Year: 2014
Paper: 2
Question Number: 2
Course: LFM Pure
Section: Integration
There were good solutions presented to all of the questions, although there was generally less success in those questions that required explanations of results or the use of diagrams and graphs to reach the solution. Algebraic manipulation was generally well done by many of the candidates although a range of common errors such as confusing differentiation and integration and simple arithmetic slips were evident. Candidates should also be advised to use the methods that are asked for in questions unless it is clear that other methods will be accepted (such as by the use of the phrase "or otherwise").
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
This question concerns the inequality
\begin{equation}
\int_0^\pi \bigl( f(x) \bigr)^2 \d x \le \int_0^\pi \bigl(
f'(x)\bigr)^2 \d x\,.\tag{$*$}
\end{equation}
\begin{questionparts}
\item Show that $(*)$ is satisfied in the case $f(x)=\sin nx$, where $n$ is a positive integer.
Show by means of counterexamples that $(*)$ is not
necessarily satisfied if either $f(0) \ne 0$ or $f(\pi)\ne0$.
\item You may now assume that $(*)$ is satisfied for any (differentiable) function $f$ for which $f(0)=f(\pi)=0$.
By setting $f(x) = ax^2 + bx +c$, where $a$, $b$ and $c$ are suitably chosen, show that
$\pi^2\le 10$.
By setting $f(x) = p \sin \frac12 x + q\cos \frac12 x +r$, where $p$, $q$ and $r$ are suitably chosen, obtain another inequality for $\pi$.
Which of these inequalities leads to a better estimate for $\pi^2\,$?
\end{questionparts}\begin{questionparts}
\item If $f(x) = \sin nx$ then $f'(x) = n \cos n x$ and so
\begin{align*}
&& LHS &= \int_0^\pi \sin^2 n x \d x \\
&&&= \left [ \frac{x+\frac1{2n}\sin 2n x}{2} \right ]_0^{\pi} \\
&&&= \frac{\pi}{2} \\
\\
&& RHS &= \int_0^{\pi} n^2 \cos^2 n x \d x \\
&&&= n^2 \left [ \frac{\frac{1}{2n}\sin 2n x + x}{2} \right]_0^{\pi} \\
&&&= n^2\frac{\pi}{2} \geq LHS
\end{align*}
[$f(0) = 0, f(\pi) \neq 0$] Suppose $f(x) = x$ then $f'(x) = 1$ and $LHS = \frac{\pi^3}{3} > \pi = RHS$.
[$f(0) \neq 0, f(\pi) = 0$] Suppose $f(x) = \pi - x$ then $f'(x) = -1$ and $LHS = \frac{\pi^3}{3} > \pi = RHS$
\item Suppose $f(x) = x(\pi - x)$ then $f'(x) = \pi - 2x$ and so
\begin{align*}
&& \int_0^\pi x^2(\pi-x)^2 \d x &\leq \int_0^\pi (\pi-2x)^2 \d x \\
\Leftrightarrow && \left [\pi^2 \frac{x^3}{3} - 2\pi \frac{x^4}{4} + \frac{x^5}{5} \right]_0^{\pi} &\leq \left [ \pi^2x - 4\pi \frac{x^2}{2} + \frac{4x^3}{3} \right]_0^{\pi} \\
\Leftrightarrow && \pi^5 \left (\frac13 - \frac12+\frac15 \right) &\leq \pi^3 \left ( 1 - 2+\frac43 \right) \\
\Leftrightarrow && \pi^2 \frac{1}{30} &\leq \frac13 \\
\Leftrightarrow && \pi^2 &\leq 10
\end{align*}
Suppose $f(x) = p\sin \tfrac12 x + q \cos \tfrac12 x + r$, so $f(0) = q + r$ and $f(\pi) = p + r$, so say $p = q = 1, r = -1$
\begin{align*}
&& LHS &= \int_0^{\pi} \left ( \sin \tfrac12 x + \cos \tfrac12 x-1\right)^2
\d x \\
&&&=\int_0^\pi \left ( \sin^2 \tfrac12 x + \cos^2 \tfrac12 x+1-2\sin \tfrac12 x - 2\cos \tfrac12 x+ \sin x \right)\\
&&&= \left [2x + 4\cos \tfrac12 x - 4\sin \tfrac12 x - \cos x \right]_0^{\pi} \\
&&&= \left ( 2\pi -4+1 \right) - \left ( 4-1 \right) \\
&&&= 2\pi -6\\
\\
&& RHS&= \int_0^{\pi} \left ( \tfrac12 \cos \tfrac12 x -\tfrac12 \sin \tfrac12 x\right)^2 \d x \\
&&&= \int_0^{\pi} \left ( \tfrac14 \cos^2 \tfrac12 x +\tfrac14 \sin^2 \tfrac12 x-\tfrac14 \sin x\right) \d x \\
&&&= \frac{\pi}{4} - \frac12 \\
\Rightarrow && 2\pi -6 &\leq \frac{\pi}{4} - \frac12 \\
\Rightarrow && \frac{7\pi}{4} &\leq \frac{11}{2} \\
\Rightarrow && \pi &\leq \frac{22}{7}
\end{align*}
$22^2/7^2 = 484/49 < 10$ therefore $\pi \leq \frac{22}{7}$ is the better estimate.
\end{questionparts}
This was one of the more popular questions of the paper. Most candidates successfully showed that the first inequality was satisfied, but when producing counterexamples, some failed to show that either f″ ≥ 0 or g″ ≥ 0 for their chosen functions. In the second part many candidates did not attempt to choose values of α, β and γ, but substituted the general form of the quadratic function into the inequality instead. In the case where the function involved trigonometric functions, many of those who attempted it were able to deduce that the required result, but several candidates made mistakes in the required integration. Those who established two inequalities were able to decide which gives the better estimate for π.