Year: 2008
Paper: 2
Question Number: 2
Course: LFM Stats And Pure
Section: Partial Fractions
There were around 850 candidates for this paper – a slight increase on the 800 of the past two years – and the scripts received covered the full range of marks (and beyond!). The questions on this paper in recent years have been designed to be a little more accessible to all top A-level students, and this has been reflected in the numbers of candidates making good attempts at more than just a couple of questions, in the numbers making decent stabs at the six questions required by the rubric, and in the total scores achieved by candidates. Most candidates made attempts at five or more questions, and most genuinely able mathematicians would have found the experience a positive one in some measure at least. With this greater emphasis on accessibility, it is more important than ever that candidates produce really strong, essentially-complete efforts to at least four questions. Around half marks are required in order to be competing for a grade 2, and around 70 for a grade 1. The range of abilities on show was still quite wide. Just over 100 candidates failed to score a total mark of at least 30, with a further 100 failing to reach a total of 40. At the other end of the scale, more than 70 candidates scored a mark in excess of 100, and there were several who produced completely (or nearly so) successful attempts at more than six questions; if more than six questions had been permitted to contribute towards their paper totals, they would have comfortably exceeded the maximum mark of 120. While on the issue of the "best-six question-scores count" rubric, almost a third of candidates produced efforts at more than six questions, and this is generally a policy not to be encouraged. In most such cases, the seventh, eighth, or even ninth, question-efforts were very low scoring and little more than a waste of time for the candidates concerned. Having said that, it was clear that, in many of these cases, these partial attempts represented an abandonment of a question after a brief start, with the candidates presumably having decided that they were unlikely to make much successful further progress on it, and this is a much better employment of resources. As in recent years, most candidates' contributing question-scores came exclusively from attempts at the pure maths questions in Section A. Attempts at the mechanics and statistics questions were very much more of a rarity, although more (and better) attempts were seen at these than in other recent papers.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1498.5
Banger Comparisons: 2
Let $a_n$ be the coefficient of $x^n$ in the series expansion,
in ascending powers of $x$, of
\[\displaystyle
\frac{1+x}{(1-x)^2(1+x^2)}
\,,
\]
where $\vert x \vert <1\,$.
Show, using partial fractions,
that either $a_n =n+1$ or $a_n = n+2$ according to the value of $n$.
Hence find a decimal approximation, to nine significant figures,
for the fraction $ \displaystyle \frac{11\,000}{8181}$.
\newline
[You are not required to justify the accuracy of your approximation.]\begin{align*}
&& \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{A}{1-x} + \frac{B}{(1-x)^2} + \frac{Cx+D}{1+x^2} \\
\Rightarrow && 1 + x &= A(1-x)(1+x^2) + B(1+x^2) + Cx(1-x)^2 + D(1-x)^2 \\
\Rightarrow && 2 &= 2B \tag{$x = 1$} \\
\Rightarrow && 1 &= B \\
\Rightarrow && 1 &= A+B+D \tag{$x = 0$}\\
\Rightarrow && A &= -D \\
\Rightarrow && 0 &= 4A+2B-4C+4D \tag{$x = -1$}\\
\Rightarrow && C &= \frac12\\
\Rightarrow && 3 &= -5A+5B+2C+D \tag{$x=2$} \\
\Rightarrow && 3 &= -6A+6 \\
\Rightarrow && A,D &=-\frac12,\frac12 \\
\Rightarrow && \frac{1+x}{(1-x)^2(1+x^2)} &= \frac{1}{(1-x)^2} +\frac{1}{2(1-x)}+ \frac{x-1}{2(1+x^2)} \\
&&&=\sum_{k=0}^{\infty}(k+1)x^k + \sum_{k=0}^{\infty}\frac12 x^k + \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k+1} - \sum_{k=0}^{\infty}\frac12 (-1)^kx^{2k}
\end{align*}
Therefore the coefficient of $x^n$ is $n+1$ or $n+2$ depending on whether the coefficients from the final series add constructively $n \equiv 1, 2 \pmod{4}$ or destructively.
\begin{align*}
\frac{11\, 000}{8181} &= \frac{(10+1) \cdot 1000}{(10-1)^2 \cdot (100+1)} \\
&= \frac{(1+\frac{1}{10})}{(1-\frac{1}{10})^2 \cdot (1+\frac1{10})} \\
&= 1 + \frac3{10} + \frac4{10^2} + \frac{4}{10^3}+\frac{5}{10^4} + \frac{7}{10^5} + \frac{8}{10^6} + \cdots \\
& \quad \quad \cdots + \frac{8}{10^7} + \frac{9}{10^8} + \frac{11}{10^9} + \frac{12}{10^{10}} + \cdots \\
&= 1.34457890 + \frac{12}{10^{10}} + \cdots
\end{align*}
\begin{align*}
&& \sum_{k=m}^{\infty} (k+2)x^k &= x^m \sum_{k=0}^{\infty} (k+m+2)x^{k} \\
&& &= \frac{x^k}{(1-x)^2} + \frac{(m+2)x^k}{1-x} \\
\Rightarrow && |\sum_{k=m}^{\infty} a_k \left ( \frac1{10} \right )^k | &\leq \frac{1}{10^m}\left ( \frac{1}{(1-\frac1{10})^2} + \frac{m+2}{1-\frac1{10}} \right) \\
&&&= \frac{1}{10^{m-1}} \left ( \frac{9m+28}{81}\right )
\end{align*}
Therefore for this will be less than $10^{-9}$, when $m = 11$, so our approximation is valid to 9sf
Noticeably less popular than Q1 – with only around 500 "hits" – and with a very much poorer mean mark of about 8, it was rather obvious that many candidates were very unsure as to what constituted the best partial fraction form for the given algebraic fraction to begin with. Then, with very little direct guidance being given in the question, candidates' confidence seemed to ebb visibly as they proceeded, being required to turn the resulting collection of single algebraic fractions into series, using the Binomial Theorem, and then into a consideration of general terms. There was much fudging of these general terms in order to get the given answers of either n + 1 or n + 2 for the general term's coefficients; even amongst those who did spot which one occurred when, there was often little visible justification to support the conclusions. As a result of all the hurdles to be cleared, those who managed to get to the numerical ending successfully were very few in number.