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# Maths

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Stephen Siklos

Fourth edition, October 2008

This booklet is intended to help you to prepare for STEP examinations. It should also be useful as preparation for any undergraduate mathematics course, even if you do not plan to take STEP.
The questions are all based on recent STEP questions. I chose the questions either because they are
‘nice’ – in the sense that you should get a lot of pleasure from tackling them – or because I felt I had something interesting to say about them. In this booklet, I have restricted myself (reluctantly) to the syllabus for Papers I and II, which is the A-level core (i.e. C1 to C4) with a few additions.
This material should be familiar to you if you are taking the International Baccalaureate, Scottish
Advanced Highers or other similar courses.
The ﬁrst two questions (the sample worked questions) are in a ‘stream of consciousness’ format.
They are intended to give you an idea how a trained mathematician would think when tackling them. This approach is much too long-winded to sustain, but it should help you to see what sort of questions you should be asking yourself as you work through the later questions.
I have given each of the subsequent questions a diﬃculty rating ranging from (∗) to (∗ ∗ ∗). A question labelled (∗) might be found on STEP I; a question labelled (∗∗) might be found on STEP
II; a question labelled (∗ ∗ ∗) might be found on STEP III. But diﬃculty in mathematics is in the eye of the beholder: you might ﬁnd a question diﬃcult simply because you overlooked some key step, which on another day you would not have hesitated over. You should not therefore be discouraged if you are stuck on a one-star question; though you should probably be encouraged if you get through three-star questions without mishap.
Each question is followed by a comment and a full solution. The comment will contain hints and/or direct your attention to key points; or it might put the question in its true mathematical context.
The solution should give suﬃcient working for you to read it through and pick up the gist of the method. It is of course just one way of producing the required result: there may be many other equally good or better ways.
Calculators are not required for any of the problems in this booklet; and calculators are not permitted in STEP examinations. In the early days of STEP, calculators were permitted but they were not required for any question. It was found that candidates who tried to use calculators sometimes ended up missing the point of the question or getting a silly answer. My advice is to remove the battery so that you are not tempted.
I hope that you will use the comments and solutions as springboards rather than feather beds. You will only really beneﬁt from this booklet if you have a good go at each question before looking at the comment and another good go before looking at the solution. You should gain by reading both even if you have completed the question on your own.
If you ﬁnd errors or have suggestions for ways in which this booklet can be improved, please e-mail me: stcs@cam.ac.uk. I update it regularly.
I hope you enjoy using this booklet as much as I have enjoyed putting it together.

STCS

STEP MATHEMATICS
What is STEP?
STEP (Sixth Term Examination Paper) is an examination used by Cambridge colleges as the basis for conditional oﬀers. There were STEPs in many subjects but after 2001 the mathematics papers only remain. They are used not just by Cambridge, but also by Warwick, and many other university mathematics departments recommend that their appicants practise on the past papers even if they do not take the examination. In 2008, 2100 scripts were marked, only about 750 of which were written by Cambridge applicants.
There are three STEPs, called papers I, II and III. Papers I and II are based the core A-level syllabus, with some minor additions, and Paper III is based on a ‘typical’ further mathematics mathematics syllabus (for which there is no recognised core). The questions on Papers II and III are supposed to be of about the same diﬃculty, and are harder than those on Paper I. Each paper has thirteen questions, including three on mechanics and two on probability/statistics. Candidates are assessed on six questions only.

What is the purpose of STEP?
From the point of view of admissions to a university mathematics course, STEP has three purposes.
First, it acts as a hurdle: success in STEP is thought to be a good indicator of potential to do well on a diﬃcult course. Second, it acts as preparation for the course, because the style of mathematics found in STEP questions is similar to that of undergraduate mathematics. 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.

STEP vs A-level
A-level tests mathematical knowledge and technique by asking you to tackle fairly stereotyped problems. STEP asks you to apply the same knowledge and technique to problems that are, ideally, unfamiliar. Here is a quotation from Roger Porkess1 which illustrates why A-level examinations do not test satisfactorily the skills required to tackle university mathematics courses at the highest levels: Students following modular syllabuses work much harder and the system rewards them for doing so. This means that there is a new sort of A–C grade student: not necessarily a particularly intuitive mathematician but someone who has shown the ability to learn the subject successfully by dint of hard work and, quite possibly, good teaching.

No one believes that modular A-levels are uniformly bad; anything that encourages more students to study mathematics must be basically good. However, they are not good preparation for the students who plan to take a degree in mathematics from one of the top universities.
1

From Every cloud has a silver lining: Roger Porkess (1995).

Here is an A-level question:
By using the substitution u = 2x − 1, or otherwise, ﬁnd
2x
dx.
(2x − 1)2

Here is a STEP question, which requires both competence in basic mathematical techniques and mathematical intuition. Note that help is given for the ﬁrst integral, so that everyone starts at the same level. Then, for the second integral, candidates have to show that they understand why the substitution used in the ﬁrst part worked, and how it can be adapted.
Use the substitution x = 2 − cos θ to evaluate the integral
2
3/2

x−1
3−x

Show that, for a < b, q p

x−a b−x 1
2

dx =

1
2

dx.

(b − a)(π + 3 3 − 6)
,
12

where p = (3a + b)/4 and q = (a + b)/2.

The diﬀerences between STEP and A-level are:
1. There is considerable choice on STEP papers (6 out of 13). This is partly to allow for diﬀerent
A-level syllabuses.
2. STEP questions are much longer. Candidates completing four questions in three hours will almost certainly get a grade 1.
3. STEP questions are much less routine.
4. STEP questions may require considerable dexterity in performing mathematical manipulations.
5. Individual STEP questions may require knowledge of diﬀerent areas of mathematics (especially the mechanics and statistics questions, which will often require advanced pure mathematical techniques). 6. The marks available for each part of the question are not disclosed on the paper.
7. Calculators are not permitted.

Advanced Extension Awards were taken for the ﬁrst time in 2002. The last papers, in all subjects except mathematics, will be sat in 2010. Mathematics has a reprieve for at least a few years. The
AEA is not a replacement for STEP. It is aimed at the top 10% or so of the A-level cohort; i.e. at the top 7000 or soi whereas STEP is aimed at the top 2000 or so. This diﬀerence matters, because in mathematics (though not perhaps in other subjects) it is important to match the diﬃculty of the

question and the ability of the candidates exactly.2 The AEA contains questions on only the core
A-level syllabus: no statistics and (more important for us) no mechanics. Many of my colleagues believe that, apart from the resulting reduction of choice, this is undesirable because it sends out the wrong signals about the importance of these other areas of mathematics.

Setting STEP
STEP is produced under the auspices of the Cambridge Assessment examining board. The setting procedure starts 18 months before the date of the examination, when the three examiners (one for each paper, from schools or universities) are asked to produce a draft paper. The ﬁrst drafts are then vetted by the STEP coordinator, who tries to enforce uniformity of diﬃculty, checks suitability of material and style and tries to reduce overlap between the papers. Examiners then produce a second draft, based on the coordinator’s suggestions and taking into account his/her comments.
The second drafts are circulated to the three moderators (normally school teachers), and to the other examiners, who produce written comments and discuss the drafts in a two-day meeting. The examiners then produce third drafts, taking into account the consensus at the meetings. These drafts are sent to a vetter, who works through the papers, pointing out mistakes and infelicities.
The ﬁnal draft is checked by a second vetter. At each stage, the drafts are produced camera-ready, using a special mathematical word-processing package called LaTeX so no typesetting errors can arise after the second vetting.

STEP Questions
STEP questions do not fall into any one category. Typically, there will be a range of types on each papers. An excellent type involves two parts: in the ﬁrst part, candidates are asked to perform some task and are told how to do it (integration by substitution, or expressing a quadratic as the algebraic sum of two squares, for example); for the second part, candidates are expected to demonstrate that they have understood and learned from the ﬁrst part by applying the method to a new and perhaps more complicated task. Another good type of question requires candidates to do some preliminary special-case work and then guess and prove a general result. In another type, candidates have to show that they can understand and use new notation or a new theorem to perform perhaps standard tasks in disguise. Multipart questions of the A-level type are generally avoided; but if they occur, there tends to be a ‘sting in the tail’ involving putting all the parts together in some way.
As a general rule, questions on standard material tend to be harder than questions on material that candidates are not expected to have seen before. Sometimes, the ﬁrst question on the paper is of a basic nature, and on material that will be familiar to all candidates.
There is a syllabus for STEP, which is given at the end of this book.
Often there is at least one question that does not rely on any part of the syllabus: it might be a
‘common sense’ question, involving counting or seeing patterns, or it might involve some aspect of
GCSE mathematics with an unusual slant.
Some questions are devised to check that you do not simply apply routine methods blindly. Remember, for example, that a function can have a maximum value at which the derivative is non-zero
2

You could reasonably ask the question ‘Was Henry VIII a good king’ on examination papers from GCSE to PhD level. The answers would (or should) diﬀer according to the level. Mathematics does not work like this.

(it would have to be at the end of a range), so ﬁnding the maximum in such a case is not simply a question of routine diﬀerentiation.
Some questions take you through a new idea and then ask you to apply it to something else. Some questions try to test your capacity for clear thought without using much mathematical knowledge
(like the calendar question mentioned in the next section or questions concerning islands populated by toads ‘who always tell the truth’ and frogs ‘who always ﬁb’).
There are always questions speciﬁcally on integration or diﬀerentiation, and many others (including mechanics and probability) that use calculus as a means to an end, Graph-sketching is regarded by most mathematicians as a fundamental skill and there are nearly always one or more sketching questions or parts of questions. Some questions simply involve diﬃcult algebraic or trigonometric b manipulation. Basic ideas from analysis (such as 0 f (x) k ⇒ 0
(b − a)k or a f (x)dx

f (x) > 0 ⇒ f (b) > f (a) for b > a or a relationship between an integral and a sum) often come up, though knowledge of such results is never assumed — a sketch ‘proof’ of an ‘obvious’ result may be asked for.
Some areas are diﬃcult to examine and questions tend to come up rarely. Complex numbers fall into this category, and the same applies to (pure mathematical applications of) vectors.
The mechanics and probability questions often involve signiﬁcant ideas from pure mathematics.
Questions on pure statistical tests are rare, because questions that require real understanding (rather than ‘cookbook’ methods) tend to be too diﬃcult. More often, the questions in this section are about probability. The mechanics questions normally require a ﬁrm understanding of the basic principles (when to apply conservation of momentum and energy, for example) and may well involve a diﬀerential equation.
Projectile questions are often set, but are never routine.

First appearances
Your ﬁrst impression on looking at a STEP paper is likely to be that it looks very hard. Don’t be discouraged! Its diﬃcult appearance is largely due to it being very diﬀerent in style from A-levels.
If you are considering studying mathematics at Cambridge (or Oxford or Warwick), it is likely that you will manage to work through most or all of an A-level module in the time available: maybe
10 questions in 75 minutes (i.e. 7.5 minutes per question). In STEP, you are only supposed to do six questions in three hours, and you are doing very well if you manage four; that means that each question is designed to take 45 minutes. Thus although the questions are long and arduous, you have much more time to work on them.
You may be put oﬀ by the number of subjects covered on the paper. You should not be. STEP is supposed to provide suﬃcient questions for all candidates, no matter which A-level mathematics syllabus they have covered. It would be a very exceptional candidate who had the knowledge required to do all the questions.
Once you get used to the idea that STEP is very diﬀerent from A-level, it becomes much less daunting. Preparation
The best preparation for STEP is to work slowly through old papers.3 Hints and answers for the some years are available, but you should use these with discretion: doing a question with hints and answers in front of you is nothing like doing it yourself, and may well miss the whole point of the question (which is to make you think about mathematics). In general, thinking about the problem is much more important than getting the answer.
Should you try to learn up areas of mathematics that are not in your syllabus in preparation for
STEP? The important thing to know is that it is much better to be very good at your syllabus than to have have a sketchy knowledge of lots of additional topics: depth rather than breadth is what matters. It may conceivably be worth your while to round out your knowledge of a topic you have already studied to ﬁt in with the STEP syllabus; it is probably not worth your while to learn a new topic for the purposes of the exam. It is worth emphasising that there is no ‘hidden agenda’: a candidate who did two complete probability questions and one complete mechanics question will obtain the same grade as one who did four complete pure questions.
Just as the examiners have no hidden agenda concerning syllabus, so they have no hidden agenda concerning your method of answering the question. If you can get to the end of a question correctly you will get full marks whatever method you use. Some years ago one of the questions asked candidates to ﬁnd the day of the week of a given date (say, the 5th of June 1905). A candidate who simply counted backwards day by day from the date of the exam would have received full marks for that question (but would not have had time to do any other questions). You may be worried that the examiners expect some mysterious thing called rigour. Do not worry: STEP is an exam for schools, not university students, and the examiners understand the diﬀerence. Nevertheless, it is extremely important that you present ideas clearly and show working at all stages.

Presentation
You should set out your answer legibly and logically (don’t scribble down the ﬁrst thought that comes into your head) – this not only helps you to avoid silly mistakes but also signals to the examiner that you know what you are doing (which can be eﬀective even if you haven’t the foggiest idea what you are doing).

3
These and the other publications www.stepmathematics.org.uk. mentioned

below

are

obtainable

from

the

STEP

web

site

What to do if you cannot get started on a problem
Try the following, in order.
• Reread the question to check that you understand what is wanted;
• Reread the question to look for clues – the way it is phrased, or the way a formula is written, or other relevant parts of the question. (You may think that the setters are trying to set diﬃcult questions or to catch you out. Usually, nothing could be further from the truth: they are probably doing all in their power to make it easy for you by trying to tell you what to do).
• Try to work out exactly what it is that you don’t understand.
• Simplify the notation – e.g. by writing out sums explicitly.
• Look at special cases (choose special values which simplify the problem) in order to try to understand why the result is true.
• Write down your thoughts – in particular, try to express the exact reason why you are stuck.
• Go on to another question and go back later.
• If you are preparing for the examination (but not in the actual examination!) take a short break.4 • Discuss it with a friend or teacher (again, better not do this in the actual examination) or consult the hints and answers, but make sure you still think it through yourself. REMEMBER: following someone else’s solution is not remotely the same thing as doing the problem yourself.
Once you have seen someone else’s solution to an example, then you are deprived, for ever, of most of the beneﬁt that could have come from working it out yourself. Even if, ultimately, you get stuck on a particular problem, you derive vastly more beneﬁt from seeing a solution to something with which you have already struggled, than by simply following a solution to something to which you’ve given very little thought.

What if a problem isn’t coming out?
If you have got started but the answer doesn’t seem to be coming out then:
• Check your algebra. In particular, make sure that what you have written works in special cases. For example: if you have written the series for log(1 + x) as
1 − x + x2 /2 − x3 /3 + · · · then a quick check will reveal that it doesn’t work for x = 0; clearly, the 1 should not be there.5 4

Littlewood (distinguished Cambridge mathematician and author of the highly entertaining Mathematicians Miscellany) used to work seven days a week until an experiment revealed that when he took Sundays oﬀ the good ideas had a way of coming on Mondays.
5

Another check will reveal that for very small positive x, log is positive (since its argument is bigger than
1) whereas the series is negative, so there is clearly something else wrong.

• Make sure that what you have written makes sense. For example, in a problem which is dimensionally consistent, you cannot add x (with dimensions length, say) to x2 or exp x
(which itself does not make sense — the argument of exp has to be dimensionless). Even if there are no dimensions in the problem, it is often possible to mentally assign dimensions and hence enable a quick check. Be wary of applying familiar processes to unfamiliar objects (very easy to do when you are feeling at sea): for example, it is all too easy, if you are not sure where your solution is going, to solve the vector equation a.x = 1 by dividing both sides by a.
• Analyse exactly what you are being asked to do; try to understand the hints (explicit and implicit); remember to distinguish between terms such as explain/prove/deﬁne/etc. (There is essentially no diﬀerence between ‘prove’ and ‘show’: the former tends to be used in more formal situations, but if you are asked to ‘show’ something, a proper proof is required.)
• Remember that diﬀerent parts of a question are often linked (it is sometimes obvious from the notation and choice of names of variables).
• If you get irretrievably stuck in the exam, state in words what you are trying to do and move on (at A-level, you don’t get credit for merely stating intentions, but STEP examiners are generally grateful for any sign of intelligent life).

What to do after completing a question
It is a natural instinct to consider that you have ﬁnished with a question once you have done got to an answer. However this instinct should be resisted both mathematically and from the much narrower view of preparing for an examination. Instead, when you have completed a question you should stop for a few minutes and think about it. Here is a check list for you to run through.
• Look back over what you have done, checking that the arguments are correct and making sure that they work for any special cases you can think of. It is surprising how often a chain of completely spurious arguments and gross algebraic blunders leads to the given answer.
• Check that your answer is reasonable. For example if the answer is a probability p then you should check that 0 ≤ p ≤ 1. If your answer depends on an integer n does it behave as it should when n → ∞? Is it dimensionally correct?

If, in the exam, you ﬁnd that your answer is not reasonable, but you don’t have time to do anything about it, then write a brief phrase showing that you understand that you answer is unreasonable (e.g. ‘This is wrong because mass must be positive’).

• Check that you have used all the information given. In many ways the most artiﬁcial aspect of examination questions is that you are given exactly the amount of information required to answer the question. If your answer does not use everything you are given then either it is wrong or you are faced with a very unusual examination question.
• Check that you have understood the point of the question. It is, of course, true that not all exam questions have a point but many do. What idea did the examiners want you to have?
Which techniques did they want you to demonstrate? Is the result of the question interesting in some way? Does it generalise? If you can see the point of the question would your working show the point to someone who did not know it in advance?

• Make sure that you are not unthinkingly applying mathematical tools which you do not fully understand.6 • In preparation for the examination, try to see how the problem ﬁts into the wider context and see if there is a special point which it is intended to illustrate. You may need help with this.
• In preparation for the examination, make sure that you actually understand not only what you have done, but also why you have done it that way rather than some other way. This is particularly important if you have had to use a hint or solution.
• In the examination, check that you have given the detail required. There often comes a point in a question where, if we could show that A implies B, the result follows. If after a lot of thought you suddenly see that A does indeed imply B the natural thing to do is to write triumphantly
‘But A implies B so the result follows’7 . Unfortunately unscrupulous individuals who have no idea why A should imply B (apart from the fact that it would complete the question) could and do write the exactly the same thing. Go back through the major points of the question making sure that you have not made any major unexplained leaps.

6

Mathematicians should feel as insulted as Engineers by the following joke.
A mathematician, a physicist and an engineer enter a mathematics contest, the ﬁrst task of which is to prove that all odd number are prime. The mathematician has an elegant argument: ‘1’s a prime, 3’s a prime,
5’s a prime, 7’s a prime. Therefore, by mathematical induction, all odd numbers are prime. It’s the physicist’s turn: ‘1’s a prime, 3’s a prime, 5’s a prime, 7’s a prime, 11’s a prime, 13’s a prime, so, to within experimental error, all odd numbers are prime.’ The most straightforward proof is provided by the engineer: ‘1’s a prime,
3’s a prime, 5’s a prime, 7’s a prime, 9’s a prime, 11’s a prime . . . ’.
7

There is a standard anecdote about the distinguished Professor X. In the middle of a lecture she writes ‘It is obvious that A’, suddenly falls silent and after a few minutes walks out of the room. The awed students hear her pacing up and down outside. Then after twenty minutes she returns says ‘It is obvious’ and continues the lecture.

Sample worked question
Let
f (x) = ax −

x3
,
1 + x2

where a is a constant.
Show that, if a

9/8, then f ′ (x)

0

for all x.

First thoughts
When I play tennis and I see a ball that I think I can hit, I rush up to it and smack it into the net.
This is a tendency I try to overcome when I am doing mathematics. In this question, for example, even though I’m pretty sure that I can ﬁnd f ′ (x), I’m going to pause for a moment before I do so.
I am going to use the pause to think about two things. First, I’m going to think about what to do when I have found f ′ (x). Second, I’m going to try to decide what the question is really about. Of course, I may not be able to decide what to do with f ′ (x) until I actually see what it looks like; and I may not be able to see what the question is really about until I have ﬁnished it; maybe not even then. (In fact, some questions are not really about anything in particular.) I am also going to use the pause to think a bit about the best way of performing the diﬀerentiation. Should I simplify ﬁrst? Should I make some sort of substitution? Clearly, if knowing how to tackle the rest of the question might guide me in deciding the best way to do the diﬀerentiation. Or it may turn out that the diﬀerentiation is fairly straight forward, so that it doesn’t matter how I do it.
Two more points occur to me as I reread the question. I notice that the inequalities are not strict
(they are rather than >). Am I going to have to worry about the diﬀerence? I also notice that there is an ‘If ... then’, and I wonder if this is going to cause me trouble. I will need to be careful to get the implication the right way round. I mustn’t try to prove that if f ′ (x) 0 then a 9/8.
It will be interesting to see why the implication is only one way — why it is not an ‘if and only if’ question. Don’t turn over until you have spent a little time thinking along these lines.

Doing the question
Looking ahead, it is clear that the real hurdle is going to be showing that f ′ (x) 0. How am I going to do that? Two ways suggest themselves. First, if f ′ (x) turns out to be a quadratic function, or an obviously positive multiple of a quadratic function, I should be able to use some standard method: looking at the discriminant (‘b2 − 4ac′ ), etc; or, better, completing the square. But if f ′ (x) is not of this form, I will have to think of something else: maybe I will have to sketch a graph. I’m hoping that I won’t have to do this, because otherwise I could be here all night.
I’ve just noticed that f is an odd function, i.e. f (x) = −f (−x) (did you notice that?). That is is helpful, because it means that f ′ (x) is an even function.8 If it had been odd, there could have been a diﬃculty, because odd functions (at least, those with no vertical asymptotes) always cross the horizontal axis y = 0 at least once (they start in one corner of the graph paper for x large and negative and end in the diagonally opposite corner for x large and positive) and cannot therefore be positive for all x.
Now, how should I do the diﬀerentiation? I could divide out the fraction, giving x3 x
=x−
2
1+x
1 + x2

and

f (x) = (a − 1)x +

x x = bx +
.
2
1+x
1 + x2

(Is this right? I’ll just check that it works when x = 2 — yes, it does: f (2) = 2a−8/5 = 2(a−1)+2/5.)
This might save a bit of writing, but I don’t at the moment see this helping me towards a positive function. On balance, I think I’ll stick with the original form.
Another thought: should I diﬀerentiate the fraction using the quotient formula or should I write it as x3 (1 + x2 )−1 and use the product rule? I doubt if there is much in it. I never normally use the quotient rule — it’s just extra baggage to carry round — but on this occasion, since the ﬁnal form
I am looking for is a single fraction, I will.
One more thought: since I am trying to prove an inequality, I must be careful throughout not to cancel any quantity which might be negative; or at least if I do cancel a negative quantity, I must remember to reverse the inequality.
Here goes (at last): f ′ (x) = a −

a + (2a − 3)x2 + (a − 1)x4
3x2 (1 + x2 ) − 2x(x3 )
=
.
(1 + x2 )2
(1 + x2 )2

This is working out as I had hoped: the denominator is certainly positive and the numerator is a quadratic function of x2 . I can ﬁnish this oﬀ most elegantly by completing the square. There are two ways of doing this, and it is marginally easier to do it the unnatural way: a + (2a − 3)x2 + (a − 1)x4 = a(1 + (2 − 3a−1 )x2 ) + (a − 1)x4
= a 1 + 1 (2 − 3a−1 )x2
2

2

+ (a − 1) − 1 a(2 − 3a−1 )2 x4 .
4

We certainly need a 0 to make the ﬁrst term non-negative. Simplifying the second square bracket gives [2 − 9a−1 /4], so it is true that if a 9/8 then f ′ (x) 0 for all x.

8

You can see this easily from a by sketching a ‘typical’ odd function.

Post-mortem
Now I can look back and analyse what I have done.
On the technical side, it seems I was right not to use the simpliﬁed form of x3 /(1 + x2 ). This would have lead to the quadratic (b + 1) + (2b − 1)x2 + bx4 , which isn’t any easier to handle than the quadratic involving a and just gives an extra opportunity to make an algebraic error.
Actually, I see on re-reading my solution that I have made a bit of a meal out of the ending. I needn’t have completed the square at all; I could used the inequality a
9/8 immediately after ﬁnding f ′ (x), since a appears in f ′ (x) with a plus sign always: f ′ (x) =

a + (2a − 3)x2 + (a − 1)x4
(1 + x2 )2

9/8 + (9/4 − 3)x2 + (9/8 − 1)x4
= (x2 − 3)2 /8
(1 + x2 )2

0

as required.
I wonder why the examiner wanted me to investigate the sign of f ′ (x). The obvious reason is to see what the graph looks like. We can now see what this question is about. It is clear that the examiners really wanted to set the question: ‘Sketch the graphs of the function ax − x3 /(1 + x2 ) in the diﬀerent cases that arise according to the value of a’ but it was thought too long or diﬃcult. It is worth looking back over my working to see what can be said about the shape of the graph of f (x) when a < 9/8 . My unnecessarily elaborate proof, involving completing the square, makes the role of a a bit clearer than it is in the shorter alternative. (I leave that to you to think about!)

Sample worked question 2
The n positive numbers x1 , x2 , . . . , xn , where n x1 = 1 +

1
,
x2

x2 = 1 +

3, satisfy
1
, x3 and also xn = 1 +

... ,

xn−1 = 1 +

1
,
xn

1
.
x1

Show that
(i)

x1 , x2 , . . . , xn > 1,

(ii)

x 1 − x2 = −

(iii)

x 1 = x2 = · · · = x n .

x2 − x3
,
x2 x3

Hence ﬁnd the value of x1 .

First thoughts
My ﬁrst thought is that this question has an unknown number of variables: x1 , . . . , xn . That makes it seem rather complicated. I might, if necessary, try to understand the result by choosing an easy value for n (maybe n = 3). If I manage to prove some of the results in this special case, I will certainly go back to the general case: doing the special case might help me tackle the general case, but I don’t expect to get many marks in an exam if I just prove the result in one special case.
Next, I see that the question has three sub-parts, then a ﬁnal one. The ﬁnal one begins ‘Hence ...’.
This must mean that the ﬁnal part will depend on at least one of the previous parts. It is not clear from the structure of the question whether the three subparts are independent; the proof of (ii) and
(iii) may require the previous result(s). However, I expect to use the results of parts (i), (ii) and
(iii) later in the question.
Actually, I think I can see how to do the very last part. If I assume that (iii) holds, so that x1 = x2 = · · · = xn = x (say), then each of the equations given in the question is identical and each gives a simple equation for x.
I am a bit puzzled about part (i). How can an inequality help to derive the equalities in the later parts? I can think of a couple of ways in which the result xi > 1 could be used. One is that I may need to cancel, say, x1 from both sides of an equation in which case I would need to know that x1 = 0. But looking back at the question, I see I already know that x1 > 0 (it is given), so this cannot be the right answer. Maybe I need to cancel some other factor, such as (x1 − 1). Another possibility is that I get two or more solutions by putting x1 = x2 = · · · = xn = x and I need the one with x > 1. This may be the answer: looking back at the question again, I see that it asks for the value of x1 — so I am looking for a single value. I’m still puzzled, but I will remember in later parts to keep a sharp look out for ways of using part (i).
One more thing strikes me about the question. The equations satisﬁed by the xi are given on two lines (‘and also’). This could be for typographic reasons (the equations would not all ﬁt on one line) but more likely it is to make sure that I have noticed that the last equation is a bit diﬀerent: all the other equations relate xi to xi+1 , whereas the last equation relates xn to x1 . It goes back to the beginning, completing the cycle. I’m pleased that I thought of this, because this circularity must be important.

Doing the question
I think I will do the very last part ﬁrst, and see what happens.
Suppose, assuming the result of part (iii), that x1 = x2 = · · · = xn = x. Then substituting into any of the equations given in the question gives x=1+ 1 x i.e. x2 − x − 1 = 0. Using the quadratic formula gives

1± 5
,
x=
2
which does indeed give two answers (despite the fact that the question asks for just one). However,
I see that one is negative and can therefore be eliminated by the condition xi > 0 which was given in the question (not, I note, the condition x1 > 1 from part (i); I still have to ﬁnd a use for this).
I needed only part (iii) to ﬁnd x1 , so I expect that either I need both (i) and (ii) directly to prove
(iii), or I need (i) to prove (ii), and (ii) to prove (iii).
Now that I have remembered that xi > 0 for each i, I see that part (i) is obvious. Since x2 > 0 then
1/x2 > 0 and the ﬁrst equation given in the question, x1 = 1 + 1/x2 show immediately that x1 > 1 and the same applies to x2 , x3 , etc.
Now what about part (ii)? The given equation involves x1 , x2 and x3 , so clearly I must use the ﬁrst two equations given in the question: x1 = 1 +

1
,
x2

x2 = 1 +

1
.
x3

Since I want x1 − x2 , I will see what happens if I subtract the two equations: x1 − x2 = (1 +

1
1
1
1
x3 − x2
) − (1 + ) =

=
.
x2 x3 x2 x3 x2 x3

(∗)

That seems to work!
One idea that I haven’t used so far is what I earlier called the circularity of the equations: the way that xn links back to x1 . I’ll see what happens if I extend the above result. Since there is nothing special about x1 and x2 , the same result must hold if I add 1 to each of the suﬃces: x2 − x3 =

x4 − x3
.
x3 x4

I see that I can combine this with the previous result: x1 − x2 =

x3 − x2 x3 − x4
.
= x2 x3 x2 x2 x4
3

I now see where this is going. The above step can be repeated to give x1 − x 2 = and eventually I will get back to x1 − x2 : x1 − x2 =

x3 − x 4 x5 − x4
2 x = x x2 x2 x = · · · x2 x3 4
2 3 4 5

x1 − x2 x5 − x4
2 x2 x = · · · = ± x x2 x2 x2 . . . x2 x2 x x2 x3 4 5
2 3 4 5 n 1 2

i.e.
(x1 − x2 ) 1 ∓

1 x2 x2 x2 x2 x2 . . . x2 n 1 2 3 4 5

=0.

