Year: 2014
Paper: 3
Question Number: 1
Course: UFM Pure
Section: Roots of polynomials
A 10% increase in the number of candidates and the popularity of all questions ensured that all questions had a good number of attempts, though the first two questions were very much the most popular. Every question received at least one absolutely correct solution. In most cases when candidates submitted more than six solutions, the extra ones were rarely substantial attempts. Five sixths gave in at least six attempts.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1542.2
Banger Comparisons: 5
Let $a$, $b$ and $c$ be real numbers such that $a+b+c=0$ and let
\[(1+ax)(1+bx)(1+cx) = 1+qx^2 +rx^3\,\]
for all real $x$. Show that $q = bc+ca+ab$ and $r= abc$.
\begin{questionparts}
\item Show that the coefficient of $x^n$ in the series expansion (in ascending powers of $x$) of $\ln (1+qx^2+rx^3)$ is $(-1)^{n+1} S_n$ where
\[S_n = \frac{a^n+b^n+c^n}{n} \,, \ \ \ \ \ \ \ \ (n\ge1).\]
\item Find, in terms of $q$ and $r$, the coefficients of $x^2$, $x^3$ and $x^5$ in the series expansion (in ascending powers of $x$) of $\ln (1+qx^2+rx^3)$ and hence show that $S_2S_3 =S_5$.
\item Show that $S_2S_5 =S_7$.
\item Give a proof of, or find a counterexample to, the claim that $S_2S_7=S_9$.
\end{questionparts}\begin{align*}
(1+ax)(1+bx)(1+cx) &= (1+(a+b)x+abx^2)(1+cx) \\
&= 1+(a+b+c)x+(ab+bc+ca)x^2+abcx^3
\end{align*}
Therefore by comparing coefficients, $q = bc + ca + ab$ and $r = abc$ as required.
\begin{questionparts}
\item
\begin{align*}
\ln (1+qx^2 + rx^3) &= \ln(1+ax) + \ln(1+bx)+\ln(1+cx) \\
&= -\sum_{n=1}^{\infty} \frac{(-ax)^n}{n}-\sum_{n=1}^{\infty} \frac{(-bx)^n}{n}-\sum_{n=1}^{\infty} \frac{(-cx)^n}{n} \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n+1}(a^n+b^n+c^n)}{n} x^n \\
&= \sum_{n=1}^{\infty} (-1)^{n+1} S_n x^n \\
\end{align*}
\item \begin{align*}
\ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} + O(x^6) \\
&= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + O(x^6) \\
\end{align*}
Comparing coefficients we see that $S_2 = -q$ and $S_3 = r$, we also must have $S_5 = -qr = S_2S_3$ as required.
\item \begin{align*}
\ln (1 + qx^2 + rx^3) &= (qx^2+rx^3) -\frac{(qx^2+rx^3)^2}{2} +\frac{(qx^2+rx^3)^3}{3}+ O(x^8) \\
&= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \frac12 rx^6 + \frac13 q^3 x^6 + q^2r x^7 + O(x^8) \\
&= qx^2 + rx^3 - \frac12 q^2 x^4 - qr x^5 + \left ( \frac12 r+ \frac13 q^3 \right)x^6 + q^2r x^7
\end{align*}
Comparing coefficients we see that $S_2 = -q$ and $S_5 =-qr$, we also must have $S_7 = q^2r = S_2S_5$ as required.
\item Let $a = b = 1, c = -2$, then $S_2 = \frac{1^2+1^2 + (-2)^2}{2} = 3, S_7 = \frac{1^2+1^2+(-2)^7}{7} = -18, S_9 = \frac{1^1+1^2+(-2)^9}{9} = \frac{2 - 512}{9} \neq 3 \cdot (-18)$
\end{questionparts}
This was the most popular question on the paper, being attempted by approximately 14 out of every 15 candidates. It was the second most successfully attempted with a mean score of half marks. The stem of the question caused no problems, but a common mistake in part (i) was to attempt derivatives to obtain the desired result. Most candidates came unstuck in part (ii), making it much more difficult for themselves by attempting to work with expressions in a, b, and c rather than using the log series working with q and r, and as a result making sign errors, putting part (iii) beyond reach, and although they could find counterexamples for the claim in part (iv), they did so without the clear direction that working with the expressions in q and r would have made obvious.