Year: 2010
Paper: 2
Question Number: 4
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: 1516.0
Banger Comparisons: 1
\begin{questionparts}
\item Let
\[
I=\int_0^a \frac {\f(x)}{\f(x)+\f(a-x)} \, \d x\,.
\]
Use a substitution to show that
\[
I =
\int_0^a \frac {\f(a-x)}{\f(x)+\f(a-x)} \, \d x\,
\]
and hence evaluate $I$ in terms of $a$.
Use this result to evaluate the integrals
\[
\int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x
\ \ \ \ \ \ \text{ and }\ \ \ \ \
\int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x
\,.
\]
\item
Evaluate
\[
\int_{\frac12}^2 \frac {\sin x}{x \big(\sin x + \sin \frac 1 x\big)}
\, \d x\,.
\]
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& I &= \int_0^a \frac{f(x)}{f(x)+f(a-x)} \d x \\
u =a-x, \d u = - \d x: &&& \int_{u=a}^{u=0} \frac{f(a-u)}{f(a-u)+f(u)} (-1) \d u \\
&&&= \int_0^a \frac{f(a-u)}{f(u)+f(a-u)} \d u \\
&&&= \int_0^a \frac{f(a-x)}{f(x)+f(a-x)} \d x \\
\Rightarrow && 2 I &= \int_0^a \left ( \frac{f(x)}{f(x)+f(a-x)} + \frac{f(a-x)}{f(x)+f(a-x)} \right) \d x \\
&&&= \int_0^a 1 \d x \\
&&&= a \\
\Rightarrow && I &= \frac{a}{2}
\end{align*}
\begin{align*}
&& J &= \int_0^1 \frac{\ln (x+1)}{\ln (2+x-x^2)}\, \d x \\
&&&= \int_0^1 \frac{\ln (x+1)}{\ln((x+1)(2-x))} \d x \\
&&&= \int_0^1 \frac{\ln (x+1)}{\ln(x+1) + \ln ((1-x)+1)} \d x \\
&&&= \frac{1}{2} \tag{$f(x) = \ln (x+1)$} \\
\\
&& K &= \int_0^{\frac\pi 2} \frac{\sin x } {\sin(x+\frac \pi 4 )} \, \d x \\
&&&= \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4}} \\
&&&= \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x }{\sin x + \sin (\frac{\pi}{2}-x)} \d x\\
&&&= \frac{\pi}{2\sqrt{2}}
\end{align*}
\item $\,$ \begin{align*}
&&I &= \int_{\frac12}^2 \frac{\sin x }{x(\sin x + \sin \frac1x)} \d x \\
u = 1/x, \d u = -1/x^2 \d x : &&&= \int_{u = 2}^{u=\frac12} \frac{\sin \frac1u}{\frac{1}{u}(\sin \frac1u + \sin u)} (-\frac{1}{u^2} ) \d u \\
&&&= \int_{\frac12}^2 \frac{\sin \frac1u}{u (\sin u + \sin \frac1u)} \d u \\
\Rightarrow && 2I &= \int_{\frac12}^2 \left ( \frac{\sin x }{x(\sin x + \sin \frac1x)} + \frac{\sin \frac1x }{x(\sin x + \sin \frac1x)}\right) \d x \\
&&&= \int_{\frac12}^2 \frac{1}{x} \d x\\
&&&= 2\ln2 \\
\Rightarrow && I &= \ln 2
\end{align*}
\end{questionparts}
This question received about the same number of "hits" as Q1 and came out with an average mark only fractionally lower. For the majority, the introductory work was successfully completed along with the rest of (i), although a lot of candidates' working was very unclear in the first integral, involving logarithms. One or two marks were commonly lost as the correct answer of ½ could easily have been guessed from the initial result, and the working produced by the candidates failed to convince markers that it had been obtained legitimately otherwise. The fault was often little more than failing either to identify the relevant "f(x)" or to show it implicitly by careful presentation of the working of the log. function. The excellent part (ii) required candidates to mimic the method used to find the opening result rather than repeat its use in a new case, and this was only accessible to those with that extra bit of insight or determination.