I have put in a ± because each step introduces a minus sign and I’m not sure yet whether the ﬁnal sign should be (−1)n or (−1)n−1 . I can check this later (for example, by working out one simple case such as n = 3); but I may not need to.
I deduce from this last equation that either x2 x2 x2 x2 x2 . . . x2 = ±1 or x1 = x2 (which is what I n 1 2 3 4 5 want). At last I see where to use part (i): I know that x2 x2 x2 x2 x2 . . . x2 = ±1 because x1 > 1, n 1 2 3 4 5 x2 > 1, etc. Thus x1 = x2 , and since there was nothing special about x1 and x2 , I deduce further that x2 = x3 , and so on, as required.

Postmortem
There were a number of useful points in this question.
The ﬁrst point concerns using the information given in the question. The process of teasing information from what is given is fundamental to the whole of mathematics. It is very important to study what is given (especially seemingly unimportant conditions, such as xi > 0) to see why they have been given. If you ﬁnd you reach the end of a question without apparently using some condition, then you should look back over your work: it is very unlikely that a condition has been given that is not used in some way. (It may not be a necessary condition — and we will see that the condition xi > 0 is not, in a sense, necessary in this question — but it should be suﬃcient.) The other piece of information in the question which you might easily have overlooked is the use of the singular (‘ ... ﬁnd the value of ...’), implying that there is just one value, despite the fact that the ﬁnal equation is quadratic..
The second point concerns using the structure of the question. Here, the position of the word ‘Hence’ suggested strongly that none of the separate parts were stand-alone results; each had to be used for a later proof. Understanding this point made the question much easier, because I was always on the look out for an opportunity to use the earlier parts. Of course, in some problems (without that
‘hence’) some parts may be stand-alone; though this is rare in STEP questions. You may think that this is like playing a game according to hidden (STEP) rules, but that is not the case. Precision writing and precision reading is vital in mathematics and in many professions (law, for example).
Mathematicians have to be good at it, which is the reason that everyone wants to employ people with mathematical training.
The third point was the rather inconclusive speculation about the way inequalities might help to derive an equality. It turned out that what was actually required was x1 x2 . . . xn = ±1. I was a bit puzzled by this possibility in my ﬁrst thoughts, because it seemed that the result ought to hold under conditions diﬀerent from those given; for example, xi < 0 for all i (does this condition work??). Come to think of it, why are conditions given on all xi when they are all related by the given equations? This makes me think that there ought to be a better way of proving the result which would reveal exactly the conditions under which it holds.
Fourth was the idea (which I didn’t actually use) that I might try to prove the result for, say, n = 3.
This would not have counted as a proof of the result (or anything like it), but it might have given me ideas for tackling the question.
Fifth was the observation that the equations given in the question are ‘circular’. It was clear that the circularity was essential to the question and it turned out to be the key the most diﬃcult part.

Having identiﬁed it early on, I was ready to use it when the opportunity arose.

Final thoughts
It occurs to me only now, after my post-mortem, that there is another way of obtaining the ﬁnal result. Suppose I start with the idea of circularity (as indeed I might have, had I not been otherwise directed by the question) and use the given equations to ﬁnd x1 in terms of ﬁrst x2 , then x3 , then x4 and eventually in terms of x1 itself. That should give me an equation I can solve, and I should be able to ﬁnd out what conditions are needed on the xi . Try it. You may need to guess a formula for x1 in terms of xi from a few special cases, then prove it by induction. You will ﬁnd it useful to deﬁne a sequence of numbers Fi such that F0 = F1 = 1 and Fn+1 = Fn + Fn−1 . (These numbers are called Fibonacci numbers9 .)

You should ﬁnd that if xn = x1 for any n (greater than 1), then xn = 1±2 5 . No conditions are required for this result to hold, except that all the xi exist, so it was not necessary to set the condition x1 > 1. (The condition on x1 for all xi to exist is that x1 = Fk /Fk−1 for any k n, as you will see if you obtain the general formula for x1 in terms of xn .)

9
Fibonacci (short for ﬁlius Bonacci — son of Bonacci) was called the greatest European mathematician of the middle ages. He was born in Pisa (Italy) in about 1175 AD. He introduced the series of numbers named after him in his book of 1202 called Liber Abbaci (Book of the Abacus). It was the solution to the problem of the number of pairs of rabbits produced by an initial pair: A pair of rabbits are put in a ﬁeld and, if rabbits take a month to become mature and then produce a new pair every month after that, how many pairs will there be in twelve months time?

Question 1(∗)

Find all the solutions of the equation
|x + 1| − |x| + 3|x − 1| − 2|x − 2| = x + 2 .

This looks more diﬃcult than it is. There is no easy way to deal with the modulus function: you have to look at the diﬀerent cases individually (for example, for |x| you have to look at x 0 and x 0). The most straightforward approach would be to solve the equation in each of the diﬀerent regions determined by the modulus signs: x −1, −1 x 0, 0 x 1, etc. You might ﬁnd a graphical approach helps you to picture what is going on (I didn’t).

Solution to question 1
Let
f (x) = |x + 1| − |x| + 3|x − 1| − 2|x − 2| − (x + 2).
We have to solve f (x) = 0 in the ﬁve regions of the x-axis determined by the modulus functions, namely x −1 ; −1 x 0 ; 0 x 1 ; 1 x 2 ; 2 x .
In the separate regions, we have

 (x + 1) − x + 3(x − 1) − 2(x − 2) − (x + 2)

 (x + 1) − x + 3(x − 1) + 2(x − 2) − (x + 2)

(x + 1) − x − 3(x − 1) + 2(x − 2) − (x + 2) f (x) =

 (x + 1) + x − 3(x − 1) + 2(x − 2) − (x + 2)

−(x + 1) + x − 3(x − 1) + 2(x − 2) − (x + 2)

=
0
= 4x − 8
= −2x − 2
=
−2
= −2x − 4

for
2 x 0.

You have to decide ﬁrst whether the statement is true of false. If true, the justiﬁcation has to be a proof. If false, you could prove that it is false, though it is often better to ﬁnd a simple counterexample. Part (iii) might look a bit odd. I suppose it relates to the standard approximation cos x ≈ 1 − x2 /2 which holds when x is small. This result can be improved by using a polynomial of higher degree: the next term is x4 /4! . It can be proved that, given any number k , cos x can be approximated, for
−k < x < k , by a polynomial of the form
N

(−1)n n=0 x2n
.
(2n)!

You have to use more terms of the approximation (i.e. a larger value of N ) if you want either greater accuracy or a larger value of k . In Part (iii) of this question, you are being asked if there is a polynomial such that the approximation is good for all values of x .

Solution to question 8
(i) True. The easiest way to see this is to log both sides. For the left hand side, we have ln(aln b ) = (ln b)(ln a) and for the right hand side we have ln(aln b ) = (ln a)(ln b) , which agree.
Note that we have to be a bit careful with this sort of argument. The argument used is that A = B because ln A = ln B . This requires the property of the ln function that ln A = ln B ⇒ A = B . You can easily see that this property holds because ln is a strictly increasing function; if A > B, then ln A > ln B . The same would not hold for (say) sin (i.e. sin A = sin B ⇒ A = B ).
(ii) False. θ = π/2 is an easy counterexample.

(iii) False. Roughly speaking, any polynomial can be made as large as you like by taking x to be very large (provided it is of degree greater than zero), whereas | cos x| 1 . There is obviously no polynomial of degree zero (i.e. no constant number) for which the statement holds.
(iv) True: x4 + 3 + x−4 = (x2 − x−2 )2 + 5

5.

Postmortem
The important point here is that if you want to show a statement is true, you have to give a formal proof, whereas if you want to show that it is false, you only need give one counterexample. It does not have to be an elaborate counterexample — in fact, the simpler the better.
There may still be a question mark in your mind over part (iii). Given a polynomial P(x), how do you actually prove that it is possible to ﬁnd x such that |P(x)| > 1 + 10−6 ? This is one of those many mathematical results that seems very obvious but is irritatingly diﬃcult to prove. (Often, the most irritatingly-diﬃcult-to-prove obvious results are false, though this one is true.)
You should try to cobble together a proof for yourself. I would start by choosing a general polynomial of degree N . Let
N

an xn ,

P(x) = n=0 where aN = 0 . Then I would assume that aN = 1 since if I could prove the result for this case, I’m sure I could easily modify the proof to deal with the general case. Now what?
A simple way forward would be to make use of the fundamental theorem of algebra, which says that
P(x) can be written in the form
P(x) ≡ (x − α1 )(x − α2 ) · · · (x − αn ) where α1 , α2 , . . . , αn are the roots of the equation P(x) = 0 . (The theorem says that this equation has n roots.) Then
|P(x)| ≡ |x − α1 ||x − α2 | · · · |x − αn | which can be made larger than any given number, M say, by choosing x so that |x − αi | > M 1/n for each i.

Actually, one one or two things have been brushed under the carpet in this proof (what if some of the roots are complex?), but I’m conﬁdent that there is nothing wrong that couldn’t be easily ﬁxed,

Question 9(∗)

Prove that the rectangle of greatest perimeter which can be inscribed in a given circle is a square.
The result changes if, instead of maximising the sum of lengths of sides of the rectangle, we seek to maximise the sum of nth powers of the lengths of those sides for n 2. What happens if n = 2?

Obviously, the perimeter of a general rectangle has no maximum (it can be as long as you like), so the key to this question is to use the constraint that the rectangle lies in the circle.
The word ‘greatest’ in the question immediately suggests diﬀerentiating something. The perimeter of a rectangle is expressed in terms of two variables, length x and breadth y, so you must ﬁnd a way of using the constraint to eliminate one variable. This could be done in two ways: express y in terms of x, or express both x and y in terms of another variable (an angle, say).
Finally, there is the matter of deciding whether your solution is the greatest or least value (or neither). This can be done either by considering second derivatives, or by trying to understand the diﬀerent situations. Often the latter method, or a combination, is preferable.
Finding stationary values of functions subject to constraints is very important: it has widespread applications (for example, in theoretical physics, ﬁnancial mathematics and in fact in almost any area to which mathematics is applied). It forms a whole branch of mathematics called optimisation.
Normally, the methods described above (eliminating one variable) are not used: a clever idea, the methods of Lagrange multipliers, is used instead,

Solution to question 9
Let the circle have diameter d and let the length of one side of the rectangle be x and the length of the adjacent side be y.
Then, by Pythagoras’s theorem, y= d2 − x2

(†)

and the perimeter P is given by
P = 2x + 2 d2 − x2 .
We can ﬁnd the largest possible value of P as x varies by calculus. We have dP x
,
= 2 − 2√
2 − x2 dx d so for a stationary point, we require (cancelling the factor of 2 and squaring)
1=

x2 d2 − x2

i.e.

2x2 = d2 .

Thus x = d/ 2 (ignoring the negative root for obvious reasons). Substituting this into (†) gives

√ y = d/ 2, so the rectangle is indeed a square, with perimeter 2 2 d .
But is it the maximum perimeter? The easiest way to investigate is to calculate the second derivative,

which is easily seen to be negative for all values of x, and in particular when x = d/ 2 . The stationary point is certainly a maximum.
For the second part, we consider f (x) = xn + (d2 − x2 )n/2 .
The ﬁrst thing to notice is that f is constant if n = 2, so in this case, the largest (and smallest) value is d2 . For n = 3 , we have f ′ (x) = 3x2 − 3x(d2 − x2 )1/2 ,

so f (x) is stationary when x4 = x2√ 2 − x2 ), i.e. when 2x2 = d2 as before or when x = 0. The
(d
corresponding stationary values are 2d3 and 2d3 , so this time the largest value occurs when x = 0.

Postmortem
You will almost certainly want to investigate the situation for other values of n yourself, just to see what happens.
Were you happy with the proof given above that the square has the largest perimeter? And isn’t it a bit odd that we did not discover a stationary point corresponding to the smallest value (which can easily be seen to occur at x = 0 or x = d)? To convince yourself that the maximum point gives the largest value, it is a good idea to sketch a graph: you can check that dP/dx is positive for

0 < x < d/ 2 and negative for d/ 2 < x < d. The smallest values of the perimeter occur at the endpoints of the interval 0 x d and therefore do not have to be turning points.
A much better way of tacking the problem is to set x = d cos θ and y = d sin θ and ﬁnd the perimeter as a function of θ. Because θ can take any value (there are no end points such as x = 0 to consider), the largest and smallest perimeters correspond to stationary points. Try it.

Question 10(∗)

Use the ﬁrst four terms of the binomial expansion of (1 − 1/50)1/2 , to derive the approximation

2 ≈ 1.414214.

Calculate similarly an approximation to the cube root of 2 to six decimal places by considering
(1 + N/125)1/3 , where N is a suitable number.

[You need not justify the accuracy of your approximations.]

Although you do not have to justify your approximations, you do need to think carefully about the number of terms required in the expansions. You ﬁrst have to decide how many you will need to obtain the given number of decimal places; then you have to think about whether the next term is likely to aﬀect the value of the last decimal.

It is not at all obvious where the 2 in the ﬁrst part comes from until you write (1 − 1/50)1/2 =

(49/50) .
In the second part, you have to choose N in such a way that 125 + N has something to do with a power of 2.
It is a bit easier to do the binomial expansions by making the denominator a power of 10 (so that the expansion in the ﬁrst part becomes (1 − 2/100)1/2 ).

Solution to question 10
First we expand binomially:
2
1−
100

1
2

1
1
1
2
1

+

2
100
2!
2
2
0.5
0.5
1

− 6 = 0.9899495 .
= 1−
100 104
10
≈ 1+

2

2
100

1
3!

1
2

1
2

3
2

2
100

3

+ ···

It is clear that the next term in the expansion would introduce the seventh and eighth places of decimals, which it seems we do not need. Of course, after further manipulations we might ﬁnd that the above calculation does not supply the 6 decimal places we need, in which case we will work out the next term in the expansion.

1

98 2
7 2
But
=
2 ≈ 9.899495/7 ≈ 1.414214 .
, so
100
10
Second part:
3
1+
125

1
3

1
3
1
1
2
+

3
125
2!
3
3
3
8
64
5 8
= 1−
− 6+
1000 10
3 109
256 1
= 1.007936 +
= 1.007937 .
3 108
≈ 1+

3
125

2

+

1
3!

1
3

2
3

5
3

3
125

3

+ ···

Successive terms in the expansion decrease by a factor of about 1000, so this should give the right number of decimal places.

1

432
832
128 3
3
=
=
so
But
2 ≈ 10.0793/8 ≈ 1.259921 .
125
5
10
Postmortem
This question required a bit of intuition, and some accurate arithmetic. You don’t have to be brilliant at arithmetic to be a good mathematician, but most mathematicians aren’t bad at it. There have in the past been children who were able to perform extroardinary feats of arithmetic. For example,
Zerah Colburn, a 19th century American, toured Europe at the age of 8. He was able to multiply instantly any two four digit numbers given to him by the audiences. George Parker Bidder (the
Calculating Boy) could perform similar feats, though unlike Colburn he became a distinguished mathematician and scientist. One of his brothers knew the bible by heart.
The error in the approximation is the weak point of this question. Although it is clear that the next term in the expansion is too small to aﬀect the accuracy, it is not obvious that the sum of all the next hundred (say) terms of the expansion is negligible (though in fact it is). What is needed is an estimate of the truncation error in the binomial expansion. Such an estimate is not hard to obtain
(ﬁrst year university work) and is typically of the same order of magnitude as the ﬁrst neglected term in the expansion.

Question 11(∗)

How many integers greater than or equal to zero and less than 1000 are not divisible by 2 or 5?
What is the average value of these integers?
How many integers greater than or equal to zero and less than 9261 are not divisible by 3 or 7?
What is the average value of these integers?

There are a number of diﬀerent ways of tackling this problem, but it should be clear that whatever way you choose for the ﬁrst part will also work for the second part (especially when you realise the signiﬁcance of the number 9261). A key idea for the ﬁrst part is to think in terms of blocks of 10 numbers, realising that all blocks of 10 are the same for the purposes of the problem.

Solution to question 11
Only integers ending in 1, 3, 7, or 9 are not divisible by 2 or 5. This is 4/10 of the possible integers, so the total number of such integers is 4/10 of 1,000, i.e. 400.
The integers can be added in pairs: sum = (1 + 999) + (3 + 997) + · · · + (499 + 501) .
There are 200 such pairs, so the sum is 1, 000 × 200 and the average is 500. A simpler argument would be to say that this is obvious by symmetry: there is nothing in the problem that favours an answer greater (or smaller) than 500.
Alternatively, we can ﬁnd the number of integers divisible by both 2 and 5 by adding the number divisible by 2 (i.e. 500) to the number divisible by 5 (i.e. 200), and subtracting the number divisible by both 2 and 5 (i.e. 100) since these have been counted twice. To ﬁnd the sum, we can sum those divisible by 2 (using the formula for a geometric progression), add the sum of those divisible by 5 and subtract the sum of those divisible by 10.
For the second part, either of these methods will work. In the ﬁrst method you consider integers in blocks of 21 (essentially arithmetic to base 21): there are 12 integers in each such block that are not divisible by 3 or 7 (namely 1, 2, 4, 5, 8, 10, 11, 13, 16, 17, 19, 20) so the total number is 9261×12/21 =
5292 . The average is 9261/2 as can be seen using the pairing argument (1 + 9260) + (2 + 9259) + · · · or the symmetry argument.
Postmortem
I was in year 7 at school when I had my ﬁrst encounter with lateral thinking in mathematics. The problem was to work out how many houses a postman delivers to in a street of houses numbered from 1 to 1000, given that he refuses to deliver to houses with the digit 9 in the number. It seemed impossible to do it systematically, until the idea of counting in base 9 occurs; and then it seemed brilliantly simple. I didn’t know I was counting in base 9; in those days, school mathematics was very traditional and base 9 would have been thought very advanced. All that was required was the idea of working out how many numbers can be made out of nine digits (i.e. 93 ) instead of 10; and of course it doesn’t matter which nine.
Did you spot the signiﬁcance of the number 9261? It is 213 , i.e. the number written as 1000 in base
21. Note how carefully the question is written and laid out to suggest a connection between the two paragraphs (and to highlight the diﬀerence); there is not even a part (i) and Part (ii) to break the symmetry. For the examination, the number 1,000,000 was used instead of 1,000 to discourage candidates from spending the ﬁrst hour of the examination writing down the numbers from 1 to
1000.
Having realised the that the ﬁrst part involves (2 × 5)3 and the second part involves (3 × 7)3 you are probably now wondering what the general result is, i.e:
How many integers greater than or equal to zero and less than (pq)3 are not divisible by p or q?
What is the average value of these integers?
I leave this to you, with the hint that you need to think about only pq, rather than (pq)3 , to start with. Question 12(∗)

Sketch the following subsets of the x-y plane:
(i)

|x| + |y|

1;

(ii)

|x − 1| + |y − 1|

1;

(iii)

|x − 1| − |y + 1|

1;

(iv)

|x| |y − 2|

1.

Often with modulus signs, it is easiest to consider the cases separately, so for example is part (i), you would ﬁrst work out the case x > 0 and y > 0, then x > 0 and y < 0, and so on. Here there is a simple geometric understanding of the diﬀerent cases: once you have worked out the ﬁrst case, the other three can be deduced by symmetry.
Another geometric idea should be in your mind when tackling this question, namely the idea of translations in the plane.

Solution to question 12
The way to deal with the modulus signs in this question is to consider ﬁrst the case when the things inside the modulus signs are positive, and then get the full picture by symmetry, shifting the origin as appropriate.
For part (i), consider the the ﬁrst quadrant x
0 and y
0. In this d quadrant, the inequality is x + y
1 . Draw the line x + y = 1 and d then decide which side of the line is described by the inequality. It is d d obviously (since x has to be smaller than something) the region to the d left of the line; or (since y also has to be smaller than something) the region below the line, which is the same region.
Similar arguments could be used in the other quadrants, but it is easier to note that the inequality |x| + |y| 1 unchanged when x is replaced by −x , or y is replaced by −y , so the sketch should have reﬂection symmetry in both axes.

Part (ii) is the same as part (i), except that the origin is translated to
(1, 1).

d d d d d d d d d d d d d d d d d d d d

For part (iii), consider ﬁrst x − y 1, where x > 0 and y > 0 , which give an area that is inﬁnite in extent in the positive y direction. Then reﬂect this in both axes and translate one unit down the y axis and 1 unit along the positive x axis.

d d d

d d d

d d d

d d d

Part (iv) is the rectangular hyperbola xy = 1 reﬂected in both axes and the region required is enclosed by the four hyperbolas. The four hyperbolas are then translated 2 units up the y axis.

Question 13(∗∗)

Show that x2 − y 2 + x + 3y − 2 = (x − y + 2)(x + y − 1) and hence, or otherwise, indicate by means of a sketch the region of the x-y plane for which x2 − y 2 + x + 3y > 2.
Sketch also the region of the x-y plane for which x2 − 4y 2 + 3x − 2y < −2.
Give the coordinates of a point for which both inequalities are satisﬁed or explain why no such point exists. Comments
This question gets a two star diﬃculty rating because inequalities always need to be handled with care. For the very ﬁrst part, you could either factorise the left hand side to obtain the right hand side, or multiply out the right hand side to get the left hand side. Obviously, it is much easier to do the multiplication than the factorisation. But is that ‘cheating’ or taking a short cut that might lose marks? The answer to this is no: it doesn’t matter if you start from the given answer and work backwards — it is still a mathematical proof and any proof will get the marks. (But note that if there is a ‘hence’ in the question, you will lose marks if you do not do it using the thing you have just proved.)
In the second part, you have to do the factorisation yourself, so you should look carefully at where the terms in the (very similar) ﬁrst part come from when you do the multiplication. It will help to spot the similarities bewteen x2 − y 2 + x + 3y − 2 and x2 − 4y 2 + 3x − 2y + 2

Since no indication is given as to what detail should appear on the sketch, you are have to use your judgement: it is clearly important to know where the regions lie relative to the coordinate axes.

Solution to question 13
To do the multiplication, it pays to be systematic and to set out the algebra nicely:
(x − y + 2)(x + y − 1) = x(x + y − 1) − y(x + y − 1) + 2(x + y − 1)
= x2 + xy − x − yx − y 2 + y + 2x + 2y − 2

= x2 + x − y 2 + 3y − 2 as required.

For the ﬁrst inequality, we need x − y + 2 and x + y − 1 both to be either positive or negative. To sort out the inequalities
(or inequations as they are sometimes horribly called), the ﬁrst thing to do is to draw the lines corresponding to the corresponding equalities. These lines divide the plane into four regions and we then have to decide which regions are relevant. The diagonal lines x − y = −2 and x + y = 1 intersect at
1
(− 2 , 3 ) ; this is the important point to mark in on the sketch.
2
The required regions are the left and right quadrants formed by these diagonal lines (since the inequalities mean that the regions are to the right of both lines or to the left of both lines). y = 1−x

d = x+2 y d d d (− 1 , 3 ) d 2 2 d d d d d d d d

For the second part, the ﬁrst thing to do is to factorise x2 −4y 2 +3x−2y +2 . The key similarity with the ﬁrst part is the absence of ‘cross’ terms of the form xy . This allows a diﬀerence-of-two-squares factorisation of the ﬁrst two terms: x2 − 4y 2 = (x − 2y)(x + 2y) . Following the pattern of the ﬁrst part, we can then try a factorisation of the form x2 − 4y 2 + 3x − 2y + 2 = (x − 2y + a)(x + 2y + b) where ab = 2 . Considering the terms linear in x and y gives a + b = 3 and 2a − 2b = −2 which quickly leads to a = 1 and b = 2 . Note that altogether there were three equations for a and b, so we had no right to expect a consistent solution (except for the fact that this is a STEP question for which we had every right to believe that the ﬁrst part would guide us through the second part).
Alternatively, we could have completed the square in x and in y and then used diﬀerence of two squares: x2 − 4y 2 + 3x − 2y + 2 = (x + 3 )2 −
2
= (x +

3 2
2)

9
4

− (2y + 1 )2 +
2

− (2y +

1 2
2)

1
4

= (x +

+2
3
2

− 2y − 1 )(x +
2

3
2

+ 2y + 1 ) .
2

2y = x + 1

¨
¨
¨¨
¨¨
¨¨
¨¨
rr ¨¨ r ¨¨r
¨
rr
3
(− 2 , 1 ) rr
4
rr rr rr r 2y = −2 − x

As in the previous case, the required area is formed by two
1
intersecting lines; this time, they intersect at (− 3 , − 4 ) and
2
the upper and lower regions are required, since the inequality is the other way round.
It is easy to see from the sketches that there are points that satisfy both inequalities: for example (1, 2) .

Question 14(∗∗)

Solve the inequalities
(i)

1 + 2x − x2 > 2/x

(ii)

(3x + 10) > 2 +

(x = 0) ,

(x + 4)

(x

−10/3) .

The two parts are unrelated (unusually for STEP questions), except that they deal with inequalities.
Both parts need care, having traps for the unwary. In the ﬁrst part you have to watch out when you multiply an inequality: if the thing you muliply is negative then the inequality reverses. In the second part you have to consider the possibility that your algebraic manipulations have created extra spurious solutions. In the ﬁrst part, you might ﬁnd sketching a graph helpful.
For the second part, it is worth (after you have ﬁnished the question) sketching the graphs of

(3x + 10) , 2 + (x + 4) and 2 − (x + 4) . You can get them by translations of x (which you can get by reﬂection y = x2 in the line y = x . I would have drawn them for you in the postmortem comment overleaf, but there wasn’t room on the page.

Solution to question 14
(i) We would like to multiply both sides of the inequality by x in order to obtain a nice cubic expression. However, we have to allow for the possibility that x is negative (which would reverse the inequality). One way is to consider the cases x > 0 and x < 0 separately.
For x > 0, we multiply the whole equation by x without changing the inequality:
1 + 2x − x2 > 2/x ⇒ x + 2x2 − x3 > 2

⇒ x3 − 2x2 − x + 2 < 0

⇒ (x − 1)(x + 1)(x − 2) < 0
⇒ 1 < x < 2,

(∗)

discarding the possibility x < −1, since we have assumed that x > 0. The easiest way of obtaining the result (∗) from the previous line is to sketch the graph of (x − 1)(x + 1)(x − 2) < 0 .
For x < 0, we must reverse the inequality when we multiply by x so in this case, (x − 1)(x + 1)(x − 2) > 0, which gives −1 < x < 0.

The smart way to do x > 0 and x < 0 in one step is to multiply by x2 (which is never negative and hence never changes the direction of the inequality) and analyse x(x−1)(x+1)(x−2) < 0. Again, a sketch is useful.

(ii) First square both sides the inequality

(3x + 10) > 2 +

1

3x + 10 > 4 + 4(x + 4) 2 + (x + 4)

i.e.

(x + 4) :
1

x + 1 > 2(x + 4) 2 .

Note that the both sides of the original inequality are positive or zero (i.e. non-negative), so the direction of the inequality is not changed by squaring. Now consider the new inequality x + 1 >
1
2(x + 4) 2 . If both sides are non-negative, that is if x > −1 , we can square both sides again without changing the direction of the inequality. But, if x < −1 , the inequality cannot be satisﬁed since the right hand side is always non-negative. Squaring gives x2 + 2x + 1 > 4(x + 4)

i.e.

(x − 5)(x + 3) > 0 .

Thus, x > 5 or x < −3. However, we must reject x = −3 because of the condition x > −1 .
Therefore the inequality holds for x > 5 .
Postmortem
The spurious result x < −3 at the end of the second part arises through loss of information in the process of squaring. (If you square an expression, you lose its sign.) After squaring twice, the

√ resulting inequality is the same as would have resulted from (3x + 10) > 2 − (x + 4) and it is to this inequality that x < −3 is the solution.

Question 15(∗∗)

Sketch, without calculating the stationary points, the graph of the function f (x) given by f (x) = (x − p)(x − q)(x − r) , where p < q < r. By considering the quadratic equation f ′ (x) = 0, or otherwise, show that
(p + q + r)2 > 3(qr + rp + pq) .
By considering (x2 + gx + h)(x − k), or otherwise, show that g2 > 4h is a suﬃcient condition but not a necessary condition for the inequality
(g − k)2 > 3(h − gk) to hold.

