final part of the question very few candidates were able to identify 
the part of the reasoning that led to v not being equal to zero in all of the cases identified. 

Question 12 

Many solutions to this question did not include sufficient explanation to gain full credit. In the first 
part, marks were not awarded simply for stating that the value of ¼ could be achieved by multiplying 
½ by ½ (often with an additional multiplication by 1) – an explanation of where this calculation 
comes from was also required. In the second part a number of candidates stated that it was 
symmetric and so the answer must be ¼ but with insufficient explanation why. In part (iii), some 
candidates obtained a geometric series which was then summed to get the probability of C winning 
if the first two tosses are TT. In the final part some correct answers were offered, but without 
explanation of the method. A number of candidates made incorrect assumptions such as that p+q=1, 
or p+q+r=1. When finding the probability that C wins a lot of candidates were able to achieve some 
of the marks by working out the probability in terms of q. 

Question 13 

This was not a popular question and those solutions that were offered generally showed a limited 
understanding of continuous probability distributions. The integration that was required was also 
generally quite poorly carried out. Often these mistakes made it difficult to answer the final section 
of part (i). Part (ii) was only attempted by a small proportion of candidates. 

675
STEP 3 2015  Examiners’ report 

A very similar number of candidates to 2014 once again ensured that all questions received a decent 
number of attempts, with seven questions being very popular rather than five being so in 2014, but 
the most popular questions were attempted by percentages in the 80s rather than 90s.  All but one 
question was answered perfectly at least once, the one exception receiving a number of very close 
to perfect solutions.  About 70% attempted at least six questions, and in those cases where more 
than six were attempted, the extra attempts were usually fairly superficial 

1.  This was the most popular question, being attempted by 85% of candidates, it was however 
only moderately successful although a number achieved full marks.  Quite often, candidates ignored 
the helpful approach suggested by the LHS of the first required result, though, of course, it was 
possible to start from the first defined integral and achieve the same result.  Many needlessly lost 
marks through omitting fairly straightforward steps such as the final evaluation in the last part of the 
question and failing to substantiate the simplified form of the result of part (i).  Some got very 
carried away with tan or sinh substitutions in part (i), usually unsuccessfully and leading to 
monstrous amounts of algebraic working.  A few failed to change the limits of integration in [part (ii). 

2.  Nearly three quarters attempted this, though again with moderate success as the main 
feature of the question was proof, and this was frequently handled cavalierly.  Whilst it was not a 
crucial aspect of the question, ignoring the fact that the question deals with sequences of positive 
numbers was careless.  Answers to the first part suffered at times from lack of argument or 
backwards logic.  Part (ii) was generally well answered, although there were some silly counter‐
examples.  This part suffered from those who completely missed the point of what the question was 
all about, forgetting the initial definition.  Whilst most appreciated that part (iv) was true, there were 
many different methods used to attempt to prove it, and often unsuccessfully.  Whilst induction 
using algebra is fairly straightforward, differentiation with or without logarithms and graphical 
methods frequently came to grief. 

3.  Under 20% attempted this, making it the least popular Pure question on the paper, and it 
was the least successfully attempted of all questions on the paper.  Candidates seemed to find it 
intimidating, and many gave up before part (ii).  They often got confused when dealing with separate 
cases and did not seem to understand what was required to show  sec 0  in part (i).  Those that 
did make a stab at (ii) usually omitted a factor of two and most failed to find the correct limit to use. 

4.  Along with questions 5 and 7, attempted by just over three quarters, this was the third most 
popular question, though a little less successful than the most popular question 1. The first part was 
frequently not well attempted, but the second part was usually mastered.  Attempts at the third part 
suffered from arguments with poor logical structure, though many did not get a start on this part. 

5.  Marginally less successful than question 2, a lot of candidates earned about half of the 
marks.  Unfortunately, many candidates approached this on the basis of their knowledge of the 
standard irrationality proof for root two employing rational numbers expressed in lowest terms 
rather than observing the specified argument.  In part (i), proving step 5 was frequently beset with 
omissions, and simple steps like  0 √2 1 1  were not acknowledged let alone justified.  The 
first result of part (ii) caused few problems except to those that did not appreciate ‘if and only if’, but 
defining a suitable set in order to construct a similar argument to prove the irrationality of the cube 
roots of 2 and 2 squared was beyond most leading to mostly spurious logic. 

6.  About three fifths of the candidates attempted this question but without great success.  The 
first part tripped up many through needing to prove ‘if and only if’.  The first part of (ii) yielded good 

676
scoring opportunities for those that did make progress on this question, though some fell by the 
wayside when it came to the situation that would not generate there possible values.  Some 
attempts at the last result failed as the counter‐example was not always shown to be a counter‐
example. 

7.  This was as successful as question 2 and so was third equal most popular and second equal 
most successful.  Usually the very first result was comfortably answered, but there were many flaws 
in part (i) as many could not carry out a proper formal induction.  In part (ii), which saw a lot fall by 
the wayside, some candidates thought that     commutes with    , and often, candidates invented 
random formulae for   1   from looking at the first few cases.  Not surprisingly, working 
towards a given result, many came up with the correct result, but through spurious working such as 
substituting   1  in   1  before using the differential operator. 

8.  This was a little less popular than question 1, but still was attempted by more than 80% of 
candidates, and was the question with highest scores.  Many managed part (i) although several 
candidates did not realise that     was a function of    .  Part (ii) was generally fine as far as the 
transformed differential equation but then the correct use of partial fractions to integrate having 
separated variables was less frequent than it should have been.  A surprising number made no 
attempt to sketch any solutions despite doing the rest of the question either well or perfectly.  
Nobody realised that the constant     was truly arbitrary in part (ii) because of the modulus signs 
appearing in the log terms form the integral.  The sketches tested all but the best. 

9.  Just over 20% attempted this, making it the most popular non‐Pure question, and attempts 
at it were slightly less successful than question 1.  Quite a few found the first required equation from 
applying Newton’s 2nd Law, when some made sign errors through not being careful with directions, 
and then integrating rather than from conserving energy.     was found easily by the majority, and 
the expression for the acceleration was commonly still by Newton, though a lot of marks were lost 
by not substituting in   .  The last part was poorly done in general with few getting more than 
an opening line, and if they made a sensible substitution, very few expanded it correctly.  Not many 
even attempted the last part of the question. 

10.  Whilst this was the least popular question with just over 8% attempting it, it was only slightly 
less successfully attempted than question 5, though those making substantial attempts at it 
invariably scored half to two thirds of the marks comfortably.  Most successfully wrote the position 
vector of one of the particles and then differentiated with respect to time to obtain the velocity 
correctly, though a few succeeded by adding velocities.  The second displayed equation was almost 
always correctly derived, though many did far too much work obtaining the corresponding equations 
for the other particle when it could just be written down.  Deducing    and     was fine, but  θ  
frequently wasn’t.  At this point, finding initial values for   and     caused some issues, if it was 
realised that these were needed, and although many wrote the uniform acceleration equation for 
the displacement of the midpoint of the rod, the final result eluded many. 

11.  Marginally more popular than the Probability and Statistics questions, this was attempted by 
just over 10% with, on average, slightly less success than question 4, though students either got 
almost full marks or virtually none.  Most did part (i) correctly except the final part, identifying the 
force from the hinge.  The most common mistake was in part (ii) by those that assumed that there 
were no perpendicular forces acting on    and   .  Students that correctly considered the total 
moment for part (ii) obtained the answer.  Some students got the direction of centripetal force 
wrong. 

677
12.  The two Probability and Statistics questions were equally popular being attempted by about 
10% of the candidates with, overall, this one achieving the same sort of scores as question 6.  About 
a fifth of the candidates attempting it got right through the question.  Most however did not seem to 
know what a probability generating function was, and it was often confused with the probability 
density function.  Equally there was confusion between the labels of the random variables and of the 
PGFs.  However most were happy working with the arithmetic congruent to moduli. 

13.  The large majority of attempts got almost no marks, and as a consequence this was the 
second worst scoring question.  A lot failed to draw the right sort of graph to attempt the first part of 
(i) or, if they did, frequently miscalculated the area to find     in the case  1 2 .  
The next major problem was an inability to see how to find the cumulative distribution function of  
.   A surprisingly large number failed to multiply by    before integrating to find the 
expectation.  A handful of candidates got most of the question right although only one made it clear 
with a symmetry argument why they could write down   . 

678
Explanation of Results STEP 2015

All STEP questions are marked out of 20. The mark scheme for each question is designed to reward
candidates who make good progress towards a solution. A candidate reaching the correct answer will
receive full marks, regardless of the method used to answer the question.

All the questions that are attempted by a student are marked. However, only the 6 best answers are
used in the calculation of the final grade for the paper.

There are five grades for STEP Mathematics which are:

S – Outstanding
1 – Very Good
2 – Good
3 – Satisfactory
U – Unclassified

The rest of this document presents, for each paper, the grade boundaries (minimum scores required
to achieve each grade), cumulative percentage of candidates achieving each grade, and a graph
showing the score distribution (percentage of candidates on each mark).

STEP Mathematics I (9465)

Grade boundaries
Maximum Mark S 1 2 3 U
120 96 65 45 28 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 3.5 18.6 42.6 73.1 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics I

www.admissionstestingservice.org

679
STEP Mathematics II (9470)

Grade boundaries
Maximum Mark S 1 2 3 U
120 94 68 60 30 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 7.9 27.9 37.9 85.5 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics II

STEP Mathematics III (9475)

Grade boundaries
Maximum Mark S 1 2 3 U
120 88 65 54 29 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 11.8 37.3 51.4 85.5 100.0

Distribution of scores

2.5

2.0

1.5
Percent

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics III

www.admissionstestingservice.org
680


681


STEP Mark Scheme 2015

Mathematics

STEP 9465/9470/9475

October 2015

682
This mark scheme is published as an aid to teachers and students, to indicate the
requirements of the examination. It shows the basis on which marks were awarded
by the Examiners and shows the main valid approaches to each question. It is
recognised that there may be other approaches and if a different approach was
taken in the exam these were marked accordingly after discussion by the marking
team. These adaptations are not recorded here.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the relevant
knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question papers and
the Report on the Examination.

The Admissions Testing Service will not enter into any discussion or correspondence
in connection with this mark scheme.

© UCLES 2015

More information about STEP can be found at:

www.stepmathematics.org.uk

683
Contents

STEP Mathematics (9465, 9470, 9475)

Mark Scheme Page

STEP Mathematics I 4
STEP Mathematics II 18
STEP Mathematics III 45

684
SI-2015/Q1
(i) y = ex(2x – 1)(x – 2) B1 Correct factorisation of quadratic term (or formula, etc.)
( 12 , 0) & (2, 0) B1 Noted or shown on sketch
dy
= ex(2x2 – x – 3) M1 Derivative attempted and equated to zero for TPs
dx
= ex(2x – 3)(x + 1)
( 32 , – e1.5) & (–1, 9e – 1) A1 A1 Noted or shown on sketch
(if y-coords. missing, allow one A1 for 2 correct x-coords.)

G1 Generally correct shape

G1 for (0, 2) noted or shown on sketch

G1 for negative-x-axis asymptote

(penalise curves that clearly turn up away from axis
or that do not actually seem to approach it)

⑧
Give M1 for either 0, 1, 2 or 3 solutions OR clear indication they know these arise from where a
horizontal line meets the curve (e.g. by a line on their diagram) – implied by any correct answer(s)
Then y = k has NO solutions for k < – e1.5 A1
ONE solution for k = – e1.5 and k > 9e – 1 A1 A1
TWO solutions for – e1.5 < k ≤ 0 and k = 9e – 1 A1 A1
THREE solutions for 0< k < 9e – 1 A1
FT from their y-coords.of the Max. &Min. points.

(ii) G1 Any curve clearly symmetric in y-axis

G1 Shape correct
G1 A Max. TP at (0, 2) FT

G1 Min. TPs at (  3
2 , – e1.5) FT

G1 Zeroes at x =  1
2 ,  2 FT

685
SI-2015/Q2
(i) M1 Use of cos(A – B) formula with A = 60o, B = 45o OR A = 45o, B = 30o
or 2 cos2 15o – 1 etc.
3 1
A1 Exact trig.values used (visibly) to gain cos 15o = legitimately (Given Answer)
2 2
M1 Similar method OR sin =  1  cos 2 (as 15o is acute, no requirement to justify +vesq.rt.)
3 1
A1 sin 15o = (however legitimately obtained) ④
2 2

(ii) M1 Use of cos(A + B) formula and double-angle formulae OR de Moivre’sThm. (etc.)

A1 cos3α  4cos3α – 3cosα
A1 Justifying/noting that x = cosα is thus a root of 4 x 3  3 x  cos 3  0
M1  
For serious attempt to factorise 4 x 3  c 3  3 x  c  as linear × quadratic factors
or via Vieta’s Theorem (roots/coefficients)
A1   
( x  c) 4 x 2  cx  c 2  3
M1 Solving 4 x  4cx  4c  3  0 FT their quadratic factor
2 2

Remaining roots are x =  c  c  4c  31

2
2 2

M1 Use of s  1  c 2 to simplify sq.rt. term

A1 x= 1
2
 cos   3 sin  
⑧
1 3 3 2
(iii) M1 y  y 0
2 2 2
3
1  1  2
A1 4 y   3 y  0
2  2  2
2
M1 cos3α = = cos 45o
2
A1 α = 15o
M1
1
2
y  cos  , 1
2
 cos     cos  
3 sin  , 1
2

3 sin  with their α

3 1
A1 y = 2 cos 15o =
2
3 1
A1 3 sin 15 o  cos15 o = 
2
A1  3 sin 15  cos15 =  2
o o

686
SI-2015/Q3
1
2
(b  a) B1 For correct lengths in smaller
C P
M1 By similar s (OR trig.ORcoord.geom.)
1
(b  a)
2
PQ 12 (b  a) b(b  a)
A1  1  PQ =
b 2
(b  a ) ba
b a Q M1 so a guard at a corner can see 2(b + PQ)
4b 2 4ba
A1 = (might be given as all but or as a
ba ba

⑤
fraction of the perimeter)
 

Q P Lengths 1
a and 12 (b  a) in smaller 

2

M1 By similar s (OR trig.ORcoord.geom.)

1
1
b 1
a b(b  a )
2
b A1 2
1 2  PQ =
PQ 2 (b  a) 2a
 
M 
M1 so a guard at a midpoint can see b + 2PQ
b2 b( 4a  b)
A1 = (might be given as all but or as a
a a
fraction of the perimeter) ④
 

Lengths 1
2
a and 12 (b  a) in smaller 
P M1 By similar s (OR trig. OR coord.geom.)
1
PQ a ba
A1  1 2  PQ =
b 2 (b  a ) ba
M Q M1 so a guard at a midpoint can see 4b – 2PQ
2b(2b  3a ) 2ba
A1 = (might be given as all but or as a
ba ba
fraction of the perimeter)
④
B1 Recognition that b = 3a is the case when guard at M / C equally preferable
(P at corner in the two M cases)
4b 2 b 2 b2
M1A1 Relevant algebra for comparison of one case   3a  b 
b  a a a(b  a)
A1 Correct conclusion: Guard stands at C for b< 3a and at M for b> 3a
4b 2 2b(2b  3a )
M1A1 Relevant algebra  
2ba
3a  b 
ba ba (b  a )(b  a )
A1 Correct conclusion: Guard stands at C for b< 3a and at M for b> 3a

⑦
Overall, I am anticipating that most attempts will do the Corner scenario and one of the Middle scenarios. This
will allow for a maximum of 12 = 5 (for the Corner work) + 4 (for the Middle work) + 3 (for the comparison).
In this circumstance, it won’t generally be suitable to give the B1 for the b = 3a observation.

687
SI-2015/Q4
M1 When P is at (x, 1
4 x 2 ) ... and makes an angle of  with the positive x-axis

A1 ... the lower end, Q, is at x  b cos  , 1

4 x 2  b sin  
dy 1
M1 Also, y  14 x 2   x = tan
dx 2
A1 x = 2 tan 
i.e. P = 2 tan  , tan 2 

A1A1 so thatQ = 2 tan   b cos  , tan   b sin  2
 obtained legitimately (Given Answer)

⑥
2 tan 
M1A1 When x = 0, 2 tan   b cos  b 
cos 
sin 
M1A1 Substg. into y-coordinate  yA = tan 2  2 tan    tan 2 
cos 

M1A1 Eqn. of line AP is y  x tan   tan 2 

M1A1 Area between curve and line is  1

4

x 2  x tan   tan 2    dx
B1 Correct limits (0, 2tan α )
A1A1 =  1
12 x 3  12 x 2 tan   x tan 2   (Any 2 correct terms; all 3)

A1A1 = 2
3 tan 3   2 tan 3   2 tan 3    (Any 2 correct terms; all 3 FT)

A1 = 2
3 tan 3  obtained legitimately (Given Answer) ⑩

P
ALTERNATIVE

O B C

A 

M1 A1 for obtaining the “conversion factor” bcos = 2 tan or tan2 = 12 b sin 

M1 A1 for distances OB = BC (= 12 b cos  ) and so PC = OA = tan2
M1 A1 giving OABCPB
A1  Area is  1
4 x 2 dx
B1 Correct limits (0, 2 tan ) used
A1 A1 Correct integration; correct Given Answer

b cos
ALTERNATIVE Translate whole thing up by tan2 and calculate 
0
1
4 
x 2  tan 2  dx – 

688
SI-2015/Q5
 (t  1) x  3
(i) M1A1 f(x) =  
 x 1

2x
A1 = ③
x

M1 Differentiating by use of Quotient RuleOR taking logs.anddiffg. implicitly)

B1 for
d x
dx
2   2 x. ln 2 seen at any stage
dy x .2 x . ln 2  2 x
A1 =
dx x2
 1 
A1 TP at  , e ln 2  (y-coordinate not required)
 ln 2 

B1 Jusitfying that the TP is a minimum ⑤

G1 Generally correct -shape

G1 Asymptotic to y-axis
and TP in FT correct position

(ii) M1 Let u 2  1  x 2  2 xt
A1 2u du = – 2xdt
B1 t: (–1, 1) u: ( | 1 + x |, | 1 – x | ) Correct limits seen at any stage
1
x 
M1A1 Full substn. attempt; correct g(x) = 1 . du

g(x) = 1  x  1  x 
1
A1 n. may be done directly, but be strict on the limits
x
 2
 x x  1


org(x) =  2  1  x  1


 2 x 1
 x
(Must have completely correct three intervals: x< –1,  1  x  1 , x> 1)

M1 Graph split into two or three regions

A1 A1 Reciprocal graphs on LHS & RHS (must be asymptotic to x-axis)
(Allow even if they approach y-axis also)
A1 Horizontal line for middle segment

689
SI-2015/Q6

Let P, Q, R and S be the midpoints of sides (as shown)

Then P
M1A1 p= 12 a  12 b , q = 12 b  12 a  ,
r= 12 a   12 b  , s = 12 b   12 a Q
and
 
M1A1 PQ  SR  a  a
1
2 S
A1 QR  PS  12 b   b 
A1 so that PQSR is a //gm. R
(opposite sides // and equal)

⑥
M1 PQ  QR  PQ  QS  1
2 a  a  12 b  b  for use of the scalar product
A1 = 1
4
a  b  a  b  a  b  a  b  Do not accept a  b  etc.
M1 Use of perpendicularity of OA, OB and OA, OB
=  14 a  b  a  b 
M1 AOB =   AOB = 180o–  ; and cos(180o –  ) = – cos
A1 = 0 since a  b  ab cos and a  b  a b cos
and we are given that a = b and a = b
A1 so that PQRS is a rectangle (adjacent sides perpendicular)

⑥
B1 PQ 2  SR 2  PQ  PQ = 1
4 a 2
 (a ) 2  2a  a  
B1 QR 2  PS 2 = 1
4 b 2
 (b ) 2  2b  b  
M1 
Since a = b, a = b and a  a  aa  cos 90 o   , b  ab   bb cos 90 o     
A1 it follows that PQRS is a square (adjacent sides equal)

④
M1A1 Area PQRS = 1
4 a 2
 ( a ) 2  2aa  cos 90 o    
M1 ... which is maximal when cos 90 o    1  
A1 i.e. when   90 o

690
SI-2015/Q7
M1 f (x) = 6ax – 18x2
= 6x(a – 3x)
A1A1 = 0 for x = 0 and x = 13 a
A1A1 f(0) = 0 f( 13 a) = 19 a3
A1 (Min. TP) (Max. TP) since f(x) is a ‘negative’ cubic
(f(0) = 0 and the TPs may be shown on a sketch – award the marks here if necessary)

⑥
M1 Evaluating at the endpoints
A1A1 f(– 13 ) = 19 (3a + 2); f(1) = 3a – 6

③
M1 1
9 (3a + 2)  19 a3  a3 – 3a – 2  0
M1  (a + 1)2(a – 2)  0
A1 and since a  0, a  2

M1 1
9 a3  3a – 6  a3 – 27a + 54  0
M1  (a – 3)2(a + 6)  0
A1 which holds for all a  0

M1 1
9 (3a + 2)  3a – 6  3a + 2  27a – 54
 8(3a – 7)  0
A1  a  73 (which, actually, affects nothing, but working should appear)

Thus
 19 (3a  2) 0  a  2
B1B1B1

M(a) =  19 a 3 2a3 (Ignore ‘non-unique’ allocation of endpoints) ③
 3a  6 a3

(Do not award marks for correct answers unsupported or from incorrect working)

691
SI-2015/Q8
(i) S = 1 + 2 + 3 + … + (n – 2) + (n – 1) + n
M1 S = n + (n – 1) + (n – 2) + … + 3 + 2 + 1 Method
M1 2S = n (n + 1) Adding
A1 S = 12 n(n + 1) obtained legitimately (Given Answer)
(Allow alternatives using induction or the Method of Differences, for instance, but NOT by stating
that it is an AP and just quoting a formula; ditto -number formula)

③
(ii) (N – m)k + mk (k odd)
k  k   k  k 1
M1A1 = N k   mN k  1   m 2 N k  2  ...   m N  m k  m k
1  2  k  1
E1 which is clearly divisible by N (since each term has a factor of N)
(Allow alternatives using induction, for instance)

③
Let S = 1k + 2k + … + nk an odd no. of terms
M1 = 0k + 1k + 2k + … + nk an even no. of terms
M1     
= (n  0) k  0 k  (n  1) k  1k  ...  ( 12 n  12 ) k  ( 12 n  12 ) k 
(no need to demonstrate final pairing but must explain fully the pairing up or the single extra nk term)
E1 and, by (ii), each term is divisible by n.

③
For S = 1k + 2k + … + nk an even no. of terms
M1 = 0k + 1k + 2k + … + nk an odd no. of terms

M1     
= (n  0) k  0 k  (n  1) k  1k  ...  ( 12 n  1) k  ( 12 n  1) k   12 n  k

(no need to demonstrate final pairing but must explain the pairing and note the separate, single term)

and, by (ii), each paired term is divisible by n

E1 and the final single term is divisible by 12 n  required result

③
M1 By the above result … for n even, so that (n + 1) is odd
A1 (n + 1) | 1k + 2k + … + nk + (n + 1)k
E1 (n + 1) |S + (n + 1)k  (n + 1) |S

M1 By the above result … for n odd, so that (n + 1) is even

1 k k k k
A1 2 (n + 1) | 1 + 2 + … + n + (n + 1)

E1 1
2 (n + 1) | S + (n + 1)k  1
2 (n + 1) |S (as 1
2 (n + 1) is an integer)

E1 Since hcf(n, n + 1) = 1  hcf( 12 n, n + 1) = 1 for n even

1
E1 and hcf(n, 2 (n + 1)) = 1 for n odd

So it follows that 1
2 n(n + 1) | S for all positive integers n ⑧
692
SI/15/Q9
M1 Time taken to land (at the level of the projection) (from y = utsin – 1
2 gt2 , y = 0, t  0)
2u sin 
A1 is t  (may be implicit)
g
  
M1 Bullet fired at time t  0  t   lands at time
 6 
2u   
A1 TL  t  sin   t 
g 3 
dTL 2 u   1  
M1A1  1 cos   t  = k  cos   t 
dt g 3  k 3 
 
A1 = 0 when k  cos   t 
3 
2u 2 sin  cos 
M1A1 Horizontal range is R  (from y = utsin – 1
2 gt2 with above time)
g

A1  RL 
2u 2
k 1 k 2 obtained legitimately (Given Answer) ⑩
g

d 2TL 22 u   
M1A1  sin   t   0  maximum distance
3 
2
dt g
 
M1A1 0t 
 1
in k  cos  t    k 
3
④
6 3  2 2

dTL
M1 If k < 1
2 then  0 throughout the gun’s firing …
dt
A1 … and TL is a (strictly) decreasing function.
M1 Then TL max. occurs at t = 0

A1 i.e.  
3

⑥
2
2u 1 3 u2 3
M1A1 and RL    
g 2 2 2g

693
SI-2015/Q10
B1 Speed of rain relative to bus is vcos – u (or u – vcos if negative)
M1A1 When u = 0, A  hvcos + avsin (width of bus and time units may be included as factors)
E1 When vcos – u > 0, rain hitting top of bus is the same, and rain hits back of bus
as before, but with vcos – u instead of vcos
E1 When vcos – u < 0, rain hitting top of bus is the same, and rain hits front of bus
as before, but with u – vcos instead of vcos
A1 Together, A  h |vcos – u| + avsin Fully justified (Given Answer)

⑥
1
M1 Journey time  so we need to minimise
u
av sin  h | v cos  u |
A1 J  (Ignore additional constant-of-proportionality factors)
u u
M1 For vcos – u > 0,
av sin  hv cos
if w  vcos , we minimise J   h
u u
E1 and this decreases as u increases
E1 and this is done by choosing u as large as possible; i.e. u = w
M1 For u – vcos > 0,
av sin  hv cos 
we minimise J   h
u u
E1 and this decreases as u increases if a sin > hcos
E1 so we again choose u as large as possible; i.e. u = w
[Note: minimisation may be justified by calculus in either case or both.]

⑧
M1 If a sin < hcos , then J increases with u when u exceeds vcos
A1 so we choose u = vcos in this case

M1A1 If a sin = hcos then J is independent of u, so we may as well take u = w ②

av sin  hv cos 
M1 Replacing  by 180o –  gives J   h
u u
A1 Which always decreases as u increases, so take u = w again

694
SI-2015/Q11

(i) B1 ↺O1: F = F1
O1 
↺O2: F = F2 (Both, with reason)
O2 
①

(ii) B1 Resg. ||plane (for C1): F1  R  W1 sin  

B1 Resg.r. plane (for C1): R1  F  W1 cos  

B1 Resg.||plane (for C2): F2  R  W2 sin  

B1 Resg.r. plane (for C2): R2  F  W2 cos  

Max 4 marks to be given for four independent statements (though only 3 are required).
One or other of
Resg.||plane (for system) : F1  F2  W1  W2 sin 

Resg.r. plane (for system) : R1  R2  W1  W2  cos 

may also appear instead of one or more of the above.

(F1 and F2 may or may not appear in these statements as F, but should do so below) ④

FR FR
M1A1 Equating for sin :  using  and 
W1 W2

 W  W2 
M1A1 Re-arranging for F in terms of R: F   1  R
 W1  W2 
M1 Use of the Friction Law, F  R

W1  W2
A1   obtained legitimately (Given Answer) ⑥
W1  W2

695
 FR
M1A1 (e.g.)   tan  
R1  F
 W  W2 
F  F  1 
 W1  W2   W  W2 
M1A1 Substg. for R = using R   1  F
R1  F  1
W  W2 

 2W1 
F  

=  1 2 
W W
A1
R1  F1
F1
M1A1 Substg. for R1 (correct inequality) using Friction Law F1  1 R1  R1 
1
 2W1 
F  
 W1  W2 

F1
 F1
1
 2W1 
F  
 W1  W2 
M1 Tidying-up algebra =
 1  1 
F  
 1 
2 1W1
A1  tan  obtained legitimately (Given Answer)
1  1 W1  W2 
⑨

696
SI-2015/Q12
nr
n  1   3 
r

(i) M1A1 P(exactly r out of n need surgery) =       (A binomial prob. term; correct)
r 4  4

②
nr
e 8 8 n
 r
n! 1 3
(ii) M1 P(S = r) = 
nr n!
     Attempt at sum of appropriate product terms
r ! (n  r ) !  4   4 
B1B1A1 Limits All internal terms correct; allow nCr for the A mark
nr
e 8  r
8n 1 3
M1 =
r!
    
n  r (n  r ) !  4   4 
Factoring out these two terms

e 8 
8n 3n  r
M1 =
r!

n  r (n  r ) !

4n
Attempting to deal with the powers of 3 and 4

e 8 
2 n  3n  r
A1 =
r!

n  r (n  r ) !
Correctly

e 8  2 r 
6n  r
M1 =
r!

n  r (n  r ) !
Splitting off the extra powers of 2 ready to …

e 8  2 r 
6m
M1 =
r!

m0 m!
… adjust the lower limit (i.e. using m = n – r)

e 8  2 r e 2  2 r
A1 =  e6 i.e.
r! r!

A1 … which is Poisson with mean 2 (Give B1 for noting this without the working) ⑪

(iii) M1 P(M = 8 |M + T = 12) Identifying correct conditional probability outcome

e 2  28 e 2  2 4

A1A1A1 = 8! 4! One A mark for each correct term (& no extras for 3rd A mark)
e 4  412
12 !
212  12 !
A1A1 = Powers of e cancelled; factorials in correct part of the fraction –
412  8 !  4 !
(unsimplified is okay at this stage)

A1 =
495
⑦
4096

697
SI-2015/Q13
Reminder
A : the 1st6 arises on the nth throw
B : at least one 5 arises before the 1st6
C : at least one 4 arises before the 1st6
D : exactly one 5 arises before the 1st6
E : exactly one 4 arises before the 1st6

(i) M1A1 P(A) =  56 n  1  16  ②

(ii) M1A1 By symmetry (either a 5 or a 6 arises before the other), P(B) = 1

2 ②

(iii) M1 The first 4s, 5s, 6s can arise in the orders 456, 465, 546, 564, 645, 654
A1  P(BC) = 1
3 (i.e. “by symmetry” but with three pairs)

 16  16     16  64  16     16  64 2  16   ...

2 3
(iv) M1A1A1 P(D) =
1 1
M1 for infinite series with 1stterm ; A1 for 2nd term ; A1 for 3rd term and following pattern 

M1 =  361 1  2 23   3 23 2  ... For factorisation and an infinite series

M1 =  361 1  23 2 Use of the given series result

1
A1 = 4

(v) M1 P(DE) = P(D) + P(E) – P(DE) Stated or used

B1 P(E) = P(D) = answer to (iv) Stated or used anywhere

 62  16  16     63  62  16  16     63 2  62  16  16   ...

3 4
M1A1A1 P(DE) =
1  2
M1 for infinite series with 1 term ; A1 for 2nd term ; A1 for 3rd term and following pattern 
st

M1 = 1081 1  3 12   6 12 2  ... For factorisation and an infinite series

M1 = 1081 1  12 3 Use of the given series result

A1 P(DE) = 1
2  272  23
54

698
Question 1 

(i)  1
ln 1 1   B1 
1
  For  0,  1  M1 
  Therefore  ln 1 0 for  0  A1 
  If  0,  ln 1 0   
  Therefore  ln 1  is positive for all positive .  B1 
     
  Therefore  ln 1 0 for all positive .   
  So, 
1 1 B1 
ln 1  

  1 1
ln 1 ln ln 1 ln   M1 
  So, 
1 M1 
ln 1 ln 1 ln ln 1 ln 1 

  Therefore, 
1 A1 
ln 1  

     
(ii)  1
ln 1 1   B1 
1
  For 0 1,  1  M1 
  Therefore  ln 1 0 for 0 1.  A1 
  If  0,  ln 1 0   
  Therefore  ln 1  is negative for 0 1.  B1 
     
  Therefore  ln 1 0 for all  1.   
  So, 
1 1 B1 
ln 1  

  1 1 M1 M1 
ln 1 ln ln 1 2 ln ln 1  
A1 
  So, 
1 M1 A1 
ln 1 ln 2 ln ln 1  

  As  → ∞, ln ln 1 → 0  B1 

  Therefore, 
1 1 1 A1 
1 ln 2 
1
 

   

699
Question 1 

Note that the statement of the question requires the use of a particular method in both parts. 

(i)   
B1  Correct differentiation of the expression. 
M1  Consideration of the sign of the derivative for positive values of  . 
A1  Deduction that the derivative is positive for all positive values of  . 
B1  Clear explanation that  ln 1  is positive for all positive . 
Note that answer is given in the question. 
B1  Use of   and summation. 
M1  Manipulation of logarithmic expression to form difference. 
M1  Attempt to simplify the sum (some pairs cancelled out within sum). 
A1  Clear explanation of result. 
Note that answer is given in the question. 
   
(ii)   
B1  Correct differentiation of the expression. 
M1  Consideration of the sign of the derivative for 0 1. 
A1  Deduction that the derivative is negative for this range of values. 
B1  Deduction that  ln 1  is negative for this range of values. 
B1  Use of   and summation. 
M1  Expression within logarithm as a single fraction and numerator simplified. 
M1  Logarithm split to change at least one product to a sum of logarithms or one quotient as a 
difference of logarithms. 
A1  Complete split of logarithm to required form. 
M1  Use of differences to simplify sum. 
A1  ln 2 correct. 
B1  Correctly dealing with limit as  → ∞. 
Note that answers which use ∞ as the upper limit on the sum from the beginning must 
have clear justification of the limit. Those beginning with   as the upper limit must have 
ln ln 1  correct in simplified sum. 
A1  Inclusion of  1 to the sum to reach the final answer. 
Note that answer is given in the question. 
 

   

700
Question 2 

  ∠ 3 B1 
 
  M1 A1 
sin 3 sin
  sin 3 sin 3   B1 
  sin 3
∴    
sin
  sin cos 2 cos sin 2
  M1 
sin
  sin 1 2 sin cos 2 sin cos
  M1 M1 
sin
  3 4 sin   A1 
     
   (or  )  B1 
     
  3 4 sin cos 2   B1 B1 
2
  3 4 sin 2 1 2 sin   M1 M1 
2 2 A1 
     
  sin ∠ , so ∠   B1  
M1 A1 
  Therefore ∠ 3 2   M1 M1 
  1 1
∠ 3 ∠  
3 3 A1 
So   trisects the angle   
 

   

701
Question 2 

B1  Expression for ∠  (may be implied by later working). 

M1  Application of the sine rule. 
A1  Correct statement. 
B1  Does not need to be stated as long as implied in working. 
M1  Use of sin  formula. 
M1  Use of double angle formula for sin. 
M1  Use of double angle formula for cos.  
A1  Simplification of expression. 
Note that answer is given in the question. 
   
B1  Identification of this relationship between distances. (just   is sufficient) 
B1  Correct expression substituted for the length of  . 
B1  Correct expression substituted for the length of  . 
M1  Use of double angle formula for cos. 
M1  Simplification of expression obtained. 
A1  Correct expression independent of  . 
   
B1  Identification of a right angled triangle to calculate sin ∠ . 
M1  Deduction that one of the lengths in sine of this angle is equal to  . 
A1  Value of the angle (Degrees or radians are both acceptable). 
M1  Obtaining ∠ 2  
M1  Use of ∠ ∠ ∠ ∠ . 
A1  Expression to show that ∠ ∠  and conclusion stated. 
 

   

702
Question 3 

   is the number of triangles that can be made using a rod of length 8 and 
M1 
two other, shorter rods. 
  If the middle length rod has length 7 then the other rod can be 1, 2, 3, 4, 5 or 6.  M1 
  If the middle length rod has length 6 then the other rod can be 2, 3, 4 or 5.   
  If the middle length rod has length 5 then the other rod can be 3 or 4.  M1 
  2 4 6.  A1 
     
  Assume that the longest of the three rods has length 7:  M1 
  If the middle length rod has length 6 then the other rod can be 1, 2, 3, 4 or 5.  M1 
  If the middle length rod has length 5 then the other rod can be 2, 3 or 4.   
  If the middle length rod has length 4 then the other rod must be 3.  M1 
  Therefore  1 3 5.  A1 
  1 2 3 4 5 6.  A1 
     
  2 4 ⋯ 2 1 B1 
  1 2 3 ⋯ 2 1 B1 
     
  3. (The possibilities are  1, 2, 3 , 1, 3, 4  and  2, 3, 4 .)  B1 
  Substituting  2 into the equation gives  2 2 1 4 2 1 3.   
  Therefore the formula is correct in the case  2.  B1 
  Assume that the formula is correct in the case  : 
M1 
 

  1 4 1 2 1    M1 
  4 3 1 12 6 4 1 1 , which is a 
M1 
statement of the formula where  1. 
  Therefore, by induction,  1 4 1   A1 
     
  2 4 ⋯ 2 1 1 .  M1 A1 
  Therefore  1 4 1 1 .   
  1 4 5 . (Or  1 4 1 )  A1 
 

   

703
Question 3 

M1  Appreciation of the meaning of  . 
M1  Identify the number of possibilities for the length of the third rod in one case. 
M1  Identify the set of possible cases and find numbers of possibilities for each. 
A1  Clear explanation of the result. 
Note that answer is given in the question. 
M1  An attempt to work out  . 
M1  Correct calculation for any one defined case. 
M1  Identification of a complete set of cases. 
A1  Correct value for  . 
A1  Correct deduction of expression for  . 
B1  Correct expression. No justification is needed for this mark. 
B1  Correct expression. No justification is needed for this mark. 
B1  Correct justification that  3. Requires sight of possibilities or other justification. 
B1  Evidence of checking a base case. (Accept confirmation that  1 gives  0 here.) 
M1  Application of the previously deduced result. 
M1  Substitution of formula for   and the formula for the sum. 
M1  Taking common factor to give a single product. 
A1  Re‐arrangement to show that it is a statement of the required formula when  1 
and conclusion stated. 
M1  Use of result from start of question. 
A1  Correct summation of 2 4 ⋯ 2 1 . 
A1  Correct formula reached (any equivalent expression is acceptable). 
 

   

704
Question 4 

(i) 

B1 
B1 

 
   

(ii)  , so stationary points when  1. 

B1 
B1 
M1 
A1 
A1 

 
   

B1 
B1 
B1 
B1 

 
   

(iii) 

B1 
B1 
B1 
B1 

 
   

B1 
B1 
B1 
B1 
B1 

705
 

Question 4 

Penalise additional sections to graphs (vertical translations by  ) only on the first occasion 
providing that the correct section is present in later parts. 

B1  Correct shape. 
B1  Asymptotes   and   shown. 
   
B1  Rotational symmetry about the point  0,0 . 
B1  Correct shape. 
M1  Differentiation to find stationary points. 
A1  Correct stationary points ‐  1, . ( ‐coordinates) 
A1  Correct  ‐coordinates. 
   
B1  Rotational symmetry about the point  0, . 
B1  Correct shape. 
B1  Stationary points have same  ‐coordinates as previous graph. (Follow through incorrect 
stationary points in previous graph if consistent here). 
B1  Correct co‐ordinates for stationary points ‐  1, arctan  
   
B1  Correct asymptotes  1. 
B1  ‐axis as an asymptote. 
B1  Middle section correct shape. 
B1  Outside sections correct shape. 
   
B1  Section for  1 1 correct shape. 
B1  1 . 
B1  1 . 
B1  Section for  1 correct with asymptote  2 . 
B1  Section for  1 correct with asymptote  0 or a rotation of  1 section about 
0, . 
 

   

706
Question 5 

(i)  tan tan arctan , so the formula is correct for  1.  B1 

  Assume that tan :   
  arctan .  M1 
 
tan    M1 

  tan , which simplifies to 
M1 A1 
tan . 
  Hence, by induction tan .  A1 
     
  Clearly,  arctan .  B1 
  Suppose that it is not true that  arctan  for all values of  .  
 
Then there is a smallest positive value,   such that  arctan . 
  Since  ,  arctan  and tan , but  arctan  
M1 M1 
. 
  However,  arctan , so this is not possible.  A1 
  Therefore  arctan .  A1 
     
(ii)  tan 2 .  M1 A1 
  So,    B1 
  Which simplifies to 2 tan 1 4 tan 2 0  M1 A1 
  tan 2 2 tan 1 0 A1 
  Since   must be acute, tan  cannot equal  2 .  B1 
  Therefore  arctan .   
 
lim arctan 1 .  M1 A1 
→ 4
 

   

707
Question 5 

B1  Confirmation that the formula is correct for  1. 

M1  Expression of   in terms of  . 
M1  Use of tan  formula. 
M1  Simplification of fraction. 
A1  Expression of   to show that it matches result. 
A1  Conclusion stated. 
   
B1  Confirmation for  1. 
M1  Observation that  0 
M1  Evidence of understanding that successive values of   with the same value of tan  must 
differ by  . 
A1  Evidence of understanding that   cannot be sufficiently large for   to be of the 
form arctan  if   is. 
A1  Clear justification. 
   
M1  Identification of the relevant sides of the triangle (diagram is sufficient). 
A1  Correct expression for tan 2 . 
B1  Use of double angle formula. 
M1  Rearrangement to remove fractions. 
A1  Correct quadratic reached. 
A1  Quadratic factorised. 
B1  Irrelevant case eliminated (must be justified). 
M1  Sum expressed as limit of   
A1  Correct value justified. 
Note that answer is given in the question. 
 

   

708
Question 6 

(i)  1
sec   B1 
cos
  1
sec   B1 
cos cos sin sin
  2
sec    
cos sin
  cos sin ≡ cos 2 sin cos sin 1 sin    M1 
  Therefore, sec   M1 A1 
     
  Therefore,  tan .  M1 A1 
     
(ii)  → 0
Limits:     
0→
  1   B1 
  sin sin   B1 
  Therefore,  sin sin 1    
  sin sin     
  So, 2 sin sin   M1 
  sin sin    A1 
     
  , and applying the result from part (i):   
  tan tan tan 2.  B1 
  2    B1 
     
(iii)  Consider  sin . Making the substitution  :   
  sin sin 1    M1 A1 
  So,  sin sin    
  Therefore,  2 3 sin 3 sin   B1 
  sec   
M1 
sec tan sec   
  2 tan tan    A1 
  And so,    B1 
  Therefore,  3 .  B1 
   

709
Question 6 

B1 
Expression of sec  in terms of any other trigonometric functions. 
B1 
Correct use of a formula such as that for cos  to obtain expression with 
trigonometric functions of  . 
M1  Expanding the squared brackets. 
M1  Use of sin ≡ 2 sin cos  and sin cos ≡ 1 
A1  Fully justified answer. 
Note that answer is given in the question. 
   
M1  Any multiple of tan . 
A1  Correct answer 
B1  Deals with change of limits correctly. AND 
Correctly deals with change to integral with respect to  . 
Note that both these steps need to be seen – the correct result reached without evidence 
of these steps should not score this mark. 
B1  Use of sin sin  (may be just seen within working) 
M1  Grouping similar integrals. 
A1  Fully justified answer. 
Note that answer is given in the question. 
B1  Evaluation of the integral from (i) with the appropriate limits. 
B1  Use of result from (ii) to evaluate required integral. 
   
M1  Attempt to make the substitution. 
A1  Substitution all completed correctly. 
B1  Rearrange to give something that can represent the required integral on one side. 
M1  Use of sec ≡ 1 tan  within integral. 
A1  Correct evaluation of this integral. 
B1  Correct use of result from part (i). 
B1  Correct application of result deduced earlier to reach final answer. 
 

   

710
Question 7 

(i)  Most likely examples:   M1 M1 

√  and  √   A1 
  If   then there cannot be two points on the circle that are a distance of 2  
apart and any two diametrically opposite points on   must be a distance of 2   B1 
apart. 
  If   then the circle must be the same as  , so there is not exactly 2 points 
B1 
of intersection. 
     
(ii)  The distances of the centre of   from the centres of   and   are    M1 A1 
and  .  B1 
  If the  ‐coordinate of the centre of   is  , then the  ‐coordinate is given by 
B1 B1 
 and   
  Therefore,    M1 
  4  and so  .  M1 A1 
     
  Therefore, the  ‐coordinate of the centre of   satisfies 
B1 
 and   
 
So 2 2    
 
2 2 2 2     
 
So,     
     
 
Therefore,  0  B1 
  16 8 8 16 0 M1 M1 
  16 16 8 8  
  16 4 M1 
  16 4 4 M1 
  16 4 2 4 2  
  16 4 4 A1 
 

   

711
Question 7 

M1  Calculation that the distance between the centres of the circles must be √ . 
M1  An example which shows that it is possible for at least one value of  . 
A1  Example showing that it is possible for all  . 
B1  Statement that the two intersection points must be a distance 2  apart. 
B1  Explanation that in the case   it would have to be the same circle. 
   
M1  The line joining the centre of   (or  ) and the radii to a point of intersection form a right 
angled triangle in each case. (one case) 
A1  Use of this to find the distance between centres of circles. 
B1  Applying the same to the other circle. 
B1  Expression relating the co‐ordinates and radii obtained from considering  . 
B1  Expression relating the co‐ordinates and radii obtained from considering  . 
M1  Elimination of   from the equations. 
M1  Either expansion of squared terms or rearrangement to apply difference of two squares. 
A1  Expression for   reached. 
Note that answer is given in the question. 
   
B1  Substitution to find expression for  ‐coordinate. 
Note that any expression for   in terms of  ,  ,   and   is sufficient, but it must be 
expressed as  ⋯, not  ⋯. 
B1  Observation that   must be positive. 
  Alternative mark scheme for this may be required once some solutions seen. 
M1  Attempt to rearrange the inequality to get 16  on the left. 
M1  Reach a point symmetric in   and  . 
M1  Reach a combination of squared terms. 
M1  Apply difference of two squares to simplify. 
A1  Reach the required inequality. 
Note that answer is given in the question. 
 

   

712
Question 8 

(i)  Let   be the vector from the centre of   to  .   

  Using similar triangles, the vector from the centre of   to   is  .  M1 A1 
  Therefore  , since these are both expressions for the vector 
M1 
from the centre of   to the centre of  . 
  So    A1 
  The position vector of   is    M1 A1 
     
(ii)  The position vectors of   and   will be   and  .  B1 
  Therefore, 
M1 A1 
 
     M1 A1 
  Similarly,    M1 A1 
M1 A1 
  Since they are multiples of each other the points  ,   and   must lie on the 
B1 
same straight line. 
     
(iii)   lies halfway between   and   if    B1 
  Therefore    M1 
  So,     
  Which simplifies to  2   M1 A1 
 

   

713
Question 8 

M1  Identification of similar triangles within the diagram. 
A1  Relationship between the two vectors to  . 
M1  Equating two expressions for the vector between the centres of the circles. 
A1  Correct simplified expression. 
M1  Calculation of vector from centre of one circle to  . 
A1  Correct position vector for  . 
Note that answer is given in the question. 
   
B1  Identifying the correct vectors for the foci of the other pairs of circles. 
M1  Expression for vector between any two of the foci. 
A1  Terms grouped by vector. 
M1  Simplification of grouped terms. 
A1  Extraction of common factor. 
M1  Expression for a vector between a different pair of foci. 
A1  Award marks as same scheme for previous example, but award all four marks for the 
M1  correct answer written down as it can be obtained by rotating 1, 2 and 3 in the previous 
A1  answer. 
B1  Statement that they lie on a straight line. 
   
B1  Statement that the two vectors must be equal. 
M1  Reduction to statement involving only   terms. 
M1  Attempt to simplify expression obtained (if necessary). 
A1  Any simplified form. 
 

   

714
Question 9 

(i)  Taking moments about  :   
  3 sin 30   M1 A1 
  5 sin 30   M1 A1 
    B1 
  5 cos 30 sin cos sin 30 3 cos 30 sin cos sin 30   M1 
  √3 1 √3 1
5 sin cos 3 sin cos   A1 
2 2 2 2
  Therefore 
A1 
4√3 sin cos  
  Either 
Use sin cos ≡ 1 and justify choice of positive square root. 
Or  M1 
Draw right angled triangle such that tan  and calculate the length of the 
√
hypotenuse. 
  sin    A1 
     
(ii)  Let   be the vertical distance of   below  . 
 
Let   be the vertical distance of   below  . 
  sin    M1 M1 
A1 
  sin    M1 A1 
  If   is the centre of mass of the triangle: 
M1 A1 
  
  Conservation of energy: 
M1 A1 
4 8 . 2  for complete revolutions. 
 
Therefore  .  A1 
 

   

715
Question 9 

M1  Attempt to find the moment of   about  . 

A1  Correct expression for moment (sin 30  may be replaced by cos 60 ). 
M1  Attempt to find the moment of   about  . 
A1  Correct expression for moment (sin 30  may be replaced by cos 60 ). 
B1  Correct statement that these must be equal. 
M1  Use of sin  or cos  formulae. 
A1  Correct values used for sin 30 and cos 30. 
A1  Correctly simplified. 
M1  Use of a correct method to find the value of sin . 
A1  Fully justified solution. If using right angled triangle method then choice of positive root not 
needed, if choice of positive root not given when applying sin cos ≡ 1 method 
then M1 A0 should be awarded. 
Note that answer is given in the question. 
   
M1  Attempt to find  . 
M1  Correctly deal with sine or cosine term. 
A1  Correct value. 
M1  Attempt to find  . 
A1  Correct value. 
M1  Combine two values to obtain distance of centre of mass from  . 
A1  Correct value 
M1  Apply conservation of energy. 
A1  Correct inequality. 
A1  Correct minimum value. 
 

   

716
Question 9 Alternative part (i) 

(i)  Let   be the centre of mass of the triangle and let the distance   be  .   

  Taking moments about  : 
M1 A1 
5 cos 3 cos
  Therefore 5 3 , so  .  A1 
   must lie on   and ∠ 30 .  B1 
  cos
sin 30   M1 
  sin 30 cos cos 30 sin cos   M1 
  √
.  A1 
  Therefore cos 4√3 sin  and so cos 48 sin   M1 
  sin , and so (since   is acute) sin .  M1 A1 
 

M1  Taking moments. 
A1  Correct equation. 
A1  Correct relationship between   and  . 
B1  Identification that   lies on   and calculation of ∠ . 
M1  Use of sine of identified angle. 
M1  Use of sin  formula. 
A1  Direct relationship between sin  and cos . 
M1  Rearrangement and squaring both sides. 
M1  Applying sin cos ≡ 1. 
A1  Final answer (choice of positive root must be explained). 
Note that answer is given in the question. 
 

   

717
Question 10 

  If the length of string from the hole at any moment is  , then  .  B1 

  The distance,  , from the point beneath the hole satisfies,  .  B1 
  Therefore  2 .  M1 A1 
  , and  cosec   M1 
  Therefore, the speed of the particle is  cosec .  A1 
     
  Acceleration:  cosec cosec cot   M1 A1 
  sin , so cos   M1 
  Therefore  cot   A1 
  The acceleration is  cot   M1 
  Since  sec , the acceleration can be written as  cot .  M1 A1 
     
  Horizontally:  M1 M1 
sin cot , so  cot cosec   A1 
     
  The particle will leave the floor when  cos   M1 A1 
  cot  and so tan   M1 A1 
 

   

718
Question 10 

B1  An interpretation of   in terms of other variables (including any newly defined ones). 

B1  Any valid relationship between the variables. 
M1  Differentiation to find horizontal velocity. 
A1  Correct differentiation. 
M1  Attempt to eliminate any introduced variables. 
A1  Correct result. 
Answers which make clear reference to the speed of the particle in the direction of the 
string being V. 
   
M1  Differentiation of speed found in first part. 
A1  Correct answer. 
M1  Attempt to differentiate to find an expression for  . 
A1  Correct answer. 
M1  Substitution to find expression for acceleration. 
M1  Relationship between required variables and any extra variables identified. 
A1  Substitution to give answer in terms of correct variables. 
   
M1  Horizontal component of tension. 
M1  Application of Newton’s second law. 
A1  Correct answer. 
   
M1  Vertical component of tension found. 
A1  Identification that particle leaves ground when tension is equal to the mass. 
M1  Substitution of their value for  . 
A1  Rearrangement to give required result. 
Note that answer is given in the question. 
 

   

719
Question 11 

(i)  cos , sin   B1 B1 

  Differentiating:  sin , cos   M1 
  Since   is moving with velocity   and is at the point  , 0  at time  ,  :   
  Velocity of   is  sin , cos .  A1 
     
(ii)  Initial momentum was   (horizontally).  M1 
  Horizontal velocity of   will be the same as that of  , so horizontally the total 
M1 
momentum is given by  2 sin  
  Therefore 3 2 sin .  A1 
     
  Initial energy was    M1 
  Total energy is  2 sin cos   M1 A1 
  Therefore  2 2 sin sin cos   M1 
  So  3 4 sin 2    
  Substituting   gives 
M1 
3 2 sin 4 sin 2 sin 6  
   6 3 4 sin 4 sin 4 sin 8 sin    
  6 4 sin 2    
  So,  .  A1 
     
(iii)  0, so there can only be an instantaneous change of direction in which   
B1 
varies at a collision. Since the first collision will be when  0, the second 
B1 
collision must be when  . 
     
(iv)  Since horizontal momentum must be  ,  0 ⟹ 2 sin .  B1 
  The KE of   must be  , so    B1 
    sin    
  sin , so   or    M1 A1 
   is only 0 when   takes these values and   is positive as   would need a non‐
zero value to satisfy 3 2 sin  if   is negative. (The relationship is still  B1 
true since collisions are elastic). 
 

   

720
Question 11 

B1  Horizontal component. 
B1  Vertical component. 
M1  Differentiation. 
A1  Complete justification, including clear explanation that  . 
Note that answer is given in the question. 
   
M1  Statement that momentum will be conserved. 
M1  Identification that horizontal momentum of   and   will be equal. 
A1  Correct equation reached. 
Note that answer is given in the question. 
M1  Statement that energy will be conserved. 
M1  Use of symmetry to obtain energy of   (accept answers which simply double the energy of 
 rather than stating the vertical velocity in opposite direction). 
A1  Correct relationship. 
M1  Use of sin cos ≡ 1. 
M1  Substituting the other relationship to eliminate  . 
A1  Correct equation reached. 
Note that answer is given in the question. 
   
B1  Correct value of  . 
B1  Answer justified. 
   
B1  First equation identified. 
B1  Second equation identified. 
M1  Solving simultaneously to find  . 
A1  Correct values for  . 
B1  Justified answer that   is not always 0 when   takes these values. 
 

   

721
Question 12 

(i)  If a tail occurs then player   must always win before   can achieve the 

sequence required. Therefore the only way for   to win is if both of the first  B1 
two tosses are heads. 
  After the first two tosses are heads it does not matter if more tosses result in 
B1 
heads as the first time tails occurs   will win. 
  The probability that   wins is therefore    B1 
     
(ii)  As before, after  , only   can win.  B1 
  Similarly, after  , only   can win.  B1 
  In all other cases for the first two tosses only   and   will be able to win.  M1 
  The probabilities for   and   to win must be equal.  M1 
  The probability of winning is   for all of the players.  A1 
     
(iii)  If the first two tosses are   then   must win (as soon as a   occurs), so the 
B1 
probability is 1. 
  After  : 
 must win if the next toss is a   as   needs two  s to win, but   will win the  M1 
next time an   occurs. 
  If the next toss is  , then the position is as if the first two tosses had been  , 
M1 
and so the probability that   wins from this point is  . 
  Therefore,  1   A1 
     
  After  : 
 
If the next toss is   then   will win with probability  . 
  If the next toss is   then   will win with probability  .  M1 
  Therefore  , and so  .  A1 
     
  After  : 
 
If the next toss is   then player   wins immediately. 
  If the next toss is   then   will win with probability  .  M1 
  Therefore  .  A1 
     
  Solving the two equations in   and  , gives  ,     
  From the third equation    M1 A1 
     
  The probability that   wins is  1   M1 A1 
 

   

722
Question 12 

B1  Identifying that   cannot win once a tail has been tossed. 

