Year: 2019
Paper: 2
Question Number: 3
Course: LFM Stats And Pure
Section: Polynomials
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
For any two real numbers $x_1$ and $x_2$, show that
$$|x_1 + x_2| \leq |x_1| + |x_2|.$$
Show further that, for any real numbers $x_1, x_2, \ldots, x_n$,
$$|x_1 + x_2 + \cdots + x_n| \leq |x_1| + |x_2| + \cdots + |x_n|.$$
\begin{questionparts}
\item The polynomial f is defined by
$$f(x) = 1 + a_1 x + a_2 x^2 + \cdots + a_{n-1} x^{n-1} + x^n$$
where the coefficients are real and satisfy $|a_i| \leq A$ for $i = 1, 2, \ldots, n-1$, where $A \geq 1$.
\begin{enumerate}
\item If $|x| < 1$, show that
$$|f(x) - 1| \leq \frac{A|x|}{1 - |x|}.$$
\item Let $\omega$ be a real root of f, so that $f(\omega) = 0$. In the case $|\omega| < 1$, show that
$$\frac{1}{1 + A} \leq |\omega| \leq 1 + A. \quad (*)$$
\item Show further that the inequalities $(*)$ also hold if $|\omega| \geq 1$.
\end{enumerate}
\item Find the integer root or roots of the quintic equation
$$135x^5 - 135x^4 - 100x^3 - 91x^2 - 126x + 135 = 0.$$
\end{questionparts}Claim: $|x_1 + x_2| \leq |x_1| + |x_2|$
Proof: Case 1: $x_1, x_2 \geq 0$. The inequality is equivalent to $|x_1 + x_2| = x_1 + x_2 = |x_1|+|x_2|$ so it's an equality.
Case 2: $x_1, x_2 \leq 0$. The inequality is equivalent to $|x_1+x_2| = -x_1-x_2 = |x_1|+|x_2$, so it's also an equality in this case.
Case 3: (wlog) $|x_1| \geq |x_2| > 0$ and $x_1x_2 < 0$ then
$|x_1+x_2| = x_1-x_2 \leq x_1 \leq |x_1|+|x_2|$
We can prove this by induction, we've already proven the base case and:
$|x_1+x_2 + \cdots + x_n| \leq |x_1 + x_2 + \cdots x_{n-1}| + |x_n| \leq |x_1| + |x_2| + \cdots + |x_n|$
\begin{questionparts}
\item $\,$ \begin{align*}
&& |f(x) - 1| &= |a_1 x + a_2x^2 + \cdots + a_{n-1}x^{n-1} + x^n| \\
&&&\leq |a_1x| + |a_2x^2| + \cdots + |a_{n-1}x^{n-1}| + |x^n| \\
&&&\leq |a_1||x| + |a_2||x|^2 + \cdots + |a_{n-1}||x|^{n-1} + |x|^n \\
&&&\leq A|x| + A|x|^2 + \cdots + A|x|^{n-1} + |x|^n \\
&&&=A|x| \frac{1-|x|^{n-1}}{1-|x|} + |x|^n \\
&&&= \frac{A|x|-A|x|^{n}+|x|^{n+1}-|x|^n}{1-|x|} \\
&&&= \frac{A|x|-|x|^n(\underbrace{A-|x|+1}_{\geq0})}{1-|x|} \\
&&&\leq \frac{A|x|}{1-|x|}
\end{align*}
\item If $f(\omega) = 0$ then \begin{align*}
&& 1 & \leq \frac{A|\omega|}{1-|\omega|} \\
\Leftrightarrow && 1-|\omega| &\leq A |\omega| \\
\Leftrightarrow && 1 &\leq (1+A) |\omega| \\
\Leftrightarrow && \frac{1}{1+A} &\leq |\omega| \\
\end{align*}
We also know $\omega \leq 1 < 1 + A$
\item If $\omega$ is a root of $f(x)$ then $1/\omega$ is a root of $1 + a_{n-1}x + a_{n-2}x^2 + \cdots + a_1x^{n-1}+x^n$ and so $1/\omega$ satisfies that inequality, ie
\begin{align*}
&& \frac{1}{1+A} && \leq &&|1/\omega| && \leq &&1 + A \\
\Leftrightarrow &&1+A && \geq&& |\omega| && \geq&& \frac{1}{1 + A}
\end{align*}
\item First notice that it's equivalent to:
$0 = x^5 - 1x^4 - \frac{100}{135}x^3-\frac{91}{135}x^2-\frac{126}{135} + 1$
therefore all integer roots must be between $-2,-1$ and $1$ and $2$.
$1$ doesn't work. $-1$ works. Clearly $2$ cannot work by parity argument, therefore the only integer root is $-1$.
\end{questionparts}
While this was a popular question it was also the one where the average mark achieved by candidates was the lowest. In this question many of the results to be reached were given in the question. Students therefore need to recognise that it is necessary for solutions to be presented very clearly, and it is for this reason that many solutions in the first parts did not achieve full marks. For example, justifications of the generalised result for a set of n real numbers expressed in the form of an inductive proof were the most successful. For most candidates the majority of marks were scored in the sections up to and including part (i)(b). Many candidates were then unable to see how to work in the cases where |x| > 1 for part (i)(c). In the final part, candidates were often unable to put the equation into the form that had been used in the earlier parts of the questions and therefore did not manage to reduce the possible values of the integer roots to a sufficiently small set.