Year: 2008
Paper: 2
Question Number: 6
Course: LFM Pure
Section: Trigonometry 2
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: 1484.0
Banger Comparisons: 1
A curve has the equation $y=\f(x)$, where
\[
\f(x) = \cos \Big( 2x+ \frac \pi 3\Big) + \sin \Big ( \frac{3x}2 - \frac
\pi 4\Big).
\]
\begin{questionparts}
\item Find the period of $\f(x)$.
\item Determine all values of $x$ in the interval $-\pi\le x \le \pi$ for which $\f(x)=0$.
Find a value of $x$ in this interval at which
the curve touches the $x$-axis
without crossing it.
\item Find the value or values
of $x$ in the interval $0\le x \le 2\pi$ for which
$\f(x)=2\,$.
\end{questionparts}\begin{align*}
&& f(x) &= \cos \left( 2x+ \frac \pi 3\right) + \sin \left ( \frac{3x}2 - \frac\pi 4\right) \\
&&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{\pi}{2} - \left ( \frac{3x}2 - \frac\pi 4\right) \right)\\
&&&= \cos \left( 2x+ \frac \pi 3\right) + \cos\left (\frac{3\pi}{4} - \frac{3x}2 \right)\\
&&&= 2 \cos \left (\frac{2x+ \frac \pi 3+\frac{3\pi}{4} - \frac{3x}2}{2} \right) \cos \left ( \frac{\left (2x+ \frac \pi 3 \right) - \left (\frac{3\pi}{4} - \frac{3x}2 \right)}{2} \right)\\
&&&= 2 \cos \left (\frac{\frac{x}{2}+ \frac {13\pi}{12}}{2} \right) \cos \left ( \frac{\frac{7x}{2}- \frac {5\pi}{12}}{2} \right)\\
&&&= 2 \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \cos \left ( \frac{7x}{4}- \frac {5\pi}{24} \right)\\
\end{align*}
\begin{questionparts}
\item The period of $f$ will be the LCM of $\frac{2\pi}{\pi}$ and $\frac{2\pi}{\frac32} = \frac{4\pi}{3}$ which is $4\pi$. (This is also clear from the factorised form).
\item $f(x) = 0$ means one of those two factors is zero, ie
\begin{align*}
\text{first factor}: && 0 &= \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \\
&&n\pi + \frac{\pi}{2}&= \frac{x}{4}+ \frac {13\pi}{24} \\
\Rightarrow && x &= 4n\pi - \frac{\pi}{6} \\
\Rightarrow && x &= -\frac{\pi}{6} \\
\\
\text{second factor}: && 0 &= \cos \left ( \frac{7x}{4}- \frac {5\pi}{24} \right) \\
&& n\pi + \frac{\pi}{2} &= \frac{7x}{4}- \frac {5\pi}{24} \\
\Rightarrow && 7x &= 4n\pi + \frac{17}{6}\pi \\
\Rightarrow && x &= \frac{4n}7\pi + \frac{17}{42}\pi \\
\Rightarrow && x &= -\frac{31}{42} \pi, -\frac16\pi, \frac{17}{42}\pi, \frac{41}{42}\pi
\end{align*}
Therefore all solutions are $-\frac{31}{42} \pi, -\frac16\pi, \frac{17}{42}\pi, \frac{41}{42}\pi$
We can see that $-\frac{\pi}{6}$ is a repeated root, therefore it touches the axis and does not cross.
\item $f(x) = 2$ requires both factors to be $1$ or $-1$.
\begin{align*}
\text{first factor}: && \pm1 &= \cos \left (\frac{x}{4}+ \frac {13\pi}{24} \right) \\
&&n\pi &= \frac{x}{4}+ \frac {13\pi}{24} \\
\Rightarrow && x &= 4n\pi - \frac{13\pi}{6} \\
\Rightarrow && x &= \frac{11}{6}\pi \\
\end{align*}
We only need to test this value, where it's $-1$, so we look at $ \cos \left ( \frac{77\pi}{24}- \frac {5\pi}{24} \right) = \cos (3\pi) = -1$, so the only value is $\frac{11}{6}\pi$
\end{questionparts}
This was the least popular of the pure maths questions. Although there were 300 starts to the question, most of these barely got into the very opening part before the attempt was abandoned in favour of another question. Most attempts failed to show that f(x) has a period of 4π. As mentioned, few proceeded further. Of those who did, efforts were generally very poor indeed – as testified to by the very low mean mark of 4 – with the necessary comfort in handling even the most basic of trig. identities being very conspicuous by its absence. Part (iii) was my personal favourite amongst the pure questions, as it contained a very uncommon – yet remarkably simple – idea in order to get started on the road to a solution. The idea is simply this: f(x), being the sum of a cosine term and sine term, is equal to 2 if and only if each of these separate terms is simultaneously at its maximum of 1. That is, the question is actually two very easy trig. equations disguised as one very complicated-looking one. Once realised, the whole thing becomes very straightforward indeed, but only a few candidates had persevered this far.