B1  Identifying that   must win once the first two tosses have been heads. 
B1  Showing the calculation to reach the answer. 
Note that answer is given in the question. 
   
B1  Recognising that the situation is unchanged for player  . 
B1  Recognising that the same logic applies to player  . 
M1  All other cases lead to wins for one of the remaining players. 
M1  Recognising that the probabilities must be equal. 
A1  Correct statement of the probabilities. 
If no marks possible by this scheme award one mark for each probability correctly 
calculated with supporting working. All four calculated scores 5 marks. 
   
B1  Explanation that probability must be 1. 
M1  Explanation of the case that the next toss is  . 
This mark and the next could be awarded for an appropriate tree diagram. 
M1  Explanation of the case that the next toss is  . 
A1  Justification of the relationship between   and  . 
Note that answer is given in the question. 
   
M1  Consideration of one case following  . 
This mark could be awarded for an appropriate tree diagram. 
A1  Establishment of the relationship. 
   
M1  Consideration of one case following  . 
This mark could be awarded for an appropriate tree diagram. 
A1  Establishment of the relationship. 
   
M1  Attempt to solve the simultaneous equations. 
A1  Correct values for  ,   and . 
M1  Attempt to combine probabilities to obtain overall probability of win. 
A1  Correct probability. 
 

   

723
Question 13 

(i)  for
  B1 
for
  M1 M1 
 
A1 
  Use the substitution   in the integral:   
 
  B1 

 
  M1 
  Therefore  .  A1 
     
  , so the stationary point occurs when  ln .  M1 A1 
  If  1 then choose  ln  as it is positive.   
  If  1 then choose  0 as the minimum occurs at a negative value of  .  B1 
     
(ii) 
2   M1 A1 

  Use the substitution   in the integral:   
 
2   B1 
  Applying integration done before: 
2  
2  
  Using integration by parts: 
M1 A1 
  
  and, applying the integration already completed, 
 
. 
  Therefore  .  A1 
  Var   M1 
 
Var .   
  Var 2 .  A1 
     
  2
Var 1   M1 
  For  0,  Var 0, so the variance decreases as   increases.  A1 
 

   

724
Question 13 

B1  Statement of random variable. 
M1  Any correct term in expectation (allow   multiplied by an attempt at the probability for 
not needing any extra costs). 
M1  Correct integral stated (allow   missing). 
A1  Fully correct statement. 
  May be altered to accommodate other methods once solutions seen. 
B1  Substitution performed correctly. 
M1  Integration by parts used to calculate integral. 
A1  Correctly justified solution. 
Note that answer is given in the question. 
M1  Differentiation to find minimum point. 
A1  Correct identification of point. 
B1  Both cases identified with the solutions stated. 
   
M1  Attempt at   (at least two terms correct). 
A1  Correct statement of  . 
B1  Substitution performed correctly. 
M1  Applying integration by parts. 
A1  Correct integration. 
A1  Correct expression for  . 
M1  Use of Var  
A1  Correct simplified form for Var  
M1  Differentiation of Var . 
A1  Correct interpretation. 
Note that answer is given in the question. 
 

725
  1.  (i)  

1 1 1 1
 
1 1 1

 
1
     
B1 
1 1
 
1 2 1 2 1

        integrating by parts    M1 A1 

1 1 1
0  
2 1 2

                A1*    (4) 
               M1 

…
              M1 
…

tan           B1 

… … ! !
      M1 
… … ! !

! !
Thus               A1*      (5) 
! !

(ii)  

.  

using the substitution    ,    and then the substitution    ,  1    M1A1* 

2 1  

So   1           M1A1* 

Using the substitution    ,  1 , 

726
 1      M1A1* (6) 

(iii)  
    M1A1 

              M1 

So   
         M1 

! !
          A1  (5) 
! !

   

727
2.     (i)   True.          B1 

   1000          B1 

If   1000 , then  1000  , so  1000  , i.e.   1000     M1A1 (4) 

(ii)  False.          B1 

E. G. Let   1  and   2  for     odd, and   2  and   1  for     even.    B1 

Then  ∄   for which for     ,    , nor           M1 

So it is not the case that    , but nor is it the case that         A1 (4) 

(iii) True.          B1 

  means that there exists a positive integer, say    , for which  for     ,    .  

E1 

  means that there exists a positive integer, say    , for which  for     ,    . 

E1 

Then if   max ,  ,         B1 

for    ,    , and so     A1 (5) 

(iv)  True.          B1 

   4          B1 

Assume   2   for some value   4 .    B1 

Then   1 1 1 2 2 2 2  
          M1A1 

4 2           B1 

so by the principle of mathematical induction,  2   for   4 , and thus   2   A1 (7) 

728
 3.   (i)    

 
  Symmetry about initial line    G1 
  Two branches        G1 
  Shape and labelling      G1 (3) 
If     | sec |   , then   sec   or   sec  

So     sec    or    sec     M1A1 

If   sec 0  ,   sec 0   as      and   sec 0   as     and     

are both positive, and thus in both cases,   0  which is not permitted.   B1 

If   sec 0 ,    sec 0  and   sec 0  giving   0 

so  sec 0  as required.      B1  (4) 

So    sec   , thus points satisfying (*) lie on a certain conchoid of Nicomedes with A being 

the pole (origin) ,        B1 

 d being b,           B1 

and L being the line   sec  .    B1 (3) 

(ii) 

Symmetry about initial line    G1 
Two branches        G1 
Loop, shape and labelling    G1  

729
 

If      , then the curve has two branches,   sec   with   sec 0  and   sec   

with   sec 0  , the endpoints of the loop corresponding to  sec   .  B1 (4) 
 
In the case    1   and    2  ,  sec 2  so    

Area of loop   

2 sec 2       M1A1 

sec 4 sec 4 tan 4 ln|sec tan | 4   M1A1 

4 √3 4 ln 2 √3 √3 4 ln 2 √3   M1A1 (6) 

   

730
4.  (i)     is continuous. 

For   → ∞ , → ∞  and for   → ∞ , → ∞  .    B1 

So the sketch of this graph must be one of the following:‐ 

B1 

Hence, it must intersect the     axis at least once, and so there is at least one real root of 

   0        B1 (3) 

(ii)    M1 

Thus         A1 

     A1 

and, as   0 ,  0 , 0 

adding these three equations we have,  

3 0   M1 

(Alternatively,  

   

3 6  

  ) 

So   3 0          M1 

Thus   6 3 2           A1* (6) 

731
(iii)  Let   cos sin   for   1, 2, 3      M1 

Then   cos 2 sin 2   and   cos 3 sin 3   by de Moivre  M1 

As  

sin 0 

sin 2 0 

sin 3 0 

0 

0 

0 

and so    ,    , and    are real ,    M1 

and therefore so are    ,   , and       A1 

Hence,  as    ,   , and     are the roots of   0  with    ,   , and     real, by part 

(i), at least one of    ,   , and    is real.    M1  

So for at least one value of    ,  cos sin   is real and thus,  sin 0 ,   

and as    ,   0  as required.  A1 (6) 

If   0  then    is real.     and    are the roots of   0 

which is   0  (say   0 ) 

  and     and so the quadratic of which    and    are the 

roots has real coefficients.  Thus  ,  .  (  0  because   0)  B1 

If   4 0 ,              M1 

Thus  cos cos , and so    , as   .   

But sin sin  and so    .        M1 A1 

If   4 0 , then     and    are real roots, so  sin sin 0 , and thus   0 , so  

 .              B1 (5) 

732
   

733
5. (i) Having assumed that  √2  is rational (step 1),  √2   , where   , ∈ , 0  B1 

Thus from the definition of   (step 2), as  ∈  and √2 ∈  , so  ∈  proving step 3. 

B1 (2) 

If   ∈  , then    is an integer and  √2 is an integer.  B1 

So   √2 1 √2  is an integer,     B1 

and  √2 1 √2 2 √2  which is an integer and so  √2 1 ∈   proving step 5.  B1 (3) 

1 √2 2   and so     M1 

0 √2 1 1, and thus 0 √2 1     A1 

and thus this contradicts step 4 that   is the smallest positive integer in   as  √2 1  has been 

shown to be a smaller positive integer and is in   .    A1 (3) 

  (ii)  If  2   is rational, then 2   , where   , ∈ , 0 

So   2   , that is  2  , which can be written  2 2    M1 

and hence 2  proving that  2  is rational.  A1 

2

If  2   is rational, then 2   , where   , ∈ , 0     M1 

and so 2   proving that 2   is rational and that  2   is rational only if  2   is rational.  

A1 (4) 

Assume that 2   is rational. 

Define the set    to be the set of positive integers with the following property:    is in   if and only if 

2   and   2  are integers.      B1 

The set   contains at least one positive integer as if  2   , where   , ∈ , 0 , then 

2 ∈    and   2 ∈   , so   ∈  .  M1A1 

Define     to be the smallest positive integer in   .  Then    2   and   2  are integers.  B1 

Consider   2 1 .      2 1 2   which is the difference of two integers and so is 

itself an integer.    2 1 2 2 2  which is an integer,  

and   2 1 2 2 2 2 2 2   which is an integer. 

734
Thus   2 1   is in  .    M1A1 

1 2 2   and so  0 2 1 1, and thus 0 2 1 , and thus this contradicts 

that   is the smallest positive integer in   as  2 1  has been shown to be a smaller positive 

integer and is in   .      M1A1 (8) 

   

735
6.  (i)  , ∈ ⟹ , ∈     B1 

For   , ∈  , we require to solve    ,    M1 

2 2 0 

2 √4 8 8 √2
 
4 2

∓ √2
 
2

              M1A1 

So for , ∈  , as   must be real,   must be real, and  2 0   

i.e.   2       B1* (5) 

(ii)   ⟹ 2   so if    ,  then   2  

so  3 1          M1A1 

3 1  

      M1A1 

Thus if    ,   1 1       M1A1 

Thus   1 1 0 ,    M1 

1 0      M1A1 

√
Thus   1  or    

So as   ∈   and   0  , the values of    are real.  B1 

√
There are three distinct values of    unless   1  in which case   √1 4 3 , i.e.   2 

            M1A1 (12) 

For   , ∈   , from (i) we require   ∈   which it is,   ∈  which it is, and   2   

in other words    .        M1 

So     and     need not be real.  A counterexample would be   1    B1 

for then   1 ,    , so   1  ,  i.e.   2 2 0 in which case 

the discriminant is  0  so  ∉  .      B1 (3) 

736
7.      M1 

      M1A1 (3) 

(i)  Suppose     is a polynomial of degree     i.e.   ⋯   

for some integer   .        B1 

Then   ⋯ ⋯  

1 ⋯ 1 ⋯   

which is a polynomial of degree    .      M1A1 

Suppose   ⋯  , then 

  ⋯ 1 ⋯   so the result 
is true for  1 ,          M1A1 

and we have shown that if it is true for    , it is true for   1 .  Hence by induction, it is 

true for any positive integer  .        B1 (6) 

(ii)   Suppose   1   is divisible by  1    i.e.   1 1   for 

some integer   , with  1.        B1 

Then   1 1 1  

1 1   

1 1    which is divisible by   1 .  M1A1 

1 1 1   so result is true for  1.   M1A1 

We have shown that if it is true for    , it is true for   1 .  Hence by induction, it is true 

for any positive integer   .    B1 (6) 

(iii) 
1 ∑ ∑ 1   M1 

So   
1 ∑ 1 ∑ 1      M1A1 

But by (ii),   1  is divisible by  1   and so   1 1  , and 

thus if   1 ,    1 0 ,  and hence 

 
∑ 1 0    M1A1* (5) 

737
8.   (i)    cos ⟹ sin cos   M1A1 

and   sin ⟹ cos sin     M1A1 

Thus       becomes    sin cos sin cos   M1 

That is   sin cos cos sin sin cos sin cos    

as   0, 0 

Multiplying out and collecting like terms gives 

cos sin sin cos 0 

which is    0  .          M1A1* (7) 

So    0        M1 

and thus      ,      A1 

         A1 

G1 (4) 

(or alternatively     M1 so   ln| |   A1 and hence   A1) 

(ii)      

becomes    sin cos cos sin cos sin  

that is   

  sin cos cos cos sin sin cos sin sin cos  

738
Multiplying out and collecting like terms gives 

cos sin cos sin sin cos 0  M1 

which is     0  .    A1 

  

  M1 

So       A1 

ln      A1 

So   
1
 

with   0 

1
 
1∓

that is 

  A1* 

G1 G1 G1 (9) 

   

739
9.  If the initial position of   is  , then at time   ,   , so conserving energy, 

1 1
 
2 2 2

                M1 A1 A1 

Thus,  

                M1 

i.e.  
 

            A1* (5) 

The greatest value,   , attained by   , occurs when  0.  M1 

Thus    

So      (negative root discounted as all quantities are positive) 

Thus  
2
 

and  

2
 

            M1 A1 (3) 

As 
 

differentiating with respect to   

1
2 2 2  
2

                M1 A1 

Thus 
√
 
√

                A1 

740
So when    , the acceleration of   is  

2 √ 2
 

                  M1 A1 (5) 

That is  

and thus 

1
 
√

where    is the period.    M1 A1 

So  
1
4  
√

4 1
 
√
1

Let  
√
 

                B1 

then 

and so 
2
 

741
 √2 1 √2  
2

as   ≪  

Thus 
1
√2  

                M1A1 

and so  

4 1 1 32 1
√2  
√1 √1

as required.              M1 A1* (7) 

   

742
10.  The position vector of the upper particle is 
sin
 
cos

              B1 B1 

so differentiating with respect to time, its velocity is 

θcos
 
θsin

              E1* (3) 

Its acceleration, by differentiating with respect to time, is thus 

θcos sin
 
θsin cos

              M1 A1 A1 

so by Newton’s second law resolving horizontally and vertically 

sin θcos sin

 
cos θsin cos

              M1 A1 

That is 

θcos sin sin 0

 
θsin cos cos 1

The other particle’s equation is 

θcos sin sin 0

 
θsin cos cos 1

            B1 (6) 

Adding these two equations we find 

0
2 2  
1

i.e.   0  and           M1 A1*  

Thus  
θcos sin sin
 
θsin cos cos

i.e.    θcos sin sin   and    θsin cos cos  

Multiplying the second of these by  sin  and the first by  cos  and subtracting, 

743
 θ 0  and so  θ 0 .      M1A1* (4) 

Thus     and as initially   2  ,     M1 A1 

Therefore the time to rotate by      is given by    , so     A1 

As     and initially    , at time  ,  , and so     as the centre of 

the rod is initially     above the table.        M1 A1 

Hence, given the condition that the particles hit the table simultaneously,  
0 / 1/2 /  
. 

Hence    0 2 2  , or  2 2   as required.  M1 A1* (7) 

   

744
11.  (i)  Suppose that the force exerted by   on the rod has components    perpendicular to the rod 
and     parallel to the rod.  Then taking moments for the rod about the hinge,   0 ,  M1 

which as   0  yields   0  and hence the force exerted on the rod by    is parallel to the rod. 

A1* (2) 

Resolving perpendicular to the rod for    ,   sin sin cos     M1 A1 

Dividing by   sin  ,  sin cot  

That is   cot cos   or in other words   cot cos   as required.  M1 A1* (4) 

The force exerted by the hinge on the rod is along the rod towards    ,   B1 

and if that force is    , then resolving vertically for    ,   cos     M1 A1 

so   sec  .            A1 (4) 

(ii)   Suppose that the force exerted by   on the rod has component    perpendicular to the rod 

towards the axis, that the force exerted by   on the rod has component    perpendicular to the 
rod towards the axis,      B1 

then resolving perpendicular to the rod for    ,   sin sin cos  

M1A1 

and similarly for    ,   sin sin cos  

M1A1 

Taking moments for the rod about the hinge,   0     M1A1 

So multiplying the first equation by    , the second by     and adding we have 

sin sin sin cos sin cos  

Dividing by   sin  ,  cot cos   M1A1 

That is   cot cos  , where         A1 (10) 

   

745
12.  (i)  The probability distribution function of    is 

  1  2  3  4  5  6 

  1   1   1   1   1   1  
6 6 6 6 6 6

so the probability distribution function of    is 

  0  1  2  3  4  5 

  1   1   1   1   1   1  
6 6 6 6 6 6

and thus   1  .    B1 

The probability distribution function of    is 

  2  3  4  5  6  7  8  9  10  11  12 

  1 2 3 4 5 6 5 4 3 2 1
36  36  36  36  36  36  36  36  36  36  36 

M1 

so the probability distribution function of    is 

  0  1  2  3  4  5 

  6 6 6 6 6 6
36  36  36  36  36  36 

A1 

which is the same as for    and hence its probability generating function is also    .  A1* 

Therefore, the probability generating function of    is also      B1 

and thus the probability that    is divisible by 6 is  1 6  .    B1 (6) 

(ii)  The probability distribution function of    is 

  0  1  2  3  4 

  1   2   1   1   1  
6 6 6 6 6

746
and thus    1 2      M1 A1 

   would be      except that the powers must be multiplied congruent to modulus 5. 

1 2 1    B1 

Thus      would be       

except   1 1      M1A1 

and   1 1 1
1 1 1 1
5                  A1 

So   2 2 5 7   M1A1* (8) 

1 1 1
7 7 7  
6 6 6

That is 

1 1 1
7 7 7 35 43  
6 6 6

We notice that the coefficient of     inside the bracket in    is   1 6 6 ⋯6  

This can be shown simply by induction.  It is true for  1 trivially. 

Consider   1 6 6 ⋯ 6 1 6 6 ⋯ 6
1 6 6 ⋯ 6  

1 6 6 ⋯6 1 6 6 ⋯ 6
1 6 6 ⋯ 6 5 1 6 6 ⋯ 6  

5 1 6 6 ⋯6 6 1 1 6 6 ⋯6 6 1 

So   1 6 6 ⋯ 6 5 1 6 6 ⋯ 6 1 6 6 ⋯6  

as required.                  M1 

However, this coefficient is the sum of a GP and so     where     is an 

integer such that  0 5 4 .            M1 A1 

So if     is not divisible by 5, the probability that    is divisible by 5 will be the coefficient of     

which in turn is the coefficient of     , namely    1  as required.  B1* 

If     is divisible by 5, the probability that    is divisible by 5 will be   1   as    

747
That is    1       M1A1 (6) 

   

748
13.   (i) 

G1 

     if  0 1    B1 

G1 

and   1 2   if  1 2  B1 

0 if   0  and   1  if   2 

0 0
0 1
So    
1 2 1 2
1 2

        B1 (5) 

Thus   1  

1
1 1
2
1 1 1
2 1 
2 2
1
0
2

            M1 A1 

So as    , 

749
 1
0
2
1 1 1
2 1 
2
1
1

as required.          M1A1* (4) 

1
2 . 2 ln  

1
1 2 ln 2 1 2 ln 2 
2

              M1 A1 (2) 

(ii)   

G1 

1
0 1
2  
1
1 1
2

                B1 (2) 

Thus 

   1 1 1 1  

So 

1 1 1
1 1 1
2 2
 
1 1 1
1 0
2 2

i.e. 

1 1
3 1
2 2  
1 1
0
2 1 2

750
              M1A1 

So as    , 

1 1
1
2 2  
1 1
1 0
2 2

              M1A1 (4) 

   because, by symmetry,      

and   1 1      B1 

1 1
1  
2 2

1 1 1
1 1 ln  
2 2 2

1 1 1 1
ln 1 ln  
2 2 2 2

1 1 1 1 1 1
ln ln  
2 2 2 2 2 2

as required.    M1A1 (3) 

751


752


STEP Solutions 2015

Mathematics

STEP 9465/9470/9475

October 2015

753
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754
Contents

STEP Mathematics (9465, 9470, 9475)

Hints and Solutions Page

STEP Mathematics I 4
STEP Mathematics II 10
STEP Mathematics III 15

755
 SI-2015 Hints and Solutions
(See the marking scheme for full details of the expected solution approach)

Q1 This question is intended to be a relatively straightforward entrée into the paper, and thus its
demands are fairly routine in nature. That does not mean that it is easy, merely that the appropriate courses
of action should be readily accessible to all candidates of a suitable standard. To begin with, the demand for
a sketch (of any function) should lead you to consider things such as
 key points (such as where the curve meets either of the coordinate axes);
 asymptotes (note the information given in italics at the end of part (i) regarding what happens as
x –, which indicates that the negative x-axis is an asymptote in this case);
 turning points of the curve, which are clearly flagged as being of significance when considering what
happens when y = k; i.e. when the curve meets a horizontal line;
 long-term behaviour (you already have sorted for you the “x –” side of things, so there is only a
quick decision to be made about what happens as x +).
For the key points, first set x = 0 and then y = 0; the asymptote is effectively given; the TPs come from
setting the first derivative to zero and solving for x again (noting, of course, that ex is always positive); and
the curve clearly grows exponentially as x increases positively. The rest of (i) then simply requires a bit of
thought as to how many times a horizontal line will cut, or touch, the curve depending upon the value of k.
In part (ii), it is clear that the x in part (i) has now been replaced with an x2, and this second curve
must therefore have reflection symmetry in the y-axis, as all negative values of x are being squared to give
the positive counterpart. Previously, when x was equal to zero, we now havex2 = 0, and so each previous
crossing-point on the positive x–axis leads to two, one on each side of the y-axis (and at the square-root of
its former value). However, the previous y-intercept is unchanged, but must now appear at a TP of the curve
(otherwise the symmetry of the gradient would be compromised). Also, the previous TP with positive x-
value (the negative one has gone) occurs at the square-roots of the previous value, but again with unchanged
y-coordinate.

Q2 It is clear that (i) is an introductory part that requires the use of the cos(A – B) formula with suitably-
chosen values of A andB. Using the sin(A – B) formula then leads to the second result, although there are
alternative trig. identities that could be used in both cases, such as a double-angle formula. Repeated use of
these, or the double-angle formulae (or de Moivre’s Theorem for those from a further maths background)
lead to the (relatively) well-known ‘triple-angle formula’ cos3α  4cos3α – 3cosα, which gives x = cosα as a
root of the equation 4 x 3  3 x  cos 3  0 . Although there are several possible methods here, a simple
  
division/factorisation leads to ( x  c) 4 x 2  cx  c 2  3 = 0, and the quadratic formula leads to the
remaining two roots which, for their simplest form, requires the use of the most elementary of trig.
identities, s 2  1  c 2 .
A bit of insight is needed in part (iii), where one should first realise that the constant term is intended
be a cosine value (of 3  some angle), and the most obvious candidate is 1
2 2 = cos 45o, so that  is 15o.
1
From here, it is now clear that “x” is 2 y , and that the three roots are those from part (ii) with the exact
numerical forms of the sine and cosine of 15o from part (i) waiting to be deployed in order to find the surd
forms requested.

756
Q3 In this question, it is important to draw suitable diagrams in order to visualise what is going on, and
these are not difficult to manage, with the guard either at a corner of the yard (C) or at its middle (M).
However, this second case has two possible sub-cases to consider, depending upon whether the ‘far’ corners
of the yard are visible to him/her (which, in fact, turns out to be the b = 3a case which separates the two
cases that the question invites you to consider), and it is the extra length of the opposite wall that is visible
4b 2 b2
that makes for different working. These lengths are (from C), (from M, with the ‘far’ corners not
ba a
2b( 2b  3a )
visible) and (from M, with the ‘far’ corners visible). Once obtained, these should be compared
ba
in order to find that the guard should stand at C for b < 3a and at M for b > 3a (and at either when b = 3a).

Q4 A quick read-through of the question should make it clear that it is the lower end of the rod that is
being referred to (as do the subtractions within the given coordinates x and y). The fact that the rod is
tangent to the given parabola means that its direction is tan   12 x , which gives the coordinates of the rod’s
midpoint as (tan, tan2); a simple right-angled triangle and some accompanying basic trig. then leads to the
given answer. The second part of the question is equally straightforward once one realises that when xA = 0,
2 tan 
2 tan   b cos  b   yA=  tan 2  . There are several ways to attack the area between two
cos 
curves – e.g. as  y
1  y 2  dx, or by translating the bit below the x-axis up by tan2 and calculating the
difference between area under the “new” curve and a triangle; the key is to eliminate the “b” and then the
given answer follows.

Q5 Once one realises that the x within the integral is not the variable, then both integrations are
2x
, while in (ii) g(x) = 1  x  1  x  … remember to use
1
relatively straightforward. In (i), we get f(x) =
x x
the modulus function when taking square-roots (although one could, alternatively, work out a piece-wise
definition for g; that is, in bits). The sketch of f should prompt the solver to differentiate in order to identify
 1 
the turning point at  , e ln 2  . Noting that y + as x 0 and that y + as x + gives all else that
 ln 2 
is needful to draw the graph in (i). In (ii), the piece-wise definition of g is certainly more useful now since its
graph is made up of two ‘reciprocal’ curve bits joined by a horizontal straight-line in the middle.

Q6 The best way to start any geometrically-inclined question is to

have a good diagram that doesn’t make the shape of the quadrilateral
look too specialised in any way (square, rectangle, parallelogram, …).
Next, label the midpoints sensibly (see diagram) and then write down P
their position vectors in terms of a and b.
It is relatively easy to prove that the opposite sides of Q
this quadrilateral are equal and opposite, but you must then
also show that adjacent sides are equal in length and that they  
are perpendicular. This last outcome is going to follow from the S
use of the scalar product.
For the final part, you should label one of the angles at the R

757
centre (say) and note that the fourth angle at O is thus 180o– .
Having already calculated the squares of the lengths of the square’s sides in the form
1 2
 o

4 a  ( a )  2 aa cos 90   
 2
the required result follows from noting that this is maximal when cos90 o

   1 ; i.e. when   90 o .

Q7 The crucial observation here is that a (continuous) function takes its maximum value on a finite
interval either at a maximum turning-point or at an endpoint. Differentiating (a ‘negative’ cubic – so we
know what its shape is) gives a MIN. TP at (0, 0) and a MAX. TP at ( 13 a, 19 a3), and evaluating at the
endpoints gives f(– 13 ) = 19 (3a + 2) and f(1) = 3a – 6.
Now, a comparison of these possible values for f then yields that 19 (3a + 2)  19 a3  a  0, a  2; and
that 19 a3  3a – 6 holds for all a  0; and also that 19 (3a + 2)  3a – 6  a  7
3 (which, actually, affects
 19 (3a  2) 0  a  2

nothing, but the working should be done anyhow). Thus M(a) =  19 a 3 2a3 .
 3a  6 a3


Q8 The standard “bookwork” approach to this opening part is to write the sum (S) both forwards and
backwards, add the terms in pairs (n pairs, each of value n + 1) and then to half this to get S = 12 n(n + 1). As
with any such invitation to establish a result, one should not simply seek to quote a result and thus merely
“write down” the given answer. When looking at part (ii)’s question, the binomial theorem should really be
screaming at you from the page, and all that is needed is to observe that the binomial expansion of (N – m)k
consists of k + 1 terms, the first k of which contain a factor of (at least one) N. The final term, since k is odd
must be –mk which then conveniently cancels with the + mk term to leave something that is clearly divisible
by N.
In the next part of the question, you are invited to explore the cases n odd and n even separately
(indeed the results that follow are slightly different). To begin with,
S = 1k + 2k + … + nk (an odd no. of terms) = 0k + 1k + 2k + … + nk (an even no. of terms)
So these terms can now be paired up:
n with 0, n – 1 with 1, …, ( 12 n  12 ) with ( 12 n  12 ) ,
so that all pairs are of the form (n – m)k+ mk , which was just established as being divisible by n. Next, in the
case when
S = 1k + 2k + … + nk (an even no. of terms) = 0k + 1k + 2k + … + nk (an odd no. of terms),
the pairs are now
n with 0, n – 1 with 1, …, ( 12 n  1) with ( 12 n  1) ,
but with an odd term, ( 12 n) k , left over. This gives us (from the same previous result as before) a sum
1
consisting of terms divisible by n and one that is divisible by 2 n, giving the second result.
Then, for n even, so that (n + 1) is odd, S + (n + 1)k is divisible by n + 1 (by the previous result)  S
is divisible by n + 1; and for n odd, so that (n + 1) is even, S + (n + 1)k is divisible by 12 (n + 1). Thus, since
hcf(n, n + 1) = 1  hcf( 12 n, n + 1) = 1 for n even, and hcf(n, 1
2 (n + 1)) = 1 for n odd, it follows that S is
1
divisible by 2 n(n + 1) for all positive integers n.

758
 2u sin 
Q9 The standard time taken to land (at the level of the projection) of a projectile is t  . Thus, a
g
 2u   
bullet fired at time t, 0  t  , lands at time TL  t  sin   t  . Differentiating this w.r.t. t and
6 g 3 
  2u 2 sin  cos
setting it equal to zero, gives k  cos   t  . The horizontal range is then given by R  and
3  g

this gives the required answer. Moreover, substituting the endpoints of the given time interval 0  t 
6
  1 3 dTL
into k  cos  t  gives  k  . However, if k < 12 , then one sees that  0 throughout the
3  2 2 dt
gun’s firing, so that TL is a (strictly) decreasing function. Hence its maximum value occurs at t = 0, i.e.
 2u 2 1 3 u2 3
 , whence R     .
3 g 2 2 2g

Q10 The difficulty in this question lies in ignoring unnecessary information (not given). Firstly, then, note
that the speed of the rain relative to the bus is vcos – u (or u – vcos if negative), and when u = 0, the area
of the bus getting wet, A, is such thatAhvcos + avsin . Now the given result follows from observing that
when vcos – u> 0, the rain hitting the top of the bus is the same, whilethe rain hits the back of the bus as
before, but with speed vcos – u instead of vcos ; and whenvcos – u< 0, the rain hitting the top of the bus
is the same, while the rain hits the front of the bus as before, but with u – vcos instead of vcos.
1 av sin  h | v cos  u |
Next, as the journey time  , we need to minimise J   . For vcos – u > 0
u u u
av sin  hv cos
and w  vcos , we minimise J    h , and this decreases as u increases, and this is done
u u
av sin  hv cos
by choosing u as large as possible; i.e. u = w. For u – vcos > 0, we minimise J   h,
u u
and this decreases as u increases if a sin > h cos , so we again choose u as large as possible; i.e. u = w.
Next, if a sin < h cos , then J increases with u when u exceeds vcos , so we choose u = vcos in this
case. Finally, if a sin = hcos then J is independent of u, so we may as well take u = w.
av sin  hv cos 
For the return journey, simply replace by 180o –  to give J    h , which
u u
always decreases as u increases, so take u = w again.

Q11 As with all statics problems, the key to getting

a good start, and to making life as easy as possible
for the working that follows, is to have a good, clear O1 
diagram with all relevant forces, in appropriate
directions, marked on it (see alongside). O2 
To begin with, take moments about the
respective cylinders’ axes yields F = F1 = F2, as
required. Next, write down the four equations
that arise from resolving for each cylinder in the
directions parallel and perpendicular to the plane.

759
These are:-
F1  R  W1 sin  ; R1  F  W1 cos ; F2  R  W2 sin   and R2  F  W2 cos .
(Note that one could replace some of these with equivalent equations gained from resolving for the whole
system.) Replacing F1 and F2 by F, equating for sin , re-arranging for F in terms of R and using the
Friction Law, F  R , appropriately leads to the first given answer in (ii). A bit more determination is
needed to gain the second given answer, however. Firstly, tan can be gained by division in at least two
ways, and both F and R must be eliminated from any equations being used. Thereafter, it is simply a matter
of forcing the working through correctly and, hopefully, concisely.

Q12 Here, you are given the relevant Poisson result at the outset, and this is intended to guide your
thinking later on in the question. To begin with, though, part (i) is actually a Binomial situation … requiring
just a single general term. In part (ii), you were asked to prove algebraically a result that you might usually
be required to quote and use. This requires a good understanding of the use of the sigma-notation and a clear
grasp as to which of the various terms are constant relative to the summation, and then combining the
remaining terms together appropriately to give the requested Poisson answer. Most important of all, of
course, it is essential to have the first line of working correct; this is
nr
e 8 8 n
 r
n! 1 3
P(S = r) =      
nr n! r ! (n  r ) !  4   4 
e 8  2 r  6 n  r
and one follows this through to the point where the result 
r ! n  r (n  r ) !
is obtained. At this stage

another simple trick is required – effectively a re-labelling of the starting-point, using m = n – r to re-write
e 8  2 r  6 m
this as 
r ! m0 m!
. The required result follows immediately since the infinite sum is just e6.

Having established this, the final part of the question is relatively straightforward, requiring only the
use of the conditional probability formula applied to P(M = 8 |M + T = 12).

Q13 The first three parts of this question are very easy indeed, if looked at in the right way. In part (i) it is
not necessary at all that you recognise the Geometric Distribution (indeed, some of you may not have
encountered it at all), but the result asked for is simply “(n – 1) failures followed by 1 success”, and one can
write down immediately, and without explanation, the answer P(A) =  56 n  1  16  . In (ii), you have a situation
in which one can apply the principle of symmetry: either a 5 arises before a 6, or vice versa, so the required
probability is just P(B) = 12 . Part (iii) can be approached similarly, in that the first 4s, 5s, 6s can arise in the
orders 456, 465, 546, 564, 645, 654  P(BC) = 1
3 (i.e. also “by symmetry”, but with three pairings to
consider).
Parts (iv) and (v), however, each turn out to require the use of the result given at the end of the
question, as the outcomes (theoretically) stretch off to infinity. For (iv), it is best to consider only on which
throw the first 6 occurs (since we stop at that point). It cannot occur on the first throw, so we have the sum
of the situations:
a 5 occurs on the first throw, followed by a 6 on the second;
one 5 and a 1-4 occur, in either order, followed by the 6 on the third;
one 5 and two 1-4s occur, in any of three possible orders, followed by a 6 on the fourth;
etc.

760
  16  16     16  64  16     16  64 2  16   ... , and this factorises as  361 1  2 23   3 23 2  ..., and the
2 3
Thus P(D) =
1 1
2
big bracket is just the given result with x = 3 and n = 2.
Before getting too deeply into part (v), a couple of simple results should be noted. Firstly, we use the
fact that P(E) = P(D) = 14 , the answer to (iv); and then that we will need to use the basic probability result
1
P(DE) = P(D) + P(E) – P(DE) = 2 – P(DE). Turning this around, since it is far easier to calculate the
probability, P(DE), that both one 4 and one 5 occur before the first 6. Again, looking at this from the
viewpoint of finishing after the first 6 is thrown, we see that
 3  4 2

P(DE) =  62  16  16     63  62  16  16     63   62  16  16   ... = 108
1
 1  3 12   6 12 2  ...
1  2
1 23
and the big bracket is the given result with x = 2 and n = 3 , leading to the answer P(DE) = 54 .

761
STEP 2 2015 Hints and Solutions 

Question 1 

For the first result, show that the gradient of the function is positive for all positive values of   (by 
differentiating) and also that  0 0. Once this result has been established sum a set of the terms, 
using  , note that ln 1  can be written as ln 1 ln  and then the required result 
follows. 

For the second part, first show that  ln 1  is negative for 0 1 and then use the 

substitution  , noting that ln 1  can be written as ln 1 2 ln ln 1 . Deal 
with the sum starting with  2 and then add the initial 1 afterwards. 

Question 2 

As with all geometric questions a good diagram of the information given makes the solution to this 
question much easier to reach. The first result in this question follows from an application of the sine 
rule with applications of the relevant formulae for sin  and the double angle formulae. From 
a diagram of the triangle it should then be an easy application of trigonometry to show that  . 
There are a number of different methods for establishing that   trisects the angle   – one 
method is to show that sin ∠ , following which it is relatively straightforward to work out 
the sizes of angles   and   in terms of   and show that they must satisfy the correct 
relationship. 

Question 3 

For the first part note that   can be interpreted as the triangles that can be made using the 
rod of length 8 and two other, shorter rods. These can then be counted by noting that there are 6 
possibilities if the length 7 rod is used, 4 possibilities if the length 6 (but not the length 7) rod is used 
and 2 possibilities if the length 5 (but not 6 or 7) rod is used. It is clear that at least one rod longer 
than length 4 must be used. To evaluate   note that it is equal to   and 
then evaluate   in a similar manner to  . Similar reasoning easily gives formulae for 
 and  . 

For the induction, the rule for   deduced in the previous part can be used to show the 
inductive step, while the easiest way to show the base case is to list the possibilities. The easiest way 
to establish the result for an odd number of rods is to use the formula for   and the 
formula for   that was just proven. 

   

762
Question 4 

For the first part, note that the graph of arctan  satisfies the requirement of being continuous, but 
does not satisfy  0 . Since tan tan , a translation of the graph of  arctan  
vertically by a distance of   gives the required graph. 

It should be clear that the graph of   has no vertical asymptotes, approaches the  ‐axis as 

→ ∞ and passes through the origin. Identifying the stationary points should be the next task 
after which a graph should be easy to sketch. The graph of   should then be easy to sketch 
by considering the fact that   is an increasing function and   is obtained by composing the 
two functions already sketched. 

To sketch the graph of   first note that there must be two vertical asymptotes. Once 
stationary points have been checked for it should be straightforward to complete the sketch. In this 
case, the asymptotes need to be considered to deduce the shape of the graph for   as the 
composition with   will lead to discontinuities. Noting again that tan tan  the 
discontinuities can be resolved by translating sections of that graph vertically by a distance of  . 

Question 5 

The initial proof by induction is a straightforward application of the tan  formula. The final 
part of section (i) requires recognition that there are many possible values of   to give a particular 
value of tan , but only one of them is the value that would be obtained by applying the arctan 
function. The result can therefore be shown by establishing that the difference between consecutive 
terms of the sequence is never more than  . 

For the second part of the question a diagram of the triangle and application of the tan 2  formula 
shows that the value of   must be of the form used in the first part of the question. All that 
remains is then to show that the limit of the sum must give the required value. 

Question 6 

The first part of the question requires use of the cos  formula. Following this the integral 
should be easy to evaluate given that  sec tan . In the second part, apply the 
substitution and note that the limits of the integral are reversed, which is equivalent to multiplying 
by ‐1. Following this a simple rearrangement (noting that the variable that the integration is taken 
over can be changed from   to  ) should establish the required result. The integral at the end of this 
part can then be evaluated simply by applying this result along with the integral evaluated in part (i). 

In the final part of the question it is tempting to make repeated applications of the result proven in 
part (ii). However, this is not valid as it would require the use of a function satisfying  sin , 
which is not possible on the interval over which the integral is defined. Instead, application of a 
similar substitution to part (ii) to  sin  will simplify to allow this integral to be evaluated 
based on the integration of  . An application of the result from part (ii) will also be required. 

   

763
Question 7 

For part (i) note that the lines joining the centres of the two circles and one of the points where the 
bisection occurs form a right‐angled triangle, so the radius of the new circle can be calculated. To 
show that no such circle can exist when   note that the diametrically opposite points on   must 
be a distance of 2  apart, and no two points on a circle of radius   can be that far apart. For the case 
 note that the new circle would be the same as   (and so would have more than two 
intersection points). 

For part (ii) a similar method can be used to deduce the distances between the centre of the new 
circle and each of   and  . From these distances equations can be formed relating the   and   
coordinates of the centre of the new circle. It is then an easy task to eliminate the  ‐coordinate of 
the centre of the circle from the equations to get the given value of the  ‐coordinate. 

The expression for   can easily be found by substituting back into the equations obtained from the 
distance between the centres of two of the circles. Once this is done, note that  0 to obtain the 
final inequality. 

Question 8 

The first part of the question follows from consideration of similar triangles in the diagram if the line 
through   and the centres of the circles is added. For the second part, expressions can be written 
down for the position vectors of   and   by noting that the same method as in part (i) will still apply. 
The vectors   and   can then be compared to show that one is a multiple of the other. 

For the final part of the question, note that   will lie halfway between   and   if  . 

Question 9 

A diagram to represent this situation will show the angles that will be required to calculate the 
moments of each of the particles about   in terms of  . Following this, simple trigonometric 
manipulation should lead to a relationship between sin  and cos . From this, either a right‐angled 
triangle or one of the basic trigonometric identities can be used to reach the required result. 

For the second part of the question the amount of potential energy that needs to be gained by the 
system should be easy to calculate and this must be equal to the initial kinetic energy of the system. 

   

764
Question 10 

The component of the velocity of the particle in the direction of the string at any moment must be 
equal to  , which leads to  cosec  as the speed of the particle along the floor. Alternatively, 
introduce a variable to represent the length of string still in the room or the height of the room and 
then differentiate  , the distance of the particle from the point directly beneath the hole, with 
respect to time. The length of the string (to the hole in the ceiling) is decreasing at a rate of  , 
which then allows the introduced variable to be eliminated to reach an expression for the speed of 
the particle. 

Differentiation of the speed of the particle allows the acceleration to be calculated. Finally, note that 
the particle will remain on the floor as long as the vertical component of the tension is less than the 
weight of the particle and then the point at which the particle leaves the floor can be identified. 

Question 11 

For the first part, the coordinates of   are found by applying simple trigonometric ratios and 
differentiation with respect to time gives the velocity of  . In the second part, the first equation 
results from consideration of conservation of momentum and the second results from conservation 
of energy (with a substitution based on the first equation made to eliminate one variable). 

Since no energy is lost in any collisions the relationships from part (ii) must continue to hold and this 
shows that   cannot be 0 which means that the direction in which   changes remains the same 
unless there is a collision. Since the first collision occurs when  0, the second one must be when 
. 

For the final part, note that the equations in part (ii) must still hold, and if  0, the kinetic energy 
of   must be 0. Since the kinetic energies of   and   must be equal (by symmetry) it must be the 
case that the kinetic energy of   is   and can also be calculated from the expression for the 
velocity of   shown in part (i). Since  0, this can then be used to find the values of  . Finally, 
note that given these values of  ,   will only be 0 on the occasions when   is positive. 

Question 12 

For the first part, note that   can only win the game if the first two tosses result in heads, since once 
there has been a tail,   will win as soon as two consecutive heads have been tossed and   cannot 
win until there have been two consecutive heads and one further toss. In the second part, note that 
this logic still applies to the game for   and similar reasoning can be applied to the game for  . For 
the other two players switching heads and tails in any sequence that results in a win for   will give a 
sequence that results in a win for  , and vice versa, so the probabilities must be equal. Since only 
sequences which alternate between heads and tails forever (and the probabilities of such sequences 
tend to zero as the lengths of the sequences increase) the probabilities must both also be ¼ . 

For the final part, note that   must win if the first two tosses are TT. Since only the previous two 
tosses are important in determining what could happen on the next toss, each case can be analysed 
by a tree diagram which shows the outcomes after one further toss. 

For example, following HT: 

765
  H gives the position if the last two tosses were TH, and so a probability of winning of  , 
 T gives the position if the last two tosses were TT and so a probability of winning of 1. 

The total probability is therefore  , but this must also be equal to  . 

This yields three equations in the three unknowns which allows all of the individual probabilities to 
be calculated. Once this is done the overall probability can be calculated. 

Question 13 

To calculate the expected value of the total cost, note that there is a constant component of   and 
then the expected value of the   given that   must be added, which can be calculated 
by integration of   with respect to  , between   and ∞. Differentiating the expression 
for   with respect to   allows the position of the stationary point to be found. If this is at a 
negative value then   should be chosen to be 0 and otherwise the value of   for the stationary point 
should be used. 

A slightly more complicated integration is needed to establish the formula for   and then 
differentiation of this gives a value that is clearly negative for positive values of  , which shows that 
the variance is decreasing as   increases. 

766
STEP 3 2015  Hints and solutions 

1.  The first result can be obtained by simplifying the LHS and then writing it as  
  and integrating this by parts.  To obtain the evaluation of    ,  the first result 
can be re‐arranged to make     the subject, and then iterating the result to express it in terms of  
  which is a standard integral.  The expression can be tidied by multiplying numerator and 
denominator by   2 2 2 … 2  .  The first result for (ii) is obtained by means of the 
substitution    , the second by adding the two versions of    , and the third by the substitution  
 , being careful with limits of integration and employing symmetry.  Part (iii) is solved by 
expressing the integrand as     and then employing first part (ii) then part (i) to obtain  
!
, which is    . 
!

2.  Part (ii) is the only false statement, and a simple counter‐example is  1  and   2  for  

  odd, and   2  and   1  for     even.  Part (i)   1000  is a suitable value, then  1000  
and as     is positive, the inequality can be multiplied by it giving the required result.  Part (iii) 
requires the use of the definition twice with values    and     say, and then using  
max ,  .  For part (iv), we can choose   4 , and an inductive argument such as 

   1 1 1 2 2 2 2   works. 

3.  The part (i) inequality for  sec   can be obtained by making     the subject of the formula as  

sec   and invoking     remembering that   0  is not permitted.   

Then the points lie on a conchoid of Nicomedes with A being the pole (origin),  d being b, and L being 
the line   sec    (" " ).  A sketch is 

 
 

In part (ii), the extra feature is the loop as specified with end‐points at the pole corresponding to 
sec  .  A sketch is  

767
   

So in the given case, the area is given by  2 sec 2  which is    √3 4 ln 2 √3  . 

4.  Part (i) is imply shown by considering the image of the function     
as   → ∞  and then observing that the function is continuous and exhibits a sign‐change.  Part (ii) 
can be approached by writing        giving    , 
 , which can be obtained by considering    and the required result for  6   which 
can be neatly obtained by considering   0 .  

 Writing   cos sin   for   1, 2, 3 , employing de Moivre’s theorem, the three 

sums imply the reality of    ,    , and    , and hence    ,   , and    which by virtue of the result of 
part (i) yields the reality of    ,   , or    and hence the required result.  The final result can be 
considered as two cases, the trivial one of all three roots being real, and the one where the other 
two are complex.  The latter can be shown to give the required result by considering the real and 
imaginary parts of the roots of a real quadratic. 

5.  (i) Step 3 is straightforward on the basis of steps 1 and 2, noting that no lowest terms 
restriction need be made in part 1.  Step 5 requires that the given expression is a positive integer as 
well as well as being integer when multiplied by root two.  Step 6 requires justification that         
√2 1 1 .  

  (ii) The rationality of  2  on the basis of   2  being rational is simply obtained by squaring 

the latter, and the opposite implication can be made by squaring the former or dividing 2 by the 
former.  To construct the similar argument, let the set    be the set of positive integers with the 
following property:    is in   if and only if  2   and   2  are integers, and taking    to be the 
smallest positive integer in that set, consider   2 1  to produce the argument. 

6.  Treating the equations for    and    as simultaneous equations for     and    , one finds that  

√ ∓√
  and     which demonstrates that if   ∈   and   2  , i.e.  ∈ ,     
and    are real.  If     and    are real, then    and    are (trivially) and  2 0 . 

In (ii),  the first two equations yield  3 1 , making it possible to write the third equation as  
√
1 1   which has an obvious factor of   1   leading to   1  or     
from the quadratic equation.  If one of the solutions of the quadratic equation gives the same root  
1 , then there are not three possible values, i.e. if   2 .  From the first part of the question, 
for     and    to be real, we would want    to be real,     to be real, and   2  , in 

768
other words    .  So a counter‐example could be   1  giving  2 2 0  which has a 
negative discriminant. 

7.  The opening result is simply achieved by following the given explanation for   with  

 .  Parts (i) and (ii) can both be shown usual the principle of mathematical induction with 
initial statements 

“Suppose     is a polynomial of degree     i.e.   ⋯   for 

some integer   .” and “Suppose   1   is divisible by  1    i.e.   1
1   for some integer   , with  1.”   Part (iii) is obtained by expressing  
1   in sigma notation (by the binomial theorem), then carrying out   1  using the 
idea in the stem, and finally invoking the result of part (ii) and then substituting   1 . 

8.  Transforming the differential equation in part (i) is made by substituting for    and     as 

given, for     using    cos sin   and a similar result for     , and then simplifying the 
algebra by multiplying out and collecting like terms bearing in mind that a factor     can be cancelled 
as   0 .  The transformed equation can be solved by separating variables or using an integrating 
factor, to give    , the sketch of which is  

 
 

The same techniques for part (ii) yields a differential equation   0  which is solved by 
separating the variables and then employing partial fractions giving a variety of possible solution 
sketches 

 
( 1 but it is possible to consider  1 in which case  0 ) 

9.  Whilst the first part can be obtained otherwise, the simplest approach is by conserving 
energy, when   √   leads to the required answer simply.     is 

found by setting   0  leading to    .  The acceleration can be found by applying 

769
Newton’s 2nd Law or by differentiating the equation found in the first part, and substituting leads to 
√
the result     for the acceleration when    .  Treating the equation found in the first 
part as a differential equation for     in terms of  , the expression for the period is  
√
  4  . Making the substitution     , leads to  
√

√2 1   , which making a binomial expansion and using the given condition to 
approximate   √2   results in the final given expression. 

sin
10.  The position vector of the upper particle is    so differentiating with respect to 
cos
time yields the correct velocity and acceleration which gives the second result when used in 
Newton’s second law resolving horizontally and vertically.  The corresponding equations are  
θcos sin sin 0
  for the other particle by merely swapping the 
θsin cos cos 1
direction of the tension and the displacement from the midpoint.  The deductions are obtained by 
adding the two equations of motion, and in the case of  θ , subtracting the two equations and then 
θcos
eliminating     between the equations for each component.  Using       and a 
θsin
similar equation for the lower particle, initial values of   and     can be found and then the time for 
the rod to rotate by    can be obtained and substituted in the displacement equation under 
uniform acceleration to obtain the final result. 

11.  (i)  The first result is obtained, as the question prompts, by considering a component of force 
on the rod due to P, and taking moments about the hinge to find that that component is zero with 
the consequence that any force exerted on the rod by P must be parallel to the rod.  Bearing in mind 
the horizontal acceleration of P towards the centre of the circle it describes, resolving perpendicular 
to the rod and writing the equation of motion for P leads directly to the given equation with the 
stated substitution being made.  The force exerted by the hinge on the rod is along the rod towards 
P and resolving vertically for forces on P and rearranging gives   sec  . 

  (ii)  Taking moments for the whole system about the hinge gives  

sin sin sin cos sin cos  

which can be rearranged into the required form with    . 

12.  (i)  The required probability generating function is  1  
and it is simple to write down the probability distribution function of    and hence of    and arrive 
at the same pgf.  As a consequence, it can be argued that the pgf for   is also    and so the 
1
required probability is   6 . 

  (ii)   1 2   .       would be      except 

that the powers must be multiplied congruent to modulus 5 , and it can be shown that   and 
5  so  obtaining the required result for   .  Obtaining      

770
where     is an integer such that  0 5 4 , and the probability that    is divisible by 5 will 
be the coefficient of     which in turn is the coefficient of     as required.  If     is divisible by 5, the 
probability that    is divisible by 5 will be   1   as    . 

13.  (i)   

lead to     if  0 1 ,  1 2   if  1 2 , 

0 if   0  and   1  if   2 . 

From this, the cumulative distribution function of    by means of the logic 
1   and the required probability density 
function can be found by differentiation.  From that, by integration,  2 ln 2 

(ii)    

 
lead to 
1
0 1
2  
1
1 1
2

That   1 1 1  and differentiation leads to  

1
the probability density function      . 
1 0

  can be written down because by symmetry,      

and   1 1 .  This is simply verified by integration using the 
probability density function found. 

771
[An earlier potential version of the question had   ,   , and   independently uniformly distributed 
on  0,1  , considered the distribution of  ln  ,  

went on to find the pdf of   , where  ln  and finished by showing that   is also 

uniformly distributed on  0,1  . ] 

772


773
 







STEP Examiners’ Report 2016

Mathematics

STEP 9465/9470/9475

November 2016

774
 
The Admissions Testing Service is a department of Cambridge English
Language Assessment, which is part of Cambridge Assessment, a not-for-profit
department of the University of Cambridge. Cambridge English Language Assessment offers
the world’s leading qualifications for learners and teachers of the English language. Over 4
million people from 130 countries take Cambridge English exams every year.

775
Contents

STEP Mathematics (9465, 9470, 9475)

Report Page
STEP Mathematics I 5
STEP Mathematics II 10
STEP Mathematics III 13
Explanation of Results 17


 

776
 
 

777
 STEPI2016REPORT

GeneralComments

Thisyear,morethan2000candidatessigneduptositthispaper,thoughjustunder2000
actuallysatit.Thisfigureisabout thesameastheentryfigurefor2015,thoughthenumber of
candidatesoptingtositSTEPIhasrisensignificantlyoverrecentyears;forinstance,itwasaround
1500in2013.

ThereisnodoubtthatthepurposeoftheSTEPsistolearnwhichstudentscangenuinelyusetheir
mathematical knowledge, skills and techniques in an arena that demands of them a level of
performance that exceeds anything they will have encountered within the standard AͲlevel (or
equivalent) assessments. The ability to work at an extended piece of mathematical work, often
withtheminimumofspecificguidance,alliedtotheneedforbothdeterminationandtheabilityto
“makeconnections”atspeedandunderconsiderabletimepressure,arecharacteristicsthatonly
follow from careful preparation, and there is a great benefit to be had from an early encounter
with,andsubsequentprolongedexposureto,thesekindsofquestions.

Itisnotalwayseasytosaywhatlevelofpreparationhasbeenundertakenbycandidates,butthe
minimum expected requirement is the ability to undertake routine AͲlevelͲstandard tasks and
procedureswithspeedandaccuracy.Atthetopendofthescale,almost100candidatesproduced
a threeͲfigure score to the paper, which is a phenomenal achievement; and around 250 others
scoredamarkof70+,whichisalsoexceptionallyimpressive.Attheotherendofthescale,over
400candidatesfailedtoreachatotalof40marksoutofthe120available.

ForSTEPI,themostapproachablequestionsarealwayssetasQs.1&2onthepaper,withQ1in
particularintendedtoaffordeverycandidatetheopportunitytogetsomethingdonesuccessfully.
Soitisperfectlyreasonableforacandidate,uponopeningthepaper,tomakeanimmediatestart
atthefirstand/orsecondquestion(s)beforelookingaroundtodecidewhichoftheremaining10
or11questionstheyfeeltheycantackle.Itisveryimportantthatcandidatesspendafewminutes
–possiblyat thebeginning–readingthroughthequestionstodecidewhichsixtheyintendtowork,
sincetheywillultimatelyonlybecreditedwiththeirbestsixquestionmarks.Manystudentsspend
time attempting seven, eight, or more questions and find themselves giving up too easily on a
question the moment the going gets tough, and this is a great pity, since they are not allowing
themselvesthinkingtime,eitheronthepaperasawholeoronindividualquestions.

Theothersidetothenotionofstrategyisthatmostcandidatesclearlybelievethattheyneedto
attempt (at least) six questions when, in fact, four questions (almost) completely done would
guaranteeaGrade1(Distinction),especiallyiftheirscoreonthesefirstfourquestionswerethen
tobesupplementedbyacoupleofearlyattemptsatthestartingpartsofacouplemorequestions
(forthefirstfiveorsixmarks);attemptswhichneednottakelongerthan,say,tenminutesoftheir
time.Itisthusperfectlyreasonabletosuggesttocandidates,intheirpreparations,thattheycan
spendmore than 30minuteson aquestion,butonlyIFtheythinkthey aregoingtofinishitoff
satisfactorily,althoughitmightbebestiftheywereadvisedtospendabsolutelynomorethan40Ͳ
45minutesonanysinglequestion;iftheyhaven’tfinishedbythen,itreallyistimetomoveon.





