Year: 2011
Paper: 3
Question Number: 4
Course: UFM Pure
Section: Integration using inverse trig and hyperbolic functions
No solution available for this problem.
The percentages attempting larger numbers of questions were higher this year than formerly. More than 90% attempted at least five questions and there were 30% that didn't attempt at least six questions. About 25% made substantive attempts at more than six questions, of which a very small number indeed were high scoring candidates that had perhaps done extra questions (well) for fun, but mostly these were cases of candidates not being able to complete six good solutions.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1516.0
Banger Comparisons: 1
The following result applies to any
function $\f$
which is continuous, has positive gradient and satisfies
$\f(0)=0\,$:
\[
ab\le \int_0^a\f(x)\, \d x + \int_0^b \f^{-1}(y)\, \d y\,,
\tag{$*$}\]
where $\f^{-1}$ denotes the inverse function of $\f$, and
$a\ge 0$ and $b\ge 0$.
\begin{questionparts}
\item
By considering the graph of $y=\f(x)$, explain briefly why
the inequality $(*)$ holds.
In the case $a>0$ and $b>0$,
state a condition on $a$ and $b$ under which equality holds.
\item
By taking $\f(x) = x^{p-1}$ in $(*)$, where $p>1$, show that if
$\displaystyle \frac 1p + \frac 1q =1$ then
\[
ab \le \frac{a^p}p + \frac{b^q}q\,.
\]
Verify that equality holds under the condition you stated above.
\item
Show that, for $0\le a \le \frac12 \pi$ and $0\le b \le 1$,
\[
ab \le b\arcsin b + \sqrt{1-b^2} \, - \cos a\,.
\]
Deduce that,
for $t\ge1$,
\[
\arcsin (t^{-1}) \ge t - \sqrt{t^2-1}\,.
\]
\end{questionparts}
About two thirds of the candidates tried this, with very slightly greater success than questions 2 and 3. They found part (i) tricky, especially understanding the integral of the inverse function. Also, commonly, they thought the condition was that f = f⁻¹. However, part (ii) was done better, most errors being due to taking the inverse incorrectly, and of course, the verification frequently went wrong due to the false condition. Most realised the function to use in part (iii) but there was plenty of inaccuracy in working this part, though the final deduction caused few worries.