Year: 2018
Paper: 1
Question Number: 7
Course: UFM Pure
Section: Roots of 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: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1516.7
Banger Comparisons: 3
\begin{questionparts}
\item
In the cubic equation $x^3-3pqx+pq(p+q)=0\,$, where $p$ and $q$ are distinct real numbers, use the substitution
\[
x=\frac{pz+q}{z+1}
\]
to show that the equation reduces to $az^3+b = 0\,$, where $a$ and $b$ are to be expressed in terms of $p$ and $q$.
\item Show further that the equation $x^3 - 3cx + d = 0\,$, where $c$ and $d$ are non-zero real numbers, can be written in the form $x^3-3pqx+pq(p+q)=0\,$, where $p$ and $q$ are distinct real numbers, provided $d^2 > 4c^3\,$.
\item Find the real root of the cubic equation $x^3+6x-2=0\,$.
\item Find the roots of the equation $x^3 - 3p^2x +2p^3=0\,$,
and hence show how the equation $x^3 - 3cx + d = 0$
can be solved in the case $d^2 = 4c^3\,$.
\end{questionparts}\begin{questionparts}
\item Let $x = \frac{pz+q}{z+1}$ then
\begin{align*}
&& 0 &= x^3-3pqx+pq(p+q) \\
&&&= \left ( \frac{pz+q}{z+1} \right)^3 - 3pq \left ( \frac{pz+q}{z+1} \right) + pq(p+q) \\
&&&= \frac{(pz+q)^3-3pq(pz+q)(z+1)^2+pq(p+q)(z+1)^3}{(z+1)^3} \\
&&&= \frac{1}{(z+1)^3} \Big ((p^3+pq(p+q)-3p^2q)z^3 + (3p^2q-6p^2q+3pq^2+3p^2q+3pq^2)z^2 + \\
&&&\qquad \qquad\quad\quad +(3pq^2-3p^2q-6pq^2+3p^2q+3qp^2)z+(q^3-3pq^2+p^2q+pq^2) \Big ) \\
&&&= \frac{(p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2)}{(z+1)^3} \\
\Rightarrow && 0 &= (p^3+pq^2-2p^2q)z^3+(q^3+p^2q-2pq^2) \\
&&&= p(p-q)^2z^3 + q(p-q)^2 \\
\Rightarrow && 0 &= pz^3 + q
\end{align*}
\item We would like to find $pq = c$ and $pq(p+q) = d$, so $p$ and $q$ are roots of the quadratic $x^2-\frac{d}{c}x + c = 0$, which has distinct real roots if $\Delta = \frac{d^2}{c^2}-4c > 0 \Rightarrow d^2>4c^3$
\item Note that $c = -2, d = -2$ so \begin{align*}
&& 0 &= x^3+6x-2 \\
\text{consider} && 0 &= X^2-X-2 \\
&& &= (X+1)(X-2) \\
\Rightarrow && p = -1, &q = 2\\
\Rightarrow && 0 &= x^3-3\cdot 2 \cdot(-1) x + 2\cdot(-1) \cdot(-2+1) \\
\Rightarrow && 0 &= -z^3+2 \\
\Rightarrow && z &= \sqrt[3]{2} \\
\Rightarrow && \frac{-z+2}{z+1} &= \sqrt[3]{2} \\
\Rightarrow && -z+2 &= \sqrt[3]{2} z + \sqrt[3]{2} \\
\Rightarrow && z &= \frac{2-\sqrt[3]{2}}{\sqrt[3]{2}+1}
\end{align*}
\item $\,$ \begin{align*}
&& 0 &= x^3 - 3p^2x + 2p^3 \\
&&&= (x-p)(x^2+px-2p^2) \\
&&&=(x-p)^2(x+2p)\\
\Rightarrow && x &= p, p, -2p
\end{align*}
Therefore if we have a repeated root to our associated quadratic we can find a cubic of the form $x^3-3p^2x+2p^3$, but we know this has roots we can find.
\end{questionparts}
As with most of Qs. 3-8, this was a very popular question, but with a low mean score; in this case, of around 6 out of 20. Part (i) was generally done well although quite a lot of candidates made an algebraic error here. It was then possible to obtain full marks in the remaining sections by follow through, but few candidates followed through correctly. There were also a few candidates who found the and constant terms, but didn't actually check that the other terms cancelled. A rare alternative method to (i) was to write in terms of and substitute into 0; where this method was used, it was generally well-executed. A very large majority of candidates attempting part (ii) did so "backwards", showing instead that 4 follows from the relations between , and , . Those who correctly interpreted this part of the question often failed to justify strictness of the inequality. There were more essentially correct responses to part (iii) as candidates were able to identify appropriate values of and , although arithmetic errors were common. Few candidates made progress with part (iv), but those who did often got 3 marks out of 4 because they did not write the solutions in terms of and/or .