For the sketch it is not necessary to do more than think about the behaviour for x large and positive, and for x large and negative, and the points at which the graph crosses the x-axis.
The argument required for the second part is very similar to that of the ﬁrst part, except that it also tests understanding of the meaning of the terms necessary and suﬃcient.

Solution to question 15

From the sketch, we see that f (x) has two turning points, so the equation f ′ (x) = 0 has two real roots.

p

q

r

Now f (x) = (x − p)(x − q)(x − r) = x3 − (p + q + r)x2 + (qr + rp + pq)x − pqr so f ′ (x) = 3x2 − 2(p + q + r)x + (qr + rp + pq) = 0 .
Using the condition ‘b2 > 4ac’ for the quadratic to have two real root gives
4(p + q + r)2 > 12(qr + rp + pq) , which is the required result.
For the second part, we can use a similar argument.
In this case, the cubic expression is factorised into a linear factor x − k and a quadratic factor x2 + gx + h . If the quadratic equation x2 + gx + h = 0 has two distinct real roots (i.e. if g2 > 4h), then the function (x2 + gx + h)(x − k) has three zeros and therefore two turning points. Thus if g2 > 4h the equation f ′ (x) = 0 has two real roots.
Expanding f (x) gives f (x) = (x2 + gx + h)(x − k) = x3 + (g − k)x2 + (h − gk) and diﬀerentiating gives f ′ (x) = 3x2 + 2(g − k)x + (h − gk) = 0 .
The condition for two real roots is 4(g − k)2 > 12(h − gk). Thus g2 > 4h is a suﬃcient condition for this inequality to hold.

Why is it not necessary? That is to say, why is it too strong a condition? How could the condition for two turning points hold without f (x) having three zeros. The answer is that both turning points could be above the x–axis or both could be below. Saying this would get all the marks. Or you could give a counterexample: the easy counterexample is g2 = 4h ; if this holds, there are still two turning points so (g − k)2 > 3(h − gk) . (Careful of this counterexample: it will not work if the double root of the quadratic (at x = −g/2 coincides with the root of the linear factor (at x = k), so we need the extra condition that g/2 + k = 0 .)
Postmortem
I got considerable satisfaction from the way that non-obvious inequalities are derived from understanding simple graphs and the quadratic formula when I set this question. But are the inequalities really obscure? The answer is unfortunately no.
The inequality of the ﬁrst part is equivalent to (q − r)2 + (r − p)2 + (p − q)2 > 0 (the inequality coming from the fact the squares of real numbers are non-negative and in this case the given condition p < q < r means that the expression is strictly positive (not equal to zero).
If we rewrite the second inequality as (k + g/2)2 + 3(g2 /4 − h) > 0 we see that it is certainly holds if (g2 /4 − h) > 0 . Clearly, this is not a necessary condition: there is the obvious counterexample
(g2 /4 − h) = 0 and (k + g/2) = 0 . But there are lots of other easy counterexamples — for example, g = 0 and k2 > 3h > 0 .

Question 16(∗∗)

In a cosmological model, the radius R of the universe is a function of the age t of the universe. The function R satisﬁes the three conditions:
R(0) = 0,

R′ (t) > 0 for t > 0,

R′′ (t) < 0 for t > 0,

(∗)

where R′′ denotes the second derivative of R. The function H is deﬁned by
H(t) =

R′ (t)
.
R(t)

(i)

Sketch a graph of R(t). By considering a tangent to the graph, show that t < 1/H(t).

(ii)

Observations reveal that H(t) = a/t, where a is constant. Derive an expression for R(t).
What range of values of a is consistent with the three conditions (∗)?

(iii)

Suppose, instead, that observations reveal that H(t) = bt−2 , where b is constant. Show that this is not consistent with conditions (∗) for any value of b.

The sketch just means any graph starting at the origin and increasing, but with decreasing gradient.
The second part of (i) needs a bit of thought (where does the tangent intersect the x-axis?) so don’t despair if you don’t see it immediately. Parts (ii) and (iii) are perhaps easier than (i).

Solution to question 16
(i) The height of the right-angled triangle in the ﬁgure is R(t) and the slope of the hypotenuse is R′ (t). The length of the base is therefore R(t)/R′ (t), i.e.1/H(t).
The ﬁgure shows that the intersection of the tangent to the graph for t > 0 intercepts the negative t–axis, so 1/H(t) > t.

R(t)

1/H(t)

t

(ii) One way to proceed is to integrate the diﬀerential equation:
H(t) =

a
R′ (t) a ⇒
= ⇒ t R(t) t R′ (t) dt =
R(t)

a dt ⇒ ln R(t) = a ln t + constant ⇒ R(t) = Ata . t The ﬁrst two conditions (∗) are satisﬁed if a > 0 and A > 0. For the third condition, we have
R′′ (t) = a(a − 1)Ata−1 which is negative provided a < 1. The range of a is therefore 0 < a < 1,

(iii) The obvious way to do this just follows part (ii). This time, we have
H(t) =

b
R′ (t) b = 2 ⇒

2 t R(t) t R′ (t) dt =
R(t)

b b dt ⇒ ln R(t) = − + constant ⇒ R(t) = Ae−b/t .
2
t t Thus R′ (t) = H(t)R(t) = Abt−2 e−b/t . Clearly A > 0 since R(t) > 0 for t > 0 so R′ (t) > 0 .
Furthermore R(0) = 0 (think about this!), so only the condition R′′ (t) < 0 remains to be checked.
Diﬀerentiating R′ (t) gives
R′′ (t) = Ab(−2t−3 )e−b/t + Abt−2 e−b/t (bt−2 ) = Abt−4 e−b/t (−2t + b)
This is positive when t < b/2, which contradicts the condition R′′ < 0 .
Instead of solving the diﬀerential equation, we could proceed as follows. We have
R′
= bt−2 ⇒ R′ = bRt−2 ⇒ R′′ = bR′ t−2 − 2bRt−3 = b2 t−4 R − 2bRt−3 = b(b − 2t)t−4 R .
R
This is positive when t < b/2, which contradicts the condition R′′ < 0.
Postmortem
The interpretation of the conditions is as follows. The ﬁrst condition R(0) = 0 says that the universe started from zero radius — in fact, from the ‘big bang’. The second condition says that the universe is expanding. This was one of the key discoveries in cosmology in the last century. The quantity called here H(t) is called Hubble’s constant (though it varies with time). It measures the rate of expansion of the universe. Its present day value is a matter of great debate (it is of the order of
75 megaparsecs per century). The last condition says that the expansion of the universe is slowing down. This is expected on physical grounds because of the gravitational attraction of the galaxies on each other.
Another observational uncertainly is the value of the deceleration. It is not clear yet whether the universe will expand for ever or recollapse.

Question 17(∗)

Show that the equation of any circle passing through the points of intersection of the ellipse
(x + 2)2 + 2y 2 = 18 and the ellipse
9(x − 1)2 + 16y 2 = 25 can be written in the form x2 − 2ax + y 2 = 5 − 4a .

When this question was set (as question 1 on STEP I) it seemed too easy. But quite a high proportion of the candidates made no real progress. Of course, is is not easy to keep a cool head under examination conditions. But surely it is obvious that either x or y has to be eliminated from the two equations for the ellipses; and having decided that, it is obvious which one to eliminate.
It is worth thinking about how many points of intersection of the ellipses we are expecting: a sketch might help.

Solution to question 17
First we have to ﬁnd the intersections of the two ellipses by solving the simultaneous equations
(x + 2)2 + 2y 2 = 18
9(x − 1)2 + 16y 2 = 25 .
We can eliminate y by multiplying the ﬁrst equation by 8 and subtracting the two equations:
8(x + 2)2 − 9(x − 1)2 = 144 − 25

i.e.

x2 − 50x + 96 = 0

i.e.

(x − 2)(x − 48) = 0 .

The two possible values for x at the intersections are therefore 2 and 48.
Next we ﬁnd the values of y at the intersection. Taking x = 2 and substituting into the ﬁrst simultaneous equation, we have 16 + 2y 2 = 18, so y = ±1. Taking x = 48 gives 502 + 2y 2 = 18 which has no (real) roots. Thus there are two points of intersection, at (2, 1) and (2, −1).
Suppose a circle through these points has centre (p, q) and radius R. Then the equation of the circle is (x − p)2 + (y − q)2 = R2 . Setting (x, y) = (2, 1) and (x, y) = (2, −1) gives two equations:.
(2 − p)2 + (1 − q)2 = R2 ,

(2 − p)2 + (−1 − q)2 = R2 .

Subtracting gives q = 0 (it is obvious anyway that the centre of the circle must lie on the y axis).
Thus the equation of any circle passing through the intersections is
(x − p)2 + y 2 = (2 − p)2 + 1 , which simpliﬁes to the given result with p = a .

Question 18(∗ ∗ ∗)
Let f (x) = xm (x − 1)n , where m and n are both integers greater than 1. Show that f ′ (x) =

m n − x 1−x

f (x) .

Show that the curve y = f (x) has a stationary point with 0 < x < 1. By considering f ′′ (x), show that this stationary point is a maximum if n is even and a minimum if n is odd.
Sketch the graphs of f (x) in the four cases that arise according to the values of m and n.

There is quite a lot in this question, but it is one of the best STEP questions on basic material that
I came across (I looked through all 15 years of STEP’s history).
First, you have to ﬁnd f ′ (x) . The very ﬁrst part (giving a form of f ′ (x)) did not occur in the original
STEP question. This particular expression is extremely helpful when if comes to ﬁnding the value of f ′′ (x) at the stationary point. In the actual exam, a handful of candidates successfully found a general expression for f ′′ (x) in terms of x then evaluated it at the stationary point: ﬁrst class work.
You should ﬁnd that the value of f ′′ (x) at the stationary point can be expressed in the form
(· · · )f (x) , where the factor in the brackets is quite simple and always negative, so that the nature of the stationary point depends only on the sign of f (x) . (Actually, this is intuitively obvious: f (0) = f (1) = 0 so the one stationary point must be, for example, a maximum if f (x) > 0 for
0 < x < 1 .)
For the sketches, you really have only to think about the behaviour of f (x) when |x| is large and the sign of f (x) between x = 0 and x = 1. That allows you to piece together the graph, knowing that there is only one stationary point between x = 0 and x = 1 .
Note that, very close to x = 0 , f (x) ≈ xm (−1)n so it is easy to see how the graph there depends on n and m ; you can use this to check that you have got the graphs right.
Postmortem
[This didn’t ﬁt on the next page: read it after you have done the question.]
You may have noticed a little carelessness at the top of the next page: what happens in the logarithmic diﬀerentiation if any of f (x) or x or x − 1 are negative? The answer is that it doesn’t matter. One way to deal with the problem of logs with negative arguments is to put modulus signs everywhere using the correct result f ′ (x) d ln |f (x)| =
.
dx f (x)
If you are not sure of this, try it on f (x) = x taking the two cases x > 0 and x < 0 separately.
A more sophisticated way of dealing with logs with negative arguments is to note that ln(−f (x)) = ln f (x) + ln(−1). We don’t have to worry about ln(−1) because it is a constant (of some sort) and so won’t aﬀect the diﬀerentiation. Actually, a value of ln(−1) can be obtained by taking logs of
Euler’s famous formula eiπ = −1 giving ln(−1) = iπ .

Solution to question 18
The ﬁrst result can be established by diﬀerentiating f (x) directly, but the neat way to do it is to start with ln f (x) : ln f (x) = m ln x + n ln(x − 1) =⇒

f ′ (x) m n
=
+
,
f (x) x x−1

as required.
Thus f (x) has stationary points at x = 0 and x = 1 (since m − 1 > 0 and n − 1 > 0) and when n m

=0, x 1−x
i.e. when m(1 − x) − nx = 0 . Solving this last equation for x gives x = m/(m + n) , which lies between 0 and 1 since m and n are positive.
Next we calculate f ′′ (x) . Starting with n m

x
1−x

f ′ (x) = we obtain f ′′ (x) =

n m −
2
x
(1 − x)2

f (x) ,

f (x) +

m n − x 1−x

f ′ (x) .

At the stationary point, the second of these two terms is zero because f ′ (x) = 0 . Thus f ′′ (x) =

m n − x2 (x − 1)2

f (x) .

The bracketed expression is negative so at the stationary point f ′′ (x) < 0 if f (x) > 0 and f ′′ (x) > 0 if f (x) < 0.
The sign of f (x) for 0 < x < 1 is the same as the sign of (x − 1)n , since xm > 0 when x > 0. The stationary point is between 0 and 1, so x − 1 < 0 and (x − 1)n > 0 if n is even and (x − 1)n < 0 if n is odd. Thus f ′′ (x) < 0 if n is even and f ′′ (x) > 0 if n is odd, which is the required result.
The four cases to sketch are determined by whether m and n are even or odd. The easiest way to understand what is going on is to consider the graphs of xm and (x − 1)n separately, then try to join them up at the stationary point between 0 and 1. If m is odd, then x = 0 is a point of inﬂection, but if m is even it is a maximum or minimum according to the sign of (x − 1)n . You should also think about the behaviour for large |x|. All the various bits of information (including the nature of the turning point investigated above) should all piece neatly together.

m even, n even

m odd, n even

m even, n odd

m odd, n odd

Question 19(∗ ∗ ∗)
1
, for x 0.
1 + x2
Find the equation of the line that intersects the curve at x = 0 and is tangent to the curve at some point with x > 0 . Prove that there are no further intersections between the line and the curve.
Draw the line on your sketch.
Give a sketch of the curve y =

By considering the area under the curve for 0

x

1, show that π > 3 .

Show also, by considering the volume formed by rotating the curve about the y axis, that ln 2 > 2/3 .
1

[ Note:
0

π
1
dx = . ]
2
1+x
4

There is quite a lot to this question, so three stars even though none of it is particularly diﬃcult.
The most common mistake in ﬁnding the equation of the tangent is to muddle the y and x that occur in the equation of the line (y = mx+ c) with the coordinates of the point at which the tangent meets the curve, getting ‘constants’ m and c that depend on x . I’m sure you wouldn’t make this rather elementary mistake normally, but it is surprising what people do under examination conditions.
The reason for sketching the curve lies in the last parts: the shape of the curve relative to the straight line provides the inequality.
The note at the end of the question had to be given because integrals of that form (giving inverse trigonometric functions) are not in the core A-level syllabus.

Solution to question 19
The sketch should show a graph which has gradient 0 at (0, 1) and which asymptotes to the x-axis for large x. The line is the tangent to the graph from the point (0, 1) which is required later. y 1







 x 

1





First we ﬁnd the equation of the tangent to the curve y = (1 + x2 )−1 at the point (p, q) . The gradient of the curve at the point (p, q) is −2p(1 + p2 )−2 = −2pq 2 , so the equation of the tangent is y = −2pq 2 x + c where c is given by q = −2pq 2 p + c . This line is supposed to pass through the point (0, 1) , so 1 = c .
Thus
2p2
1
+
1=
1 + p2 (1 + p2 )2 which simpliﬁes to
(1 + p2 )2 = 1 + 3p2

p4 = p2 .

i.e.

The only positive solution is p = 1, so the equation of the line is y = −x/2 + 1 .

We can check that this line does not meet the curve again by solving the equation

1 x +1=
.
2
1 + x2

Multiplying by (1 + x)2 gives
(1 − x/2)(1 + x2 ) = 1

i.e.

x3 − 2x2 + x = 0 .

Factorising shows that x = 1 and x = 0 satisfy this equation (these are the known roots, x = 1 being a double root corresponding to the tangency) and that there are no more roots.
The area under the curve for 0 x 1 is π/4 as given. The sketch shows that this area is greater
1
1 than the area under the tangent line for 0 x 1, which is 2 + 4 (the area of the rectangle plus the area of the triangle above it). Comparing with with π/4 gives the required result.
The volume formed by rotating the curve about the y axis is
1

1

2πyxdx = π
0

0

2x dx = π ln 2 .
1 + x2

This is greater than the volume formed by rotating the line about the y axis, which is
1

1

2πyxdx = 2π
0

Comparison gives the required result.

0

(1 − x/2)xdx = 2π/3 .

Question 20(∗∗)

Let

a

I=
0

where 0
Find also:

cos x dx sin x + cos x

a

and

J=
0

sin x dx , sin x + cos x

a < 3π/4 . By considering I + J and I − J, show that 2I = a + ln(sin a + cos a) . π/2 cos x dx , where p and q are positive numbers; p sin x + q cos x

π/2

cos x + 4 dx .
3 sin x + 4 cos x + 25

(i)
0

(ii)
0

I love this question. When I ﬁrst saw it (one of my colleagues proposed the ﬁrst part for STEP
Paper I a few years ago) I was completely taken by surprise. You might like to think how you would evaluate these integrals without using this method.
In order to use it again for STEP in 2002, I added parts (i) and (ii). It took me some time to think of a suitable extension. I was disappointed to ﬁnd that the basic idea is more or less a one-oﬀ: there are very few denominators, besides the ones given, that lead to integrands amenable to this trick. But I was pleased with what I came up with. The question leads you through the opening paragraph and the extra parts depend very much on your having understood why the opening paragraph works.
In the examination, candidates who were successful in parts (i) and (ii) nearly always started oﬀ with the statement ‘Now let I = · · · and J = · · · .’ The choice of signiﬁcant numbers in part (ii) (3,
4 and 25) should be a clue.
Finally, you will have noticed that 0 a < 3π/4 was given in the question. Understanding the reason for this restriction does not help to do the question, but you should try to work out its purpose. Solution to question 20
We have a I +J

=
0

I −J

a

=
0

a cos x + sin x dx = dx = a sin x + cos x
0
cos x − sin x dx = ln(cos a + sin a) sin x + cos x

(in the second integral, the numerator is the derivative of the denominator). Adding these equations gives the required expression for 2I . π/2 (i) Similarly, let I =
0

cos x dx and J = p sin x + q cos x

π/2
0

sin x dx . Then p sin x + q cos x

π/2

π q cos x + p sin x dx = p sin x + q cos x
2

π/2

qI + pJ

p cos x − q sin x dx = ln(p sin π/2 + q cos π/2) − ln(p sin 0 + q cos 0) = ln(p/q) . p sin x + q cos x

=
0

pI − qJ

=
0

Now we multiply the ﬁrst of these equations by q and the second by p and add the resulting equations:
(p2 + q 2 )I = π/2 (ii) This time, let I =
0

+ p ln(p/q) .
2

cos x + 4 dx and J =
3 sin x + 4 cos x + 25

π/2
0

sin x + 3 dx . Then
3 sin x + 4 cos x + 25

π/2

π
4 cos x + 3 sin x + 25 dx =
3 sin x + 4 cos x + 25
2

π/2

4I + 3J

3 cos x − 4 sin x dx = ln(28/29) .
3 sin x + 4 cos x + 25

=
0

3I − 4J

=
0

Now we multiply the ﬁrst of these equations by 4 and the second by 3 and add the resulting equations:
25I = 2π + 3 ln(28/29) .
Postmortem
The reason for restriction 0 a < 3π/4 is that the denominator should not be 0 for any value of x in

the range of integration — otherwise, the integral is undeﬁned. Writing sin x+cos x = 2 sin(x+π/4) shows that the denominator is ﬁrst zero when x = 3π/4 .
Perhaps the easiest way of doing this sort of integral is to use the substitution t = tan(x/2) which converts the denominator to a quadratic in x .

Question 21(∗∗)

In this question, you may assume that if k1 , . . . , kn are distinct positive real numbers, then
1
n

n

kr > r=1 1 n n

kr

,

r=1

i.e. their arithmetic mean is greater than their geometric mean.
Suppose that a, b, c and d are positive real numbers such that the polynomial f(x) = x4 − 4ax3 + 6b2 x2 − 4c3 x + d4 has four distinct positive roots.
(i) By considering the relationship between the coeﬃcients of f and its roots, show that c > d.
(ii) By diﬀerentiating f, show that b > c.
(iii) Show that a > b.

This result is both surprising and pleasing. I’m sorry to say that I don’t understand it. I’ve looked at it from various points of view, and asked a few selected colleagues who are usually good at this sort of thing, but so far no one has come up with a good explanation. Maybe the penny will drop in time for the next edition. (I haven’t even worked out if the converse of this result holds, namely that the above inequalities are suﬃcient for the roots to lie in the given interval; I don’t think this is hard, but I thought I would leave it until I undersood it better.)
There are general theorems on the number of roots in any given interval. This problem was studied by Descartes (1596–1650), who came up with his rule of signs. This relates the number of roots to the signs of terms in the polynomial, but it only gives an upper bound for the number of roots. The problem was ﬁnally cracked by Sturm in about 1835, but his solution is quite complicated, and does not seem to throw any light on the special case we are looking at here (where the interval is (0, ∞)).
It will be clear from the proof that it can be generalised to any equation of the form
N

x +

N −1 k=0 N
(−ak )N −k xk = 0 , k where the numbers ak are distinct and positive, which has positive distinct roots.
The question looks diﬃcult, but you don’t have to go very far before you come across something to substitute into the arithmetic/geometric inequality. Watch out for the condition in the arithmetic/geometric inequality that the numbers are distinct: you will have to show that any numbers
(or algebraic expressions) you use for the inequality are distinct.
To obtain the relationship between the coeﬃcients and the roots, you need to write the quartic equation in the form (x − p)(x − q)(x − r)(x − s) = 0 .

Solution to question 21
(i) First write f (x) ≡ (x − p)(x − q)(x − r)(x − s) where p , q , r and s are the four roots of the equation (known to be real, positive and distinct).
Multiplying out the brackets and comparing with x4 − 4ax3 + 6b2 x2 − 4c3 x + d4 shows that pqrs = d4 and pqr + qrs + rsp + spq = 4c3 .
The required result, c > d , follows immediately by applying the arithmetic/geometric mean inequality to the positive real numbers pqr, qrs, rsp and spq : c3 =

pqr + qrs + rsp + spq
> [(pqr)(qrs)(rsp)(spq)]1/4 = [p3 q 3 r 3 s3 ]1/4 = [d12 ]1/4 .
4

Taking the cube root (c and d are real) preserves the inequality.
The arithmetic/geometric mean inequality at the beginning of the question is stated only for the case when the numbers are distinct (though in fact it holds provided at least two of the numbers are distinct). To use the inequality, we have therefore to show that no two of pqr , qrs , rsp and spq are equal. This follows immediately from the fact that the roots are distinct and non-zero. For if, for example, pqr = qrs then qr(p − s) = 0 which means that q = 0 , r = 0 or p = s any of which is a contradiction.
(ii) The polynomial f ′ (x) is cubic so it has three zeros (roots). These are at the turning points of f (x) (which lie between the zeros of f (x)) and are therefore distinct and positive.
Now
f ′ (x) = 4x3 − 12ax2 + 12b2 x − 4c3
At the turning points, x3 − 3ax2 + 3b2 x − c3 = 0 .
Suppose that the roots of this cubic equation are u , v and w all real, distinct and positive. Then comparing x3 −3ax2 +3b2 x−c3 = 0 and (x−u)(x−v)(x−w) shows that c3 = uvw and vw+wu+uv =
3b2 .
The required result, b > c , follows immediately by applying the arithmetic/geometric mean inequality to the positive real numbers vw , wu and uv .
(iii) Apply similar arguments to f ′′ (x)/12.

Question 22(∗∗)

The nth Fermat number, Fn , is deﬁned by n Fn = 22 + 1 ,

n = 0, 1, 2, . . . ,

n

where 22 means 2 raised to the power 2n . Calculate F0 , F1 , F2 and F3 . Show that, for k = 1, k = 2 and k = 3 ,
F0 F1 . . . Fk−1 = Fk − 2 .
(∗)
Prove, by induction, or otherwise, that (∗) holds for all k have a common factor greater than 1.

1. Deduce that no two Fermat numbers

Hence show that there are inﬁnitely many prime numbers.

Comments n Fermat (1601—1665) conjectured that every number of the form 22 + 1 is a prime number. F0 to
F4 are indeed prime, but Euler showed in 1732 that F5 (4294967297) is divisible by 641. As can be seen, Fermat numbers get very big and not many more have been investigated; but those that have been investigated have been found not to be prime numbers. It is now conjectured that in fact only a ﬁnite number of Fermat numbers are prime numbers.
The Fermat numbers have a geometrical signiﬁcance as well: Gauss proved that a regular polygon of n sides can be inscribed in a circle using a Euclidean construction (i.e. only a straight edge and a compass) if and only if n is a power of 2 times a product of distinct Fermat primes.
This rather nice proof that there are inﬁnitely many prime numbers comes from Proofs from the
Book11 , a set of proofs thought by Paul Erd¨s to be heaven-sent. (Most of them are much less o elementary this one.) Erd¨s was an extraordinarily proliﬁc mathematician. He had almost no o personal belongings and no home12 . He spent his life visiting other mathematicians and proving theorems with them. He collaborated with so many people that every mathematician is (jokingly) assigned an Erd¨s number; for example, if you wrote a paper with someone who wrote a paper with o Erd¨s, you are given the Erd¨s number 2. It is conjectured that no mathematician has an Erd¨s o o o number greater than 7. Mine is 8, but that is because I am more a theoretical physicist than a mathematician.13 11

See The man who loved only numbers by Paul Hoﬀman published by Little Brown and Company in 1999 – a wonderfully readable and interesting book.
13
Since I wrote this, I discovered ‘collaborator distance’ on http://www.ams.org/mathscinet, which gives the distance between any two researchers. I found that I am not after all counterexample to the conjecture: my Erd¨s is 5 (via o Einstein). Sorry to bore you with all this personal stuﬀ.
12

Solution to question 22
To begin with, we have by direct calculation that F0 = 3 , F1 = 5 , F2 = 17 and F3 = 257, so it is easy to verify the displayed result for k = 1, 2, 3.
For the induction, we start by assuming that the result holds for k = m so that
F0 F1 . . . Fm−1 = Fm − 2 .
We need to show that this implies that that the result holds for k = m + 1, i.e. that
F0 F1 . . . Fm = Fm+1 − 2 .
Starting with the left hand side of this equation, we have m m

m

F0 F1 . . . Fm−1 Fm = (Fm − 2)Fm = (22 − 1)(22 + 1) = 22

2

m+1

− 1 = 22

− 1 = Fm+1 − 2,

as required. Since we know that the result holds for k = 1, the induction is complete.
For the rest of the question: if p divides Fl and Fm , where l < m, then p divides F0 F1 . . . Fl . . . Fm−1 and therefore p divides Fm − 2, by (∗). But all Fermat numbers are odd so no number other than 1 divides both Fm − 2 and Fm , which gives the contradiction. Hence no two Fermat numbers have a common factor greater than 1.
For the last part, note that every number can be written uniquely as a product of prime factors and that because the Fermat numbers are co-prime, each prime can appear in at most one Fermat number. Thus, since there are inﬁnitely many Fermat numbers, there must be inﬁnitely many primes. Postmortem
This question is largely about proof. The above solution has a proof by induction and a proof by contradiction. The last part could have been written as a proof by contradiction too.
The diﬃculty is in presenting the proof. It is not enough to understand your own proof: you have to be able to set it out so that it is clear to the reader. This is a vital part of mathematics and its most useful transferable skill.

Question 23(∗∗)

Show that the sum SN of the ﬁrst N terms of the series
1
3
5
2n − 1
+
+
+ ··· +
+ ···
1.2.3 2.3.4 3.4.5 n(n + 1)(n + 2) is 1
2

3
1
5
+

2 N +1 N +2

.

What is the limit of SN as N → ∞?
The numbers an are such that

Find an expression for

an
(n − 1)(2n − 1)
=
. an−1 (n + 2)(2n − 3)

an and hence, or otherwise, evaluate a1 ∞

n=1

an when a1 =

2
.
9

If you haven’t the faintest idea how to do the sum, then look at the ﬁrst line of the solution; but don’t do this without ﬁrst having a long think about it, and a long look at the form of the answer given in the question.
Limits form an important part of ﬁrst year university mathematics. The deﬁnition of a limit is one of the basic ideas in analysis which is the rigorous study of calculus. Here no deﬁnition is needed: you just see what happens when N gets very large (some terms get very small and eventually go away). The second part of the question looks as if it might be some new idea. Since this is STEP, you will probably realise that the series in the second part must be closely related to the series in the ﬁrst part. The peculiar choice for a1 (= 2/9) should make you suspect that the sum will come out to some nice round number (not in fact round in this case, but straight and thin).
Having decided how to do the ﬁrst part, please don’t use the ‘cover up’ rule unless you understand why it works: mathematics at this level is not a matter of applying learned recipes.

Solution to question 23
The ﬁrst thing to do is to put the general term of the series into partial fractions in the hope that something good might then happen. (The form of the answer suggests partial fractions have been used; and it is diﬃcult to think of anything else to do.)
The general term of the series is
2n − 1
A
B
C
≡ +
+
. n(n + 1)(n + 2) n n+1 n+2
(The ‘equivalent’ sign, ≡ , indicates an identity (something that holds for all values of n) rather than an equation to solve for n . The easiest way to deal with this is to multiply both sides of the identity by one of the terms in the denominator and then set this term equal to zero (since the identity must hold for all values of n). For example, multiplying by n gives
2n − 1
Bn
Cn
≡A+
+
.
(n + 1)(n + 2) n+1 n+2
1
and setting n = 0 shows immediately that A = − 2 . Similarly, multiplying by n + 1 and n + 2 respectively and setting n = −1 and n = −2 respectively gives B = 3 and C = − 5 . Thus
2

2n − 1
1/2
3
5/2
=−
+
− n(n + 1)(n + 2) n n+1 n+2
The series can now be written
1/2 3 5/2
1/2 3 5/2
1/2 3 5/2
+ −
+ −
+ −
+ −
+ −
+

1
2
3
2
3
4
3
4
5
1/2
3
5/2
3
5/2
3
5/2
1/2
1/2
+ ··· + −
+

+

+

+ −
+ −
N −2 N −1
N
N −1 N
N +1
N
N +1 N +2

.