778
CurveͲsketchingskillsareusuallyacommonweakness,butwereonlytestedonthispaperinQ3.
Theothercommonareaofweakness–algebra–wastestedrelativelyfrequently,andprovedto
beastestingasusual.Calculusskillsweregenerally“okay”althoughtheintegrationoffirstͲorder
differential equations by the separation of variables, as appearing repeatedly in Q4, was found
challenging by many of the candidates who attempted this question. The most noticeable
deficiency,however,wasinthewidespreadinabilitytoconstructanargument,particularlyinQs.
5,7&8.Vectorsareoftenpoorlyhandled,andthisyearprovednoexception.

Commentsonindividualquestions

[Examiner’snote:inordertoextractthemaximumamountofprofitfromthisReport,itisfirmly
recommend that the reader studies it alongside the Solutions and/or Mark Scheme supplied
separately.]

Question1
Marginallythemostpopularquestiononthepaper,andthehighestͲscoring,thiswasarelatively
carefully signposted question; thus, even though the demands were entirely algebraic, it was a
goodstarterquestionandgaveallcandidatessomethingtogettheirteethinto.Itwasstilloften
thecasethatcandidatesspentalotmoretimedoingfairlysimplethingsthantheyshouldhave;
forinstance,anawfullotofattemptsproducedoverandoveragain(effectively)thesameworkto
x 2 n1  1
show that { x 2 n  x 2 n1  ...  x 2  x  1  in each of the four given cases. And a similar
x 1
amount of unnecessary effort was expended on what should have been some fairly simple
binomialexpansions.Nevertheless,mostcandidatesmadegoodprogressforthen=1,2,3cases.
Toshowthatp4andq4arenotidentical,itsufficestochooseanyonevalueof‫ݔ‬forwhichthey
yield different outputs, but most approaches preferred to deal with the full polynomial
expansions,whichisfinebut(again)notanoptimalapproach.
Inpart(ii),mostcandidatesrealisedthattheyweretousetheresultsofpart(i),andtheywere
generallyabletodosoforatleast(a).Relativelyfewrealisedthatthesameidea(theuseofthe
differenceoftwosquaresfactorisation)wastobedeployedin(b),presumablyputoffbythelarge
numbersinvolvedandthenotionthattheanswercouldbeleftintermsofpowersof7.

Question2
Thiswasthesecondmostpopularquestionofall,attemptedbyover80%ofcandidates,though
pulling in a mean score of just under halfͲmarks. Most candidates managed the tricky
differentiation; tricky, since it required the repeated use of the Chain and Product rules of
differentiation.Thebigkeytolaterprogressinthisquestionwasthesuccess,orotherwise,ofthe
1 x2
simplifying that was done. Those who realised that 1  x 2 needed to be reͲwritten as 
1 x2
flew throughthesimplification,andthismadetheideabehindthethreelater partsmuchmore
transparent. In fact, careful differentiation and a sound simplification made choosing the
appropriatevaluesofthegivenconstantsatoestraightforward.







779
Question3
Itwasveryencouraging,onthisquestion,toseeboththenumberofattempts(onlyafewbehind
those for Qs.1 & 2) and the quality of them. The introduction of a new function is frequently a
guarantee of poor answers, especially when it is made to act on another function (the sine
function here), but many candidates seemed to cope quite well with the “int” function and the
properties of its “step” graphs. Although reminded (at the end of the question) to make the
endpointsofthestepscompletelyclear,therewasalotofvaguenessincandidates’answerson
thesepoints,andthiswasoneoftheprincipalcausesoflostmarks.Anoverallmeanscoreofjust
over12½outof20istestimonytotheoverallsuccessonthispotentiallytrickyquestion.

Question4
Around60%ofcandidatesattemptedthisquestion,butthemeanscoreachievedonitwasunder
3½ out of 20. The first three marks were awarded for a correct differentiation and subsequent
simplification (of a very similar kind to that in Q2 but much, much easier), but it is clear that
almost noͲone was able to proceed further into the question, due almost entirely to difficulties
dy d2 y
working with  as an entity in its own right (along with its derivative, 2 ). This is especially
dx dx
disappointing,sincethisislittlemorethanalabellingmatter.Evenforthosewhodidseewhatto
do in that respect, spotting that one then had a separableͲvariable 1stͲorder d.e. to deal with
proveddifficult.

Question5
Attemptsatthisquestionweredowntoaroundathirdofthecandidatureand,overall,itproved
to be the lowest scoring question on the paper with a mean score of just 3 out of 20. The big
problem here was the widespread inability to draw some suitable lines onto the diagram, or to
realise that the lines joining the centres necessarily passed through the points of contact. Once
thisisdone,thereisanobviousrightͲangledtriangletobefoundembeddedwithinthisgeometric
arrangementthatthenrequiresonlytheuseofPythagoras’Theoremtomakeastart.
Althoughtherewererelativelyfewcompleteeffortstobefound,manywhodidmanagetomakea
good start found aspects of the algebra difficult: for instance, turningresult (*) into (**).  Here,
thereareseveralgoodapproachesprovidedoneuses(*)sensiblytoeliminate(say)b.
Admittedly,thelastparttothequestionwasmoredemandingtodealwith,butitwasclearfrom
responsesthatthemajorityofeventhosecandidateswhogotthisfarstruggledtoseethelogical
differencebetweenprovingAŸBandprovingBŸA.
 
Question6
This vectors question proved both unpopular and lowͲscoring, being attempted by only 30% of
candidatesandelicitingameanscoreofonly5outof20.Onceagain,itwasclearthatcandidates
were,ingeneral,doinglittlemorethanmakingahalfͲheartedattemptattheveryfirstpartofthe
question.Therestofthequestionreliedonlyontheabilitytowritethevectorequationsofpairs
of lines and then solve simultaneously by considering the coefficients of a and b. Overall
thisquestionsuggeststhatvectorsarenotwellunderstood.








780
Question7
This was a rather splendid reasoning question, and many candidates responded very well to its
demands.Almosthalfofallcandidatesmadeanattemptatit,achievingameanscoreofover7
outof20.Whilstmanyattemptswentlittlebeyondthefirsttwoparts,manycandidatesactually
didwellonthequestionasawhole.
Part (iii) required a mixture of a proof by contradiction based on at least an informal
understandingofthemethodofproofbyinduction,soitwasnosurprisethatmanygaveupatthis
stage. The “fun” part was (v), in which candidates had to choose some numbers that
demonstratedtherequiredproperties;unfortunately,itisclearthissortofrequestisnotalways
wellanswered.

Question8
Generating functions generally appear very late on in statistics modules whose entry numbers
seldom reach three figures. Nevertheless, a mean score of almost 9 out of 20 on this question,
obtainedbymorethanaquarterofthecandidates(veryfew,ifany,ofwhomwillhaveseenthe
ideabefore),suggeststhatitiseasytointroducethetopicandthatgoodcandidateswillpickitup
quicklyandsuccessfully.
Apartfromalgebraicdifficulties,therealhurdletomakingacompleteattemptlayinpart(ii)(a);
although most of the candidates who got this far went on to earn the 4 marks allocated to this
part, they tended to do so by ignoring the guidance of the question. Although there are two or
threeperfectlygoodalternativeapproachestotheoneintended,theydon’tnecessarilypointin
thedirectionthathelpsthesolvercopewiththemoreimaginativelyconstructedpart(ii)(b);and
thiswaswhymostcandidatesstruggledtoknowquitewhattodowithit.

Questions9,10&11
The three mechanics questions each received a healthy number of “hits” this year, and marks
werefairlyrespectablealso.

Question9
Inthisquestions–astaticsquestion–candidateswerehelpedenormouslybythegivendiagram,
which allowed most takers the opportunity to get most of the preliminary “force” statements
downcorrectly:resolvingtwiceandtakingmoments(choosingsensibledirectionsandasuitable
axialpoint),alongwiththestraightforwarduseofthe“F= PR”frictionlaw,meant that7marks
could be obtained simply for noting all the correct ingredients. Suitable choices for eliminating
termswouldthenleadtotheprintedanswerwithlittledifficultybeyondthealgebraic.Itisfairto
say,however,thattheprospectofdoingitallagainforthesecondconfigurationprovedtoomuch
formanycandidates,althoughitactuallyonlyinvolvedtherealisationthatcertainforceschanged
directionsandcouldthusbereplacedbytheirnegatives.












781
Question10
Thisquestionwasarelativelyundemandingcollisionsquestion,afactthatwasapparenttoalotof
thecandidates,almosthalf ofwhommadeanattemptat thequestion.Itturned outtobeonly
one of three questions on the paper for which the mean achieved mark was over 10 out of 20
(after Qs. 1 & 3), and this is no doubt partly due to the fact that there are generally only three
physical“laws”or“principles”tobeappliedonthistypeofmechanicsproblem.Onthisoccasion,
noenergyconsiderationswereinvolved,whichmeantthatcarefulapplicationofConservationof
Linear Momentum and Newton’s (Experimental) Law of Restitution pretty much saw the solver
throughthewholething.Theonlyminorhurdleisthatcandidatesoftengetconfusedaboutwhich
directions are being taken as positive, and this is usually down to the lack of a suitablyͲmarked
diagram.Otherthanthat,itwasonlyafewalgebraicslipshereandtherethatpreventedfulland
successfulattemptsfrombeingmade.
 
Question11
This question was also a rather pleasant question of its kind, dealing with projectile motion. It
provedtobeaverypopularquestion(with40%ofallcandidatesattemptingit,almostaspopular
asQ10)butwithameanscoreofalmost8outof20.Thereweretwomainobstaclestoprogressin
this question: first, rather a lot of candidates failed to appreciate that finding the greatest
somethingͲorͲother might require calculus; secondly, and more importantly, putting everything
togethertocreateaquadraticinc=cos2Dprovedchallenging.

Question12 
Thisquestionprovedtobeanespeciallypopularchoice(almost45%uptake)andahighͲscoring
oneatthat(meanscoreofover9outof20).However,mostofthesemediumͲsuccessfulmarks
came from attempts at the first two (specific) cases, in which Bob threw 2 coins, then 3. There
were 6 marks for successful attempts at each of these preliminary parts, and candidates were
happytoworkthemthroughreasonablysuccessfully,althoughsomeoftheseattemptswerevery
poorlyexplained;andoccasionallyitwasthecasethattherewasnoexplanationatall.Although
they may have been getting the correct answers, it is imperative to ensure that answers are
sufficiently coherent for the reader to be able to follow their reasoning and/or structure and
henceunderstandhowtheanswershavebeenarrivedat.
ThefinalpartofQ12elicitedlittlemorethanafewhalfͲeffortsatthegeneralsituation,butthe
general reasoning required proved beyond most candidates. A small number of candidates
thoughttheywerebeingaskedforaninductiveproofofsomekind,usingparts(i)and(ii)asbaseͲ
linecases.

Question13
Thisattractedthesmallestamountofattention:83attempts,withameanscoreof5outof20.
Candidates should recognise the (negative) exponential distribution here, and many candidates
didso.Itisthusaquestionaboutpdfsandcdfsandmultiplyingtheprobabilitiesofindependent
events.Theonlycommonlyscoredmarkswerethosefortheexpectation,gainedusingintegration
byparts.Almostnocandidatesmadeittopart(ii).


782
 STEPII2016REPORT

GeneralComments

AsinpreviousyearsthePurequestionswerethemostpopularofthepaperwithquestions1,3
and7themostpopularofthese.Theleastpopularquestionsofthepaperwerequestions10,11,
12and13withfewerthan400attemptsforeachofthem.Thereweremanyexamplesofsolutions
inthispaperthatwereinsufficientlywellexplained,giventhattheanswertobereachedhadbeen
providedinthequestion.

Commentsonindividualquestions

Question1
Thiswasapopularquestionandmanyverygoodsolutionswereseen.Thefirstpartofthe
questionwasarelativelystraightforwardapplicationofdifferentiationandparametricequations
andwassuccessfullycompletedbymanyofthecandidates.Thesketchesproducedweregenerally
thecorrectshapefortheparabola,althoughinsomecasesitwasnotinthecorrectposition.The
othercurvecausedmoreproblemswithsomecandidatesdrawinganotherparabolaorsimilar
shape.

Question2
Thisquestionreceivedpoorresponsesintermsoftheaveragemarkperattempt.Applicationof
thefactortheoremtothefirstpartoftenproducedaneatsolution.However,anumberof
candidatesinthefirstpartdidnotusethefactortheoremasrequestedandsoproducedvery
complicatedalgebraicexpressionsthatweremoreofachallengetosimplify.Wheretheresulthad
beensuccessfullyshownandtheexpressionfactorised,manycandidateswereabletoseethe
relevancetosolvingtheequationinpart(i).Thosecandidateswhowereabletofollowthrough
themethodtosuccessfullyfactorisethesecondexpressionwereoftenabletocompletethe
question.

Question3
Thiswasthemostpopularquestiononthepaper,andcandidatesgenerallyscoredwellhere.The
firsttwopartswererelativelystraightforwardandwereansweredwell.Thefinalpartrequired
carefulexplanationfromcandidatesandmanywerenotabletotaketheinitialstepofrewriting
therelationshipbetweenthefunctionanditsderivativeinausefulway.Thosewhodid
successfullycompletethispartwereoftenabletoidentifythenumberofrootsineachofthetwo
cases.

Question4
Thisquestionattractedmanygoodresponseswhichoftensuccessfullyaccomplishedthefirstand
thirdresultsofpart(i)ofthequestion,althoughinmanycasesmarkswerelostthrough
incompleteexplanationsofsomeofthestepstaken.Part(ii)ofthequestionwasgenerally
answeredpoorlywithsomecandidatessimplystatingthatthesquarerootofoneexpressionin
termsofsurdswasequaltotheotheronethattheyhadachieved.Onlyasmallnumberof
candidatessuccessfullycompletedthelastsectionofthisquestion.






783
Question5
ThiswastheleastattemptedofallofthePurequestionsandonethatgenerallyproducedlow
marksforcandidates,whogenerallyappearedtohavedifficultyinexpressingthecoefficientsof
thebinomialexpansionintherequiredform.Ineachcasethedesiredanswerwasgiveninthe
question,andsosuccessfulsolutionsalsoneededtobeveryclearaboutthereasoningusedto
reachtheanswer.

Question6
Thisquestionwasansweredrelativelypoorlycomparedtotheothers,asmanycandidatesdidnot
appeartoreadthequestioncarefullyenoughandsoattemptedtosolvethedifferentialequations
inparts(i)and(ii)ratherthansimplyverifyingtheresultsgiven.Wherecandidatesmovedonto
parts(iii)and(iv)theyweregenerallysuccessfuliftheywereabletocompletethedifferentiation
ofthenewfunction.

Question7
ThiswasthesecondmostpopularandoneofthebestͲansweredquestionsonthepaper,with
manycandidatesscoringveryhighmarks.Inmanycasestheinitialresultwasexplainedclearlyand
thenappliedsuccessfullytothefirstexample.Candidatesweregenerallyabletofollowthrough
thecalculationswheretheywereabletoseethewayinwhichtheresultcouldbeachieved,andso
mostoftheattemptsthatfollowedacorrectmethodonlylostmarksthroughoccasionalerrorsin
calculation.

Question8
Thefirsttaskinthisquestionwasgenerallywellanswered,althoughsketcheswereoftenunclear
ordifficulttointerpret.Inpart(i)manycandidateswereabletoobtainthefirstapproximation,
butthencouldnotseehowtoachievetheothertwo,oftenofferingsumsofanonͲintegernumber
oftermswhichgivesthecorrectapproximationwhensubstitutedintotheformula.Thosewho
successfullycompletedpart(i)wereoftenabletoapproximatetheerrorinpart(ii)andseehowit
appliestothefinalsum.

Question9
Thisquestionwasthemostpopularofthemechanicsquestions,andmanycandidateswereable
tocompletethefirstpartofthequestionsuccessfully.Inmanycasesthiswasasfarastheygot,as
alargenumberofcandidateswereunabletomakesignificantprogressonpart(ii).Where
candidateswereabletoidentifyacorrectstrategyforsolvingtheproblemtheywereoften
successfulinreachingexpressions,onlylosingmarksthrougherrorsinthealgebra.

Question10
Thisquestionreceivedanumberofverygood,andoftenconcise,answers.However,therewasa
significantnumberofcandidateswhodidnotcalculatethecentreofmass,ormistakenlyassumed
thattheformulaforthecentreofmassforanequilateraltrianglecouldbeapplied.Many
candidateschosetoconsiderthelimitingcasefirstandthendeducetheinequalityinthefinal
step,butdidnotjustifythedirectionoftheinequalityclearly.Therewereanumberofcases
wheretherequiredangleswherenotcalculatedcorrectlywhenresolvingtheforces.








784
Question11
Althoughnotapopularquestion,thiswasoneofthebetterͲansweredquestionsintermsofthe
averagenumberofmarksachievedpercandidate.Manycandidateswhoattemptedthisquestion
ୠ
wereabletogainmanyofthemarksforpart(i),oftenbysubstituting–ƒ Ʌfor intothe
ୟ
simultaneousequationsandtheneliminating–.Somecandidates,however,lostsomemarksfor
assumingthatƒ ൌ …‘• Ʌand„ ൌ •‹ Ʌ.Part(ii)waswellansweredbymanycandidates,butvery
fewsolutionssuccessfullyexplainedthelinkbetweenthetwopartsofthequestion.

Question12
Thefirstpartofthequestionrequiredtheprooftofollowfromtheresultquotedinthequestion.
Forthisreason,solutionsthatexplainedtheresultbydrawingaVenndiagramwerenotawarded
fullmarks.Inmanycases,setsneededtobemoreclearlydefined(forexample ‫) ׫  ת‬.This
wasagainanexampleofaquestionwithagivenanswerwheremanycandidatesdidnotfully
explainallofthestepsintheproof.Theresultfortheunionoffoursetswasgenerallywell
answered,althoughtherewereseveralanswersinwhichnotallofthepairwiseintersectionswere
identified.
Manycandidateswereabletocalculatetheprobabilitiesrequiredinparts(i),(ii)and(iii),butfew
wereabletoapplytheresultsfromthestartofthequestiontothefinalcalculations.

Question13
Thisquestionreceivedonlyasmallnumberofattempts,withasignificantnumberofcandidates
notidentifyingthatarectanglecouldbeusedtoapproximatetheareaineachofthecasesandso
unabletomakemuchprogressonthequestion.Theaveragemarkpercandidateforthisquestion
wasthelowestonthepaper,withasignificantnumberofattemptsnotprogressingbeyondthe
firststep.



















785
 STEPIII2016REPORT

GeneralComments

Asubstantiallylargernumberofcandidatestookthepaperthisyear:14%morethanin2015.
However,themeanscorewasvirtuallyidenticaltothatin2015.Fivequestionswerevery
popular,withtwobeingattemptedbyinexcessof90%ofthecandidates,butonceagain,all
questionswereattemptedbysignificantnumbers,withonlyonedippingunder10%attempting
it,andeveryquestionwasansweredperfectlybyatleastonecandidate.Mostcandidateskept
tosixsensibleattempts,althoughsomedidseveralmorescoringweaklyoverall,exceptinsix
outstandingcasesthatearnedveryhighmarks.

Commentsonindividualquestions

Question1
Thiswasthemostfrequentlyattemptedquestionwithmorethan93%ofcandidatesattempting
it.Itwasalsosuccessfullyattempted,thesecondmostinthisrespectandonlybyasmallmargin,
earningtwothirdsmarksonaverage.Themajorityofcandidatescompletedparts(i)and(iii)with
littletrouble,withvariousalgebraicmistakesoccurringthroughout,andafewcandidates
forgettingtosubstituteforthelimitsinpart(i).Manygotstartedonpart(ii)usingthe
substitutionfrompart(i),andthengetstuckfacedwiththeconsequentintegral.

Question2
Thisquestionwasquitepopular,beingattemptedbyjustoverthreeͲquartersofthecandidates,
butsuccesswasmoderate.Mostgotunderwaydifferentiatingimplicitly,ratherthan
parametrically,andwereabletofindequationsoftangents,normalandchords,butnotalways
simplifyingbyfactorisingtomaketheirliveseasier;thosewhocouldfactorisemadegood
progress,whilstthosewhodidnotstruggledtofind(i).Otherweaknesseswerenotappreciating
thattheycouldfindr+qandrq,whichthenledtonotfindingthecertainpointin(ii).Inthefinal
part,squarerootingtheinequalityandonlyconsideringthepositivecasewasnotuncommon.

Question3
Marginallylesspopularthanquestion2,thiswasveryslightlybetterattempted.Inbothparts,
candidatessuccessfullyequatedthedifferentialoftheexpressionsontherightoftheequations
totheexpressiontobeintegrated,withtheexponentialfunctioncancelled.Inthefirstpart,
manyobtainedP(x)inthegivencaseofQ(x),buttheattemptedproofsthatthedegreeofP(x)is
onemorethanthatofQ(x)andforpart(ii)thatnosuchpolynomialsexistledtomanyillogical
steps.Manywouldhavebenefitedbymultiplyingupbydenominatorsandusingthe
remainder/factortheorem,ratherthanattemptingargumentsbasedondegreesofrational
expressions.Somesubvertedpart(i)bysuccessfullyintegratingthefirstexpression.











786
Question4
Veryslightlymorepopularthanquestion2withfourfifthsattemptingit,theydidsowithslightly
lesssuccess.Thefirstpartofthequestion,beingwellsignposted,wasprettywellattempted,
althoughtherewassomeverypoornotationforlimitargumentsasNwascommonlytakento
equalinfinity.Inpart(ii),itwasquitecommonforcandidatestowritetheexpressionforsech(ry)
intermsofpositivepowersofexponentialswhichmadeattemptstoapply(i)invalid.Few
candidatesfullysimplifiedthefinalresult,andpriortothat,manydidnothandlethepositiveand
negativepartsofthesumcorrectlywithsomejustdoublingthesumfrom1toinfinity.

Question5
Thiswasattemptedbyhalfthecandidature,scoringsimilarlytoquestion2.Thebinomial
expansionanditssymmetrywerewellͲhandledinthefirstpart,aswasapplyingtheresultof
(iii)inpart(iv).Part(ii)wasnotwellͲansweredastheretendedtobecavalierstatements
regardingdivisibility,andinpart(iii),fewnoticedthatthecondition ൅ ͳ ൑ ʹwas
requiredinordertouse(ii).

Question6
Justunder40%attemptedthisquestion,anddidsowithoutgreatsuccess,scoringaboutoneͲ
thirdmarks.Themajorityknewtheformulaforcoshofasum,orifnot,couldusedefinitionsand
comparecoefficientstoobtainthefirstresultinthestem.However,theywereveryweakonthe
 ൌ case.
Explanationforpart(i)waspoor,andtheplusorminuswasfrequentlynotproperlyunderstood;
asaconsequence,aspeciousplusorminusoftenappearedinpart(ii).Arguingnecessaryand
sufficientconditionsin(iii)and(iv)wasweak.However,asmallnumberofgoodcandidatesdid
completethisquestion.

Question7
ThiswastheleastpopularquestioninthePuresectionofthepaperbeingattemptedbyless
than30%ofthecandidates,andwhilstthereweresomeverygoodsolutions,thestandardof
attemptswasgenerallynotgood,andinfact,onlyoneoftheMechanicsquestionswaslesswellͲ
answeredinthewholepaper.Ingeneral,thestemwasprettywellanswered,butevenhere,
havingappreciatedthateachfactorontheLHSwasafactoroftheRHS,frequentlytherewasno
considerationgiventothe,admittedlysimpletoobtain,scalarfactorbeingone.Inspiteofthe
stem,mostdidnotappreciatehowtoproceedwithpart(i),andsowentlittlefurther.

 



787
Question8
Attemptedbynearlyasmanyasattemptedquestion1,itwasmarginallymoresuccessful,anda
goodnumberachievedfullmarks.Generally,theideaofrepeatedlyapplyingafunctiontocreate
acyclewaswellͲspotted.However,candidatesdidsometimesfalldowntryingtofind‰ሺšሻin(ii)
andsomesubstitutedthegiven‰ሺšሻratherthanfindingit.Inpart(iii),somestoppedhaving
madethefirstsubstitutionandsocouldnotfindthesolution.Also,someguessedthesolution
forpart(iii)but,ofcourse,thisdidnotdothefulljob.

Question9
Althoughnotoverlypopular,beingattemptedbylessthanafifthofcandidates,thisquestion
wasmoderatelysuccessful,alittlebetterthanquestion3.Findingtheextensionsandtensionin
thefirsttwopartsofthequestionwascompletedbymostcandidates,buthavinggenerally
writtendowntheequationofmotion,veryfewthoughtofapplyingthebinomialexpansionand
socouldnotproceedtothefinalresult.

Question10
Oneoftheleastpopularquestions,beingattemptedbyaboutaseventhofthecandidates,itwas
theleastsuccessfullyansweredwithjustunderquartermarksscored.Asever,thereweresome
verystrongsolutions.Generally,thefirstresultofthequestionwasdonewellastheycould
resolveaccuratelyandidentifythecondition,൐ Ͳ.Forthefinalresult,notmanyidentifiedthe
condition ൐ Ͳ,andsocouldnotproceed.Generally,energywasconservedwell,thoughsome
omittedoneorotherofthekineticenergyterms.Asfarascircularmotionwasconcerned,some
treatedtheradiusasƒratherthan…‘• Ⱦ.

Question11
ThemostpopularofthenonͲPurequestions,itwasattemptedbyathirdofthecandidates,but
generally,onlyaboutaquarterofthemarkswerescored.Alargenumberattemptedtouse
constantaccelerationformulae,oriftheyrealisedthatcalculuswasrequired,failedto
appreciatethattheyneededtouse“ƒ ൌ ˜†˜Ȁ†š̶.Forthosethatseparatedvariables,the
integrationscausedfewproblems.Part(iii)causeddifficultiesascandidateswerenot
comfortableusingthebounds;hadtheyconsideredɉଵ െ ɉଶ theymighthaveencountered
fewerproblems.


Question12
Whilstaspopular(orratherunpopular)asquestion10,attemptsatquestion12weremore
successfulthanallbutquestions1and8withonaveragehalfmarksbeingscored.Part(i)was
generallywelldoneusingthebinomialdistribution,andthosethatspottedtheyshouldusethe
Poissondistributioninpart(ii)usuallydidwelltoo.However,candidateswereoftensloppyin
theirexplanationsoftherearrangementsofChebyshev,andalsoquiteoftencandidateshada
fixationwiththeNormaldistributionwhichdidnothelp.

 



788
Question13
Thiswastheleastpopularquestionbeingattemptedbyonlyhalfthenumberattempting
question12,andwithslightlylesssuccessthanforquestion11.Mostcandidatespickedupa
fewmarksatthestartofthequestionandthenasmallnumberusedintegrationbypartsin(i),
butothersattemptedthisunsuccessfullytryingintegrationbychangeofvariable.The
multiplicationofͶwassurprisinglybadlydone,andtheexpectationofaconstantbeingzero
wassimilarlysurprisinglycommon.











789
Explanation of Results STEP 2016

All STEP questions are marked out of 20. The mark scheme for each question is designed to reward
candidates who make good progress towards a solution. A candidate reaching the correct answer will
receive full marks, regardless of the method used to answer the question.

All the questions that are attempted by a student are marked. However, only the 6 best answers are
used in the calculation of the final grade for the paper.

There are five grades for STEP Mathematics which are:

S – Outstanding
1 – Very Good
2 – Good
3 – Satisfactory
U – Unclassified

The rest of this document presents, for each paper, the grade boundaries (minimum scores required
to achieve each grade), cumulative percentage of candidates achieving each grade, and a graph
showing the score distribution (percentage of candidates on each mark).

STEP Mathematics I (9465)

Grade boundaries
Maximum Mark S 1 2 3 U
120 102 75 60 41 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 3.7 17.2 37.0 75.7 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics I

 790
STEP Mathematics II (9470)

Grade boundaries
Maximum Mark S 1 2 3 U
120 95 74 65 31 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 8.0 26.5 37.3 81.6 100.0

Distribution of scores

3.0

2.5

2.0
Percent

1.5

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics II

STEP Mathematics III (9475)

Grade boundaries
Maximum Mark S 1 2 3 U
120 88 64 55 32 0

Cumulative percentage achieving each grade

Maximum Mark S 1 2 3 U
120 11.0 36.6 50.1 85.1 100.0

Distribution of scores

2.5

2.0

1.5
Percent

1.0

0.5

0.0
0 10 20 30 40 50 60 70 80 90 100 110 120

Score on STEP Mathematics III

 791
 792
.

The Admissions Testing Service is part of Cambridge English Language Assessment, a not-for-profit department of
the University of Cambridge. We offer a range of tests and tailored assessment services to support selection and
recruitment for educational institutions, professional organisations and governments around the world. Underpinned
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Admissions Testing Service

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Language Assessment
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Admissions tests support:

www.admissionstestingservice.org/help

793
STEP Mark Schemes 2016

Mathematics

STEP 9465/9470/9475

November 2016

794
The Admissions Testing Service is a department of Cambridge English
Language Assessment, which is part of Cambridge Assessment, a
not-for-profit department of the University of Cambridge.

Cambridge English Language Assessment offers the world’s leading

qualifications for learners and teachers of the English language. Over 4 million
people from 130 countries take Cambridge English exams every year.

795
Contents

STEP Mathematics (9465, 9470, 9475)

Mark Schemes Page

Introduction 4
STEP Mathematics I 5
STEP Mathematics II 18
STEP Mathematics III 45

796
Introduction

These mark schemes are published as an aid to teachers and students, to indicate
the requirements of the examination. It shows the basis on which marks were
awarded by the Examiners and shows the main valid approaches to each question. It
is recognised that there may be other approaches and if a different approach was
taken in the exam these were marked accordingly after discussion by the marking
team. These adaptations are not recorded here.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the relevant
knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question papers and
the Report on the Examination.

The Admissions Testing Service will not enter into any discussion or correspondence
in connection with this mark scheme.

797
 STEP I 2016 MARK SCHEME

Question 1 (i)
x3  1 x5  1 x7  1 x9  1
B1 for at least 3 of q1(x) = , q2(x) = , q3(x) = , q4(x) = correct
x 1 x 1 x 1 x 1

M1 A1 for p1(x) = ( x 2  2 x  1)  3x(1) ; = x 2  x  1 { q1(x)

M1 A1 for p2(x) = ( x 4  4 x 3  6 x 2  4 x  1)  5x( x 2  x  1) ; = x 4  x 3  x 2  x  1 { q2(x)

M1 for attempt at binomial expansion of (x + 1)6 and squaring ( x 2  x  1 )

A1 for (x + 1)6 = x 6  6 x 5  15 x 4  20 x 3  15 x 2  6 x  1
A1 for ( x 2  x  1 )2 = x 4  2 x 3  3x 2  2 x  1
A1 for p3(x) = x 6  x 5  x 4  x 3  x 2  x  1 { q3(x) shown legitimately ླྀ

M1 for valid method to show p4(x) ี q4(x)

Method I: p4(x) = x 8  x 7  x 6  2 x 5  7 x 4  2 x 3  x 2  x  1
while q4(x) = x 8  x 7  x 6  x 5  x 4  x 3  x 2  x  1
Method II: partial expansion showing one pair of coefficients not equal
19  1
Method III: e.g. p4(1) = 28 – 9.1.33 = 13 z q4(1) = 1
11
A1 A1 for correct/valid partial working; completely and correctly concluded ི

Question 1 (ii) (a)

M1 M1 A1 for use of p1(300) = q1(300); use of difference-of-two-squares factorisation; 271 u 331 ི

Question 1 (ii) (b)

M1   
for use of p3 77 = q3 77

for identifying squares: >7 @  

3 2 2
M1 1  7 8 714  7 7  1
7

M1 for use of difference-of-two-squares factorisation

A1 A1 >7 7 3
@ > 3
@
 1  718  711  7 4  u 7 7  1  718  711  7 4 
  
or 7 21  3.714  3.7 7  1  718  711  7 4 u 7 21  3.714  3.7 7  1  718  711  7 4  ུ

798
Question 2
For y ax 2
 
 bx  c ln x  1  x 2  dx  e  1  x 2 
M1 use of Product Rule twice
M1 A1 use of Chain Rule in 1st product for the log. term (allow correct unsimplified here)
dy
dx

ax 2  bx  c
1
x  1 x2 ©

u §¨1  12 1  x 2 2 .2 x ·¸  2ax  b  ln x  1  x 2
1

¹
> @  
M1 A1 use of Chain Rule in 2nd product (allow correct unsimplified here)
+ dx  e §¨ 12 >1  x 2 @ 2 .2 x ·¸  d 1  x 2
1

© ¹

ax  bx c
2
> 1 x  x@ +  2

  xdx  e
>x  1 x @ u
dy
2ax  b  ln x  1  x 2 +  d 1 x2
dx 2
1 x 2
1 x 2

M1 cancelling the [-] terms

A1 A1 one mark for each term, correct and simplified
dy ( a  2 d ) x 2  (b  e) x  (c  d )
+ 2ax  b  ln x  1  x 2   ཷ
dx 1 x 2

Question 2 (i)
M1 A1 A1 for choosing a = d = 0, b = 1, e = –1 and c = 0 so that
dy (0) x 2  (0) x  (0)
+ 0  1 ln x  1  x 2  
dx 1 x 2

A1 and ³ ln x  
1  x 2 dx = x ln x  1  x 2   1  x 2 (+ C) clearly stated ཱི

Question 2 (ii)
1
M1 A1 A1 for choosing a = b = e = 0 and c = d = 2 so that
dy (0  1) x 2  (0) x  (1)

+ 0  0  ln x  1  x 2 
dx 1 x 2

A1 and ³ 1  x 2 dx = 1
2
 
ln x  1  x 2  12 x 1  x 2 (+ C) clearly stated ཱི

Question 2 (iii)
M1 A1 A1 for choosing a = 1
2 , b = e = 0 and c = 1
4 and d =  14 so that
dy
dx
( 12  12 ) x 2  (0) x  ( 14  14 )
1 x 2

+  x  0  ln x  1  x 2 
A1 and ³ x ln x  1  x 2 dx =   1
2 x2  1
4
ln x  
1  x 2  14 x 1  x 2 (+ C) clearly stated ཱི

Alternative: results for (i) and (ii) enable (iii) to be done using Integration by Parts:
I3 = ³ x . ln x   1  x 2 dx

= x^x ln x  1  x   1  x ` ³ 1 . ^ln x 
2 2

1 x2  1 x2 ` M1 A1
= x ln x  1  x   x 1  x – I + (ii)
2 2 2
3

M1 for turning it round, collecting I3’s etc. A1 for final answer (FT (ii))


799
Question 3 (i)
M1 for steps
A1 y-values change at integer x-values
A1 y-values at unit heights
A1 LH Ɣs and RH Ⴜs correct
(ignoring 2 at ends)
A1 for very LH & RH bits correct

Question 3 (ii)
M1 for steps
A1 y-values change at integer x-values
A1 y-values at sin(k’s), k  ള
A1 LH Ɣs and RH Ⴜs correct
(ignoring 2 at ends)
A1 for very LH & RH bits correct

Question 3 (iii)
M1 A1 for two main steps; endpoints in right places
A1 for all endpoints correct in these two lines
B1 for Ɣ at  12 S , 1 with clear Ⴜ in line below
B1 for Ɣ at  S , 0 

Question 3 (iv)
M1 for steps at integer y-values
A1 essentially correct domains (ignoring Ɣs and Ⴜs)
A1 for all lines’ endpoints correct
B1 for Ɣ at  12 S , 2 with clear Ⴜ in line below
B1 for Ɣ at  S , 0 

800
Question 4 (i)
z
M1 use of Quotient Rule (or equivalent) on y
1 z2
A1 for correct use of Chain Rule for the diffl. of the denominator
dy 
1  z 2 .1  z. 12 1  z 2  1
2
.2 z
 1 z 
= 2
dz 2

1
A1 all correct and simplified: ི
1  z 
3
2 2

Question 4 (ii)
§ d2 y · dz
¨¨ 2 ¸¸
M1 for using z
dy
in © dx ¹ N to get dx N
1  z 
3 3
dx ­° § dy · 2 ½° 2
2 2

®1  ¨ ¸ ¾
°̄ © dx ¹ °¿
dz
M1 A1 for separating variables; correctly: ³ ³ N dx
1  z 
3
2 2

z
A1 for correct integration using (i)’s result: N ( x  c) (+ c in any form)
1 z2
M1 for re-arranging for z or z2 : z 2 N 2 ( x  c) 2 ( z 2  1) Ÿ …
u
A1 correct: z r , u = N(x + c), any correct form (ignore lack of r throughout) ཱུ
1 u2

dy dy du
M1 for attempt at .
dx du dx
du dy u
M1 A1 for use of the Chain Rule (e.g.) with N ; correct diffl. eqn. N r ི
dx du 1 u2

u
M1 for separating variables: ³ N dy r³
1  u2
du

u
M1 M1 A1 for method to integrate ³ 1 u 2
du =  1  u 2

(by “recognition”, “reverse chain rule” or substitution) ཱི

M1 for integrating and substituting for u : N y  d # 1  N 2 ( x  c)2

2 2

1  N 2 ( x  c) 2 or §¨ y 
d· §1·
M1 A1 for working towards a circle eqn. : (Ny  d ) 2 ¸  ( x  c)
2
¨ ¸
© N¹ ©N ¹
B1 for noting that radius of circle is the reciprocal of the curvature ཱི

801
Question 5 (i)
M1 for attempt at any of PR, PQ, QR using Pythagoras’ Theorem
PR = PQ + QR Ÿ (a  c) 2  (a  c) 2 = (b  a ) 2  (b  a ) 2 + (c  b ) 2  (c  b ) 2

A1 A1 A1 for correct, simplified lengths: 4ac = 4ab + 4bc

1 1 1
A1 given answer legitimately obtained by dividing by 4abc :  ུ
b c a

M1 M1 for working suitably on RHS of (*); substituting for b, e.g.

2
§ 1 ­1 1½ 1 ·
2
§ 1 1 1· ¨ ® 
2
¨   ¸ ¨a  ¾  ¸¸
©a b c¹ © ¯a ac c ¿ c ¹
§ 1 3 1 2 2 ·
A1 = 4¨¨ 2   2   ¸¸ any form suitable for comparison
© a ac c a ac c ac ¹
M1 for working suitably on LHS of (*) and substituting for b2, e.g.
§ 1 1 1 · 2 2 § 1 4 6 4 1 ·
A1 for correct b2 in 2¨ 2  2  2 ¸  2  2¨¨ 2     2 ¸¸
©a c ¹ 2
b a c ©a a ac ac c ac c ¹
4 12 4 8 8
A1 shown equal to RHS: = 2   2  
a ac c a ac c ac
ཱུ

1 1 1 1 1 2 1
Alternative:  Ÿ   M1 squaring
b c a b a ac c
2
§1 1 1· 4
Ÿ ¨   ¸ M1 M1 rearranging and squaring again
©b a c¹ ac
1 1 1 2 2 2 4
Ÿ 2  2  2    A1 correct LHS
a b c ab bc ac ac
2
§ 1 1 1 · 1 1 1 2 2 2 §1 1 1·
Ÿ 2¨ 2  2  2 ¸ = 2  2  2    = ¨   ¸ M1 A1
©a b c ¹ a b c ab bc ca © a b c ¹
Question 5 (ii)
2
§ 1 1 1 · §1 1 1· 1 1 1 2 2 2
M1 If 2¨ 2  2  2 ¸ = ¨   ¸ then 2  2  2  
©a b c ¹ ©a b c¹ a b c ab bc ca
1 1 1
M1 Let x , y , z with or without actual substitution
a b c
so that x 4  y 4  z 4 2x 2 y 2  2 y 2 z 2  2z 2 x 2
M1 for recognition of conditions b < c < a Ÿ y > z > x
for completing the square: x 2  z 2  y 2 
2
M1 A1 4x2 z 2
A1 œ x2  z 2  y2 r 2 xz
œ z # x 
2
y2
A1 for the four cases y = x – z , y = z – x , y = x + z or y = – x – z
E1 for use of conditions to show that only y = x + z is suitable
1 1 1
A1 for legitimately obtaining given answer:  ླྀ
b c a


802
Question 6
E1 for explanation that x = ma since OX || OA
B1 for 0 < m < 1 (since X between O and A): don’t penalise any equality interval endpoints
E1 for explanation that BC || OA Ÿ c – b = ka and so c = ka + b
B1 for k < 0 (since BC in opposite direction to OA) ཱི

B1 for correct set-up for D = OB ˆ AC: a + D (c – a) = E b

§ 1 ·
M1 for equating coefficients: 1 – D + Dk = 0 and D = E ¨ ¸
© 1 k ¹
1
A1 for d = b ི
1 k

§ 1 ·
B1 for correct set-up for Y = XD ˆ BC: ma + D ¨ b  ma ¸ = b + Ek a
©1 k ¹
D
M1 for equating coefficients: m – Dm – Ek = 0 and 1
1 k
A1 for y = kma + b from D = 1 – k, E = m ི

B1 for correct set-up for Z = OY ˆ AB: (1 – D)a + Db = E km a  b 

§ 1 ·
M1 for equating coefficients: 1 – D – kmE = 0 and D = E ¨ ¸
© 1  km ¹
§ km · § 1 ·
A1 for z = ¨ ¸a + ¨ ¸b (Given Answer) ི
© 1  km ¹ © 1  km ¹

1 § km 1 1 ·
B1 for correct set-up for T = DZ ˆ OA: D a = b + E¨ a b b¸
1 k © 1  km 1  km 1  k ¹
Ekm 1 E E
M1 for equating coefficients: D = and 0 = 
1  km 1  k 1  km
§ m · m 1  km
A1 for t = ¨ ¸a from D , E ི
©1 m ¹ 1 m k (1  m)

§ m ·
M1 A1 for setting up all lengths: OA = a , OX = ma , OT = ¨ ¸a ,
©1  m ¹
§ m2 ·
¸¸a , TA = §¨
1 ·
TX = ¨¨ ¸a , XA = (1 – m)a
© 1 m ¹ ©1 m ¹
where |a| = a, which may (w.l.o.g.) be taken to be 1
1 1§ 1· 1 1
A1 for 1st correctly derived result: ¨1  ¸ 
OT a © m ¹ OA OX
§ m · 2 § 1 ·
A1 for 2nd correctly derived result: OT . OA = ¨ ¸a ( ma ) . ¨ ¸ a = OX . TA ཱི
©1 m ¹ ©1 m ¹



803
Question 7 (i)
B1 B1 for S ˆ T = I ; S ‰ T = the set of positive odd numbers ཱ

Question 7 (ii)
M1 A1 for (4a + 1)(4b + 1) = 4(4ab + a + b) + 1 (which is in S)
M1 A1 for (4a + 3)(4b + 3) = 4(4ab + 3a +3b + 2) + 1 (not necessarily as shown here)
A1 for clearly demonstrating this is not in T ུ

Question 7 (iii)
M1 M1 for attempting a proof by contradiction; method for establishing contradiction
Suppose all of t’s prime factors are in S
B1 for no even factors
t = (4a + 1) (4b + 1) (4c + 1) …(4n + 1)
A1 Then t = 4{ … … …} + 1
E1 for convincing explanation that this is always in S
(may appeal inductively to (ii)’s result) ུ

Question 7 (iv) (a)

B1 for writing an element of T as products of T-primes
M1 for noting that every pair of factors in T multiply to give an element of S (by (ii))
A1 so there must be an odd number of them ི

Question 7 (iv) (b)

M1 for recognisable method to find composites in S whose prime-factors are in T
M1 for recognition of the regrouping process
M1 A1 for correct example demonstrated:
e.g. 9 u 77 = 21 u 33 (= 693) where 9, 21, 33, 77 are in S
and 9 = 3 u 3, 21 = 3 u7, 33 = 3 u 11, 77 = 7 u 11 with 3, 7, 11 in T
B1 for correctly-chosen S-primes ུ



804
Question 8 (i)
B1 for f(x) = 0  x  2 x 2  3x 3  ...  nx n  ...
M1 for use of (1  x) 2 1  2 x  3x 2  4 x 3  ...  nx n  1  ... (forwards or backwards)
A1 for given result correctly shown: f(x) = x (1  x) 2 ི

M1 A1 for x(1  x) 3 
x 1  3 x  6 x 2  10 x 3  ...  12 n( n  1) x n  1  ... 
= 0  x  3 x 2  6 x 3  ...  12 n(n  1) x n  ...
A1 for un = 12 n 2  12 n ི

2x x
M1 A1 for use of first two results: 2 u (2nd) – (1st) gives  with un = n2 ཱ
(1  x) (1  x) 2
3

Question 8 (ii) (a)

Method I: B1 for f(x) a  (ka) x  (k 2 a) x 2  (k 3a) x 3  ...  (k n a) x n  ...
§ 1 ·
M1 A1 for a u sum-to-infinity of a GP with common ratio kx : f(x) a¨ ¸
© 1  kx ¹
B1 for showing (retrospectively) that f(x) = a + kx f(x)

Method II: B1 for f(x) = a  akx  ak 2 x 2  ak 3 x 3  ...  ak n x n  ...

M1 
= a  kx a  akx  ak 2 x 2  ak 3 x 3  ...  ak n x n  ... 
A1 = a + kx f(x)
§ 1 ·
A1 for f(x) a¨ ¸ ཱི
© 1  kx ¹

Question 8 (ii) (b)

f
M1 A1 for summing, and splitting off initial terms: f(x) = 0  x  ¦ u n x n
n 2
f
M1 for use of given recurrence relation: = 0  x  ¦ u n  1  u n  2 x n
n 2
f f
M1 for dealing with limits: = x  x ¦ un 1 x n 1  x 2 ¦ un  2 x n  2
n 2 n 2
f f
A1 for re-creating f(x)’s: = x  x¦ u n x n  x 2 ¦ u n x n
n 1 n 0

A1 for correctly expressing all terms in f(x): = x  x^f ( x)  0` x 2 f ( x)

x
M1 A1 for re-arranging to get f(x) = ཷ
1 x  x2




805
Question 9
RP
B Diagram for Case 1:
FA G b P FP G between walls;
T a rod about to slip down LH wall
A RA

B1 for both FA = O RA and FP = P RP noted or used somewhere

M1 for resolving in one direction (with correct number of forces)

A1 e.g. Res.n W = FA + RP sinT + FP cosT
M1 for eliminating the F’s (e.g.): W = O RA + RP sinT + P RP cosT

M1 for resolving in second direction (with correct number of forces)

A1 e.g. Res.o RA = RP cosT – FP sinT
M1 for eliminating the F’s (e.g.): RA = RP cosT – P RP sinT

M1 for taking moments (with correct number of forces)

A1 e.g. ඣA W a sinT = RP (a + b)
M1 for correct introduction of d: W a sin2T = RP d or other suitable distance

M1 A1 for getting W in terms of one other force: e.g. W = RP O cos T  OP sin T  sin T  P cos T 
M1 for eliminating W and that force from two relevant equations: e.g. these last two
A1 for legitimately obtaining given result: dcosec2T a[O  P ] cosT  [1  OP ]sinT  ཽ

For Case 2: G the other side of P; rod about to slide up LH wall …

M1 M1 M1 FA o – FA ; FP o – FP ; a + b o a – b (or switching signs of O and P)

A1 W = RP  O cos T  OP sin T  sin T  P cos T  and W a sin2T = RP d (e.g.)
M1 A1 for obtaining dcosec2T a [O  P ] cosT  [1  OP ]sinT  ཱུ



806
Question 10 (i)

ou o0 0m um

A B C D

o vA o vB m vC m vD

For collision A/B For collision C/D

B1 B1 for CLM statements: m(Ou = OvA +vB) m(u = vC +vD)
B1 B1 for NEL/NLR statements: eu = vB – vA eu = vC – vD
Watch out for different signs from alternative choices of directions
M1 solving for at least vB and vC
O (1  e) O e
A1 A1 for vB u , vC 12 (1  e)u NB v A u and vD 12 (1  e)u not needed ྲྀ
O 1 O 1

o vB vC m

B C

o0 o wC
M1 A1 A1 for CLM and NEL/NLR statements: m(vB – vC) = m wC and e(vB + vC) = wC
M1 for substituting previous answers in terms of e and u
O 1
M1 A1 for identifying e : e Given Answer legitimately obtained
3O  1
E1 for justifying that e < 1
3 (can’t just show that e o 13 ) ྲྀ

Question 10 (ii)
(1  e)(O  1)
NB wC = u correct from previous bit of work
2(O  1)
M1 for setting wC = vD in whatever forms they have (not just saying they are equal)
(1  e)(O  1)
A1 correct to here: u = 12 (1  e)u FT previous answers
2(O  1)
M1 for substituting for e (e.g.)
M1 A1 A1 for solving for O and e : O 5 2, e = 52 ཱུ



807
Question 11
gx 2
M1 A1 for stating, or obtaining, the Trajectory Equation: y x tan D 
2u 2 cos 2 D
M1 for setting y = –h and re-arranging
gx 2
2
2h cos 2 D  2 x sin D cosD
u
A1 for legitimately obtaining given answer from use of double-angle formulae:
gx 2
h(1  cos 2D )  x sin 2D ཱི
u2

d § gx 2 · § dx ·
M1 A1 for differentiating w.r.t. D : ¨¨ 2 ¸¸ h 2 sin 2D   ¨ x.2 cos 2D  sin 2D . ¸
dD © u ¹ © dD ¹
M1 for using both derivatives = 0
A1 for legitimately obtaining given answer x = h tan2D ཱི

gh 2 tan 2 2D
M1 for substituting back: h(1  cos 2D )  h tan 2D sin 2D
u2
M1 cancelling one h and (e.g.) writing all trig. terms in c = cos2D

A1

gh 1  c 2  1 c 
1 c
2
Ÿ gh  ghc2 
u 2 c 2  c3  c  c3 
u 2c 2 c
M1 A1 for a quadratic in c : 0 u  ghc  u c  gh
2 2 2

M1 for solving attempt: 0 >u  gh c  gh@(c  1)

2

gh
A1 for cos2D = ྲྀ
u  gh
2

M1 for substituting x = h tan2D and y = –h in '2 x2  y2

M1 A1 for use of relevant trig. result(s) = h2 sec22D i.e. ' h sec 2D
u  gh
2
M1 for use of previous result: ' h.
gh
u2
A1  h correct given answer legitimately obtained ུ
g



808
Question 12 (i)
M1 for some systematic approach to counting cases
A1 A1 A1 for correct cases: e.g. p(A=0).p(B=1,2,3) + p(A=1).p(B =2,3) + p(A=2).p(B=3)
M1 for some correct probabilities: 14 u 78  2 u 14 u 84  14 u 18
1
A1 for correctly obtained answer, 2

If no other marks scored, B1 for 32 outcomes ཱུ

Question 12 (ii)
M1 for some systematic approach to counting cases
A1 A1 A1 for identifying the correct cases and/or probabilities
e.g. 18 u  461641   83 u  61641   83 u  4161   18 u 161 
M1 for all cases/probabilities correct: 14 u 78  2 u 14 u 84  14 u 18
1
A1 for correctly obtained answer, 2

If no other marks scored, B1 for 128 outcomes ཱུ

Question 12 (iii)
B1 for stating that, when each tosses n coins, p(B has more Hs) = p(A has more Hs) = p2
B1 for stating that p(AH = BH) = p1
B1 for statement (explained or not) that p1 + 2p2 = 1

M1 for considering what happens when B tosses the extra coin

A1 p(B has more Hs) = p(B already has more Hs) u p(B gets T)
A1 + p(B already has more, or equal Hs) u p(B gets H)
A1 correct probs. used = p2 u 12  ( p1  p2 ) u 12
A1 for correct answer, fully justified: 1
2 ( p1  2 p2 ) 1
2 ཷ



809
Question 13 (i)
For the i-th e-mail,
M1 for integrating f i (t ) Oe  Ot
A1 for Fi (t )  e  Ot + C
M1 A1 for justifying or noting that C = 1 (from F(0) = 0)

For n e-mails sent simultaneously,

M1 A1 for F(t) = P(T d t) = 1 – P(all n take longer than t)
B1 for  n
= 1  e Ot i.e. the product of n independent probabilities
A1 for = 1  Oe Ont
M1 A1 for differentiating this: f(t) = nOe Ont ཹ

f
Ont
M1 for attempt at E(T) = ³0 t u nOe dt

f
M1 A1 A1 for use of integration by parts: E(T) =  te Ont > @ + ³ nOe O
f
0
 nt
dt
0

ª  e Ont º f
A1 = 0 + « »
¬ On ¼ 0
1
A1 for E(T) =
nO
NB – anyone able to identify this as the Exponential Distribution can quote the
Expectation (or from the Formula Book) and get 6 marks for little effort ཱུ

Question 13 (ii)
M1 for observing that 2nd email is simply the 1st from the remaining (n – 1) …
1
A1 … with expected arrival time
(n  1)O
E1 for careful explanation of the result
1 1 1 §1 1 ·
A1 for a legitimately obtained given answer  ¨¨  ¸ ཱི
nO (n  1)O O © n (n  1) ¸¹



810
 STEPII2016 MARKSCHEME

Question1

 Ifthevalueoftheparameteratܲis‫݌‬andthevalueoftheparameteratܳis‫ݍ‬: 
 ௣య ି଴ M1
Gradientoflineܱܲis ൌ ‫݌‬andsimilarlythegradientofܱܳis‫ݍ‬.
௣మ ି଴
 Iftheangleatܱisarightangle,then‫ ݍ݌‬ൌ െͳ A1
 
ௗ௫ ௗ௬
 ൌ ʹ‫ݐ‬, ൌ ͵‫ ݐ‬ଶ  M1
ௗ௧ ௗ௧
 ௗ௬ ଷ A1
Therefore ൌ ‫ݐ‬
ௗ௫ ଶ
 Equationoftangentatthepointሺ‫ ݐ‬ଶ ǡ ‫ ݐ‬ଷ ሻǣ M1
͵ A1
‫ ݕ‬െ ‫ ݐ‬ଷ ൌ ‫ݐ‬ሺ‫ ݔ‬െ ‫ ݐ‬ଶ ሻ
ʹ
 ͵ ͵ M1
‫݌‬ሺ‫ ݔ‬െ ‫ ݌‬൅ ‫ ݌‬ൌ ‫ݍ‬ሺ‫ ݔ‬െ ‫ ݍ‬ଶ ሻ ൅ ‫ ݍ‬ଷ 
ଶሻ ଷ
ʹ ʹ
 ͵‫ ݔ݌‬െ ͵‫݌‬ଷ ൅ ʹ‫݌‬ଷ ൌ ͵‫ ݔݍ‬െ ͵‫ ݍ‬ଷ ൅ ʹ‫ ݍ‬ଷ 
 ‫݌‬ଷ െ ‫ ݍ‬ଷ ͳ A1
‫ݔ‬ൌ ൌ ሺ‫݌‬ଶ ൅ ‫ ݍ݌‬൅ ‫ ݍ‬ଶ ሻ
͵ሺ‫ ݌‬െ ‫ݍ‬ሻ ͵
 Substitutefor‫ݕ‬: M1
͵ ͳ
‫ ݕ‬െ ‫݌‬ଷ ൌ ‫ ݌‬൬ ሺ‫݌‬ଶ ൅ ‫ ݍ݌‬൅ ‫ ݍ‬ଶ ሻ െ ‫݌‬ଶ ൰ ൅ ‫݌‬ଷ 
ʹ ͵
 ͳ A1
‫ ݕ‬ൌ ‫ݍ݌‬ሺ‫ ݌‬൅ ‫ݍ‬ሻ
ʹ
  
 Use‫ ݍ݌‬ൌ െͳ:
‫݌‬ସ െ ‫݌‬ଶ ൅ ͳ
‫ݔ‬ൌ 
͵‫݌‬ଶ
 ‫݌‬ଶ െ ͳ B1
‫ݕ‬ൌെ 
ʹ‫݌‬
  
 ‫݌‬ସ െ ʹ‫݌‬ଶ ൅ ͳ M1
Ͷ‫ ݕ‬ଶ ൌ ൌ ͵‫ ݔ‬െ ͳ ሺ‫כ‬ሻ
‫݌‬ଶ A1
  
 If‫ܥ‬ଵ and‫ܥ‬ଶ meetthentheremustbeavalueof‫ ݐ‬suchthat: B1
Ͷ‫ ଺ ݐ‬ൌ ͵‫ ݐ‬ଶ െ ͳ
 Ͷ‫ ଺ ݐ‬െ ͵‫ ݐ‬ଶ ൅ ͳ ൌ Ͳ 
 ሺʹ‫ ݐ‬െ ͳሻሺʹ‫ ݐ‬ସ ൅ ‫ ݐ‬ଶ െ ͳሻ ൌ Ͳ
ଶ M1
 ሺʹ‫ ݐ‬ଶ െ ͳሻଶ ሺ‫ ݐ‬ଶ ൅ ͳሻ ൌ Ͳ A1
 ξଶ B1
Therefore,pointsofintersectiononlywhen‫ ݐ‬ൌ േ 
ଶ
 Graph: B1
B1
B1



811
M1 Anexpressionforthegradientofthelinefromtheorigintoapointonthecurve.
IfapplyingPythagorastoshowthattheangleisarightangle,theawardM1foracorrect
expressionforthedistancefromtheorigintoapointonthecurve.
A1 Correctlydeducingthat‫ ݍ݌‬ൌ െͳ
M1 Differentiationofbothfunctions.
A1 Divisiontoobtaincorrectgradientfunction.
M1 Attempttofindtheequationofatangenttothecurveatoneofthepoints.
A1 Correctequation.
M1 Eliminationofonevariablebetweenthetwotangentequations.
A1 Correctexpressionforeither‫ݔ‬or‫ݕ‬found.
M1 Substitutiontofindtheothervariable.
A1 Correctexpressionsfoundforbothvariables.
B1 Usingtherelationship‫ ݍ݌‬ൌ െͳfoundearlier.
M1 Anattempttoeliminatetheparameter
A1 Fullycorrectreasoningleadingtotheequationgiveninthequestion.
B1 Conditionforcurvestomeetidentified.
M1 Attempttofactorisetheequation.
A1 Correctlyfactorised.
B1 Pointsofintersectionidentified.
B1 Correctshapefor‫ ݔ‬ൌ ‫ ݐ‬ଶ ǡ ‫ ݕ‬ൌ ‫ ݐ‬ଷ .
B1 CorrectshapeforͶ‫ ݕ‬ଶ ൌ ͵‫ ݔ‬െ ͳ.
B1 Graphsjusttouchattwopoints.
 



