Year: 2021
Paper: 3
Question Number: 3
Course: UFM Additional Further Pure
Section: Reduction Formulae
The total entry was a marginal increase from that of 2019, that of 2020 having been artificially reduced. Comfortably more than 90% attempted one of the questions, four others were very popular, and a sixth was attempted by 70%. Every question was attempted by at least 10% of the candidature. 85% of candidates attempted no more than 7 questions, though very nearly all the candidates made genuine attempts on at most six questions (the extra attempts being at times no more than labelling a page or writing only the first line or two). Generally, candidates should be aware that when asked to "Show that" they must provide enough working to fully substantiate their working, and that they should follow the instructions in a question, so if it says "Hence", they should be using the previous work in the question in order to complete the next part. Likewise, candidates should be careful when dividing or multiplying, that things are positive, or at other times non-zero.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Let $\displaystyle I_n = \int_0^{\beta} (\sec x + \tan x)^n \, dx$, where $n$ is a non-negative integer and $0 < \beta < \dfrac{\pi}{2}$.
For $n \geqslant 1$, show that
\[
\tfrac{1}{2}(I_{n+1} + I_{n-1}) = \frac{1}{n}\bigl[(\sec\beta + \tan\beta)^n - 1\bigr].
\]
Show also that
\[
I_n < \frac{1}{n}\bigl[(\sec\beta + \tan\beta)^n - 1\bigr].
\]
\item Let $\displaystyle J_n = \int_0^{\beta} (\sec x \cos\beta + \tan x)^n \, dx$, where $n$ is a non-negative integer and $0 < \beta < \dfrac{\pi}{2}$.
For $n \geqslant 1$, show that
\[
J_n < \frac{1}{n}\bigl[(1 + \tan\beta)^n - \cos^n\beta\bigr].
\]
\end{questionparts}
\begin{questionparts}
\item $\,$ \begin{align*}
&& I_n &= \int_0^{\beta} (\sec x + \tan x)^n \, \d x \\
&& \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} \left ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}\right) \, \d x \\
&& \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( (\sec x + \tan x)^{2}+1\right) \, \d x \\
&& \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( \sec^2 x + \tan^2 x + 2\sec x \tan x + 1\right) \, \d x \\
&& \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( 2\sec x \tan x +2\sec^2 x \right) \, \d x \\
&&& = \left [\frac1n(\sec x + \tan x)^{n} \right]_0^{\beta} \\
&&&= \frac1n[(\sec \beta + \tan \beta)^n - 1]
\end{align*}
Notice that by AM-GM $\tfrac12( ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}) \geq (\sec x + \tan x)^{n}$ with equality not holding most of the time. Integrating we obtain our result.
\item $\,$ \begin{align*}
&& J_n &= \int_0^{\beta} (\sec x \cos \beta + \tan x )^n \d x \\
&& \tfrac12( J_{n+1} + J_{n-1}) &= \tfrac12 \int_0^{\beta} \left ( (\sec x \cos \beta + \tan x )^{n+1} +(\sec x \cos \beta + \tan x )^{n-1}\right ) \d x \\
&& &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( (\sec x \cos \beta + \tan x )^{2} + \right ) \d x \\
&& &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec^2 x \cos^2 \beta + \tan^2 x+ 2\sec x \tan x \cos \beta +1 \right ) \d x \\
&& &= \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\tfrac12(\cos^2 \beta +1)\sec^2 x \right ) \d x \\
&& &< \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\sec^2 x \right ) \d x \\
&&&= \left [\frac1n (\sec x \cos \beta + \tan x)^{n} \right]_0^{\beta} \\
&&&= \frac1n[ (1 + \tan \beta)^n - \cos^n \beta]
\end{align*}
But notice we can use the same AM-GM argument from before to show that $J_n < \tfrac12( J_{n+1} + J_{n-1}) < \frac1n[ (1 + \tan \beta)^n - \cos^n \beta]$
Whilst this was the second most popular question, being attempted by 84%, it was the fifth most successful with a mean mark a little below 8/20. Most candidates scored full marks for successfully obtaining the first result of part (i), and many gained nearly full credit for obtaining the second result of that part. As is nearly always true, the rule of thumb that it is usually easier to prove that something is greater than (or less than) zero applied here, and so those that considered ½(I_{n+1} + I_{n−1}) − I_n (and a similar expression for part (ii)) generally fared better. A small, but not insignificant number of candidates solved part (ii) by a direct method and were generally successful if they did so. Common errors when considering inequalities were failure to fully justify positivity of integrals in both parts, incorrect flows of logic, obtaining weak rather than strict inequalities, and stating inequalities that were inconsistent with the claimed ranges of validity. Otherwise, use of induction or integration by parts caused difficulties, and a number expected, when replicating the first part of working in (ii) from part (i), that there would again be an equation, and overlooked the extra term that arose in (ii). Some did not understand that sin x cos β ≤ 1 in (ii).