The ﬁrst term has 3 as the last number in the denominator so the N th term has N + 2 . All the terms in the series cancel, except those with denominators 1, 2, N + 1 and N + 2. The sum of these exceptions is the sum of the series.
The limit as N → ∞ is

3
4

since the other two terms obviously tend to zero.

For the second paragraph, we have ﬁrst part of the question. Thus bn an
=
a1 b1 bn an =
, where bn is the general term of the series in the an−1 bn−1

and

1

a1 an = b1 ∞
1

bn =

2/9 3
× = 1.
1/6 4

Alternatively, an =
=
=

(n − 1)(2n − 1)
(n − 1)(2n − 1) (n − 2)(2n − 3) an−1 = an−2 (n + 2)(2n − 3)
(n + 2)(2n − 3) (n + 1)(2n − 5)
5 × 11 4 × 9 3 × 7 2 × 5 1 × 3
(n − 1)(2n − 1) (n − 2)(2n − 3) (n − 3)(2n − 5)
···
a1
(n + 2)(2n − 3) (n + 1)(2n − 5) (n)(2n − 7)
8×9 7×7 6×5 5×3 4×1
2n − 1
3×2×1
2n − 1
12
a1 =
,
n(n + 1)(n + 2)
1
9 n(n + 1)(n + 2)

all other terms cancelling.
Now using the result of the ﬁrst part gives

1

an = 1.

Question 24(∗ ∗ ∗)
(i) Show that, if m is an integer such that
(m − 3)3 + m3 = (m + 3)3 ,

(∗)

then m2 is a factor of 54. Deduce that there is no integer m which satisﬁes the equation (∗).
(ii) Show that, if n is an integer such that
(n − 6)3 + n3 = (n + 6)3 ,

(∗∗)

then n is even. By writing n = 2m deduce that there is no integer n which satisﬁes the equation
(∗∗).
(iii) Show that, if n is an integer such that
(n − a)3 + n3 = (n + a)3 ,

(∗ ∗ ∗)

where a is a non-zero integer, then n is even and a is even. Deduce that there is no integer n which satisﬁes the equation (∗ ∗ ∗).

I slightly simpliﬁed the ﬁrst two parts of the question, which comprised the whole of the original
STEP question, and added the third part. This last part is conceptually tricky and very interesting: hence the three star rating.
It is an example of a method of proof by contradiction, called the method of inﬁnite descent. You will understand why it is called inﬁnite descent when you have ﬁnished the question (in fact, the name may serve as a hint). It was used by Fermat (1601 – 1665) to prove special cases of his
Last Theorem14 , which of course is exactly what you are doing in this question. The method was probably invented by him and his faith in it sometimes led him astray. It is even possible that he thought he could use it to prove his Last Theorem in full. In fact, the proof of this theorem was only given in 1994 by Andrew Wiles; and it is 150 pages long.

14
The theorem says that the equation xn + y n = z n , where x , y , z and n are positive integers, can only hold if n = 1 or 2

Solution to question 24
(i) Simplifying gives m2 (m − 18) = 54.

Both m2 and (m − 18) must divide 54, which is impossible since the only squares that divide 54 are
1 and 9, and neither m = 1 nor m = 3 satisﬁes m2 (m − 18) = 54.
You could also argue that m − 18 must be positive so m

19 and m2

(ii) Suppose n is odd. The two terms on the left hand side are both odd, which means that that the left hand side is even. But the right hand side is odd so the equation cannot balance.
Setting n = 2m gives
(2m − 6)3 + (2m)3 = (2m + 6)3 .

Taking a factor of 23 out of each term leaves (m − 3)3 + m3 = (m + 3)3 which is the same as the equation that was shown in part (i) to have no solutions.
(iii) First we show that n is even, dealing with the cases a odd and a even separately. The ﬁrst case is n odd and a odd. In that case, (n − a)3 is even, n3 is odd and (n + a)3 is even, so the equation does not balance. In the second case, n is odd and a is even, the three terms are all odd and again the equation does not balance. Therefore n cannot be odd.
Next, we investigate the case n is even and a odd. This time the three terms are odd, even and odd respectively so there is no contradiction. But multiplying out the brackets and simplifying gives n2 (n − 6a) = 2a3 .
Since n is even the left hand side is divisible by 4, because of the factor n2 . That means that a3 is divisible by 2 and hence a is even.
Now that we know n and a are both even, we can follow the method used in part (ii) and set n = 2m and a = 2b . This gives
(2m − 2b)3 + (2m)3 = (2m + 2b)3

from which a factor of 23 can be cancelled from each term. Thus m and b satisfy the same equation as n and a. They are therefore both even and we can repeat the process.

Repeating the process again and again will eventually result in an integer that is odd which will therefore not satisfy the equation that it is supposed to satisfy: a contradiction. There is therefore no integer n that satisﬁes (∗ ∗ ∗).

Question 25(∗ ∗ ∗)
The function f satisﬁes 0 why f (t)

K when 0

t

1. Explain by means of a sketch, or otherwise,

1

f (t) dt

0

K.

0
1

By considering
0

t dt , or otherwise, show that, if n > 1, n(n − t)
0

ln

n n−1 and deduce that

N n=1 1
1
− n−1 n

N

0

Deduce that

1 n −

ln N −

1 n n=2

1.

1
→ ∞ as N → ∞ . n N

Noting that 210 = 1024, show also that if N < 1030 then n=1 1
Quite a lot of diﬀerent ideas are required for this question; hence three stars. The ﬁrst hurdle is to decide what the constant K in the ﬁrst part has to be to make the second part work. You might try to maximise the integrand using calculus, but that would be the wrong thing to do (you should ﬁnd if you do this that the integrand has no turning point in the given range: it increases throughout the range. The next hurdle is the third displayed equation, which follows from the preceding result.
Then there is still more work to do.
N

1
(which is called the harmonic series tends n 1 to inﬁnity as N increases, though there are easier ways of proving this. This way tells us two very interesting things; First it tells us that the sum increases very slowly indeed: the ﬁrst 1030 terms only get to 100. Second it tells us that
It is not at all obvious at ﬁrst sight that the series

N

0
1

1
− ln N n 1

for all N . (This is the third displayed equation in the question slightly rewritten.) Not only does the sum diverge, it does so logarithmically. In fact, in the limit N → ∞,
N
1

1
− ln N → γ n where γ is Euler’s constant. Its value is about 1/2.

Solution to question 25
Any sketch showing a squiggly curve all of which lies beneath the line x = K and above the x-axis will do (area under curve is less than the area of the rectangle).
Evaluate the integral, noting that t/(n − t) = n/(n − t) − 1:
1
0

t dt = n(n − t)

1

1
1
− n−t n

0

dt = ln

n n−1 −

1
.
n

Now note that the largest value of the integrand in the interval [0, 1] occurs at t = 1, since the numerator increases as t increases and the denominator decreases as t increases (remember that n > 1) so
1
1
1
1 n =
− .

0 ln n−1 n n(n − 1) n−1 n
Summing both sides from 2 to N and cancelling lots of terms in pairs gives
N

0

ln N −

2

1 n 1−

1
.
N
N

Note that 1−1/N < 1. Since ln N → ∞ as N → ∞, so also must

2

1
(it diﬀers from the logarithm n by less than 1).
N

N

1
1
ln N , i.e. n n
1
2 and using the identity given and also the inequality e > 2 gives:
Finally, rearranging the inequality gives

1030
1

1 n ln N + 1 . Setting N = 1030 ,

ln(1030 ) + 1 < ln(102410 ) + 1 = 100 ln 2 + 1 < 100 ln e + 1 = 100 + 1 .

Postmortem
This question seems very daunting at ﬁrst, because you are asked to prove a sequence of completely unfamiliar results. However, you should learn from this question that if you keep cool and follow the implicit hints, you can achieve some surprisingly sophisticated results. You might think that this situation is very artiﬁcial: in real life, you do not receive hints to guide you. But often in mathematical research, hints are buried deep in the problem if only you can recognise them.

Question 26(∗ ∗ ∗)

Find

dy if dx y= ax + b
.
cx + d

(∗)

By using changes of variable of the form (∗), or otherwise, show that
1
0

1 ln (x + 3)2

x+1 x+3 dx =

1
1
1 ln 3 − ln 2 −
,
6
4
12

and evaluate the integrals
1
0

1 ln (x + 3)2

x2 + 3x + 2
(x + 3)2

1

dx

and
0

1 ln (x + 3)2

x+1 x+2 dx .

——————————————————————————————————————————
By changing to the variable y deﬁned by y= evaluate the integral

4
2

Evaluate the integral

25
9

2x − 3
,
x+1

2x − 3
2x − 3 ln 3
(x + 1) x+1 dx .

2z −3/2 − 5z −2 ln 2 − 5z −1/2 dz .

If you are thinking that this question looks extremely long, you are right. It is two questions — in fact, two versions of one question — separated by the horizontal line. Beneath the line is the ﬁrst draft of a STEP question; above the line is the question as it was actually set. I thought you would be interested to see the evolution of a question. Note in particular the way that the ideas in the ﬁnal draft are closely knit and better structured: the ﬁrst change of variable is strongly signalled, but the remaining two parts, though based on the same idea, require increasing ingenuity. The ﬁrst draft just required two unrelated changes of variable. Note also that the integral in the last part of the ﬁrst draft has an unpleasant contrived appearance, whereas the the integrals of the ﬁnal draft are rather pleasing: beauty matters to mathematicians.
A change of variable of the form (∗) is called a linear fractional or M¨bius transformation. It is of o great importance in the theory of the geometry of the complex plane.

Solution to question 26 dy a(cx + d) − c(ax + b) ad − bc
=
=
.
2 dx (cx + d)
(cx + d)2 x+1 For the ﬁrst integral, set y =
:
x+3

Diﬀerentiating gives

1
0

1 ln (x + 3)2

x+1 x+3 dx =

1
2

1/2

ln y dy =
1/3

1 y ln y − y
2

1/2
1/3

,

which gives the required answer. (The integral of ln y is a standard integral; it can be done by parts, ﬁrst substituting z = ln y, if you like.)
The second integral can be expressed as the sum of two integrals of the same form as the ﬁrst integral, since ln x2 + 3x + 2
(x + 3)2

(x + 1)(x + 2)
(x + 3)2

= ln

x+1 x+3 = ln

+ ln

We have already done the ﬁrst of these integrals. Using the substitution y = these integrals gives
3/4

ln y dy =
2/3

The required integral is therefore

x+2 x+3 .

x+2 in the second of x+3 13
1
17 ln 3 − ln 2 −
.
12
6
12

29
1
19 ln 3 − ln 2 − .
12
12
6

For the next integral, note that ln x+1 x+2 x+1 x+3 = ln

− ln

x+2 x+3 ,

so the required integral is the diﬀerence of the two integrals that we summed in the previous part,
i.e.
23
5
ln 2 .
− ln 3 +
4
12
——————————————————————————————————–
The change of variable gives
1
5

1

y ln y dy
1/3

2
1
ln 3 −
.
90
45
For the last part, you have to guess the substitution. There are plenty of clues, but the most obvious place to start is the log. The argument of the log can be written as

which can be integrated by parts. The result is

2(x + 1) − 5
5
=2− x+1 x+1 so it looks as if we should take z = (x + 1)2 . Making this transformation gives the previous integral almost exactly (note especially the the limits transform as they should). The only diﬀerence is a dz so the answer is twice the previous answer. factor of 2 which comes from dx Question 27(∗)

The curve C has equation y=√ (x2

x
,
− 2x + a)

where the square root is positive. Show that, if a > 1, then C has exactly one stationary point.
Sketch C when (i) a = 2 and (ii) a = 1.

There is something you must know in order to do this question, and that is the deﬁnition of
√ 2
(x − 2x + a) . You might think that this is ambiguous, because the square root could be positive

of negative, but by convention it means the positive square root. Thus (x2 ) = |x| (not x ).

For the sketches, you just need the position of any stationary points, any other points that the graph passes through, behaviour as x → ±∞ and any vertical asymptotes. It should not be necessary in such a simple case to establish the nature of the stationary point(s).
You should at some stage think about the condition given on a . In fact, I doubt if many of you will want to leave it there. It is clear that the examiner would really have liked you to do is to sketch
C in the diﬀerent cases that arise according to the value of a but was told that this would be to long and/or diﬃcult for the exam; I am sure that this is what you will do (or maybe what you have already done by the time you read this).
Please do not use your calculator for these sketches (except perhaps to check your answers): there is no point — the graph will either be wrong, in which case you will be confused, or it will be right and the value that you might have gained from thinking about the question yourself will be lost to you for ever.

Solution to question 27
First diﬀerentiate to ﬁnd the stationary points: dy a−x
(x2 − 2x + a) − x(x − 1)
= 2
=
2 − 2x + a)3/2 dx (x
(x − 2x + a)3/2 which is only zero when x = a.

15

(i) When a = 2, the stationary point is at (2, 2) (which can be seen to be a maximum by evaluating the second derivative at x = a , though this is not necessary). The curve passes through (0, 0). As x → ∞, y → 1; as x → −∞, y → −1. x (ii) When a = 1, we can rewrite y as
. The modulus
|x − 1| signs arise because the square root is taken to be positive
(or zero — but that can be discounted here since it is in the denominator). The graph is as before, except that the maximum point has been stretched into a vertical asymptote
(like a volcanic plug) at x = 1 .
Postmortem
The ﬁrst thing to do after ﬁnishing the question is to try to understand the condition on a . Part
(ii) above is the key. Writing f (x) = x2 − 2x + a = (x − 1)2 + a − 1 we see that f (x) > 0 for all x if a > 1 but if a < 1 there are values of x for which f (x) < 0 . is negative. The borderline value is a = 1, for which f (x) 0 and f (1) = 0 .

If a < 1 , we ﬁnd f (x) = 0 when x = 1 ± (1 − a) and f (x) < 0 between these two values of x . The signiﬁcance of this is that the square root in the denominator of the function we are trying to sketch

√ is imaginary, so there is a gap in the graph between x = 1 − (1 − a) and x = 1 + (1 − a) and there are vertical asymptotes at these values. The point x = a lies between those values for a < 1 .
Is that the whole story? Well, no. We now have to decide whether the two signiﬁcant points of our graph, namely x = 0 where the curve crosses the axis and x = a where the curve has a stationary

√ point, lie inside or outside the forbidden zone 1 − (1 − a) x 1 + (1 − a) . Since f (0) < 0 if a < 0 and f (a) < 0 if 0 < a < 1 we see that a has another critical value besides a = 1; if a < 0 the graph is going to look qualitatively diﬀerent.

15
My drawing package is absolutely useless, so I’m not going to attempt to improve these sketches, or label them fully (which you should certainly do). Maybe in the next draft ...

Question 28(∗ ∗ ∗)
Prove that

n

sin kθ = k=0 (i)

Deduce that, for n

1,

cos 1 θ − cos(n + 1 )θ
2
2
.
1
2 sin 2 θ

n

sin k=0 (ii)

kπ n = cot

π
2n

(∗)

.

By diﬀerentiating (∗) with respect to θ, or otherwise, show that, for n n k sin2 k=0 kπ
2n

=

π
(n + 1)2 1
+ cot2
4
4
2n

1,

.

1
The very ﬁrst part can be done by multiplying both sides of (∗) by sin 2 θ and using the identity

2 sin A sin B = cos(B − A) − cos(B + A) .
It can also be done by induction (worth a try even if you do it by the above method) or by considering the imaginary part of exp(ikθ) (summing this as a geometric progression), if you know about de
Moivre’s theorem.
Before setting pen to paper for part (ii), it pays to think very hard about simpliﬁcations of (∗) that might make the diﬀerentiation more tractable – perhaps bearing in mind the calculations involved with part (i).
In the original question, you were asked to show that for large n n sin

kπ n kπ
2n

k=0

and

n

k sin2 k=0 ≈

2n π 1
1
+ 2
4 π

n2 ,

1 using the approximations, valid for small θ, sin θ ≈ θ and cos θ ≈ 1 − 2 θ 2 . Of course, these follow quickly from the exact results; but if you only want approximate results you can save yourself a bit of work by approximating early to avoid doing some of the trigonometric calculations. I thought that the exact result was nicer than the approximate result (though approximations are an important part of mathematics).

Solution to question 28
For the ﬁrst part, we will show that n k=0

1
1
1
2 sin kθ sin 2 θ = cos 2 θ − cos(n + 2 )θ .

Starting with the left hand side, we have n n
1
2 sin kθ sin 2 θ =

k=0

k=0

1
− cos(k + 1 )θ + cos(k − 2 )θ
2

which gives the result immediately, since almost all the terms in the sum cancel. (Write our a few terms if you are not certain of this.)
(i) Let θ = π/n. Then n sin k=0 kπ n =

1 cos 2 (π/n) + cos 1 (π/n)) cos 1 (π/n) − cos(π + 1 (π/n))
2
2
2
=
= cot 1 (π/n)
2
1
2 sin 2 ( π )
2 sin 1 (π/n) n 2

as required.
(ii) Diﬀerentiating the left hand side of (∗) and using a double angle trig. formula gives k cos kθ = k=0 n

n

n

k=0

1 k 1 − 2 sin2 2 kθ = 1 n(n + 1) − 2
2

1 k sin2 2 kθ .

(†)

k=0

Before attempting to diﬀerentiate the right hand side of (∗) it is a good idea to write it in a form that gets rid of some of the fractions. Omitting for the moment the factor 1/2, we have
RHS of (∗) = (1 − cos nθ) cot 1 θ + sin nθ
2

1 and diﬀerentiating gives − 1 (1 − cos nθ) cosec2 1 θ + n sin nθ cot 2 θ + n cos nθ . Now setting θ = π/n
2
2 gives (‡)
− cosec2 1 (π/n) − n .
2

Setting θ = π/n in equation (†) and comparing with (‡) (remembering that there is a factor of 1/2 missing) gives the required result.
Postmortem
Now that I have had another go at this question it does not seem terribly interesting. At ﬁrst, I thought I might ditch it. Then I thought that it perhaps was worthwhile: keeping cool under the pressure of the diﬀerentiation for part (ii) – it should just be a few careful lines – and getting it out is a good conﬁdence booster. Anyway, now that I have slogged16 through it, I am going to leave it in. (Next day) I recall now that the reason that I included this question in the ﬁrst place was because in its original form (with approximate answers), the result for part (ii) can be obtained very quickly from part (i) by diﬀerentiating with respect to π. The snag with this is that it is not obviously going to work, because the derivatives of terms such as sin π that are omitted from (i) might feature in (ii).
It is disappointing but not surprising that this trick cannot be used to obtain the exact formulae.
16

I should come clean at this point: I just diﬀerentiated (∗) on autopilot, without considering whether I could simplify the task by rewriting the formula as suggested in my hint on the previous page. It was hard work. Serves me right for not following my own general advice.

Question 29(∗ ∗ ∗)
Consider the cubic equation x3 − px2 + qx − r = 0 , where p = 0 and r = 0.
(i)

If the three roots can be written in the form ak−1 , a and ak for some constants a and k, show that one root is q/p and that q 3 − rp3 = 0 .

(ii)

If r = q 3 /p3 , show that q/p is a root and that the product of the other two roots is (q/p)2 .
Deduce that the roots are in geometric progression.

(iii)

Find a necessary and suﬃcient condition involving p, q and r for the roots to be in arithmetic progression. Comments
The Fundamental Theorem of Algebra says that a polynomial of degree n can be written as the product of n linear factors, so we can write x3 − px2 + qx − r = (x − α)(x − β)(x − γ) ,

(∗)

where α , β and γ are the roots of the equation x3 − px2 + qx − r = 0 . The basis of this question is the comparison between the left hand side of (∗) and the right hand side, multiplied out, of (∗).
Some of the roots may not be real, but you don’t have to worry about that here.
The ‘necessary and suﬃcient’ in part (iii) looks a bit forbidding, but if you just repeat the steps of parts (i) and (ii), it is straightforward.

Solution to question 29
(i) We have
(x − ak−1 )(x − a)(x − ak) ≡ x3 − px2 + qx − r
i.e.
x3 − a(k−1 + 1 + k)x2 + a2 (k−1 + 1 + k)x − a3 = x3 − px2 + qx − r .

Thus p = a(k−1 + 1 + k) , q = a2 (k−1 + 1 + k) , and r = a3 . Dividing gives q/p = a , which is a root, and q 3 /p3 = a3 = r as required.
(ii) Set r = q 3 /p3 . Substituting x = q/p into the cubic shows that it is a root:
(q/p)3 − p(q/p)2 + q(q/p) − (q/p)3 = 0 .
Since q/p is one root, and the product of the three roots is q 3 /p3 (= r in the original equation), the product of the other two roots must be q 2 /p2 . The two roots can therefore be written in the form k−1 (q/p) and k(q/p) for some number k , which shows that they are in geometric progression.
(iii) The roots are in arithmetic progression if and only if they are of the form (a − d) , a and (a + d) .
If the roots are in this form, then the equation is

x − (a − d) x − a x − (a + d) ≡ x3 − px2 + qx − r
i.e.
x3 − 3ax2 + (3a2 − d2 )x − a(a2 − d2 ) ≡ x3 − px2 + qx − r .

Thus p = 3a , q = 3a2 − d2 , r = a(a2 − d2 ) . A necessary condition is therefore r = (p/3)(q − 2p2 /9).
Note that one root is p/3.
Conversely, if r = (p/3)(q − 2p2 /9) , the equation is x3 − px2 + qx − (p/3)(q − 2p2 /9) = 0.
We can verify that p/3 is a root by substitution. Since the roots sum to p and one of them is p/3, the others must be of the form p/3 − d and p/3 + d for some d. They are therefore in arithmetic progression. A necessary and suﬃcient condition is therefore r = (p/3)(q − 2p2 /9).
Postmortem
As usual, it is a good idea to give a bit of thought to the conditions given, namely p = 0 and r = 0 .
Clearly, if the roots are in geometric progression, we cannot have a zero root. We therefore need r = 0 . However, we don’t need the condition p = 0 . If the roots are in geometric progression with p = 0, then q = 0 , but there is no contradiction: the roots are be−2πi/3 , b and be2πi/3 where b3 = r , and these are certainly in geometric progression. So the necessary and suﬃcient conditions are r = 0 and p3 r = q 3 .
Neither of these conditions is required if the roots are in arithmetic progression.
Therefore, the condition p = 0 is only there as a convenience — one thing less for you to worry about. The condition r = 0 should really have been given only for the ﬁrst part. It should be said that this sort of question is incredibly diﬃcult to word, which is why the examiner was a bit heavy handed with the conditions.

Question 30(∗ ∗ ∗)
(i)
Let f (x) = (1 + x2 )ex − k , where k is a given constant. Show that f ′ (x) graph of f (x) in the cases k < 0 , 0 < k < 1 and k > 1 .

0 and sketch the

Hence, or otherwise, show that the equation
(1 + x2 )ex − k = 0 , has exactly one real root if k > 0 and no real roots if k
(ii)

0.

Determine the number of real roots of the equation
(ex − 1) − k tan−1 x = 0

in the diﬀerent cases that arise according to the value of the constant k . dy 1
Note: If y = tan−1 x, then
=
. dx 1 + x2

In good STEP style, this question has two related parts. In this case, the ﬁrst part not only gives you guidance for doing the second part, but also provides a result that is useful for the second part.
In the original question you were asked to determine the number of real roots in the cases (a)
0 < k
2/π and (b) 2/π < k < 1 . I thought that you would like to work out all the diﬀerent cases for yourself — but it makes the question considerably harder, especially if you think about the special values of k as well as the ranges of k .
For part (i) you need to know that xex → 0 as x → −∞ . This is just a special case of the result that exponentials go to zero faster than any power of x .
For part (ii), it is not really necessary to do all the sketches, but I think you should (though you perhaps wouldn’t under examination conditions) because it gives you a complete understanding of the way the function depends on the parameter k . For the graphs, you may want to consider the sign of f ′ (0) ; this will give you an important clue. Remember that, by deﬁnition, −π/2 < tan−1 x < π/2 .

Solution to question 30
(i) Diﬀerentiating gives f ′ (x) = 2xex + (1 + x2 )ex = (1 + x)2 ex which is non-negative (because the square is non-negative and the exponential is positive.
There is one stationary point, at x = −1 , which is a point of inﬂection, since the gradient on either side is positive.
Also, f (x) → −k as x → −∞ , f (0) = 1 − k and f (x) → ∞ as x → +∞ . The graph is essentially exponential, with a hiccup at x = −1 . The diﬀerences between the three cases are the positions of the horizontal asymptote (x → −∞) and the place where the graph cuts the y axis (above or below the x axis?).
Since the graph is increasing and 0 < f (x) < ∞, the given equation has one real root if k > 0 and no real roots if k 0. k 0).

Question 32(∗ ∗ ∗)
To nine decimal places, log10 2 = 0.301029996 and log10 3 = 0.477121255 .
(i)

Calculate log10 5 and log10 6 to three decimal places. By taking logs, or otherwise, show that 5 × 1047 < 3100 < 6 × 1047 .
Hence write down the ﬁrst digit of 3100 .

(ii)

Find the ﬁrst digit of each of the following numbers: 21000 ; 210 000 ; and 2100 000 .

This nice little question shows why it is a good idea to ban calculators from some mathematics examinations. When I was at school, before calculators were invented, we had to spend quite a lot of time in year
8 (I think) doing extremely tedious calculations by logarithms. We were provided with a book of tables of four-ﬁgure logarithms and the book also had tables of values of trigonometric functions.
When it came to antilogging, to get the answer, we had to use the tables backwards, since there were no tables of inverse logarithms (exponentials).
A logarithm consists of two parts: the characteristic which is the number before the decimal point and the mantissa which is the number after the decimal point. It is the mantissa that gives the signiﬁcant ﬁgures of the number that has been logged; the characteristic tells you where to put the decimal point. This question is all about calculating mantissas.
The characteristic of a number greater than 1 is non-negative but the characteristic of a number less than 1 is negative. The rules for what to do in the case of a negative characteristic were rather complicated: you couldn’t do ordinary arithmetic because the logarithm consisted of a negative characteristic and a positive mantissa. In ordinary arithmetic, the number −3.4 means −3 − 0.4 whereas the corresponding situation in logarithms, normally written ¯ , means −3 + 0.4 . Instead
3.4
of explaining this, the teacher gave a complicated set of rules, which just had to be learned — not the right way to do mathematics.
There are no negative characteristics in this question, I’m happy to say.

Solution to question 32
(i) For the very ﬁrst part, we have log 2 + log 5 = log 10 = 1 so log 5 = 1 − 0.301029996 = 0.699 to 3 d.p. log 6 = log 2 + log 3 = 0.778 to 3 d.p.

Taking logs preserves the inequalities, so we need to show that
47 + log 5 < 100 log 3 < 47 + log 6
i.e. that
47 + 0.699 < 47.7121 < 47 + 0.778 which is true. The ﬁrst digit of 3100 is therefore 5 .
(ii) To ﬁnd the ﬁrst digit of these numbers, we use the method of part (i).
We have log 21000 = 1000 log 2 = 301.030 = 301 + 0.030 (to 3 d.p.). But log 1 < 0.030 < log 2, so the ﬁrst digit of 21 000 is 1 .
Similarly, log 210 000 = 10 000 log 2 = 3010.29996 . But log 1 < 0.29996 < log 2, so the ﬁrst digit of
210 000 is 1 .
Finally, log 2100 000 = 30102 + 0.9996 (4 d.p.) and log 9 = 2 log 3 = 0.95 (2 d.p.), so the ﬁrst digit of
2100 000 is 9 .
Postmortem
Although the ideas in this question are really quite elementary, you needed to understand them deeply. You should feel pleased with yourself it you got this one out.

Question 33(∗∗)

For any number x , the largest integer less than or equal to x is denoted by [x] . For example,
[3.7] = 3 and [4] = 4 .
(i)

Sketch the graph of y = [x] for 0

x < 5 and evaluate
5

[x] dx .
0

(ii)

Sketch the graph of y = [ex ] for 0

x < ln n , where n is an integer, and show that ln n

0

Hence show that n!

[ex ] dx = n ln n − ln(n!) .

nn e1−n .

Again, I have had to add a bit to the original question because it was all dressed up with nowhere to go. The question is clearly about estimating n! so I added in the last line (which makes the question a bit longer but not much more diﬃcult). I could have added another part, but I thought that you would probably add it yourself: you will easily spot, once you have drawn the graphs, that a similar result for n! , with the inequality reversed, can be obtained by considering rectangles the tops of which are above the graph of ex instead of below it. You therefore end up with a nice sandwich inequality for n! .