812
Question2

 Letܿ ൌ ܽ ൅ ܾ: M1
ሺʹܽ ൅ ʹܾሻଷ െ ͸ሺʹܽ ൅ ʹܾሻሺܽଶ ൅ ܾ ଶ ൅ ሺܽ ൅ ܾሻଶ ሻ ൅ ͺሺܽଷ ൅ ܾଷ ൅ ሺܽ ൅ ܾሻଷ ሻ
 ൌ ͺሺܽ ൅ ܾሻଷ െ ʹͶሺܽ ൅ ܾሻሺܽଶ ൅ ܾܽ ൅ ܾ ଶ ሻ ൅ ͺሺʹܽଷ ൅ ͵ܽଶ ܾ ൅ ͵ܾܽ ଶ ൅ ʹܾ ଷ ሻ
 ൌͲ M1
 Thereforeሺܽ ൅ ܾ െ ܿሻisafactorof(*) A1
 
 Bysymmetry,ሺܾ ൅ ܿ െ ܽሻandሺܿ ൅ ܽ െ ܾሻmustalsobefactors. B1
 So(*)mustfactoriseto݇ሺܽ ൅ ܾ െ ܿሻሺܾ ൅ ܿ െ ܽሻሺܿ ൅ ܽ െ ܾሻ M1
 Toobtainthecorrectcoefficientofܽଷ ,݇ ൌ െ͵ M1
 (*)factorisestoെ͵ሺܽ ൅ ܾ െ ܿሻሺܾ ൅ ܿ െ ܽሻሺܿ ൅ ܽ െ ܾሻ A1
 
(i) Tomatchtheequationgiven,weneed M1
ହ ଵଷ
ܽ ൅ ܾ ൅ ܿ ൌ ‫ ݔ‬൅ ͳ,ܽଶ ൅ ܾ ଶ ൅ ܿ ଶ ൌ andܽଷ ൅ ܾ ଷ ൅ ܿ ଷ ൌ .
ଶ ସ
 ͵ ͳ A1
ܽ ൌ ‫ݔ‬ǡ ܾ ൌ ǡ ܿ ൌ െ 
ʹ ʹ
 Theequationthereforefactorisesto M1
െ͵ሺ‫ ݔ‬൅ ʹሻሺͳ െ ‫ݔ‬ሻሺ‫ ݔ‬െ ʹሻ ൌ Ͳ
 ‫ ݔ‬ൌ െʹǡ ͳ ‫ʹ ݎ݋‬ A1
 
(ii) Let݀ ൅ ݁ ൌ ܿin(*): 
ܽ ൅ ܾ െ ݀ െ ݁isafactorof
ሺܽ ൅ ܾ ൅ ݀ ൅ ݁ሻଶ െ ͸ሺܽ ൅ ܾ ൅ ݀ ൅ ݁ሻሺܽଶ ൅ ܾ ଶ ൅ ሺ݀ ൅ ݁ሻଶ ሻ ൅ ͺሺܽଷ ൅ ܾ ଷ ൅ ሺ݀ ൅ ݁ሻଷ ሻ
 Whichis: M1
ሺܽ ൅ ܾ ൅ ݀ ൅ ݁ሻଶ െ ͸ሺܽ ൅ ܾ ൅ ݀ ൅ ݁ሻሺܽଶ ൅ ܾ ଶ ൅ ݀ ଶ ൅ ݁ ଶ ሻ ൅ ͺሺܽଷ ൅ ܾ ଷ ൅ ݀ ଷ ൅ ݁ ଷ ሻ
െ͸ሺܽ ൅ ܾ ൅ ݀ ൅ ݁ሻሺʹ݀݁ሻ ൅ ͺሺ͵݀ଶ ݁ ൅ ͵݀݁ ଶ ሻ
 െ͸ሺܽ ൅ ܾ ൅ ݀ ൅ ݁ሻሺʹ݀݁ሻ ൅ ͺሺ͵݀ ଶ ݁ ൅ ͵݀݁ ଶ ሻ ൌ െͳʹܽ݀݁ െ ͳʹܾ݀݁ ൅ ͳʹ݀ଶ ݁ ൅ ͳʹ݀݁ ଶ  M1
 Whichisെͳʹሺܽ ൅ ܾ െ ݀ െ ݁ሻሺ݀݁ሻ.Thereforeሺܽ ൅ ܾ െ ݀ െ ݁ሻ isafactorof: A1
ሺܽ ൅ ܾ ൅ ݀ ൅ ݁ሻଶ െ ͸ሺܽ ൅ ܾ ൅ ݀ ൅ ݁ሻሺܽଶ ൅ ܾ ଶ ൅ ݀ ଶ ൅ ݁ ଶ ሻ ൅ ͺሺܽଷ ൅ ܾ ଷ ൅ ݀ ଷ ൅ ݁ ଷ ሻ
  
 Bysymmetry,ܽ െ ܾ െ ݀ ൅ ݁andܽ െ ܾ ൅ ݀ െ ݁ mustalsobefactors,soitmustfactoriseto: M1
݇ሺܽ ൅ ܾ െ ݀ െ ݁ሻሺܽ െ ܾ െ ܿ ൅ ݀ሻሺܽ െ ܾ ൅ ܿ െ ݀ሻ
 Toobtainthecorrectcoefficientwerequire݇ ൌ ͵. A1
 
 Tomatchtheequationgivenweneed M1
ܽ ൅ ܾ ൅ ܿ ൅ ݀ ൌ ‫ ݔ‬൅ ͸,ܽଶ ൅ ܾ ଶ ൅ ܿ ଶ ൅ ݀ ଶ ൌ ‫ ݔ‬ଶ ൅ ͳͶandܽଷ ൅ ܾ ଷ ൅ ܿ ଷ ൅ ݀ ଷ ൌ ‫ ݔ‬ଷ ൅ ͵͸
 ܽ ൌ ‫ݔ‬ǡ ܾ ൌ ͳǡ ܿ ൌ ʹǡ ݀ ൌ ͵ A1
 Theequationthereforefactorisesto M1
͵‫ݔ‬ሺ‫ ݔ‬െ Ͷሻሺ‫ ݔ‬െ ʹሻ
 ‫ ݔ‬ൌ Ͳǡ ʹ ‫ ݎ݋‬Ͷ A1


 



813
M1 Substitutionofܿ ൌ ܽ ൅ ܾ.
M1 Clearalgebraicstepstoshowthatthevalueofthefunctionis0.
A1 Conclusionthatthismeansthatሺܽ ൅ ܾ െ ܿሻisafactor.
B1 Identificationoftheotherfactors.
M1 Correctformofthefactorisationstated.
M1 Considerationofanyonecoefficienttofindthevalueof݇.
A1 Correctfactorisation.
M1 Identificationoftheequationsthatܽ,ܾandܿmustsatisfy.
A1 Correctselectionofܽ,ܾandܿ.
M1 Correctfactorisation.
A1 Solutionsoftheequation.
M1 Substitutionintotheequationandrearrangementintotheexpressionofthequestionand
anextraterm.
M1 Simplificationoftheextratermandfactorisation.
A1 Conclusion.
M1 Identificationoftheotherfactors.
A1 Correctcoefficientfound.
M1 Identificationoftheequationsthatܽ,ܾ,ܿ and݀ mustsatisfy.
A1 Correctselectionofܽ,ܾ,ܿand݀.
M1 Factorisationoftheequation.
A1 Solutionsfound.
 



814
Question3

(i) ‫ݔ‬ଶ ‫ ݔ‬௡ିଵ B1

݂௡ᇱ ሺ‫ݔ‬ሻ ൌ ͳ ൅ ‫ ݔ‬൅ ൅ ‫ڮ‬൅ 
ʹǨ ሺ݊ െ ͳሻǨ
  
(ii) Ifܽisarootoftheequationthen݂௡ ሺܽሻ ൌ Ͳ B1
 Eachofthetermsof݂ሺܽሻwillbepositiveifܽ ൐ Ͳ. M1
 Therefore݂௡ ሺܽሻ ൐ Ͳ A1
  
(iii) ௔೙ ௔೙ M1
݂௡ᇱ ሺܽሻ ൌ ݂௡ ሺܽሻ െ ൌെ ,andsimilarlyforܾ.
௡Ǩ ௡Ǩ
A1
 Sinceܽandܾarebothnegative,݂௡ᇱ ሺܽሻand݂௡ Ԣሺܾሻmusthavethesamesign. M1
M1
 Therefore݂௡ᇱ ሺܽሻ݂௡ᇱ ሺܾሻ ൐ Ͳ A1
 Twocases(positiveandnegativegradients) B1
Sketchneededforeach B1
 Sincethegraphiscontinuous,theremustbeanadditionalrootbetweenܽandܾ. M1
A1
 Thiswouldimplyinfinitelymanyroots. M1
 But݂௡ ሺ‫ݔ‬ሻisapolynomialofdegree݊,sohasatmost݊roots M1
 Thereforethereisatmostoneroot. A1
  
 If݊isoddthen݂௡ ሺ‫ݔ‬ሻ ՜ െλas‫ ݔ‬՜ െλ and݂௡ ሺ‫ݔ‬ሻ ՜ λ as‫ ݔ‬՜ λ M1
Thereisonerealroot. A1
 If݊iseventhen݂௡ ሺ‫ݔ‬ሻ ՜ λas‫ ݔ‬՜ െλ and݂௡ ሺ‫ݔ‬ሻ ՜ λ as‫ ݔ‬՜ λ M1
Therearenorealroots. A1
 



815
B1 Someexplanationofthegeneraltermisrequiredforthismark.
B1 Statedorimpliedelsewhereintheanswer(suchaswhendrawingconclusion).
M1 Clearstatementabouttheindividualterms.
A1 Clearlystatedconclusion.
M1 Attempttorelatefunctiontoitsderivative
A1 Correctrelationship
M1 Statementthatthesignsmustbethesame.
M1 Considerationofthedifferentcasesfor݊.
A1 Conclusionthattheproductispositive.
B1 Sketchofgraphwithtworootswiththecurvepassingthroughwithpositivegradienteach
time.
B1 Sketchofgraphwithtworootswiththecurvepassingthroughwithnegativegradienteach
time.SecondB1maybegivenifonlyonegraphsketchedwithaclearexplanationofthe
similaritiesthattheothergraphwouldhave.
M1 Anattempttoexplainthattheremustbearootbetweenthetwo.
A1 Clearexplanationincludingreferencetocontinuity.
M1 Statementthatthiswouldimplyinfinitelymanyroots.
OR
Statementthatthegradientwouldbenegativeor0atthatrootiftheothertworootshad
positivegradients.
M1 Statementthatthereareatmost݊roots.
OR
Statementthatanegativeorzerogradientattherootinbetweenwouldgiveapairofroots
contradictingtheearlierconclusion.
A1 Conclusion.
M1 Correctidentificationoftheoutcomefor݊ odd.
A1 Acorrectjustificationfortheconclusion.
M1 Correctidentificationoftheoutcomefor݊even.
A1 Acorrectjustificationfortheconclusion.


 



816
Question4

(i) ሺ‫ ݔ‬ଶ ൅ ‫ ߠ ‹• ݔ‬൅ ͳሻ …‘• ߠ െ ሺ‫ ݔ‬ଶ ൅ ‫ ߠ •‘… ݔ‬൅ ͳሻ •‹ ߠ M1
‫ ߠ •‘… ݕ‬െ •‹ ߠ ൌ 
‫ ݔ‬ଶ ൅ ‫ ߠ •‘… ݔ‬൅ ͳ A1
ଶ
ሺ‫ ݔ‬െ ͳሻሺ…‘• ߠ െ •‹ ߠሻ
ൌ 
‫ ݔ‬ଶ ൅ ‫ ߠ •‘… ݔ‬൅ ͳ
 ‫ݔ‬ሺ•‹ ߠ െ …‘• ߠሻ B1
‫ݕ‬െͳൌ ଶ 
‫ ݔ‬൅ ‫ ߠ •‘… ݔ‬൅ ͳ
  
ଶ ଶ ሺ•‹ ଶ
 ሺ‫ ݔ‬െ ͳሻ ߠ െ …‘• ߠሻ M1
ሺ‫ ߠ •‘… ݕ‬െ •‹ ߠሻଶ ൌ
ሺ‫ ݔ‬ଶ ൅ ‫ ߠ •‘… ݔ‬൅ ͳሻଶ
ሺ‫ ݔ‬ଶ െ ͳሻଶ
ൌ ሺ‫ ݕ‬െ ͳሻଶ ൈ 
‫ݔ‬ଶ
 ଶ
ሺ‫ ݔ‬െ ͳሻ ଶ
ͳ ଶ M1
ൌ ൬‫ݔ‬ െ ൰ 
‫ݔ‬ଶ ‫ݔ‬
 ଵ ଶ M1
Minimumvalueofቀ‫ ݔ‬െ ቁ is4,thereforeሺ‫ ߠ •‘… ݕ‬െ •‹ ߠሻଶ ൒ Ͷሺ‫ ݕ‬െ ͳሻଶ ሺ‫כ‬ሻ
௫ A1
  
 ‫ ߠ •‘… ݕ‬െ •‹ ߠcanbewrittenasඥ‫ ݕ‬ଶ ൅ ͳ …‘•ሺߠ ൅ ߙሻforsomevalueofߙ. M1
A1

 Therefore‫ ݕ‬ଶ ൅ ͳ ൒ ሺ‫ ߠ •‘… ݕ‬െ •‹ ߠሻଶ ൒ Ͷሺ‫ ݕ‬െ ͳሻଶ  A1
  
 ‫ ݕ‬ଶ ൅ ͳ ൒ Ͷ‫ ݕ‬ଶ െ ͺ‫ ݕ‬൅ Ͷ 
 ͵‫ ݕ‬ଶ െ ͺ‫ ݕ‬൅ ͵ ൑ Ͳ M1
 ଼േඥሺ଼ሻమ ିସሺଷሻሺଷሻ A1
Criticalvaluesare:‫ ݕ‬ൌ 
ଶሺଷሻ
Ͷ െ ξ͹ Ͷ ൅ ξ͹
൑‫ݕ‬൑ 
͵ ͵
  
(ii) ସାξ଻ ଵ଺ା଼ξ଻ା଻ ଷଶା଼ξ଻
If‫ ݕ‬ൌ ,thenඥ‫ ݕ‬ଶ ൅ ͳ ൌ ට ൅ͳൌට 
ଷ ଽ ଽ
 ଶାଶξ଻ M1
ʹሺ‫ ݕ‬െ ͳሻ ൌ 
ଷ
 ଶ A1
ଶାଶξ଻ ସା଼ξ଻ାଶ଼
ቀ ቁ ൌ ,soඥ‫ ݕ‬ଶ ൅ ͳ ൌ ʹሺ‫ ݕ‬െ ͳሻ
ଷ ଽ
  
 Sinceඥ‫ ݕ‬ଶ ൅ ͳ ൌ ʹሺ‫ ݕ‬െ ͳሻ,thevalueofߠmustbethevalueofߙwhen B1
‫ ߠ •‘… ݕ‬െ •‹ ߠiswrittenasඥ‫ ݕ‬ଶ ൅ ͳ …‘•ሺߠ ൅ ߙሻ.
 ଵ ସିξ଻ M1
Therefore–ƒ ߠ ൌ ൌ 
௬ ଷ A1
 
 Tofind‫ݔ‬: M1
‫ ݔ‬ଶ ‫ ݕ‬൅ ‫ ߠ •‘… ݕݔ‬൅ ‫ ݕ‬ൌ ‫ ݔ‬ଶ ൅ ‫ ߠ ‹• ݔ‬൅ ͳ
ଶ ሺ‫ݕ‬
 ‫ݔ‬ െ ͳሻ ൅ ‫ݔ‬ሺ‫ ߠ •‘… ݕ‬െ •‹ ߠሻ ൅ ‫ ݕ‬െ ͳ ൌ Ͳ 
 Since‫ ߠ •‘… ݕ‬െ •‹ ߠ ൌ േʹሺ‫ ݕ‬െ ͳሻ,and‫ ݕ‬െ ͳ ് Ͳthissimplifiesto: M1
‫ ݔ‬ଶ േ ʹ‫ ݔ‬൅ ͳ ൌ Ͳ
 Sowehaveeither‫ ݔ‬ൌ ͳor‫ ݔ‬ൌ െͳ A1
 



817
M1 Substitutionfor‫ݕ‬into‫ ߠ •‘… ݕ‬െ •‹ ߠ.
A1 Correctlysimplified.
B1 Correctsimplificationof‫ ݕ‬െ ͳ.
M1 Relationshipbetween‫ ߠ •‘… ݕ‬െ •‹ ߠand‫ ݕ‬െ ͳ.
M1 Simplificationofthemultiplier.
M1 Justificationthattheminimumvalueis4.
A1 Conclusionthatthegivenstatementiscorrect.
M1 Calculationoftheamplitudeof‫ ߠ •‘… ݕ‬െ •‹ ߠ.
A1 Correctvalue.
A1 Usetodemonstratetherequiredresult.
M1 Rearrangementtogivequadraticinequality.
A1 Solveinequalityandconcludetherangegiven.
M1 Substitutionof‫ݕ‬intothetwoexpressions.
A1 Demonstrationthattheequationholds.
B1 Statementthatthismustbeanoccasionwhere‫ ߠ •‘… ݕ‬െ •‹ ߠtakesitsmaximumvalue.
M1 Calculationofthevalueof–ƒ ߠ.
A1 Correctsimplification.
M1 Substitutiontofind‫ݔ‬.
M1 Simplificationoftheequationtoeliminateߠ.
A1 Valuesof‫ݔ‬found.


 



818
Question5

(i) ିேሺିேିଵሻǥሺିேି௡ାଵሻ ேሺேାଵሻǥሺேା௡ିଵሻ M1

Coefficientof‫ ݔ‬௡ is ሺെͳሻ௡ ൌ 
௡Ǩ ௡Ǩ
ܰ൅݊െͳ ܰ൅݊െͳ M1
orቀ ቁorቀ ቁ A1
ܰെͳ ݊
 Expansionistherefore: B1
σஶ
ேሺேାଵሻǥሺேା௥ିଵሻ ܰ൅‫ݎ‬െͳ ௥
௥ୀ଴ ‫ ݔ‬௥ orσஶ
௥ୀ଴ ቀ ቁ‫ ݔ‬
௥Ǩ ܰെͳ
 
 ሺͳ െ ‫ݔ‬ሻିଵ ൌ ͳ ൅ ‫ ݔ‬൅ ‫ ݔ‬ଶ ൅ ‫ڮ‬ B1
 Thereforethecoefficientof‫ ݔ‬௡ intheexpansionofሺͳ െ ‫ݔ‬ሻିଵ ሺͳ െ ‫ݔ‬ሻିே isthesumof M1
thecoefficientsofthetermsupto‫ ݔ‬௡ intheexpansionofሺͳ െ ‫ݔ‬ሻିே . A1
௡
ܰ൅݆െͳ ሺܰ ൅ ͳሻ ൅ ݊ െ ͳ ܰ൅݊
෍൬ ൰ൌቀ ቁൌቀ ቁ ሺ‫כ‬ሻ
݆ ݊ ݊
௝ୀ଴
 
(ii) Writeሺͳ ൅ ‫ݔ‬ሻ௠ା௡ asሺͳ ൅ ‫ݔ‬ሻ௠ ሺͳ ൅ ‫ݔ‬ሻ௡ . B1
 Whenmultiplyingthetwoexpansions,termsin‫ ݔ‬௥ willbeobtainedbymultiplyingthe M1
termin‫ ݔ‬௝ fromoneexpansionbythetermin‫ ݔ‬௥ି௝ intheotherexpansion. 
 ݉൅݊ M1
Thecoefficientof‫ ݔ‬௥ intheexpansionofሺͳ ൅ ‫ݔ‬ሻ௠ା௡ isቀ ቁ
‫ݎ‬
 ݉ M1
Thecoefficientof‫ ݔ‬௝ intheexpansionofሺͳ ൅ ‫ݔ‬ሻ௠ isቀ ݆ ቁ
 ݊ M1
Thecoefficientof‫ ݔ‬௥ି௝ intheexpansionofሺͳ ൅ ‫ݔ‬ሻ௡ isቀ‫ ݎ‬െ ݆ቁ
 Therefore,summingoverallpossibilities: A1
௡
݉൅݊ ݉ ݊
ቀ ቁ ൌ ෍ ቀ ݆ ቁ ቀ‫ ݎ‬െ ݆ቁ ሺ‫כ‬ሻ
‫ݎ‬
௝ୀ଴
 
(iii) Writeሺͳ െ ‫ݔ‬ሻே asሺͳ െ ‫ݔ‬ሻேା௠ ሺͳ െ ‫ݔ‬ሻି௠  B1
 ܰ M1
Thecoefficientof‫ ݔ‬௡ intheexpansionofሺͳ െ ‫ݔ‬ሻே isሺെͳሻ௡ ቀ ቁ
݊
 ௡ି௝ ேା௠ ܰ൅݉ ௡ି௝ M1
Thecoefficientof‫ ݔ‬inሺͳ െ ‫ݔ‬ሻ is൬ ൰ ሺെͳሻ 
݊െ݆ A1
 ݉൅݆െͳ M1
Thecoefficientof‫ ݔ‬௝ inሺͳ െ ‫ݔ‬ሻି௠ is൬ ൰
݆
 Therefore M1
௡
ܰ൅݉ ݉൅݆െͳ ܰ
෍൬ ൰ ሺെͳሻ௡ି௝ ൬ ൰ ൌ ሺെͳሻ௡ ቀ ቁ
݊െ݆ ݆ ݊
௝ୀ଴
 Andso, A1
௡
ܰ൅݉ ݉൅݆െͳ ܰ
෍൬ ൰ ሺെͳሻ௝ ൬ ൰ൌቀ ቁ ሺ‫כ‬ሻ
݊െ݆ ݆ ݊
௝ୀ଴
 



819
M1 Fullcalculationwrittendown.
M1 ሺെͳሻfactorsinalltermsdealtwith.
A1 Correctexpression.
B1 Expansionwrittenusingsummationnotation.
B1 Expansionofሺͳ െ ‫ݔ‬ሻିଵ .
M1 Sumthatwillmakeupthecoefficientof‫ ݔ‬௡ identified.
A1 Fullexplanationofgivenresult.
B1 Correctsplittingoftheexpression.
M1 Identificationofthepairsthataretobemultipliedtogether.
M1 Correctstatementofthecoefficientintheexpansionofሺͳ ൅ ‫ݔ‬ሻ௠ା௡ 
M1 Correctstatementofthecoefficientintheexpansionofሺͳ ൅ ‫ݔ‬ሻ௠ 
M1 Correctstatementofthecoefficientintheexpansionofሺͳ ൅ ‫ݔ‬ሻ௡ 
A1 Correctconclusion.
Notethattheanswerisgiven,soeachstepmustbeexplainedclearlytoreceivethemark.
B1 Correctsplittingoftheexpression.
M1 Correctstatementofthecoefficientintheexpansionofሺͳ െ ‫ݔ‬ሻே .
M1 Attempttogetthecoefficientintheexpansionofሺͳ െ ‫ݔ‬ሻேା௠ –awardthemarkifnegative
signisincorrect.
A1 Correctcoefficient.
M1 Correctstatementofthecoefficientintheexpansionofሺͳ െ ‫ݔ‬ሻି௠ .
M1 Combinationofalloftheaboveintothesum.
A1 Correctsimplification.


 



820
Question6

(i) ݀‫ ݕ‬ଶ 
ሺͳ െ ‫ ݔ‬ଶ ሻ ൬
൰ ൅ ‫ ݕ‬ଶ ൌ ͳ
݀‫ݔ‬
 ௗ௬ B1
If‫ ݕ‬ൌ ‫ݔ‬,then ൌ ͳandsoLHSbecomes
ௗ௫
ሺͳ െ ‫ ݔ‬ଶ ሻሺͳሻଶ ൅ ሺ‫ݔ‬ሻଶ ൌ ͳ ൌ ܴ‫ܵܪ‬
 ‫ݕ‬ଵ ሺͳሻ ൌ ͳ,sotheboundaryconditionisalsosatisfied. B1
  
(ii) ݀‫ ݕ‬ଶ
ሺͳ െ ‫ ݔ‬ଶ ሻ ൬ ൰ ൅ Ͷ‫ ݕ‬ଶ ൌ Ͷ
݀‫ݔ‬
 ଶ ௗ௬ M1
If‫ ݕ‬ൌ ʹ‫ ݔ‬െ ͳ,then ൌ Ͷ‫ݔ‬andsoLHSbecomes
ௗ௫
ሺͳ െ ‫ ݔ‬ଶ ሻሺͶ‫ݔ‬ሻଶ ൅ Ͷሺʹ‫ ݔ‬ଶ െ ͳሻଶ ൌ ͳ͸‫ ݔ‬ଶ െ ͳ͸‫ ݔ‬ସ ൅ ͶሺͶ‫ ݔ‬ସ െ Ͷ‫ ݔ‬ଶ ൅ ͳሻ A1
ൌ Ͷ ൌ ܴ‫ܵܪ‬
 ‫ݕ‬ଶ ሺͳሻ ൌ ʹሺͳሻଶ െ ͳ ൌ ͳ,sotheboundaryconditionisalsosatisfied. B1
  
(iii) ଶ ௗ௭ ௗ௬೙ M1
If‫ݖ‬ሺ‫ݔ‬ሻ ൌ ʹ൫‫ݕ‬௡ ሺ‫ݔ‬ሻ൯ െ ͳ,then ൌ Ͷ‫ݕ‬௡ ሺ‫ݔ‬ሻ 
ௗ௫ ௗ௫
A1
 SubstitutingintotheLHSofthedifferentialequation: M1
݀‫ݕ‬௡ ଶ
ሺͳ െ ‫ ݔ‬ଶ ሻ ൬Ͷ‫ݕ‬௡ ൰ ൅ Ͷ݊ଶ ሺʹሺ‫ݕ‬௡ ሻଶ െ ͳሻଶ 
݀‫ݔ‬
 ݀‫ݕ‬௡ ଶ M1
ൌ ͳ͸‫ݕ‬௡ଶ ቈሺͳ െ ‫ ݔ‬ଶ ሻ ൬ ൰ ൅ ݊ଶ ‫ݕ‬௡ଶ െ ݊ଶ ቉ ൅ Ͷ݊ଶ  A1
݀‫ݔ‬
 Since‫ݕ‬௡ isasolutionof(*)when݇ ൌ ݊: A1
ൌ Ͷ݊ଶ
ଶ
 Since‫ݖ‬ሺͳሻ ൌ ʹሺͳሻ െ ͳ ൌ ͳ,‫ݖ‬isasolutionto(*)when݇ ൌ ʹ݊Ǥ M1
 ଶ A1
Therefore‫ݕ‬ଶ௡ ሺ‫ݔ‬ሻ ൌ ʹ൫‫ݕ‬௡ ሺ‫ݔ‬ሻ൯ െ ͳ
 
(iv) ݀‫ݕ݀ ݒ‬௡ ݀‫ݕ‬௠ B1
ൌ ൫‫ݕ‬௠ ሺ‫ݔ‬ሻ൯ ሺ‫ݔ‬ሻ
݀‫ݔ‬ ݀‫ݔ‬ ݀‫ݔ‬
 SubstitutingintoLHSof(*)with݇ ൌ ݉݊ǣ M1
ଶ
݀‫ݕ‬௡ ݀‫ݕ‬௠ ଶ
ሺͳ െ ‫ ݔ‬ଶ ሻ ൬ ൫‫ݕ‬௠ ሺ‫ݔ‬ሻ൯ ሺ‫ݔ‬ሻ൰ ൅ ሺ݉݊ሻଶ ቀ‫ݕ‬௡ ൫‫ݕ‬௠ ሺ‫ݔ‬ሻ൯ቁ 
݀‫ݔ‬ ݀‫ݔ‬
 ଶ M1
݀‫ݕ‬௡ ݀‫ݕ‬௠ ଶ
ൌ ൫‫ݕ‬௠ ሺ‫ݔ‬ሻ൯ ൮ሺͳ െ ‫ ݔ‬ଶ ሻ ൭ ሺ‫ݔ‬ሻ൱ ൲ ൅ ݉ଶ ݊ଶ ቀ‫ݕ‬௡ ൫‫ݕ‬௠ ሺ‫ݔ‬ሻ൯ቁ 
݀‫ݔ‬ ݀‫ݔ‬

 ଶ M1
ௗ௬೘
From(*),ሺͳ െ ‫ ݔ‬ଶ ሻ ൬ ሺ‫ݔ‬ሻ൰ ൌ ݉ଶ െ ݉ଶ ‫ݕ‬௠ ሺ‫ݔ‬ሻଶ 
ௗ௫
 Therefore,wehave: 
݀‫ݕ‬௡ ଶ
൫‫ݕ‬௠ ሺ‫ݔ‬ሻ൯ሺ݉ଶ െ ݉ଶ ‫ݕ‬௠ ሺ‫ݔ‬ሻଶ ሻ ൅ ݉ଶ ݊ଶ ቀ‫ݕ‬௡ ൫‫ݕ‬௠ ሺ‫ݔ‬ሻ൯ቁ 
݀‫ݔ‬
 Let‫ ݑ‬ൌ ‫ݕ‬௠ ሺ‫ݔ‬ሻ,thenthissimplifiesto M1
݀‫ݕ‬௡
݉ଶ ሾሺͳ െ ‫ݑ‬ଶ ሻ ሺ‫ݑ‬ሻ ൅ ݊ଶ ‫ݕ‬௡ ሺ‫ݑ‬ሻଶ ሿ
݀‫ݔ‬
 Andbyapplying(*)when݇ ൌ ݊,thissimplifiesto݉ଶ ݊ଶ ,so‫ݒ‬satisfies(*)when A1
݇ ൌ ݉݊.
 ‫ݒ‬ሺͳሻ ൌ ‫ݕ‬௡ ൫‫ݕ‬௠ ሺͳሻ൯ ൌ ‫ݕ‬௡ ሺͳሻ ൌ ͳ,so‫ݒ‬ሺ‫ݔ‬ሻ ൌ ‫ݕ‬௠௡ ሺ‫ݔ‬ሻ A1
 



821
 B1 Checkthatthefunctionsatisfies thedifferentialequation.
B1 Checkthattheboundaryconditionsaresatisfied.
M1 Differentiationandsubstitution.
A1 Confirmthatthefunctionsatisfiesthedifferentialequation.
B1 Checkthattheboundaryconditionsaresatisfied.
M1 Differentiationof‫ݖ‬.
A1 Fullycorrectderivative.
M1 SubstitutionintoLHSofthedifferentialequation.
M1 Appropriategrouping.
A1 Expressedtoshowtherelationshipwiththegeneraldifferentialequation.
A1 Useofthefactthat‫ݕ‬௡ isasolutionofthedifferentialequationtosimplifytotheRHS.
M1 Checktheboundarycondition.
A1 Concludetherequiredrelationship.
B1 Differentiationof‫ݒ‬.
M1 Substitutionintothecorrectversionofthedifferentialequation.
M1 Groupingoftermstoapplythefactthat‫ݕ‬௠ isasolutionofadifferentialequation.
M1 Useofthedifferentialequation.
M1 Simplificationoftheresultingexpression.
A1 Conclusionthatthissimplifiedto݉ଶ ݊ଶ 
A1 Checkofboundaryconditionandconclusion.


 



822
Question7

 Let‫ ݕ‬ൌ ܽ െ ‫ݔ‬: 

 Limits: B1
‫ ݔ‬ൌ ܽ ՜ ‫ ݕ‬ൌ Ͳ
‫ ݔ‬ൌ Ͳ ՜ ‫ ݕ‬ൌ ܽ
 ݀‫ݕ‬ B1
ൌ െͳ
݀‫ݔ‬
 ௔ ଴ 
න ݂ሺ‫ݔ‬ሻ ݀‫ ݔ‬ൌ െ න ݂ሺܽ െ ‫ݕ‬ሻ ݀‫ݕ‬
଴ ௔
 Swappinglimitsoftheintegralchangesthesign(andwecanreplace‫ ݕ‬by‫ݔ‬inthe B1
integralontheright:
௔ ௔
න ݂ሺ‫ݔ‬ሻ ݀‫ ݔ‬ൌ න ݂ሺܽ െ ‫ݔ‬ሻ ݀‫ݔ‬
଴ ଴
  
(i) Using(*): M1
ଵ ଵ ͳ
ଶ
గ
•‹ ‫ݔ‬ ଶ
గ •‹ሺ ߨ െ ‫ݔ‬ሻ
න ݀‫ ݔ‬ൌ න ʹ ݀‫ ݔ‬
଴ …‘• ‫ ݔ‬൅ •‹ ‫ݔ‬ ͳ ͳ
଴ …‘•ሺ ߨ െ ‫ݔ‬ሻ ൅ •‹ሺ ߨ െ ‫ݔ‬ሻ
ʹ ʹ
 ଵ ଵ A1
గ గ
ଶ •‹ ‫ݔ‬ ଶ …‘• ‫ݔ‬
න ݀‫ ݔ‬ൌ න ݀‫ ݔ‬
଴ …‘• ‫ ݔ‬൅ •‹ ‫ݔ‬ ଴ …‘• ‫ ݔ‬൅ •‹ ‫ݔ‬
 Therefore M1
ଵ ଵ A1
గ గ
ଶ •‹ ‫ݔ‬ ଶ •‹ ‫ ݔ‬൅ …‘• ‫ݔ‬
ʹන ݀‫ ݔ‬ൌ න ݀‫ ݔ‬
଴ …‘• ‫ ݔ‬൅ •‹ ‫ݔ‬ ଴ …‘• ‫ ݔ‬൅ •‹ ‫ݔ‬
ଵ
గ
ଶ
ൌ න ͳ݀‫ ݔ‬
଴
ͳ
ൌ ߨ
ʹ
 ଵ A1
గ
ଶ •‹ ‫ݔ‬ ͳ
න ݀‫ ݔ‬ൌ ߨ
଴ …‘• ‫ ݔ‬൅ •‹ ‫ݔ‬ Ͷ
  
(ii) Using(*): 
ଵ ଵ ͳ
ସ
గ
•‹ ‫ݔ‬ ସ
గ •‹ሺ ߨ െ ‫ݔ‬ሻ
න ݀‫ ݔ‬ൌ න Ͷ ݀‫ ݔ‬
଴ …‘• ‫ݔ‬ ൅ •‹ ‫ݔ‬ ͳ ͳ
଴ …‘•ሺ ߨ െ ‫ݔ‬ሻ ൅ •‹ሺ ߨ െ ‫ݔ‬ሻ
Ͷ Ͷ
 ͳ ξʹ M1
•‹ ቀ ߨ െ ‫ݔ‬ቁ ሺ…‘• ‫ ݔ‬െ •‹ ‫ݔ‬ሻ
Ͷ ൌ ʹ  M1
ͳ ͳ
…‘• ቀ ߨ െ ‫ݔ‬ቁ ൅ •‹ ቀ ߨ െ ‫ݔ‬ቁ ξʹ ሺ…‘• ‫ ݔ‬൅ •‹ ‫ ݔ‬൅ …‘• ‫ ݔ‬െ •‹ ‫ݔ‬ሻ A1
Ͷ Ͷ ʹ
ͳ
ൌ ሺͳ െ –ƒ ‫ݔ‬ሻ
ʹ
 ଵ M1
గ ଵ
ͳ ସ ͳ గ
න ͳ െ –ƒ ‫ ݔ݀ ݔ‬ൌ ሾ‫ ݔ‬െ Žȁ•‡… ‫ݔ‬ȁሿସ଴ 
ʹ ଴ ʹ
 ͳ ͳ A1
ൌ ߨ െ Ž ʹ
ͺ Ͷ
 



823
(iii) Using(*): M1
ଵ ଵ
గ గ
ସ ସ ͳ
න Žሺͳ ൅ –ƒ ‫ݔ‬ሻ ݀‫ ݔ‬ൌ න Ž ൬ͳ ൅ –ƒ ൬ ߨ െ ‫ݔ‬൰൰ ݀‫ ݔ‬
଴ ଴ Ͷ
ଵ
గ
ସ ͳ െ –ƒ ‫ݔ‬
ൌන Ž ൬ͳ ൅ ൰ ݀‫ ݔ‬
଴ ͳ ൅ –ƒ ‫ݔ‬
ଵ
గ
ସ ʹ
ൌන Ž ൬ ൰ ݀‫ ݔ‬
଴ ͳ ൅ –ƒ ‫ݔ‬
భ
 గ ଵ M1
Therefore,if‫ ܫ‬ൌ ‫׬‬଴ Žሺͳ ൅ –ƒ ‫ݔ‬ሻ ݀‫ ݔ‬,then‫ ܫ‬ൌ ߨ Ž ʹ െ ‫ܫ‬
ర
ସ
 ଵ A1
గ
ସ ͳ
න Žሺͳ ൅ –ƒ ‫ݔ‬ሻ ݀‫ ݔ‬ൌ ߨ Ž ʹ
଴ ͺ
  
(iv) Using(*): M1
ଵ ଵ ͳ
ସ
గ
‫ݔ‬ ସ
గ ߨെ‫ݔ‬
‫ܫ‬ൌන ݀‫ ݔ‬ൌ න Ͷ ݀‫ ݔ‬
଴ …‘• ‫ ݔ‬ሺ…‘• ‫ ݔ‬൅ •‹ ‫ݔ‬ሻ ଴ ξʹ
ሺ…‘• ‫ ݔ‬൅ •‹ ‫ݔ‬ሻξʹ …‘• ‫ݔ‬
ʹ
ଵ
గ
ͳ ସ ͳ
ൌ ߨන ݀‫ ݔ‬െ ‫ܫ‬
Ͷ ଴ ሺ…‘• ‫ ݔ‬൅ •‹ ‫ݔ‬ሻ …‘• ‫ݔ‬
 ଵ ଵ M1
గ గ ଵ
ସ ͳ ସ •‡… ଶ ‫ݔ‬ గ
න ݀‫ ݔ‬ൌ න ݀‫ ݔ‬ൌ ሾŽሺͳ ൅ –ƒ ‫ݔ‬ሻሿସ଴  A1
଴ ሺ…‘• ‫ ݔ‬൅ •‹ ‫ݔ‬ሻ …‘• ‫ݔ‬ ଴ ͳ ൅ –ƒ ‫ݔ‬
 Therefore A1
ͳ
ʹ‫ ܫ‬ൌ ߨ Ž ʹ
Ͷ
ͳ
‫ ܫ‬ൌ ߨ Ž ʹ
ͺ
 



824
B1 Considerationofthelimitsoftheintegral.
B1 Completionofthesubstitution.
B1 Clearexplanationaboutchangingthesignwhenswitchinglimits.
M1 Applicationofthegivenresult.
A1 Simplificationofthetrigonometricratios.
M1 Useoftherelationshipbetweenthetwointegrals.
A1 Integrationcompleted.
A1 Finalanswer.
M1 ଵ ଵ
Correctreplacementofeither•‹ ቀ ߨ െ ‫ݔ‬ቁor…‘• ቀ ߨ െ ‫ݔ‬ቁ
ସ ସ
M1 Allfunctionsofଵ ߨ െ ‫ݔ‬replaced.
ସ
A1 Expressionwrittenintermsof–ƒ ‫ݔ‬.
M1 Integrationcompleted.
A1 Limitssubstitutedandintegralsimplified.
M1 Simplificationoftheintegral.
M1 Useofpropertiesoflogarithmstoreachanequationin‫ܫ‬.
A1 Correctvalue.
M1 Rearrangementandsplitintotwointegrals.
M1 ௙ ᇲ ሺ௫ሻ
Rearrangetowriteintheform .
௙ሺ௫ሻ
A1 Correctintegration.
A1 Correctvaluefortheoriginalintegral.


 



825
Question8

 ஶ
ͳ ͳஶ ʹ M1
න ݀‫ݔ‬ ൌ ൤െ ൨ ଵൌ 
ଵ ‫ݔ‬ଶ ‫ ݔ‬௠ି ʹ݉ െ ͳ A1
௠ି ଶ
ଶ
 ଵ B1
Sketchof‫ ݕ‬ൌ మ
௫
 ଵ ଵ ଵ B1
Rectangledrawnwithheight andwidthgoingfrom݉ െ to݉ ൅ 
௠మ ଶ ଶ
 ଵ ଵ ଵ B1
Rectangledrawnwithheight మ andwidthgoingfrom ݊ െ to݊ ൅ 
௡ ଶ ଶ
 Atleastoneotherrectangleinbetween,showingthatnogapsareleftbetweenthe B1
rectangles.
 Anexplanationthattherectangleareasmatchthesum. B1
 
(i) ଶ M1
Taking݉ ൌ ͳandaverylargevalueof݊,theapproximationsfor‫ܧ‬isʹ െ 
ଶ௡ାଵ
 Thereforewith݉ ൌ ͳ,‫ ܧ‬՜ ʹas݊ ՜ λ A1
 ଶ M1
If݉ ൌ ʹ,‫ ܧ‬՜ as݊ ՜ λ
ଷ
 ஶ ଵ ହ A1
Thereforeanapproximationfor‫ܧ‬isͳ ൅ ‫׬‬య ݀‫ ݔ‬ൌ 
మ ௫మ ଷ
 ଶ 
Similarly,if݉ ൌ ͵,‫ ܧ‬՜ as݊ ՜ λ
ହ
 ଵ ஶ ଵ ହ ଶ ଷଷ A1
Thereforeanapproximationfor‫ܧ‬isͳ ൅ ൅ ‫׬‬ఱ ݀‫ ݔ‬ൌ ൅ ൌ 
ସ మ ௫మ ସ ହ ଶ଴
  
(ii) ଵ ଵ M1
௥ା ௥ା
ଶ ͳ ͳ ʹ ʹ ଶ Ͷ
න ݀‫ ݔ‬ൌ ൤െ ൨ ଵ ൌ െ ൌ  A1
௥ି
ଵ ‫ݔ‬ଶ ‫ ݔ‬௥ି ʹ‫ ݎ‬െ ͳ ʹ‫ ݎ‬൅ ͳ Ͷ‫ ݎ‬ଶ െ ͳ
ଶ ଶ
 ସ ଵ ଵ ଵ M1
Theerroris െ ൌ ሺସ௥ మ ൎ forlargevaluesof‫ݎ‬.
ସ௥ మ ିଵ ௥మ ିଵሻ௥ మ ସ௥ ర
A1
  
 Theerrorintheestimatefor‫ ܧ‬isapproximately B1
ஶ
ͳ
෍ 
‫ݎ‬ସ
௥ୀଵ
 ଷଷ M1
Using‫ ܧ‬ൎ ,
ଶ଴
ஶ
ͳ ͵͵
෍ ସ
ൎ െ ͳǤ͸Ͷͷ ൌ ͲǤͲͲͷ
Ͷ‫ݎ‬ ʹͲ
௥ୀଷ
 Therefore: M1
ஶ
ͳ A1
෍ ൎ ͳ ൅ ͲǤͲ͸ʹͷ ൅ ͶሺͲǤͲͲͷሻ ൌ ͳǤͲͺ͵
‫ݎ‬ସ
௥ୀଵ
 



826
M1 Functionintegratedcorrectly.
A1 Limitsapplied.
B1 Sketchonlyrequiredforpositive‫ݔ‬.
B1 Rectanglemusthavecorrectheight.
B1 Rectanglemusthavecorrectheight.
B1 Itmustbeclearthattherearenogapsbetweentherectangles.
B1 Clearjustification.
M1 Correctexpressionforlarge݊.Awardthismarkifthefirstintegralfromthequestionisused
inthesubsequentestimates.
A1 Correctexplanationoftheestimateinthiscase.
M1 Valueofintegralforthecase݉ ൌ ʹ.
A1 Addthefirstvalue.
A1 Applythesameprocessfor݉ ൌ ͵.
M1 Evaluationoftheintegralwithappropriatelimits.
A1 Correctexpression.
M1 Calculationoftheerror.
A1 Clearexplanationthatthegivenvalueisthecorrectapproximation.
B1 Expressionoftheerrorasasum.
M1 Useofmostaccurateestimatefrompart(i)
M1 Rearrangementtomakethesumthesubject.
A1 Correctanswer.


 



827
Question9
ଵ
(i) Kineticenergylostbybulletis ݉‫ݑ‬ଶ  M1
ଶ
 Workdoneagainstresistancesisܴܽ M1
 Energylost=Workdone M1
 ௠௨మ A1
Thereforeܽ ൌ .
ଶோ
  
(ii) Let‫ݒ‬bethevelocityofthecombinedblock andbullet oncethebullethasstopped M1
movingrelativetotheblock. A1
Momentumisconserved,so݉‫ ݑ‬ൌ ሺ‫ ܯ‬൅ ݉ሻ‫ݒ‬
 Inthecasewheretheblockwasstationary,thebulletcomestorestoveradistanceof M1
௨మ A1
ܽ,soitsaccelerationisെ .
ଶ௔
 Considerthemotionofthebulletuntilitcomestorestrelativetotheblock: M1
‫ݑ‬ଶ A1
‫ ݒ‬ଶ ൌ ‫ݑ‬ଶ ൅ ʹ ቆെ ቇ ሺܾ ൅ ܿሻ
ʹܽ
௠௨
 Since‫ ݒ‬ൌ ǣ M1
ெା௠
݉‫ ݑ‬ଶ ‫ݑ‬ଶ
ቀ ቁ ൌ ‫ݑ‬ଶ െ ሺܾ ൅ ܿሻ
‫ܯ‬൅݉ ܽ
 Andso: A1
݉ ଶ
ܽቀ ቁ ൌ ܽ െ ܾ െ ܿ
‫ܯ‬൅݉
௠
 Theaccelerationoftheblockmustbe timestheaccelerationofthebulletinthe M1
ெ
casewheretheblockwasfixed.
௠௨
 Therefore,theblockacceleratesfromresttoaspeedof overadistanceofܿ. M1
ெା௠
 Ԣ‫ ݒ‬ଶ ൌ ‫ݑ‬ଶ ൅ ʹܽ‫ ݏ‬ᇱ ǣ M1
݉‫ ݑ‬ଶ ݉‫ݑ‬ଶ ܿ A1
ቀ ቁ ൌͲ൅ 
‫ܯ‬൅݉ ‫ܽܯ‬
 Therefore: A1
݉ ଶ ݉ܿ
ቀ ቁ ൌ 
‫ܯ‬൅݉ ‫ܽܯ‬
andso
݉‫ܽܯ‬
ܿൌ 
ሺ‫ ܯ‬൅ ݉ሻଶ
 Substitutingtogetܾ: M1
݉ ଶ ݉‫ܽܯ‬
ܽቀ ቁ ൌܽെܾെ 
‫ܯ‬൅݉ ሺ‫ ܯ‬൅ ݉ሻଶ
ଶ
 ݉‫ܯ‬ ݉ M1
ܾ ൌ ܽ ቆͳ െ െ ቇ
ሺ‫ ܯ‬൅ ݉ሻ ଶ ሺ‫ ܯ‬൅ ݉ሻଶ
 ‫ܽܯ‬ A1
ܾൌ 
ሺ‫ ܯ‬൅ ݉ሻଶ
 



828
M1 CalculationoftheKineticEnergy.
M1 Calculationoftheworkdone.
M1 Statementthatthetwoareequal.
A1 Rearrangementtogiveexpressionforܽ.
M1 Considerationofmomentum.
A1 Correctrelationshipstated.
M1 Attempttofindtheaccelerationofthebullet.
A1 Correctexpressionfound.
M1 Applicationoftheaccelerationfoundtothemotionofthebulletwhentheblockmoves.
A1 Correctrelationshipfound.
M1 Useoftherelationshipfoundfrommomentumconsiderations.
A1 Eliminationof‫ݑ‬fromtheequation.
M1 Statementoftherelationshipbetweenthetwoaccelerations.
M1 Correctidentificationoftheotherinformationrelatingtotheuniformaccelerationofthe
block.
M1 Useof‫ ݒ‬ଶ ൌ ‫ݑ‬ଶ ൅ ʹܽ‫ݏ‬
A1 Relationshipfound.
A1 Simplificationtogetexpressionforܿ.
M1 Substitutionintootherequation.
M1 Rearrangementtomakeܾthesubject.
A1 Correctexpression.


 



829
Question10

 Findthecentreofmassofthetriangle: M1
Letthetwosidesofthetrianglewithequallengthhavelengthܾandtheotherside
havelengthʹܽ.
Let‫ݔ‬ҧ bethedistanceofthecentreofmassfromtheside‫ܥܤ‬andalongthelineof
symmetry.
 ͳ M1
ሺʹܽ ൅ ʹܾሻ‫ݔ‬ҧ ൌ ʹܾ ൬ ܾ …‘• ߠ൰
ʹ M1
 ܾ ଶ …‘• ߠ A1
‫ݔ‬ҧ ൌ 
ʹሺܽ ൅ ܾሻ
  
 Letthepointofcontactbetweenthetriangleandthepegbeadistance‫ ݕ‬fromthe B1
midpointof‫ܥܤ‬. B1
Lettheweightofthetrianglebeܹ,thereactionforceatthepegbeܴandthe B1
frictionalforceatthepegbe‫ܨ‬.
Lettheanglebetween‫ܥܤ‬andthehorizontalbeߙ.
 Resolvingparallelto‫ܥܤ‬: M1
‫ ܨ‬ൌ ܹ •‹ ߙ A1
 Resolvingperpendicularto‫ܥܤ‬: M1
ܴ ൌ ܹ …‘• ߙ A1
 ‫ݕ‬ B1
–ƒ ߙ ൌ 
‫ݔ‬ҧ
 Topreventslipping: M1
‫ ܨ‬൑ ߤܴ
 ߤ ൒ –ƒ ߙ A1
  
 Therefore M1
ʹ‫ݕ‬ሺܽ ൅ ܾሻ A1
ߤ൒ ଶ 
ܾ …‘• ߠ
 and‫ݕ‬cantakeanyvalueuptoܽ,sothelimitonߤiswhen‫ ݕ‬ൌ ܽ. M1
 ʹܽሺܽ ൅ ܾሻ
ߤ൒ ଶ 
ܾ …‘• ߠ
 Sinceܽ ൌ ܾ •‹ ߠǣ M1
 ʹ •‹ ߠ ሺ•‹ ߠ ൅ ͳሻ M1
ߤ൒ ൌ ʹ –ƒ ߠ ሺͳ ൅ •‹ ߠሻ
…‘• ߠ A1
 



830
M1 Notationsdevisedtoallowcalculationstobecompleted.Maybeseenonadiagram.
M1 Correctpositionsofcentresofmassesforindividualpieces.
M1 Correctequationwrittendown.
A1 Positionofcentreofmassfound.
B1 Specificationofavariabletorepresentthepositionofthecentreofmass.
B1 Notationsforalloftheforces.
B1 Anappropriateangleidentified.(Allthreeofthesemarksmaybeawardedforsightofthe
featuresonadiagram).
M1 Resolvinginonedirection.
A1 Correctequationstated.Mustuseangleߠ.
M1 Resolvinginanotherdirection.
A1 Correctequationstated.Mustuseangleߠ.
B1 Statementofthevalueof–ƒ ߙ.
M1 Useofcoefficientoffriction.
A1 Correctconclusion.
M1 Substitutionfortheangle.
A1 Correctinequality.
M1 Identificationofthelimitingcase.
M1 Eliminationofthesidelengths.
M1 Inequalityonlyintermsofߠfound.
A1 Correctanswer.


