Year: 2010
Paper: 2
Question Number: 2
Course: LFM Pure
Section: Integration
There were just under 1000 entries for paper II this year, almost exactly the same number as last year. Of this number, more than 60 scored over 90% while, at the other end of the scale, almost 200 failed to score more than 40 marks. In hindsight, many of the pure maths questions were a little too accessible and lacked a sufficiently tough 'difficulty gradient', so that scores were slightly higher than anticipated. This was reflected in the grade boundaries for the "1" and the "2" (around ten marks higher than is generally planned) in particular. Next year's questions may be expected to be a little bit more demanding, but only in the sense that the final 5 or 6 marks on each question should have rather more bite to them: it should certainly not be the case that all questions are tougher to get into at the outset. Most candidates attempted the requisite number of questions (six), although many of the weaker brethren made seven or eight attempts, most of which were feeble at best and they generally only picked up a maximum of 5 or 6 marks per question. It is a truth universally acknowledged that practice maketh if not perfect then at least a whole lot better prepared, and choosing to waste time on a couple of extra questions is not a good strategy on the STEPs. The major down-side of the present modular examination system is that students are not naturally prepared to approach the subject holistically; ally this to the current practice of setting highly-structured, fully-guided questions requiring no imagination, insight, depth or planning from A-level candidates in a system that fails almost nobody and rewards even the most modestly able with high grades in a manner reminiscent of a dentist giving lollipops to kids who have done little more than been brave and seen the course through, it is even more important to ensure a full and thorough preparation for these papers. The 20% of the entry who seem to be either unprepared for the rigours of a STEP, or unwittingly possessed of only a smattering of basic advanced-level skills, seems to be remarkably steady year-on-year, even in a year when their more suitably prepared compatriots found the paper appreciably easier than usual. 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.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1506.8
Banger Comparisons: 4
Prove that
\[
\cos 3x = 4 \cos^3 x - 3 \cos x \,.
\]
Find and prove a similar result for $\sin 3x$ in terms of $\sin x$.
\begin{questionparts}
\item Let
\[
{\rm I}(\alpha) = \int_0^\alpha \big(7\sin x - 8 \sin^3 x\big) \d x\,.
\]
Show that
\[
{\rm I}(\alpha) = -\tfrac 8 3 c^3 + c +\tfrac5 3\,,
\]
where $c = \cos \alpha$.
Write down one value of $c$ for which ${\rm I}(\alpha) =0$.
\item Useless Eustace believes that
\[
\int \sin^n x \, \d x =\dfrac {\sin^{n+1}x}{n+1}\,
\]
for $n=1, \ 2, \ 3, \ldots\, $.
Show that Eustace would obtain the correct value of ${\rm I}(\beta)\,$, where $\cos \beta= -\frac16$.
Find all values of $\alpha$ for which he would obtain the correct value of ${\rm I}(\alpha)$.
\end{questionparts}\begin{align*}
\cos 3x &\equiv \cos (2x + x) \\
&\equiv \cos 2x \cos x - \sin 2x \sin x \\
&\equiv (2\cos^2 x - 1) \cos x - 2 \sin x \cos x \sin x \\
&\equiv 2 \cos^3 x - \cos x - 2\cos x (\sin^2 x) \\
&\equiv 2 \cos^3 x - \cos x - 2\cos x (1- \cos^2 x) \\
&\equiv 4\cos^3 x - 3\cos x
\end{align*}
Similarly,
\begin{align*}
\sin 3x &\equiv \sin (2x + x) \\
&\equiv \sin 2x \cos x + \cos 2x \sin x \\
&\equiv 2 \sin x \cos x \cos x + (1-2\sin^2 x) \sin x \\
&\equiv 2 \sin x (1-\sin^2 x) + \sin x - 2 \sin^3 x \\
&\equiv 3 \sin x -4 \sin ^3 x
\end{align*}
\begin{questionparts}
\item \begin{align*}
I(\alpha) &= \int_0^{\alpha} (7 \sin x - 8 \sin^3 x) \d x \\
&= \int_0^{\alpha} (7 \sin x - (6\sin x-2 \sin 3x) ) \d x \\
&= \int_0^{\alpha} (\sin x +2 \sin 3x ) \d x \\
&= -\cos \alpha - \frac23 \cos 3\alpha +1+\frac23 \\
&= -c - \frac23 (4c^3-3c) + \frac53 \\
&= -\frac83 c^3 +c + \frac53
\end{align*} as required.
When $c = -1$ this value is $0$.
Eustace will obtain the value $\frac{7}{2} \sin^2 \beta - 2 \sin^4 \beta = \frac72 (1-\cos^2 \beta) - 2(1-\cos^2 \beta)^2 = \frac32 + \frac12\cos^2 \beta -2\cos^4 \beta$
So if $\cos \beta = -\frac16$ he will obtain $\frac32 + \frac{1}{2\cdot36} - \frac{2}{6^4}$ and he should obtain $\frac{8}{3} \frac{1}{6^3} - \frac{1}{6} + \frac{5}{3}$ which are equal.
We want to find all roots of:
\begin{align*}
&& \frac32 + \frac12 c^2 - 2c^4 &= -\frac83 c^3+ c + \frac53 \\
\Rightarrow && 0 &=2c^4-\frac83c^3-\frac12 c^2+c +\frac{1}{6} \\
&&&= 12c^4-16c^3-3c^2+6c+1\\
&&&= (6c+1)(2c^3-3c^2+1) \\
&&&= (6c+1)(2c+1)(c-1)^2
\end{align*}
Therefore $\cos \alpha = - \frac16, -\frac12, 1$ will give the correct answers.
\end{questionparts}
This was the most popular question on the paper, drawing an attempt from almost every candidate. There were several proofs of the initial trigonometric identities using de Moivre's Theorem but most settled for the more standard cosine and sine of (2x + x). Personally, I was against the inclusion of the given answer of cos⁻¹ 16 in (ii) as it led to what struck me as an unwelcome dichotomy of approaches. Most candidates opted to verify that the two polynomials in "c" that arose gave the same numerical answer, and this working was not entirely straightforward – in the event, lots of candidates failed to show the markers that they had done the working correctly for both expressions – whereas my original intention had been that they should collect terms up into a single polynomial equation and factorise it by first spotting the (repeated) factor (c – 1) hinted at in (i). There was one important mathematical oversight that many candidates made during this question, and it was due to not reading the question sufficiently carefully. The wording of the question in (ii) clearly states that Eustace's misunderstanding of the integration of powers of the sine function was for n = 1, 2, 3, … . Unfortunately, rather a lot of candidates thought that he would then have integrated sin x (i.e. the case n = 1) correctly as –cos x. We concocted a mark-scheme for this eventuality which allowed candidates 'follow-through' for 6 out of the 10 marks allocated here, but the self-imposed penalty of four marks could not be avoided as it was just no longer possible to get, for instance, the given answer. Finally, there is a bit of an apology to make: at some final stage of the printing process, the bit of the question that identified α as lying in the range 0 to π got removed; this left candidates having to think about general solutions rather than just the two decently small ones that had been looked-for when the question was first written. Nevertheless, not only was this the most popular question for number of attempts, it was also the most successful for candidates with a mean score of almost 15.