Year: 2019
Paper: 1
Question Number: 4
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; all of which are available from the STEP and Cambridge Examinations Board websites. 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 for several months beforehand. 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 it is that a candidate should never think that they are simply required to 'go through the motions' but must expect, sooner or later, to 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. When you read through the report and look at the solutions (either in the mark-scheme or the Hints & 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's paper produced the usual sorts of outcomes, with far too many candidates wasting valuable time by attempting more than six questions, and with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. Around one candidate in eight failed to hit the 30 mark overall, though this is an improvement on last year. Most candidates were able to produce good attempts at two or more questions. At the top end of the scale, around a hundred candidates scored 100 or more out of 120, with four hitting the maximum of 120 and many others not far behind. 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, although under two-thirds of the entry attempted it this year, and it also turned out to be the most successful question on the paper with a mean score of about 12 out of 20. In order of popularity, Q1 was followed by Qs.3, 4 and 2. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the applied questions combined scoring fewer 'hits' than any one of the first four questions on its own. Though slightly more popular than the applied questions, the least successful question of all was Q5, on vectors. This question was attempted by almost 750 candidates, but 70% of these scored no more than 2 marks, leaving it with a mean score of just over 3 out of 20. Q9 (a statics question) was found only marginally more appetising, with a mean score of almost 3½ out of 20. In general, it was found that explanations were poorly supplied, with many candidates happy to overlook completely any requests for such details.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Find integers $m$ and $n$ such that
$$\sqrt{3+2\sqrt{2}} = m + n\sqrt{2}.$$
\item Let $f(x) = x^4 - 10x^2 + 12x - 2$. Given that the equation $f(x) = 0$ has four real roots, explain why $f(x)$ can be written in the form
$$f(x)=(x^2 + sx + p)(x^2 - sx + q)$$
for some real constants $s$, $p$ and $q$, and find three equations for $s$, $p$ and $q$.
Show that
$$s^2(s^2 - 10)^2 + 8s^2 - 144 = 0$$
and find the three possible values of $s^2$.
Use the smallest of these values of $s^2$ to solve completely the equation $f(x) = 0$, simplifying your answers as far as you can.
\end{questionparts}\begin{questionparts}
\item $(1+\sqrt{2})^2 = 3 + 2\sqrt{2}$ so $\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}$
\item We can always factorise any quartic in the form $(x^2+ax+b)(x^2+cx+d)$, since $x^3$ has a coefficient of $a+b$ we must have $a = -b$, ie the form in the question.
\begin{align*}
&& 0 &= (x^2+sx+p)(x^2-sx+q) \\
&&&= x^4+(p+q-s^2)x^2+s(q-p)x+pq \\
\Rightarrow && pq &= -2 \\
&& s(q-p) &= 12 \\
&& p+q-s^2 &= -10 \\
\\
&& p+q &= s^2-10 \\
&& (p+q)^2 &= (s^2-10)^2 \\
&& (q-p)^2 &= \frac{12}{s^2} \\
\Rightarrow && (s^2-10)^2 &= \frac{12}{s^2} + 4pq \\
&& (s^2-10)^2 &= \frac{144}{s^2} -8 \\
&& 0 &= s^2(s^2-10)^2+8s^2-144 \\
&&&= s^6-20s^4+108s^2-144 \\
&&&= (s^2-2)(s^2-6)(s^2-12)
\end{align*}
Suppose $s = \sqrt{2}$, and we have
\begin{align*}
&& q-p &= 6\sqrt{2} \\
&& p+q &= -8 \\
\Rightarrow && q &= 3\sqrt{2}-4 \\
&& p &= -4-3\sqrt{2}
\end{align*}
Solving our quadratic equations, we have
\begin{align*}
&& 0 &= x^2-\sqrt{2}x-4+3\sqrt{2} \\
\Rightarrow && x &= \frac{\sqrt{2}\pm \sqrt{2-4\cdot(-4+3\sqrt{2})}}{2} \\
&&&= \frac{\sqrt{2}\pm \sqrt{18-12\sqrt{2}}}{2} \\
&&&= \frac{\sqrt{2}\pm (2\sqrt{3}-\sqrt{6})}{2} \\
\\
&& 0 &= x^2+\sqrt{2}x-3\sqrt{2}-4 \\
&& x &= \frac{-\sqrt{2} \pm \sqrt{2-4\cdot(3\sqrt{3}-4)}}{2}\\
&& &= \frac{-\sqrt{2} \pm \sqrt{18+12\sqrt{2}}}{2}\\
&& &= \frac{-\sqrt{2} \pm (\sqrt{6}+2\sqrt{3})}{2}\\
\end{align*}
\end{questionparts}
This question was also attempted by many candidates, being the last of the "big four" early questions, and – along with questions 1 and 2 – one of the only questions for which scores exceeded 10/20 on average. Most students did part (i) well, although it was quite common to see negative or non-integer values of m and n included in the answer, despite the question's clear wording. It was also unfortunate that many candidates went about it in the longest way imaginable, squaring m + n√2 and then comparing the result with the intended answer … then solving for m and n from a pair of simultaneous equations (one linear and one quadratic) when the small numbers involved, and the fact that the question stated that they would be integers, required a careful evaluation of the situation. In part (ii), most students correctly got three equations for p, q and s by expanding the factorisation given and comparing coefficients. However, relatively few gave a clear justification that such a factorisation must exist; there were some wordy but vague attempts at this, with the logic commonly being reversed. Some candidates found it difficult to manipulate the simultaneous non-linear equations to obtain useful expressions, but there were also many very good derivations of the desired equation, following several different routes. Most students, including those who had been unable to derive it, reduced the given equation to a cubic equation for s and solved this with no problems. A good number simply spotted the roots with no working shown. In using the value s = √2 to obtain two quadratic equations, students often obtained p, q = 4 ± 3√2 but confused which variable took which sign, or even used both possibilities in each case, leading either to incorrect signs on all four roots or to eight roots of which four were spurious. We also saw a significant number incorrectly taking s = 2. Most people who got to the end were very good at getting rid of the nested square roots using the method of part (i).