Year: 2009
Paper: 1
Question Number: 7
Course: LFM Pure
Section: Integration
There were significantly more candidates attempting this paper again this year (over 900 in total), and the scores were pleasing: fewer than 5% of candidates failed to get at least 20 marks, and the median mark was 48. The pure questions were the most popular as usual; about two-thirds of candidates attempted each of the pure questions, with the exceptions of question 2 (attempted by about 90%) and question 5 (attempted by about one third). The mechanics questions were only marginally more popular than the probability and statistics questions this year; about one quarter of the candidates attempted each of the mechanics questions, while the statistics questions were attempted by about one fifth of the candidates. A significant number of candidates ignored the advice on the front cover and attempted more than six questions. In general, those candidates who submitted answers to eight or more questions did fairly poorly; very few people who tackled nine or more questions gained more than 60 marks overall (as only the best six questions are taken for the final mark). This suggests that a skill lacking in many students attempting STEP is the ability to pick questions effectively. This is not required for A-levels, so must become an important part of STEP preparation. Another "rubric"-type error was failing to follow the instructions in the question. In particular, when a question says "Hence", the candidate must make (significant) use of the preceding result(s) in their answer if they wish to gain any credit. In some questions (such as question 2), many candidates gained no marks for the final part (which was worth 10 marks) as they simply quoted an answer without using any of their earlier work. There were a number of common errors which appeared across the whole paper. These included a noticeable weakness in algebraic manipulations, sometimes indicating a serious lack of understanding of the mathematics involved. As examples, one candidate tried to use the misremembered identity cos β = sin √(1 − β²), while numerous candidates made deductions of the form "if a² + b² = c², then a + b = c" at some point in their work. Fraction manipulations are also notorious in the school classroom; the effects of this weakness were felt here, too. Another common problem was a lack of direction; writing a whole page of algebraic manipulations with no sense of purpose was unlikely to either reach the requested answer or gain the candidate any marks. It is a good idea when faced with a STEP question to ask oneself, "What is the point of this (part of the) question?" or "Why has this (part of the) question been asked?" Thinking about this can be a helpful guide. One aspect of this is evidenced by pages of formulæ and equations with no explanation. It is very good practice to explain why you are doing the calculation you are, and to write sentences in English to achieve this. It also forces one to focus on the purpose of the calculations, and may help avoid some dead ends. Finally, there is a tendency among some candidates when short of time to write what they would do at this point, rather than using the limited time to actually try doing it. Such comments gain no credit; marks are only awarded for making progress in a question. STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1484.0
Banger Comparisons: 1
Show that, for any integer $m$,
\[
\int_0^{2\pi} \e^x \cos mx \, \d x = \frac
{1}{m^2+1}\big(\e^{2\pi}-1\big)\,.
\]
\begin{questionparts}
\item
Expand
$\cos(A+B) +\cos(A-B)$. Hence show that
\[\displaystyle
\int_0^{2\pi} \e^x \cos x \cos 6x \, \d x\,
= \tfrac{19}{650}\big( \e^{2\pi}-1\big)\,.
\]
\item Evaluate $\displaystyle
\int_0^{2\pi} \e^x \sin 2x \sin 4x \cos x \, \d x\,$.
\end{questionparts}\begin{align*}
&& I &= \int_0^{2 \pi} e^{x} \cos m x \d x \\
&&&= \left [e^x \cos m x \right]_0^{2 \pi}-\int_0^{2 \pi} e^x m (-\sin mx) \d x\\
&&&= e^{2\pi}-1 + m\int_0^{2\pi}e^x \sin m x \d x \\
&&&= e^{2\pi}-1 + m\left [e^x \sin m x \right]_0^{2\pi} - m \int_0^{2\pi} e^x m \cos x \d x \\
&&&= e^{2\pi}-1+0 - m^2 I\\
\Rightarrow && (m^2+1)I &= e^{2\pi}-1 \\
\Rightarrow && I &= \frac{1}{m^2+1} (e^{2\pi}-1)
\end{align*}
\begin{questionparts}
\item $\,$
\begin{align*}
&& \cos(A+B) + \cos(A-B) &= \cos A\cos B - \sin A \sin B + \cos A \cos B + \sin A \sin B \\
&&&= 2 \cos A \cos B
\end{align*}
Therefore
\begin{align*}
&& I &= \int_0^{2\pi} e^x \cos x \cos 6x \d x \\
&&&= \int_0^{2\pi} e^x \frac12\left (\cos 7x + \cos 5x \right) \d x\\
&&&= \left ( \frac{1}{2(1+7^2)} + \frac1{2(1+5^2)}\right)(e^{2\pi}-1) \\
&&&= \left (\frac{1}{100}+\frac{1}{52} \right) (e^{2\pi}-1) \\
&&&= \frac{19}{650}(e^{2\pi}-1)
\end{align*}
\item $\,$
\begin{align*}
&& I &= \int_0^{2\pi} e^x \sin 2x \sin 4x \cos x \d x\\
&&&= \int_0^{2\pi} e^x \tfrac12(\cos2x-\cos 6x) \cos x \d x\\
&&&= \frac12 \int_0^{2\pi} e^x \left (\cos 2x \cos x -\cos 6x \cos x \right) \d x \\
&&&= \frac14 \int_0^{2\pi} e^x \left (\cos 3x + \cos x-\cos 7x-\cos 5x \right) \d x \\
&&&= \frac14 \left (\frac{1}{1+3^2}+\frac{1}{1+1^2}-\frac{1}{1+7^2} - \frac{1}{1+5^2} \right)(e^{2\pi}-1) \\
&&&= \frac14 \left (\frac{1}{10}+\frac{1}{2}-\frac{1}{50} - \frac{1}{26} \right)(e^{2\pi}-1) \\
&&&= \frac{44}{325}(e^{2\pi}-1)
\end{align*}
\end{questionparts}
Candidates found this question very approachable, and was one of the highest scoring questions on the paper. The standard integral at the start was generally well done, and about three-quarters of candidates gained full marks for it. Common mistakes included incorrect notation for integration by parts, forgetting to use the [···]₀²π notation when using parts with a definite integral, and of course making sign errors. The trigonometrical identity in part (i) was done perfectly by most candidates who tried it. The following integral was usually performed successfully, and over half of all attempts got this far. The errors which occurred were similar to above, along with the standard inability to add and simplify fractions. The integral in part (ii) caused significantly more problems. Most candidates wrote down the product formula for sin A sin B, but then many got stuck: they could not see how to proceed. It is very common in STEP questions for the later parts of a question to depend on the ideas developed earlier parts. In this case, we end up after expansion with two integrals like the one seen in part (i), but it seems as though this was not noticed. The other common problem which occurred in this part was missing or losing the factor of ½ in this part; this is another example of algebraic carelessness costing marks.