Year: 2008
Paper: 2
Question Number: 1
Course: LFM Pure
Section: Proof
No solution available for this problem.
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: 1500.7
Banger Comparisons: 2
A sequence of points $(x_1,y_1)$, $(x_2,y_2)$, $\ldots$ in the
cartesian plane is generated by first choosing $(x_1,y_1)$ then
applying the rule, for $n=1$, $2$, $\ldots$,
\[
(x_{n+1}, y_{n+1}) = (x_n^2-y_n^2 +a, \; 2x_ny_n+b+2)\,,
\]
where $a$ and $b$ are given real constants.
\begin{questionparts}
\item In the case $a=1$ and $b=-1$, find the values
of $(x_1,y_1)$ for which the sequence is constant.
\item Given that $(x_1,y_1) = (-1,1)$, find the values
of $a$ and $b$ for which the sequence has period
2.
\end{questionparts}
The first question is invariably intended to be a gentle introduction to the paper, and to allow all candidates to gain some marks without making great demands on either memory or technical skills. As such, most candidates traditionally tend to begin with question 1, and this proved to be the case here. Almost 700 candidates attempted this question, making it (marginally) the second most popular question on the paper; and it gained the highest mean score of about 14 marks. There were still several places where marks were commonly lost. In (i), setting (x₂, y₂) = (x₁, y₁) and eliminating y (for instance) leads to a quartic equation in x. There were two straightforward linear factors easily found to the quartic expression, leaving a quadratic factor which could yield no real roots. Many candidates failed to explain why, or show that, this was so. In (ii), the algebra again leads to two solutions, gained by setting (x₃, y₃) = (x₁, y₁). However, one of them corresponds to one of the solutions already found in (i), where the sequence is constant, and most candidates omitted either to notice this or to discover it by checking. Another very common oversight – although far less important in the sense that candidates could still gain all the marks by going the long way round – was that the algebra in (ii) was exactly the same as that in (i), but with a = –x and b = –y. For the very few who noticed this, the working for the second half of the question was remarkably swift.