 



831
Question11

(i) Sincetheparticlescollidethereisavalueof‫ݐ‬suchthat M1
ܽ ൅ ‫ ߙ •‘… ݐݑ‬ൌ ‫ߚ •‘… ݐݒ‬
‫ ߙ ‹• ݐݑ‬ൌ ܾ ൅ ‫ߚ ‹• ݐݒ‬
 Multiplythefirstequationbyܾandmakeܾܽthesubject: M1
ܾܽ ൌ ܾ‫ ߚ •‘… ݐݒ‬െ ܾ‫ߙ •‘… ݐݑ‬
 Multiplythesecondequationbyܽandmakeܾܽthesubject: M1
ܾܽ ൌ ܽ‫ ߙ ‹• ݐݑ‬െ ܽ‫ߚ ‹• ݐݒ‬
 Equating: M1
ܾ‫ ߚ •‘… ݐݒ‬െ ܾ‫ ߙ •‘… ݐݑ‬ൌ ܽ‫ ߙ ‹• ݐݑ‬െ ܽ‫ߚ ‹• ݐݒ‬
andso:
ܽ‫ ߙ ‹• ݐݑ‬൅ ܾ‫ ߙ •‘… ݐݑ‬ൌ ܾ‫ ߚ •‘… ݐݒ‬൅ ܽ‫ߚ ‹• ݐݒ‬
 ܽ‫ ߙ ‹• ݐݑ‬൅ ܾ‫ ߙ •‘… ݐݑ‬ൌ ܴଵ •‹ሺߙ ൅ ߠଵ ሻ M1
whereܴଵଶ ൌ ሺܽ‫ݐݑ‬ሻଶ ൅ ሺܾ‫ݐݑ‬ሻଶ  A1
௕ A1
and–ƒ ߠଵ ൌ 
௔
 ܾ‫ ߚ •‘… ݐݒ‬൅ ܽ‫ ߚ ‹• ݐݒ‬ൌ ܴଶ •‹ሺߚ ൅ ߠଶ ሻ M1
whereܴଶଶ ൌ ሺܽ‫ݐݒ‬ሻଶ ൅ ሺܾ‫ݐݒ‬ሻଶ  A1
௕ A1
and–ƒ ߠଶ ൌ 
௔
 Sinceߠଵ ൌ ߠଶ : M1
ܴଵ •‹ሺߠ ൅ ߙሻ ൌ ܴଶ •‹ሺߠ ൅ ߚሻ A1
andsince‫ܴݒ‬ଵ ൌ ‫ܴݑ‬ଶ ǣ
‫‹• ݑ‬ሺߠ ൅ ߙሻ ൌ ‫‹• ݒ‬ሺߠ ൅ ߚሻ ሺ‫כ‬ሻ
  
(ii) Vertically: M1
ଵ
Bullet’sheightabovethegroundattime‫ݐ‬isܾ ൅ ‫ ߚ ‹• ݐݒ‬െ ݃‫ ݐ‬ଶ  M1
ଶ
ଵ A1
Target’sheightabovethegroundattime‫ݐ‬is‫ ߙ ‹• ݐݑ‬െ ݃‫ ݐ‬ଶ 
ଶ
௕
Thereforethecollisionmustoccurwhen‫ ݐ‬ൌ 
௨ ୱ୧୬ ఈି௩ ୱ୧୬ ఉ
 ௕௨ ୱ୧୬ ఈ ଵ ௕ ଶ 
Theverticalheightofthetargetatthistimeis െ ݃ቀ ቁ 
௨ ୱ୧୬ ఈି௩ ୱ୧୬ ఉ ଶ ௨ ୱ୧୬ ఈି௩ ୱ୧୬ ఉ A1
 Ifthisisbeforeitreachestheground: M1
ଶ
ܾ‫ߙ ‹• ݑ‬ ͳ ܾ
െ ݃൬ ൰ ൐ Ͳ
‫ ߙ ‹• ݑ‬െ ‫ ߙ ‹• ݑ ʹ ߚ ‹• ݒ‬െ ‫ߚ ‹• ݒ‬
 Therefore:
ʹܾ‫ ߙ ‹• ݑ‬ሺ‫ ߙ ‹• ݑ‬െ ‫ߚ ‹• ݒ‬ሻ െ ܾ ଶ ݃ ൐ Ͳ
 ʹ‫ ߙ ‹• ݑ‬ሺ‫ ߙ ‹• ݑ‬െ ‫ߚ ‹• ݒ‬ሻ ൐ ܾ݃ A1
  
 Boththebulletandtargetareaffectedequallybygravity,soanycollisionwould B1
correspondtothetimeforthestraightlinemotioninpart(i)
 Inpart(i)therecanclearlyonlybeacollisionifߙ ൐ ߚ B1
 



832
M1 Pairofequationsstated.
M1 Makeܾܽthesubjectofthefirstequation.
M1 Makeܾܽthesubjectofthesecondequation.
M1 Putthetwotogether.
M1 Rewriteintheformܴ •‹ሺߙ ൅ ߠሻ.
A1 Correctvalueofܴ.
A1 Correctvalueof–ƒ ߠ.
M1 Rewriteintheformܴ •‹ሺߚ ൅ ߠሻ.
A1 Correctvalueofܴ.
A1 Correctvalueof–ƒ ߠ.
M1 Identifythatthetwovaluesofߠareequal.
A1 Usetherelationshipbetweenthevaluesofܴ toreachthecorrectanswer.
M1 Considerthemotionofthebulletvertically.
M1 Considerthemotionofthetargetvertically.
A1 Findthevalueof‫ݐ‬forwhichthecollisionoccurs.
A1 Substitutethevalueof‫ݐ‬intooneoftheexpressionsfortheheight.
M1 Stateasaninequality.
A1 Rearrangetoreachtherequiredinequality.
B1 Relationshipwithpart(i)identified.
B1 Requiredconditionforacollisiontotakeplacein(i)identified.


 



833
Question12

 ܲሺ‫ܥ ׫ ܤ ׫ ܣ‬ሻ ൌ ܲ൫ሺ‫ܤ ׫ ܣ‬ሻ ‫ܥ ׫‬൯ ൌ ܲሺ‫ܤ ׫ ܣ‬ሻ ൅ ܲሺ‫ܥ‬ሻ െ ܲሺሺ‫ܤ ׫ ܣ‬ሻ ‫ܥ ת‬ሻ M1
 ܲ൫ሺ‫ ܤ ׫ ܣ‬ሻ ‫ ܥ ת‬൯ ൌ ܲ ሺ ሺ ‫ ܥ ת ܣ‬ሻ ‫ ׫‬ሺ ‫ ) ܥ ת ܤ‬ሻ M1
ൌ ܲሺ‫ܥ ת ܣ‬ሻ ൅ ܲሺ‫ܥ ת ܤ‬ሻ െ ܲሺሺ‫ܥ ת ܣ‬ሻ ‫ ת‬ሺ‫ܥ ת ܤ‬ሻሻ
 ܲ൫ሺ‫ܥ ת ܣ‬ሻ ‫ ת‬ሺ‫ܥ ת ܤ‬ሻ൯ ൌ ܲሺ‫ܥ ת ܤ ת ܣ‬ሻ M1
 Therefore: A1
ܲሺ‫ܥ ׫ ܤ ׫ ܣ‬ሻ ൌ ܲሺ‫ܣ‬ሻ ൅ ܲሺ‫ܤ‬ሻ ൅ ܲሺ‫ܥ‬ሻ െ ܲሺ‫ܤ ת ܣ‬ሻ െ ܲሺ‫ܥ ת ܤ‬ሻ െ ܲሺ‫ܣ ת ܥ‬ሻ ൅
ܲሺ‫ܥ ת ܤ ת ܣ‬ሻ
  
 ܲሺ‫ܦ ׫ ܥ ׫ ܤ ׫ ܣ‬ሻ ൌ ܲሺ‫ܣ‬ሻ ൅ ܲሺ‫ܤ‬ሻ ൅ ܲሺ‫ܥ‬ሻ ൅ ܲሺ‫ܦ‬ሻ B1
െܲሺ‫ܤ ת ܣ‬ሻ െ ܲሺ‫ܥ ת ܣ‬ሻ െ ܲሺ‫ܦ ת ܣ‬ሻ B1
െܲሺ‫ܥ ת ܤ‬ሻ െ ܲሺ‫ܦ ת ܤ‬ሻ െ ܲሺ‫ܦ ת ܥ‬ሻ
൅ܲሺ‫ܥ ת ܤ ת ܣ‬ሻ ൅ ܲሺ‫ܦ ת ܤ ת ܣ‬ሻ ൅ ܲሺ‫ܦ ת ܥ ת ܣ‬ሻ ൅ ܲሺ‫ܦ ת ܥ ת ܤ‬ሻ
െܲሺ‫ܦ ת ܥ ת ܤ ת ܣ‬ሻ
 
ଵ
(i) ܲሺ‫ܧ‬௜ ሻ ൌ  B1
௡
  
(ii) Thereareatotalof݊Ǩarrangementspossible. M1
 ሺ݊ െ ʹሻǨofthesewillhavethe݅thand݆thinthecorrectposition. M1
 ଵ A1
ܲ൫‫ܧ‬௜ ‫ܧ ת‬௝ ൯ ൌ 
௡ሺ௡ିଵሻ
 
(iii) ଵ M1
Bysimilarreasoningto(ii)theprobabilitywillbe 
௡ሺ௡ିଵሻሺ௡ିଶሻ
  M1
  A1
  
 Atleastonecardisinthepositionasthenumberitbearsistheunionofallofthe‫ܧ‬௜ s B1
 M1
ܲ ൭ ራ ‫ܧ‬௜ ൱ ൌ ෍ ܲሺ‫ܧ‬௜ ሻ െ ෍ ܲ൫‫ܧ‬௜ ‫ܧ ת‬௝ ൯ ൅ ෍ ܲ൫‫ܧ‬௜ ‫ܧ ת‬௝ ‫ܧ ת‬௞ ൯ െ ‫ڮ‬
ଵஸ௜ஸ௡ ଵஸ௜ஸ௡ ଵஸ௜ழ௝ஸ௡ ଵஸ௜ழ௝ழ௞ஸ௡
൅ ሺെͳሻ௡ାଵ ܲሺ‫ܧ‬ଵ ‫ܧ ת‬ଶ ‫ ת‬ǥ ‫ܧ ת‬௡ ሻ
 ͳ ݊ ͳ ݊ ͳ M1
ܲ ൭ ራ ‫ܧ‬௜ ൱ ൌ ݊ ൈ െ ቀ ቁ ൈ ൅ቀ ቁൈ െ ‫ڮ‬ M1
݊ ʹ ݊ሺ݊ െ ͳሻ ͵ ݊ሺ݊ െ ͳሻሺ݊ െ ʹሻ
ଵஸ௜ஸ௡
ͳ
൅ሺെͳሻ௡ାଵ ൈ 
݊ሺ݊ െ ͳሻሺ݊ െ ʹሻ ǥ ʹ ൈ ͳ
 ͳ ͳ ͳ A1
ܲ ൭ ራ ‫ܧ‬௜ ൱ ൌ ͳ െ ൅ െ ‫ ڮ‬൅ ሺെͳሻ௡ାଵ 
ʹǨ ͵Ǩ ݊Ǩ
ଵஸ௜ஸ௡
 
 Theprobabilitythatnocardsareinthesamepositionasthenumbertheybearis M1
ͳ ͳ ͳ
െ ൅ ‫ ڮ‬൅ ሺെͳሻ௡ 
ʹǨ ͵Ǩ ݊Ǩ
 Thereforetheprobabilitythatexactlyonecardisinthesamepositionasthenumber
itbearsis݊ ൈ ܲሺ‫ܧ‬ଵ ሻ ൈtheprobabilitythatnocardfromasetofሺ݊ െ ͳሻisinthe
samepositionasthenumberitbears.
 ͳ ͳ ͳ A1
െ ൅ ‫ ڮ‬൅ ሺെͳሻ௡ିଵ 
ʹǨ ͵Ǩ ሺ݊ െ ͳሻǨ
 



834
M1 Applicationofthegivenresultappliedforsomesplittingof‫ ܥ ׫ ܤ ׫ ܣ‬intotwosets.
M1 Correcthandlingoftheintersectionterminpreviousline.
M1 Intersectionscorrectlyinterpreted.
A1 Fullycorrectstatement.
B1 Allpairwiseintersectionsincluded.
B1 Allothertermsincluded.
B1 Correctanswer.
M1 Totalnumberofarrangementsfound.
OR
Atreediagramdrawn.
M1 Numberofarrangementswheretwoareintherightplacefound.
OR
Correctprobabilitiesonthetreediagram.
A1 Correctprobability.
M1 Totalnumberofarrangementsfound.
OR
Atreediagramdrawn.
M1 Numberofarrangementswheretwoareintherightplacefound.
OR
Correctprobabilitiesonthetreediagram.
A1 Correctprobability.
B1 Identificationoftherequiredeventintermsoftheindividual‫ܧ‬௜ s.
M1 Useofthegeneralisationoftheformulafromthestartofthequestion(precisenotationnot
required).
M1 Atleastoneoftheindividualsumsworkedoutcorrectly.
M1 Allofthepartsofthesumworkedoutcorrectly.
A1 Correctanswer.
M1 Probabilityofnocardincorrectpositionfound.
A1 Correctanswer.


 



835
Question13

(i) ଵ ଵହ ଵ଻ B1
̱ܺ‫ܤ‬ሺͳ͸ǡ ሻisapproximatedbyܻ̱ܰሺͺǡͶሻ,soܲሺܺ ൌ ͺሻ ൎ ܲሺ ൏ܻ൏ ሻ
ଶ ଶ ଶ
B1
 ଵ ଵ A1
Intermsofܼ̱ܰሺͲǡͳሻ,thisisܲሺെ ൏ ܼ ൏ ሻ
ସ ସ
 Theprobabilityisthereforegivenby M1
ଵ
ସ
ͳ ଵ మ
න ݁ ିଶ௫ ݀‫ݔ‬
ଵ
ξʹߨ
ି
ସ
 ଵ ଵ M1
Thiscanbeapproximatedasarectanglewithawidthof andaheightof .
ଶ ξଶగ
ଵ A1
Theareaistherefore 
ଶξଶగ
ͳ
ܲሺܺ ൌ ͺሻ ൎ 
ʹξʹߨ
 
(ii) ଵ ௡ ଶ௡ିଵ ଶ௡ାଵ B1
̱ܺ‫ܤ‬ሺʹ݊ǡ ሻcanbeapproximatedbyܻ̱ܰሺ݊ǡ ሻ,soܲሺܺ ൌ ݊ሻ ൎ ܲሺ ൏ܻ൏ ሻ
ଶ ଶ ଶ ଶ
B1
 Inthesamewayaspart(i)ܲሺܺ ൌ ݊ሻcanbeapproximatedbyarectangleofheight M1
ଵ ଶ A1
.Thewidthwillnowbeට .
ξଶగ ௡
 Therefore: M1
ሺʹ݊ሻǨ ͳ ଶ௡ ͳ A1
ܲሺܺ ൌ ݊ሻ ൌ ൬ ൰ ൎ  A1
݊Ǩ ݊Ǩ ʹ ξ݊ߨ
 Rearranginggives: B1
ʹଶ௡ ሺ݊Ǩሻଶ
ሺʹ݊ሻǨ ൎ ሺ‫כ‬ሻ
ξ݊ߨ
  
(iii) ଶ௡ିଵ ଶ௡ାଵ B1
̱ܺܲ‫݋‬ሺ݊ሻcanbeapproximatedbyܻ̱ܰሺ݊ǡ ݊ሻ,soܲሺܺ ൌ ݊ሻ ൎ ܲሺ ൏ܻ൏ ሻ
ଶ ଶ
 Inthesamewayaspart(i)ܲሺܺ ൌ ݊ሻ canbeapproximatedbyarectangleofheight M1
ଵ ଵ ଵ A1
.Thewidthwillnowbeට .Theareaistherefore .
ξଶగ ௡ ξଶగ௡ 
 Therefore:
 ݁ ି௡ ݊௡ ͳ M1
ൎ 
݊Ǩ ξʹߨ݊ A1
 Whichsimplifiesto: A1
݊Ǩ ൎ ξʹߨ݊݁ ି௡ ݊௡
 



836
 B1 Correctapproximation.
B1 Probabilitywithcontinuitycorrectionapplied.
A1 Convertedtostandardnormaldistribution.
M1 Expressionoftheprobabilityasanintegral.
M1 Useofarectangletoapproximatethearea.
A1 Correctanswer.
B1 Correctapproximation.
B1 Probabilitywithcontinuitycorrectionapplied.
M1 Useofarectangletoapproximatethearea.
A1 Correctdimensionsintermsof݊.
M1 UseofformulaforBinomialprobability.
A1 Correctsubstitution.
A1 Correctvalueforapproximation.
B1 Rearrangetogiveanswerfromthequestion.
B1 Correctapproximation.
M1 Onedimensionfortheapproximatingrectanglecorrect.
A1 Correctapproximation.
M1 UseofformulaforPoissonprobability.
A1 Correctsubstitution.
A1 Simplificationtotherequiredform.




837
 ^dW///ϮϬϭϲDARK^CHEME

1. (i)
ஶ
ͳ
‫ܫ‬ଵ ൌ න ݀‫ݔ‬
‫ݔ‬ଶ ൅ ʹܽ‫ ݔ‬൅ ܾ
ିஶ

‫ ݔ‬൅ ܽ ൌ ඥܾ െ ܽଶ –ƒ ‫ݑ‬

݀‫ݔ‬
ൌ ඥܾ െ ܽଶ •‡… ଶ ‫ݑ‬
݀‫ݑ‬
M1A1
గ గ
ଶ ଶ
ξܾ െ ܽଶ •‡… ଶ ‫ݑ‬ ξܾ െ ܽଶ •‡… ଶ ‫ݑ‬
‫ܫ‬ଵ ൌ න ݀‫ݑ‬ ൌ න ݀‫ ݑ‬
ሺܾ െ ܽଶ ሻ –ƒଶ ‫ ݑ‬൅ ሺܾ െ ܽଶ ሻ ሺܾ െ ܽଶ ሻ •‡… ଶ ‫ݑ‬
ିగ ିగ
ଶ ଶ
M1A1 M1
గ
ଶ
ͳ ߨ
‫ܫ‬ଵ ൌ න ͳ݀‫ ݑ‬ൌ
ξܾ െ ܽଶ ିగ ξܾ െ ܽଶ
ଶ
A1*  (6)
(ii)
ஶ ஶ
ͳ ‫ݔ‬ ஶ െʹ݊‫ݔ‬ሺ‫ ݔ‬൅ ܽሻ
‫ܫ‬௡ ൌ න ଶ ݀‫ݔ‬ ൌ ൤ ൨ െ න ଶ ݀‫ ݔ‬
ሺ‫ ݔ‬൅ ʹܽ‫ ݔ‬൅ ܾሻ ௡ ሺ‫ ݔ‬൅ ʹܽ‫ ݔ‬൅ ܾሻ ିஶ
ଶ ௡ ሺ‫ ݔ‬൅ ʹܽ‫ ݔ‬൅ ܾሻ௡ାଵ
ିஶ ିஶ
M1A1

ஶ ஶ
‫ ݔ‬ଶ ൅ ܽ‫ݔ‬ ‫ ݔ‬ଶ ൅ ʹܽ‫ ݔ‬൅ ܾ ܽ‫ ݔ‬൅ ܾ
‫ܫ‬௡ ൌ ʹ݊ න ଶ ݀‫ݔ‬ ൌ ʹ݊ න െ ଶ ݀‫ݔ‬
ሺ‫ ݔ‬൅ ʹܽ‫ ݔ‬൅ ܾሻ ௡ାଵ ሺ‫ ݔ‬൅ ʹܽ‫ ݔ‬൅ ܾሻ
ଶ ௡ାଵ ሺ‫ ݔ‬൅ ʹܽ‫ ݔ‬൅ ܾሻ௡ାଵ
ିஶ ିஶ
M1A1
ܽ ஶ
ሺʹ‫ ݔ‬൅ ʹܽሻ ሺܾ െ ܽଶ ሻ
‫ܫ‬௡ ൌ ʹ݊‫ܫ‬௡ െ ʹ݊ න ଶ ʹ ൅ ݀‫ݔ‬
ሺ‫ ݔ‬൅ ʹܽ‫ ݔ‬൅ ܾሻ௡ାଵ ሺ‫ ݔ‬ଶ ൅ ʹܽ‫ ݔ‬൅ ܾሻ௡ାଵ
ିஶ
M1
െܽ ஶ

‫ܫ‬௡ ൌ ʹ݊‫ܫ‬௡ െ ʹ݊ ቎ ଶ ʹ݊ ቏ െ ʹ݊ሺܾ െ ܽଶ ሻ‫ܫ‬௡ାଵ 

ሺ‫ ݔ‬൅ ʹܽ‫ ݔ‬൅ ܾሻ௡
ିஶ
A1

ʹ݊ሺܾ െ ܽଶ ሻ‫ܫ‬௡ାଵ ൌ ሺʹ݊ െ ͳሻ‫ܫ‬௡

A1*(7)

గ ʹ݇ െ ʹ
(iii)Suppose‫ܫ‬௞ ൌ భ ቀ ቁforsomeintegerk,݇ ൒ ͳ
ೖష
ଶమೖషమ ሺ௕ି௔మ ሻ మ ݇െͳ
B1
ଶ௞ିଵ గ ʹ݇ െ ʹ గ ଶሺଶ௞ିଵሻ ʹ݇ െ ʹ
Then‫ܫ‬௞ାଵ ൌ ቀ ቁൌ భ ൈ ቀ ቁ
ଶ௞ሺ௕ି௔మ ሻ మೖషమ ሺ௕ି௔మ ሻೖషభమ ݇ െ ͳ ೖశ ௞ ݇െͳ
ଶ ଶమೖ ሺ௕ି௔మ ሻ మ
M1

 838
 ʹሺʹ݇ െ ͳሻ ʹ݇ െ ʹ ʹሺʹ݇ െ ͳሻ ሺʹ݇ െ ʹሻǨ ʹ݇ሺʹ݇ െ ͳሻ ሺʹ݇ െ ʹሻǨ ʹ݇
ቀ ቁൌ ൌ ൌ ቀ ቁ
݇ ݇െͳ ݇ ሺ݇ െ ͳሻǨ ሺ݇ െ ͳሻǨ ݇݇ ሺ݇ െ ͳሻǨ ሺ݇ െ ͳሻǨ ݇

soresulttruefork+1.M1A1

For݊ ൌ ͳ,

ߨ ʹ݊ െ ʹ ߨ Ͳ ߨ
ଵቀ ݊െͳ
ቁൌ ଵ ቀͲቁ ൌ ଵ
ʹଶ௡ିଶ ሺܾ െ ܽଶ ሻ௡ିଶ ሺܾ െ ܽ ଶ ሻଶ ሺܾ െ ܽଶ ሻଶ

  M1A1


whichisthecorrectresult.

Soresulthasbeenprovedby(principleof)(mathematical)induction.dB1(7)

 

 839
 ௗ௫
2.(i) ‫ ݔ‬ൌ ܽ‫ ݐ‬ଶ ֜ ൌ ʹܽ‫ݐ‬
ௗ௧

ௗ௬
 ‫ ݕ‬ൌ ʹܽ‫֜ ݐ‬ ൌ ʹܽ
ௗ௧
ௗ௬ ଶ௔ ଵ
So ൌ ൌ 
ௗ௫ ଶ௔௧ ௧

Thus,thegradientofthenormalatQisെ‫ݍ‬.M1A1
ଶ௔௣ିଶ௔௤ ଶ௔ሺ௣ି௤ሻ ଶ
ButthisnormalisthechordPQandsohasgradient ൌ ൌ M1A1
௔௣మ ି௔௤ మ ௔ሺ௣ି௤ሻሺ௣ା௤ሻ ௣ା௤

ଶ
Soെ‫ ݍ‬ൌ whichrearrangesto‫ ݍ‬ଶ ൅ ‫ ݌ݍ‬൅ ʹ ൌ ͲA1*(5)
௣ା௤

(ii) Similarly,‫ ݎ‬ଶ ൅ ‫ ݌ݎ‬൅ ʹ ൌ Ͳ

 B1
ି൫ଶା௤ మ ൯ ି൫ଶା௥ మ ൯
Makingpthesubjectofeachresult,‫ ݌‬ൌ ൌ 
௤ ௥

Soʹ‫ ݎ‬െ ʹ‫ ݍ‬൅ ‫ ݍ‬ଶ ‫ ݎ‬െ ‫ ݎݍ‬ଶ ൌ Ͳ

ʹሺ‫ ݎ‬െ ‫ݍ‬ሻ െ ‫ݎݍ‬ሺ‫ ݎ‬െ ‫ݍ‬ሻ ൌ Ͳ

Asሺ‫ ݎ‬െ ‫ݍ‬ሻ ് Ͳ,‫ ݎݍ‬ൌ ʹM1A1
௬ିଶ௔௤ ௬ିଶ௔௥
ThelineQRis ൌ 
௫ି௔௤ మ ௫ି௔௥ మ

Thatis‫ ݕݔ‬െ ʹܽ‫ ݔݍ‬െ ܽ‫ ݎ‬ଶ ‫ ݕ‬൅ ʹܽଶ ‫ ݎݍ‬ଶ ൌ ‫ ݕݔ‬െ ʹܽ‫ ݔݎ‬െ ܽ‫ ݍ‬ଶ ‫ ݕ‬൅ ʹܽଶ ‫ ݍ‬ଶ ‫ݎ‬

ʹܽ‫ݔ‬ሺ‫ ݎ‬െ ‫ݍ‬ሻ െ ܽሺ‫ ݎ‬ଶ െ ‫ ݍ‬ଶ ሻ‫ ݕ‬൅ ʹܽଶ ‫ݎݍ‬ሺ‫ ݎ‬െ ‫ݍ‬ሻ ൌ Ͳ

Again,asሺ‫ ݎ‬െ ‫ݍ‬ሻ ് Ͳ,and് Ͳ,ʹ‫ ݔ‬െ ሺ‫ ݎ‬൅ ‫ݍ‬ሻ‫ ݕ‬൅ ʹܽ‫ ݎݍ‬ൌ ͲM1A1

Because‫ ݎݍ‬ൌ ʹ,QRisʹ‫ ݔ‬െ ሺ‫ ݎ‬൅ ‫ݍ‬ሻ‫ ݕ‬൅ Ͷܽ ൌ Ͳ

Sowhen‫ ݕ‬ൌ Ͳ,‫ ݔ‬ൌ െʹܽandthusሺെʹܽǡ Ͳሻisasuitablefixedpoint.B1(6)

(iii)Because‫ ݍ‬ଶ ൅ ‫ ݌ݍ‬൅ ʹ ൌ Ͳand‫ ݎ‬ଶ ൅ ‫ ݌ݎ‬൅ ʹ ൌ Ͳsubtractinggives

‫ ݍ‬ଶ െ ‫ ݎ‬ଶ ൅ ‫ ݌ݍ‬െ ‫ ݌ݎ‬ൌ Ͳ

Again,asሺ‫ ݎ‬െ ‫ݍ‬ሻ ് Ͳ,‫ ݍ‬൅ ‫ ݎ‬൅ ‫ ݌‬ൌ ͲM1A1

SothelineQRisʹ‫ ݔ‬൅ ‫ ݕ݌‬൅ Ͷܽ ൌ ͲM1

ଶ௔௣ ଶ
ThelineOPis‫ ݕ‬ൌ ‫ݔ‬i.e.‫ ݕ‬ൌ ‫ݔ‬B1
௔௣మ ௣

ିଶ௔
ThustheintersectionofQRandOPisatቀെܽǡ ቁ,whichlieson‫ ݔ‬ൌ െܽ.B1(5)
௣

ିଶ ଶ௤ ଶ௥
ൌ ൌ 
௣ ଶା௤ మ ଶା௥ మ

ିଶ
Suppose݇ ൌ ,then݇‫ ݍ‬ଶ െ ʹ‫ ݍ‬൅ ʹ݇ ൌ Ͳ M1
௣

 840
Asthisequationhastwodistinctrealroots,qandr,thenthediscriminantispositiveM1

andso
ଵ ଵ ଵ
Ͷ െ ͺ݇ ଶ ൐ ͲA1so݇ ଶ ൏ thatisെ ൏݇൏ whichmeansthatthedistanceofthatpointof
ଶ ξଶ ξଶ
௔
intersectionislessthan fromthexaxis.A1*(4)
ξଶ

 

 841
3.(i)
݀ ܲ݁ ௫ ܳሺܲ݁ ௫ ൅ ܲԢ݁ ௫ ሻ െ ܲ݁ ௫ ܳԢ
ቆ ቇൌ 
݀‫ܳ ݔ‬ ܳଶ

ௗ ௉௘ ೣ ௫ య ିଶ
Itisrequiredthat ቀ ቁ ൌ ሺ௫ାଵሻమ ݁ ௫ soitfollowsthat
ௗ௫ ொ

ሾܳሺܲ݁ ௫ ൅ ܲԢ݁ ௫ ሻ െ ܲ݁ ௫ ܳԢሿሺ‫ ݔ‬൅ ͳሻଶ ൌ ሺ‫ ݔ‬ଷ െ ʹሻ݁ ௫ ܳ ଶ 

    M1A1

Thus,

ሾܳሺܲ ൅ ܲԢሻ െ ܲܳԢሿሺ‫ ݔ‬൅ ͳሻଶ ൌ ሺ‫ ݔ‬ଷ െ ʹሻܳ ଶ 

Letting‫ ݔ‬ൌ െͳ,Ͳ ൌ െ͵ሾܳሺെͳሻሿଶ soܳሺെͳሻ ൌ ͲandthusQhasafactorሺ‫ ݔ‬൅ ͳሻasrequired.

   M1A1(4)

Supposethatthedegreeofܲሺ‫ݔ‬ሻispandthatofܳሺ‫ݔ‬ሻisq.   M1

ThenthedegreeofܲԢሺ‫ݔ‬ሻis‫ ݌‬െ ͳandofܳԢሺ‫ݔ‬ሻis‫ ݍ‬െ ͳ

Soܲ ൅ ܲԢhasdegreep,ܳሺܲ ൅ ܲԢሻhasdegree‫ ݌‬൅ ‫ݍ‬,ܲܳԢhasdegree‫ ݌‬൅ ‫ ݍ‬െ ͳ,
ሾܳሺܲ ൅ ܲԢሻ െ ܲܳԢሿhasdegree‫ ݌‬൅ ‫ݍ‬,andthusሾܳሺܲ ൅ ܲԢሻ െ ܲܳԢሿሺ‫ ݔ‬൅ ͳሻଶ hasdegree‫ ݌‬൅ ‫ ݍ‬൅ ʹ

      A1

ሺ‫ ݔ‬ଷ െ ʹሻܳଶ hasdegreeʹ‫ ݍ‬൅ ͵

Thus‫ ݌‬൅ ‫ ݍ‬൅ ʹ ൌ ʹ‫ ݍ‬൅ ͵whichmeansthat‫ ݌‬ൌ ‫ ݍ‬൅ ͳasrequired.  A1(3)

Ifܳሺ‫ݔ‬ሻ ൌ ‫ ݔ‬൅ ͳ,ܳԢሺ‫ݔ‬ሻ ൌ ͳ,andso

ሺ‫ ݔ‬൅ ͳሻሺܲ ൅ ܲԢሻ െ ܲ ൌ ሺ‫ ݔ‬ଷ െ ʹሻ

    M1A1

Thatis‫ ܲݔ‬൅ ሺ‫ ݔ‬൅ ͳሻܲᇱ ൌ ሺ‫ ݔ‬ଷ െ ʹሻ

ܲሺ‫ݔ‬ሻ ൌ ܽ‫ ݔ‬ଶ ൅ ܾ‫ ݔ‬൅ ܿandsoܲԢሺ‫ݔ‬ሻ ൌ ʹܽ‫ ݔ‬൅ ܾ B1

Therefore‫ݔ‬ሺܽ‫ ݔ‬ଶ ൅ ܾ‫ ݔ‬൅ ܿሻ ൅ ሺ‫ ݔ‬൅ ͳሻሺʹܽ‫ ݔ‬൅ ܾሻ ൌ ሺ‫ ݔ‬ଷ െ ʹሻ M1

andequatingcoefficients

ܽ ൌ ͳǡ ܾ ൅ ʹܽ ൌ Ͳǡ ܿ ൅ ʹܽ ൅ ܾ ൌ Ͳǡ ܾ ൌ െʹ

     A1

Theseequationsareconsistent,withܽ ൌ ͳǡ ܾ ൌ െʹǡ ܿ ൌ Ͳsoܲሺ‫ݔ‬ሻ ൌ ‫ ݔ‬ଶ െ ʹ‫ݔ‬ A1(6)

ௗ ௉௘ ೣ ଵ
(ii)ForsuchPandQtoexist, ቀ ቁൌ ݁௫
ௗ௫ ொ ௫ାଵ

 842
andso

ሾܳሺܲ݁ ௫ ൅ ܲԢ݁ ௫ ሻ െ ܲ݁ ௫ ܳԢሿሺ‫ ݔ‬൅ ͳሻ ൌ ݁ ௫ ܳ ଶ 

     M1

and
ሾܳሺܲ ൅ ܲԢሻ െ ܲܳԢሿሺ‫ ݔ‬൅ ͳሻ ൌ ܳଶ 

     A1

Letting‫ ݔ‬ൌ െͳ,Ͳ ൌ ሾܳሺെͳሻሿଶ soܳሺെͳሻ ൌ ͲandthusQhasafactorሺ‫ ݔ‬൅ ͳሻasbeforein(i).

However,lettingܳሺ‫ݔ‬ሻ ൌ ሺ‫ ݔ‬൅ ͳሻܴሺ‫ݔ‬ሻ,then   M1

ሾሺ‫ ݔ‬൅ ͳሻܴሺܲ ൅ ܲԢሻ െ ܲሺܴ ൅ ሺ‫ ݔ‬൅ ͳሻܴԢሻሿሺ‫ ݔ‬൅ ͳሻ ൌ ሺ‫ ݔ‬൅ ͳሻଶ ܴ ଶ 

andso

ሺ‫ ݔ‬൅ ͳሻሺܴܲ ൅ ܴܲᇱ െ ܴܲԢሻ െ ܴܲ ൌ ሺ‫ ݔ‬൅ ͳሻܴ ଶ 

Letting‫ ݔ‬ൌ െͳ,ܲሺെͳሻܴሺെͳሻ ൌ Ͳ,butܲሺെͳሻ ് ͲasPandQhavenocommonfactors,andso

ܴሺെͳሻ ൌ ͲwhichmeansthatRinturnhasafactorሺ‫ ݔ‬൅ ͳሻ.  A1

ThusQmusthaveafactorofሺ‫ ݔ‬൅ ͳሻଶ .

Supposeܳሺ‫ݔ‬ሻ ൌ ሺ‫ ݔ‬൅ ͳሻ௡ ܵሺ‫ݔ‬ሻ,where݊ ൒ ʹandܵሺെͳሻ ് Ͳ M1

Then
ሾܳሺܲ ൅ ܲԢሻ െ ܲܳԢሿሺ‫ ݔ‬൅ ͳሻ ൌ ܳ ଶ 

becomes

ሺ‫ ݔ‬൅ ͳሻሾሺ‫ ݔ‬൅ ͳሻ௡ ܵሺܲ ൅ ܲԢሻ െ ܲሺ݊ሺ‫ ݔ‬൅ ͳሻ௡ିଵ ܵ ൅ ሺ‫ ݔ‬൅ ͳሻ௡ ܵԢሻሿ ൌ ሺ‫ ݔ‬൅ ͳሻଶ௡ ܵ ଶ 

Dividingbythefactorሺ‫ ݔ‬൅ ͳሻ௡ gives,

ሾሺ‫ ݔ‬൅ ͳሻܵሺܲ ൅ ܲԢሻ െ ܲሺ݊ܵ ൅ ሺ‫ ݔ‬൅ ͳሻܵԢሻሿ ൌ ሺ‫ ݔ‬൅ ͳሻ௡ ܵ ଶ 

     A1

Letting‫ ݔ‬ൌ െͳ,݊ܲሺെͳሻܵሺെͳሻ ൌ Ͳ,but݊ ് Ͳ,ܲሺെͳሻ ് Ͳandܵሺെͳሻ ് Ͳgivinga

contradictionandhencenosuchPandQcanexist.  E1

 

 843
4.(i)
ͳ ͳ ሺͳ ൅ ‫ ݔ‬௥ାଵ ሻ െ ሺͳ ൅ ‫ ݔ‬௥ ሻ ሺ‫ ݔ‬െ ͳሻ‫ ݔ‬௥
െ ൌ ൌ 
ͳ ൅ ‫ ݔ‬௥ ͳ ൅ ‫ ݔ‬௥ାଵ ሺͳ ൅ ‫ ݔ‬௥ ሻሺͳ ൅ ‫ ݔ‬௥ାଵ ሻ ሺͳ ൅ ‫ ݔ‬௥ ሻሺͳ ൅ ‫ ݔ‬௥ାଵ ሻ

     B1

Therefore
ே ே
‫ݔ‬௥ ͳ ͳ ͳ ͳ ͳ ͳ
෍ ൌ ෍൬ െ ൰ൌ ൤ െ ൨
ሺͳ ൅ ‫ ݔ‬ሻሺͳ ൅ ‫ ݔ‬ሻ ሺ‫ ݔ‬െ ͳሻ
௥ ௥ାଵ ͳ൅‫ݔ‬ ௥ ͳ൅‫ݔ‬ ௥ାଵ ሺ‫ ݔ‬െ ͳሻ ͳ ൅ ‫ ͳ ݔ‬൅ ‫ ݔ‬ேାଵ
௥ୀଵ ௥ୀଵ

   M1    M1A1
ଵ
Asܰ ՜ λ,asȁ‫ݔ‬ȁ ൏ ͳ, ՜ ͳ  M1
ଵା௫ ಿశభ

So
ஶ
‫ݔ‬௥ ͳ ͳ ͳ ͳെͳെ‫ݔ‬ െ‫ݔ‬ ‫ݔ‬
෍ ൌ ൤ െ ͳ൨ ൌ ൤ ൨ൌ ଶ ൌ 
ሺͳ ൅ ‫ ݔ‬ሻሺͳ ൅ ‫ ݔ‬ሻ ሺ‫ ݔ‬െ ͳሻ ͳ ൅ ‫ݔ‬
௥ ௥ାଵ ሺ‫ ݔ‬െ ͳሻ ͳ ൅ ‫ݔ‬ ‫ ݔ‬െ ͳ ͳ െ ‫ݔ‬ଶ
௥ୀଵ

         A1*(6)

(ii)
ͳ ʹ ʹ݁ ି௥௬
•‡…Šሺ‫ݕݎ‬ሻ ൌ ൌ ௥௬ ൌ 
…‘•Šሺ‫ݕݎ‬ሻ ݁ ൅ ݁ ି௥௬ ͳ ൅ ݁ ିଶ௥௬

ʹ݁ ିሺ௥ାଵሻ௬
•‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯ ൌ 
ͳ ൅ ݁ ିଶሺ௥ାଵሻ௬

     M1

Thus
Ͷ݁ ି௬ ݁ ିଶ௥௬
•‡…Šሺ‫ݕݎ‬ሻ •‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯ ൌ 
ሺͳ ൅ ݁ ିଶ௥௬ ሻሺͳ ൅ ݁ ିଶሺ௥ାଵሻ௬ ሻ

      A1

Soif‫ ݔ‬ൌ ݁ ିଶ௬ ,M1

ஶ ஶ
ି௬
‫ݔ‬௥ ‫ݔ‬
෍ •‡…Šሺ‫ݕݎ‬ሻ •‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯ ൌ Ͷ݁ ෍ ൌ Ͷ݁ ି௬ 
ሺͳ ൅ ‫ ݔ‬ሻሺͳ ൅ ‫ ݔ‬ሻ
௥ ௥ାଵ ͳ െ ‫ݔ‬ଶ
௥ୀଵ ௥ୀଵ

      A1  A1

Thus
ஶ
݁ ିଶ௬ ʹ
෍ •‡…Šሺ‫ݕݎ‬ሻ •‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯ ൌ Ͷ݁ ି௬ ିସ௬
ൌ ʹ݁ ି௬ ଶ௬ 
ͳെ݁ ݁ െ ݁ ିଶ௬
௥ୀଵ

       

 844
 ஶ

෍ •‡…Šሺ‫ݕݎ‬ሻ •‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯ ൌ ʹ݁ ି௬ …•…Šሺʹ‫ݕ‬ሻ

௥ୀଵ

      M1A1*(7)

(iii)
ஶ ஶ

෍ •‡…Šሺ‫ݕݎ‬ሻ •‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯ ൌ ʹ ൥෍ •‡…Šሺ‫ݕݎ‬ሻ •‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯ ൅ •‡…Š ‫ݕ‬൩
௥ୀିஶ ௥ୀଵ

     M1A1

ʹ݁ ି௬ ͳ ݁ ି௬ ͳ
ൌ ʹሾʹ݁ ି௬ …•…Šሺʹ‫ݕ‬ሻ ൅ •‡…Š ‫ݕ‬ሿ ൌ ʹ ൤ ൅ ൨ ൌ ʹ൤ ൅ ൨
•‹Š ʹ‫ݕ Š•‘… ݕ‬ •‹Š ‫ݕ Š•‘… ݕ Š•‘… ݕ‬

 A1       M1A1

ʹ ʹ݁ ି௬ ൅ ݁ ௬ െ ݁ ି௬ ʹ ʹ …‘•Š ‫ݕ‬
ൌ ൤ ൨ൌ ൤ ൨ ൌ ʹ …•…Š ‫ݕ‬
…‘•Š ‫ݕ‬ ʹ •‹Š ‫ݕ‬ …‘•Š ‫ݕ Š‹• ʹ ݕ‬

     M1   A1(7)

 

 845
5.(i)
ʹ݉ ൅ ͳ ௠ାଵ
ሺͳ ൅ ‫ݔ‬ሻଶ௠ାଵ ൌ ͳ ൅ ቀʹ݉ ൅ ͳቁ ‫ ݔ‬൅ ‫ ڮ‬൅ ቀʹ݉ ൅ ͳቁ ‫ ݔ‬௠ ൅ ቀ ቁ‫ݔ‬ ൅ ‫ ڮ‬൅ ‫ ݔ‬ଶ௠ାଵ 
ͳ ݉ ݉൅ͳ

       B1

ʹ݉ ൅ ͳ ʹ݉ ൅ ͳ ௠ ʹ݉ ൅ ͳ ௠ାଵ
ൌͳ൅ቀ ቁ‫ ݔ‬൅ ‫ڮ‬൅ ቀ ቁ‫ ݔ‬൅ ቀ ቁ‫ݔ‬ ൅ ‫ ڮ‬൅ ‫ ݔ‬ଶ௠ାଵ 
ͳ ݉ ݉

      M1

ʹ݉ ൅ ͳ ʹ݉ ൅ ͳ ʹ݉ ൅ ͳ
‫ ݔ‬ൌ ͳ ֜ ʹଶ௠ାଵ ൌ ʹ ቂͳ ൅ ቀ ቁ ൅ ‫ڮ‬൅ ቀ ቁቃ ൐ ʹ ቀ ቁ
ͳ ݉ ݉

      M1

ʹ݉ ൅ ͳ
andhenceቀ ቁ ൏ ʹଶ௠     A1*(4)
݉
ʹ݉ ൅ ͳ ሺଶ௠ାଵሻǨ
(ii)ቀ ቁ ൌ ሺ௠ାଵሻǨ௠Ǩisaninteger.  E1
݉

Ifpisaprimegreaterthan݉ ൅ ͳandlessthanorequaltoʹ݉ ൅ ͳ,thenpisafactorofሺʹ݉ ൅ ͳሻǨ

          E1

ʹ݉ ൅ ͳ
andisnotafactorofሺ݉ ൅ ͳሻǨ ݉Ǩ,E1andsoitisafactorofቀ ቁ. E1
݉
ʹ݉ ൅ ͳ
Therefore,ܲ௠ାଵǡଶ௠ାଵ ,whichistheproductofsuchprimes,dividesቀ ቁ.  E1
݉
ʹ݉ ൅ ͳ
Hence,݇ܲ௠ାଵǡଶ௠ାଵ ൌ ቀ ቁwhere݇ ൒ ͳisaninteger,M1andhence
݉
ଵ ʹ݉ ൅ ͳ ଵ
ܲ௠ାଵǡଶ௠ାଵ ൌ ቀ ቁ ൏ ʹଶ௠ ,i.e.ܲ௠ାଵǡଶ௠ାଵ ൏ ʹଶ௠  A1*(7)
௞ ݉ ௞

(iii)ܲଵǡଶ௠ାଵ ൌ ܲଵǡ௠ାଵ ܲ௠ାଵǡଶ௠ାଵ  M1

݉ ൒ ͳ ֜ ݉ ൅ ݉ ൒ ݉ ൅ ͳi.e.݉ ൅ ͳ ൑ ʹ݉andsoܲଵǡ௠ାଵ ൏ Ͷ௠ାଵ applyinggivenconditionE1

By(ii),ܲ௠ାଵǡଶ௠ାଵ ൏ ʹଶ௠ ൌ Ͷ௠  M1

Thus,ܲଵǡଶ௠ାଵ ൏ Ͷ௠ାଵ Ͷ௠ ൌ Ͷଶ௠ାଵ asrequired. A1*(4)

(iv)Supposeܲଵǡ௠ ൏ Ͷ௠ forall݉ ൑ ݇forsomeparticular݇ ൒ ʹ.  E1

Thenif݇ ൌ ʹ݉,ܲଵǡ௞ାଵ ൏ Ͷ௞ାଵ by(iii).    E1

ܲଵǡଶ௠ାଶ ൌ ܲଵǡଶ௠ାଵ ൏ Ͷଶ௠ାଵ ൏ Ͷଶ௠ାଶ (equalityasʹ݉ ൅ ʹisnotprime)using(iii). E1

Soif݇ ൌ ʹ݉ ൅ ͳ,ܲଵǡ௞ାଵ ൏ Ͷ௞ାଵ . E1

ܲଵǡଶ ൌ ʹ ൏ Ͷଶ andhencerequiredresultistruebyprincipleofmathematicalinduction.dE1(5)

 

 846
6.
ܴ …‘•Šሺ‫ ݔ‬൅ ߛሻ ൌ ܴሺ…‘•Š ‫ ߛ Š•‘… ݔ‬൅ •‹Š ‫ߛ Š‹• ݔ‬ሻ

Sowerequire ൌ •‹Š ߛand ൌ …‘•Š ߛwhichispossibleif‫ ܤ‬൐ ‫ ܣ‬൐ Ͳ

஺
Thusܴ ൌ ξ‫ܤ‬ଶ െ ‫ܣ‬ଶ andߛ ൌ –ƒŠିଵ . B1
஻

If‫ ܤ‬ൌ ‫ܣ‬,then‫ ݔ Š‹• ܣ‬൅ ‫ ݔ Š•‘… ܤ‬ൌ ‫ ݁ܣ‬௫  B1

Ifെ‫ ܣ‬൏ ‫ ܤ‬൏ ‫ܣ‬,then‫ ݔ Š‹• ܣ‬൅ ‫ݔ Š•‘… ܤ‬canbewritten

•‹Šሺ‫ ݔ‬൅ ߛሻ ൌ ܴሺ•‹Š ‫ ߛ Š•‘… ݔ‬൅ …‘•Š ‫ߛ Š‹• ݔ‬ሻ

requiring ൌ …‘•Š ߛand ൌ •‹Š ߛ.

஻
Soܴ ൌ ξ‫ܣ‬ଶ െ ‫ܤ‬ଶ andߛ ൌ –ƒŠିଵ   B1
஺

If‫ ܤ‬ൌ െ‫ܣ‬,then‫ ݔ Š‹• ܣ‬൅ ‫ ݔ Š•‘… ܤ‬ൌ െ‫ି ݁ܣ‬௫  B1

IF‫ ܤ‬൏ െ‫ܣ‬,then‫ ݔ Š‹• ܣ‬൅ ‫ݔ Š•‘… ܤ‬canbewrittenܴ …‘•Šሺ‫ ݔ‬൅ ߛሻͲ
஺
requiring ൌ •‹Š ߛand ൌ …‘•Š ߛ,soܴ ൌ െξ‫ܤ‬ଶ െ ‫ܣ‬ଶ andߛ ൌ –ƒŠିଵ B1(5)
஻

(i)‫ ݕ‬ൌ ܽ –ƒŠ ‫ ݔ‬൅ ܾ ൌ •‡…Š ‫ ݔ‬ M1

Thusܽ •‹Š ‫ ݔ‬൅ ܾ …‘•Š ‫ ݔ‬ൌ ͳ  A1

௔
Soξܾ ଶ െ ܽଶ …‘•Š ቀ‫ ݔ‬൅ –ƒŠିଵ ቁ ൌ ͳusingfirstresultofquestion M1
௕

ܽ ͳ
…‘•Š ቀ‫ ݔ‬൅ –ƒŠିଵ ቁ ൌ
ܾ ξܾ െ ܽଶ
ଶ

ܽ ͳ
‫ ݔ‬൅ –ƒŠିଵ ൌ േ …‘•Šିଵ ൬ ൰
ܾ  ξܾ െ ܽଶ
ଶ

M1

andso
ͳ ܽ
‫ ݔ‬ൌ േ …‘•Šିଵ ൬ ൰ െ –ƒŠିଵ
ξܾ ଶ െ ܽଶ ܾ

A1*(5)

(ii)

ͳ ܾ
‫ ݔ‬ൌ •‹Šିଵ ൬ ൰ െ –ƒŠିଵ
ξܽଶ െ ܾ ଶ ܽ

 M1A1(2)

 847
(iii)Forintersectiontooccurattwodistinctpoints,werequiretwosolutionstoexisttothe
equationsconsideredsimultaneously.Consideringthetwographs,therecanbeatmostonlyone
intersection,whichwouldoccurfor‫ ݔ‬൐ Ͳ,ifܾ ൑ Ͳ.

ଵ
Thuswerequireܾ ൐ ܽandቀ ቁ ൐ ͳ  M1
ξ௕ మ ି௔మ

Thatisܽ ൏ ܾ ൏ ξܽଶ ൅ ͳ.  A1

Similarlyviceversa,iftheseconditionsapply,thentherearetwosolutionsandhencetwo
intersections.  E1(3)

ଵ
(iv)Totouch,werequiretwocoincidentsolutions.i.e.ቀ ቁ ൌ ͳ
ξ௕ మ ି௔మ

Thatisܾ ൌ ξܽଶ ൅ ͳ,andequally,ifthisappliesthentheywilltouch,  E1

so
ܽ
‫ ݔ‬ൌ െ –ƒŠିଵ 
ξܽଶ ൅ ͳ

     M1

௔ ௔మ ଵ
andthus‫ ݕ‬ൌ ܽ –ƒŠ ቀെ –ƒŠିଵ ቁ ൅ ξܽଶ ൅ ͳ ൌ െ ൅ ξܽଶ ൅ ͳ ൌ 
ξ௔మ ାଵ ξ௔మ ାଵ ξ௔మ ାଵ

   A1    M1 A1(5)

 

 848
7.If
ଶగ௜
߱ൌ݁ ௡ 

thenifͲ ൑ ‫ ݎ‬൑ ݊ െ ͳ,

ଶగ௜௥௡ ௥
ሺ߱௥ ሻ௡ ൌ ݁ ௡ ൌ ൫݁ ଶగ௜ ൯ ൌ ͳ௥ ൌ ͳ

      M1

Soͳǡ ߱ǡ ߱ଶ ǡ ǥǡ ߱௡ିଵ arethenrootsof‫ ݖ‬௡ ൌ ͳ,thatisof‫ ݖ‬௡ െ ͳ ൌ Ͳ. A1

Thusሺ‫ ݖ‬െ ߱௥ ሻisafactorof‫ ݖ‬௡ െ ͳ     B1

Hence‫ ݖ‬௡ െ ͳ ൌ ݇ሺ‫ ݖ‬െ ͳሻሺ‫ ݖ‬െ ߱ሻሺ‫ ݖ‬െ ߱ଶ ሻ ǥ ሺ‫ ݖ‬െ ߱௡ିଵ ሻandcomparingcoefficientsof‫ ݖ‬௡ ,݇ ൌ ͳ

          M1

Soasrequiredሺ‫ ݖ‬െ ͳሻሺ‫ ݖ‬െ ߱ሻሺ‫ ݖ‬െ ߱ଶ ሻ ǥ ሺ‫ ݖ‬െ ߱௡ିଵ ሻ ൌ ‫ ݖ‬௡ െ ͳ  A1*(5)

(i)Withoutlossofgenerality,letܺ௥ berepresentedby߱௥    M1

ഏ೔ ഏ
ቀ ାగቁ௜
Thenܲwillberepresentedeitherby‫ ݁ݎ‬೙ ൌ ‫ݖ‬,or‫݁ݎ‬ ೙ ൌ ‫ݖ‬Ԣwithȁܱܲȁ ൌ ‫ݎ‬ M1

గ௜ గ௜ గ௜
ȁܲܺ଴ ȁ ൈ ȁܲܺଵ ȁ ൈ ǥ ൈ ȁܲܺ௡ିଵ ȁ ൌ ฬͳ െ  ‫ ݁ݎ‬௡ ฬ ฬ߱ െ  ‫ ݁ݎ‬௡ ฬ ǥ ฬ߱௡ିଵ െ  ‫ ݁ݎ‬௡ ฬ

          M1

ൌ ȁሺ‫ ݖ‬െ ͳሻሺ‫ ݖ‬െ ߱ሻሺ‫ ݖ‬െ ߱ଶ ሻ ǥ ሺ‫ ݖ‬െ ߱௡ିଵ ሻȁ ൌ ȁ‫ ݖ‬௡ െ ͳȁ ൌ ห‫ ݎ‬௡ ݁ గ௜ െ ͳห ൌ ȁെ‫ ݎ‬௡ െ ͳȁ ൌ ‫ ݎ‬௡ ൅ ͳ

  A1    A1      A1*

orȁ‫ݖ‬Ԣ௡ െ ͳȁ ൌ ห‫ ݎ‬௡ ݁ ሺ௡ାଵሻగ௜ െ ͳห ൌ ห‫ ݎ‬௡ ݁ గ௜ ݁ ௡గ௜ െ ͳห ൌ ȁെ‫ ݎ‬௡ െ ͳȁ ൌ ‫ ݎ‬௡ ൅ ͳas݁ ௡గ௜ ൌ ͳbecausen
iseven.  E1(7)

Soȁܲܺ଴ ȁ ൈ ȁܲܺଵ ȁ ൈ ǥ ൈ ȁܲܺ௡ିଵ ȁ ൌ ȁܱܲȁ௡ ൅ ͳasrequired.

