Year: 2012
Paper: 2
Question Number: 8
Course: UFM Pure
Section: Sequences and series, recurrence and convergence
There were just over 1000 entries for paper II this year, almost exactly the same number as last year. Overall, the paper was found marginally easier than its predecessor, which means that it was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50 candidates scored 100 marks or more, with more than 400 gaining at least half marks on the paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40 marks. It was pleasing to note that the advice of recent years, encouraging students not to make attempts at lots of early parts to questions but rather to spend their time getting to grips with the six that can count towards their paper total, was more obviously being heeded in 2012 than I can recall being the case previously. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were the most popular questions, although each drew only around 800 "hits" – fewer than usual. Questions 3 – 5 & 8 were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most successfully attempted question on the paper – with very little interest shown in the rest of Sections B or C.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1485.7
Banger Comparisons: 1
The positive numbers $\alpha$, $\beta$ and $q$ satisfy
$\beta-\alpha >q$. Show that
\[
\frac{\alpha^2+\beta^2 -q^2}{\alpha\beta}-2> 0\,.
\]
The sequence $u_0$, $u_1$, $\ldots$ is defined by $u_0=\alpha$,
$u_1=\beta$ and
\[
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
u_{n+1} = \frac {u_{n}^2 -q^2}{u_{n-1}}
\ \ \ \ \ \ \ \ \ \ \ (n\ge1),
\]
where $\alpha$, $\beta$ and $q$ are given positive numbers (and $\alpha$ and $\beta$ are such that no term in the sequence is zero).
Prove that $u_n(u_n+u_{n+2}) = u_{n+1}(u_{n-1}+u_{n+1})\,$.
Prove also that
\[
u_{n+1} -pu_n + u_{n-1}=0
\]
for some number $p$ which you should express in terms of $\alpha$, $\beta$ and $q$. Hence, or otherwise, show that if $\beta> \alpha+q$, the sequence is strictly increasing (that is, $u_{n+1}-u_n > 0$ for all $n$). Comment on the case $\beta =\alpha +q$.\begin{align*}
&& \beta - \alpha &> q \\
\Rightarrow &&(\beta - \alpha)^2 &> q^2 \\
\Rightarrow && \beta^2 +\alpha^2 - 2\beta \alpha &> q^2 \\
\Rightarrow && \alpha^2+\beta^2-q^2 -2 \beta \alpha &> 0 \\
\Rightarrow && \frac{\alpha^2+\beta^2-q^2}{\alpha\beta} - 2 &> 0
\end{align*}
\begin{align*}
&& u_n(u_n+u_{n+2}) &= u_n \cdot \left (u_n + \frac {u_{n+1}^2 -q^2}{u_{n}}\right) \\
&&&= u_n^2 + u_{n+1}^2-q^2 \\
&&&= u_n^2 + u_{n+1}^2 - (u_n^2-u_{n-1}u_{n+1}) \\
&&&= u_{n+1}^2 + u_{n+1}u_{n-1} \\
&&&= u_{n+1}(u_{n-1}+u_{n+1}) \\
\\
&& u_{n+1}-pu_n+u_{n-1} &= -pu_n+\frac{u_{n}(u_{n-2}+u_n)}{u_{n-1}} \\
&&&= \frac{u_n(u_{n}-pu_{n-1}+u_{n-2})}{u_{n-1}}
\end{align*}
Therefore if $u_2 -pu_1 + u_0 = 0$ it is always zero, ie if
\begin{align*}
&& u_2 &= p\beta - \alpha \\
&& u_2 &= \frac{\beta^2-q^2}{\alpha} \\
\Rightarrow && \frac{\beta^2-q^2}{\alpha} &= p\beta - \alpha \\
\Rightarrow && p &= \frac{\alpha^2+\beta^2-q^2}{\alpha\beta}
\end{align*}
If $\beta > \alpha + q$ we must have that $p > 2$, and so $u_{n+1}-u_n = (p-1)u_n - u_{n-1} > u_n-u_{n-1} > 0$, therefore the sequence is strictly increasing.
If $\beta = \alpha + q$ the sequence follows $u_{n+1} - 2u_n + u_{n-1} =0$ and so $u_{n+1}-u_n = u_n - u_{n-1}$ for all $n$ (which is still increasing - it's an arithmetic progression with common difference $\beta - \alpha$).
Well over a half of all candidates attempted this question but, on average, it proved to be the least well scoring. The initial inequality was usually well handled, but most of the remaining parts of the question were poorly handled in very circuitous ways, with few candidates being very clear in either what they were trying to prove or how. The first, given, result simply follows from equating for q² in two successive cases of the given recurrence definition. This result was then supposed to help with the following result, but almost no-one seemed to realise this, and attempts at inductive proofs were quite common at this stage (usually unsuccessfully). Candidates' confidence had clearly ebbed away well before the final part of the question, and so there were very few attempts at the two cases of the final paragraph.