Year: 2024
Paper: 3
Question Number: 3
Course: LFM Pure
Section: Differentiation
No solution available for this problem.
The total entry was an increase on that of 2023 by more than 10%. One question was attempted by more than 98% of candidates, another two by about 80%, and another five by between 50% and 70%. The remaining four questions were attempted by between 5% and 30% of candidates, these being from Section B: Mechanics, and Section C: Probability and Statistics, though the Statistics questions were in general attempted more often and more successfully. All questions were perfectly solved by some candidates. About 84% of candidates attempted no more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Throughout this question, consider only $x > 0$.
\begin{questionparts}
\item Let
\[\mathrm{g}(x) = \ln\left(1 + \frac{1}{x}\right) - \frac{x+c}{x(x+1)}\]
where $c \geqslant 0$.
\begin{enumerate}
\item Show that $y = \mathrm{g}(x)$ has positive gradient for all $x > 0$ when $c \geqslant \frac{1}{2}$.
\item Find the values of $x$ for which $y = \mathrm{g}(x)$ has negative gradient when $0 \leqslant c < \frac{1}{2}$.
\end{enumerate}
\item It is given that, for all $c > 0$, $\mathrm{g}(x) \to -\infty$ as $x \to 0$.
Sketch, for $x > 0$, the graphs of
\[y = \mathrm{g}(x)\]
in the cases
\begin{enumerate}
\item $c = \frac{3}{4}$,
\item $c = \frac{1}{4}$.
\end{enumerate}
\item The function $\mathrm{f}$ is defined as
\[\mathrm{f}(x) = \left(1 + \frac{1}{x}\right)^{x+c}.\]
Show that, for $x > 0$,
\begin{enumerate}
\item $\mathrm{f}$ is a decreasing function when $c \geqslant \frac{1}{2}$;
\item $\mathrm{f}$ has a turning point when $0 < c < \frac{1}{2}$;
\item $\mathrm{f}$ is an increasing function when $c = 0$.
\end{enumerate}
\end{questionparts}
The second most popular question, it was the eighth most successful with a mean score of a little under 9/20. Whilst some candidates did not make progress with differentiating f in (iii), most differentiated well in (i) and (iii). However in (i), sufficient justification for the positive gradient for c ≥ 1/2 was often missing in (a), and some occasionally forgot that inequalities reverse when divided by a negative number in (b). In part (ii), both sketch graphs were mostly drawn correctly. However, in part (a), many did not justify the positive gradient or asymptote for large x. In part (b), whilst most found the turning point correctly, few justified the positive gradient before the turning point. The justifications, or otherwise, in (iii) varied a lot in the level of detail. Forgetting to mention that f > 0 was a common way that candidates did not achieve full marks.