Fornodd,

ȁܲܺ଴ ȁ ൈ ȁܲܺଵ ȁ ൈ ǥ ൈ ȁܲܺ௡ିଵ ȁ ൌ ȁ‫ ݖ‬௡ െ ͳȁ ൌ ห‫ ݎ‬௡ ݁ గ௜ െ ͳห ൌ ȁെ‫ ݎ‬௡ െ ͳȁ ൌ ‫ ݎ‬௡ ൅ ͳ ൌ ȁܱܲȁ௡ ൅ ͳ

           M1A1

or

ȁܲܺ଴ ȁ ൈ ȁܲܺଵ ȁ ൈ ǥ ൈ ȁܲܺ௡ିଵ ȁ ൌ ห‫ ݖ‬ᇱ ௡ െ ͳห ൌ ห‫ ݎ‬௡ ݁ ሺ௡ାଵሻగ௜ െ ͳห ൌ ȁ‫ ݎ‬௡ െ ͳȁ ൌ ȁܱܲȁ௡ െ ͳ

         B1

ifȁܱܲȁ ൒ ͳ,andൌ ͳ െ ȁܱܲȁ௡ ifȁܱܲȁ ൏ ͳA1(4)

 849
(ii)
ȁܺ଴ ܺଵ ȁ ൈ ȁܺ଴ ܺଶ ȁ ൈ ǥ ൈ ȁܺ଴ ܺ௡ିଵ ȁ ൌ ȁሺͳ െ ߱ሻሺͳ െ ߱ଶ ሻ ǥ ሺͳ െ ߱௡ିଵ ሻȁ

         M1

But

ሺ‫ ݖ‬െ ͳሻሺ‫ ݖ‬െ ߱ሻሺ‫ ݖ‬െ ߱ଶ ሻ ǥ ሺ‫ ݖ‬െ ߱௡ିଵ ሻ ൌ ‫ ݖ‬௡ െ ͳ

andso

‫ݖ‬௡ െ ͳ
ሺ‫ ݖ‬െ ߱ሻሺ‫ ݖ‬െ ߱ଶ ሻ ǥ ሺ‫ ݖ‬െ ߱௡ିଵ ሻ ൌ ൌ ‫ ݖ‬௡ିଵ ൅ ‫ ݖ‬௡ିଶ ൅ ‫ ڮ‬൅ ͳ
‫ݖ‬െͳ

          M1

ሺ‫ ݖ‬െ ߱ሻሺ‫ ݖ‬െ ߱ଶ ሻ ǥ ሺ‫ ݖ‬െ ߱௡ିଵ ሻ ൌ ‫ ݖ‬௡ିଵ ൅ ‫ ݖ‬௡ିଶ ൅ ‫ ڮ‬൅ ͳ

          A1

istrueforall‫ݖ‬sofor‫ ݖ‬ൌ ͳ,ሺͳ െ ߱ሻሺͳ െ ߱ଶ ሻ ǥ ሺͳ െ ߱௡ିଵ ሻ ൌ ͳ ൅ ͳ ൅ ‫ ڮ‬൅ ͳ ൌ ݊A1*(4)

 

 850
8.(i) ݂ሺ‫ݔ‬ሻ ൅ ሺͳ െ ‫ݔ‬ሻ݂ሺെ‫ݔ‬ሻ ൌ ‫ ݔ‬ଶ 

Let‫ ݔ‬ൌ െ‫ݑ‬,then݂ሺെ‫ݑ‬ሻ ൅ ሺͳ െ െ‫ݑ‬ሻ݂ሺെ െ ‫ݑ‬ሻ ൌ ሺെ‫ݑ‬ሻଶ 

i.e.݂ሺെ‫ݑ‬ሻ ൅ ሺͳ ൅ ‫ݑ‬ሻ݂ሺ‫ݑ‬ሻ ൌ ‫ݑ‬ଶ 

Let‫ ݑ‬ൌ ‫ݔ‬,then݂ሺെ‫ݔ‬ሻ ൅ ሺͳ ൅ ‫ݔ‬ሻ݂ሺ‫ݔ‬ሻ ൌ ‫ ݔ‬ଶ asrequired.  E1

Substitutingfor݂ሺെ‫ݔ‬ሻfromtheequationjustobtainedintheoriginal, M1

݂ሺ‫ݔ‬ሻ ൅ ሺͳ െ ‫ݔ‬ሻ൫‫ ݔ‬ଶ െ ሺͳ ൅ ‫ݔ‬ሻ݂ሺ‫ݔ‬ሻ൯ ൌ ‫ ݔ‬ଶ 

Thus‫ ݔ‬ଶ ݂ሺ‫ݔ‬ሻ ൌ ‫ ݔ‬ଷ ,andhence݂ሺ‫ݔ‬ሻ ൌ ‫ݔ‬  M1A1

Verification:Ͳ‫ ݔ‬൅ ሺͳ െ ‫ݔ‬ሻ ൈ െ‫ ݔ‬ൌ ‫ ݔ‬െ ‫ ݔ‬൅ ‫ ݔ‬ଶ ൌ ‫ ݔ‬ଶ asrequired. B1(5)

(ii) 
‫ݔ‬൅ͳ
‫ݔ‬൅ͳ ቀ ቁ ൅ ͳ ‫ ݔ‬൅ ͳ ൅ ‫ ݔ‬െ ͳ ʹ‫ݔ‬
‫ܭ‬൫‫ܭ‬ሺ‫ݔ‬ሻ൯ ൌ ‫ ܭ‬൬ ൰ൌ ‫ݔ‬െͳ ൌ ൌ ൌ ‫ݔ‬
‫ݔ‬െͳ ‫ݔ‬൅ͳ ‫ݔ‬ ൅ ͳ െ ‫ݔ‬ ൅ ͳ ʹ
ቀ ቁെͳ
‫ݔ‬െͳ

    M1  M1    A1*(3)

asrequired.

‫ݔ‬൅ͳ
݃ሺ‫ݔ‬ሻ ൅ ‫ ݃ݔ‬൬ ൰ ൌ ‫ݔ‬
‫ݔ‬െͳ

So
‫ݔ‬൅ͳ
‫ݔ‬൅ͳ ‫ݔ‬൅ͳ ቀ ቁ൅ͳ ‫ݔ‬൅ͳ
݃൬ ൰൅൬ ൰݃ቌ ‫ ݔ‬െ ͳ ቍൌ൬ ൰
‫ݔ‬െͳ ‫ݔ‬െͳ ‫ݔ‬൅ͳ ‫ݔ‬ െ ͳ
ቀ ቁെͳ
‫ݔ‬െͳ

         M1

Thatis
‫ݔ‬൅ͳ ‫ݔ‬൅ͳ ‫ݔ‬൅ͳ
݃൬ ൰൅൬ ൰ ݃ሺ‫ݔ‬ሻ ൌ ൬ ൰
‫ݔ‬െͳ ‫ݔ‬െͳ ‫ݔ‬െͳ

         A1

௫ାଵ
Sosubstitutingfor݃ ቀ ቁfromtheequationjustobtainedintheinitialequationM1
௫ିଵ

‫ݔ‬൅ͳ ‫ݔ‬൅ͳ
݃ሺ‫ݔ‬ሻ ൅ ‫ ݔ‬ቆ൬ ൰െ൬ ൰ ݃ሺ‫ݔ‬ሻቇ ൌ ‫ݔ‬
‫ݔ‬െͳ ‫ݔ‬െͳ

ሾሺ‫ ݔ‬െ ͳሻ െ ‫ݔ‬ሺ‫ ݔ‬൅ ͳሻሿ݃ሺ‫ݔ‬ሻ ൅ ‫ݔ‬ሺ‫ ݔ‬൅ ͳሻ ൌ ‫ݔ‬ሺ‫ ݔ‬െ ͳሻ

ሺെ‫ ݔ‬ଶ െ ͳሻ݃ሺ‫ݔ‬ሻ ൌ െʹ‫ݔ‬

 851
 ʹ‫ݔ‬
݃ሺ‫ݔ‬ሻ ൌ 
ሺ‫ ݔ‬ଶ ൅ ͳሻ

    M1A1*(5)

NotrequiredͲverification:Ͳ
‫ݔ‬൅ͳ
ʹ‫ݔ‬ ʹቀ ቁ ʹ‫ݔ‬ ʹሺ‫ ݔ‬൅ ͳሻሺ‫ ݔ‬െ ͳሻ ʹ‫ݔ‬ ʹሺ‫ ݔ‬ଶ െ ͳሻ
൅ ‫ݔ‬ ‫ݔ‬െͳ ൌ ൅ ‫ݔ‬ ቆ ቇ ൌ ൅ ‫ݔ‬ 
ሺ‫ ݔ‬ଶ ൅ ͳሻ ‫ݔ‬൅ͳ ଶ ሺ‫ ݔ‬ଶ ൅ ͳሻ ሺ‫ ݔ‬൅ ͳሻଶ ൅ ሺ‫ ݔ‬െ ͳሻଶ ሺ‫ ݔ‬ଶ ൅ ͳሻ ʹሺ‫ ݔ‬ଶ ൅ ͳሻ
ቆቀ ቁ ൅ ͳቇ
‫ݔ‬െͳ

ʹ‫ ݔ‬൅ ‫ݔ‬ሺ‫ ݔ‬ଶ െ ͳሻ ‫ݔ‬ሺʹ ൅ ‫ ݔ‬ଶ െ ͳሻ

ൌ ൌ ൌ ‫ݔ‬
ሺ‫ ݔ‬ଶ ൅ ͳሻ ሺ‫ ݔ‬ଶ ൅ ͳሻ

asexpected.

(iii)
ͳ ͳ
݄ሺ‫ݔ‬ሻ ൅ ݄ ൬ ൰ൌͳെ‫ݔ‬െ 
ͳെ‫ݔ‬ ͳെ‫ݔ‬

   (EquationA)

ͳ ͳ ͳ ͳ
݄൬ ൰൅݄ቌ ቍൌͳെ൬ ൰െ 
ͳെ‫ݔ‬ ͳ ͳെ‫ݔ‬ ͳ
ͳെቀ ቁ ͳെቀ ቁ
ͳെ‫ݔ‬ ͳെ‫ݔ‬

      M1A1

Thus
ͳ ‫ݔ‬െͳ ͳ ͳെ‫ݔ‬
݄൬ ൰൅݄൬ ൰ൌͳെ൬ ൰൅൬ ൰
ͳെ‫ݔ‬ ‫ݔ‬ ͳെ‫ݔ‬ ‫ݔ‬

   (EquationB)

Then
‫ݔ‬െͳ ͳ ‫ݔ‬െͳ ͳ
݄൬ ൰൅݄ቌ ቍൌͳെ൬ ൰െ 
‫ݔ‬ ‫ݔ‬ െ ͳ ‫ݔ‬ ‫ݔ‬െͳ
ͳെቀ ቁ ͳെቀ ቁ
‫ݔ‬ ‫ݔ‬

      M1A1

Thatis
‫ݔ‬െͳ ‫ݔ‬െͳ
݄൬ ൰ ൅ ݄ሺ‫ݔ‬ሻ ൌ ͳ െ ൬ ൰ െ ‫ݔ‬
‫ݔ‬ ‫ݔ‬

   (EquationC)

 852
A+CͲBgives     M1
ͳ ‫ݔ‬െͳ ͳ ͳെ‫ݔ‬
ʹ݄ሺ‫ݔ‬ሻ ൌ ͳ െ ‫ ݔ‬െ ൅ͳെ൬ ൰ െ ‫ ݔ‬െ ቆͳ െ ൬ ൰൅൬ ൰ቇ
ͳെ‫ݔ‬ ‫ݔ‬ ͳെ‫ݔ‬ ‫ݔ‬

     A1

ʹ݄ሺ‫ݔ‬ሻ ൌ ͳ െ ʹ‫ݔ‬

So
ͳ
݄ሺ‫ݔ‬ሻ ൌ െ ‫ݔ‬
ʹ

     A1(7)

NotrequiredͲverification:Ͳ
ͳ ͳ ͳ ͳ
െ‫ݔ‬൅ െ ൌͳെ‫ݔ‬െ 
ʹ ʹ ͳെ‫ݔ‬ ͳെ‫ݔ‬

asexpected.

 

 853
 ଶ ଶ గ ଶ
9.ܲܺ ൌ ξ͵ܽ ൌ ܽoralternativelyܲܺ ൌ ܽ •‡… ൌ ܽ M1A1
ଷ ξଷ ଺ ξଷ

ଶ
Sotheextensionis ܽ െ ݈.     A1*(3)
ξଷ

ଵ ଶ
Displacingܺadistance‫ݔ‬towardsܲ,ܴܺwillbeටܽଶ ൅ ቀ ܽ ൅ ‫ݔ‬ቁ M1A1
ξଷ

andthusthetensioninܴܺwillbe
ଶ
ߣ ͳ ߣ Ͷ ʹ
ቌඨܽଶ ൅ ൬ ܽ ൅ ‫ݔ‬൰ െ ݈ቍ ൌ ቌඨ ܽଶ ൅ ܽ‫ ݔ‬൅ ‫ ݔ‬ଶ െ ݈ቍ
݈ ξ͵ ݈ ͵ ξ͵

       M1A1*(4)

Thecosineoftheanglebetweenܴܺandܲܺproducedwillbe
ͳ
ܽ൅‫ݔ‬
ξ͵ 
ଶ
ͳ
ඨܽଶ ൅ ൬ ܽ ൅ ‫ݔ‬൰
ξ͵

    B1

sotheequationofmotionforܺ,resolvinginthedirectionܺܲis

ͳ
ܽ൅‫ݔ‬
ߣ ʹ ߣ Ͷ ʹ ξ͵
൬ ܽ െ ݈ െ ‫ݔ‬൰ െ ʹ ቌඨ ܽଶ ൅ ܽ‫ ݔ‬൅ ‫ ݔ‬ଶ െ ݈ቍ ൌ ݉‫ݔ‬ሷ 
݈ ξ͵ ݈ ͵ ξ͵ ଶ
ͳ
ඨܽଶ ൅ ൬ ܽ ൅ ‫ݔ‬൰
ξ͵

       M1A1A1(4)

ͳ ͳ
ܽ൅‫ݔ‬ ݈൬ ܽ ൅ ‫ݔ‬൰
Ͷ ʹ ξ͵ ͳ ξ͵
ቌඨ ܽଶ ൅ ܽ‫ ݔ‬൅ ‫ ݔ‬ଶ െ ݈ቍ ൌ ܽ൅‫ݔ‬െ 
͵ ξ͵ ଶ ξ͵ ଶ
ͳ ͳ
ඨܽଶ ൅ ൬ ܽ ൅ ‫ݔ‬൰ ඨܽ ଶ ൅ ൬ ܽ ൅ ‫ݔ‬൰
ξ͵ ξ͵

so

ͳ
ܽ൅‫ݔ‬
ߣ ʹ ߣ Ͷ ʹ ξ͵
൬ ܽ െ ݈ െ ‫ݔ‬൰ െ ʹ ቌඨ ܽଶ ൅ ܽ‫ ݔ‬൅ ‫ ݔ‬ଶ െ ݈ቍ 
݈ ξ͵ ݈ ͵ ξ͵ ଶ
ͳ
ඨܽଶ ൅ ൬ ܽ ൅ ‫ݔ‬൰
ξ͵

    M1A1

 854
 ͳ
‫ۇ‬ ʹߣ ൬ ܽ ൅ ‫ݔ‬൰ ‫ۊ‬
ʹߣ ߣ ʹ ߣ ʹߣ ξ͵
ൌ ൬ ܽ െ ߣ െ ‫ݔ‬൰ െ ‫ ܽ ۈ‬൅ ‫ ݔ‬െ ‫ۋ‬
ξ͵ ݈ ݈ ‫ۈ‬ξ͵ ݈ ݈ ଶ‫ۋ‬
ͳ
ඨܽ ଶ ൅ ൬ ܽ ൅ ‫ݔ‬൰
‫ۉ‬ ξ͵ ‫ی‬
ିଵ
͵ߣ ͳ Ͷ ʹ ଶ
ൌ െߣ െ ‫ ݔ‬൅ ʹߣ ൬ ܽ ൅ ‫ݔ‬൰ ൬ ܽଶ ൅ ܽ‫ ݔ‬൅ ‫ ݔ‬ଶ ൰ 
݈ ξ͵ ͵ ξ͵
ିଵ
͵ߣ ͳ ξ͵ ξ͵ ‫ ݔ͵ ݔ‬ଶ ଶ
ൌ െߣ െ ‫ ݔ‬൅ ʹߣ ൬ ܽ ൅ ‫ݔ‬൰ ቆͳ ൅ ൅ ቇ 
݈ ξ͵ ʹܽ ʹ ܽ Ͷܽଶ

      A1

͵ߣ ͳ ξ͵ ξ͵ ‫ݔ‬
ൎ െߣ െ ‫ ݔ‬൅ ߣ ൬ ܽ ൅ ‫ݔ‬൰ ቆͳ െ ቇ
݈ ξ͵ ܽ Ͷ ܽ

      M1A1

͵ߣ ξ͵ߣ‫ ݔ‬ξ͵ ߣ‫ݔ‬

ൎ െߣ െ ‫ݔ‬൅ߣ൅ െ 
݈ ܽ Ͷ ܽ

      M1

͵ߣ ͵ξ͵ ߣ‫ݔ‬
ൌെ ‫ݔ‬൅ 
݈ Ͷ ܽ
͵ߣ
ൌെ ൫Ͷܽ െ ξ͵݈൯‫ݔ‬
Ͷ݈ܽ

     A1

Thisisapproximatelytheequationofsimpleharmonicmotionwithperiod

ʹߨ Ͷ݈݉ܽ
ൌ ʹߨඨ 
͵൫Ͷܽ െ ξ͵݈൯ߣ
ට ͵ߣ ൫Ͷܽ െ ξ͵݈൯
Ͷ݈݉ܽ

asrequired.M1A1*(9)

 

 855
10.Resolvingupwardsalongalineofgreatestslopeinitially,ifthetensioninthestringisܶ,
‫ݑ‬ଶ
ܶ …‘• ߚ െ ݉݃ •‹ ߙ ൌ ݉ 
ܽ …‘• ߚ

   M1  M1B1A1(4)

Resolvingperpendiculartotheslope,ifthenormalcontactforceisܴ,
ܴ ൅ ܶ •‹ ߚ െ ݉݃ …‘• ߙ ൌ Ͳ

     M1A1(2)

Theparticlewillnotimmediatelyleavetheplaneifܴ ൐ Ͳ.  M1

Thisis
݉݃ …‘• ߙ ൐ ܶ •‹ ߚ

     A1

So
‫ݑ‬ଶ
݉ ൅ ݉݃ •‹ ߙ
ܽ …‘• ߚ
݉݃ …‘• ߙ ൐ •‹ ߚ
…‘• ߚ

     M1

Thatis
‫ݑ‬ଶ
݃ …‘• ߙ …‘• ߚ ൐ ݃ •‹ ߙ •‹ ߚ ൅ –ƒ ߚ
ܽ

whichbecomesܽ݃ሺ…‘• ߙ …‘• ߚ െ •‹ ߙ •‹ ߚሻ ൐ ‫ݑ‬ଶ –ƒ ߚ  M1

or,asrequired,ܽ݃ …‘•ሺߙ ൅ ߚሻ ൐ ‫ݑ‬ଶ –ƒ ߚ    A1*(5)

Anecessaryconditionfortheparticletoperformacompletecirclewhilstincontactwiththeplaneis
thatthestringremainsintensionwhentheparticleisatitshighestpointinthemotion.E1

Ifthespeedoftheparticleatthatmomentisv,thenconservingenergy,
ͳ ͳ
݉‫ݑ‬ଶ ൌ ݉‫ ݒ‬ଶ ൅ ݉݃ʹܽ …‘• ߚ •‹ ߙ
ʹ ʹ

      M1A1

andthus‫ ݒ‬ଶ ൌ ‫ݑ‬ଶ െ Ͷܽ݃ …‘• ߚ •‹ ߙ

Resolvingdownwardsalongalineofgreatestslope,ifthetensioninthestringisnowܶԢ,
‫ݒ‬ଶ
ܶ ᇱ …‘• ߚ ൅ ݉݃ •‹ ߙ ൌ ݉ 
ܽ …‘• ߚ

      B1

 856
 ‫ݒ‬ଶ
ܶ ᇱ  ൐ Ͳ ֜ ݉ ቆ െ ݃ •‹ ߙቇ ൐ Ͳ
ܽ …‘• ߚ

    M1

whichmeansthat
‫ݑ‬ଶ െ Ͷܽ݃ …‘• ߚ •‹ ߙ
െ ݃ •‹ ߙ ൐ Ͳ
ܽ …‘• ߚ

Thus‫ݑ‬ଶ ൐ ͷܽ݃ …‘• ߚ •‹ ߙ       A1(6)

Aswealreadyhaveܽ݃ሺ…‘• ߙ …‘• ߚ െ •‹ ߙ •‹ ߚሻ ൐ ‫ݑ‬ଶ –ƒ ߚ

ͷܽ݃ …‘• ߚ •‹ ߙ –ƒ ߚ ൏ ܽ݃ሺ…‘• ߙ …‘• ߚ െ •‹ ߙ •‹ ߚሻ

       M1

Soͷ •‹ ߙ •‹ ߚ ൏ …‘• ߙ …‘• ߚ െ •‹ ߙ •‹ ߚ

i.e.͸ •‹ ߙ •‹ ߚ ൏ …‘• ߙ …‘• ߚor,asisrequired,͸ –ƒ ߙ –ƒ ߚ ൏ ͳ  M1A1*(3)

 

 857
11.(i)Supposeܴ ൌ ݇‫ݒ‬forsomeconstant݇
௉ ௉ ௉
Thenas െ ܴ ൌ ݉ܽ, െ Ͷܷ݇ ൌ Ͳgiving݇ ൌ  B1
௩ ସ௎ ଵ଺௎ మ

௉ ௗ௩ ௉
As݉ܽ ൌ െ ܴ,݉‫ݒ‬ ൌ െ ݇‫ݒ‬    M1
௩ ௗ௫ ௩

Separatingvariables,
݉‫ ݒ‬ଶ
න ݀‫ ݒ‬ൌ න ݀‫ݔ‬
ܲ െ ݇‫ ݒ‬ଶ

      M1

So
݉ ͳ͸ܷ ଶ ‫ ݒ‬ଶ
න ݀‫ ݒ‬ൌ න ݀‫ݔ‬
ܲ ͳ͸ܷ ଶ െ ‫ ݒ‬ଶ

ͳ͸ܷ ଶ ‫ ݒ‬ଶ ଶ
ͳ͸ܷ ଶ ʹܷ ʹܷ
ଶ ଶ
ൌ ͳ͸ܷ ቆ ଶ ଶ
െ ͳቇ ൌ ͳ͸ܷ ଶ ൬ ൅ െ ͳ൰
ͳ͸ܷ െ ‫ݒ‬ ͳ͸ܷ െ ‫ݒ‬ Ͷܷ െ ‫ ݒ‬Ͷܷ ൅ ‫ݒ‬

      M1A1

So
݉ ଶ௎
ቂͳ͸ܷ ଶ ሺെʹܷ ŽሺͶܷ െ ‫ݒ‬ሻ ൅ ʹܷ ŽሺͶܷ ൅ ‫ݒ‬ሻ െ ‫ݒ‬ሻቃ ൌ ܺଵ 
ܲ ௎

      M1A1

ͳ͸ܷ݉ ଷ
ܺଵ ൌ ሺെʹ Ž ʹܷ ൅ ʹ Ž ͸ܷ െ ʹ ൅ ʹ Ž ͵ܷ െ ʹ Ž ͷܷ ൅ ͳሻ
ܲ

Thus
͸ܷ ൈ ͵ܷ ͻ
ߣܺଵ ൌ ʹ Ž ൬ ൰ െ ͳ ൌ ʹ Ž െ ͳ
ʹܷ ൈ ͷܷ ͷ

      M1A1(9)

(ii)Supposeܴ ൌ ߤ‫ ݒ‬ଶ forsomeconstantߤ

௉ ௉
Then െ ͳ͸ߤܷ ଶ ൌ Ͳgivingߤ ൌ   B1
ସ௎ ଺ସ௎ య

௉
Again,as݉ܽ ൌ െ ܴ,
௩
݀‫ܲ ݒ‬ ܲ െ ߤ‫ ݒ‬ଷ
݉‫ݒ‬ ൌ െ ߤ‫ ݒ‬ଶ ൌ 
݀‫ݒ ݔ‬ ‫ݒ‬

݉‫ ݒ‬ଶ
න ݀‫ ݒ‬ൌ න ݀‫ݔ‬
ܲ െ ߤ‫ ݒ‬ଷ

      M1

 858
So
െ݉ ଶ௎
൤ Žሺܲ െ ߤ‫ ݒ‬ଷ ሻ൨ ൌ ܺଶ 
͵ߤ ௎

     M1A1

െ͸Ͷܷ ଷ ݉ ͹ ͸͵ െ͸Ͷܷ ଷ ݉ ͺ
ܺଶ ൌ ൬Ž ܲ െ Ž ܲ൰ ൌ Ž 
͵ܲ ͺ ͸Ͷ ͵ܲ ͻ
ସ ଽ
Thusߣܺଶ ൌ Ž      M1A1(6)
ଷ ଼

ଽ ସ ଽ ଼
(iii)ߣܺଵ െ ߣܺଶ =ʹ Ž െ ͳ െ Ž ൌ Ͷ Ž ͵ െ ʹ Ž ͷ െ ͳ െ Ž ͵ ൅ Ͷ Ž ʹ
ହ ଷ ଼ ଷ

      M1

Ͷ ͳ ͳ
ൌ Ž ʹͶ െ ʹ Ž ͷ െ ͳ ൐ ሺͶ ൈ ͵Ǥͳ͹ െ ͸ ൈ ͳǤ͸ͳ െ ͵ሻ ൌ ሺͳʹǤ͸ͺ െ ͻǤ͸͸ െ ͵ሻ ൐ Ͳ
͵ ͵ ͵

 A1   M1   A1

Soܺଵ islargerthanܺଶ    A1(5)

 

 859
12.(i)ܺ ‫ܤ ׽‬ሺͳͲͲ݊ǡ ͲǤʹሻ  B1

Soߤ ൌ ͳͲͲ݊ ൈ ͲǤʹ ൌ ʹͲ݊M1A1andߪ ଶ ൌ ͳͲͲ݊ ൈ ͲǤʹ ൈ ͲǤͺ ൌ ͳ͸݊M1A1

Soܲሺͳ͸݊ ൑ ܺ ൑ ʹͶ݊ሻ ൌ ܲሺȁܺ െ ʹͲ݊ȁ ൑ Ͷ݊ሻ ൌ ܲ൫ȁܺ െ ʹͲ݊ȁ ൑ ξ݊ ൈ ξͳ͸݊൯

   M1   M1A1

ଵ ଶ ଵ
SobyChebyshev,ܲሺͳ͸݊ ൑ ܺ ൑ ʹͶ݊ሻ ൒ ͳ െ ቀ ቁ ൌ ͳ െ asrequired.A1*(9)
ξ௡ ௡

(ii)Supposeܺ ‫݋ܲ ׽‬ሺ݊ሻ B1

Thenߤ ൌ ݊B1andߪ ଶ ൌ ݊ B1

ଵ
ByChebyshev,ܲሺȁܺ െ ߤȁ ൐ ݇ߪሻ ൑ 
௞మ

ଵ
solet݇ ൌ ξ݊andhenceܲሺȁܺ െ ݊ȁ ൐ ݊ሻ ൑ 
௡

M1   A1

݊ଶ ݁ ି௡ ݊ଶ௡ ݁ ି௡
ܲሺȁܺ െ ݊ȁ ൐ ݊ሻ ൌ ܲሺܺ ൏ Ͳ‫ ܺݎ݋‬൐ ʹ݊ሻ ൌ ܲሺܺ ൐ ʹ݊ሻ ൌ ͳ െ ݁ ି௡ െ ݊݁ ି௡ െ െ ‫ڮ‬െ 
ʹǨ ʹ݊Ǩ

 M1   A1   A1

௡మ ௡మ೙ ଵ
Soͳ െ ݁ ି௡ ቀͳ ൅ ݊ ൅ ൅‫ڮ‬൅ ቁ൑   M1A1
ଶǨ ଶ௡Ǩ ௡

௡మ ௡మ೙ ଵ
andhenceͳ ൅ ݊ ൅ ൅‫ڮ‬൅ ൒ ቀͳ െ ቁ ݁ ௡  A1*(11)
ଶǨ ଶ௡Ǩ ௡

 

 860
13.Letܻ ൌ ܺ െ ܽ,thenߤ௒ ൌ ‫ܧ‬ሺܻሻ ൌ ‫ܧ‬ሺܺ െ ܽሻ ൌ ‫ܧ‬ሺܺሻ െ ܽ ൌ ߤ െ ܽ B1

‫ܧ‬ሺሺܻ െ ߤ௒ ሻସ ሻ ൌ ‫ܧ‬ሺሺܺ െ ܽ െ ߤ ൅ ܽሻସ ሻ ൌ ‫ܧ‬ሺሺܺ െ ߤሻସ ሻ

       B1

ߪ௒ ଶ ൌ ‫ܧ‬ሺሺܻ െ ߤ௒ ሻଶ ሻ ൌ ‫ܧ‬ሺሺܺ െ ܽ െ ߤ ൅ ܽሻଶ ሻ ൌ ‫ܧ‬ሺሺܺ െ ߤሻଶ ሻ ൌ ߪ ଶ 

       B1

sothekurtosisofܺ െ ܽis
‫ܧ‬ሺሺܻ െ ߤ௒ ሻସ ሻ ‫ܧ‬ሺሺܺ െ ߤሻସ ሻ
െ ͵ ൌ െ ͵
ߪ௒ ସ ߪସ

whichisthesameasthatforܺ     B1*(4)

(i)Ifܺ ‫ܰ ׽‬ሺͲǡ ߪ ଶ ሻthenithaspdf

ͳ ିଵ ௫ మ
ቀ ቁ
݁ ଶ ఙ 
ߪξʹߨ

So
ஶ ିଵ ௫ మ ஶ ିଵ ௫ మ
ͳ ቀ ቁ ͳ ቀ ቁ
‫ܧ‬ሺሺܺ െ ߤሻସ ሻ ൌ න ‫ݔ‬ସ ݁ ଶ ఙ ݀‫ݔ‬ ൌන ‫ݔ‬ଷ‫ݔ‬ ݁ ଶ ఙ ݀‫ ݔ‬
ିஶ ߪξʹߨ ିஶ ߪξʹߨ

     M1A1

Byparts,

ஶ ஶ ஶ
ͳ ିଵ ௫ మ ͳ ିଵ ௫ మ ͳ ିଵ ௫ మ
ଷ ቀ ቁ ቀ ቁ ቀ ቁ
න ‫ݔ ݔ‬ ݁ ଶ ఙ ݀‫ݔ‬ ൌ ቈ‫ ݔ‬ଷ ൈ െߪ ଶ
݁ଶ ఙ ቉ െන ͵‫ ݔ‬ଶ ൈ െߪ ଶ ݁ ଶ ఙ ݀‫ ݔ‬
ିஶ ߪξʹߨ ߪξʹߨ ିஶ ିஶ ߪξʹߨ

    M1A1

ൌ Ͳ ൅ ͵ߪ ଶ ߪ ଶ ൌ ͵ߪ ସ 

    A1

Sothekurtosisis
͵ߪ ସ
െ ͵ ൌ Ͳ
ߪସ


asrequired.     (5)

 861
(ii)

௡ ସ
ܶ ସ ൌ ൬෍ ܻ௥ ൰ ൌ ෍൫ܻ௥ ସ ൅ Ͷܻ௥ ଷ ܻ௦ ൅ ͸ܻ௥ ଶ ܻ௦ ଶ ൅ ͳʹܻ௦ ܻ௧ ܻ௥ ଶ ൅ ʹͶܻ௥ ܻ௦ ܻ௧ ܻ௨ ൯
௥ୀଵ

     M1A1

wherethesummationisoverallvalueswithoutrepetition.
AstheYsareindependent,theexpectationofproductsareproductsofexpectationsandas

‫ܧ‬ሺܻሻ ൌ Ͳ,
௡ ௡ିଵ ௡

‫ܧ‬ሺܶ ସሻ
ൌ ‫ ܧ‬ቀ෍൫ܻ௥ ൅ ͸ܻ௥ ܻ௦ ൯ቁ ൌ ‫ ܧ‬൭෍ ܻ௥ ൱ ൅ ‫ ܧ‬൭෍ ෍ ͸ܻ௥ ଶ ܻ௦ ଶ ൱
ସ ଶ ଶ ସ

௥ୀଵ ௥ୀଵ ௦ୀ௥ାଵ

     M1
௡ ௡ିଵ ௡

ൌ ෍ ‫ܧ‬൫ܻ௥ ൯ ൅ ͸ ෍ ෍ ‫ܧ‬൫ܻ௥ ଶ ൯‫ܧ‬൫ܻ௦ ଶ ൯

ସ

௥ୀଵ ௥ୀଵ ௦ୀ௥ାଵ

     A1*(4)

(iii)
‫ܧ‬ሺሺܺ௜ െ ߤሻସ ሻ
െ ͵ ൌ ߢ
ߪସ

Letܻ௜ ൌ ܺ௜ െ ߤthenbythefirstresult,thekurtosisofܻ௜ is,

i.e.
‫ܧ‬൫ܻ௜ ସ ൯
െ ͵ ൌ ߢ
ߪସ

so‫ܧ‬൫ܻ௜ ସ ൯ ൌ ሺ͵ ൅ ߢሻߪ ସ   M1A1

௡

‫ ܧ‬൭෍ ܺ௜ ൱ ൌ ݊ߤ
௜ୀଵ

and
௡

ܸܽ‫ ݎ‬൭෍ ܺ௜ ൱ ൌ ݊ߪ ଶ 
௜ୀଵ

    B1

 862
sothekurtosisof
௡

෍ ܺ௜ 
௜ୀଵ
is
‫ܧ‬ሺሺσ௡௜ୀଵ ܺ௜ െ ݊ߤሻସ ሻ ‫ܧ‬ሺሺσ௡௜ୀଵ ܻ௜ ሻସ ሻ
െ ͵ ൌ െ ͵
ሺ݊ߪ ଶ ሻଶ ݊ଶ ߪ ସ

     M1

Let
௡
ܶൌ෍ ܻ௥ 
௥ୀଵ

Thenwerequire
‫ܧ‬ሺܶ ସ ሻ
െ ͵
݊ଶ ߪ ସ

whichby(ii)is
σ௡௥ୀଵ ‫ܧ‬൫ܻ௥ ସ ൯ ൅ ͸ σ௡ିଵ ௡ ଶ ଶ
௥ୀଵ σ௦ୀ௥ାଵ ‫ܧ‬൫ܻ௥ ൯‫ܧ‬൫ܻ௦ ൯
െ ͵
݊ଶ ߪ ସ

     M1A1

݊ሺ͵ ൅ ߢሻߪ ସ ൅ ͵݊ሺ݊ െ ͳሻߪ ଶ ߪ ଶ ͵ ൅ ߢ ൅ ͵ሺ݊ െ ͳሻ െ ͵݊ ߢ

ൌ ଶ ସ
െ͵ൌ ൌ 
݊ ߪ ݊ ݊

     A1*(7)

 863
 864
.

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865
STEP Solutions 2016

Mathematics

STEP 9465/9470/9475

November 2016

866
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Language Assessment, which is part of Cambridge Assessment,
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867
Contents

STEP Mathematics (9465, 9470, 9475)

Solutions Page
STEP Mathematics I 5
STEP Mathematics II 19
STEP Mathematics III 24

868
4

869
 STEPI2016Solutions

Question1

Asastartertothepaper,thisisastraightforwardquestionintermsofitsearlydemandsandinvolveslittle
morethantheneedtosortoutsomeclearlyͲsignpostedalgebra.Tobeginwith,itisclearthat,whenevern
isodd,theexpression(xn+1)has(x+1)asafactor(bytheFactorTheorem),sothat
qn(x)= x 2 n  x 2 n  1  x 2 n  2  ...  x 3  x 2  x  1 .
Examiningthepn(x)’sinturn,usingthebinomialtheorem(andPascal’sTriangleforthecoefficients),gives
 
p1(x)= x 2  2 x  1  3x(1) = x 2  x  1 ,
p2(x)= x  4 x  6 x
4 3 2

 4 x  1  5 x( x 2  x  1) = x 4  x 3  x 2  x  1 ,and
p3(x)= x  6 x  15x
6 5 4 3

 20 x  15x 2  6 x  1 –7x( x 2  x  1 )2.
Expanding( x 2  x  1 )2isrelativelystraightforward,anditisrelativelyeasytoobtaintherequiredresults.

There are several ways to demonstrate that two given expressions of the given kind are not identically
equal.Oneistoexpandthembothaspolynomialsandshowthattheyarenotthesame.Inthiscase,
p4(x)= x 8  x 7  x 6  2 x 5  7 x 4  2 x 3  x 2  x  1 whileq4(x)= x 8  x 7  x 6  x 5  x 4  x 3  x 2  x  1 .
Alternatively,oneneedonlyshowthatonecorrespondingpairofcoefficientsarenotthesame–here,the
coefficientsof(say)x5arenotequal.However,thesimplestthingistofindanyonevalueofxforwhichthe
twoexpressionsgivedifferentvalues.Itturnsout,infact,thatonlyx=0actuallydoesgiveequaloutputs,
soalmostanychosenvalueofxwouldsuffice,andthekeyisthentochooseoneforwhichtheworking
19  1
involvestheminimumofeffort,suchasp4(1)=28–9.1.33=13zq4(1)= 1 .
11

In(ii)(a),thegivennumericalexpressionisclearlythatforq1(x)withx=300.Sincepandqarethesame
thingwhenn=1,weinsteadexaminep1(300),anditbecomesclearthatifxis3timesaperfectsquare(in
thiscase3u102)thenwecanusethedifferenceͲofͲtwoͲsquaresfactorisationon(301)2–(3u10)2togetthe
answer271u331.

Part(b)hasasimilarthinggoingon,buthereweneedxtobe7timesaperfectsquare,andwefindthat

>
we have 77  1 @
3 2
 78 714  77  1 , which again requires the use of the differenceͲofͲtwoͲsquares
2

>   3
@ >  
factorisationandyields 77  1  718  711  74 u 77  1  718  711  74 or
3
@
7 21
 3.7  3.7  1  7  7
14 7 18 11
 7  u 7  3.7
4 21 14
 3.7  1  718  711  7 4 ,
7

eitherofwhichanswerswouldsuffice.
              








5

870

Question2

Onceagain,thisquestionbeginswithaverysimpleinstruction,todifferentiate afunctionofx,andwill
clearlyinvolvetheuseofboththeProductRule(twice)andtheChainRule(inordertodealwiththelog.
termandthesquareͲroot).Inprinciple,thislooksverystraightforward,thoughthekeyistobecarefulnot
tooverlooksomeaspectofthevariousprocessesatplay,andthentosimplifytheresultingexpressionsina
suitableway.Tobeginwith,onewillobtainsomethingthatlooksquitemessy:

dy
dx
ax 2
 bx  c  1
x  1 x ©
2
u §¨1  12 1  x 2 > @ 1
2
¹

.2 x ·¸  2ax  b  ln x  1  x 2  
+ dx  e §¨ 12 1  x 2 2 .2 x ·¸  d 1  x 2 
©
1

¹
> @
anditiseasytobeputoff;itisespeciallyimportantnottoattempttoomuch“inyourhead”.Youshould
findthissimplifiesto
dy ax 2  bx  c > 1  x  x@+ 2ax  blnx 
2
 xdx  e
dx >x  1 x 2
@ u
1 x 2
1  x 2 +
1 x 2
 d 1 x2 

1 1 x 2
andcollectinguptermssuitably,andnotingthat { leadstoanexpressionwhichcontains
1 x 2 1 x2

onlysimplemultiplesof
1
1 x 2

and ln x  1  x 2 ;namely 
dy
dx
( a  2 d ) x 2  (b  e ) x  ( c  d )
1 x 2

+ 2ax  b  ln x  1  x 2 . 

Alltheresultsoftheremainingpartsofthequestioncannowbededucedbychoosingsuitablevaluesfor
theconstantsatoe.

In(i),choosinga=d=0,b=1,e=–1andc=0gives
dy (0) x 2  (0) x  (0)
dx 1 x 2
+ 0  1 ln x  1  x 2 ,so  
  
that ³ ln x  1  x 2 dx= x ln x  1  x 2  1  x 2 (+C). 
In(ii),choosinga=b=e=0andc=d= 12 gives
dy
dx
(0  1) x 2  (0) x  (1)
1 x 2
 
+ 0  0 ln x  1  x 2 ,sothat

³  
1  x 2 dx= 12 ln x  1  x 2  12 x 1  x 2 (+C).
Andin(iii),choosinga= 12 ,b=e=0,c= 14 andd=  14 gives
dy
dx
( 12  12 ) x 2  (0) x  ( 14  14 )
1 x 2

+  x  0 ln x  1  x 2  

andhence ³ x ln x  1  x 2 dx= 12 x 2   1
4 ln x  
1  x 2  14 x 1  x 2 (+C).
              




6

871
 
Question3

Ifyouhavenotseenthissortoffunctionbefore,thenitisworthwhileplayingaroundwithsuchthingsas
partofyourpreparationfortheSTEPs,whichfrequentlytestperfectlysimpleideasincontextsthatarenot
astandardpartofAͲlevel(orequivalent)courses.Beingabletothinkthingsthroughcalmlyandcarefully
underexaminationconditionsisanespeciallyhighͲlevelskill,butonethatcanbepractised.

In this case, the “integer part” function is a relatively simple one to deal with, as it only changes values
whenthefunctionitactsonhitsanintegervalue.Beforecommencingworkonthisquestion,notethatthe
“integerpart”ofanegativenumberistheonetotheleftofit(ifitliesbetweenintegers,ofcourse),and
manyfunctionͲplottingpackagesaresetto“goright”fornegativenumbers,whichisunfortunate.Thereis
also the small matter of how to illustrate the “y” values at those points when the “step” occurs … the
traditionistoemploya“filled”circle(භ)forinclusionandan“open”circle(ӑ)forexclusion.Withabitof
careyoushouldfindthatthefourgraphsrequiredherelookasfollows.

(i)         (ii)











(iii)         (iv)









              







872
Question4

AswithQ2,thisbeginswithasimpleinstructiontodifferentiate;again,theideaisthatyouwilltidyupthe
z
finalanswerforlaterreference.UsingtheQuotientandChainRuleson y gives
1 z2
1  z 2 .1  z. 12 1  z 2  2 .2 z
1
dy
 
= 2

dz 1 z2
1
whichsimplifiesto .
1  z 
3
2 2

The given expression in part (ii) initially appears to be quite awful, until you realise that writing, for
§ d2 y · dz
¨¨ 2 ¸¸
instance, z
dy
turns © dx ¹ N into dx N ,andthiscannowbeseentobeastandard
1  z 
3 3
dx ­° § dy · 2 ½° 2 2 2

®1  ¨ ¸ ¾
°̄ © dx ¹ °¿
dz
“separable variables” firstͲorder differential equation:  ³ ³ N dx . Using (i)’s result then gives
1  z 
3
2 2

z
N ( x  c) (wheretheusual“+c”hasbeenincorporatedintoaslightlymorehelpfulformhere).
1 z2
u
ReͲarranging this for z or z2 leads to  z 2 N 2 ( x  c) 2 ( z 2  1)  Ÿ z r
 where (again) the more
1 u2
complicatedͲlookingtermhasbeengivenanewlabel,whichisasimplebuteffectivedevicetomakewhat
todonextmoreobvious:here,u=N(x+c).

dy d y d u du
Wenowsubstitutebackforz= . andusetheChainRule(e.g.)with N toobtainanother
d x d u dx dx
dy u u
“separablevariables”firstͲorderdifferentialequation, N r or ³ N dy r ³ du.Atthis
du 1 u2 1 u2
u
point,youshouldbeabletoseethat ³ ducanbeintegrated(by“recognition”,“reversechainrule”
1 u2
orasubstitution) to give  1  u 2 . Substituting for u then gives ky  d # 1  N 2 ( x  c ) 2  and
squaring both sides leads towards a circle equation (Ny  d ) 2 1  N 2 ( x  c) 2  or
2 2
§ d· §1· § d·
( x  c) 2  ¨ y  ¸ ¨ ¸ ,whichistheequationofacircle,withcentre ¨  c,  ¸ andaradiuswhichis
© N ¹ ©N ¹ © N¹
thereciprocalofthecurvature .
              






873
Question5

Thissortofsituationisrelativelycommonin‘MathsChallenges’andtheusualapproachistojoinallthe
circles’ centres to the points of tangential contact and then form some rightͲangled triangles by
considering(here)thehorizontallinethroughB’scentre.BecauseofthewellͲknown(GCSE)CircleTheorem
“tangent perpendicularto radius” result, it is the case that each of AB, BC and CA is astraight line. This
enablesustousePythagoras’Theorem:
PR=PQ+QRŸ ( a  c ) 2  ( a  c ) 2 = (b  a ) 2  (b  a ) 2 + (c  b ) 2  (c  b ) 2 
which simplifies to 4ac = 4ab + 4bc  and, upon division throughout by 4abc , gives the required
1 1 1
answer  .
b c a

There are many ways to approach the next result, but it should be clear that each will, at some stage,
require the replacement of the b’s with a’s and c’s (or equivalent). The most direct route would be to
examinetheLHSandRHSof(**)separately,andthenshowthattheymatchup.Thiswouldlooklike:
§ 1 1 1 · 2 2 § 1 4 6 4 1 · 4 12 4 8 8
LHS= 2¨ 2  2  2 ¸  2  2¨¨ 2     2 ¸¸ = 2   2   .
©a c ¹ 2
b a c ©a a ac ac c ac c ¹ a ac c a ac c ac 
2
§ 1 ­1 1½ 1 ·
2 2 2
§ 1 1 1· 2 §2 2 2· §1 1 1·
RHS= ¨   ¸ ¨ ® 
¨a  ¾  ¸¸ ¨   ¸ 4¨   ¸
©a b c¹ © ¯a ac c ¿ c ¹ ©a ac c ¹ ©a ac c ¹ 
§ 1 3 1 2 2 ·
4¨ 2   2   ¸ ,andtheseareclearlythesame.
 © a ac c a ac c ac ¹

Working in the other direction is trickier, but not much more so, and it is again helpful to reͲlabel the
variables to make things look simpler, especially if we can somehow remove the need for everything to
appearasafraction.So,followinganinitialobservationthat
2
§ 1 1 1 · §1 1 1· 1 1 1 2 2 2
2¨ 2  2  2 ¸ = ¨   ¸ Ÿ 2  2  2   
©a b c ¹ ©a b c¹ a b c ab bc ca
1 1 1
wecouldwrite x ,y ,z ,sothatwearenowtryingtoprovethat
a b c
x4  y4  z4 2x2 y 2  2 y 2 z 2  2z 2 x2 . 
(Althoughitisnotessentialtodothisatthisstage,itisoftenthecasethatfolksforgettodoitattheendif
theydon’t;andthatistoconsiderthegivenconditionsb<c<a,whichtranslatetoy>z>x.)

Now, completing the square: x 2  z 2  y 2 2
4 x 2 z 2  œ x 2  z 2  y 2 r 2 xz  œ z # x 
2
y 2 , and there
arethefourcasestoconsider:y=x–z,y=z–x,y=x+zory=–x–z.Considerationoftheabove
1 1 1
conditionsonx,y,zthenshowsthatonlyy=x+zissuitable,andso  ,asrequired.
b c a
              






874
Question6

This is a fairly straightforward vectors question that involves little beyond working with the vector
equationofaline.Tobeginwith,youarerequiredtoexplainacoupleofintroductoryresultsthatrelyon
thefactthattwo(nonͲzero)vectorsaremultiplesofeachotherifandonlyiftheyareparallel.Thus,OX||
OAŸx=ma,with0<m<1,sinceXisbetweenOandA;andBC||OAŸc–b=kaandsoc=ka+b,with
k<0sinceBCisintheoppositedirectiontoOA.

Then, lines OB and AC have vector equations r = E b and r = a + D (c – a) respectively, for some scalar
parametersDandE.Replacingcbyka+bandequatingthetwor’sforthepointofintersectionthengives
Eb=a+D(ka+b–a).Sinceaandbarenotparallel,wecanequatetermstofindthat1–D+Dk=0and
1 1
D=E.SolvingleadstoD=E ,sothatd= b.
1 k 1 k

§ 1 ·
Inanexactlysimilarway,wethenhaveY=XDˆBCŸma+D ¨ b  ma ¸ =b+Eka.(Notethatthereis
©1 k ¹
no reason why we have to use differentsymbols for the scalar parameterseach time, as they are of no
D
actualsignificanceinrelationtotheproblem.)Equatingcoefficients Ÿm– Dm– Ek=0and 1 ,so
1 k
thaty=kma+b.

§ 1 ·
Next,Z=OYˆABŸ(1–D)a+Db=E kma  b  ,andso1–D–kmE=0andD=E ¨ ¸ ,givingz=
© 1  km ¹
§ km · § 1 · 1 § km 1 1 ·
¨ ¸ a+ ¨ ¸ b;andT=DZ ˆOA Ÿ Da= b+ E ¨ a b b ¸ ,whence D=
© 1  km ¹ © 1  km ¹ 1  k © 1  km 1  km 1  k ¹
Ekm 1 E E § m ·
and0=  ,sothatt= ¨ ¸a .
1  km 1  k 1  km ©1 m ¹

Noticethatallthathasbeendonesofaristoworkoutthepositionvectorsofthepointscreatedasthe
intersectionsofvariouspairsoflines.Ifthisisdifficulttovisualize,thenadiagramshouldbedrawnfirst.

Allthatremainsistosetupthelengthsofthevariouslinesegmentsofinterest.IfwecallOA=a,thenit
§ m · § m2 ·
¸¸a , TA = §¨
1 ·
follows that OX = ma, OT = ¨ ¸ a , TX = ¨¨ ¸a and XA = (1 – m)a. (Note that the
© 1  m ¹ © 1  m ¹ © m¹
1 
shrewd solver would simply take, w.l.o.g., the value of a to be 1, as it isan arbitrary length and cancels
throughoutanyoftheworkingthatfollowsinordertoobtainthetwogivenanswers,
1 1§ 1· 1 1 § m · 2 § 1 ·
¨1  ¸  andOT.OA= ¨ ¸a (ma ) . ¨ ¸a =OX.TA.
OT a © m ¹ OA OX  ©1 m ¹ ©1 m ¹
              





10

875
Question7

Firstly,S‰T=thesetofallpositiveoddnumbers;andSˆT=I.

Next,wemustshowthattheproductoftwodifferentelementsofSisalsoanelementofS,andthendoa
similarthingfortwodifferentelementsofT.Thisinvolvesnotingthat
(4a+1)(4b+1)=4(4ab+a+b)+1,
whichisanelementofS;and
(4a+3)(4b+3)=4(4ab+3a+3b+2)+1,
whichis,infact,inSratherthanT.

Theresultofpart(iii)essentiallyrequiresproofbycontradiction,sowefirstsupposethattisanelementof
T,andthatalloft’sprimefactorsareinS.Notingthattherearenoevenfactors,wecanwrite
t=(4a+1)(4b+1)(4c+1)…(4n+1).
ButwehavealreadynotedthattheproductofanypairofelementsofSwillalwaysyieldanotherelement
ofS,andhence(inductively,Isuppose),t=4{………}+1isalwaysinS.Whichcontradictstheassumption
thattisanelementofT.
 
Forpart(iv)(a),wenotethatanelementofTiseitherTͲprimeorTͲcomposite.Ifitisthelatter,thenitcan
beexpressedasaproductofTͲprimes.However,wehavealreadyestablishedthateverypairoffactorsinT
multiply to give an element of S, as do every pair of elements of T. So, after every pairingͲup of this
element’s factors, there must be an odd one left over to multiply by in order to give an element of T.
Hence,altogether,thereisanoddnumberofthem.

Forthefinalpartofthequestion,wearerequiredtofindnonͲprimesinSthatareproductsofelementsof
T that can be “reͲgrouped” suitably. One such example involves the numbers 9, 21, 33 and 77, each of
whichisbothSͲprimeandaproductofelementsinT:
9=3u3,21=3u7,33=3u11and77=7u11;with3,7and11inT.
This leads to the example 9 u 77 = 21 u 33 (= 693). Of course, the main purpose of the question is to
demonstrate the existence of perfectly reasonable numberͲsets (in this case, S), having the standard
properties of multiplication, yet for which the “unique factorisation” principle no longer holds. This is a
very important principle relating to prime numbers within the set of positive integers, which you have
beentaught(quiterightly)toassume,butthatIimagineyouhaveneverhadreasontothinkthatitmight
notnecessarilyholdinthiscase,orinothersimilarsituations.
              










11

876
Question8

Thisisasimpleideainvolvingtermsofseries. Themostimportantthing here istobesurethatyoucan
justifytheformofthenthtermforanygivenseries.Tobeginwith,itishelpfultorealisethattheBinomial
Theoremappliedtothenegativeintegers,givesthefollowingresults:
(1  x) 1 1  x  x 2  x 3  ...  x n  ... ,withthecoefficientofxnbeing1;
(1  x) 2 1  2 x  3x 2  4 x 3  ...  (n  1) x n  ... ,withthecoefficientofxnbeing(n+1);and
(1  x) 3 1  3x  6 x 2  10 x 3  ...  12 (n  1)(n  2) x n  ... ,havingthetriangularnumbersascoefficients;
etc.
Havingestablishedtheseresults,itiseasytoshowthat
0  x  2 x 2  3x 3  ...  nx n  ... =x (1  x) 2 
andthat
x(1  x) 3  
x 1  3x  6 x 2  10 x 3  ...  12 n(n  1) x n  1  ... = 0  x  3x 2  6 x 3  ...  12 n(n  1) x n  ... 
hascoefficientofx equaltoun= 12 n  12 n  .
n 2


2x x
Usingthesefirsttworesults:2u(2nd)–(1st)gives  withun=n2. 
(1  x) (1  x)
3 2


Thereareseveralwaystoproceedwithpart(ii)(a);thesimplestbeingtonotethat

a  akx  ak 2 x 2  ak 3 x 3  ...  ak n x n  ... = a  kx a  akx  ak 2 x 2  ak 3 x 3  ...  ak n x n  ...  
=a+kxf(x)
§ 1 ·
andsof(x) a¨ ¸ .
© 1  kx ¹
Forpart(ii)(b),youshouldnotethatthegivensecondͲorderrecurrencerelation(i.e.twoprecedingterms
areinvolved)requirestwostartingtermsbeforethingsgetgoingsystematically,soitisbesttosplitoffthe
firsttwotermsoftheseriesbeforeattemptingtomakeuseofthisdefiningfeature.
f f f f
Writingf(x)= 0  x  ¦ u n x n = 0  x  ¦ u n  1  u n  2 x n = x  x ¦ u n  1 x n  1  x 2 ¦ u n  2 x n  2 
n 2 n 2 n 2 n 2

Note:wearetryingtoreͲcreatef(x)ontherightͲhandside    
f f
  = x  x ¦ u n x n  x 2 ¦ u n x n = x  x^f ( x)  0` x 2 f ( x) 
n 1 n 0

x
sothatf(x)= .
1 x  x2
              









12

877
Question9

AswithallStaticsproblems,itissoimportanttohaveacleardiagram,suitablymarkedwithalltherelevant
forces(andpreferablyinthecorrectdirections).Thegivendiagramappearstogivethesituationwhenthe
horizontalrailismorethanhalfͲwayalongtherod;inthisconfiguration,therodwillslipsothattheendA
slidesdownthewall,andtotheleftrelativetothecontactatP(asshownbelow).

      RP
        B 
   FAGbPFP  
    Ta     
   A  RA


      W
      
     d

Thenextessentialpartofanysolutionistowritedownallthekeystatementsbeforeattemptingtoprocess
them in some way. The guidance “resolve twice and take moments” is surely a stock part of every
mechanics teacher’s repertoire, and it is sound advice. It is customary to resolve in two perpendicular
directions (vertically and horizontally here) and to find some point about which to take moments that
minimisesthe“clutter”ofsubsequentalgebraͲandͲtrigonometry(Ahasbeenchosenhere).Theadditional
useoftheFrictionLawfor,onthisoccasion,thecaseoflimitingequilibriumatthetwopointsofcontactis
alsoneeded.Thisgivesusthefollowing“ingredients”foruseinasolution.
FrictionLaw FA=ORAandFP=PRP.
Res.n W=FA+RPsinT+FPcosT
Res.o RA=RPcosT–FPsinT
օA  WasinT=RP(a+b)

EliminatingtheF’sfromthetworesolvingstatementsgives
W=ORA+RPsinT+PRPcosTandRA=RPcosT–PRPsinT
andintroducingdinthemomentsstatement(notingthatitneedstoappearintheanswer)gives
Wasin2T=RPd.
SincethislastequationinvolvesjusttheforcesWandRPitmakessensetoeliminateRAnext,toget
W RP O cosT  OP sin T  sin T  P cos T . 
Finally,dividingtheselasttwoleadstothegivenanswer,
dcosec2T a[O  P ] cosT  [1  OP ]sin T  .

