Year: 2014
Paper: 3
Question Number: 6
Course: UFM Pure
Section: Hyperbolic functions
No solution available for this problem.
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: 1516.0
Banger Comparisons: 1
Starting from the result that
\[
\.h(t) >0\ \mathrm{for}\ 0< t < x \Longrightarrow \int_0^x \.h(t)\ud t
> 0 \,,
\]
show that, if $\.f''(t) > 0$ for $0 < t < x_0$ and $\.f(0)=\.f'(0) =0$,
then $\.f(t)>0$ for $0 < t < x_0$.
\begin{questionparts}
\item Show that,
for $0 < x < \frac12\pi$,
\[
\cos x \cosh x <1
\,.
\]
\item Show that, for $0 < x < \frac12\pi$,
\[
\frac 1 {\cosh x} < \frac {\sin x} x < \frac x {\sinh x} \,.
\]
% \item Show that, for $0 < x < \frac12\pi$, $\tanh x < \tan x$.
\end{questionparts}
The third most popular question, as well as the third most successful being only marginally behind question one in marks, having been attempted by about 70%. Many did not use the required starting point, instead resorting to monotonicity or drawing pictures (graphs) which were not proofs. In parts (i) especially and (ii) as well, candidates failed to use the result that f″(x) ≥ 0, cavalierly using f′(x) ≥ 0, or even f(x) ≥ 1 without justification. Many made complicated choices of functions for (i) and (ii), and then got lost in their differentiations, and finally there was frequent lack of care to ensure that quantities dividing inequalities were positive.