Year: 2016
Paper: 2
Question Number: 4
Course: LFM Stats And Pure
Section: Quadratics & Inequalities
As in previous years the Pure questions were the most popular of the paper with questions 1, 3 and 7 the most popular of these. The least popular questions of the paper were questions 10, 11, 12 and 13 with fewer than 400 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained, given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Let
\[
y=\dfrac{x^2+x\sin\theta+1}{x^2+x\cos\theta+1}
\,.\]
\begin{questionparts}
\item
Given that
$x$ is real, show that
\[
(y\cos\theta -\sin\theta)^2 \ge 4 (y-1)^2
\,.
\]
Deduce that
\[
y^2+1
\ge
4(y-1)^2
\,,
\]
and hence that
\[
\dfrac {4-\sqrt7}3
\le y \le
\dfrac {4+\sqrt7}3 \,.
\]
\item
In the case $y=
\dfrac {4+\sqrt7}3 \,$, show that \[\sqrt{y^2+1}=2(y-1)\]
and find the corresponding values of $x$ and
$\tan\theta$.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& y&=\frac{x^2+x\sin\theta+1}{x^2+x\cos\theta+1} \\
\Leftrightarrow && 0 &= x^2(y-1) + x(y \cos \theta - \sin \theta) + y-1 \\
\Leftrightarrow && 0 &\leq \Delta = (y\cos \theta - \sin \theta)^2 - 4(y-1)^2 \\
\Leftrightarrow && (y\cos \theta - \sin \theta)^2 &\geq 4(y-1)^2
\end{align*}
[Assuming that $y \neq 1$, if $y = 1$ then the RHS is $0$ and it is automatically satisfied].
Notice that $(y\cos \theta - \sin \theta)^2 \leq (y^2+1)(\cos^2 \theta + \sin^2 \theta)$ by Cauchy-Schwarz, so $y^2 + 1 \geq 4(y-1)^2$.
\begin{align*}
&& y^2 + 1 &\geq 4(y-1)^2 \\
\Leftrightarrow && 0 &\geq 3y^2-8y+3 \\
\text{c.v.} && y&= \frac{8 \pm \sqrt{64-4\cdot3 \cdot 3}}{6} \\
&&&= \frac{4 \pm \sqrt{16-9}}{3} = \frac{4 \pm \sqrt{7}}3
\end{align*}
so $\frac{4-\sqrt{7}}3 \leq y \leq \frac{4+\sqrt7}3$.
\item If $y = \frac{4+\sqrt7}3$ then $y - 1 = \frac{1+\sqrt7}3$ and since $y^2+1 = 4(y-1)^2$ taking square roots we obtain $\sqrt{y^2+1} = 2(y-1)$.
Since equality must hold in our C-S identity, we must have $\langle y, -1 \rangle$ parallel to $ \langle \cos \theta , \sin \theta \rangle$, ie $\tan \theta = -\frac{3}{4+\sqrt{7}}$ and
\begin{align*}
&& x & = \frac{-(y \cos \theta - \sin \theta) \pm \sqrt{\Delta}}{2(y-1)} \\
&&&= \frac{\pm2(y-1)}{2(y-1)} \\
&&&= \pm1
\end{align*}
\end{questionparts}
This question attracted many good responses which often successfully accomplished the first and third results of part (i) of the question, although in many cases marks were lost through incomplete explanations of some of the steps taken. Part (ii) of the question was generally answered poorly with some candidates simply stating that the square root of one expression in terms of surds was equal to the other one that they had achieved. Only a small number of candidates successfully completed the last section of this question.