ForthecasewhenPislessthanhalfͲwayalongtherod,whichwillnowslideupthewallatA(etc.),wecan
simplywriteFA o–FA;FP o–FP;a+b oa–b,orjustswitchthesignsof Oand P,aseverythingelse
remainsunchanged.Thecorrespondingresultisthusshowntobe
dcosec2T a [O  P ] cosT  [1  OP ]sin T  .
              

13

878
   
Question10

When it comes to it, collisions questions involve only the use of two (sometimes three, if energy
considerations are involved) principles. In this case, the principles of Conservation of Linear Momentum
(CLM)andNewton’s(Experimental)LawofRestitution(NELorNLR).Diagramsarealsoquiteimportantin
thesetypesofquestions,butlargelytoenablethesolvertobeclearaboutwhatdirectionsarebeingtaken
aspositive:rememberthatvelocityandmomentumarevectorialinnature.Ifyouareeverunsureabout
whichdirectionanyoftheobjects(particlesorspheres,etc.)willbemovinginafterthecollisionhastaken
place,thenjusthavethemallgoingthesameway;thismakesitmucheasiertointerpretnegativesignsin
anylateranswers.

Tobeginwith,therearetwoseparatecollisions,betweenA/BandC/D,asshownbelow.

  ou  o0  0m   um


    A  B    C  D


    ovA ovB   mvC  mvD

ForcollisionA/B  CLM:m(Ou=OvA+vB)andNEL:eu=vB–vA
andforcollisionC/D  CLM:m(u=vC+vD)andNEL:eu=vC–vD
Solvingeachpairofequations,separately,gives
O e O (1  e)
vA u , v B u , vC 12 (1  e)u and vD 12 (1  e)u ;
O 1 O 1
eventhoughsomeoftheseturnoutnottoberequired(thoughtheymaybelateron,ofcourse).

ThereisthenafurthercollisionbetweenB/C:

  ovB  vCm  

    B  C    


    o0  owC   
CLM:m(vB–vC)=mwCandNEL:e(vB+vC)=wC

Now,substitutingpreviousanswersintermsofeandu,andidentifyingfore,leadstotherequiredanswer
O 1
e .Tojustifythefollowingconditionone,notethat
3O  1
O  1 1 § 3O  3 · 1 § 3O  1  4 · 1 4
e ¨ ¸ ¨ ¸  3
 1
3O  1 © 3O  1 ¹ © 3O  1 ¹ 3O  1 3
3 3 3

sincethetermbeingsubtractedispositive.(Itisnotsufficientsimplytoshowthateo 13 .)  


14 879
 (1  e)(O  1)
Finally,usingwC= u frompreviouswork,andequatingthistovD,gives
2(O  1)
(1  e)(O  1)
u = 12 (1  e)u ;
2(O  1)
andsubstitutingfore(e.g.)enablesustosolveforOandthenfinde: O 5  2 ,e= 5  2 .
              

 

15

880
Question11

Projectilesquestionsareoftenstraightforwardinprinciple,butcanoftenrequirebothcarefulthoughtand
sometrickytrigonometricmanipulation.Occasionally,thereistheadditionalrequirementtomaximiseor
minimisesomethingͲorͲotherusingcalculus.
Tobeginwith,itisusefultoknow(ortobeabletoderivequickly)theTrajectoryEquationoftheparabola:
gx 2 gx 2
y x tan D  .Settingy=–handreͲarrangingthengives 2h cos 2 D  2 x sin D cosD ,andit
2u cos D
2 2
u 2

shouldbefairlyclearfromthegivenanswerthatalittlebitofworkusingthedoubleͲangleformulaewill
gx 2
leadtotherequiredresult h(1  cos 2D )  x sin 2D .
u2

d § gx 2 · § dx ·
Differentiatingw.r.t.DŸ ¨¨ 2 ¸¸ h 2 sin 2D   ¨ x.2 cos 2D  sin 2D . ¸ .Noticethatthereisactually
dD © u ¹ © dD ¹

noneedtoreͲarrangefor
d
... sincewerequirebothderivativetermstodisappear.Thisimmediately
dD
gh 2 tan 2 2D
leads to x = h tan2D ; and, substituting back, we get  h(1  cos 2D )  h tan 2D sin 2D . By
u2
cancellingoneoftheh’sand(e.g.)writingalltrig.termsinc=cos2Dthenyields

gh 1  c 2  1 c 
1 c
2
Ÿ gh  ghc 2  
u 2 c 2  c 3  c  c 3 .
u 2c 2 c
Asaquadraticinc: 0 u 2
 >  @
 gh c 2  u 2 c  gh = u 2  gh c  gh (c  1) …notethatitisalwaysworthtrying
tofactorisebeforedeployingthemessierquadraticformula.
gh
Wenowhavecos2D= 2 ,andsubstitutingx=htan2Dandy=–hin '2 x 2  y 2 ,thengives
u  gh
'2 h2sec22Di.e. ' h sec 2D 
u 2  gh u2
sothat ' h.  h ,asrequired.
gh g
              















16

881
Question12

Bizarrely,thisprobabilityquestionconcerningtossingfaircoinshasallofitsanswersequalto 12 .Asone
mightimagine,suchananswercanbeobtainedinverymanywaysindeedandsothekeyistomakeyour
workingreallyclearastohowtheanswerisarrivedat.Merelywritingdownawholeloadoffractionson
thepageisreallynotveryenlightening.Thereneedstobesome(visible)systematicapproachtocounting
cases,followedbythenumericalworkthatgoeswithit.Forinstance,inpart(i),wecouldbreakdownthe
answerintoeachpossiblevalueforAandthevaluesofBthatcouldthengowithit.
e.g. p(A=0).p(B=1,2or3)+p(A=1).p(B=2or3)+p(A=2).p(B=3)= 14 u 78  2 u 14 u 84  14 u 18 = 12 .

Forpart(ii),oneshouldbynowhavebeenableto“see”howtheresultsarise,andcanappealtoa“similar”
process;e.g. 18 u  461641   83 u  61641   83 u  4161   18 u 161  …noticetheappearanceofthebinomialcoefficients
1
forcountingtherelevantnumbersofB’spossiblevalues.Thisgives 128 15  3 u11  3 u 5  1 1
128 64 = 12 .

Inpart(iii),youshouldnotethat,wheneachofthemhastossedncoins,
p(BhasmoreHs)=p(AhasmoreHs)=p2
andthatp(AH=BH)=p1.Thusp1+2p2=1.

Next,consideringwhathappenswhenBtossestheextracoin,
p(BhasmoreHs)=p(BalreadyhasmoreHs)up(BgetsT)
+p(Balreadyhasmore,orequal,Hs)up(BgetsH)
= p2 u 2  ( p1  p2 ) u 12 = 12 ( p1  2 p2 ) 12 .
1

              

 

17

882
Question13

FortheiͲtheͲmail,thepdfis f i (t ) Oe Ot .Integratingthisgivesthecdf: Fi (t )  e  Ot +C,andF(0)=0Ÿ
C=1.

Then,forneͲmailssentsimultaneously,
 n
F(t)=P(Tdt)=1–P(allntakelongerthant)= 1  e Ot ,
(usingtheproductofnindependentprobabilities)
= 1  Oe Ont .
DifferentiatingthisthengivestherequiredpdfofT,f(t)= nOe Ont .

f
Findinganexpectedvalueisastandardintegrationprocess:E(T)= ³ t u nOe Ont dt,andthisrequiresthe
0

f
ª  e Ont º f 1
>
useofintegrationbyparts:E(T)=  te Ont @ + ³ nOe O
f  nt
dt=[0]+ « » = .
¬ O n ¼ 0 nO
0
0


Fortheveryfinalpart,onecouldgoagainthroughtherouteofpdfsandcdfs,butitshouldbeobviousthat
thewaitingͲtimeforthe2ndemailissimplythe1stfromtheremaining(n–1)…withexpectedarrivaltime
1 1 1 1 §1 1 ·
,givingatotalwaitingtimeof  ¨¨  ¸ .
(n  1)O nO (n  1)O O © n (n  1) ¸¹
              

18

883
 STEP//2016Solutions
Question1
Use‫ݐ‬ଵ and‫ݐ‬ଶ torepresentthevalueoftheparameter‫ݐ‬ateachofthepointsPandQ.The
equationsofthetwotangentscanthereforebefoundintermsof‫ݐ‬ଵ and‫ݐ‬ଶ andthefactthat
POQisarightanglecanbeusedtofindarelationshipbetween‫ݐ‬ଵ and‫ݐ‬ଶ .Thepointof
intersectionofthetwotangentscanthereforebefoundintermsofjust‫ݐ‬ଵ andthisisapair
ofparametricequationsforthecurvethatthepointofintersectionmakes.
Substitutingtheparametricequationsfor‫ܥ‬ଵ intotheequationfor‫ܥ‬ଶ givesacubicequation
in‫ ݐ‬ଶ whichcanbesolvedbyinspectiontoshowthattherearejusttwointersectionsandso
thetwocurvesjusttouch,butdonotcross.

Question2
Substituteܿ ൌ ܽ ൅ ܾintotheexpressiontoshowthatܽ ൅ ܾ െ ܿisafactor.Oncethisis
done,thesymmetryshowsthatܾ ൅ ܿ െ ܽandܿ ൅ ܽ െ ܾmustalsobefactorsandtherefore
thereisjustaconstantmultiplierthatneedstobededucedtoobtainthefullfactorisationof
(*).
Forpart(i),choicesofܽ,ܾandܿneedtobemadesothat
ܽ ൅ ܾ ൅ ܿ ൌ ‫ ݔ‬൅ ͳ
ʹ‫ ݔ‬ଶ ൅ ͷ
ܽଶ ൅ ܾ ଶ ൅ ܿ ଶ ൌ 
ʹ
Ͷ‫ ݔ‬ଷ ൅ ͳ͵
ܽଷ ൅ ܾଷ ൅ ܿ ଷ ൌ 
Ͷ
Oncethesehavebeenidentifiedthesolutionstotheequationfollowfromthefactorisation
alreadydeduced.
Oncethesubstitution݀ ൅ ݁ ൌ ܿhasbeenmadeitisonlynecessarytoidentifythepartsof
theexpressionwhichdifferfrom(*)inthefirstpartofthequestion(whicharisefromtheܿ ଶ 
andܿ ଷ terms).Thefactorisationandsolutionoftheequationthenfollowasimilarprocess
tothefirstpartofthequestion.


 

19

884
Question3
Thedifferentiationtoshowtheresultinpart(i)shouldnotpresentmuchdifficulty,although
itisimportanttoshowthatalloftheterms(andnoothers)arepresent.
Forpart(ii)observethateachindividualtermof݂௡ ሺ‫ݔ‬ሻhasapositivecoefficient,soforany
positivevalueof‫ݔ‬thevalueof݂௡ ሺ‫ݔ‬ሻmustbepositive.
Forpart(iii),usetheresultinpart(i)torewrite݂௡ Ԣሺ‫ݔ‬ሻintermsof݂௡ ሺ‫ݔ‬ሻandnotethat݂௡ ሺܽሻ
and݂௡ ሺܾሻmustbe0.Thismeansthatanypairofrootsmusthaveagradientofthesame
sign,whichleadstoanargumentthattheremustbeanotherrootbetweenthetwo.Asthis
wouldleadtoaninfinitenumberofrootstoapolynomial,therecannotbemorethanone
root.
Toestablishthenumberofrootsinthetwocasesconsiderthebehaviourofthegraphas
‫ ݔ‬՜ λandas‫ ݔ‬՜ െλ

Question4
Theequationgivencanberewrittenasaquadraticin‫ݔ‬.Thediscriminantthenestablishes
therequiredresult.Toshowthesecondresult,showthat‫ݕ‬ଶ ൅ ͳ ൒ ሺ‫ ߠ •‘… ݕ‬െ •‹ ߠሻଶ ,
whichcanbeshownbywriting‫ ߠ •‘… ݕ‬െ •‹ ߠintheformܴ …‘•ሺߠ ൅ ߙሻandthenthisresult
isaquadraticinequalitythatleadsdirectlytothenextresult.
ସାξ଻
Inthecase‫ ݕ‬ൌ ,carefulmanipulationofsurdsshowstherequiredresultandsothe
ଷ
valueofߠmustbethevalueofߙobtainedintheprevioussection.Finally,thevalueof‫ݔ‬
canbeobtainedbyreturningtotheoriginalequationandsubstitutinginthevaluesthatare
known.

 

20

885
Question5
Thebinomialexpansionforሺͳ െ ‫ݔ‬ሻିே shouldbeeasyenough,itisthenrequiredtowrite
‫݌‬
theproductintermsoffactorialssothattheexpressioncanbewrittenintermsofቀ‫ ݍ‬ቁ.

Sincetheexpansionofሺͳ െ ‫ݔ‬ሻିଵ involvesacoefficientof1foreveryterm,thecoefficientof

‫ ݔ‬௡ intheexpansionofሺͳ െ ‫ݔ‬ሻିଵ ሺͳ െ ‫ݔ‬ሻିே issimplythesumofthecoefficientsofallofthe
termsintheexpansionofሺͳ െ ‫ݔ‬ሻିே uptoandincludingthetermin‫ ݔ‬௡ .
TheproductsinthesumontherightͲhandsideoftheresultinpart(ii)shouldbe
recognisableasbinomialcoefficientsinthecasewherethepowerisapositiveinteger,so
use
ሺͳ ൅ ‫ݔ‬ሻ௣ ሺͳ ൅ ‫ݔ‬ሻ௤ ‫ ؠ‬ሺͳ ൅ ‫ݔ‬ሻ௣ା௤ 
andcomparecoefficientsasinpart(i).
Similarlyforpart(iii),identifythattheresultwillcomefromconsiderationof
ሺͳ ൅ ‫ݔ‬ሻேା௠ ሺͳ ൅ ‫ݔ‬ሻି௠ ‫ ؠ‬ሺͳ ൅ ‫ݔ‬ሻே Ǥ

Question6
Parts(i)and(ii)onlyrequireverificationineachofthecases,sosimplydifferentiatethe
functionsgivenandsubstituteintothedifferentialequationtoconfirmthattheyare
solutions.Remembertocheckaswellthattheboundaryconditionsaresatisfied.
Forpart(iii),differentiatethegivenformulafor‫ݖ‬andsubstituteintothedifferential
equation.Byobservingthatthenewdifferentialequationisofthesameformas(*),butfor
ʹ݊insteadof݊,theexpressionfor‫ݕ‬ଶ௡ ሺ‫ݔ‬ሻcanbeestablished.
Forpart(iv),againdifferentiatethegivenformula,beingcarefulabouttheapplicationofthe
chainruleandsubstitute.Again,bycomparingwith(*)thefinalresultshouldbeclear.

Question7
Thefirstresultcanbeshownbyusingasubstitutionintotheintegral,beingcarefulto
explainthechangeofsignwhenthelimitsoftheintegralareswitched.
Simpleapplicationofknowledgeoftrigonometricgraphsoncethesubstitutionhasbeen
madecanbeusedtoshowthattwicetheintegralisequivalenttointegratingthefunction1
overtheinterval.
Similarly,theremainingintegralscanallberearrangedusingstandardtrigonometric
identitiesandknowledgeoflogarithmsintoformsthatcanbeintegratedfromstandard
resultsoncethesubstitutionfrom(*)hasbeenmade.


21

886

Question8
Theintegralrequiredatthestartofthequestionshouldbeastraightforwardoneto
evaluate.Whenmakingasketchtoillustratetheresultinthesecondpart,ensurethatthe
sumisindicatedbyaseriesofrectangles,withthegraphofthecurvepassingthroughthe
midpointsofthetops.
Inpart(i),theintegralthatwouldmatchthesumgivenresultsinananswerof2,sothisis
thefirstoftheestimates.Theremainingestimatesarisefromusingtheintegraltoestimate
mostofthesum,buttakingthefirstfewtermsastheexactvalues(soineachcasethe
integrationistakenfromadifferentlowerlimit).
Forpart(ii),evaluatetheintegralforoneparticulartermofthesumandnotethatitis
ଵ ଷଷ
approximately .Finally,usingthemostaccurateestimatefor‫ܧ‬ቀ ቁthesumfrom‫ ݎ‬ൌ ͵
ସ௥ ర ଶ଴
ଵ
onwardscanbecalculatedandthenthefirsttwovaluesof రcanbeaddedtoachievethe
௥
desiredresult.

Question9
Theresultinpart(i)followsfromconsiderationofkineticenergylostandworkdone.
Inpart(ii)applyconservationofmomentumtothecombinedblockandbulletafterthe
bullethitstheblock.Bycomparingtothecaseinpart(i)themotionofthebulletuntilitisat
restrelativetotheblockcanbeanalysed.Oncealloftherelevantequationsofmotionhave
beenwrittendown,aseriesofsimultaneousequationswillhavebeenfoundfromwhichthe
valuesofܾandܿcanbefound.

Question10
Thefirstrequirementwillbetofindthecentreofmassofthetriangle.Oncethisisdonea
diagramwillbeveryusefulandnotationswillneedtobeaddedforvariousdistances,angles
andthefrictionalforce.Fromthisdiagramtheforcescanberesolvedintwoperpendicular
directionsandmomentscanbetaken.Thisleadstoaseriesofequationswhichcanbe
solvedtoworkoutthevaluethatthefrictionalforcewouldhavetotaketopreventslipping.
Fromthistherequiredresultcanbeestablished.

 

22

887
Question11
Theparticlesmustcollideiftheywouldbeinthesamepositionforoneparticularvalueof‫ݐ‬.
Therefore,writingouttheequationsofmotionforthetwoparticlesandeliminatingthe
variablesthatarenotneededtherequiredresultcanbereached.
Forthesecondpart,thetimeofthecollisioncanbefoundbyconsideringtheheightsofthe
bulletandtargetattime‫ݐ‬andnotingthatthesemustbeequal.Oncethevalueof‫ݐ‬has
beenfound,thefactthatthismustbepositiveleadstotheinequalitythatisrequiredforthe
firstresult.
Forthefinalpart,notethatgravityaffectsboththebulletandtargetinthesameway,soifit
isignoredthenthetimeofcollision(ifthereisone)willbethesameandthisisasituationas
inpart(i).Clearly,inpart(i)thetwoobjectsmustbemovingtowardseachotherifthereis
tobeacollision.

Question12
Replace‫ܤ‬withሺ‫ܥ ׫ ܤ‬ሻintheresultthatyoumuststartwithandthenobservethat
‫ ת ܣ‬ሺ‫ܥ ׫ ܤ‬ሻisthesameasሺ‫ܤ ת ܣ‬ሻ ‫ ׫‬ሺ‫ܥ ת ܣ‬ሻ.Thecorrespondingresultforfourevents
shouldbeclear,butcaremustbetakentoincludeallofthepossiblepairs.
Theresultsforparts(i),(ii)and(iii)shouldbeclearfromconsiderationofarrangementsin
eachcaseandtheresultrequiredfollowsfromthegeneralisationoftheresultfromthestart
ofthequestion.
Theprobabilitythatthefirstcardisinthecorrectpositionandnoneoftheothersiscanbe
establishedandthereforetheprobabilitythatexactlyonecardisinthecorrectpositionwill
be݊timesthat.

Question13
Forpart(i)theapproximationofthebinomialdistributionbyanormaldistributionshould
beknownandtheareaunderthecurve(applyingacontinuitycorrection)canthenbe
approximatedbyarectangle.
Thesecondresultfollowsfromasimilarapproximationandtheuseoftheformulafora
probabilityfromthebinomialdistribution.
Part(iii)followsfromanapproximationofaPoissondistributionwithanormaldistribution
andagainapproximatingtherequiredareabyarectangle.

23

888
^dW///ϮϬϭϲ^ŽůƵƚŝŽŶƐ

1. Part(i)ismostsimplydealtwithbythesuggestedmethod,changeofvariable,anditis
worthcompletingthesquareinthedenominatortosimplifythealgebraleadingtoatrivialintegral.
Part(ii)caneitherbeattemptedimmediatelyusingintegrationbyparts,startingfrom‫ܫ‬௡ and
ஶ ଶ௡௫ሺ௫ା௔ሻ
obtaining‫ି׬‬ஶ ሺ௫ మ ݀‫ ݔ‬andthenwritingthenumeratoras
ାଶ௔௫ା௕ሻ೙శభ
ʹ݊ሺ‫ ݔ‬ଶ ൅ ʹܽ‫ ݔ‬൅ ܾሻ െ ݊ܽሺʹ‫ ݔ‬൅ ʹܽሻ െ ʹ݊ሺܾ െ ܽଶ ሻ.Alternatively,useofthesamesubstitutionin
‫ܫ‬௡ାଵ asinpart(i)leadstotheneedtointegrate…‘• ଶ௡ ‫ݑ‬,whichinturncanbewrittenas
…‘• ଶ௡ିଶ ‫ݑ‬ሺͳ െ •‹ଶ ‫ݑ‬ሻ ൌ …‘• ଶ௡ିଶ ‫ ݑ‬െ …‘• ଶ௡ିଶ ‫ݑ ‹• ݑ‬Ǥ •‹ ‫ݑ‬,withthesecondtermbeingsusceptible
tointegrationbyparts.Part(iii)followsfromthepreviouspartsbyinductionusingpart(ii)to
achievetheinductivestepand(i)thebasecase.

2. Therearenumerouscorrectwaysthroughthisquestion.Workingparametricallywith
‫ ݔ‬ൌ ܽ‫ ݐ‬,‫ ݕ‬ൌ ʹܽ‫ݐ‬givesanormalas‫ ݔݐ‬൅ ‫ ݕ‬ൌ ܽ‫ ݐ‬ଷ ൅ ʹܽ‫ݐ‬andimposingthatthispassesthrough
ଶ

ሺܽ‫݌‬ଶ ǡ ʹܽ‫݌‬ሻyields‫ ݐ‬ଶ ൅ ‫ ݌ݐ‬൅ ʹ ൌ Ͳ(*)whichhasroots‫ݍ‬and‫ݎ‬,theformergiving(i).Asa

consequence,‫ ݎ‬൅ ‫ ݍ‬ൌ െ‫݌‬and‫ ݍݎ‬ൌ ʹ,sothatQR,ʹ‫ ݔ‬െ ሺ‫ ݎ‬൅ ‫ݍ‬ሻ‫ ݕ‬൅ ʹܽ‫ ݎݍ‬ൌ Ͳ,simplifiesto
ିଶ௔
ʹ‫ ݔ‬൅ ‫ ݕ݌‬൅ Ͷܽ ൌ Ͳ,andthuspassesthroughሺെʹܽǡ Ͳሻfor(ii).Tcanbeshowntobeቀെܽǡ ቁ,
௣
which,ofcourse,lieson‫ ݔ‬ൌ െܽ,andas(*)hadtworealdistinctroots,‫ݍ‬and‫ݎ‬,‫݌‬ଶ െ ͺ ൐ Ͳ,which
ିଶ௔ ௔
yieldsቚ ቚ൏ .
௣ ξଶ

3. Differentiating,multiplyingbydenominatorsanddividingbytheexponentialfunction,gives
ሾܳሺܲ ൅ ܲԢሻ െ ܲܳԢሿሺ‫ ݔ‬൅ ͳሻଶ ൌ ሺ‫ ݔ‬ଷ െ ʹሻܳଶ which,invokingthefactortheorem,givesthefirst
requiredresult.DenotingthedegreeofPbypandthatofQbyqinthisexpressionyields
‫ ݌‬൅ ‫ ݍ‬൅ ʹ ൌ ʹ‫ ݍ‬൅ ͵andhencethedesiredresultin(i).Furthermore,inthegivencase,substitution
inthesameresultandpostulatingܲሺ‫ݔ‬ሻ ൌ ܽ‫ ݔ‬ଶ ൅ ܾ‫ ݔ‬൅ ܿyieldsconsistentequationsforܽ,ܾandܿ
andthusܲሺ‫ݔ‬ሻ ൌ ‫ ݔ‬ଶ െ ʹ‫ݔ‬.
Forpart(ii),commencingasinpart(i)demonstratesagainthatQhasafactorሺ‫ ݔ‬൅ ͳሻas
ሾܳሺܲ ൅ ܲԢሻ െ ܲܳԢሿሺ‫ ݔ‬൅ ͳሻ ൌ ܳଶ .Supposingܳሺ‫ݔ‬ሻ ൌ ሺ‫ ݔ‬൅ ͳሻ௡ ܵሺ‫ݔ‬ሻ,where݊ ൒ ʹand
ܵሺെͳሻ ് Ͳ,withܲሺെͳሻ ് Ͳandsubstitutingintheexpressionalreadyderivedleadstoa
contradiction.
ሺ௫ିଵሻ௫ ೝ
4. Theconsideredexpressionequatesto ሺଵା௫ ೝ ሻሺଵା௫ ೝశభ ሻandso,bythemethodofdifferences,

௫ೝ ଵ ଵ ଵ
σே
௥ୀଵ ൌ ሺ௫ିଵሻ ቂ െ ቃ,andlettingܰ ՜ λ,thedesiredresultisobtained.
ሺଵା௫ ೝ ሻሺଵା௫ ೝశభ ሻ ଵା௫ ଵା௫ ಿశభ
ଶ௘ షೝ೤
Writing•‡…Šሺ‫ݕݎ‬ሻas andsimilarly•‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯,theresultofpart(i)with‫ ݔ‬ൌ ݁ ିଶ௬ can
ଵା௘ షమೝ೤
beusedtoobtaintheresult.Careneedstobetakentowriteσஶ
௥ୀିஶ •‡…Šሺ‫ݕݎ‬ሻ •‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯as

ʹൣσஶ
௥ୀଵ •‡…Šሺ‫ݕݎ‬ሻ •‡…Š൫ሺ‫ ݎ‬൅ ͳሻ‫ݕ‬൯ ൅ •‡…Š ‫ݕ‬൧whichwiththepreviousdeductionof(ii)canbe
simplifiedtoʹ …‘•‡…Š ‫ݕ‬.

5. Thebinomialexpansion,evaluatedfor‫ ݔ‬ൌ ͳ,appreciatingthattermsaresymmetrical

containstwotermsequaltotheLHSoftheinequality,andsotruncatingtothemgivesdoublethe
ሺଶ௠ାଵሻǨ
requiredresultin(i).Appreciatingthatሺ௠ାଵሻǨ௠Ǩisanintegerandthatif݉ ൅ ͳ ൏ ‫ ݌‬൑ ʹ݉ ൅ ͳ,
withpaprime,impliespdividesthenumeratorandnotthedenominatorofthisexpressionand
hencedividestheintegerthencanbeextendedforallsuchprimesyieldingtheresult,withthe
deductionfollowingfrom(i).For(iii),itcanbeshownthat݉ ൅ ͳ ൑ ʹ݉andwritingܲଵǡଶ௠ାଵ as

24 889
ܲଵǡ௠ାଵ ܲ௠ାଵǡଶ௠ାଵ ,combiningthegivenresultand(ii),thedesiredresultisobtained.Part(iv)is
obtainedbyuseofstronginductionwiththesupposition,ܲଵǡ௠ ൏ Ͷ௠ forall݉ ൑ ݇forsome
particular݇ ൒ ʹ,andconsideringthecaseskevenandoddseparatelyandmakinguseof(iii).

6. Usingܴ …‘•Šሺ‫ ݔ‬൅ ߛሻ ൌ ܴሺ…‘•Š ‫ ߛ Š•‘… ݔ‬൅ •‹Š ‫ߛ Š‹• ݔ‬ሻ,ܴ ൌ ξ‫ܤ‬ଶ െ ‫ܣ‬ଶ and
஺
ߛ ൌ –ƒŠିଵ if‫ ܤ‬൐ ‫ ܣ‬൐ Ͳ.If‫ ܤ‬ൌ ‫ܣ‬,then‫ ݔ Š‹• ܣ‬൅ ‫ ݔ Š•‘… ܤ‬ൌ ‫ ݁ܣ‬௫ .Ifെ‫ ܣ‬൏ ‫ ܤ‬൏ ‫ܣ‬,the
஻
஻
expressioncanbewrittenas•‹Šሺ‫ ݔ‬൅ ߛሻwithܴ ൌ ξ‫ܣ‬ଶ െ ‫ܤ‬ଶ andߛ ൌ –ƒŠିଵ .If‫ ܤ‬ൌ െ‫ܣ‬,
஺
then‫ ݔ Š‹• ܣ‬൅ ‫ ݔ Š•‘… ܤ‬ൌ െ‫ି ݁ܣ‬௫ ,andif‫ ܤ‬൏ െ‫ܣ‬,theexpressioncanbewrittenas
஺
ܴ …‘•Šሺ‫ ݔ‬൅ ߛሻwithܴ ൌ െξ‫ܤ‬ଶ െ ‫ܣ‬ଶ andߛ ൌ –ƒŠିଵ .Forpart(i),solvingsimultaneouslygives
஻
ܽ •‹Š ‫ ݔ‬൅ ܾ …‘•Š ‫ ݔ‬ൌ ͳ,whichgivesthedesiredsolutionsusingthefirstresultofthequestion.
ଵ ௕
Similarlyforpart(ii)usingtheappropriateresult,‫ ݔ‬ൌ •‹Šିଵ ቀ ቁ െ –ƒŠିଵ .For(iii),we
ξ௔మ ି௕ మ ௔
ଵ
requirethattheconditionsfor(i)givetwosolutions,i.e.thatܾ ൐ ܽandቀ ቁ ൐ ͳ,andso
ξ௕ మ ି௔మ
ܽ ൏ ܾ ൏ ξܽଶ ൅ ͳ,andviceversa,ifthisappliesthereareindeedtwosolutions.For(iv),werequire
௔
case(i)togivecoincidentsolutions,i.e.ܾ ൌ ξܽଶ ൅ ͳandhence‫ ݔ‬ൌ െ –ƒŠିଵ మ ,andso
ξ௔ ାଵ
ଵ
‫ݕ‬ൌ .Thereverseargumentalsoapplies.
ξ௔మ ାଵ

7. Consideringሺ߱௥ ሻ௡ establishesbythefactortheoremthateachfactorontheLHSisafactor
oftheRHS,andcomparingcoefficientsof‫ ݖ‬௡ betweenthetwosidesestablishesthatnonumerical
factorisrequired.Forpart(i),representingܺ௥ by߱௥ ,thentherearetwocasestoconsider,ܲwill
ഏ೔ ഏ
ቀ ାగቁ௜
berepresentedeitherby‫ ݁ݎ‬೙ ,or‫ ݁ݎ‬೙ .Theproductofmoduliisthemodulioftheproductof
factors,andtheproductofthefactorscanbesimplifiedusingthestemandchoosing‫ݖ‬inturnasthe
representationsofܲtogivetherequiredresultinbothcases.Proceedingsimilarlyfornodd,the
firstcaseyieldsȁܱܲȁ௡ ൅ ͳ,andthesecond,ȁܱܲȁ௡ െ ͳ,ifȁܱܲȁ ൒ ͳ,andͳ െ ȁܱܲȁ௡ ifȁܱܲȁ ൏ ͳ.
Usingthesamerepresentationsfortheܺ௥ inpart(ii),andthesametechniquewiththemoduli,the
stemcanbedividedbyሺ‫ ݖ‬െ ͳሻtogiveሺ‫ ݖ‬െ ߱ሻሺ‫ ݖ‬െ ߱ଶ ሻ ǥ ሺ‫ ݖ‬െ ߱௡ିଵ ሻ ൌ ‫ ݖ‬௡ିଵ ൅ ‫ ݖ‬௡ିଶ ൅ ‫ ڮ‬൅ ͳ
whichthengivesthedesiredresultwhen‫ ݖ‬ൌ ͳ.

8. Thefirstresultin(i)isobtainedbythesubstitution‫ ݔ‬ൌ െ‫(ݑ‬followedbyasecond‫ ݑ‬ൌ ‫)!ݔ‬.

Substitutingfor݂ሺെ‫ݔ‬ሻintheinitialstatementusingtheresultobtainedreadilyleadsto݂ሺ‫ݔ‬ሻ ൌ ‫ݔ‬
whichissimplyverified.Alternatively,subtractingtheresultfromtheinitialequationleadsto
݂ሺ‫ݔ‬ሻ ൌ ݂ሺെ‫ݔ‬ሻwhichsubstitutinggivestherequiredresultagain.Inpart(ii),substituting‫ܭ‬ሺ‫ݔ‬ሻfor‫ݔ‬
௫ାଵ
intheequationfor݃ሺ‫ݔ‬ሻgivesanequationfor݃ ቀ ቁwhichcanbesubstitutedintheequationto
௫ିଵ
ଵ
besolvedtogivethedesiredresult.Similarly,inpart(iii),substituting for‫ݔ‬givesanequationfor
ଵି௫
ଵ ௫ିଵ
݄ቀ ቁand݄ ቀ ቁ,andthenrepeatingthissubstitutionintheequationjustobtainedgivesan
ଵି௫ ௫
௫ିଵ
equationfor݄ ቀ ቁand݄ሺ‫ݔ‬ሻ.Addingthegivenandlastequationsandsubtractingthatfirstfound
௫
ଵ
leadsto݄ሺ‫ݔ‬ሻ ൌ െ ‫ݔ‬.
ଶ
ଶ
9. Therearenumerouswaystoobtainܲܺ ൌ ܽviae.g.knowledgeofthecentroidofa
ξଷ
triangle,PythagorasΖtheorem,trigonometryoracombinationofthese,leadingtotheinitial
extensionresult.PythagorasΖtheoremcanbeusedtofindܴܺandhencethegiventension.The
equationofmotioninthedirectionܺܲcombinesthetensioninܲܺandtheresolvedpartsofthe
tensionsintheothertwosprings.Writingthecosineoftheanglebetweenܴܺandܲܺproducedas

25 890
 భ షభ
௔ା௫ ଷఒ ଵ ξଷ ξଷ ௫ ଷ௫ మ మ
ξయ
,leadstoെߣ െ ‫ ݔ‬൅ ʹߣ ቀ ܽ൅ ‫ݔ‬ቁ ቀͳ ൅ ൅ ቁ ൌ ݉‫ݔ‬ሷ whichmakingan
మ ௟ ξଷ ଶ௔ ଶ ௔ ସ௔మ
ට௔మ ାቀ భ ௔ା௫ቁ
ξయ
ଷఒ
approximationforsmall‫ݔ‬andthebinomialexpansionleadstoെ ൫Ͷܽ െ ξ͵݈൯‫ ݔ‬ൌ ݉‫ݔ‬ሷ ,andhence
ସ௟௔
thefinalresult.

10. Resolvingalongalineofgreatestslopeinitially,bearinginmindtheaccelerationdueto
circularmotion,givesanexpressionfortheinitialtensioninthestringwhichcanbesubstitutedin
theexpressionobtainedfornormalcontactforceobtainedbyresolvingperpendiculartotheslope.
Requiringapositivenormalcontactforcethengivesthedesiredresult.Tocompletecircles,there
mustbeatensioninthestringwhentheparticleisatthehighestpointitcanreachontheplane.
Conservingenergygives‫ ݒ‬ଶ ൌ ‫ݑ‬ଶ െ Ͷܽ݃ …‘• ߚ •‹ ߙandresolvingdowntheplaneyields
௩మ
ܶ'…‘• ߚ ൅ ݉݃ •‹ ߙ ൌ ݉ resultingin‫ݑ‬ଶ ൐ ͷܽ݃ …‘• ߚ •‹ ߙ;thiscombinedwiththefirst
௔ ୡ୭ୱ ఉ
resultwillgivethefinaldesiredresult.(Thefirstresultcanbefoundelegantlybyresolving
perpendicularlytothestring.)

11. Inpart(i),expressingtheresistanceas݇‫ݒ‬,thenthezeroaccelerationconditiongives
௉ ௗ௩
݇ൌ .Writingtheequationofmotionusingܽ ൌ ‫ݒ‬ ,andsolvingthedifferentialequationby
ଵ଺௎ మ ௗ௫
௠ ସ௎ା௩ ଶ௎
separatingvariables,theintegrationgivesܺଵ ൌ ቂ ቀʹܷ Ž ቀ ቁ െ ‫ݒ‬ቁቃ whichevaluatedand
௞ ସ௎ି௩ ௎
rearrangedistherequiredresult.Part(ii)followsasimilarroute,insteadexpressingtheresistance
௉
as݇‫ ݒ‬ଶ ,with݇ ൌ .Thesametechniqueforthedifferentialequationgivesaslightlysimpler
଺ସ௎ య
ସ
integrationtoyieldtheresult.ߣܺଵ െ ߣܺଶ canbemanipulatedtobe Ž ʹͶ െ ʹ Ž ͷ െ ͳwhichcan
ଷ
beshowntobepositiveusingtheappropriateboundsandsoansweringpart(iii)thatܺଵ isthe
larger.

12. Usingthebinomialdistribution,ߤ ൌ ʹͲ݊,ߪ ଶ ൌ ͳ͸݊,writingͳ͸݊ ൑ ܺ ൑ ʹͶ݊as

ȁܺ െ ʹͲ݊ȁ ൑ Ͷ݊enablesChebyshevtobeappliedwith݇ ൌ ξ݊leadingtotherequiredresultin(i).
Similarly,inpart(ii),consideringaPoissondistributionwithmean݊,andappreciatingthat
ȁܺ െ ݊ȁ ൐ ݊impliesܺ ൐ ʹ݊inthesecircumstances,thesamevalueof݇asinpart(i)with
Chebyshevleadstothedesiredresult.

13. Showingthatܺ െ ܽhasthesamekurtosisasܺrequirestheexpectationsofܺ െ ܽ,

ሺܺ െ ܽ െ ߤ ൅ ܽሻଶ ,andሺܺ െ ܽ െ ߤ ൅ ܽሻସ tobeobtainedandsubstituted.Forpart(i),the
numeratorcanbeobtainedbyanintegrationbypartsreducingtheintegraltotheonethatgivesthe
variance.Expandingܶ ସ asσ൫ܻ௥ ସ ൅ Ͷܻ௥ ଷ ܻ௦ ൅ ͸ܻ௥ ଶ ܻ௦ ଶ ൅ ͳʹܻ௦ ܻ௧ ܻ௥ ଶ ൅ ʹͶܻ௥ ܻ௦ ܻ௧ ܻ௨ ൯,wherethe
summationisoverallvalueswithoutrepetition,andtakingtheexpectationofthesetermsgivesthe
requestedresultinpart(ii).Definingܻ௜ ൌ ܺ௜ െ ߤ,thekurtosisofܻ௜ bythefirstresultgives
‫ܧ‬൫ܻ௜ ସ ൯ ൌ ሺ͵ ൅ ߢሻߪ ସ anddefiningܶasin(ii),thekurtosisofσ௡௜ୀଵ ܺ௜ is,usingtheresultof(ii),
௡ሺଷା఑ሻఙ ర ାଷ௡ሺ௡ିଵሻఙ మ ఙ మ
െ ͵givingtherequiredanswer.
௡మ ఙ ర

26 891
27 892
.

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893
STEP Mark Schemes 2017
Mathematics

STEP 9465/9470/9475

November 2017

894
Admissions Testing is a department of Cambridge Assessment English, which is
part of Cambridge Assessment, a not-for-profit department of the University of
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Cambridge Assessment English offers the world’s leading qualifications for

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2 895
Contents

STEP Mathematics (9465, 9470, 9475)

Mark Schemes Page

Introduction 4
Marking Notation 5
STEP Mathematics I 6
STEP Mathematics II 32
STEP Mathematics III 54

3 
 
896
Introduction

These mark schemes are published as an aid for teachers and students, and
indicate the requirements of the examination. It shows the basis on which marks
were awarded by the Examiners and shows the main valid approaches to each
question. It is recognised that there may be other approaches; if a different approach
was taken by a candidate, their solution was marked accordingly after discussion by
the marking team. These adaptations are not recorded here.

All Examiners are instructed that alternative correct answers and unexpected
approaches in candidates’ scripts must be given marks that fairly reflect the relevant
knowledge and skills demonstrated.

Mark schemes should be read in conjunction with the published question papers and
the Report on the Examination.

Admissions Testing will not enter into any discussion or correspondence

in connection with this mark scheme.

4 897
Marking notation

NOTATION MEANING NOTES

M Method mark For correct application of a Method.
dM or m Dependent method mark This cannot be earned unless the
preceding M mark has been earned.
A Answer mark M0 ⇒ A0
B Independently earned Stand alone for “right or wrong”.
mark
E B mark for an explanation
G B mark for a graph
ft Follow through To highlight where incorrect answers
should be marked as if they were correct.
CAO or CSO Correct Answer/Solution To emphasise that ft does not apply.
Sometimes Only
written as
A*
AG Answer Given Indicates answer is given in question.

5 898
 STEP I 2017 Mark Scheme 

Question 1 

(i)    sin cos  

  sin cos sin M1 A1 
  cos  
      M1 A1 

       
    ln| |   A1 
   ∴ ln| sin cos |   M1 
     
      M1 
  Let  cos sin    
  cos sin cos  
  M1 A1 
  sin  
    ln| |    

   ∴ ln| cos sin |  A1 

     
(ii)  Let  sec tan   M1 A1 
  sec 2 sec sec tan 2 sec tan    A1 

  sec 2 sec tan sec 2 sec tan    A1 

  So  ln| |    

  ln| sec tan |    A1 

     
     M2 A1 

Using same substitution as previous integral:
 
  A1 
  

     A1 
 

   

6 899
 

Question 1 

An A1 should be lost if modulus signs or   are omitted in the final answer for any section, but only 
on the first occasion. 

M1  Calculation of   

A1  Correct expression 
M1  Use of tan  
A1  Integral simplified and in terms of   
A1  Integration completed correctly in terms of   
M1  Integral rewritten in terms of 
Subtotal: 6
M1  Rewriting integral in a form ready for substitution 
M1  Correct choice of substitution 
A1  Correctly differentiated 
A1  Correct final answer. 
Subtotal: 4
M1  Choice of a sensible substitution, based on the denominator.
A1  Correct choice 
A1  Differentiation of sec  
A1  Correct   
A1  Correct final answer. 
Subtotal: 5
M2  Transformation of the integral so that the denominator is similar to the first part. 
A1  Correctly transformed. 
A1  Correct integral in terms of   
A1  Correct final answer. 
Subtotal: 5
 

7 900
Question 2 

(i)  ln| | ln    M1 

  1 1    
  Therefore, ln 1  A1 AG 
     
  Over the interval  1,  1  M1 
Taking the integral over the range  1 gives the inequality 
   
ln 1
  Therefore ln 1  A1 AG 
     
(ii)  1    M1 A1 
Therefore, integrating both sides gives 
  1 ln    M1 A1 
and so ln 1  for  1 
For 0 1, integrating   over the interval  1 gives: 
  M1 
1 ln   
  So ln 1  for 0 1 as well and so (**) is true for  0.  A1 
     
  ALTERNATIVE   
  1  and  ln   M1 A1 
  When  1, 1 0 ln   dM1 
   for  1  B1 
Therefore, since the two sides of the inequality are equal when  1, the LHS 
  grows more rapidly for  1 and the RHS grows more rapidly for  1, the  G1 E1 
inequality is true. 
     
(iii)  ln 1 ln   
ln 1 M1 
  

  ln ln 1 ln 1   A1 
Integrating (*): 
  For  1:  M1 
ln 1   
  Therefore 2 ln 1  and so        (since 2 1 0   A1 
For 0 1: 
  M1 
ln 1   
  Therefore 2 ln  and so        (since 2 1 0   A1 
     
Integrating (**): 
  For  1:  M1 
ln 1 ln 1
  Therefore  1 ln 2 1  and so    (since  1 1 0   A1 
For 0 1: 
  M1 
ln 1 ln 1
  Therefore  1 ln 2 1  and so    (since  1 1 0   A1 
 
8 901
Question 2 

M1  Integration of one of the sides of the inequality (indefinite integration OK) 
A1  Integration of both sides of the inequality and conclusion reached. (In the case of the RHS 
an alternative would be a clear explanation in terms of area of rectangle) 
M1  Statement of the inequality for this range of values for  .
A1  Correctly drawn conclusion. 
Subtotal: 4
M1  Integration of LHS of inequality. (indefinite integration OK)
A1  Integration completed correctly.
M1  Inequality formed by integrating both sides of inequality 
A1  Correct deduction of (**) for  1 
Marks up to this point can be awarded if there is no consideration of which values of   (**) is 
shown for. 
M1  Integration of correct inequality for 0 1
A1  Conclusion of (**) including clear explanation of how it is shown for whole range. 
Note that substituting   for   in (*) gives ln 1, which leads to  ln 1 and then to 
(**) directly, but no marks for this as question requires starting from a different inequality. 
ALTERNATIVE 
M1  Two differentiations. 
A1  Correctly completed. 
dM1  Consideration of  1 
B1  Correct inequality. 
G1  Graph to illustrate that the inequality holds. 
E1  Explanation (award the G1 also for a good explanation without the graph sketched. 
 Subtotal: 6
M1  Use of integration by parts to integrate ln  (indefinite integration OK) 
A1  Correct integral 
M1  Integration of both sides of inequality in case  1.
A1  Inequality deduced for case  1. 
M1  Integration of both sides of inequality in case 0 1 . 
A1  Inequality deduced for case 0 1. 
M1  Integration of both sides of inequality in case  1.
A1  Inequality deduced for case  1. 
M1  Integration of both sides of inequality in case 0 1 . 
A1  Inequality deduced for case 0 1. 
Subtotal: 10
 

9 902
Question 3 

  2 4    M1 

  At P,    A1 
Therefore the equation of the tangent is:
  M1 
2   

     A1 
     
  Similarly, the equation of the tangent at Q is    B1 
Coordinates of R: 
  
M1 
     A1 A1 
  
Therefore  . 
  [   ]   
  The coordinates of R are  ,    
Other coordinates are: 
  B1 
0,  and  0,  
Area of RST: 
Using the edge along the  ‐axis as the base: 
Base length    M1 A1 
 
Height    A1 
Therefore the area is   
     
  Area of OPQ:   
Trapezium formed by adding horizontals to  ‐axis from P and Q:
  M2 A1 
Area  2 2  
Triangles to be removed: 
  (P): Area  2   A1 A1 
(Q): Area  2  
  Therefore area of OPQ    
M1 A1 
  Area   
B1 
  Therefore the area of OPQ is twice the area of RST   
 

   

10 903
Question 3 

M1  Differentiation. 
A1  Gradient correct. 
M1  Equation of line through P with this gradient. 
A1  Establish given equation. 
Subtotal: 4
B1  Tangent at Q. 
Subtotal: 1
M1  Equating   coordinates. 
A1  Solve for   (allow mark if signs are incorrect).
A1  Correct signs. 
Subtotal: 3
B1  Coordinates of other points. 
M1  Valid method for area of triangle.
A1  Area correct. 
A1  Sign correct. 
Subtotal: 4
M2  Valid method for calculation of the area.
A1  Correct area for one shape that is needed. 
A1  Correct area for at least one other shape that is needed. 
A1  Correct areas found for all shapes needed. 
M1  Combine all areas correctly. 
A1  Correct area of the triangle. 
B1  All correct and conclusion reached. 
Subtotal: 8
 

11 904
Question 4 

(i) 
1 ≡  
Therefore minimum point is  ,  
B1 
M1 
A1 
B1 
B1 

Since | | 1, upper bound for  is the value when  1.

  B1 
Therefore   3 
Since | | 1, there is only one value of   that corresponds to each value of   in 
  the range 1 3 (as can be seen on the graph). Therefore values of   in that  B1 
range determine   (and hence   uniquely. 
  If  1 then there are two values of   that satisfy 1 .   
  Rearranging   gives  1   M1 
Substituting: 
  M1 
1 1 1   

  So 3  and therefore  3 3 1 0   A1 AG 

     
(ii) 
1 2 3 ≡3    M1 A1 
At  1, 1 2 3 2 
  B1 
At  1, 1 2 3 6 
  Therefore values of   in the range 2 6 determine , and hence   uniquely.  B1 
  If   then the value of   and hence   is also determined uniquely.  B1 
     
  For  2, there are two values of  .  B1 
  1  , so  1   M1 
√
  Therefore  1 2 1 3 1   A1 
√ √
  6   and so  6 3 8√   M1 
√
Squaring: 
  6 6 6 9 64    A1 
6 28 6 9 0
 

   

12 905
Question 4 

B1  Correct shape (must include range | |<1). 

M1  Completing the square (or differentiating) to find minimum.
A1  Correct minimum point. 
B1  ‐intercept at (0,1). 
B1  No intercept with  ‐axis. 
B1  Upper bound for   justified. 
B1  Any valid explanation. 
M1  Rearrangement of the formula for S. 
M1  Substitution into the formula for p.
A1  Correct solution. 
Subtotal: 10
M1  Completing the square (or differentiation to find minimum) 
A1  Completion of square done correctly.
B1  Calculation of the value at the two endpoints. 
B1  Interval 2 6 identified. 
B1   identified. 
Subtotal: 5
B1  Correct interval identified. 
M1  Rearrangement of the formula for T. 
A1  Substitution into the formula for q. 
M1  Simplification into a three term quadratic in √ . 
A1  Quadratic in T found. 
Subtotal: 5
 

13 906
Question 5 

  The width of the rectangle is     
  The height of the rectangle is  tan  
  Therefore the area is  tan   B1 AG 
  The coordinates of   must be  , √   B1 
  The coordinates of Q must be  , tan    
  Therefore  tan √  since both must have the same  ‐coordinate  M1 
  So  tan ,  tan   A1 
     
  2 2 tan    M1 
  tan 1 tan    M1 A1 

Substituting for  : 
  M1 
tan tan 1 tan   

  2 tan tan    A1 AG 

     
  ALTERNATIVE (from  tan ,  tan )   
  tan tan tan    dM1 
  tan tan 2 tan    M1 A1 

  Since  tan ,  2 tan tan   M1 A1 

     
Stationary points when 
   
2 tan tan 0  
  2 tan 0   M1 
By quadratic formula: 
  M1 
1 tan  (since  )  
M1 A1 
  Therefore  1 sec  
AG 
     
1 sec   
  M1 
tan 1 sec
  1 2 sec sec tan  
  2 sec 1 sec    A1 
     
  sec tan    M1 
     A1 

      

     M1 

  tan    A1 AG 
     
  tan ∠     

  tan ∠ tan  so ∠   B1 AG 

 

   
14 907
Question 5 

B1  Clear explanation of the formula for the area of the rectangle. 
B1  Coordinates of R deduced (use of Pythagoras / radius of circle).
M1  Equating y‐coordinates for Q and R. 
A1  Expression for   in terms of  ,   and  . 
M1  Differentiation to find   
M1  Differentiation of the expression for  . 
A1  Correct differentiated expression. 
M1  Substitution for   
A1  Rearrangement to required form. 
ALTERNATIVE 
dM1  Substitution. 
M1  Differentiation with respect to  (must either treat tan as constant or have an expression 
involving  .) 
A1  Correct derivative. 
M1  Use of other formula. 
A1  Correctly deduced relationship. 
Subtotal: 9
M1  Use of  0. 

M1  Use of quadratic formula. 
M1  Use of 1 tan ≡ sec  
A1  Correct value for  determined, including justification of which root of the quadratic. 
Subtotal: 4
M1  Attempt to eliminate   from equation found in the first part of the question. 
A1  Correct relationship. 
M1  Substitution into the formula for the area. 
A1  Correct formula obtained. 
M1  Use of double angle formulae. 
A1  Correct expression for the area.
B1  Justification for the size of angle ROS. 
Subtotal: 7
 

15 908
Question 6 

(i)  Suppose that  0 for all values of   in the interval. Then  0 

B1 
(since  0 for some value of   in the interval). 
Similarly, if  0 for all values of   in the interval. Then  0 
   
(since  0 for some value of   in the interval). 
Therefore if  0,   must take both positive and negative values in 
  B1 
the interval 0 1. 
     
(ii)  2    M1 
  2 0   A1 
Since  1,   must be non‐zero for some value of   in the interval 
  M1 
0 1. 
Therefore, by (i),   takes both positive and negative values in the interval 
  M1 
0 1. 
  Therefore,   takes the value 0 at some point in the interval 0 1.  A1 
     
If  , then: 
   
  
  Therefore,  1  B1 
      
  Therefore,    B1 
      
  Therefore,    B1 
From the first two equations: 
   
4 6 ,  12 6 
Substituting into the third equation: 
  M1 
4 6 12 6   
  0    
√
  Therefore    A1 
  So  1 √3 2 √3 or  1 √3 2 √3  A1 
Therefore the values of  0  and  1  are 1 √3 and 1 √3 and so  0 
  E1 
for some value of   in the interval 0 1. 
     
(iii)  ′ 1 0 1   B1 
′   
  M1 A1 
So  ′ 1  
′ 2   
  M1 A1 
So  ′ 1 2 1  
  Therefore, the conditions of (*) are met ( ′ ,  1 ).   
  Therefore  0 for some value of   in the interval 0 1.  B1 
 

   

16 909
Question 6 

B1  Correct statement considering either   always positive or always negative. 

B1  Statement that corresponding result is true in the other case and conclusion of proof by 
contradiction. 
Subtotal: 2
M1  Expansion of   and split of integral. 
A1  Establish that the value of the integral is 0. 
M1  Identify one of the conditions required to apply the result from (i) 
M1  Apply result from (i) 
A1  Draw the required conclusion. 
Subtotal: 5
B1  Relationship between   and  . 
B1  Relationship between  ,   and  .
B1  Relationship between  ,  and  . 
M1  Elimination of two of the variables. 
A1  Value of one of the variables found. 
A1  Correct choice of  . 
B1  Verification that there is a root in the interval. 
Subtotal: 7
B1  Confirm that first condition is satisfied. 
M1  Use of integration by parts. 
A1  Confirm that second condition is satisfied. 
M1  Use of integration by parts. 
A1  Confirm that the third condition is satisfied.
B1  Apply the previous result to draw the conclusion. 
Subtotal: 6
 

17 910
Question 7 

(i)   is an isosceles triangle, with | |  and | | | |   

  Angle  120°  B1 
  2| | 2| | cos 120 M1 
  | |     

  | |    A1 AG 
√

  | |    B1 
√
     
(ii)  | | | | | | 2| || | cos ∠   
M1 
| | cos ∠   
| | | | | | 2| || | cos ∠   
  M1 
2 cos ∠
  ∠ ∠ 60°   M1 
√
  Therefore cos ∠ cos ∠ sin ∠   M1 
  | | cos ∠ √3 sin ∠    M1 
√
  | | sin ∠     
M1 A1 
  6| | 4√3∆  
AG 
A similar argument will show that 6| | 6| | 4√3∆, so 
  B1 
 is an equilateral triangle. 
√
  The area of   is  | |    
√ √
  If  4√3∆, then the area of   is  ∆ ∆  B1 
If the area of   is equal to the area of  : 
  √  
4√3∆ ∆  
  √3 12∆ 24∆  
  4√3∆   B1 
     
(iii)  If  2
cos 1
60° , then 
 
2 2 cos 2 60°
  cos √3 sin M1 
  ∆ sin , and by the cosine rule  2 cos   M1 
  2√3∆   A1 
  All steps are reversible, so the conditions are equivalent.  B1 
     
The areas of the triangles are equal if and only if  4√3∆, which is 
   
equivalent to  2 1 cos 60° . 
Since  0 and  2 1 cos 60° 0 this can only be satisfied if 
  M1 
both sides are equal to 0. 
Therefore   and cos 60° 1, so  60°. 
  A1 
So   is an equilateral triangle. 
 