Solution to question 33
(i)
y

y=x

The graph of [x] is an ascending staircase with 5 stairs, the lowest at height 0, each of width 1 unit and rising 1 unit.
The integral is the sum of the areas of the rectangles shown in the ﬁgure: area = 0 + 1 + 2 + 3 + 4 = 10 . x (ii)
The graph of [ex ] is a also staircase: the height of each stair is 1 unit and the width decreases as x increases (because the gradient of ex increases). It starts at height 1 and ends at height
(n − 1) .
The value of [ex ] changes from k to k +1 when ex = k i.e. when x = ln(k + 1) . Thus [ex ] = k when ln k x < ln(k + 1) and the area of the corresponding rectangle is k(ln(k + 1) − ln k) .
The total area under the curve is therefore

y

y = ex

1(ln 2 − ln 1) + 2(ln 3 − ln 2) + · · · + (n − 1)(ln n − ln(n − 1))

x

i.e. n ln n − ln n − ln(n − 1) − ln(n − 2) − · · · − ln 1 which sums to n ln n − ln(n!) as required.

For the last part, note that [ex ] ln n
0

ex (this is clear from the deﬁnition) so ln n

[ex ] dx
0

ex dx = n − 1

i.e.

n ln n − ln(n!)

n−1.

Taking exponentials gives the required result.
Postmortem
Considering [ex + 1] gives rectangles above the graph of y = ex rather than below. The calculations are roughly the same, so you should easily arrive at n! nn+1 e1−n . We have therefore proved that nn e1−n

n!

nn+1 e1−n .

This is a pretty good (given the rather elementary method at our disposal) approximation for n!
1

−n
for large n . Stirling (1692 – 1770) proved in 1730 that n! ≈ 2π nn+ 2 e√ for large n. This was a brilliant result — even reading his book, it is hard to see where he got 2π from (especially as he wrote in Latin) — but he went on to obtain the approximation in terms of an inﬁnite series. The expression above is just the ﬁrst term.

Question 34(∗ ∗ ∗)
Given that tan π = 1 show that tan π =
4
8
Let

2 − 1.

1

I=
−1

(1 + x) +

1

(1 − x) + 2

dx .

Show, by using the change of variable x = sin 4t , that π/8 I=
0

Hence show that

2 cos 4t dt . cos2 t

I =4 2−π−2.

This tests trigonometric manipulation and integration skills. You will certainly need tan 2θ in terms of tan θ , cos 2θ in √ terms of cos θ, and maybe other formulae. A good idea is required in the ﬁrst part to deal with 1 ± sin 4t — look away if you want to ﬁnd it for yourself — try writing 1 as cos2 2t + sin2 2t.
Both parts of the question are what is called multistep: there are half a dozen consecutive steps, each diﬀerent in nature, with no guidance. This sort of thing is unusual at A-level but normal in university course mathematics. In this question, the most steps require trigonometric manipulation similar to that appearing in earlier parts of the question.
I checked the answer on www.integrals.com, which turned out to be very good indeed at doing this sort of thing. I asked it to do the indeﬁnite integral and, in less than a second, it came up with

1 − x −1 − ( x + 1 + 1)−1 − 2 arcsin (x + 1)/2 + [ x + 1 + 1]−1 .
Not a pretty sight and not in its neatest form by a long way (the arcsin term reduces, after a bit of algebra, to −π/2 + arcsin x(!)). I also asked it to do the same integral with the 2 replaced by a which it did in less than four seconds. The answer was about 20 times longer than the above result.
It seemed to enjoy the problem, as far as I could tell.
I think that it does it by multiplying top and bottom of the fraction by suitable expressions (maybe

(1+x)− (1−x)+2 to start with) until the square roots have been removed from the denominator.

Solution to question 34
Let t = tan π . Then
8
−2 ±
2t
= 1 ⇒ t2 + 2t − 1 = 0 ⇒ t =
2
1−t
2

8

= −1 ±

2.

We take the root with the + sign since we know that t is positive.
Now the integral. Substituting x = sin 4t gives: π/8 I=
−π/8

4 cos 4t

dt = 2
1 + sin 4t + 1 − sin 4t + 2

π/8
0

4 cos 4t dt ,
(cos 2t + sin 2t) + (cos 2t − sin 2t) + 2

using
1 ± sin 4t = 1 ± 2 sin 2t cos 2t = cos2 2t + sin2 2t ± 2 sin(t/2) cos 2t = (cos 2t ± sin 2t)2 and remembering to take the positive square root (cos 2t sin 2t for values of t in the range of integration). One more trig. manipulation gives the required result.
Now
cos 4t = 2 cos2 t − 1 = 2(2 cos2 t − 1)2 + 1 = 8 cos4 t − 8 cos2 t + 1 so π/8

I=2
0

π/8

8 cos2 t − 4 + sec2 t)dt = 4 sin 2t − 8t + tan t

,
0

which gives the required result.
Postmortem
Manipulating the integral after the change of variable was really quite demanding. If you didn’t

think of the idea referred to in the comments for dealing with 1 ± sin 4t you could of course have

√ veriﬁed that 1 + sin 4t + 1 ± sin 4t + 2 = 4 cos2 t, though even this would require some thought: you would have to subtract 2 from both sides before you squared them, and then you would have to worry about signs, since square-rooting the squares would lead to a ±.
It occurs to me that there ought to be a better way of doing the integral, without at any stage spoiling the x → −x symmetry of the integrand; maybe I’ll think again about this one day.

Question 35(∗∗)
(i)

Show that, for 0

x

1, the largest value of

x6
1
. is 2 + 1)4
(x
16

What is the smallest value?
(ii)

Find constants A , B , C and D such that, for all x ,
Ax5 + Bx3 + Cx
(x2 + 1)3

d
1

2 + 1)4
(x
dx
(iii)

+

Dx6
.
(x2 + 1)4

Hence, or otherwise, prove that
11
24

1
0

(x2

1 dx + 1)4

1
11
+
.
24 16

You should think about part (i) graphically, though it is not necessary to draw the graph: just set about it as if you were going to (starting point, ﬁnishing point, turning points, etc).
The equivalence sign in part (ii) indicates an equality that holds for all x — you are not being asked to solve the equation for x. After doing the diﬀerentiation, you obtain four linear equations for A ,
B , C and D by equating coeﬃcients of powers of x . You could instead put four carefully chosen values of x into the equation; one good choice would be x = i to eliminate terms with factors of x2 + 1 and setting x = 0 would be useful; after that, it becomes more diﬃcult.
Note that there are only odd powers of x inside the derivative. You could include even powers as well, but you would ﬁnd that their coeﬃcients would be zero: the derivative has to be an even function (even powers of x) so the function being diﬀerentiated must be odd.
For part (iii), you need to know that inequalities can be integrated: this is ‘obvious’ if you think about integration in terms of area, though a formal proof requires a formal deﬁnition of integration and this is the sort of thing you would do in a ﬁrst university course in mathematical analysis.
This question is about estimating an integral. Although the idea is good, the ﬁnal result is a bit feeble. It only gives the value of the integral to an accuracy of about (1/16)(11/24) which is about 15%. The actual value of the integral can be found fairly easily using the substitution x = tan t and is 11/48 + 5π/64 so the inequalities give (not particularly good) estimates for π namely (2.933 π 3.733 ).

Solution to question 35
(i) The derivative of

x6 is (x2 + 1)4
8x7
2(3 − x2 )x5
6x5
− 2
=
(x2 + 1)4 (x + 1)5
(x2 + 1)5

which is positive for 0 < x < 1. The function therefore increases in value from 0 to 1. The largest value occurs when x = 1 and is 1/16 and the smallest value occurs at x = 0 and is 0.
(ii) Doing the diﬀerentiation gives
1
5Ax4 + 3Bx2 + C
6x(Ax5 + Bx3 + Cx)
Dx6

+ 2
(x2 + 1)4
(x2 + 1)3
(x2 + 1)4
(x + 1)4 and multiplying by (x2 + 1)4 gives the following identity:
1 ≡ (5Ax4 + 3Bx2 + C)(x2 + 1) − 6x(Ax5 + Bx3 + Cx) + Dx6
i.e.
1 ≡ (D − A)x6 + (5A − 3B)x4 + (3B − 5C)x2 + C .
This holds for all values of x so we can equate coeﬃcients of the diﬀerent powers of x on each side of the equivalence sign. This gives
1 = C, 0 = 3B − 5C, 0 = 5A − 3B, 0 = D − A, so A = 1, B = 5/3, C = 1 and D = 1.

With these values of A, B, C and D the identity in the question holds (i.e. the these values are not merely necessary, they are also suﬃcient).
(iii) Using the results of parts (i) and (ii), we see that for 0 d dx

x5 + (5/3)x3 + x
(x2 + 1)3

1
2 + 1)4
(x

d dx x

1

x5 + (5/3)x3 + x
(x2 + 1)3

+

1
.
16

Inequalities can be integrated, so
1
0

i.e.

d dx x5 + (5/3)x3 + x
(x2 + 1)3

1

dx

x5 + (5/3)x3 + x
(x2 + 1)3

0

1 dx 2 + 1)4
(x

1
0

1
0

1 dx 2 + 1)4
(x

from which the required result follows immediately.

1
0

d dx x5 + (5/3)x3 + x
(x2 + 1)3

x5 + (5/3)x3 + x
(x2 + 1)3

1

+
0

1

dx +
0

x
16

1
0

1 dx 16

Question 36(∗∗)

Show that

1
−1

| xex |dx = −

0
−1

1

xex dx +

xex dx

0

and hence evaluate the integral.
Evaluate the following integrals:
4

(i)
0

| x3 − 2x2 − x + 2 | dx ;

π

(ii)
−π

| sin x + cos x | dx .

The very ﬁrst part shows you how to do this sort of integral (with mod signs) by splitting up the range of integration at the places where the integrand changes sign. In parts (i) and (ii) you have to use the technique on diﬀerent examples. This learning process and the nice mixture of ideas make it an ideal STEP question.

Solution to question 36
For the ﬁrst part, note that |xex | = −xex if x < 0 . Then integrate by parts to evaluate the integrals:
0

1

xex dx +
0

−1

0

xex dx = −

0

xex
−1

1

ex dx

+

xex

−1

0

1

ex dx

0

= 2 − 2e−1 .

For the next parts, we have to ﬁnd out where the integrand is positive and where it is negative.
(i) x3 − 2x2 − x + 2 = (x − 1)(x + 1)(x − 2) (spotting the factors), so the integrand is positive for
0 x < 1 , negative for 1 < x < 2 and positive for 2 < x < 4 . Splitting the range of integration into these ranges and integrating gives
4
0

x3 − 2x2 − x + 2 dx
1

=
0

(x3 − 2x2 − x + 2) dx −

2
1

1

=

1 4
4x

2
− 3 x3 − 1 x2 + 2x
2

0

4

(x3 − 2x2 − x + 2) dx +

2

(x3 − 2x2 − x + 2) dx

2

1 4
4x

2
1
− 3 x3 − 2 x2 + 2x

4

+
1

1 4
4x

1
− 2 x3 − 2 x2 + 2x
3

2

1
= 22 6 .

(ii) sin x + cos x changes sign when tan x = −1, i.e. when x = −π/4 and x = 3π/4. Splitting the range of integration into these ranges and integrating gives π −π

sin x + cos x dx = −

(sin x + cos x) dx +

−π

= − − cos x + sin x

=4 2

π

3π/4

−π/4

−π/4
−π

Alternatively, start by writing sin x + cos x = easier to spot and the integrals easier to do.

(sin x + cos x) dx −

−π/4

+ − cos x + sin x

(sin x + cos x) dx
3π/4
π

3π/4
−π/4

− − cos x + sin x

3π/4

2 cos(x − π/4) which makes the changes of sign

Postmortem
If you give a bit more thought to part (ii), you will see easier ways of doing it. Since the trigonometric functions are periodic with period 2π the integrand is also periodic. Write the integrand in the form

2 cos(x − π/4) . Integrating this over any 2π interval gives the same result. Indeed, we may as

well integrate 2 sin x ; from 0 to π and double the answer.

Question 37(∗)

A number of the form 1/N , where N is an integer greater than 1, is called a unit fraction.
Noting that

1
1 1
1
1
1
= + and = + ,
2
3 6
3
4 12

guess a general result of the form

1 1
1
= +
(∗)
N a b and hence prove that any unit fraction can be expressed as the sum of two distinct unit fractions.
By writing (∗) in the form
(a − N )(b − N ) = N 2

and by considering the factors of N 2 , show that if N is prime, then there is only one way of expressing
1/N as the sum of two distinct unit fractions.
Prove similarly that any fraction of the form 2/N , where N is prime number greater than 2, can be expressed uniquely as the sum of two distinct unit fractions.

Fractions written as the sum of unit fractions are called Egyptian fractions: they were used by
Egyptians. The earliest record of such use is 1900BC. The Rhind papyrus in the British Museum gives a table of representations of fractions of the form 2/n as sums of unit fractions for all odd integers n between 5 and 101 — a superb achievement when you consider that algebra was 3.500 years away.
It is not clear why Egyptians represented fractions this way; maybe it just seemed a good idea at the time. Certainly the notation they used, in which for example 1/n was denoted by n with an oval on top, does not lend itself to generalisation to fractions that are not unit. One of the rules
2
was that all the unit fractions in an Egyptian fraction should be distinct, so had to be expressed
7
1
1
as +
, which seems pretty daft.
4 28
It is not obvious that every fraction can be express in Egyptian form; this was proved by Fibonacci in 1202. There are however still many unsolved problems relating to Egyptian fractions.
Egyptian fractions have been called a wrong turn in the history of mathematics; if so, it was a wrong turn that favoured style over utility which is no bad thing, in my opinion.

Solution to question 37
1
1 is . In that case, the
N
N +1
. That proves the result that every unit fraction can be

A good guess would be that the ﬁrst term of the decomposition of
1
1
1

i.e.
N
N +1
N (N + 1) expressed as the sum of two unit fractions.

other term is

The only factors of N 2 (since N is prime) are 1 , N and N 2 . The possible factorisations of N 2 are therefore N 2 = 1 × N 2 or N 2 = N × N so, since a = b, a − N = 1 and b − N = N 2 (or the other way round). Thus N + 1 and N 2 + N are the only possible values for a and b and the decomposition is unique. For the second half, set

1 1
2
= +
N
a b
(a + b)N where a = b . Proceeding as before, we have ab −
= 0 which we write as
2
a−

N
2

b−

N
2

=

N2
4

i.e.

(2a − N ) (2b − N ) = N 2 .

Thus 2a − N = N 2 and 2b − N = 1 (or the other way round). The decomposition is therefore unique and given by
2
1
1
=
+
.
N
(N 2 + N )/2 (N + 1)/2
The only possible ﬂy in the ointment is the 1/2 in the denominators: a and b are supposed to be integers. However, all prime numbers greater than 2 are odd, so N + 1 and N 2 + N are both even and the denominators are indeed integers.
Postmortem
Another way of getting the last part (less systematically) would have been to notice that
1
1
2
1
1
1
=
+
=⇒
=
+
N
(N + 1) N (N + 1)
N
(N + 1)/2 N (N + 1)/2 which gives the result immediately. How might you have noticed this? Well, I noticed by trying to work out some examples, starting with what is given at the very beginning of the question. If
1
1
2
2
2
1
= + then = + which works.
3
4 12
3
4 12
Of course, the advantage of the systematic approach is that it allows generalisation: what happens if N is odd but not prime; what happens if the numerator is 3 not 2?

Question 38(∗)

Prove that if (x − a)2 is a factor of the polynomial p(x), then p′ (a) = 0 . Prove a corresponding result if (x − a)4 is a factor of p(x) .
Given that the polynomial

x6 + 4x5 − 5x4 − 40x3 − 40x2 + kx + m has a factor of the form (x − a)4 , ﬁnd k and m .

The very ﬁrst part has an ‘obvious’ feel to it but the diﬃculty lies in setting out a clear mathematical proof. The key is the factor theorem: if (x − a) is a factor of a polynomial, then it can be written as the product of polynomial and (x − a) .
Please take some care over the presentation of next part.

There are a few numerical calculations to do. Since calculators are neither permitted nor needed in
STEP, it is probably best if you don’t use yours.
In the original question, k was given (it is 32). But this seemed rather unsatisfactory because it meant that not all the information available is required to solve the problem. Even my new improved version is not perfect in this respect because p′′ (a) = 0 is used only to distinguish between three possibilities for a .

Solution to question 38
If (x − a)2 is a factor of p(x), we can write p(x) = (x − a)2 q(x), where q(x) is a polynomial.
Diﬀerentiating gives p′ (x) = 2(x − a)q(x) + (x − a)2 q′ (x) and p′ (a) = 0.
Similarly, if (x − a)4 is a factor, then p(x) = (x − a)4 q(x) for some polynomial q(x) . Following the above argument, we have p′ (x) = 4(x − a)3 q(x) + (x − a)4 q′ (x)

so p′ (x) has a factor (x − a)3 and can be expressed in the form p′ (x) = (x − a)3 r(x) for some polynomial. Similarly, p′′ (x) has a factor (x − a)2 and p′′′ (x) has a factor (x − a) . A necessary condition for p(x) to have a factor (x − a)4 is therefore p(a) = p′ (a) = p′′ (a) = p′′′ (a) = 0 .

For the given polynomial,

p′′′ (a) = 0 ⇒ 120a3 + 240a2 − 120a − 240 = 0 ⇒ a3 + 2a2 − a − 2 = 0 ⇒ (a − 1)(a + 1)(a + 2) = 0
(spotting factors), so the possible values of a are 1, −1 and −2. But p′′ (1) = 0 and p′′ (−1) = 0, so a = −2.
Now we have two remaining equations.

p′ (−2) = 0 ⇒ 6(−2)5 + 20(−2)4 − 20(−2)3 − 120(−2)2 − 80(−2) + k = 0 so k = 32 . Similarly p(−2) = 0 ⇒ m = 48 .
Postmortem
Another way of doing the question would be simply to write x6 + 4x5 − 5x4 − 40x3 − 40x2 + kx + m ≡ (x − a)4 (bx2 + cx + d) and equation coeﬃcients of xn on either side of the equation. That gives 7 equations for 6 unknowns.
In general, the system would be overdetermined (too many equations for the number of unknowns) but here we know that a solution exists.

Question 39(∗ ∗ ∗)
Show that sin θ =

2t
,
1 + t2

1 − t2
,
1 + t2

cos θ =

1 + cos θ
= tan(π/2 − θ/2) , sin θ

where t = tan(θ/2) .
Let

π
2

I=
0

1 dθ .
1 + cos α sin θ

Use the substitution t = tan(θ/2) to show that, for 0 < α < π/2 ,
1

I=
0

2 dt .
(t + cos α)2 + sin2 α

By means of the further substitution t + cos α = sin α tan u show that
I=
Deduce a similar result for

π
2

0

α
.
sin α

1 dφ .
1 + sin α cos φ

The ﬁrst of the two substitutions is familiarly known as the ‘t substitution’. It would have been very standard fare 30 years ago, but seems to have gone out of fashion now. The second is the normal substitution for integrals with quadratic denominators and is only just over the edge of the core
A-level syllabus.
For the last part, ‘deduce’ implies that you don’t have to do any further integration. This can only really mean a change of variable to make the integrand into the one you want; you have to hope that the limits then transform to the one given. Note that the change of variable is signalled by the use of the variable φ in the last integral; since it is a deﬁnite integral, it doesn’t matter what the variable is called, and could have been called θ as I ; the use of a diﬀerent variable was just a kindness on the part of the examiner (me). Of course, you can change α without any such complications.
The ‘similar result’ result should include the conditions under which it is true. It is worth thinking about why the condition 0 < α π/2 is required – or indeed, if it is required.

Solution to question 39
The three identities just require use of the half-angle formulae cos θ = cos2 (θ/2) − sin2 (θ/2) and sin θ = 2 sin(θ/2) cos(θ/2) . Remember, for the last one, that cot x = tan(π/2 − x).
For the ﬁrst change of variable, we have dt = and 1, so π 2

1
2

sec2 (θ/2) dθ = 1 (1 + t2 )dθ and the new limits are 0
2

0

1

1 dθ =
1 + cos α sin θ

1
2
dt
2 ) 1 + t2
1 + (cos α)2t/(1 + t

1

I=

2 dt =
1 + 2t cos α + t2

0

=
0

1
0

2 dt .
(t + cos α)2 + sin2 α

For the second change of variable, we have dt = sin α sec2 u du . When t = 0 , tan u = cot α so u = π/2 − α . When t = 1 , sin α tan u = 1 + cos α so u = π/2 − α/2 . Thus π/2−α/2 I = π/2−α π/2−α/2

= π/2−α 2 sin α sec2 udu sin α(1 + tan2 u)
2

2
2
u du = sin α sin α

π/2−α/2 π/2−α =

α
.
sin α

For the last part, we want to make a change of variable the changes the cosine in the denominator to a sine. One possibility is to set φ = π/2 − θ . This will swaps the limits but also and introduces a minus sign since dθ = −dφ . Thus α =I =− sin α

0

π
2

1 dφ =
1 + cos α cos φ

π
2

0

1 dφ .
1 + cos α cos φ

This is almost the integral we want: we still need to replace cos α in the denominator by sin α .
Remembering what we did a couple of lines back, we just replace α by π/2 − α in the integral and in the answer, giving π 2
1
π/2 − α dφ =
.
cos α
0 1 + cos α cos φ
Postmortem
The restriction on α in the integral might prevent the denominator being zero for some value of θ ; a zero in the denominator usually means that the integral is undeﬁned. However, the only value of α for which cos α sin θ could possibly be as small as −1 (for 0 θ π/2) is π (and of course 3π, etc). From the point of view of the integral, the only restriction should be α = π (etc). There is a slight awkwardness in the answer when α = 0 but this can be overcome by taking limits: α =1 lim α→0 sin α which you can easily verify is the correct answer in this case.
You may by now have noticed a curious thing: increasing α by 2π does not change I but does change the answer. If you work through the solution with this in mind you see that the only step where it can make a diﬀerence is in the line where you work out tan−1 u , which by deﬁnition lies in the range −π/2 to π/2 . It is this that determines the given restriction. (Note that negative values of α are not required because setting α → −α changes neither the integral nor the given solution.)

Question 40(∗∗)

The line ℓ has vector equation r = λs , where s = (cos θ +

3 ) i + ( 2 sin θ) j + (cos θ − 3 ) k

and λ is a scalar parameter. Find an expression for the angle between ℓ and the line r = µ(a i + b j + c k) . Show that there is a line m through the origin such that, whatever the value of θ , the acute angle between ℓ and m is π/6 .

A plane has equation x − z = 4 3 . The line ℓ meets this plane at P . Show that, as θ varies, P describes a circle, with its centre on m. Find the radius of this circle.

It is not easy to set vector questions at this level: they tend to become merely complicated (rather than diﬃcult in an interesting way). Usually, there is some underlying geometry and it pays to try to understand what this is. Here, the question is about the geometrical object traced out by ℓ as θ varies. You will need to know about scalar products of vectors for this question, but otherwise it is really just coordinate geometry.
Vectors form an extremely important part of almost every branch of mathematics (maybe every branch of mathematics) and will probably be one of the ﬁrst topics you tackle on your university course. Solution to question 40
Both ℓ and the line r = µ(a i + b j + c k) pass through the origin, so the angle α between the lines is given by the scalar product of the unit vectors, i.e. the scalar product between the given vectors divided by the product of the lengths of the two vectors:

a(cos θ + 3) + 2b sin θ + (cos θ − 3)c cos α = √

[(cos θ + 3)2 + 2 sin2 θ + (cos θ − 3)2 ] [a2 + b2 + c2 ]

(a + c) cos θ + 2 b sin θ + (a − c) 3
√ √
=
.
2 2 [a2 + b2 + c2 ]
Now we want to show that there is some choice of a , b and c such that cos α does not depend on the value of θ . By inspection, this requires a = −c and b = 0 . For these values, the terms depending

on θ disappear from the expression for cos α leaving cos α = ± 1 3 according to the whether a is
2
1√ positive or negative. If we choose a = 1 then cos α = + 2 3 and α = π/6 as required. The line m is given by r = µ(i − k) .
Note that as θ varies, the line ℓ generates a cone with axis m; this is what the question is about.
The coordinates of a general point on the line ℓ are x = λ(cos θ +

3)

√ y = λ 2 sin θ

z = λ(cos θ −

3) .

For a point which is also on the plane x − z = 4 3 we have λ(cos θ +

3) − λ(cos θ − 3) = 4 3

so λ = 2 . The point P on the intersection between the line and the plane has coordinates

(2 cos θ + 2 3 , 2 2 sin θ , 2 cos θ − 2 3) .

A general point on the line m has coordinates x = µ, y = 0, z = −µ . It meets the plane x− z = 4 3

when µ = 2 3 i.e. at the point O with coordinates (2 3, 0 , −2 3).
We need to check that, as θ varies, P moves round a circle with centre O, i.e. that the length OP is independent of θ. We have

OP 2 = (2 cos θ)2 + (2 2 sin θ)2 + (2 cos θ)2 = 8 ,

which is indeed independent of θ as required. The radius of the circle is therefore 2 2.

Question 41(∗∗)

Given that x4 + px2 + qx + r ≡ (x2 − ax + b)(x2 + ax + c) , express p , q and r in terms of a , b and c .
Show that a2 is a root of the cubic equation u3 + 2pu2 + (p2 − 4r)u − q 2 = 0 .
Verify that u = 9 is a root in the case p = −1, q = −6, r = 15 .

Hence, or otherwise, solve the equation

y 4 − 8y 3 + 23y 2 − 34y + 39 .

The long-sought solution of the general cubic was found, in 1535 by Niccol` Tartaglia (c. 1500-57). o He was persuaded to divulge his secret (in the form of a poem) by Girolamo Cardano (1501–76), who promised not to publish it before Cardano did. However, Cardano discovered that it had previously been discovered by del Ferro (1465–1525/6) before 1515 so he published it himself in his algebra book The Great Art. There followed an acrimonious dispute between Tartaglia and Cardano, in which the latter was championed by his student Ferrari (1522–1565). The dispute culminated in a public mathematical duel between Farrari and Tartaglia held in the church of Santa Maria in Milan in 1548, in which they attempted to solve each others’ cubics. The duel ended in a shouting match with Tartaglia storming oﬀ. It seems Ferrari was the winner. Tartaglia was sacked from his job as lecturer and Ferrari made his fortune as a tax assessor before becoming a professor of mathematics at Bologna. He was poisoned by his sister, with arsenic, in 1565.
Ferarri found the a way of reducing quartic equations to cubic equations; his method (roughly) is used in this question to solve a quartic which could probably be solved easier ‘otherwise’. But it is the method that is interesting, not the solution.
The ﬁrst step of the Ferrari method is to reduce the general quartic to a quartic equation with the cubic term missing by means of a linear transformation of the form x → x − a . Then this reduced quartic is factorised (the ﬁrst displayed equation in this question). The factorisation can be found by solving a cubic equation (the second displayed equation above) that must be satisﬁed by one of the coeﬃcients in the factorised form.
The solutions of the quartic are all complex, but don’t worry if you haven’t come across complex numbers: you will be able to do everything except perhaps write down the last line.
You will no doubt be full of admiration for this clever method of solving quartic equations. One thing you are bound to ask yourself is how the other two roots of the cubic equation ﬁt into the picture. In fact, the cubic equation gives (in general) 6 distinct values of a so there is quite a lot of explaining to do, given that the quartic has at most 4 distinct roots.

Solution to question 41
We have
(x2 − ax + b)(x2 + ax + c) = (x2 − ax)(x2 + ax) + b(x2 + ax) + c(x2 − ax) + bc
= x4 + (b + c − a2 )x2 + a(b − c)x + bc

so p = b + c − a2 ,

q = a(b − c) ,

r = bc .

(∗)

To obtain an equation for a in terms of p , q and r we eliminate b and c from (∗) using the identity
(b + c)2 = (b − c)2 + 4bc . This gives (p + a2 )2 = (q/a)2 + 4r which simpliﬁes easily to the given cubic with u replaced by a2 . For the purposes of this equation, all that is required is to verify that setting u = a2 in the given cubic equation works (on use of (∗)), but this is probably no easier than constructing the equation by elimination, as above.
We can easily verify that u = 9 satisﬁes the given cubic by direct substitution.
To solve the quartic equation, the ﬁrst task is to reduce it to the form x4 + px2 + qx + r , which has no term in x3 . This is done by means of a translation. Noting that (y − a)4 = y 4 − 4ay 3 + · · · , we set x = y − 2 . This gives
(x + 2)4 − 8(x + 2)3 + 23(x + 2)2 − 34(x + 2) + 39 . which (not surprisingly) boils down to x4 − x2 − 6x + 15 , so that p = −1, q = −6 and r = 15 . We have already shown that one root of the cubic corresponding to these values is u = 9 . Thus we can achieve the factorisation of the quartic into two quadratic factors by setting a = 3 (or a = −3 — it doesn’t matter which we use) and
1
p + a2 +
2
1 p + a2 − c= 2 b= q
=3
a q =5. a Thus x4 − x2 − 6x + 15 = (x2 − 3x + 3)(x2 + 3x + 5) .
Setting each of the two quadratic factors equal to zero gives

3±i 3
−3 ± i 11 x= and x =
,
2
2
so

7±i 3 y= 2

or

1 ± i 11 y= .
2

Postmortem
Did you work out the relation between the six possible values of a (corresponding to given values of p , q and r) and the roots of the quartic? The point is that the quartic can be written as the product of four linear factors (by the fundamental theorem of algebra) and there are six ways of grouping the linear factors into two quadratic factors. Each way corresponds to a value of a .

