Year: 2024
Paper: 2
Question Number: 8
Course: LFM Pure
Section: Integration
No solution available for this problem.
Many candidates produced good solutions to the questions, with the majority of candidates opting to focus on the pure questions of the paper. Candidates demonstrated very good ability, particularly in the area of manipulating algebra. Many candidates produced clear diagrams which in many cases meant that they were more successful in their attempts at their questions than those who did not do so. The paper also contained a number of places where the answer to be reached was given in the question. In such cases, candidates must be careful to ensure that they provide sufficient evidence of the method used to reach the result in order to gain full credit.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, the following theorem may be used without proof.
Let $u_1, u_2, \ldots$ be a sequence of real numbers. If the sequence is
\begin{itemize}
\item bounded above, so $u_n \leqslant b$ for all $n$, where $b$ is some fixed number
\item and increasing, so $u_n \leqslant u_{n+1}$ for all $n$
\end{itemize}
then there is a number $L \leqslant b$ such that $u_n \to L$ as $n \to \infty$.
For positive real numbers $x$ and $y$, define $\mathrm{a}(x,y) = \frac{1}{2}(x+y)$ and $\mathrm{g}(x,y) = \sqrt{xy}$.
Let $x_0$ and $y_0$ be two positive real numbers with $y_0 < x_0$ and define, for $n \geqslant 0$
\[ x_{n+1} = \mathrm{a}(x_n, y_n)\,, \]
\[ y_{n+1} = \mathrm{g}(x_n, y_n)\,. \]
\begin{questionparts}
\item By considering $(\sqrt{x_n} - \sqrt{y_n})^2$, show that $y_{n+1} < x_{n+1}$, for $n \geqslant 0$. Show further that, for $n \geqslant 0$
\begin{itemize}
\item $x_{n+1} < x_n$
\item $y_n < y_{n+1}$.
\end{itemize}
Deduce that there is a value $M$ such that $y_n \to M$ as $n \to \infty$.
Show that $0 < x_{n+1} - y_{n+1} < \frac{1}{2}(x_n - y_n)$ and hence that $x_n - y_n \to 0$ as $n \to \infty$.
Explain why $x_n$ also tends to $M$ as $n \to \infty$.
\item Let
\[ \mathrm{I}(p,q) = \int_0^{\infty} \frac{1}{\sqrt{(p^2 + x^2)(q^2 + x^2)}}\,\mathrm{d}x, \]
where $p$ and $q$ are positive real numbers with $q < p$.
Show, using the substitution $t = \frac{1}{2}\!\left(x - \dfrac{pq}{x}\right)$ in the integral
\[ \int_{-\infty}^{\infty} \frac{1}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}}\,\mathrm{d}t, \]
that
\[ \mathrm{I}(p,q) = \mathrm{I}\!\left(\mathrm{a}(p,q),\, \mathrm{g}(p,q)\right). \]
Hence evaluate $\mathrm{I}(x_0, y_0)$ in terms of $M$.
\end{questionparts}
This question was attempted by approximately half of the candidates. In part (i) candidates were often able to produce the necessary algebra, but some did not justify the strictness of the inequality or failed to use a full inductive structure for the proof. There was a roughly even split between candidates who identified that x₀ could be used as a bound in order to apply the given result and those who incorrectly used the fact that yₙ < xₙ. When considering the behaviour of (xₙ − yₙ) some candidates incorrectly asserted that the fact that the sequence is bounded below by 0 is sufficient to show that the sequence tends to 0. Almost all candidates were able to show that this result implies that the two sequences tend to the same limit. In part (ii) most candidates were able to apply the substitution, but some did not justify the new values of the limits or comment on the evenness of the integrand. The evaluation of the final integral was generally done well.