   

18 911
Question 7 

B1  Value of angle  . 
M1  Use of cosine rule for triangle  .
A1  Correct value reached. 
B1  Correct value stated. 
Subtotal: 4
M1  Application of cosine rule to triangle  .
M1  Application of cosine rule to triangle  . 
M1  Relationship between angles   and  . 
M1  Application of cos ≡ cos cos sin sin .
M1  Combination of the previous results. 
M1  Use of ∆ sin  
A1  Fully correct solution. 
B1  Deduction that the triangle is equilateral. 
B1  Justification that the condition implies that the areas are equal.
B1  Justification that the areas being equal implies that the condition holds. 
Subtotal: 10
M1  Expansion and rearrangement.
M1  Use of area of triangle and cosine rule.
A1  Fully correct justification. 
B1  Clear indication that the reasoning applies both ways. 
M1  Observation that inequality can only be satisfied in this case if both sides are 0. 
A1  Clear explanation that this implies that the triangle is equilateral. 
Subtotal:6 
 

19 912
Question 8 

  Check  1:  2 1 2 1 2 2 1  B1 

Assume that the result is true for  : 
   
2 1 
2
  M1 A1 
2 2 2 2 5 2 5  
  2 1  
  Therefore, by induction,  2 1  1 for all  B1 
     
(i)  From the definitions of the sequences  0 and  0 for all    B1 
  2 2 5 , so  2 5  is true in the case  1.  B1 
  Assume that  2 5  for some value  .   
  Then  2 5 2 5   M1 A1 
  Therefore , by induction,  2 5  for all  1  B1 
     
  ALTERNATIVE   
  0 for all  1   B1 
  2,  2 5 5 5 ⋯ 5   M1 A1 
  2 5 B2 
     
  From (*):  2 1   B1 
  Therefore, as  → ∞,   approaches a root of  2 1 0  M1 
  The roots of  2 1 0 are  1 √2  A1 
  Since  0,  → √2 1 as  → ∞.  A1 
     
(ii)     M1 

  Therefore     

  0   M1 
  Therefore the sequence   is decreasing. A1 
  Therefore  √2 1.  A1 AG 
  1 √2 and so  √2 1  A1 AG 
     
  ALTERNATIVE FOR  √    
  1 2    M1 A1 
  0   B1 
  √2 1   A1 
     
  1 and  2, so     

  5 and  12, so     

  29 and  70, so    M1 

  Therefore  √2 1   

  √2    A1 
 

   

20 913
Question 8 

B1  Check the case  1 
M1  Attempt to relate the case  1 to the case 
A1  Deduce that the result holds in the case  1 if it holds in the case  
B1  Conclusion of proof by induction. 
Subtotal: 4
B1  Observe that all values in both sequences are positive.
B1  Check the case  1 
M1  Attempt to relate the case  1 to the case   
A1  Deduce that the result holds in the case  1 if it holds in the case   
B1  Conclusion of proof by induction. 
ALTERNATIVE 
B1  Observe that all values of   are positive. 
M1  Inequality between consecutive values for 
A1  Repeated application of inequality. 
B2  Conclusion clearly justified. 
Subtotal: 5
B1  Deduce formula satisfied by 
M1  Find equation satisfied by limit of sequence. 
A1  Solve quadratic. 
A1  Justify choice of positive root. 
Subtotal: 4
M1  Write   in terms of   and   
M1  Find expression for   
A1  Conclude that the sequence is decreasing.
A1  Explain why this shows that  √2 1 
ALTERNATIVE 
M1  Solution of quadratic. 
A1  Choice of positive square root. 
A1  Observe that  0 
A1  Clear explanation that  √2 1 
   
A1  Conclude required inequality 
M1  Calculate   
A1  Deduce required inequality. 
Subtotal: 7
 

21 914
Question 9 

(i)  Horizontal speed =  cos , therefore the particle passes through P after   

M1 A1 
seconds. 
Vertically: 
  Initial speed =  sin , acceleration  , displacement  tan .  M1 A1 
tan sin   

  tan tan sec     

      

   will be as small as possible at a point where  0:   
M1 M1 
  2   
A1 
  2     
  Therefore  0 if  tan tan 2 tan sec 0   
  tan tan 2 tan tan 1 0 M1 
  tan 2 tan tan 1 0  
  So tan  and tan tan   A1 

Therefore  tan   M1 A1 

 
AG 
  tan cot 2    M1 
The graph of  cot  is a translation of the graph of  tan  by 90 °  
horizontally.  M1 A1 
 
 must be greater than  .   AG 
Therefore 2 90° 
     
(ii)  When the particle passes through P: 
 
Horizontal velocity is  cos  
  Vertical velocity is  sin   B1 
Therefore if the angle to the horizontal is  , then 
  M1 A1 
 tan tan sec  

  Since  tan : tan tan tan sec 1  M1 A1 

  Therefore  90° (or  90 ° below the horizontal).  A1 
 

   

22 915
Question 9 

M1  Use of constant horizontal velocity to determine time passing through P. 
A1  Correct expression for time. 
M1  Uniform acceleration formula. 
A1  Correct equation. 
M1  Attempt to differentiate either   or   with respect to   
M1  Application of quotient rule. 
A1  Correctly differentiated. 
M1  Set derivative equal to 0. 
A1  Rearrange to get formula for tan  
M1  Substitute to get expression for   
A1  Fully correct justification. 
M1  Observe relationship between tan  and cot 2  
M1  Reference to the relationship between the two functions. 
A1  Correct relationship, fully justified. 
Subtotal: 14
B1  Calculation of vertical velocity through P (must be in terms of  ) 
M1  Division of two velocities to get tan of required angle. 
A1  Simplified form. 
M1  Substitution of result from part (i)
A1  tan tan 1   
A1  Removal of tan functions. 
Subtotal: 6
 

23 916
Question 10 

(i)  Conservation of momentum: 
B1 
  
Law of Restitution: 
  B1 
  
Eliminating  : 
  M1 
  
  1 1     
     A1 AG 
For the first collision with particle   ( 1 : 
     M1 A1 
  
      

Eliminating  : 
   
  
  1 1     

  Therefore    A1 

     M1 

1   
  M1  
  

  So    A1 
     
(ii)  If   then  1, so   for every choice of   and so there cannot  M1 M1 
be any subsequent collisions.  A1 
     
(iii)  If   then all particles will have the same velocity after their second collision. 
M1 A1 
1
The Kinetic Energies of the particles after their second collisions will form a 
  M1 
geometric series with first term  1  and common ratio  . 

  Therefore the sum will approach  1   A1 

  The initial KE was  , so the fraction that has been lost approaches  .  A1 AG 

     
(iv)  If  1 then all particles stop after their second collision.  B1 
  All of the energy is lost eventually in this case.  B1 
 

   

24 917
Question 10 

B1  Correct equation. 
B1  Correct equation. 
M1  Attempt to eliminate   
A1  Reach given equation correctly. 
M1  Consideration of conservation of momentum for  collision. 
A1  Simplified form. 
A1  Correct equation for   
M1  Substitution to find   
M1  Simplification. 
A1  Adjustment to get   
Subtotal: 10
M1 
1    
M1  Relationship between velocities.
A1  Clear explanation why this implies no further collisions. 
Subtotal: 3
M1  Comment that all velocities will be equal. 
A1  Correct common velocity stated.
M1  Identify that the KEs will form a geometric series. 
A1  Sum to infinity. 
A1  Clear justification that fraction of KE lost approaches 
Subtotal: 5
B1  Observation that all particles stop. 
B1  All KE lost (fraction lost = 1) 
Subtotal: 2
 

25 918
Question 11 

Forces at A: 
  Reaction force   (perpendicular to the slope)  B1 
Frictional force   (parallel to slope, towards O) 
Forces at B: 
  Reaction force   (parallel to slope)  B1 
Frictional force   (perpendicular to slope, away from O) 
Since equilibrium is limiting at both A and B: 
  B1 
tan  and  tan  
Resolving parallel to the slope: 
  M1 A1 
tan sin   
Resolving perpendicular to the slope: 
  M1 A1 
cos tan   
Eliminating  :  M1 
sin cos cos tan cos sin tan sin    M1 
  tan tan tan tan    M1 
, so  tan   A1 
Taking moments about the centre of the rod:  M1 M1 
 
cos sin tan cos tan sin A1 
  So  tan tan tan tan   M1 M1 
     
Therefore: 
  M1 
tan tan tan tan tan tan
  tan tan 2    M1 
  Therefore  2   M1 
  Since   and   is acute,  2 .  A1 AG 
 

   

26 919
Question 11 

B1  Identification of the forces at A (may be implied by later work). 
B1  Identification of the forces at B (may be implied by later work).
B1  Use of limiting equilibrium at both points. 
M1  Resolve parallel to slope. 
A1  All correct. 
M1  Resolve perpendicular to slope.
A1  All correct. 
M1  Elimination of any one variable from equations. 
M1  Manipulation of trigonometric functions (may occur later in solution).
M1  Use of tan  (or equivalent) formula (may occur later in solution). 
A1  Correct relationship between two reaction forces. 
M1  Take moments about centre of rod (at least 2 correct) 
M1  Moments about centre of rod (at least 3 correct)
A1  Fully correct. 
M1  Cancel   from the equation. 
M1  Apply tan  
M1  Eliminate so that  ,   and   are not present in the equation. 
M1  Rearrange to apply tan  formula. 
M1  Full solutions to tan equation just reached. 
A1  Deduce relationship between  ,  and  , explaining why it can’t be any of the others. 
Subtotal: 20
 

27 920
Question 12 

(i)  The probability that any one participant will choose the correct number is  . 
M1 A1 
Therefore,  1  
  The expected amount that will need to be paid in prizes is  1 1 .  M1 

  Therefore the expected profit is  1 1 .  A1 
  Therefore the expected profit is approximately  1   A1 
If 2  then the expected profit is  1 0, therefore the organizer 
  A1 AG 
will expect to make a loss. 
     
(ii)  The probability of picking a number between 1 and N is  
M1 
  1 1 
  1   A1 
If the number that is drawn is popular then the probability that no participant 
   
will choose it is  1  
If the number that is drawn is not popular then the probability that no 
  B1 
participant will choose it is  1  
The probability that no participant chooses the winning number is therefore: 
  M1 A1 
1 1 1   
The expected profit is therefore 
  A1 
  1 1 1 1  
which can be approximated to 
  M1 A1 
1 1 1   
     
  If  , then  1. If  9 , then    M1 
     A1 
If 2 , then the profit will be: 
  M1 
7 4   
√
  7 4 7 4 4, since  3  M1 
√ √
  16, so  4 0, meaning that the organiser will expect to  M1 A1 
make a profit. 
 

   

28 921
Question 12 

M1  Identification of the probability of choosing the winning number. 
A1  Correct probability that no participant chooses the winning number.
M1  Expected amount to be paid out. 
A1  Correct expected profit. 
A1  Use of approximation. 
A1  Justification that organizer will expect to make a loss.
Subtotal: 6
M1  Consideration of probability that the number chosen is between 1 and  . 
A1  Correct relationship. 
B1  Correct probabilities of no winner for both cases. 
M1  Find probability that no participant chooses a winning ticket. 
A1  Correct probability. 
A1  Correct expected profit. 
M1  Use of approximation. 
A1  Simplification to required form. 
Subtotal: 8
M1  Substitution and attempt to solve simultaneous equations.
A1  Values of a and b correct. 
M1  Profit calculated in the case 2  
M1  Rearranged and use of  3 
M1  Attempt to show that the expected profit is positive.
A1  Fully clear explanation. 
Subtotal: 6
 

29 922
Question 13 

  0   B1 
If the   slice is to be used to make toast then either  
  the  1  slice was used as the second slice for a sandwich (probability  )  M1 
the  1  slice was used for toast (probability  ) 
The probability that the next slice is used for toast is  . 
  A1 
Therefore   
The   slice being the second slice for a sandwich is equivalent to the  1  
  slice being the first slice for a sandwich, so the probability that the  1  slice  M1 M1 
is the first slice for a sandwich is also   

Since there are only three possibilities for the use of a slice,  1 
  A1 
and so  1   
Valid for  2 as the reasoning only refers to the previous slice. 
  B1 
Formula for   is not valid for   as the final slice must be toast. 
     
  1 , so  1   M1 
  Therefore  1 1   M1 
  1 1 1 1     
  1    A1 
     
  0, which is correct.  B1 
  Assume that     

  Then  1   M1 

     A1 

  Therefore, by induction,   for 1 1  B1 

      M1 
  Therefore   for 1 1  A1 
     
  1    M1 

  1 1     

     A1 
Since the last slice must either be the second slice of a sandwich or toast: 
  M1 A1 
1   
 

   

30 923
Question 13 

B1  Correct value. 
M1  Identification of the two possibilities.
A1  Clear justification of the equation. 
M1  Identification of the probability that it is the first slice of a sandwich. 
M1  Identification of the three possibilities in general. 
A1  Clear justification of the equation.
B1  Clear justification of the ranges for which the equations are valid. 
Subtotal: 7
M1  Rearrangement. 
M1  Substitution. 
A1  Correct equation. 
B1  Check first case. 
M1  Relate case  1 to case  . 
A1  Show that the correct formula follows. 
B1  Complete proof by induction. 
M1  Use relationship between   and  . 
A1  Correct equation, including range for which it is valid.
Subtotal: 9
M1  Substitute into formula for   
A1  Correct formula. 
M1  Observe that  1  
A1  Correct formula. 
Subtotal: 4
 

31 924
 STEP II 2017 Mark Scheme 

Question 1 
1
(i) In =  arctan x . x n dx M1 Use of intgrn. by parts (parts correct way round)
0

1
 xn 1 
1
1 x n 1
n  1  0 1  x 2 n  1
= arctan x .   . dx A1 Correct to here

0

 1
1
 1 xn 1
= .  0   2 dx
 4 n 1  n 1 0 1 x

3
1
xn 1
 (n + 1)In =
4
 0 1  x 2 dx A1 Given Answer legitimately established

 1
x
Setting n = 0, I0 =
4
 1 x
0
2
dx M1 Attempt to solve this using recognition/ substitution


=
4
  ln1  x 
1
2
2
M1 Log integral involved

=

4
 12 ln 2 A1 CAO 3
(ii) n  n + 2 in given result:
 1
xn  3
(n + 3)In + 2 =
4
 0 1  x 2 dx B1 Noted or used somewhere

 1

xn 1 1  x2 
(n + 3)In + 2 + (n + 1)In =
2
 
0
1  x2
dx M1 Adding and cancelling ready to integrate

=

2

1
n2
A1 CAO 3
Setting n = 0 and then n = 2 in this result (or equivalent involving integrals):
 1  1
3I2 + I0 =  and 5I4 + 3I2 =  M1
2 2 2 4
Eliminating I2 and using value for I0 to find I4 M1 By subtracting, or equivalent
I4 = 20 1    2 ln 2 
1
A1 FT from their I0 value 3
(iii) For n = 1, 5I4 = A  12  1  12   A  14
= 1
4  14   12 ln 2 M1 Comparing formula with found I4 value
and the result is true for n = 1 provided
A = 14   12 ln 2 A1 FT from their I4 value 2

32 925
 2k
1
Assuming (4k + 1)I4k + 1 = A – 1
2  (1)
r 1
r

r
M1 For a clearly stated induction hypothesis

(or a fully explained “if … then …” at end)

1 
(4k + 5)I4k + 4 + (4k + 3)I4k + 2 =  B1
2 4k  4
 1
(4k + 3)I4k + 2 + (4k + 1)I4k =  B1
2 4k  2
Subtracting:
1 1
(4k + 5)I4k + 4 = (4k + 1)I4k   M1
4k  2 4k  4
2k
1 1 1
= A – 12  (1) r   M1 Use of assumed result
r 1 r 4k  2 4k  4
2k
1 1 1 1
=A– 1
2  (1)
r 1
r

r
 2 (1) 2 k  1
2k  1
 12 (1) 2 k  2
2k  2
2 ( k  1)
1
=A– 1
2 
r 1
(1) r
r
A1 A clear demonstration of how the two extra

terms fit must be given

6

33 926
Question 2 

 aX  1 
a  1
aX 1  X b 
Let xn = X. Then xn + 1 = and xn + 2 = M1 A1 Correct, unsimplified
X b  aX  1 
 b
 X b 
(a 2  1) X  (a  b)
i.e. xn + 2 = M1 Attempt to remove “fractions within fractions”
(a  b) X  (b 2  1)
A1 Correct, simplified 4
(i) If xn + 1 = xn then aX – 1 = X 2 + bX M1
 0 = X 2 – (a – b)X + 1 A1

If xn + 2 = xn then
(a2 – 1)X – (a + b) = (a + b) X 2 + (b2 – 1)X M1
 0 = (a + b){ X 2 – (a – b)X + 1} M1 A1 Factorisation
and so, for xn + 2 = xn but xn + 1  xn
we must have a + b = 0 A1 Given Answer fully justified & clearly stated
(No marks for setting b = –a, for instance, and showing sufficiency)
For “comparing coefficients” approach (must be all 3 terms) max. 3/4 6
(a 2  1) xn  2  (a  b)
(ii) xn + 4 = M1 Use of the two-step result from earlier
(a  b) xn  2  (b 2  1)
 (a 2  1) X  (a  b) 
(a 2  1)    ( a  b)
 (a  b) X  (b 2  1) 
= A1 Correct, unsimplified, in terms of X
 (a 2  1) X  (a  b) 
( a  b)    (b 2  1)
 (a  b) X  (b  1) 
2

If xn + 4 = xn then M1 Equating
(a2 – 1)2X – (a + b) (a2 – 1) – (a + b)2X – (a + b) (b2 – 1) A1 LHS correct
2 2 2 2 2 2 2
= (a + b) (a – 1)X – (a + b) X + (a + b) (b – 1)X + (b – 1) X A1 RHS correct
 0 = (a + b) (a2 + b2 – 2)X 2 – [(a2 – 1)2 – (b2 – 1)2]X + (a + b) (a2 + b2 – 2)
M1 Good attempt to simplify
2 2 2
 0 = (a + b) (a + b – 2) {X – (a – b)X + 1} M1 Factorisation attempt
A1 A1 Partial; complete
and the sequence has period 4 if and only if
a2 + b2 = 2, a + b  0, X 2 – (a – b)X + 1  0 B1 CAO Correct final statement

[Ignore any discussion or confusion regarding issues of necessity and sufficiency]

NB Some candidates may use the one-step result repeatedly and get to xn + 4 via xn + 3:
(a 3  2a  b) X  (a 2  ab  b 2  1) ax  1
xn + 3 = 2 and xn + 4 = n  3 starts the process; then as above.
(a   ab  b  1) X  (a  2b  b )
2 3
xn  3  b
10

34 927
ALT. Consider the two-step sequence {… , xn , xn + 2 , xn + 4 , …} given by (assuming a + b  0)
 a2 1
 X 1
 a  b  AX  1
xn +2 =  , which is clearly of exactly the same form as before.
 b  1
2
X B
X   
 ab 
Then xn + 4 = xn if and only if a + b  0, X 2 – (a – b)X + 1  0 (from xn + 4  xn + 2 and
xn + 4  xn as before), together with the condition A + B = 0 (also from previous work);
a 2 1 b2 1
i.e.   0 , which is equivalent to a 2  b 2  2  0 since a + b  0.
ab ab

Note that it is not necessary to consider xn + 4  xn + 3 since if xn + 4 = xn + 3 = X then the sequence

would be constant.

35 928
Question 3 

(i) sin y = sin x  y = n + (–1)n x

n = –1 : y=––x B1
n=0: y=x B1
n=1: y=–x B1 Withhold final B mark for any number of extra eqns.
y

 B1 n = 0 case correctly drawn

B1 n = 1 case correctly drawn

                 
– O  x

B1 n = –1 case correctly drawn

Withhold final B mark for any number of extra lines
–  Ignore “endpoint” issues
6
dy
(ii) sin y = 1
2 sin x  cos y = 1
2 cos x M1 Implicit diffn. attempt (or equivalent)
dx
dy cos x
 A1 Correct
dx 2 cos y
cos x cos x
=
2 1  14 sin x 2
or
4  sin 2 x
A1 Correct and in terms of x only
3
   
1 1
d2 y 4  sin 2 x 2 .  sin x  cos x. 12 4  sin 2 x 2 .  2 sin x cos x

dx 2 4  sin 2 x
M1 For use of the Quotient Rule (or equivalent)
M1 For use of the Chain Rule for d/dx(denominator)
A1
 sin x4  sin x   cos x. sin x
2 2
= M1 Method for getting correct denominator
4  sin 2 x 2
3

=

sin x cos 2 x  4  sin 2 x 
4  sin x 
3
2 2

 3 sin x
= A1 Given Answer correctly obtained from c2 + s2 = 1
4  sin x 2
3
2
5

36 929
 dy
Initially, = 1
2 at (0, 0) increasing to a maximum
dx
d2 y
at ( 2 , 6 ) since 0
dx 2
B1 (Gradient and coordinate details unimportant unless
graphs look silly as a result)


B1 Reflection symmetry in x = 2

B1 Rotational symmetry about O

B1 Reflection symmetry in y =  
2

4
(iii)

B1 RHS correct

B1 LHS correct

37 930
Question 4 

(i) Setting f(x) = 1 in (*) gives

2
b  b  b 
  g ( x) dx     1 dx   g ( x)2 dx  B1 Clearly stated
    
a  a  a 
2
b  b 
Let g(x) = e :   e x dx   (b  a )  e 2x dx 
x
M1
a  a 

 eb  e a  2
 
 (b  a ). 12 e 2b  e 2 a

 e b  e a  (b  a ). e b  e a e b  e a 
2
1
2

 e b  e a  (b  a )e b  e a 
1
2
A1
Choosing a = 0 and b = t gives M1
t
e 1 1

e t  1  12 t e t  1  
et  1
 2t A1 Given Answer legitimately obtained 5
(ii) Setting f(x) = x, a = 0 and b = 1 in (*) gives
2
1  1  1 
  x g ( x) dx     x 2 dx   g ( x)2 dx  B1 Clearly stated
    
0  0  0 
 1 x2
Choosing g(x) = e 4 gives M1
2
 1  14 x 2   1  12 x 2 
  xe
 dx   13 13  0 3  
e
 dx 
0  0 
2
 1 2 1
  2e  4 x    1  e  2 x dx 
1 1 2

   
 0
3   A1 A1 LHS, RHS correct
 0 
2
1
 1 x2    1 
 e
0
2 dx  3  2  e 4  1 
  
1 2
 1 x2  1 
i.e. 0
e 2 dx  121  e 4 
 
A1 Given Answer legitimately obtained
5
(iii) With f(x) = 1, g(x) = sin x , a = 0, b = 12  , M1 Correct choice for f, g (or v.v.)
(*) becomes M1 Any sensible f, g used in (*)
2
 1
2    1
2
 sin x dx   1   sin x dx 
  2    A1
0  0 
1
RHS is 1
2   cos x  = 12 
0
1
2

(and since LHS is positive) we have 
0
sin x dx 
2
A1 RH half of Given inequality obtained
from fully correct working
4
38 931
With f(x) = cos x, g(x) = 4
sin x , a = 0, b = 12  , M1 Correct choice for f, g (or v.v.)
(*) gives
2
 12    12   12  
 cos x.(sin x) 14 dx    cos 2 x dx  sin x dx  A1
     
0  0  0 
2
 1

 
LHS =   54 (sin x) 4   
5 2
16
M1 A1 By recognition/substitution integration
    25
 0

 
1 1
2 2

 cos x dx    12 cos 2 x  dx
2
and 1
2
M1
0 0

2
 1

 
=  12 x  14 sin 2 x    14 
2
A1
 0 
 
 
Giving the required LH half of the Given inequality:
 12   1
2
64
    sin x dx  i.e.  sin x dx 
6
16 1
25 4  25
0  0

Withhold the last A mark if final result is not arrived at

39 932
Question 5 
dy
dy 2a 1
(i)  dt
  M1 Finding gradt. of tgt. (or by implicit diffn.)
dx dx
dt 2at t
 Grad. nml. at P is –p A1
 Eqn. nml. to C at P is x – 2ap = –p(x – ap2) B1 FT any form, e.g. y = –px + ap(2 + p2)
Nml. meets C again when x = an2, y = 2an M1 Substd. into nml. eqn.
 2an = – pan2 + ap(2 + p2)
 0 = pn2 + 2n – p(2 + p2)
 0 = (n – p)(pn + [2 + p2]) M1 Solving attempt
2 p 2
Since n = p at P, it follows that n   at N
p
 2
i.e. n    p  
 p
A1 Given Answer legitimately obtained
6
(ii) Distance P(ap2, 2ap) to N(an2, 2an) is given by
 
PN2 = a ( p 2  n 2 )  2a( p  n)
2 2
M1
=a 2
( p  n)  ( p  n)
2 2
 4

2    2  
2 2

= a 2  2 p      4 M1 Substituting for n
 p   p  

2
 p 2  1  1  p 2  2 ( p  1)
2 3
= 16a   
3
2
 = 16a A1 Given Answer legitimately obtained
 p   p  p4

d( PN 2 ) d( p 2  3  3 p 2  p 4 )
 16a 2 M1 Differentiation directly,
dp dp
or by the Quotient Rule
3 5
= 16a (2 p  6 p  4 p )
2
A1 Correct, unsimplified
p6  3 p2  2
= 32a 2
p5

=
32a 2 2
p 5

( p  1) 2 p 2  2 
d( PN 2 )  p 4 .3( p 2  1) 2 .2 p  ( p 2  1) 3 .4 p 3 
Note that  16a 2  
dp  p8 

=
32a 2 3 2
p8
 
. p ( p  1) 2 3 p 2  2( p 2  1) by the Quotient Rule

d( PN 2 )
 0 only when p2 = 2 A1 Given Answer fully shown
dp
Justification that it is a minimum E1
2
d( PN )
(either by examining the sign of
dp
2
or by explaining that PN cannot be maximised 4
40 933
 2
(iii) Grad. PQ is B1
pq
2 2
Grad. NQ is or B1
nq 2
q p
p
Since PQN = 90 (by “ in a semi-circle”; i.e. Thales Theorem)
2 2
  1 M1
pq q p 2
p
 2 2q
 4  ( p  q ) p  q    p 2  q 2  2 
 p p
2q
 2  p2  q2 
p
A1 Given Answer legitimately obtained
4
2q
PN minimised when p2 = 2  q 2  M1 Substituted into given expression
p
2
 q = 0 or q =  2 A1
p
But q   2  q = p (which is not the case) E1 Other cases must be ruled out

Special Case: 1/3 for substg. q = 0 and verifying that p2 = 2 3

41 934
Question 6 

(i)     
When  1  Clear verification. 
1 2√1 1 B1 
Assume that the statement is true when  :  B1  Must be clear that this is 
2√ 1 assumed. 
Then    Linking   and   
1 M1 
 
√ 1
1 M1  Using assumed result 
2√ 1  
√ 1
Sufficient to prove:  M1   
1
2√ 1 2√ 1 1 
√ 1

i.e. 2 1 1 2 1   A1  Multiplying by √ 1 or 

putting over a common 
denominator 
i.e. 2 1 2 1     
i.e. 4 4 4 4 1  A1   
Which is clearly true. Therefore by induction the statement is  B1  Clear conclusion showing logic 
true for all  1.  of induction. 
  [8]   
(ii)     
Required to prove:  Squaring given inequality
4 1 1 4 3 M2 
i.e. 16 24 9 1 16 24 9   A1
which is clearly true. 
  [3]   
When  1:  M1
1
1 2  
2
So we need    A1
Prove   works using induction  M1
Assume holds when  :  M1  Allow a general c. 
1 3
2√  
2√ 2
Then   M1   
1 1 1
2√  
√ 1 2√ √ 1
Sufficient to prove:  A1
1 1 1
2√ 2√ 1  
2√ √ 1 2√ 1
i.e. 4 √ 1 √ 1 2√ 4√ 1 √   A1A1
Which simplifies to the previously proved inequality.  B1
No further restrictions on c, so the minimum value is   
  [9]
 

42 935
Question 7 

(i) For 0 < x < 1, x is positive and ln x is negative

so 0 > x ln x > ln x
 e0 > ex ln x > eln x or ln 1 > ln xx > ln x
 (1 >) f(x) > x since ln is a strictly increasing fn. B1
Again, since ln x < 0, it follows that
ln x < f(x) ln x < x ln x
 ln x < ln{g(x)} < ln{f(x)} M1 Suitably coherent justification
 x < g(x) < f(x) A1 Given Answer legitimately obtained

For x > 1, ln x > 0 and so x < f(x) < g(x) B1 No justification required 4
(ii) ln{f(x)} = x ln x M1 Taking logs and attempting implicit diffn.
Alt. Writing y = ex ln x and diffg.
1 1
.f ( x)  x.  1. ln x i.e. f (x) = 1 ln x  f(x) A1
f ( x) x
f (x) = 0 when 1 + ln x = 0, ln x = –1, x = e – 1 A1
3
(iii) _|Å f ( x) = _|Å e  = _|Å e  = 1 B1 Suitably justified
x0 x0
x ln x
x0
0

_|Å g( x) = _|Å x f ( x)  = _|Å x  = 0 B1 May just be stated

x0 x0 x0
1

Alt. _|Å g ( x )  = _|Å e

x0
f ( x ) ln x
x0
 = _|Å e ln x  = _|Å x = 0
x0 x0
2
1
(iv) For y =  ln x (x > 0),
x
dy 1 1 x 1
  2  or 2 = 0 … M1 Diffg. and equating to zero
dx x x x
… when x = 1 A1 From correct derivative
dy dy
For x = 1–,  0 and for x = 1+, 0 M1 Method for deciding
dx dx
1
(1, 1) is a MINIMUM of y =  ln x A1
x
(Since there are no other TPs or discontinuities)
y  1 for all x > 0 Conclusion must be made for all 4 marks 4
ln(g(x)) = f(x) ln x M1 Taking logs and attempting implicit diffn.
1 1
.g ( x)  f ( x).  ln xf ( x)1  ln x  A1 using f (x) from (ii)
g ( x) x
1 
 g ( x )  f ( x).g ( x)   ln x  (ln x) 2 
x 
 f ( x).g ( x)1  (ln x) 2  M1 using previous result of (iv)
> 0 since f, g > 0 from (i)
and 1 + (ln x)2  1 > 0 A1 Given Answer fully justified 4
43 936
 B1 One of f, g correct …
g f B1 Both correct …
… relative to y = x
y=x
B1 All three passing thro’ (1, 1)

1 f

g 3
O 1

44 937
Question 8 

Line thro’ A perpr. to BC is r = a +  u B1

Line thro’ B perpr. to CA is r = b +  v B1
Lines meet when (r = p =) a +  u = b +  v M1 Equated
1
 v= a  b   u  A1

Since v is perpr. to CA, a  b   u   (a  c) = 0 M1

 a  b   a  c  +  u  (a  c) = 0 A1 Correctly multiplied out

 
b  a   a  c  M1 Re-arranging for 
u  a  c 
A1 Correct (any sensible form)
 b  a   a  c  
 p  a    u A1 FT their  (if only a, b, c, u involved)
 u  a  c  
9
CP  p  c  a  c  u B1 FT their 

Attempt at CP  AB M1
= a  c  u   b  a  A1 Correct to here

= a  c   b  a   u  b  a 
Now u  (b  c)  0 since u perpr. to BC M1

 ub  uc A1

so that CP  AB = a  c   b  a   u  c  a  M1 Substituted in

= a  c   a  b  u  M1 A1 Factorisation attempt; correct

= 0 from boxed line above A1 E1 Statement; justified
 CP is perpr. to AB E1 For final, justified statement 11

Notice that the “value” of is never actually required

Any candidate who states the result is true because P is the orthocentre of ABC may be awarded B2 for
actually knowing something about triangle-geometry, but only in addition to any of the first 3 marks earned
in the above solution: i.e. a maximum of 5/11 for the second part of the question.

45 938
Question 9

(i) 

O↺ F . r = F1 . r B1  For correct moment equation. 
 F = F1 
Res.  F + F1cos = Rsin B1  For resolving horizontally for 
one cylinder. 
Together give Rsin = F(1 + cos ) AG 
Since F1   R , with  = 12 ,  M1  Use of the Friction law 
F sin  A1  Combining with previous 
it follows that  1
  1
R 2
1  cos  2 answer 
i.e. 2 sin  1 + cos AG 
Subtotal: 
4 
(ii)
Res.  for RH cylinder W = N – Rcos – F1sin B1  F1 might correctly be replaced 
with F. 
Res.  for plank kW = 2Rcos + 2Fsin B1 

Eliminating W: M1  For eliminating W 

k(N – Rcos – Fsin ) = 2Rcos + 2Fsin

2  2  A1  For correct rearrangement for 
N  R cos    1  F sin    1 N 
k  k 
 2  1  cos   M1   1  cos  
N    1 . cos   sin   F For use of  R   F
k  sin    sin  
 2  cos  cos   sin  
2 2 B1  Obtaining  cos sin  
N    1  F
 k  sin  
 2  cos   1  AG 
=   1 F
k  sin  
For no slipping at the ground, F  N M1  Using Friction equation 

 2  cos   1  A1  Using  previous part 

 F 1
  1 F
 sin  
2
k

ie. 2k sin  (k + 2)(1 + cos ) A1  Rearranging into a “useful” 

form. 

46 939
However, we already have that E1  Properly justified 
2k sin  k(1 + cos )  (k + 2)(1 + cos )
so there are no extra restrictions on  .
Subtotal:   
10 
(iii)    
4 sin 1 2cos M1  Squaring up an appropriate 
trig inequality 
4 1 cos 1 2cos cos
0 5 cos 2 cos 3 M1  Creating and simplifying  
0 5 cos 3 cos 1 quadratic inequality in one trig 
ratio 
Since cos 0 we have cos   A1
For appropriate angles cos  is decreasing and sin  is  E1  A graphical argument is 
increasing.  perfectly acceptable here. 
N.b It is possible that 
inequalities like 2 1  
are squared. If this is done 
without justifying that both 
sides are positive then 
withhold this final E1. 
Therefore sin   AG   
ra B1
sin =  
r
So 5r – 5a  4r M1 Combining with previous 
result 
r  5a AG   
Subtotal: 
6 
 

47 940
Question 10 

ma = F – (Av2 + R)   B1  Clear use of N2L 

d M1   
WD =  F dx  
0
d AG   
 ma  Av 
 R dx  
2
=
0

dv B1   
Since a  v  
dx
xd M1 Attempting to change variable of 
 ma  Av R  ddxv dv
2
WD = integration. 
x0
xd

 ma  Av  R
2 v
= dv
x0
a
 
Using v2 = u2 + 2as with v = w, u = 0, s = d  B1 Justifying limits. Ignore absence of 
w  2ad    
Therefore:  AG
vw
ma  Av 2
 Rv
WD =
v0
 a
dv  

  [5]   
(i) 
2 ad M1  Performing integration 
 R  v 2 Av 4 
WD =  m      
 a 2 4a  0

 R A1 Correct answer in terms of d.
 m  ad  Aad  
2

 a
For second half-journey, B1B1  B1 for correct limits 
0
 ma  Av 2
 Rv B1 for correct integrand 
WD = 
w
a
dv

 
= – mad + Rd + Aad2 A1   
Summing gives 2dR + 2Aad2 AG  N.b.  integrals may be combined to 
get to the same result.  
R > ma  F = Av2 + R – ma > 0 always E1   
[6]   
   

48 941
(ii)
If R < ma then F is zero when Av2 = ma – R B1  Finding an expression for the 
ma  R critical speed. 
i.e. when v = V =
A
For F to fall to zero during motion, V < w E1   
ma  R E1   
i.e. when  2ad i.e. R > ma – 2Aad
A
In this case, WD = mad + Rd + Aad2 , B1   
as before, for the first half-journey
V
 ma  Av 2
 Rv M2   
For the second half WD = 
w
a
dv

A1   
 v 2 Av 4  V
ma  R  2a  4a 
 w
2 M1  Substituting expressions for V and 
1  ma  R  A  ma  R 
= (ma  R )    – w. 
2a  A  4a  A 
1
2a
(ma  R)2ad  
A
4a

4a 2 d 2 
1 1    
= (ma  R) 2  (ma  R ) 2 – (ma – R)d +
2 Aa 4 Aa
Aad2
1 A1 CAO  Without wrong working 
= (ma  R) 2 – mad + Rd + Aad2
4 Aa
1 AG
So total WD = (ma  R) 2 + 2Rd + 2Aad2
4 Aa
[9]   
 

49 942
Question 11 

   

(i)     
At A,  ,  0  B1

At  , , 2   B1   
Conservation of energy:  M1   
5 1
2  
2 2
 
     
   A1
  [4]   
If angle at   is   and it just passes the second wall then we 
have: 
1 M1 Using   
0 sin  
2
So  sin   A1  Solving for t at second 
wall. 
Also,  cos   M1  Considering horizontal 
distance 
2 sin cos   N.b. Some candidates 
  may just quote this (or 
equivalent). Give full 
credit. 
2 sin cos A1  Combining previous 
results. 
So sin 2 1  A1   
Therefore  45°  AG Condone absence of 
domain considerations. 
  [5]
 velocity is constant so     
cos M1  Comparing   velocities 
1 A1
5 cos  
√2
1
cos  
√10
3 A1  Converting to a more 
sin , tan 3 
√10 useful ratio. 
   

50 943
Method 1:  M1 Using   
3 1
2 5  
√10 2
3 1
 
√2 2
So 
3√2 4
0 

   
2 2
2 0 

A1   
First time over the wall means that   
A1   
So  5  
√
Method 2:  M1  Using trajectory 
sec equation 
tan  
2
A1  Combining with 
2 3   previous results 
2 0    
  A1   
  [6]
If the speed at   above first wall is   then by conserving  M1   
energy,  
1 1
5 2  
2 2
 
2   B1   
 
Using trajectory equation with  origin at top of first wall and  M1 Use of trajectory 
angle   as particle moves over first wall:  equation (might be 
1 tan several kinematics 
tan   equations effectively 
2
When   we need  0:  leading to the same 
1 tan thing) 
0 tan  
2
Treating this as a quadratic in tan :  M1  Considering the 
quadratic (or 
tan tan 0  equivalently 
2 2
tan 2 tan 2 0  differentiating to find 
The discriminant is:  the max) 
4 4 2  
 
4 4 4 2 2   A1  Obtaining a clearly 
4 4 4 2 4   negative discriminant – 
8   this might take many 
0  alternative forms. 
Therefore no solution. 
  [5]   
       

51 944
Question 12 

(i)  B2   

P X r P Y n r  

e λ e μ B1   
 
r! n r !

n! M1  Attempting to manipulate 
λμ   factorials towards a binomial 
! r! n r !
coefficient 
n B1  Identifying  correct binomial 
λμ   coefficient 
! r

B1
 
!
Which is the the formula for    E1  Recognising result. Must state 
parameters 
  [7]   
(ii)     
M2 (may be implied by following 
|  
line) 
A1
! !
 
!
! A1   
 
! !
Which is a  ,  distribution.  E1 Parameters must be stated.
  [5]   
(iii) This corresponds to r=1, k=1 from (ii) M2 Can be implied by correct 
answer. 
So probability is  .  A1
(iv)  [3]
Expected waiting time given that Adam is first is waiting time  B2  Also accept  waiting time given 
for first fish plus waiting time for Eve    Eve is first. Must be clearly 
identified. 
Expected waiting time is:  M2
E(Waiting time|Adam first)P(Adam first)+E(Waiting time|Eve 
first)P(Eve first) 
=   A1
1 1 1   No need for this algebraic 
 
simplification. 
  [5]   
 

52 945
Question 13 

(i)     
1 1 M1A1 M1 for any attempt relating to 
correct key on attempt 1   the geometric distribution – 
e.g. missing first factor or 
power slightly wrong. 
    Although not strictly necessary, 
Where  , 1   you may see this substitution 
frequently 
Expected number of attempts is given by  M1  May be written in sigma 
2 3 … notation 
1 2 3 …    
1 M1  Linking to binomial expansion 
1
 
  A1   
  [5]
(ii)     
correct key on attempt  for  1. . .   B1   
Expected number of attempts is given by  M1   
1 2 3
…  
 
1 M1A1  M1 for clearly recognising sum 
 
2 of integers / arithmetic series. 
  [4]   
(iii)     
correct key on attempt   M1  M1 for an attempt at this, 
1 1 1 A1  possibly by pattern spotting the 
…   first few cases. Condone 
1 2 1
absence of checking  1 case 
explicitly. 
1 M1 M1 for attempting telescoping 
 
2 1 AG  (may be written as an 
induction) 
1 1 M2 Attempting partial fractions
1  
1 2 A1  (This may be seen later) 
  [6]
Expected number of attempts is given by  M1   

1  
2 1
1 1 2 2
1
1 1
3 3
…  
1 2
1 1 1 M1A1  M1 for attempting telescoping 
1 …  
1 1
1 1 B1
1  

In the brackets there is an infinite sum minus a finite sum, so  E1   
the result is infinite. 
  [5]   
       
53 946
 STEP III 2017 Mark Scheme 

Question 1 
! ! ! !
  (i)        M1 
! !

! !
    
!

! !
    M1 
!

1 1 1 1 1 ! !
∴  
!
! !
    A1*  (3) 
!

∑ ∑   M1 

⋯   M1 

    because   → ∞   as   → ∞  E1 

  A1  (4) 

∑ 1     M1 M1 (2) 

!
(ii)     M1 
! ! ! ! !

!
So    A1*  (2) 

20 1 5! 120 120 120

 
1 1 4

4 1 4   M1 

0   

as   3 and so denominator is positive.  E1  (2) 
!
Hence,    

 
Alternatively,  
20 1 5! 5! 5!
4 1  
1 1 4 1 4

5! 5 5!
 
5 4
as   3  and so   5 
 
54 947
 3! 1 1 1
 
2

            M1 
! ! !
So   ∑  

            M1 

And therefore  ∑  

            A1* (3) 

5! 20 1 1 1 5
20 10  
2 4

        M1          M1 
Therefore  
5! 35
 
4

1 35 1 1 7 96 12 115
 
4 5! 1 8 96 96 96 96

          M1        A1* (4) 
   

55 948
Question 2 

 (i)        M1 

Thus   1     A1*  (2) 

(ii)   1  

1 1   M1 A1 

So    

This is a rotation about     if 

1   M1 

If    2  ,   1 0 ,  so     cannot be found.  E1 

Otherwise, multiplying by    , 

  

⁄ ⁄
2 sin 2 sin 2 sin   M1 

⁄ ⁄
sin sin sin   A1* (6) 

If    2  ,     M1 

So    1    A1 

This is a translation by   1    A1 (3) 

(iii)  If RS = SR, and if   2  , then 

1 1  

         M1 

2 0 

So   , or if    ,  2 0 

        A1 
2 cos 2 0  M1 
Thus   2   A1  (4) 
If   θ 2nπ 
⁄ ⁄ ⁄ ⁄
sin sin sin sin  M1 

2 sin sin 0   A1 

So    ,   2  , or   2  

A1                     A1       A1  (5) 
   
56 949
Question 3 

  M1 
  A1 (2) 
(i) 3 40 84 0  M1 A1 
2 42 0 M1 
2 7 6 0 M1 A1 
So   7  A1  (6) 
(ii)  3 4 
                        M1       M1  A1 (3) 
0  M1 
Thus     is a root of   4 0  M1 
So   2  A1 
6 
6 
2 6 M1 
3  as     (and so   2) 
So   5 A1 (5) 
 
Alternatively,   10 ,  M1 A1  so     and     are the roots of  
7 10 0  M1 A1  and as    ,   5 (and   2 ). A1 (5) 
 
(iii)  Thus     and     are the roots of   2 5 0  and     and     are the roots of  
2 2 0 M1 A1 
So   1 , 1 2  A1 A1 (4) 
   

57 950
Question 4 

 (i)     (formula book) 
So if   log  

  E1 
and so  ln ln ln log   B1 
Therefore,  

   
        M1        M1 

Thus,      A1* (5) 

(ii)     M1 

 

   M1 

    
  M1            A1* (4) 
  (iii)  Let    , 

Then     
            M1    M1 

√   
A1        M1        A1* (5)       

(iv)    

ln ln ln    

ln ln   M1 

ln ln  M1 

ln so    M1 

Integrating  ln ln ln ln ln ln  M1 A1 
ln   
  
 A1* (6) 

58 951
Question 5 

sin  

cos sin cos sin  M1 

cos   

sin cos sin cos  M1 

   M1 A1 (4) 

1  M1 

tan tan tan tan tan tan   

sec 0   M1 
0  A1* (3) 
 
1 sin   
cos   
So    cos 1 sin 0  M1 

sec tan   A1 

ln ln sec tan ln cos ln ln   M1 A1 

1 sin   M1 A1) 

Alternatively,    M1 A1 

ln ln 1 sin ln 1 sin  M1 

and hence   1 sin  A1 

 

4 ,    so  4 2    M1 

Thus    2 1 sin   A1 (8) 

59 952
  
G1 G1 dG1 G1 G1(5)   

60 953
Question 6 

 (i)     

Let     ,   

B1 
So  

   

                                   M1             M1 
∞    A1* (4) 

(ii)   ⇔ ⇔ 1 ⇔  

M1 

0    M1 

1 1   

   A1* (3) 

Alternatively,  
1 1 1 1
⇔  
1 1 1 1 1
  M1 
1 1
 
1 1 1
    M1      A1  
 

  

                                                     M1             M1 

   A1* (3) 

As   ∞  ,   ∞  
So 

   ∞ ∞ ∞

∞  

                               M1  M1 
 
61 954
 
Thus 

2 ∞    A1* (3) 

Let    ,   , then     implies     which is     M1 

2 ∞ 2 ∞  A1* (2) 

(iii)  Using    2 ∞   with   √3  M1 

√ √
√3 2 ∞ √3   
√ √

√3 2 ∞ √3 √3   

3 √3 2 ∞ ⇔ √3 ∞    A1* (2) 

Using    ∞   with   1 , 

1 ∞ 1    and so    1 ∞   B1 

Using     with   √2 1  and    1  M1 

√
√2 1 1   
√

√
√2 1 1 1 √2 1 1   
√ √

Using    ∞   ,    √2 1 ∞ √2 1  

So    √2 1 ∞ √2 1 1  

2 √2 1 ∞ 1 ∞ ∞   

√2 1 ∞    A1* (3) 

Alternatively, using     with   √2 1   

2 √2 1 2 √2 1
√2 1 √2 1 √2 1  
1 √2 1 2 √2 1

Therefore 2 √2 1 1   and so   √2 1 1 ∞  

   

62 955
 
Question 7 

   1  B1 (1) 

(i)   1 ⇒ 0    M1 

   M1 A1 

So L is      M1 

2 1 4 1 1 1   
2 1 1 1 1 4 1   
Thus  2 1 1   M1 
and as  ,  lies on this line  2 1 1  
0 2   A1*  (6) 
For there to be two distinct lines, there need to be two values of    . 
So the discriminant must be positive,   2 4 0  M1 
4 4   
   A1* 

1   

1   so   ,   lies outside the ellipse.  B1 (3) 

 
However, if    ,   , one tangent is at  0  or  ∞  ,   a vertical line.  E1 

If   1 , then   0 .    E1  

63 956
  
G1 (3) 
(ii)     and     are the roots of  0 2  

So     and       M1 

Thus     and   2  A1 A1 (3) 

Without loss of generality   0,  lies on   2 0 
and    0,  lies on   2 0 
So   2 0 , that is   2 0  M1 
and    2 0 

As   2   ,  2   M1 

4  

4  

4  M1 

4   
64 957
2 4   
4 4   

1   

1     A1* (4) 

   

65 958
Question 8 

  

         M1 

    M1 
Hence, 

     A1* (3) 
 
(i)   Let   1 (or any constant) and   sin   ,   M1 
then  

sin 1 sin sin 1 sin sin 1 1 1  

     M1 A1 
So 

1 1
2 cos sin sin 1 sin  
2 2

  M1 A1 
and therefore
1 1 1
cos sin 1 sin csc  
2 2 2

       A1* (6) 
(ii)  Let     and   sin 1 sin   ,  M1 
then  
sin sin 1 sin 1 sin  
1 1 1 1
2 cos sin 2 cos sin  
2 2 2 2
     M1 A1 

66 959
 4 sin sin sin   M1 A1 

Thus, using the stem 

1
4 sin sin
2
1 sin sin 1 1 sin 0 sin

sin sin 1  

  M1 A1 
 
 
So  
1
4 sin sin 1 sin sin 1 sin sin sin 1  
2

       M1 A1 

1
4 sin sin 1 sin sin 1  
2

Thus 
1
sin sin sin 1 csc  
2

where 
1
 
4
     A1 
and 
1
1  
4
     A1 (11) 

Alternatively, let     and   cos  , using stem,  M1 

1 1
cos cos
2 2
1 1 1
1 cos cos cos  
2 2 2

                M1 A1 
 
 
67 960
 
So, 

1
2 sin sin
2
1 1 1 1
1 cos cos sin 1 sin csc  
2 2 2 2
    M1 A1 
1 1 1 1 1 1
csc 1 cos sin sin cos sin 1 sin  
2 2 2 2 2 2

      M1 A1 
1 1 1 1 1 1
csc 2 1 cos sin 2sin cos sin 1 sin  
2 2 2 2 2 2
1 1
csc 1 sin 1 sin sin sin 1 sin  
2 2
M1 A1 
1 1
csc sin 1 1 sin  
2 2
giving result as before. 
 
   

68 961
Question 9 

For A,      and for B,   2   where     is tension. M1 A1 A1 

Adding,   2   M1 
Integrating with respect to time,   2  
Initially,   0 ,   0 ,   0⇒ 0 

Integrating with respect to time,    2   M1 M1

Initially,   0 ,   0 ,   0⇒ 0 

So    2   A1* (7)

When     ,    so   M1 A1 

Conserving energy, at time     we have shown there is no elastic potential energy, so 
1 1
0 2  
2 2
M1 A1 A1 A1 (6) 
That is 
2 2  
  B1 

But also    2    and so    2 6   M1 A1 

Thus  2 6 2 2    M1 A1

6 4 6 4 0 

2 2
2 0 
3 3

2
0 
3

M1 

and so     A1* (7) 

Alternatively,  

M1 

69 962
Subtracting,  
2 3 2  
3
 
2
M1 
So, 
2
1 cos  
3
M1 
where  
3
 
2
As    2  ,    3 1 cos   M1 

When     , 

so   3 3 1 cos   and thus   4   ,   M1 

2 sin 6
3 0 
3

2
3

M1 A1* (7) 

70 963
Question 10 

Moment of inertia of PQ about axis through P is    3 3   B1 

Conserving energy,  0 3 sin sin   M1 A1 A1 A1 

Thus   3 3 2 sin     A1*  (6) 

Differentiating with respect to time, 

2 3 3 2 cos  
M1 
So  
2 3 3 2 cos  
  A1 (2) 
Alternatively, taking moments about axis through P 
3
3 cos  
2
M1 
So  
2 3 3 2 cos  
  A1 (2) 

Resolving perpendicular to the rod for the particle, 

cos  
  M1 A1 
Thus  
3 2
cos cos 1  
2 3

  M1 A1 

3 2 6 2 3 2 3 2
1 0 
2 3 2 3 2 3
because    2   A1 (5) 

Resolving along the rod towards P for the particle, 

sin  
  M1 A1 

71 964
Thus  
3 2 3
sin sin 1 sin  
3 3

M1 
On the point of slipping    , so   B1 
3 3 2
sin cos  
3 2 3

Thus 
2
tan  
2
  A1* (5) 
At the instant of release, the equation of rotational motion for the rod ignoring the particle 
is  
3
3  
2
and thus  
 
2
M1 
Therefore the acceleration of the point on the rod where the particle rests equals  
  if   2  , and so the rod drops away from the particle faster than the particle 
accelerates and the particle immediately loses contact.  A1 (2) 

(Alternatively, for particle to accelerate with rod from previous working   0 , M1  
meaning that it would have to be attached to so accelerate, and as it is only placed on the 
rod, this cannot happen.) A1 (2) 

72 965
Question 11 

 (i)  Conserving (linear) momentum 
0 
M1 

A1 
1 1 1 1 1
1  
2 2 2 2 2
M1        M1    A1* (5) 
as required. 
(ii)  Conserving momentum before and after r th gun fired 

1  

    M1 A1 
Therefore 
 
M1 
and so 
 

  A1* (4) 
Summing this result for  1 to    , 

⋯  
1 2 3
    M1 
Because 
0 1 
1  

1
with equality only for the term    
Thus 
⋯
1 2 3
E1 

As   0 ,    A1* (3) 

73 966
(iii)  Considering the energy of the truck and the  1  projectiles before and after 
the    projectile is fired (the other  1  already fired do not change their kinetic energy 
at this time), 
1 1 1
1  
2 2 2
M1 A1 
1 1 1
2 2 2
1 1
 
2 2
1 1
 
2 2
1 1
 
2 2
    M1 
Summing this result for  1 to    , 
1 1
2 2
    M1 
So  
1 1
2 2
A1* (5) 
Now 

so  
1 1
2 2
M1 
and thus 
1 1 1 1 1 1
1 1
2 2 2 2 2 2
 
M1 

as   1  E1 (3) 

74 967
Question 12 

(i) 

, 1 

1 

     M1 

1
1 1 
2

     M1 A1 
1 1
1 1 1 
2 2

     M1 
Therefore, 
1
 
1
     A1 (5) 

1 2 1 1 2
1  
2 2 1 2 1

             M1 A1 (2) 
1 2
 
2 1
     B1 
For     and     to be independent,   ,   M1 
So  
1 2 1 2
 
2 1 2 1 1
       M1 
1 2 1 2 4 1  
1 2 1 4 0 

1 0 

    M1 
which does not happen for e.g.     ,   1 .  (Many equally valid examples possible.) 
  and     are not independent.  E1 (5) 
75 968
 
(ii)  
1 2 1 1
 
6 2

            M1 
1 2 1 1 2 1
 
6 6
     M1 A1 (3)  
1 2
1 2 2 1 6 1 2 1
 
2 1 2 1

       M1 A1 
1 2 1 7 5
 
4 6 12
         A1 (3) 
Thus 
1 2 1 7 5 2 1 1
, 0 
6 12 144 144

  M1          E1 (2) 
   

76 969
Question 13 

 
2 2  
  M1    M1        M1     A1 (4) 

2  

         M1    A1* (2) 

If    ~ 0,1  , then     and    , so   

                                         B1  B1    M1 A1 (4) 

1 1 1
 
3 12 2
1 1
∈ ,  
12 3

1 1 1 1 1 1
12 2 2 12 2 12

1
2  
12

  M1      M1      A1 

2  ,       and  0  otherwise. 

                                                       M1       A1        A1 (6) 

1 1 1 1 1
 
12 12 12 12 12

                                                      M1        M1 

2 1 1 1 1 1 1
2  
3 12 6 12 12 12 12

  M1        A1 (4) 
as required. 
Alternatively, for final integral, 

let     , 

77 970
 1
1 2 1 1
12 2 2  
12 3 6 12

                                                   M1     M1      M1    A1 (4) 

 
or further 

let     , 

 
1
1 2 1 1
12 2  
12 3 6 12

                                                   M1     M1      M1    A1 (4) 

 

78 971
79 972
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