Question 42(∗ ∗ ∗)
3 tan θ − tan3 θ
.
1 − 3 tan2 θ

11
.
Given that θ = cos−1 (2/ 5) and 0 < θ < π/2, show that tan 3θ =
2
Hence, or otherwise, ﬁnd all solutions of the equations
11
(i) tan(3 cos−1 x) =
,
2
Show that tan 3θ =

(ii)

cos

2
1
tan−1 y = √ .
3
5

Perhaps three stars is a bit over the top for this question, but I thought it was tricky. Very few candidates got more than half marks in the examination (Paper I, 2001).
It is not hard to see what you have to do, but great care is required when writing out the argument.
The important thing to remember is that if (say) tan A = tan B then you cannot say that A = B ; rather, you must say A = B + nπ .
For the very ﬁrst part, you might wonder where to start. This did not trouble the candidates in the examination: they all quoted the formula for tan(α + β) and worked from there.

Solution to question 42
First part: tan 3θ = where t = tan θ.

tan 2θ + tan θ
2t/(1 − t2 ) + t
3t − t3
=
=
,
1 − tan 2θ tan θ
1 − 2t2 /(1 − t2 )
1 − 3t2

For θ in the given range, we can use Pythagoras’s theorem (think of a triangle with sides 1, 2 and

5) to show that
2
1
1
arccos √ = arcsin √ = arctan
2
5
5
and then set t = 1/2 in the above formula. Alternatively, we can obtain tan θ from tan2 θ = sec2 θ−1 taking the positive square root because 0 < θ < π/2 .
(i) We have tan(3 arccos x) =

11
2
= tan 3 arccos √ ,
2
5

2 so 3 arccos x = 3 arccos √ + nπ. Thus
5
2
1
nπ nπ 2 x = cos arccos( √ + nπ/3 = √ cos
− √ sin
.
3
3
5
5
5
This gives six possibilities:
2
x= √
5

=

2 x = ( √ )(−1)
5

=

2 1
1
3 x = ( √ )( ) − ( √ )(
)
5 2
5 2

3
2 1
1
x = ( √ )( ) − ( √ )(−
)
5 2
5
2

1
1
3
2
) x = ( √ )(− ) − ( √ )(
5
2
5 2

2
1
1
3
x = ( √ )(− ) − ( √ )(−
)
5
2
5
2

=
=
=
=

2

5
2
−√
5

2− 3

2 5

2+ 3

2 5

−2 − 3

2 5

−2 + 3

2 5

(n = 0)
(n = 3)
(n = 1)
(n = −1)
(n = 2)
(n = −2)

(ii) Similarly
11
1
1
arctan( ) cos( arctan y) = cos
3
3
2

1
1
11 arctan y = ± arctan( ) + 2nπ
3
3
2
11
⇒ arctan y = ± arctan( ) + 6nπ
2
11
⇒ y = tan ± arctan( ) + 6nπ
2
11
11
⇒ y = tan ± arctan( ) = ± .
2
2

Question 43(∗ ∗ ∗)
The curve C1 passes through the origin in the x–y plane and its gradient is given by dy 2
= x(1 − x2 )e−x . dx 1
Show that C1 has a minimum point at the origin and a maximum point at 1, 2 e−1 . Find the coordinates of the other stationary point. Give a rough sketch of C1 .

The curve C2 passes through the origin and its gradient is given by dy 3
= x(1 − x2 )e−x . dx Show that C2 has a minimum point at the origin and a maximum point at (1, k) , where k >
(You need not ﬁnd k .)

1
2

e−1 .

No work is required to ﬁnd the x coordinate of the stationary points, but you have to integrate the diﬀerential equation to ﬁnd the y coordinate. For the second part, you cannot integrate the equation — other than numerically, or in terms of rather obscure special functions that you almost certainly haven’t come across, such as the incomplete gamma function, deﬁned by a Γ(x, a) =

tx−1 e−t dt .

0

However, you can obtain an estimate, which is all that is required, by comparing the gradients of
C1 with C2 and thinking of the graphs for −1 x 1 . This is perhaps a bit tricky; an idea that you may well not alight on under examination conditions.

Solution to question 43 dy C1 has stationary points when
= 0, i.e. when x = 0, x = +1 or x = −1. To ﬁnd the y coordinates dx of the stationary points, we integrate the diﬀerential equation, using integration by parts:
2

(1 − x2 )xe−x dx

y =

2

1
= (1 − x2 )(− 2 e−x ) −
2

2

xe−x dx
2

1
= − 1 (1 − x2 )e−x + 2 e−x + const
2

=

1 2 −x2
2x e

,

where we have used the fact that C1 passes through the origin to evaluate the constant of integration.
1
The coordinates of the stationary points are therefore (1, 2 e−1 ), (0, 0) and (−1, 1 e−1 ).
2
One way of classifying the stationary points is to look at the second derivative: d2 y dy 2
2
2
2
= (x − x3 )e−x =⇒
= (1 − 3x2 )e−x − 2x2 (1 − x2 )e−x = (1 − 5x2 + 2x4 )e−x ,
2
dx dx which is positive when x = 0 (indicating a minimum) and negative when x = 1 and x = −1
(indicating maxima).
Since y → 0 as x → ±∞, the sketch is: y 1
2e

−1

1

x

For C2 , the stationary points are again at x = 0 and x = ±1. To classify the stationary points, we calculate the second derivative: d2 y
3
= (1 − 3x2 − 3x3 + 3x5 )e−x ,
2
dx which is positive at x = 0 (a minimum) and negative at x = 1 (a maximum).
Since we cannot integrate this diﬀerential equation explicitly, we must compare C2 with C1 to see whether the value of y at the maxima is indeed greater for C2 than for C1 .
3

2

3

2

For 0 < x < 1, x3 < x2 , so ex < ex and e−x > e−x . Thus the gradient of C2 is greater than the gradient of C1 . Since both curves pass through the origin, we deduce that C2 lies above C1 for
1
0 < x 1 and therefore the maximum point (1, k) on C2 has k > 2 e−1 .
Postmortem
You

Question 44(∗ ∗ ∗)
A spherical loaf of bread is cut into parallel slices of equal thickness. Show that, after any number of the slices have been eaten, the area of crust remaining is proportional to the number of slices remaining. A European ruling decrees that a parallel-sliced spherical loaf can only be referred to as ‘crusty’ if the ratio of volume V (in cubic metres) of bread remaining to area A (in square metres) of crust remaining after any number of slices have been eaten satisﬁes V /A < 1. Show that the radius of a crusty parallel-sliced spherical loaf must be less than 2 2 metres.
3
[The area A and volume V formed by rotating a curve in the x–y plane round the x-axis from x = −a to x = −a + t are given by
A = 2π

−a+t
−a

dy y 1+ dx 2

1
2

dx ,

V =π

−a+t

y 2 dx . ]

−a

The ﬁrst result came as a bit of a surprise to me — though no doubt it is well known. I wondered if it was the only surface of revolution with this property. You might like to think about this.
For the last part, you will need to minimise a ratio as a function of t (the ‘length’ of loaf remaining
— which is of course proportional to the number of slices remaining if the slices are all the same thickness); to ﬁnd the ratio you have to do a couple of integrals. It is this ‘multi-stepping’ that makes the problem diﬃcult (and very diﬀerent from typical A-level questions) rather than any individual step. Solution to question 44
The ﬁrst thing we need is an equation for the surface of a spherical loaf. The obvious choice, especially given the hint at the bottom of the question, is x2 + y 2 = a2 (and z = 0 ), rotated about the x-axis.
If the loaf is cut at a distance t from the end x = a, then the area remaining is

−a+t
−a

dy y 1+ dx 2

1
2

dx = 2π

−a+t
−a

= 2π

−a+t

2

2

(a − x )

1
2

1+

−x

2
1

(a2 − x2 ) 2

1
2

dx

−a

= 2πat .

This is proportional to the length t of remaining loaf, so proportional to the number of slices remaining (if the loaf is evenly sliced).
The remaining volume is π −a+t
−a

y 2 dx = π

−a+t
−a
2

(a2 − x2 )dx = π(a2 x − x3 /3)
3

2

−a+t
−a

= π[(a (−a + t) − (−a + t) /3] − π[(a (−a) − (−a)3 /3] = π(at2 − t3 /3).

As a quick check on the algebra, notice that this is zero when t = 0 and 4πa3 /3 when t = 2a.
Thus V /A = (3at − t2 )/(6a). This is a quadratic curve with zeros at t = 0 and t = 3a, so it has a maximum at t = 3a/2 (by diﬀerentiating or otherwise), and where V /A = 3a/8. Since we require this ratio to be less than one, we must have a < 8/3 metres.

Question 45(∗∗)

A tortoise and a hare have a race to the vegetable patch, a distance X kilometres from the starting post, and back. The tortoise sets oﬀ immediately, at a steady v kilometers per hour. The hare goes to sleep for half an hour and then sets oﬀ at a steady speed V kilometres per hour. The hare overtakes the tortoise half a kilometre from the starting post, and continues on to the vegetable patch, where she has another half an hour’s sleep before setting oﬀ for the return journey at her previous pace. One and quarter kilometres from the vegetable patch, she passes the tortoise, still plodding gallantly and steadily towards the vegetable patch. Show that
V =

10
4X − 9

and ﬁnd v in terms of X.
Find X if the hare arrives back at the starting post one and a half hours after the start of the race.
How long does it take the tortoise to reach the vegetable patch?

You might ﬁnd it helpful to draw a distance–time diagram. You might also ﬁnd it useful to let the times at which the two animals meet be T1 and T2 .

Solution to question 45 x 

L 

L 
L  Tortoise

Hare
 L 

L 
L


L


Et
T

Let the times of the ﬁrst and second meetings be T1 and T2 . Then vT1 = 1 ,
2

1
V (T1 − 2 ) = 1 ,
2

5 vT2 = (X − 4 ),

5
V (T2 − 1) = X + 4 .

The ﬁrst pair and the second pair of equations give, respectively,
1
1

= 1, v V

X − 5/4
X + 5/4
=
+ 1, v V

and hence (ﬁrst eliminating v):
V =

10
,
4X − 9

v=

10
.
4X + 1

The total distance travelled by the hare in 1/2 hour is 2X, so
2X =

10
1
× ,
4X − 9 2

i.e.

8X 2 − 18X − 5 = 0.

Factorizing the quadratic gives two roots −1/4 and 5/2, and we obviously need the positive root.

The speed of the tortoise is given by

v=

10
10
=
4X + 1
11

so the time taken to travel 5/2 kilometres is 11/4 hours.

Question 46(∗)

A single stream of cars, each of width a and exactly in line, is passing along a straight road of breadth b with speed V . The distance between successive cars (i.e. the distance between back of one car and the front of the following car) is c.
A chicken crosses the road in safety at a constant speed u in a straight line making an angle θ with the direction of traﬃc. Show that
Va
u
.
(∗) c sin θ + a cos θ
Show also that if the chicken chooses θ and u so that it crosses the road at the least possible
(constant) speed, it crosses in time b c a
.
+
V a c

I like this question because it relates to (an idealized version) of a situation we have probably all thought about. Once you have visualized it, there are no great diﬃculties. As usual, you have to be careful with the inequalities, though it turns out here that there is no danger of dividing by a negative quantity.
When I arrived at (∗), I noticed that there is another inequality that might be relevant (did you?); but when I looked at the rest of the question I realized I had worried unnecessarily.
Is the time given in the ﬁnal part (which corresponds to the least possible speed) the longest time that the chicken can spend crossing the road?

Solution to question 46
Let t be the time taken to cross the distance a in which the chicken is at risk. Then a = ut sin θ.
For safety, the chicken must choose ut cos θ
V t − c: this is the case where the chicken starts at the near-side rear of one car and just avoids being hit by the far-side front of the next car.
Eliminating t from these two equations gives the required inequality: ut cos θ
=⇒ (u cos θ − V )t a =⇒ (u cos θ − V ) u sin θ
=⇒ au cos θ − aV

=⇒ u(c sin θ + a cos θ)

Vt−c

−c

−c
−cu sin θ

aV

which is the required result
For a given value of θ, the minimum speed satisﬁes u(c sin θ + a cos θ) = aV.
The smallest value of this u is therefore obtained when c sin θ + a cos θ is largest. This can be found by calculus (regard it as a function of θ and diﬀerentiate: the maximum occurs when tan θ = c/a) or by trigonometry: c sin θ + a cos θ = a2 + c2 cos (θ − arctan(c/a))

so the maximum value is a2 + c2 and it occurs when tan θ = c/a.
The time of crossing is b b(c sin θ + a cos θ) b(c + a cot θ) b(c + a2 /c)
=
=
=
. u sin θ
V a sin θ
Va
Va
Postmortem
There is another inequality besides (∗) that you should have noticed: u cos θt

Vt+c

otherwise the chicken will run into the back of the car ahead. This leads to (changing the sign of c and reversing the inequality in (∗)) u(−c sin θ + a cos θ)

aV .

This looks potentially as if it might also give a lower bound on u (if (−c sin θ + a cos θ) < 0). But it turns out that this is not relevant.
Of course, the chicken can take as long as it wants to cross the road just by going at an extremely acute angle at a speed marginally greater than V . (For maximum time, the minimum value of u sin θ, not of u) is required.)

Question 47(∗)

I

2d

  B

A
 

B

B
B
B
B
B
B
B
G
B
EP
B
B f

B f  B


B
B

 f 
B

B


f  B  w f

B

f
 f  f
I

B

f
A fh f
B  f d f  f  f  f  x f
 α

A lorry of weight W stands on a plane inclined at an angle α to the horizontal. Its wheels are a distance 2d apart, and its centre of gravity G is at a distance h from the plane, and halfway between the sides of the lorry. A horizontal force P acts on the lorry through G , as shown.
(i)

If the normal reactions on the lower and higher wheels of the lorry are equal, show that the sum of the frictional forces between the wheels and the ground is zero.

(ii)

If P is such that the lorry does not tip over (but the normal reactions on the lower and higher wheels of the lorry need not be equal), show that
W tan(α − β)

P

W tan(α + β) ,

where tan β = d/h .

There is not much more to this than just resolving forces and taking moments about a suitable point. You might think it a bit odd to have a force that acts horizontally through the centre of gravity of the lorry: it is supposed to be centrifugal. The ﬁrst draft of the question was intended as a model of a lorry going round a bend on a cambered road. The idea was to relate the speed of the lorry to the angle of camber: the speed should be chosen so that there is be no tendency to skid. However, the wording was rather diﬃcult, because motion in a circle is not included in the STEP syllabus, and the modelling part of the question was eventually abandoned.
You can do the second part algebraically (using equations derived from resolving forces and taking moment), but there is a more direct approach.
There was a slight mistake in the wording of the question, which I have not corrected here. Maybe you will spot it in the course of your solution. The diagram is correct.

Solution to question 47
Let the normal reactions at the lower and upper wheels be N1 and N2 , respectively, and let the frictional forces at the lower and upper wheels be F1 and F2 , respectively, (both up the plane).
(i) Taking moments about G, we have
(N1 − N2 )d = (F1 + F2 )h

(∗)

so if N1 = N2 the sum of the frictional forces is zero.
(ii) The condition for the lorry not tip over down the plane is N2 0 , which is the same as saying that the total moment of W and P about the point of contact between the lower wheel and the plane in the clockwise sense is positive. This gives one of the inequalities directly (after a bit of geometry to ﬁnd the distance between the lie of action of the forces and the points of contact of the wheels), and the other follows similarly.
Alternatively, resolving forces perpendicular to the plane and parallel to the plane gives
N1 + N2 = W cos α + P sin α
F1 + F2 = W sin α − P cos α
Multiplying the ﬁrst equation by d and the second by h and adding (using (∗) gives
2dN1 = W (h sin α + d cos α) + P (−h cos α + d sin α)

0,

i.e.
W sin(α + β)

P cos(α + β) .

(∗)

Similarly, the ﬁrst equation by d and the second by h and subtracting (using (∗) gives
2dN2 = W (−h sin α + d cos α) + P (h cos α + d sin α)

0,

i.e
P cos(α − β)

W sin(α − β) .

(∗∗)

Postmortem
To obtain the required inequalities from (∗) and (∗∗) above, you have to divide by a cosine. As always with inequalities, you have to worry about the sign of anything you divide by. In this case, cos(α − β) > 0, since both angles are acute. But what about cos(α + β)? If this is negative, then
(∗∗) is satisﬁed with no constraint and it would be wrong to divide through the inequality without changing the direction of the inequality. Can cos(α + β) be negative? The answer, which was overlooked by the setter (me), is that it is negative if α + β > π/2. In this case, the line of action of P passes between the wheels of the lorry (not, as shown in the diagram, higher than the higher wheel). Thus increasing P will not cause to the lorry to tip up the slope and there is, in this case, no upper limit to P .
An alternative way of obtaining the inequalities of the second part is to work with the resultant of
P and W . The direction of this resultant force (given by tan−1 (W/P )) must be such that the line of action of the force passes between the wheels. The line of action makes an angle of tan−1 (P/W ) with the vertical through G. The lines through G to points of contact of the upper wheel and the lower wheel with the plane make angles of β + α and α − β, respectively, with the vertical through
G. Hence the inequalities in part (ii).

Question 48(∗∗)

In a game of cricket, a ﬁelder is perfectly placed to catch a ball. She watches the ball in ﬂight takes the catch just in front of her eye. The angle between the horizontal and her line of sight at a time t d after the ball is struck is θ. Show that
(tan θ) is constant during the ﬂight. dt The next ball is also struck in the direction of the ﬁelder but at a diﬀerent velocity. In order to be perfectly placed to catch the ball, the ﬁelder runs at constant speed towards the batsman. Assuming d (tan θ) is constant during the ﬂight. that the ground is horizontal, show that again dt Comments
I came across this while looking through some forty year old Entrance Examination papers. Candidates sat these papers in December and were called for interview if they did well enough. This sounds more eﬃcient in terms of everyone’s time than the current system in which all applicants are interviewed, but the disadvantage of implementing it now would be that students would have to sit the examination after only four terms in the sixth form. Forty years ago the situation was diﬀerent because nearly all applicants to Oxford and Cambridge were post A-level.
As with all the very best questions, 9/10 of this question is submerged below the surface. It uses the deepest properties of Newtonian dynamics, and a good understanding of the subject makes the question completely transparent. However, it can still be done without too much trouble by a straightforward approach: take the height above the ﬁelder’s eye-level at which the ball is struck to be h, and later eliminate h by writing it in terms of the ﬂight time, the speed and angle of elevation at which the ball is hit.
The cleverness of this question lies in its use of two fundamental invariances of Newton’s law d2 x
= g. dt2 (†)

The ﬁrst is invariance under what is called time reﬂection symmetry. The equation (†) is invariant under the transformation t → −t, because replacing t by −t does not aﬀect the equation. This means that any given solution can be replaced by one where the projectile goes back along the trajectory. The second is invariance under what are called Galilean transformations. Equation (†) is also invariant under the transformation x → x + vt, where v is an arbitrary constant velocity. This means that we can solve the equation in a frame that moves with constant speed.

Solution to question 48
We take a straightforward approach. Let the height above the ﬁelder’s eye-level at which the ball is struck be h. Let the speed at which the ball is struck be u and the angle which the ball makes with the horizontal be α.
Then, taking x to be the horizontal distance of the ball at time t from the point at which the ball was struck and y to be the height at of the ball at time t above the ﬁelder’s eye-level, we have x = (u cos α)t ,

1 y = h + (u sin α)t − 2 gt2 .

Let d be the horizontal distance of the ﬁelder from the point at which the ball is struck, and let T be the time of ﬂight of the ball. Then d = (u cos α)T ,

0 = h + (u sin α)T − 1 gT 2
2

(∗)

and y d−x h + (u sin α)t − 1 gt2
2
= d − (u cos α)t
1
1
−(u sin α)T + 2 gT 2 + (u sin α)t − 2 gt2
=
(using ∗)
(u cos α)T − (u cos α)t
1
−(u sin α)(T − t) + 2 g(T 2 − t2 )
=
(u cos α)(T − t)
−u sin α + 1 g(T + t)
2
.
(cancelling the factor (T − t))
=
u cos α
This last expression is a polynomial of degree one in t, so its derivative is constant, as required. tan θ =

For the second part, let l be the distance from the ﬁelder’s original position to the point at which she catches the ball. Then l = vT and
1
−u sin α + 2 g(T + t) y y
=
= tan θ =
(l − vt) + d − x v(T − t) + d − x v + u cos α cancelling the factor of (T − t) as before. This again has constant derivative.

Postmortem
The invariance mentioned above can be used to answer the question almost without calculation.
Using time reﬂection symmetry to reverse the trajectory shows that the batsman is completely irrelevant: it only matters that the ﬁelder caught a ball. We just think of the ball being projected from the ﬁelder’s hands (the time-reverse of a catch). Taking her hands as the origin of coordinates, and using u to denote the ﬁnal speed of the ball and α to be the ﬁnal value of θ, we have y =
(u sin α)t − 1 gt2 , x = (u cos α)t and tan θ = (u sin α − 1 gt)/u cos α. The ﬁrst derivative of this
2
2 expression is constant, as before.
We use the Galilean transformation for the second part of the question. Instead of thinking of the ﬁelder running with constant speed v towards the batsman, we can think of the ﬁelder being stationary and the ball having an additional horizontal speed of v. The situation is therefore not changed from that of the ﬁrst part of the question, except that u cos θ should be replaced by v + u cos θ.

Question 49(∗∗)

A rigid straight rod AB has length l and weight W . Its weight per unit length at a distance x from
B is αW l−1 (x/l)α−1 , where α is a constant greater than 1. Show that the centre of mass of the rod is at a distance αl/(α + 1) from B.
The rod is placed with the end A on a rough horizontal ﬂoor and the end B resting against a rough vertical wall. The rod is in a vertical plane at right angles to the plane of the wall and makes an angle of θ with the ﬂoor. The coeﬃcient of friction between the ﬂoor and the rod is µ and the coeﬃcient of friction between the wall and the rod is also µ . Show that, if the equilibrium is limiting at both A and B, then
1 − αµ2 tan θ =
.
(1 + α)µ
Given that α = 3/2 and given also that the rod slides for any θ < π/4 ﬁnd the greatest possible value of µ .
This is a pretty standard situation: a rod leaning against a wall, prevented from slipping by friction at both ends. The only slight variation is that the rod is not uniform; the only eﬀect of this is to alter the position of the centre of gravity through which the weight of the rod acts.
The question is not general, since the coeﬃcient of friction is the same at both ends of the rod. I think that it is a good idea (at least, if you are not working under examination conditions) to set out the general equations (with µA and µB ), and at ﬁrst no limiting friction (with frictional forces
FA and FB ). This will help understand the structure of the equations and highlight the symmetry between the ends of the rod.
You might wonder why the weight per unit length is given in such a complicated way; why not simply kxα ? The reason is to keep the dimensions honest. The factor of W/l appears so that the dimension is clearly weight divided by length. Then x appears divided by l to make it dimensionless.
Finally, the factor of α is required for the total weight of the rod is W
Postmortem
In the original version, the question spoke about limiting friction without saying that it was limiting at both ends. We then wondered whether the friction could be limiting at one end only. The answer is yes. The three equations that govern the system do not determine uniquely the four unknown forces (normal reaction and frictional forces at each end of the rod); extra information is required.
If the angle is also to be determined, two extra pieces of information are required; in this case the information is that the friction is limiting at both ends.
The same lack of determinism occurs in many statics problems. For example, the forces on the legs of a four-legged table cannot all be determined from the three available equations (vertical forces, moment about the x-axis and moments about the y-axis) — only three legs are required for the table to stand (at least,in the case of a three-legged table with an added fourth leg), so extra information is needed to determine all the forces. This extra information might come from considerations of the internal structure of the table; otherwise, the way the table arrived at its current state might matter; for example, whether all four legs were placed on the ﬂoor at the same time.

Solution to question 49
The distance, x, of the centre of mass from the end B of the rod is given by
¯
x=
¯

l
0
l
0

αW l−α xα dx

αW l−α xα−1 dx

=

l
0
l
0

xα dx

xα−1 dx

=

αl
(α + 1)−1 lα+1
=
.
−1 lα
(α)
α+1

The response to the rest of this question should be preceded by an annotated diagram showing all relevant forces on the rod and with G nearer to A than B. This will help to clarify ideas.
FB

T
B ERB
Z
Z
Z
Z
Z
Z
Z G
Z
Z
RA
Z
Z T θ Z
'
FA A c W

The equations that determine the equilibrium are
FB + RA = W

(vertical forces)

FA − RB = 0

(horizontal forces)

RB x sin θ + FB x cos θ = RA (l − x) cos θ − FA (l − x) sin θ
¯
¯
¯
¯

If the rod is about to slip then the frictional and normal forces at A can be speciﬁed as FA = µRA and FB = µRB . Substituting in ﬁrst two of the above questions gives
RA =

W µW µW µ2 W
, RB = 2
, FA = 2
, FB = 2
.
µ2 + 1 µ +1 µ +1 µ +1

Substituting into the third (moments) equation gives the required result.
For the ﬁnal part, setting α = 3/2 gives tan θ =

2 − 3µ2

If the rod slips for any angle less than θ = π/4, then the angle at which limiting friction occurs at both ends must be at least π/4. Therefore
2 − 3µ2

i.e. 3µ2 + 5µ − 2

1

0, or (3µ − 1)(µ + 2) < 0. The greatest possible value of µ is 1/3.

Question 50(∗∗)

N particles P1 , P2 , P3 , . . ., PN with masses m, qm, q 2 m, . . . , q N −1 m, respectively, are at rest at distinct points along a straight line in gravity-free space. The particle P1 is set in motion towards P2 with velocity V and in every subsequent impact the coeﬃcient of restitution is e, where 0 < e < 1.
Show that after the ﬁrst impact the velocities of P1 and P2 are
1 − eq
V
1+q

and

1+e
V,
1+q

respectively.
Show that if q e, then there are exactly N − 1 impacts and that if q = e, then the total loss of kinetic energy after all impacts have occurred is equal to
1
me 1 − eN −1 V 2 .
2

This situation models the toy called ‘Newton’s Cradle’ which consists of four or more heavy metal balls suspended from a frame so that they can swing. At rest, they are in contact in a line. When the ﬁrst ball is raised and let swing, there follows a rather pleasing pattern of impacts. In this case, the coeﬃcient of restitution is nearly 1 and the balls all have the same mass, so, as the ﬁrst displayed formula shows, the impacting ball is reduced to rest by the impact. At the ﬁrst swing of the ball, nothing happens except that the ﬁrst ball is reduced to rest and the last ball swings away.
Note that this is consistent with the balls being separated by a very small amount; what actually happens is that the ball undergoes a small elastic deformation at the impact, and the impulse takes a small amount of time to be transmitted across the ball to the next ball. You can order a Newton’s
Cradle (and lots of other interesting toys and games) online from www.Hawkin.com.

Solution to question 50
Using conservation of momentum and Newton’s law of impact at the ﬁrst collision results in the equations mV = mV1 + mqV2 ,
V1 − V2 = −eV.
Solving these for V1 and V2 gives for the speeds of P1 and P2 after the collision
V1 =
Note that V1 is positive since eq

1 − eq
V
1+q

and

1+e
V.
1+q

V2 =

e2 < 1 , and that V1 < V , and that V < V2 since (1 + e) > (1 + q) .

Similarly, the speeds of P2 and P3 after the next collision are
1 − eq
V2 =
1+q

1 − eq
1+q

1+e
V
1+q

and

1+e
V2 =
1+q

1+e
1+q

2

V.

Note that the speed of P2 is positive and greater than V1 (the speed of P1 ) so there is no further collision between P1 and P2 . Applying this argument at each collision shows that there are exactly
N − 1 collisions: P1 with P2 ; P2 with P3 ; ... , PN −1 with PN .
The speed of Pk after it has hit Pk+1 is similarly
1 − eq
1+q

k−1

1+e
1+q

V,

and the speed of Pk+1 after this collision and before it hits Pk+2 is
1+e
1+q
If q = e,
V1 =

k

V.

1 − e2
V = (1 − e)V
1+e

and similarly the ﬁnal speeds of P2 , . . . , PN −1 are (1 − e)V . The ﬁnal speed of PN is V . Thus the total kinetic energy is
1
1
(m + me + · · · + meN −2 )(1 − e)2 V 2 + meN −1 V 2 =
2
2
=

mV 2 (1 − eN −1 ) mV 2 N −1
(1 − e)2 + e 2
1−e
2 mV 2
(1 − e + eN ),
2

(replacing all the q’s with e’s and summing the geometric progression). Thus the loss of kinetic energy is mV 2
1
mV 2 −
(1 − e + eN ),
2
2 as required.

