Year: 2018
Paper: 2
Question Number: 1
Course: LFM Stats And Pure
Section: Polynomials
The pure questions were again the most popular of the paper, with only two of those questions being attempted by fewer than half of the candidates (none of the other questions was attempted by more than half of the candidates). Good responses were seen to all of the questions, but in many cases, explanations lacked sufficient detail to be awarded full marks. Candidates should ensure that they are demonstrating that the results that they are attempting to apply are valid in the cases being considered. In several of the questions, later parts involve finding solutions to situations that are similar to earlier parts of the question. In general candidates struggled to recognise these similarities and therefore spent a lot of time repeating work that had already been done, rather than simply observing what the result must be.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
Show that, if $k$ is a root of the quartic equation
\[
x^4 + ax^3 + bx^2 + ax + 1 = 0\,,
\tag{$*$}
\]
then $k^{-1}$ is a root.
You are now given that $a$ and $b$ in $(*)$ are both
real and are such that the roots are all real.
\begin{questionparts}
\item Write down all the values of $a$ and $b$ for which $(*)$ has only one distinct root.
\item
Given that $(*)$ has exactly three distinct roots, show that either $b=2a-2$ or $b=-2a-2\,$.
\item
Solve $(*)$ in the case $b= 2 a -2\,$, giving your solutions in terms of $a$.
\end{questionparts}
Given that $a$ and $b$ are both real and that the roots of $(*)$ are all real, find necessary and sufficient conditions, in terms of $a$ and $b$, for $(*)$ to have exactly three distinct real roots.Let $f(x) = x^4 + ax^3 + bx^2 + ax + 1$, and suppose $f(k) = 0$. Since $f(0) = 1$, $k \neq 0$, therefore we can talk about $k^{-1}$.
\begin{align*}
&& f(k^{-1}) &= k^{-4} + ak^{-3} + bk^{-2} + ak^{-1} + 1 \\
&&&= k^{-4}(1 + ak + bk^2 + ak^3 + k^4) \\
&&&= k^{-4}(k^4+ak^3+bk^2+ak+1) \\
&&&= k^{-4}f(k) = 0
\end{align*}
Therefore $k^{-1}$ is also a root of $f$
\begin{questionparts}
\item If $f$ has only on distinct root, we must have $f(x) = (x+k)^4$ therefore $k = k^{-1} \Rightarrow k^2 = 1 \Rightarrow k = \pm1$, or $a = 4, b = 6$ or $a = -4, b = 6$
\item If $f$ has exactly three distinct roots then one of the roots must be a repeated $1$ or $-1$, ie $0 = f(1) = 1 + a + b + a + 1 = 2 + b +2a \Rightarrow b = -2a-2$ or $0 = f(-1) = 1 -a + b -a + 1 \Rightarrow b = 2a - 2$
\item If $b = 2a-2$, we have
\begin{align*}
&& f(x) &= 1 + ax + (2a-2)x^2 + ax^3 + x^4 \\
&&&= (x^2+2x+1)(1+(a-2)x+x^2) \\
\Rightarrow && x &= \frac{2-a \pm \sqrt{(a-2)^2 - 4}}{2} \\
&&&= \frac{2-a \pm \sqrt{a^2-4a}}{2}
\end{align*}
\end{questionparts}
$f$ has exactly three distinct real roots iff $b = \pm 2a - 2$ and $b \neq 6$
This was the most popular question of the paper, attempted by 93% of candidates. While many candidates scored well on this question there were very few who achieved full marks and a significant number who scored 0 for this question. The first two parts of the questions were generally done well, with the majority of marks being lost for arithmetic errors or from just considering one of the two cases. In some cases, candidates produced the reasoning for part (ii) as their answer for part (i). In these cases, the candidates generally repeated the work as their answer for part (ii). There were a number of solutions to part (ii) that were overcomplicated, involving consideration of the factorised form and comparison of coefficients. Part (iii) was also very well completed by those who attempted it, with the main cause for loss of marks being algebraic errors. A number of errors were seen in the final part of the question, particularly not considering the second case or not realising that the quadratic would have a different discriminant in this case and a failure to check that the roots of the quadratic did not duplicate the repeated root already found in this case.