2025 Paper 2 Q3
Year: 2025
Paper: 2
Question Number: 3
Course: LFM Pure
Section: Differentiation
Difficulty: 1500.0
Banger: 1515.3
curve sketching
differentiation
logarithms
inequality
stationary points
ln x / x
comparison of powers
optimisation
Problem
- Sketch a graph of \(y = \frac{\ln x}{x}\) for \(x > 0\).
- Use your graph to show the following.
- \(3^{\pi} > \pi^3\)
- \(\left(\frac{9}{4}\right)^{\sqrt{5}} > \sqrt{5}^{\frac{9}{4}}\)
- Given that \(1 < x < 2\), decide, with justification, which is the larger of \(x^{x+2}\) or \((x+2)^x\).
- Show that the inequalities \(9^{\sqrt{2}} > \sqrt{2}^9\) and \(3^{2\sqrt{2}} > (2\sqrt{2})^3\) are equivalent. Given that \(e^2 < 8\), decide, with justification, which is the larger of \(9^{\sqrt{2}}\) and \(\sqrt{2}^9\).
- Decide, with justification, which is the larger of \(8^{\sqrt[4]{3}}\) and \(\sqrt[3]{8}\).
Solution
- \begin{enumerate}
- since \(\frac{\ln x}{x}\) is decreasing on \((e, \infty)\) we must have that \(\frac{\ln 3}{3} > \frac{\ln \pi}{\pi} \Rightarrow e^\pi > \pi^3\)
- similarly, since \(\frac{\ln x}{x}\) is increasing on \((0, e)\) we must have that \(\frac{\ln \sqrt{5}}{\sqrt{5}} < \frac{\ln 9/4}{9/4} \Rightarrow \left(\frac{9}{4}\right)^{\sqrt{5}} > \sqrt{5}^{\frac{9}{4}}\)
- Since \(2^4 = 4^2\) notice also that:
from the graph we must have the green area between \(1\) and \(2\) mapping to the (higher) green area between \(3\) and \(4\). Therefore \((x+2)^x > x^{x+2}\) for \(1 < x < 2\)
- \begin{align*} && 9^{\sqrt 2} & \stackrel{?}{>} \sqrt{2}^9 \\ \Leftrightarrow && (3^2)^{\sqrt2} &\stackrel{?}{>} (\sqrt{2}^3)^3 \\ \Leftrightarrow && 3^{2 \sqrt2} &\stackrel{?}{>} (2\sqrt2)^3 \end{align*} Since \(e^2 < 8 < 9\Rightarrow e < 2\sqrt2 < 3\) therefore: \begin{align*} && \frac{\ln 2 \sqrt2}{2 \sqrt 2} &> \frac{\ln 3}{3} \\ \Leftrightarrow && (2 \sqrt{2})^3 &> 3^{2 \sqrt{2}} \\ \Leftrightarrow && \sqrt{2}^9 &> 9^{\sqrt 2} \\ \end{align*}
- \begin{align*} && 8^{\sqrt[3]{3}} & \stackrel{?}{>} \sqrt[3]{3}^8 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (\sqrt[3]{3}^4)^2 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (3\sqrt[3]{3})^2 \\ \end{align*} Since \(3\sqrt[3]{3} > 4\) we have \begin{align*} && \frac{\ln (3 \sqrt[3]3)}{3 \sqrt[3]3} &< \frac{\ln 4}{4} \\ &&&= \frac{\ln 2}{2}\\ \Rightarrow && (3 \sqrt[3]{3})^2 &< 2^{3 \sqrt[3]{3}} \\ \Rightarrow && \sqrt[3]3^8 &< 8^{\sqrt[3]3} \end{align*}
Examiner's report
— 2025 STEP 2, Question 3
Above Average
Described as 'popular'; good progress through most parts but part (v) very challenging
From the paper introduction
As is commonly the case, the vast majority of candidates focused on the Pure questions in Section A of the paper, with a good number of attempts made on all of those questions. Candidates that attempted the Mechanics questions in Section B generally answered both questions. More candidates attempted Question 11 in Section C than either Mechanics question, but very few attempted Question 12 in that section. There were a large number of good responses seen for all the questions, but a significant number of responses lacked sufficient detail in the presentation, particularly when asked to prove a given result or provide an explanation. Candidates who did well on this paper generally: gave careful explanations of each step within their solutions; indicated all points of interest on graphs and other diagrams clearly; made clear comments about the approach that needed to be taken, particularly when having to explore a number of cases as part of the solution to a question; used mathematical terminology accurately within their solutions. Candidates who did less well on this paper generally: made errors with basic algebraic manipulation, such as incorrect processing of indices; produced sketches of graphs in which significant points were difficult to see clearly because of the chosen scale; skipped important lines within lengthy sections of algebraic reasoning.
