Year: 2011
Paper: 2
Question Number: 3
Course: LFM Pure
Section: Differentiation
There were just under 1000 entries for paper II this year, almost exactly the same number as last year. After the relatively easy time candidates experienced on last year's paper, this year's questions had been toughened up significantly, with particular attention made to ensure that candidates had to be prepared to invest more thought at the start of each question – last year saw far too many attempts from the weaker brethren at little more than the first part of up to ten questions, when the idea is that they should devote 25-40 minutes on four to six complete questions in order to present work of a substantial nature. It was also the intention to toughen up the final "quarter" of questions, so that a complete, or nearly-complete, conclusion to any question represented a significant (and, hopefully, satisfying) mathematical achievement. Although such matters are always best assessed with the benefit of hindsight, our efforts in these areas seem to have proved entirely successful, with the vast majority of candidates concentrating their efforts on four to six questions, as planned. Moreover, marks really did have to be earned: only around 20 candidates managed to gain or exceed a score of 100, and only a third of the entry managed to hit the half-way mark of 60. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were attempted by almost all candidates; 3 and 4 by around three-quarters of them; 6, 7 and 9 by around half; the remaining questions were less popular, and some received almost no "hits". Overall, the highest scoring questions (averaging over half-marks) were 1, 2 and 9, along with 13 (very few attempts, but those who braved it scored very well). This at least is indicative that candidates are being careful in exercising some degree of thought when choosing (at least the first four) 'good' questions for themselves, although finding six successful questions then turned out to be a key discriminating factor of candidates' abilities from the examining team's perspective. Each of questions 4-8, 11 & 12 were rather poorly scored on, with average scores of only 5.5 to 6.6.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, you may assume without proof that
any function $\f$ for which $\f'(x)\ge 0$ is \textit{increasing}; that is,
$\f(x_2)\ge \f(x_1)$ if $x_2\ge x_1\,$.
\begin{questionparts}
\item
\begin{enumerate}
\item
Let $\f(x) =\sin x -x\cos x$.
Show that $\f(x)$ is
increasing
for $0\le x \le \frac12\pi\,$
and deduce that $\f(x)\ge 0\,$ for
$0\le x \le \frac12\pi\,$.
\item Given that $\dfrac{\d}{\d x} (\arcsin x) \ge1$ for
$0\le x< 1$,
show that
\[
\arcsin x\ge x
\quad (0\le x < 1).
\]
\item Let $\g(x)= x\cosec x\, \text{ for }0< x < \frac12\pi$. Show that
$\g$ is increasing and deduce that
\[
({\arcsin x})\, x^{-1} \ge x\,{\cosec x}
\quad (0 < x < 1).
\]
\end{enumerate}
\item
Given that $\dfrac{\d}{\d x}
(\arctan x)\le 1\text{ for }x\ge 0$, show
by considering the function $x^{-1} \tan x$
that
\[
(\tan x)( \arctan x) \ge x^2
\quad
(0< x < \tfrac12\pi).
\]
\end{questionparts}\begin{questionparts}
\begin{enumerate}
\item Let $f(x) = \sin x - x \cos x$, then $f'(x) = \cos x - \cos x + x \sin x = x \sin x$ which is clearly non-negative when both $x$ and $\sin x$ are on that interval. Since $f(0) = 0 - 0 \cdot 1 = 0$ clearly $f(x) \geq 0$ on that interval.
\item Given $\frac{\d}{\d x} (\arcsin x) \geq 1$ we must have $\frac{\d}{ \d x} (\arcsin x - x) \geq 0$ for $0 \leq x < 1$, but since $\arcsin 0 - 0 = 0$ this means that $\arcsin x - x \geq 0$ on this interval, ie $ \arcsin x \geq x$.
\item Let $g(x) = x \cosec x$ then $g'(x) = \cosec x - x \cosec x \cot x = \cosec x(1 - x \cot x) = \cosec x \cot x (\tan x - x) \geq 0$ therefore $g$ is increasing.
\begin{align*}
&& g(\arcsin x) &\geq g(x) \tag{$f$ increasing} \\
\Rightarrow&& (\arcsin x) x^{-1} &\geq x \cosec x
\end{align*}
\end{enumerate}
\item Given $\frac{\d}{\d x} (\arctan x) \leq 1$ we must have $\frac{\d}{ \d x} (x-\arctan x) \geq 0$ for $x \geq 0$, but since $ 0 - \arctan 0 = 0$ this means that $x - \arctan x \geq 0$, ie $ \arctan x \geq x$ for $x \geq 0$
$g(x) = x^{-1} \tan x \Rightarrow g'(x) = -x^{-2}\tan x +x^{-1} \sec^2 x$. If we can show $f(x) = x \sec ^2 x - \tan x$ is positive that would be great. However $f'(x) = x 2 \tan x \sec^2 x \geq 0$ and $f(0) = 0$ so $f(x)$ is positive and $g'(x)$ is positive and hence increasing, therefore $g(x) \geq g(\arctan x) \Rightarrow \frac{\tan x}{x} \geq \frac{x}{\arctan x}$ from which the result follows.
\end{questionparts}
Again, despite the obvious presence of inequalities in the question, this was another very popular question, and was generally well-handled very capably in part (i), where the structure of the question provided the necessary support for successful progress to be made here. Part (ii) was less popular and less well-handled, even though the only significant difference between this and (i)(c) was (effectively) that the direction of the inequality was reversed. Although the intervals under consideration were clearly flagged, many candidates omitted to consider that, having shown the function increasing on this interval, they still needed to show something simple such as f(0) = 0 in order to show that f(x) ≥ 0 on this interval. A few also thought that f'(x) increasing implied that f(x) was also increasing.