Year: 2014
Paper: 3
Question Number: 4
Course: LFM Pure
Section: Integration
A 10% increase in the number of candidates and the popularity of all questions ensured that all questions had a good number of attempts, though the first two questions were very much the most popular. Every question received at least one absolutely correct solution. In most cases when candidates submitted more than six solutions, the extra ones were rarely substantial attempts. Five sixths gave in at least six attempts.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Let
\[
I = \int_0^1 \bigl((y')^2 -y^2\bigr)\d x \qquad\text{and}\qquad
I_1=\int_0^1 (y'+y\tan x)^2 \d x \,,
\]
where $y$ is a given function of $x$ satisfying $y=0$ at $x=1$. Show that $I-I_1=0$ and deduce that $I\ge0$. Show further that $I=0$ only if $y=0$ for all $x$ ($0\le x \le 1$).
\item Let
\[
J = \int_0^1 \bigl((y')^2 -a^2y^2\bigr)\d x
\,,
\]
where $a$ is a given positive constant and $y$ is a given function of $x$, not identically zero, satisfying $y=0$ at $x=1$. By considering an integral of the form
\[
\int_0^1 (y'+ay\tan bx)^2 \d x \,,
\]
where $b$ is suitably chosen, show that $J\ge0$. You should state the range of values of $a$, in the form $a < k$, for which your proof is valid. In the case $a=k$, find a function $y$ (not everywhere zero)
such that $J=0$.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& I - I_1 &= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) \d x - \int_0^1 \left ( y' + y \tan x \right)^2 \d x\\
&&&= \int_0^1 \left ( \left ( y' \right)^2 - y^2 \right) - \left ( y' + y \tan x \right)^2 \d x\\
&&&= \int_0^1 \left (-y^2-2yy' \tan x - y^2 \tan^2 x \right) \d x\\
&&&= \int_0^1 \left (-2yy' \tan x - y^2(1+ \tan^2 x )\right) \d x\\
&&&= \int_0^1 \left (-2yy' \tan x - y^2 \sec^2 x\right) \d x\\
&&&= \int_0^1 -\frac{\d}{\d x} \left (y^2 \tan x \right) \d x\\
&&&= \left [-y^2 \tan x \right]_0^1 \\
&&&= 0 \\
\\
\Rightarrow && I &= I_1 = \int_0^1 \left ( y' + y \tan x \right)^2 d x \geq 0
\end{align*}
The only way $I_0 = 0$ is is $y' + y \tan x =0$, so
\begin{align*}
&& \frac{\d y}{\d x} &= - y \tan x \\
\Rightarrow && \int \frac{1}{y} &= \int -\tan x \d x \\
\Rightarrow && \ln |y| &= \ln |\cos x| + C \\
\Rightarrow && y &= A \cos x \\
\Rightarrow && A &= 0 \Rightarrow y = 0
\end{align*}
\item Let $J_1 = \int_0^1 (y'+ay\tan ax)^2 \d x$, then
\begin{align*}
&& J-J_1 &= \int_0^1 \left ( \left ( y' \right)^2 - a^2y^2 \right) - \left ( y' + ya \tan ax \right)^2 \d x\\
&&&= \int_0^1 \left (-a^2y^2-2yy' a \tan a x-y^2a^2 \tan^2 ax \right) \d x \\
&&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2(1+\tan^2 ax) \right) \d x \\
&&&= \int_0^1 \left (-2yy' a \tan ax - a^2y^2\sec^2 ax \right) \d x \\
&&&= \left [ - a y^2 \tan a x \right]_0^1 = 0
\end{align*}
This is true if $a < \frac{\pi}{2}$, since otherwise we might care about the order of the zero for $y$ at $x = 1$.
Consider $y = \cos \frac{\pi}{2} x$, then $y' = -\frac{\pi}{2} \sin^2\frac{\pi}{2} x$ and
\begin{align*}
&& \int_0^1 \frac{\pi^2}{4} \left (\sin^2 \frac{\pi}{2}x - \cos^2 \frac{\pi}{2} x \right) \d x &= -\frac{\pi^2}{4} \int_0^1 \cos(\pi x) \d x \\
&&&= 0
\end{align*}
\end{questionparts}
Two thirds of the candidature attempted this but with only moderate success earning just a third of the marks. The very first result was frequently obtained although some fell at the first hurdle through not appreciating that they needed to use sec²θ = 1 + tan²θ, or else that there was then an exact differential. The second result in part (i) was 'only if' whereas many read it, or answered it, as 'if'. In part (ii), most spotted the substitution. There were many inappropriate functions suggested for the last part of the question, many which ignored the requirement that f(0) = 0, f(1) = 1.