Question 51(∗∗)

An automated mobile dummy target for gunnery practice is moving anti-clockwise around the circumference of a large circle of radius R in a horizontal plane at a constant angular speed ω. A shell is ﬁred from O, the centre of this circle, with initial speed V and angle of elevation α. Show that if V 2 < gR, then no matter what the value of α, or what vertical plane the shell is ﬁred in, the shell cannot hit the target.
Assume now that V 2 > gR and that the shell hits the target, and let β be the (positive) angle through which the target rotates between the time at which the shell is ﬁred and the time of impact.
Show that β satisﬁes the equation g2 β 4 − 4ω 2 V 2 β 2 + 4R2 ω 4 = 0.
Deduce that there are exactly two possible values of β.
Let β1 and β2 be the possible values of β and let P1 and P2 be the corresponding points of impact. By
2
2
2 2 considering the quantities (β1 + β2 ) and β1 β2 , or otherwise, show that the linear distance between
P1 and P2 is ω√ 2
2R sin
V − Rg . g Comments
The rotation of the target is irrelevant for the ﬁrst part, which contravenes the setters’ law of not introducing information before it is required. In this case, it seemed better to describe the set-up immediately – especially as you are asked for a familiar result.
Remember, when you are considering the roots of the quartic (which is really a quadratic in β 2 ) that β > 0 .
The hint in the last paragraph (‘by considering ...’) is supposed to direct you towards the sum and product of the roots of the quadratic equation; otherwise, you get into some pretty heavy algebra.

Solution to question 51
If the shell hits the target, the range is R, so
R = (V cos α)t,

0 = (V sin α)t − 1 gt2 .
2

Eliminating t gives Rg = V 2 sin(2α) which can be satisﬁed, for some value of α, if and only if
V 2 Rg.
The time of ﬂight is R/(V cos α), and this must also equal the time for the target to rotate through β, i.e. β/ω. Thus cos α = Rω/(βV ). Substituting this into the range equation Rg = 2V 2 sin α cos α gives 1
R2 ω 2 2
2 Rω
.
Rg = 2V
1− 2 2 βV β V
Squaring both sides of the equation, and simplifying, leads to the given quadratic in β 2 . Solving the quadratic using the quadratic formula gives g2 β 2 = 2ω 2 V 2 ±

V 4 − g 2 R2 .

Since V 2 > gR, both the values of β 2 roots are real and positive, but they lead to only two relevant values of β since we only want positive values.
2
2
The linear distance between the two points of impact is 2R sin 1 (β2 − β1 ). Now β1 + β2 = 4ω 2 V 2 /g2
2
2 β 2 = 4R2 ω 4 /g 2 (sum and product of the roots of the quadratic), so and β1 2
2
2
(β2 − β1 )2 = β2 + β1 − 2β1 β2 =

which leads to the given result.

4ω 2 V 2 4Rω 2

, g2 g

Question 52(∗ ∗ ∗)
A particle of unit mass is projected vertically upwards with speed u. At any height x, while the particle is moving upwards, it is found to experience a total force F , due to gravity and air resistance, given by F = αe−βx , where α and β are positive constants. Calculate the energy expended in reaching a given height z. Show that
1
1
F = 2 βv 2 + α − 2 βu2 ,
1
where v is the speed of the particle, and explain why α = 2 βu2 + g, where g is the acceleration due to gravity.

Determine an expression, in terms of y, g and β, for the air resistance experienced by the particle on its downward journey when it is at a distance y below its highest point.

This is rather ingenious. The usual approach to this sort of problem is to say at the outset the particle experiences the force due to gravity and also a force due to air resistance of the form kv 2 .
In this question, the total force is given and you have to deduce the constituent parts of the force, making good use of the basic laws of mechanics (change of energy equals work done, for example).
1
I remember the discussion of the phrasing of the next part of the question ‘explain why α = 2 βu2 +g’.
How best to convey the idea that no real calculation is required, while not suggesting that the answer is trivially obvious? (All that is required, since we are trying to evaluate a constant, is to take a particular value of v.) I’m not sure ‘explain why’ was right; but I can’t think of anything better.

For the last part, you have to integrate the equation of motion, using the standard form of the acceleration used when there is y dependence instead of t dependence.

Solution to question 52
The energy expended in moving a small distance dx is F dx so the energy expended in reaching height z is z α(1 − e−βz ) αe−βx dx ,
i.e.
. β 0
By conservation of energy, the loss of kinetic energy is equal to the work done against the force in reaching this height, so
1 2 1 2 α(1 − e−βz ) u − v =
,
2
2
β where v is the speed at height y and e−βz = 1 −

1
1
β( 2 u2 − 2 v 2 )
.
α

The force experienced at this height is therefore given by
1
1
F ≡ αe−βz = α − β( u2 − v 2 )
2
2 as required.
To obtain an expression for the constant α we just have to evaluate the above equation at a value of v for which F is known. The obvious choice is v = 0: at maximum height the speed is zero so there is no air resistance, the only force on the particle being the force due to gravity. Thus F = −g when v = 0, which gives the required result.
Substituting for α in F gives
F = −g + 1 βv 2 ,
2

so the air resistance to the particle when it is moving at speed v (up or down) is equal to 1 βv 2 .
2
1
For the last part, we just have to express 2 βv 2 in terms of y, β and g.

For the downward journey, let v = v dy
. Then dt dv
= g − 1 βv 2
2
dy

i.e.

v g − 1 βv 2
2

dv =

dy.

Integrating, and using the initial condition that v = 0 when y = 0, gives y=− 1
1
1
1
ln(g − 2 βv 2 ) + ln(g) = − ln β β β Thus 1 βv 2 = g 1 − e−βy , which is therefore the air resistance.
2

1 g − 2 βv 2 g .

Question 53(∗∗)

Hank’s Gold Mine has a very long vertical shaft of height l. A light chain of length l passes over a small smooth light ﬁxed pulley at the top of the shaft. To one end of the chain is attached a bucket
A of negligible mass and to the other a bucket B of mass m.
The system is used to raise ore from the mine as follows. When bucket A is at the top it is ﬁlled with mass 2m of water and bucket B is ﬁlled with mass λm of ore, where 0 < λ < 1. The buckets are then released, so that bucket A descends and bucket B ascends. When bucket B reaches the top both buckets are emptied and released, so that bucket B descends and bucket A ascends. The time to ﬁll and empty the buckets is negligible. Find the time taken from the moment bucket A is released at the top until the ﬁrst time it reaches the top again.
This process goes on for a very long time. Show that, if the greatest amount of ore is to be raised in that time, then λ must satisfy the condition f ′ (λ) = 0 where f(λ) =

λ(1 − λ)1/2
.
(1 − λ)1/2 + (3 + λ)1/2

One way of working out the acceleration of a system of two masses connected by a light string passing over a pulley is to write down the equation of motion of each mass, bearing in mind that the force due to tension will be the same for each mass (it cannot vary along the string, because then the acceleration of some portion of the massless string would be inﬁnite). Then you eliminate the tension.
Alternatively, you can use the equation of motion of the system of two joined masses. The system has inertial mass equal to the sum of the masses (because both masses must accelerate equally) but gravitational mass equal to the diﬀerence of the masses (because the gravitational force on one mass cancels, partially, the gravitational force on the other), so the equation of motion is just (Newton’s law of motion)
(m1 + m2 )a = (m1 − m2 )g.
In the last part, the question says that the process goes on for a very long time. The reason for this is just so that you don’t have to worry about whether it is advantageous to stop in mid-cycle: the eﬀect of doing so after completing many cycles is insigniﬁcant. It is therefore suﬃcient to choose λ so as to maximise the rate at which ore is raised in just one cycle.

Solution to question 53
For bucket A’s downward journey, at acceleration a, the equations of motion for bucket A and bucket
B, respectively, are
−T + 2mg = 2ma,
T − (1 + λ)mg = (1 + λ)ma, where T is the tension in the rope. Eliminating T gives so a = g(1 − λ)/(3 + λ).

When bucket A ascends, the acceleration is g.

The time of descent (using l = 1 at2 ) is 2l/a and the time of ascent is
2
required for one complete cycle is therefore
2l
g

1+

3+λ
1−λ

2l/g. The total time

.

Call this t. The number of round trips in a long time tlong is tlong /t so the amount of ore lifted in time tlong is λmtlong /t.
To maximise this, we have to maximise λ/t with respect to λ, and λ/t is exactly the f (λ) given.
Postmortem
If you are feeling exceptionally energetic, you could try to ﬁnd the value of λ for which f ′ (λ) = 0.
This will eventually lead you to the rather discouraging quartic equation λ4 + 4λ3 + 2λ2 − 8λ + 3 = 0.
A quick look at the quartic equation solver http://www.1728.com/quartic.htm reveals, mysteriously, two real solutions: 0.7037579200493 and 0.5280411216973; ‘mysteriously’ because we were expecting, from the wording of the question, only one solution. In any case, there should be an odd number of turning points between λ = 0 and λ = 1, because f (λ) = 0 when λ = 0 and λ = 1 and f (λ) > 0 for 0 < λ < 1.
It turns out that in order to remove surds, it was necessary to square an expression and this introduced an extra ﬁctitious solution. The larger of the two solutions quoted above is in fact the correct one, as can be ascertained by substituting both back into the original equation.

Question 54(∗)

A smooth cylinder with circular cross-section of radius a is held with its axis horizontal. A light elastic band of unstretched length 2πa and modulus of elasticity λ is wrapped round the circumference of the cylinder, so that it forms a circle in a plane perpendicular to the axis of the cylinder. A particle of mass m is then attached to the rubber band at its lowest point and released from rest.
(i)

Given that the particle falls to a distance 2a below the below the axis of the cylinder, but no further, show that
9πmg
λ= √
.
(3 3 − π)2

(ii)

Given instead that the particle reaches its maximum speed at a distance 2a below the axis of the cylinder, ﬁnd a similar expression for λ .

This question uses the most basic ideas in mechanics, such as conservation of energy. I included it in this selection of STEP questions without realising that the properties of stretched strings are not in the syllabus (which is the syllabus for STEP I and II given in the appendix). However, I didn’t throw it out: it is a nice question and the only two things you need to know about stretched strings are, for a stretched string of natural (i.e. unstretched) length l and extended length l + x with modulus of elasticity λ:
(i) The potential energy stored in the stretched string is
(ii) The tension in the stretched string is

λx
.
l

λx2
2l

Solution to question 54

a

a
2a

m

The length of the extended band is

2πa − 2a arccos(1/2) + 2 3 a so the extension is

−2a arccos(1/2) + 2 3 a ,

i.e.

2a( 3 − π/3) .

(i) By conservation of energy (taking the initial potential energy to be zero), when the particle has fallen a distance d and has speed v,

λ [2a( 3 − π/3)]2
1
2
− mgd .
0 = mv +
2
2
2πa
At the lowest point, v = 0 and d = a, which gives the required answer.
(ii) The (vertical) acceleration (¨) of the particle when it has fallen a distance d is given by y m¨ = 2T cos θ − mg y where sin θ = a/(a + d) and T is the tension. The maximum speed occurs when y = 0 and d = a as
¨
before, i.e. when

2a( 3 − π/3)λ 3
2
− mg = 0
2πa
2

√ so that in this case λ = 3πmg/(3 3 − π) .

Question 55(∗∗)

A wedge of mass M rests on a smooth horizontal surface. The face of the wedge is a smooth plane inclined at an angle α to the horizontal. A particle of mass kM slides down the face of the wedge, starting from rest. At a later time t, the speed V of the wedge, the speed v of the particle and the angle β of the velocity of the particle below the horizontal are as shown in the diagram.

V



Z
Z
Z u
Z
A
Z
A Z
U
A Z v Z
Z
e
Z
e
Z
αZ e eβ e Let y be the vertical distance descended by the particle. Derive the following results, stating in (ii) and (iii) the mechanical principles you use:
(i)

V sin α = v sin(β − α);

(ii)

tan β = (1 + k) tan α;

(iii)

2gy = v 2 (1 + k cos2 β).

Write down a diﬀerential equation for y and hence show that
1
y = 2 g ′ t2

where g′ =

g sin2 β
(1 + k) sin2 α
=
.
(1 + k cos2 β) k + sin2 α

I was surprised how diﬃcult this problem is, compared with a particle on a ﬁxed wedge. I don’t think that there is an easier method: the numbered parts of the question (i) – (iii) seem unavoidable, since they embody the basic principles necessary to solve the problem. For a ﬁxed wedge, the horizontal component of momentum is not conserved, because of the force required to hold the wedge; but (i) and (iii) are just what you would use in the ﬁxed case.
Part (ii) is used only (apart from the very last result) to show that the angle β is constant. This is a bit surprising: it means that the particle moves in a straight line. To put it another way, the net force on the particle is parallel to the direction of motion. Why is this?
Most of the results of the important intermediate steps are given to you, but it is still very good discipline to check that they hold in special cases: for example, k = 0 corresponding to a massless particle or equivalently a ﬁxed wedge; and α = 0 or α = π/2 corresponding to a horizontal or vertical wedge face. You should check that you understand what should happen in these special cases and that your understanding is consistent with the formulae.

Solution to question 55
(i) Particle remains in contact with the plane so components of the velocities of the particle and the wedge perpendicular to the face must be equal.
(ii) Horizontal momentum is conserved, so M V = mv cos β. Substitution for V /v using (i) gives k cos β =

V sin(β − α) sin β cos α − sin α cos β sin β
=
=
=
− cos β v sin α sin α tan α

which leads immediately to the required result. Note that (rather surprisingly) the angle β remains constant in the motion.
(iii) Conservation of energy gives
1
1 mgy = M V 2 + mv 2 ,
2
2

i.e.

1
1
kgy = (kv cos β)2 + kv 2 .
2
2

Thus 2gy = (k cos2 β + 1)v 2 .
The next step is to write v in terms of y to obtain the diﬀerential equation. The vertical component of velocity is v sin β, so dy . v sin β = dt Now using the result of (iii) gives dy = v sin β = dt 1 1
2gy
sin β = (2g′ ) 2 y 2 .
1 + k cos2 β

Now, at last, we use the hard-won result that β, and hence g′ , is constant:
1

y− 2

1
1
1 dy = (2g′ ) 2 ⇒ 2y 2 = (2g′ ) 2 t + constant. dt The constant is zero, since y is the vertical distance descended by the particle. Squaring gives the required answer.
There is one ﬁnal trigonometric manipulation to perform, using (ii) to convert β (the unknown angle) to α (the given angle). The methodical way (not necessarily the quickest) is to write all the trigonometric functions in terms of tan or cot: tan2 β tan2 β sin2 β
=
=
1 + k cos2 β sec2 β + k tan2 β + k + 1
(1 + k)2 tan2 α
(1 + k) tan2 α
(1 + k) sin2 α
(1 + k) sin2 α
=
=
=
=
(1 + k)2 tan2 α + k + 1
(1 + k) tan2 α + 1
(1 + k) sin2 α + cos2 α k sin2 α + 1

g′ ≡

Postmortem
Why is β constant? There are two forces acting on the particle: gravity and the reaction of the wedge on the particle. One compenent of the reaction must exactly balance one component of gravity so as to keep the particle on the wedge; the only remaining components of both forces must be perpendicular to this direction and therefore in direction of motion of the particle. You can appreciate this argument best be assuming that the particle and wedge are instantaneously at rest, which is allowable because the argument is about forces acting at a given instant and the forces are unchanged by addition of a constant velocity (‘Galilean invariance’).

Question 56(∗)

A uniform solid sphere of radius a and mass m is drawn very slowly and without slipping from horizontal ground onto a step of height a/2 by a horizontal force of magnitude F which is always applied to the highest point of the sphere and is always perpendicular to the vertical plane which forms the face of the step. Find the maximum value of F in the motion, and prove that the coeﬃcient

of friction between the sphere and the edge of the step must exceed 1/ 3.

This is quite straightforward once you have realised that very slowly means so slowly that the sphere can be considered to be static at each position. You just have to solve a statics problem with the usual tools: resolving forces and taking moments.

Solution to question 56

F

'

'$ θ &% d d d cd f  d R mg Let the angle between the radius to the point of contact with the step and the downward vertical be θ, as shown. Taking moments about the point of contact with the step gives F a(1 + cos θ) = mga sin θ, so F = mg sin θ = mg tan(θ/2). 1 + cos θ This takes its maximum value when θ is largest, i.e. when the sphere just touches the hoqizontal √ ground. At this position, cos θ = 1/2 and sin θ = 3/2 and Fmax = mg/ 3. At the point of contact, let the frictional force, which is tangent to the sphere, be f and the reaction R (along the radius of the sphere). Taking moments about the centre of the sphere gives F = f , and resolving forces parallel to the radius at the point of contact gives R = F sin θ + mg cos θ. Now sin2 θ 1 − cos2 θ = mg = mg(1 − cos θ) 1 + cos θ 1 + cos θ so R = mg. We therefore need µmg > Fmax . F sin θ = mg Postmortem To be precise about the meaning of slow in this context, you have to compare the dynamic forces connected with the motion of the sphere with the static forces. The motion of the sphere is rotation about the ﬁxed point of contact with the step. If we assume that the centre moves with constant speed, the only extra force due to the motion is an additional reaction at the point of contact with the step. (No tangential force is required to maintain a constant speed in circular motion.) This reaction is centrifugal in nature and so is of the form kmv 2 /d where d is some number which might be hard to calculate. The force calculated above is roughly mg, so the static approximation (‘very slowly’) is valid if v 2 /d ≪ g. Question 57(∗∗) Harry the Calculating Horse will do any mathematical problem I set him, providing the answer is 1, 2, 3 or 4. When I set him a problem, he places a hoof on a large grid consisting of unit squares and his answer is the number of squares partly covered by his hoof. Harry has circular hoofs, of radius 1/4 unit. After many years of collaboration, I suspect that Harry no longer bothers to do the calculations, instead merely placing his hoof on the grid completely at random. I often ask him to divide 4 by 4, but only about 1/4 of his answers are right; I often ask him to add 2 and 2, but disappointingly only about π/16 of his answers are right. Is this consistent with my suspicions? I decide to investigate further by setting Harry many problems, the answers to which are 1, 2, 3, or 4 with equal frequency. If Harry is placing his hoof at random, ﬁnd the expected value of his answers. The average of Harry’s answers turns out to be 2. Should I get a new horse? Comments Hans von Osten, a horse, lived in Berlin around the turn of the last century. He was known far and wide for his ability to solve complex arithmetical problems. Distinguished scientists travelled to Berlin to examine Hans and test his marvellous ability. They would write an equation on a chalkboard and Hans would respond by pawing the ground with his hoof. When Hans reached the answer he would stop. Though he sometimes made errors, his success rate was far higher than would be expected if his answers were random. The accepted verdict was that Hans could do arithmetic. Hans’s reputation as a calculating horse nosedived when an astute scientist simply made sure that neither the person asking the questions nor the audience knew the answers. Hans became an instant failure. His success was based on his ability to sense any change in the audience: a lifted eyebrow, a sigh, a nodding head or the tensing of muscles was enough to stop him from pawing the ground. Anyone who knew the answer was likely to give almost imperceptible clues to the horse. But we shouldn’t overlook Hans’s talents: at least he had terriﬁc examination technique. We are investigating the situation when Harry places his hoof at random, so that the probability of the centre of his hoof lands in any given region is proportional to the area of the region. We therefore move swiftly from a question about probability to a question about areas. You just have to divide a given square into regions. each determined by the number of squares that will be partially covered by Harry’s hoof if its centre lands in the region under consideration. Remember that the areas must add to one, so the most diﬃcult area calculation can be left until last and deduced from the others. To answer the last part properly, you really need to set out a hypothesis testing argument: you will accept the null hypothesis (random hoof placing) if, using the distribution implied by the null hypothesis, the probability of obtaining the given result is greater than some predecided ﬁgure. Obviously, nothing so elaborate was intended here, since it is just the last line of an already long question: just one line would do. Solution to question 57 Harry’s hoof will be completely within exactly one square if he places the centre of his hoof in a square of side 1/2 centred on the centre of any square. The area of any such smaller square is 1/4 square units. His hoof will partially cover exactly four squares if he places the centre in a circle of radius 1/4 centred on any intersection of grid lines. The total area of any one such circle is π/16 square units, which may be thought of as four quarter circles, one in each corner of any given square. His hoof will partially cover exactly two squares if he places the centre in any one of four 1/2 by 1 1/4 rectangles, of total area 4 × in any given square. 8 Otherwise, his hoof will partially cover three squares; the area of this remaining region is 1− 1 π 1 − . 4 16 2 If Harry placed his hoof at random, the probabilities of the diﬀerent outcomes would be equal to the corresponding area calculated above (divided by the total area of the square, which is 1). Thus the data given in the question are consistent with random placement. The expected value given random placements is 1 1 1 1 π +2× +3× 1− − − 4 2 4 2 16 +4× π π =2+ 16 16 The expected value (given that Harry gets all questions right) is (1 + 2 + 3 + 4)/4 = 5/2 . Harry has a less accurate expected value even than the random expected value. He is clearly hopeless and should go. Question 58(∗) In order to get money from a cash dispenser I have to punch in an identiﬁcation number. I have forgotten my identiﬁcation number, but I do know that it is equally likely to be any one of the integers 1, 2, . . . , n. I plan to punch in integers in order until I get the right one. I can do this at the rate of r integers per minute. As soon as I punch in the ﬁrst wrong number, the police will be alerted. The probability that they will arrive within a time t minutes is 1 − e−λt , where λ is a positive constant. If I follow my plan, show that the probability of the police arriving before I get my money is n 1 − e−λ(k−1)/r . n k=1 Simplify the sum. On past experience, I know that I will be so ﬂustered that I will just punch in possible integers at random, without noticing which I have already tried. Show that the probability of the police arriving before I get my money is 1 1− . n − (n − 1)e−λ/r Comments This was originally about getting money from cash dispenser using a stolen card, but it was decided that STEP questions should not be immoral. Hence the rather more improbable scenario. The exponential distribution, here governing the police arrival time is often used for failure of equipment (light-bulbs, etc). It has the useful property (not used here) that the reliability for a light bulb (here the probability of the police not coming within a certain time period of duration t doesn’t depend on which time period you choose; i.e. given that the light bulb has survived to time T , the probability of it surviving until time T + t is independent of T (which doesn’t seem very suitable for light bulbs). This is referred to as the memoryless property. To simplify the sums, you have to recognise a geometric series where the common ratio of terms is an exponential; not diﬃcult, but easy to miss the ﬁrst time you see it. Solution to question 58 The probability that I get the right number on the kth go is 1/n . The time taken to key in k integers is (k − 1)/r . The probability that the police arrive before this is 1 − e−λ(k−1)/r so the total probability that the police arrive before I get my money is n k=1 We have n k=1 1 × 1 − e−λ(k−1)/r n 1 × 1 − e−λ(k−1)/r = 1 − n . n e−λ(k−1)/r 1 − e−λn/r , =1− n n(1 − e−λ/r ) k=1 since the last part of the sum is a geometric progression with common ratio e−λ/r . This time, the probability of getting my money on the kth go is 1/n times the probability of not having punched in the correct integer in the preceding k − 1 turns: P(money on kth go) = 1 × n n−1 n k−1 . Thus the probability of the police arriving before I get my money is k=1 1 × n n−1 n k−1 × 1 − e−λ(k−1)/r = 1 1 1 1 . × − × −λ/r (n − 1)/n n 1 − (n − 1)/n n 1 − e Postmortem For each part, you have to calculate the probablity of the ﬁrst success occuring on the kth go. You have to remember to multiply the probability of success on this go by the probability of failure on the preceding k − 1 goes — a very typical idea in this sort of probability question. After that, the question is really just algebra. STEP questions on this sort of material nearly always involve signiﬁcant algebra or calculus. Car batteries as well as light bulbs often crop up as examples of objects to which exponential reliability can be applied. Suppose your car battery has a guarantee of three years, and your car is completely destroyed after two years. Your insurance company oﬀers to give you one third of the price of the battery. What do you think they would say to your counterclaim for the full price, based on the fact that, given it had lasted two years, the exponential model says that it would have had an expected further three years life in it? Question 59(∗) Four students, Arthur, Bertha, Chandra and Delilah, exchange gossip. When Arthur hears a rumour, he tells it to one of the other three without saying who told it to him. He decides whom to tell by choosing at random amongst the other three, omitting the ones that he knows have already heard the rumour. When Bertha, Chandra or Delilah hear a rumour, they behave in exactly the same way (even if they have already heard it themselves). The rumour stops being passed round when it is heard by a student who knows that the other three already have aready heard it. Arthur starts a rumour and tells it to Chandra. By means of a tree diagram, or otherwise, show that the probability that Arthur rehears it is 3/4. Find also the probability that Bertha hears it twice and the probability that Chandra hears it twice. Comments This is pretty straightforward, though you have to marshall the information carefully. Arthur, the hero of this question, is named after one of my sons, who was born in the year this was set. I can’t remember who Bertha and co are. Next time this comes up, you can expect ﬁve students since I now have another son called Edward. But don’t hold your breath: the tree diagram is so horrible to produce in LaTeX (the mathematical word-processing language in which these papers, and the solutions, are set) that such questions are rare. Solution to question 59 The probability tree is B (1) C A (2) B (3) D B A 1 2 1 2 A D A C (4) C C (5) D (6) D (7) B D A B C (8) 1 Each of the branches has probablility 1 , so the probability of each outcome is 8 . 2 A rehears the rumour on 1, 2, 3, 5, 6, and 7, so with probability B hears the rumour twice on 1 and 3, so with probability 2 8 6 8 . . C hears the rumour twice on 2, 4, 5 and 8, so with probability 4 8 . Postmortem I’ve nothing to add here. The question has not much underlying interest. So here is the source code, mentioned in the comment on the previous page, for the tree diagram: \addtolength{\tabcolsep}{-2.1 mm} \newcommand{\lin}{\rule[1mm]{15mm}{0.2mm}} \begin{tabular}{lllllllllllllllll} &&&&&&&&&&B&\hspace{1cm}(1) \\ &&&&&&A&\lin&D&\lin&\vline \\ &&&&&&\vline&&&&C &\hspace {1cm}(2)\\ &&&&B&\lin&\vline \\ &&&&\vline&&\vline&&A&\lin&B &\hspace{1cm}(3)\\\ &&&\mbox{\hspace{1.2cm}$\frac12$} &\vline&&D&\lin&\vline\\ &&&&\vline&&&&C&&&\hspace{1cm}(4)\\ A&\lin&C&\lin&\vline\\ &&&&\vline&&&&&&C&\hspace{1cm}(5)\\ &&&\mbox{\hspace{1.2cm}$\frac12\$}&\vline&&A&\lin&B&\lin&\vline \\
&&&&\vline&&\vline&&&&D &\hspace{1cm}(6)\\
&&&&D&\lin&\vline&\\
&&&&&&\vline&&A&\lin&D&\hspace{1cm}(7)\\
&&&&&&B&\lin&\vline\\
&&&&&&&&C&&&\hspace{1cm}(8)\\
\end{tabular}

Question 60(∗)

Four students, one of whom is a mathematician, take turns at washing up over a long period of time.
The number of plates broken by any student in this time obeys a Poisson distribution, the probability of any given student breaking n plates being e−λ λn /n! for some ﬁxed constant λ, independent of the number of breakages by other students. Given that ﬁve plates are broken, ﬁnd the probability that three or more were broken by the mathematician.

The way this is set up, it is largely a counting exercise (but see postmortem). To start with, you work out the probability of ﬁve breakages, then follow that with the probability that the mathematician broke three of more plates. You need to calculate the number of ways that 5 plates can be shared amongst four students, for which you have to consider each partition of the number 5 and the number of diﬀerent ways it can arise.

Solution to question 60
First we work out the probability of ﬁve breakages.
Let P(5, 0, 0, 0) denote the probability that student A breaks 5 plates and students B, C and D break no plates. Then
P(5, 0, 0, 0) = (Prob. of breaking 5) × (Prob of breaking none)3 =

e−λ λ5 −λ e 5!

3

=

λ5 e−4λ
.
5!

The probability that one student breaks all the plates is P(5, 0, 0, 0) + P(0, 5, 0, 0) + P(0, 0, 5, 0) +
P(0, 0, 0, 5), i.e. (4/5!) × λ5 e−4λ (the factor 4 because it could be any one of the four students).
Let P(4, 1, 0, 0) denote the probability that student A breaks 4 plates. student B breaks one plate and students C and D break no plate. Then
P(4, 1, 0, 0) = (Prob. of breaking 4) × (Prob. of breaking one) × (Prob of breaking none)2 e−λ λ4 e−λ λ −λ 2 λ5 e−4λ e =
.
=
4!
1!
4!
The probability that any one student breaks four plates and any other student breaks one plate is therefore 12 × λ5 e−4λ /4! .

Considering P(3, 2, 0, 0), P(3, 1, 1, 0), P(2, 2, 1, 0) and P(2, 1, 1, 1) in turn, together with the number of diﬀerent ways these probabilities can occur, shows that the probability of ﬁve breakages is λ5 e−4λ

12
12
12
12
4
4
+
+
+
+
+
5! 4! 1! 3! 2! 3! 1! 1! 2! 2! 1! 2! 1! 1! 1!