Source: Cambridge STEP 2025 Examiner's Report
· 2025-p2.pdf
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1515.3
Banger Comparisons: 1
Show LaTeX source
Problem source
\begin{questionparts}
\item Sketch a graph of $y = \frac{\ln x}{x}$ for $x > 0$.
\item Use your graph to show the following.
\begin{enumerate}[label=(\alph*)]
\item $3^{\pi} > \pi^3$
\item $\left(\frac{9}{4}\right)^{\sqrt{5}} > \sqrt{5}^{\frac{9}{4}}$
\end{enumerate}
\item Given that $1 < x < 2$, decide, with justification, which is the larger of $x^{x+2}$ or $(x+2)^x$.
\item Show that the inequalities $9^{\sqrt{2}} > \sqrt{2}^9$ and $3^{2\sqrt{2}} > (2\sqrt{2})^3$ are equivalent. Given that $e^2 < 8$, decide, with justification, which is the larger of $9^{\sqrt{2}}$ and $\sqrt{2}^9$.
\item Decide, with justification, which is the larger of $8^{\sqrt[4]{3}}$ and $\sqrt[3]{8}$.
\end{questionparts}Solution source
\begin{questionparts}
\item
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
\def\xl{-1};
\def\xu{10};
\def\yl{-0.2};
\def\yu{0.6};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=0.1:10, samples=100]
plot (\x, {ln(\x)/\x});
\filldraw ({exp(1)}, {exp(-1)}) circle (1pt) node [above] {$(e, \tfrac1{e})$};
\filldraw ({1}, {0}) circle (1pt) node [below right] {$(1,0)$};
% \draw[thick, red, dashed, domain=-5:5, samples=100]
% plot (\x, {2*\x*\x*\x-5*\x});
% \draw[thick, blue, smooth, domain=-5:5, samples=100]
% plot (\x, {2*min(\x*\x, \x*\x*\x) - 5*\x});
% \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
% \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
% \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
\item \begin{enumerate}
\item since $\frac{\ln x}{x}$ is decreasing on $(e, \infty)$ we must have that $\frac{\ln 3}{3} > \frac{\ln \pi}{\pi} \Rightarrow e^\pi > \pi^3$
\item similarly, since $\frac{\ln x}{x}$ is increasing on $(0, e)$ we must have that $\frac{\ln \sqrt{5}}{\sqrt{5}} < \frac{\ln 9/4}{9/4} \Rightarrow \left(\frac{9}{4}\right)^{\sqrt{5}} > \sqrt{5}^{\frac{9}{4}}$
\item Since $2^4 = 4^2$ notice also that:
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
\def\xl{-1};
\def\xu{10};
\def\yl{-0.2};
\def\yu{0.6};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
\draw[thick, blue, smooth, domain=0.1:10, samples=100]
plot (\x, {ln(\x)/\x});
\draw[thick, green, smooth, domain=1:2, samples=100]
plot (\x, {ln(\x)/\x});
\draw[thick, green, smooth, domain=3:4, samples=100]
plot (\x, {ln(\x)/\x});
\filldraw ({exp(1)}, {exp(-1)}) circle (1pt) node [above] {$(e, \tfrac1{e})$};
\filldraw ({1}, {0}) circle (1pt) node [below right] {$(1,0)$};
\filldraw (2, {ln(2)/2}) circle (1pt) node [below] {$(2,\frac{\ln 2}{2})$};
\filldraw (4, {ln(2)/2}) circle (1pt) node [below] {$(4,\frac{\ln 2}{2})$};
\draw[red, dashed] (0, {ln(2)/2}) -- (10, {ln(2)/2});
% \draw[thick, red, dashed, domain=-5:5, samples=100]
% plot (\x, {2*\x*\x*\x-5*\x});
% \draw[thick, blue, smooth, domain=-5:5, samples=100]
% plot (\x, {2*min(\x*\x, \x*\x*\x) - 5*\x});
% \draw[thick, red, dashed] (-2, \yl) -- (-2, \yu) node [below] {$x = -2$};
% \draw[thick, red, dashed] (2, \yl) -- (2, \yu) node [below] {$x = 2$};
% \draw[thick, red, dashed] (-5, -10) -- (5, 10) node [below left] {$y = 2x$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
\end{tikzpicture}
\end{center}
from the graph we must have the green area between $1$ and $2$ mapping to the (higher) green area between $3$ and $4$.
