Year: 2018
Paper: 3
Question Number: 5
Course: LFM Pure and Mechanics
Section: Differentiation from first principles
The total entry was a record number, an increase of over 6% on 2017. Only question 1 was attempted by more than 90%, although question 2 was attempted by very nearly 90%. Every question was attempted by a significant number of candidates with even the two least popular questions being attempted by 9%. More than 92% restricted themselves to attempting no more than 7 questions with very few indeed attempting more than 8. As has been normal in the past, apart from a handful of very strong candidates, those attempting six questions scored better than those attempting more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
The real numbers $a_1$, $a_2$, $a_3$, $\ldots$ are all positive. For each positive integer $n$, $A_n$ and $G_n$ are defined by
\[
A_n = \frac{a_1+a_2 + \cdots + a_n}n
\ \ \ \ \ \text{and } \ \ \ \ \
G_n = \big( a_1a_2\cdots a_n\big) ^{1/n}
\,.
\]
\begin{questionparts}
\item Show that, for any given positive integer $k$,
\[
(k+1) ( A_{k+1} - G_{k+1}) \ge k (A_k-G_k)
\]
if and only if
\[\lambda^{k+1}_k -(k+1)\lambda_{{k}} +k \ge 0\,,
\]
where
$ \lambda_{{k}} = \left(\dfrac{a_{k+1}}{G_{k}}\right)^{\frac1 {k+1}}\,$.
\item
Let
\[
\f(x)=x^{k+1} -(k+1)x +k \,,
\]
where $x > 0$ and $k$ is a positive integer. Show that $\f(x)\ge0$ and that $\f(x)=0$ if and only if $x = 1\,$.
\item
Deduce that:
\begin{enumerate}
\item
$A_n \ge G_n$ for all $n$;
\\
\item
if $A_n=G_n$ for some $n$, then $a_1=a_2 = \cdots = a_n\,$.
\end{enumerate}
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& (k+1) (A_{k+1} - G_{k+1}) & \geq k(A_k - G_k) \\
\Leftrightarrow && \sum_{i=1}^{k+1} a_i - (k+1)G_{k+1} &\geq \sum_{i=1}^k a_i - kG_k \\
\Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} & \geq - k G_k \\
\Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} + k G_k& \geq 0\\
\Leftrightarrow && \frac{a_{k+1}}{G_k} -(k+1)G_k^{k/(k+1)-1}a_{k+1}^{1/(k+1)} + k & \geq 0\\
\Leftrightarrow && \lambda_k^{k+1} -(k+1)\lambda_k+ k & \geq 0\\
\end{align*}
as required.
\item \begin{align*}
&& f'(x) &= (k+1)x^k - (k+1) \\
&&&= (k+1)(x^k-1)
\end{align*}
Therefore $f(x)$ is strictly decreasing on $(0,1)$ and strictly increasing on $(1,\infty)$ and so the minimum will be $f(1) = 1 - (k+1) + k = 0$, so $f(x) \geq 0$ with equality only at $x = 1$.
\item \begin{enumerate}
\item We can proceed by induction to show since the inequality holds for $n=1$ and since if it holds for $n=k$ it will hold for $n=k+1$ as $A_{k+1}-G_{k+1}$ must have the same sign as $A_k - G_k$.
\item The only way for equality to hold is if $\lambda_k = 1$ for $k = 1, \cdots n$, ie $a_{k+1} = G_k$, but this means $a_2 = a_1, a_3 = a_1$ etc. Therefore all values are equal.
\end{enumerate}
\end{questionparts}
A little under half the candidates attempted this, scoring marginally less on average than on question 1. Some candidates used the arithmetic mean/geometric mean inequality which was what the question was proving in part (iii), and so were heavily penalised as their arguments were thus circular. Part (i) was generally well done, with most justifying that their steps were reversible to obtain 'if and only if'. Marks were sometimes lost when justifying that 0 was not used to legitimise division by and also that inequalities did not change sign. Part (ii) was well done too, though many candidates did not justify the 'only if', although some tried to use part (i) to prove (ii), which could not succeed. Part (iii) expressly required deduction, and only deduction, so those who had learned another proof of the inequality and just copied it out could not be rewarded. Many just used the results of (i) and (ii) without justifying why they could be used and some were imprecise with their induction arguments.