= λ5 e−4λ

1024
5!

.

The probability that the mathematician breaks three or more is P(5, 0, 0, 0) + 3P(4, 1, 0, 0) +
3P(3, 2, 0, 0) + 3P(3, 1, 1, 0), i.e. λ5 e−4λ

4
3
3
4
+
+
+
5! 4! 1! 3! 2! 3! 1! 1!

= λ5 e−4λ

106
5!

.

The probability that the mathematician breaks three or more, given that ﬁve are broken is therefore
106/1024 .
Postmortem
If you did the question by the method suggested above, you will probably be wondering about the answer: why is it independent of λ; and why is the denominator 45 . You will quickly decide that the
Poisson distribution was a red herring (though I promise you that it was not an intentional herring), since there is no trace of the distribution in the answer.
The denominator is pretty suggestive. A completely diﬀerent approach is as follows, It is clear that
1
the probability that the mathematician breaks any given plate is 4 . The probability that he or she breaks k plates is therefore binomial:
5
k

1
4

k

3
4

5−k

and adding the cases k = 3, k = 4 and k = 5 gives rather rapidly
90 + 15 + 1
.
1024

Question 61(∗∗)

The national lottery of Ruritania is based on the positive integers from 1 to N , where N is very large and ﬁxed. Tickets cost £1 each. For each ticket purchased, the punter (i.e. the purchaser) chooses a number from 1 to N . The winning number is chosen at random, and the jackpot is shared equally amongst those punters who chose the winning number.
A syndicate decides to buy N tickets, choosing every number once to be sure of winning a share of the jackpot. The total number of tickets purchased in this draw is 3.8N and the jackpot is £W .
Assuming that the non-syndicate punters choose their numbers independently and at random, ﬁnd the most probable number of winning tickets and show that the expected net loss of the syndicate is approximately
5 1 − e−2.8
N −
W .
14

This is a binomial distribution problem: the probability that n out of m punters choose the winning ticket is m n m−n p q n where here m = 2.8N , p = 1/N and q = 1 − p. It is clear that an approximation to the binomial distribution is expected (for example, the question uses the word ‘approximately’ and you have to think about how an approximation might arise); and the presence of the exponential in the given result gives a pretty broad hint that it should be a Poisson distribution — the use of which has to be justiﬁed. One can expect the Poisson approximation to work when the number of trials (call it
m) is large (e.g. m > 150) and when np ≈ npq, i.e. the mean is roughly equal to the variance (since these are equal for the Poisson distribution) — so q ≈ 1.
You can’t ﬁnd the most probable number of winning tickets by diﬀerentiation (unless you fancy diﬀerentiating the factorial x!); instead, you must look at the ratios of consecutive terms and see when these turn from being greater than one to less than one.

Solution to question 61
Let X be the random variable whose value is the number of winning tickets out of the 2.8N tickets purchased by the non-syndicate punters. Then X ∼ Poisson(2.8) .
Let pj = P(X = j) . Then

pj =

(2.8)j e−2.8 pj+1 2.8

=
.
j! pj j+1

This fraction is greater than 1 if j < 1.8, so the most probable number of winning tickets by non-syndicate punters is 2. Overall, the most probable number of winning tickets is 3.
The expected winnings of the syndicate is
1
1
W p0 + W p1 + W p2 + · · ·
2
3

1
1
1
1
× (2.8) + × (2.8)2 + · · ·
2 1!
3 2!
2.8 − 1 e ,
= e−2.8 W
2.8

= e−2.8 W

1+

so the expected loss is as given.
Postmortem
This is rather interesting. You might wonder whether it would be worth borrowing the money to buy every single lottery combination, in order to win a share of the jackpot. Clearly, the people who run the lottery have to think about this sort of thing.
Suppose the number of tickets sold, excluding the N that we plan to buy, is expected to be kN .
Suppose also that a fraction α of the total is paid out in the jackpot. Then the above formula becomes 1 − e−k α(k + 1) .
N 1− k If k is very small (take k = 0), we lose N (1 − α) (obviously). If k is large, k = 3 say, the exponential can be ignored and we lose N (1 − α(k + 1)/k). If k ≫ 1, this becomes N (1 − α) again. In between, there is a value of k that, for each ﬁxed α, gives a minimum loss (which may be a gain if α is close to 1).
Note how informative it is to have k rather than 2.8; the numerical value was chosen in the question to model roughly that lottery system in the UK.
Having gone back to this solution after a break, I am now wondering about the use of the Poisson approximation. Of course, the set-up (large m, small p) begs us to approximate, but did we need to? Certainly not for the ﬁrst result, since we can just as well look at the ratio of two terms of the
Binomial distribution as at two terms of the Poisson distribution. Try it; the result is of course the same. The second part is more diﬃcult. What we want is the expectation of 1/(n + 1), and this turns out to be a diﬃcult sum using the Binomial distribution (in fact, it can only be expressed in terms of a hypergeometric function, which would then have to be approximated to get a less obscure answer).
It seems to me that the solution above is therefore a bit unsatisfactory. It would surely have been better to work with the exact distribution until it was necessary to approximate, even though one knows that the approximation is so good that the answers would be the same. We are, after all, mathematicians and not engineers.

Question 62(∗)

I have k diﬀerent keys on my key ring. When I come home at night I try one key after another until
I ﬁnd the key that ﬁts my front door. What is the probability that I ﬁnd the correct key, for the ﬁrst time, on the nth attempt in each of the following three cases?
(i)

At each attempt, I choose a key that I have not tried before, each choice being equally likely. (ii)

At each attempt, I choose a key from all my keys, each of the k choices being equally likely.

(iii)

At the ﬁrst attempt, I choose from all my keys, each of the k choices being equally likely.
At each subsequent attempt, I choose from the keys that I did not try at the previous attempt, each of the k − 1 choices being equally likely.

This is very easy, and would really be far too easy if the answers were given.
You should set out your argument clearly and concisely, because if you come up with the wrong answer and an inadequate explanation you will not get many marks. Even if you write down the correct answer you may not get the marks if your explanation is inadequate.
You should of course run the usual checks on your answers. Do they lie in the range 0 p 1? If you sum over all outcomes, do you get 1? (The latter is a good and not diﬃcult exercise; I insist that you try it.)
You will probably be struck by the simplicity of the answer to part (i), after simpliﬁcation. Is there an easy way of thinking about it (or, if you did it the easy way, is there a hard way)?

Solution to question 62

(i)
P(ﬁnding correct key, for the ﬁrst time, on nth attempt)
=P(fail ﬁrst, fail second, ... , fail (n − 1)th, succeed nth)

 1

 k

k−1 k−2 k − (n − 1)
1
1
=
 k × k − 1 × · · · × k − (n − 2) × k − (n − 1) = k

 0

for

n = 1;

for

2

for

n > k.

n

k;

(ii)
P(ﬁnd correct key, for the ﬁrst time, on nth attempt)

=P(fail ﬁrst, fail second, ... , succeed nth) =

k−1 k n−1

1
.
k

(iii)
P(ﬁnd correct key, for the ﬁrst time, on nth attempt)
=P(fail ﬁrst, fail second, ... , succeed nth)

 1

for n = 1;

 k
=
 k − 2 n−2 1

 for n 2.

k−1 k Postmortem

The above solution is the ‘hard way’. The easy way is to consider the keys all laid out in a row, instead of being picked sequentially. Exactly one of the keys is the correct one, which is equally likely to be any of the keys in the row, so has a probability of 1/k of being in any given position.
Clearly, this corresponds exactly to picking them one by one.
I gave the hard solution above, because I think most people will be drawn into doing it this way by the phrasing of the question: keys are picked one by one and tried before going on to the next.

Question 63(∗∗)

Tabulated values of Φ(·), the cumulative distribution function of a standard normal variable, should not be used in this question.
Henry the commuter lives in Cambridge and his working day starts at his oﬃce in London at 0900.
He catches the 0715 train to King’s Cross with probability p, or the 0720 to Liverpool Street with probability 1 − p. Measured in minutes, journey times for the ﬁrst train are N (55, 25) and for the second are N (65, 16). Journey times from King’s Cross and Liverpool Street to his oﬃce are
N (30, 144) and N (25, 9), respectively. Show that Henry is more likely to be late for work if he catches the ﬁrst train.
Henry makes M journeys, where M is large. Writing A for 1 − Φ(20/13) and B for 1 − Φ(2), ﬁnd, in terms of A, B, M and p, the expected number, L, of times that Henry will be late and show that for all possible values of p,
BM L AM.
Henry noted that in 3/5 of the occasions when he was late, he had caught the King’s Cross train.
Obtain an estimate of p in terms of A and B.
[Note: A random variable is said to be N µ, σ 2 if it has a normal distribution with mean µ and variance σ 2 .]

This is impossible unless you know the following result:
2
If X1 and X2 are independent and normally distributed according to X1 ∼ N (µ1 , σ1 and X2 ∼
2
2
2
N (µ2 , σ2 , then X1 + X2 is also normally distributed and X1 + X2 ∼ N (µ1 + µ2 , σ1 + σ2 ).

Even if you didn’t know this result, you do now and it should not be diﬃcult to complete the ﬁrst parts question.
For the last part, you need to know something about conditional probability, namely that
P(A B) =

P(B ∩ A)
P(B)

which is intuitively and can easily be understood in terms of Venn diagrams. The denominator is essentially a normalising constant. The formula may be taken as the deﬁnition of conditional probability on the left hand side.

Solution to question 63
Let T1 be the random variable representing the total journey time via Kings Cross, so that
T1 ∼ N (55 + 30, 25 + 144) = N (85, 169), using the theorem mentioned on the previous page, and and let T2 be the random variable representing the total journey time via Liverpool Street, so that
T2 ∼ N (65 + 25, 16 + 9) = N (90, 25).
Then the probabilities of being late are, respectively, P(T1 > 105) and P(T2 > 100), i.e. 1−Φ(20/13) and 1 − Φ(2). Note that Φ(20/13) < Φ(2).
We have

L = [pA + (1 − p)B]M = [B + (A − B)p]M.
L increases as p increases, since A > B, hence given inequalities corresponding to p = 0 and p = 1.
We have
P(Kings Cross given late) =

P(Late and Kings Cross)
P(Late)

An estimate for p (all it p) is therefore given by
˜
3

p
=
,
5
A˜ + B(1 − p) p ˜ so p = 3B/(2A + 3B).
˜
Postmortem
The result mentioned in the comment on the previous page is just the sort of thing you should try to prove yourself rather than take on trust. Unfortunately, such results in probability tend to be pretty hard to prove. You can also prove it fairly easily using generating functions (which are not in the syllabus for STEP I and II). You can also prove it from ﬁrst principles, but that would be diﬃcult for you but not impossible because of the awkward integrals involved.

Question 64(∗)

A group of biologists attempts to estimate the magnitude, N , of an island population of voles
(Microtus agrestis). Accordingly, the biologists capture a random sample of 200 voles, mark them and release them. A second random sample of 200 voles is then taken of which 11 are found to be marked. Show that the probability, pN , of this occurrence is given by
2

pN

(N − 200)!
=k
,
N !(N − 389)!

where k is independent of N .
The biologists then estimate N by calculating the value of N for which pN is a maximum. Find this estimate. All unmarked voles in the second sample are marked and then the entire sample is released. Subsequently a third random sample of 200 voles is taken. Using your estimate for N , write down the probability that this sample contains exactly j marked voles, leaving your answer in terms of binomial coeﬃcients.
Deduce that

200 j=0 389 j 3247
200 − j

=

3636
200

.

This is really just an exercise in combinations. (Recall that a permutation is a reordering of a set of objects, and a combination is a selection of a subset from a set.) You assume that you are equally likely to choose any given subset of the same size, so that the probability of a set of speciﬁc composition is the number of ways of choosing a set of that composition divided by the total number of ways of choosing any set of the same size. Of course, you are assuming that the voles are indistinguishable, except for the marks made by the biologists.
Maximising a discrete (not a continuous) function of N came up on one of the previous questions: you have to compare adjacent terms.

Solution to question 64
200 out of N voles are marked, so pN is just the number of ways of choosing 11 from 200 and 189 from N − 200 divided by the number of ways of choosing 200 from N : pN =

200
11

N − 200
189
N
200

so k= =

(N − 200)!
200!
11! 189! 189! (N − 389)!
,
N!
200! (N − 200)!

(200!)2
.
11! (189!)2

At the maximum value, pN ≈ pN −1 , i.e.
(N − 200)2
≈ 1,
N (N − 389) which gives N ≈ 2002 /11 ≈ 3636 (just divide 40000 by 11).

At the third sample, there are 389 marked voles and an estimated 3636 − 389 = 3247 unmarked voles. Hence
389
3247 j 200 − j
.
P(exactly j marked voles) =
3636
200
The ﬁnal part follows immediately, using
200

P(exactly j marked voles) = 1. j=0 Postmortem
Adding the Latin name of the species was a nice touch, I thought (not my idea); it adds an air of verisimilitude to the problem.
I suppose that this might be the basis of a method of estimating the population of voles — rather clever really. I don’t know how well it works in practice though. The assumption mentioned earlier, that picking one set of voles of size 200 is just as likely as picking any other set, surely relies on perfect mixing of the marked voles, which would be rather diﬃcult to achieve (especially as female voles can be highly territorial).

Question 65()

A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio,
R, of the length of the shorter piece to the length of the longer piece is less than r, where r is a given positive number.
Find the probability density function for R, and calculate the mean and variance of R.

Solution to question 65
Let X be the distance of the break from one end of the stick, and assume that X ℓ/2, where ℓ is the length of the stick (if this assumption does not hold, then we could measure from the other end of the stick). Then x P(X < x) =
ℓ/2
Now R =

X
, by deﬁnition, so
ℓ−X

X=

ℓR
.
1+R

The cumulative distribution function for R is
P(R < r) = P

X −1/3? Find E(X) and show that Var(X) = 1/18.

The experiment is repeated independently 800 times generating the larger roots X1 , X2 , . . . , X800 .
If
Y = X1 + X2 + · · · + X800 . ﬁnd an approximate value for K such that
P(Y

K) = 0.08.

Solution to question 66
The larger root is given by X = −1 +

1 − T , so X > −1/3 if T < 5/9. Thus P(X > −1/3) = 5/9.

For the repeated experiment we use the normal distribution. First calcalate the expectation and variance: 1

1
(−1 + 1 − t)dt = − ;
E(X) =
3
0
1

E(X 2 ) =

(−1 +
0

1 − t)2 dt =

1 1
1
1
; Var(X) = − =
.
6
6 9
18

Using standard results for the mean of a sum of random variables (always equal to the sum of the means) and the variance (equal to the sum of the variances if the variables are uncorrelated; independent implies uncorrelated), we have µ = E(Y ) = −
Now P (Y

K) = P(Z

800 and 3

σ 2 = Var(Y ) =

(K − µ)/σ), where Z is standard normal.

For a standard normal distribution, we have
P (Z

−1.40) = 0.08076

so we want K to satisfy
K + 800/3
= −1.4 .
20/3
Thus K = −276.

800
.
18

STEP Syllabus
General Introduction
There are two syllabuses, one for Mathematics I and Mathematics II, the other for Mathematics
III. The syllabus for Mathematics I and Mathematics II is based on a single subject Mathematics
Advanced GCE. Questions on Mathematics II are intended to be more challenging than questions on Mathematics I. The syllabus for Mathematics III is wider.
In designing the syllabuses, the speciﬁcations for all the UK Advanced GCE examinations, the
Scottish Advanced Higher and the International Baccalauriat were consulted.
Each of Mathematics I, II and III will be a 3-hour paper divided into three sections as follows:
Section A (Pure Mathematics) eight questions
Section B (Mechanics) three questions
Section C (Probability and Statistics)three questions
All questions will carry the same weight. Candidates will be assessed on the six questions best answered; no restriction will be placed on the number of questions that may be attempted from any section. Normally, a candidate who answers at least four questions well will be awarded a grade 1.
The marking scheme for each question will be designed to reward candidates who make good progress towards a complete solution.

Syllabuses
The syllabuses given below are for the guidance of both examiners and candidates. The following points should be noted.
1. Questions may test candidates’ ability to apply mathematical knowledge in novel and unfamiliar ways.
2. Solutions will frequently require insight, ingenuity, persistence and the ability to work through substantial sequences of algebraic manipulation.
3. Some questions may be largely independent of particular syllabus topics, being designed to test general mathematical potential.
4. Questions will often require knowledge of several diﬀerent syllabus topics.
5. Examiners will aim to set questions on a wide range of topics, but it is not guaranteed that every topic will be examined every year.
6. Questions may be set that require knowledge of elementary topics, such as basic geometry, mensuration (areas of triangles, volume of sphere, etc), sine and cosine formulae for triangles, rational and irrational numbers.

Formula booklets and calculators
Candidates may use an Advanced GCE booklet of standard formulae, deﬁnitions and statistical tables. The inter-Board agreed list of mathematical notation will be followed. Calculators are not permitted (or required).

MATHEMATICS I (9465) and MATHEMATICS II (9470)
Section A: Pure Mathematics
This section comprises the Advanced GCE common core (typically, modules C1 — C4 of an Advanced GCE speciﬁcation) broadly interpreted, together with a few additional items ∗enclosed in asterisks∗. Speciﬁcation

Notes

General
Mathematical vocabulary and notation
Methods of proof

including: equivalent to; necessary and suﬃcient; if and only if; ⇒ ; ⇔ ; ≡ . including proof by contradiction and disproof by counterexample; ∗including, in simple cases, proof by induction∗.

Algebra
Indices and surds

including rationalising denominators.

including proving positivity by completing a square. including knowledge of the general term; n! n
.
notation:
=
r r! (n − r)!

The expansion for (a + b)n

Algebraic operations on polynomials and rational functions.

including factorisation, the factor theorem, the remainder theorem; including understanding that, for example, if x3 + bx2 + cx + d ≡ (x − α)(x − β)(x − γ),

Partial fractions
Sequences and series

The binomial series for (1 + x)k , where k is a rational number.
Arithmetic and geometric series
∗Inequalities∗

then d = −αβγ. including denominators with a repeated or quadratic factor. including use of, for example, an+1 = f (an ) or an+1 = f (an , an−1 ); including understanding of the terms convergent, divergent and periodic in simple cases; including use of n k to obtain related sums. k=1 including understanding of the condition |x| < 1 . including sums to inﬁnity and conditions for convergence, where appropriate. x 1 including solution of, eg, a−x > x−b ; including simple inequalities involving the modulus function; including the solution of simultaneous inequalities by graphical means.

Functions
Domain, range, composition, inverse

including use of functional notation such as y = f (ax + b) , x = f −1 (y) and z = f g(x) .

Increasing and decreasing functions

both the common usages of the term ‘increasing’
(i.e. x > y ⇒ f (x) > f (y) or x > y ⇒ f (x) f (y) ) will be acceptable. including x = ay ⇔ y = loga x , x = ey ⇔ y = ln x ;
∗including the exponential series

Exponentials and logarithms

The eﬀect of simple transformations

ex = 1 + x + · · · + xn /n! + · · · .

such as y = af (bx + c) + d.

The modulus function.
Location of roots of f (x) = 0 by considering changes of sign of f (x) .
Approximate solution of equations using simple iterative methods.
Curve sketching*
General curve sketching

including use of symmetry, tranformations, behaviour as x → ±∞, points or regions where the function is undeﬁned, turning points, asymptotes parallel to the axes.

Trigonometry
Radian measure, arc length of a circle, area of a segment.
Trigonometric functions

Double angle formulae

including knowledge of standard values, such as tan(π/4) , sin 30◦ ; including identities such as sec2 φ − tan2 φ = 1 ; including application to geometric problems in two and three dimensions. including their use in calculating, eg, tan(π/8) .

Formulae for sin(A ± B) and cos(A ± B)

including their use in solving equations such as

Inverse trigonometric functions

a cos θ + b sin θ = c. deﬁnitions including domains and ranges; notation: arctan θ , etc

Coordinate geometry
Straight lines in two-dimensions

including the equation of a line through two given points, or through a given point and parallel to a given line or through a given point and perpendicular to a given line; including ﬁnding a point which divides a segment in a given ratio.

Circles

using the general form (x − a)2 + (y − b)2 = R2 ; including points of intersection of circles and lines. Cartesian and parametric equations of curves and conversion between the two forms. Calculus
Interpretation of a derivative as a limit and as a rate of change

including knowledge of both notations f ′ (x) and dy dx .

Diﬀerentiation of standard functions

including algebraic expressions, trigonometric
(but not inverse trigonometric) functions, exponential and log functions.

Diﬀerentiation of composite functions, products and quotients and functions deﬁned implicitly.
Higher derivatives

including knowledge of both notations f ′′ (x) and d2 y
;
dx2 dn y including knowledge of the notation dxn .

Applications of diﬀerentiation to gradients, tangents and normals, stationary points, increasing and decreasing functions
Integration as reverse of diﬀerentiation

including ﬁnding maxima and minima which are not stationary points; including classiﬁcation of stationary points using the second derivative.

Integral as area under a curve

including area between two graphs; including approximation of integral by the rectangle and trapezium rules.

Volume within a surface of revolution
Knowledge and use of standard integrals

Deﬁnite integrals

Integration by parts and by substitution

including the forms f ′ (g(x))g′ (x)dx and f ′ (x)/f (x)dx ; including transformation of an integrand into standard (or some given) form; including use of partial fractions; not including knowledge of integrals involving inverse trigonometric functions.
∗including calculation, without justiﬁcation, of

simple improper integrals such as 0 e−x dx and
1 −1
2 dx (if required, information such as the
0 x behaviour of xe−x as x → ∞ or of x ln x as x → 0 will be given).∗ including understanding their relationship with diﬀerentiation of product and of a composite function; including application to ln xdx .

Formulation and solution of diﬀerential equations formulation of ﬁrst order equations; solution in the case of a separable equation or by some other method given in the question.

Vectors
Vectors in two and three dimensions
Magnitude of a vector
Vector addition and multiplication by scalars Position vectors

including use of column vector and i, j, k notation. including the idea of a unit vector. including geometrical interpretations. including application to geometrical problems.

The distance between two points.
Vector equations of lines

The scalar product

including the ﬁnding the intersection of two lines; understanding the notion of skew lines
(knowledge of shortest distance between skew lines is not required). including its use for calculating the angle between two vectors.

Section B: Mechanics
This section is roughly equivalent to two GCE modules M1 and M2; candidates who have only studied one module of mechanics are likely to lack both the knowledge and the experience of mechanics required to attempt the questions. Questions may involve any of the material in the Pure
Mathematics syllabus.
Force as a vector

Centre of mass

Equilibrium of a rigid body or several rigid bodies in contact

Kinematics of a particle in a plane

Energy (kinetic and potential), work and power Collisions of particles

Newton’s ﬁrst and second laws of motion

including resultant of several forces acting at a point and the triangle or polygon of forces; including equilibrium of a particle; forces include weight, reaction, tension and friction. including obtaining the centre of mass of a system of particles, of a simple uniform rigid body (possible composite) and, in simple cases, of non-uniform body by integration. including use of moment of a force; for example, a ladder leaning against a wall or on a movable cylinder; including investigation of whether equilibrium is broken by sliding, toppling or rolling; including use of Newton’s third law; excluding questions involving frameworks. including the case when velocity or acceleration depends on time (but excluding use of dv acceleration = v dx ) ; questions may involve the distance between two moving particles, but detailed knowledge of relative velocity is not required. including application of the principle of conservation of energy. including conservation of momentum, conservation of energy (when appropriate); coeﬃcient of restitution, Newton’s experimental law; including simple cases of oblique impact (on a plane, for example); including knowledge of the terms perfectly elastic
(e = 1) and inelastic (e = 0); questions involving successive impacts may be set. including motion of a particle in two and three dimensions and motion of connected particles, such as trains, or particles connected by means of pulleys.

Motion of a projectile under gravity

including manipulation of the equation y = x tan α −

gx2
,
2V 2 cos2 α

viewed, possibly, as a quadratic in tan α ; not including projectiles on inclined planes.

Section C: Probability and Statistics
The emphasis, in comparison with Advanced GCE and other comparable examinations is towards probability and formal proofs, and away from data analysis and use of standard statistical tests.
Questions may involve use of any of the material in the Pure Mathematics syllabus.
Probability
Permutations, combinations and arrangements Exclusive and complementary events

Conditional probability

including sampling with and without replacement. including understanding of
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) , though not necessarily in this form. informal applications, such as tree diagrams.

Distributions
Discrete and continuous probability distribution functions and cumulative distribution functions.

Binomial distribution
Poisson distribution

Normal distribution

including calculation of mean, variance, median, mode and expectations by explicit summation or integration for a given (possibly unfamiliar) distribution (eg exponential or geometric or something similarly straightforward); notation: f (x) = F′ (x). including explicit calculation of mean. including explicit calculation of mean; including use as approximation to binomial distribution where appropriate. including conversion to the standard normal distribution by translation and scaling; including use as approximation to the binomial or Poisson distributions where appropriate; notation: X ∼ N(µ, σ 2 ) .

Hypothesis testing
Basic concepts in the case of a simple null hypothesis and simple or compound alternative including knowledge of the terminology null hypothesis and alternative hypothesis, one and two tailed tests.

MATHEMATICS III (9475)
In each section knowledge of the corresponding parts of the syllabus for Mathematics I and II is assumed, and may form the basis of some questions.

Section A: Pure Mathematics
Algebra
Summation of series

Polynomial equations

Induction

including use of method of diﬀerences, recognition of Maclaurin’s series (including binomial series) and partial fractions; including investigation of convergence, by consideration of the sum to n terms. including complex solutions (and the fact that they occur in conjugate pairs if the cooeﬃcients are real); including the relationship between symmetric functions of the roots and the coeﬃcients; including formation of a new equation whose roots are related to the roots of the old equation. including application complicated situations such as divisibility problems.

Trigonometry
Formula for sin A sin B (etc) and sin A + sin B (etc)
Coordinate geometry
Ellipse, parabola, hyperbola and rectangular hyperbola

Polar coordinates

including standard Cartesian forms and parametric forms; including geometrical understanding, but not detailed knowledge of, the terms focus, directrix, eccentricity. including relation with Cartesian coordinates; including ability to sketch simple graphs; including integral formula for area.

Complex numbers
Algebra of complex numbers

including Cartesian and polar forms.

Argand diagram

including geometrical representation of sums and products; including identiﬁcation of loci and regions (eg
|z − a| = k|z − b|).

Euler’s identity: eiθ ≡ cos θ + i sin θ

de Moivre’s theorem

including the relationship between trigonometrical and exponential functions; including application to summation of trigonometric series. including application to ﬁnding nth roots.

Hyperbolic functions
Deﬁnitions and properties of the six hyperbolic functions

including integrals and derivatives of cosh, sinh, tanh and coth.

Formulae corresponding to trigonometric formulae Inverse functions

such as cosh2 x − sinh2 x = 1 and sinh(x + y) = · · · . including ability to derive logarithmic forms and graphs; including derivatives.

Calculus
Integrals resulting in inverse trig. or hyperbolic functions
Maclaurin’s series
Arc length of a curve expressed in
Cartesian coordinates
First order linear diﬀerential equations

Second order linear diﬀerential equations with constant coeﬃcients

1

including integrands such as (x2 − ax)− 2 . including knowledge of expansions for sin x, cos x, cosh x, sinh x and ln(1 + x). including curves described parametrically. including the general form dy + P(x)y = Q(x). dx including the inhomogeneous case a d2 y dy + cy = f (x),
+b
dx2 dx where f (x) is an exponential function, a sin/cos function or a polynomial (but excluding resonant cases); including equations that can be transformed into the above form; including initial or boundary conditions.

Section B: Mechanics
Motion of a particle subject to a variable force Elastic strings and springs

Motion in a circle
Simple harmonic motion

Relative motion

including use of v

dv for acceleration. dx including Hooke’s law and the potential energy; notation: T = kx = λx/l, where k is stiﬀness, λ is modulus of elasticity. including motion with non-uniform speed. including obtaining from a given model an equation of the form x + ω 2 x = a, where a is a
¨
constant, and quoting the solution.

Moment of inertia

Motion of a rigid body about a ﬁxed axis

including calculations involving integration (such as a non-uniform rod, or a lamina); including parallel and perpendicular axis theorems. including understanding of angular momentum and rotational kinetic energy and their conservation; including the understanding of couple and the work done by a couple; including calculation of forces on the axis.

Section C: Probability and Statistics
Algebra of expectations

Further theory of distribution functions

Generating functions

Discrete bivariate distributions

Sampling

for example,
E(aX + bY + c) = aE(X) + bE(Y ) + c ,
Var (X) = E(X 2 ) − E(X)2 ,
Var (aX + b) = a2 Var (X) . including, for example, use of the cumulative distribution function to calculate the distribution of a related random variable (eg X 2 ). including the use of generating functions to obtain the distribution of the sum of independent random variables. including understanding of the idea of independence; including marginal and conditional distributions; including use of Var (aX + bY ) = a2 Var (X) + b2 Var (Y ) + 2abCov(X, Y ) . including sample mean and variance; including use of the central limit theorem.…...

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