Year: 2023
Paper: 3
Question Number: 7
Course: LFM Pure
Section: Integration
No solution available for this problem.
The total entry was a marginal increase on that of 2022 (by just over 1%). Two questions were attempted by more than 90% of candidates, another two by 80%, and another two by about two thirds. The least popular questions were attempted by more than a sixth of candidates. All the questions were perfectly answered by at least three candidates (but mostly more than this), with one being perfectly answered by eighty candidates. Very nearly 90% of candidates attempted no more than 7 questions. One general comment regarding all the questions is that candidates need to make sure that they read the question carefully, paying particular attention to command words such as "hence" and "show that".
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Let $\mathrm{f}$ be a continuous function defined for $0 \leqslant x \leqslant 1$. Show that
\[\int_0^1 \mathrm{f}(\sqrt{x})\,\mathrm{d}x = 2\int_0^1 x\,\mathrm{f}(x)\,\mathrm{d}x\,.\]
\item Let $\mathrm{g}$ be a continuous function defined for $0 \leqslant x \leqslant 1$ such that
\[\int_0^1 \big(\mathrm{g}(x)\big)^2\,\mathrm{d}x = \int_0^1 \mathrm{g}(\sqrt{x})\,\mathrm{d}x - \frac{1}{3}\,.\]
Show that $\displaystyle\int_0^1 \big(\mathrm{g}(x) - x\big)^2\,\mathrm{d}x = 0$ and explain why $\mathrm{g}(x) = x$ for $0 \leqslant x \leqslant 1$.
\item Let $\mathrm{h}$ be a continuous function defined for $0 \leqslant x \leqslant 1$ with derivative $\mathrm{h}'$ such that
\[\int_0^1 \big(\mathrm{h}'(x)\big)^2\,\mathrm{d}x = 2\mathrm{h}(1) - 2\int_0^1 \mathrm{h}(x)\,\mathrm{d}x - \frac{1}{3}\,.\]
Given that $\mathrm{h}(0) = 0$, find $\mathrm{h}$.
\item Let $\mathrm{k}$ be a continuous function defined for $0 \leqslant x \leqslant 1$ and $a$ be a real number, such that
\[\int_0^1 \mathrm{e}^{ax}\big(\mathrm{k}(x)\big)^2\,\mathrm{d}x = 2\int_0^1 \mathrm{k}(x)\,\mathrm{d}x + \frac{\mathrm{e}^{-a}}{a} - \frac{1}{a^2} - \frac{1}{4}\,.\]
Show that $a$ must be equal to $2$ and find $\mathrm{k}$.
\end{questionparts}
This was the third most popular question and was only marginally less successfully attempted than questions 1 and 11 with over 9.4/20 the mean mark. It was generally answered well by candidates, with many candidates earning more than half the marks for this question and many candidates earning full, or close to full, marks. The majority of candidates successfully earned full credit in part (i). They also did well on part (ii) though quite a number did not use (g(x) − x)² ≥ 0 when justifying why g(x) = x, or incorrectly stated that (g(x) − x)² > 0. In part (iii), candidates who integrated 2∫₀¹ xh′(x)dx by parts generally went on to earn full, or close to full, marks for this part. A number of candidates began by writing down the equation from (ii) with h′(x) in place of g(x). In some cases, candidates successfully 'worked backwards', cancelling down each side to verify their initial equality, however less successful attempts simply assumed the initial equation, without justification. Many candidates observed that part (iv) could be solved by considering the integral ∫₀¹(e^(½ax)k(x) − e^(−½ax))²dx. Sadly, many candidates failed to factorise the resulting quadratic in 1/a, and another common error was simply to set the quadratic equal to zero without valid justification. Many, too, tried (unsuccessfully) to solve using integration by parts.