Year: 2018
Paper: 1
Question Number: 5
Course: LFM Stats And Pure
Section: Polynomials
In order to get the fullest picture, this document should be read in conjunction with the question paper, the marking scheme and (for comments on the underlying purpose and motivation for finding the right solution-approaches to questions) the Hints and Solutions document. The purpose of the STEPs is to learn what students are able to achieve mathematically when applying the knowledge, skills and techniques that they have learned within their standard A-level (or equivalent) courses … but seldom within the usual range of familiar settings. STEP questions require candidates to work at an extended piece of mathematics, often with the minimum of specific guidance, and to make the necessary connections. This requires a very different mind-set to that which is sufficient for success at A-level, and the requisite skills tend only to develop with prolonged and determined practice at such longer questions. One of the most crucial features of the STEPs is that the routine technical and manipulative skills are almost taken for granted; it is necessary for candidates to produce them with both speed and accuracy so that the maximum amount of time can be spent in thinking their way through the problem and the various hurdles and obstacles that have been set before them. Most STEP questions begin by asking the solver to do something relatively routine or familiar before letting them loose on the real problem. Almost always, such an opening has not been put there to allow one to pick up a few easy marks, but rather to point the solver in the right direction for what follows. Very often, the opening result or technique will need to be used, adapted or extended in the later parts of the question, with the demands increasing the further on that one goes. So a candidate should never think that they are simply required to 'go through the motions'; rather they will, sooner or later, be required to show either genuine skill or real insight in order to make a reasonably complete effort. The more successful candidates are the ones who manage to figure out how to move on from the given starting-point. Finally, reading through a finished solution is often misleading – even unhelpful – unless you have attempted the problem for yourself. This is because the thinking has been done for you. So, when you read through the report and look at the solutions (either in the mark scheme or the Hints and Solutions booklet), try to figure out how you could have arrived at the solution, learn from your mistakes and pick up as many tips as you can whilst working through past paper questions. This year far too many candidates wasted time by attempting more than six questions, with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. There were almost 2000 candidates for this SI paper. Almost one-sixth of candidates failed to reach a total of 30 and around two-thirds fell below half-marks overall. This highlights the fact that many candidates don't find this test an easy one. At the other end of the spectrum, almost one-in-ten managed a total of 84 out of 120 – these candidates usually marked out by their ability to complete whole questions – with almost 4% of the entry achieving the highly praiseworthy feat of getting into three-figures with their overall score. The paper is constructed so that question 1 is very approachable indeed, the intention being to get everyone started with some measure of success; unsurprisingly, Q1 was the most popular question of all, with almost all candidates attempting it, and it also turned out to be the most successful question on the paper with a mean score of more than 15 out of 20. Around 7% of candidates didn't make any kind of attempt at it at all. In order of popularity, Q1 was followed by Qs. 2, 7, 4 and 3. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the most popular applied question (Q9, mechanics) still getting fewer 'hits' than the least popular pure question (Q5). Questions 10, 11 and 13 proved to attract very little attention from candidates and many of the attempts were minimal.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item Write down the most general polynomial of degree 4 that leaves a remainder of 1 when divided by any of
$x-1\,$, $x-2\,$, $x-3\,$ or $x-4\,$.
\item The polynomial $\P(x)$ has degree $N$, where $N\ge1\,$,
and satisfies
\[
\P(1) = \P(2) = \cdots = \P(N) =1\,.
\]
Show that $\P(N+1) \ne 1\,$.
Given that $\P(N+1)= 2\,$, find $\P(N+r)$ where $r$ is a positive integer. Find a positive integer $r$, independent of $N,$ such that $\P(N+r) = N+r\,$.
\item The polynomial ${\rm S}(x)$ has degree 4. It has integer coefficients and the coefficient of $x^4$ is 1. It satisfies
\[
{\rm S}(a) =
{\rm S}(b) =
{\rm S}(c) =
{\rm S}(d) = 2001\,,
\]
where $a$, $b$, $c$ and $d$ are distinct (not necessarily positive)
integers.
\begin{itemize}
\item Show that there is no integer $e$ such that ${\rm S}(e) = 2018\,$.
\item Find the number of ways the (distinct) integers $a$, $b$, $c$ and $d$ can be chosen such that ${\rm S}(0) = 2017$ and $a < b< c< d\,.$
\end{itemize}
\end{questionparts}\begin{questionparts}
\item $p(x) = C(x-1)(x-2)(x-3)(x-4)+1$
\item Suppose $P(N+1) = 1$ them we could consider $f(x) = P(x) - 1$ to be a polynomial of degree $N$ with at least $N+1$ roots, which would be a contradiction. Therefore $P(N+1) \neq 1$.
Since $P(x) = C(x-1)(x-2)\cdots(x-N) + 1$ and $P(N+1) = 2$ we must have $C \cdot N! + 1 = 2 \Rightarrow C = \frac{1}{N!}$, hence $P(x) = \binom{x-1}{N} + 1$ ie $P(N+r) = \binom{N+r-1}{N}+1$ so $P(N+2) = \binom{N+1}{N} +1= N+2$, so we can take $r=2$.
\item \begin{enumerate}
\item Suppose consider $p(x) = S(x) - 2001$, then $p(x)$ has roots $a,b,c,d$ and suppose we can find $e$ such that $p(e) = 17$ then we must have $(e-a)(e-b)(e-c)(e-d) = 17$ but the only possible factors of $17$ are $-17,-1,1,17$ and we cannot have all $4$ of them. Hence this is not possible.
\item Now we have $abcd = 16$, so we can have factors $-16,-8,-4, -2, -1, 1, 2, 4,8,16$ (and we need to have $4$ of them).
If we have $0$ negatives, the smallest product is $1 \cdot 2 \cdot 4 \cdot 8 > 16$
If we have $2$ negatives we must have $1$ and $-1$ (otherwise we have the same problem of being too large. So $\{-1,1,-2,8\},\{-1,1,2,-8\},\{-1,1,-4,4\},$
If we have $4$ negatives that's the same issue as with $0$ negatives.
\end{enumerate}
\end{questionparts}
Around a third of all candidates made an attempt at this question, yet it had the second lowest mean score of around only 4 marks out of 20. To begin with, many candidates seemed to think that every polynomial begins with a coefficient of 1, and this led to serious difficulties with part (ii). Part (ii) also involved a proof by contradiction (or equivalent) and the exercise of choice … and most candidates aren't happy with finding one answer of what might be many. Those who spotted that r = 2 worked then frequently failed to verify it clearly. Part (iii) was all about factors and the biggest obstacle to success lay in the widespread inability to think of integer factors as being anything other than positive; thus, the statement that 17 couldn't have four distinct factors – whilst correct – was almost invariably given because the candidate could only think of 1 and 17. In fact, 17 does have four integer factors, but the reason the 'proof' failed was actually because +17 and –17 couldn't both be used. Most who persevered this far seemed to realise that factors of 16 were to be used but failed to give all the correct possibilities (or to show why there were no others) as a result of a poorly devised system for enumerating them.