Year: 2017
Paper: 3
Question Number: 4
Course: LFM Pure
Section: Integration
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
For any function $\f$ satisfying $\f(x) > 0$, we define the \textit{geometric mean}, F, by
\[ F(y) = e^{\frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \quad (y > 0). \]
\begin{questionparts}
\item The function f satisfies $\f(x) > 0$ and $a$ is a positive number with $a\ne1$. Prove that
\[ F(y) = a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx}. \]
\item The functions f and g satisfy $\f(x) > 0$ and $\g(x) > 0$, and the function $\h$ is defined by $\h(x) = \f(x)\g(x)$. Their geometric means are F, G and H, respectively.
Show that $H(y)= \F(y) \G(y)\,$.
\item Prove that, for any positive number $b$, the geometric mean of $b^x$ is $\sqrt{b^y}\,$.
\item Prove that, if $\f(x)>0$ and the geometric mean of $\f(x)$ is $\sqrt{\f(y)}\,$, then $\f(x) = b^x$ for some positive number $b$.
\end{questionparts}\begin{questionparts}
\item \begin{align*}
&& a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} &= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} \\
&&&= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \frac{\ln f(x)}{\ln a} \, dx} \\
&&&= e^{ \frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \\
&&&= F(y)
\end{align*}
\item $\,$ \begin{align*}
&& H(y) &= e^{\frac1y \int_0^y \ln h(x) \d x} \\
&&&= e^{\frac1y \int_0^y \ln (f(x)g(x))\d x} \\
&&&= e^{\frac1y \int_0^y \left ( \ln f(x)+\ln g(x) \right)\d x} \\
&&&= e^{\frac1y \int_0^y \ln f(x) \d x +\frac1y \int_0^y \ln g(x) \d x} \\
&&&= e^{\frac1y \int_0^y \ln f(x) \d x }e^{\frac1y \int_0^y \ln g(x) \d x} \\
&&&= F(y)G(y)
\end{align*}
\item Suppose $f(x) = b^x$, then
\begin{align*}
&& F(y) &= b^{\frac1y \int_0^y \log_b f(x) \d x} \\
&&&= b^{\frac1y \int_0^y x \d x}\\
&&&= b^{\frac1y \frac{y^2}{2}} \\
&&&= b^{\frac{y}2} = \sqrt{b^y}
\end{align*}
\item Suppose the geometric mean of $f(x)$ is $\sqrt{f(y)}$ then the geometric mean of $f(x)^2$ is $f(y)$ by the the second part.
\end{questionparts}
Three fifths of the candidates attempted question 4 with a marginally better success rate than question 3. A significant proportion of candidates struggled with changing base for part (i), but almost all completed (ii) successfully. A common strategy for part (iii) was to use the result of part (i) but very few remembered to check for 1. There were very few successful attempts for part (iv); many tried integration by parts, but rarely successfully.