Therefore $(x+2)^x > x^{x+2}$ for $1 < x < 2$
\item \begin{align*}
&& 9^{\sqrt 2} & \stackrel{?}{>} \sqrt{2}^9 \\
\Leftrightarrow && (3^2)^{\sqrt2} &\stackrel{?}{>} (\sqrt{2}^3)^3 \\
\Leftrightarrow && 3^{2 \sqrt2} &\stackrel{?}{>} (2\sqrt2)^3
\end{align*}
Since $e^2 < 8 < 9\Rightarrow e < 2\sqrt2 < 3$ therefore:
\begin{align*}
&& \frac{\ln 2 \sqrt2}{2 \sqrt 2} &> \frac{\ln 3}{3} \\
\Leftrightarrow && (2 \sqrt{2})^3 &> 3^{2 \sqrt{2}} \\
\Leftrightarrow && \sqrt{2}^9 &> 9^{\sqrt 2} \\
\end{align*}
\item \begin{align*}
&& 8^{\sqrt[3]{3}} & \stackrel{?}{>} \sqrt[3]{3}^8 \\
\Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (\sqrt[3]{3}^4)^2 \\
\Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (3\sqrt[3]{3})^2 \\
\end{align*}
Since $3\sqrt[3]{3} > 4$ we have
\begin{align*}
&& \frac{\ln (3 \sqrt[3]3)}{3 \sqrt[3]3} &< \frac{\ln 4}{4} \\
&&&= \frac{\ln 2}{2}\\
\Rightarrow && (3 \sqrt[3]{3})^2 &< 2^{3 \sqrt[3]{3}} \\
\Rightarrow && \sqrt[3]3^8 &< 8^{\sqrt[3]3}
\end{align*}
\end{questionparts}
This was a popular question with many candidates able to make good progress through most parts of the question. Part (i) was answered well by the vast majority of candidates, with the maximum point clearly labelled in most cases. In a small number of cases the asymptotes were not sufficiently clear, although in a small number of cases the behaviour near x = 0 was not correct. Part (ii) (a) was generally completed well, with most candidates recognising the relationship between the graph and the required results. In some cases, solutions were presented showing that the given result implied the correct ordering of the numbers, but did not present the logic correctly to show that the ordering of the numbers implies the given result. In part (ii) (b) a significant number of candidates did not justify the order of the three numbers within their solution. Many candidates were able to produce good solutions to part (iii). Some chose to consider a translation of the curve from part (i) and looked for the point of intersection between y = ln x / x and y = ln(x+2) / (x+2) as the method to justify the inequality. Part (iv) was well answered by most candidates who attempted it. Most were able to show the equivalence of the two inequalities, but some only showed the logic in one direction. A large proportion of candidates were then able to see how to apply the equivalence between the two inequalities to determine which of the given values was the larger. Part (v) was found to be very challenging, with some candidates making no written attempt. Those candidates that made progress deduced that there was a need to find an equivalent inequality to allow a similar process to part (iv) to be carried out. Several candidates made mistakes with their manipulation of indices within this part of the question. A good proportion of those who identified the equivalent inequality were then able to recognise that an approach similar to part (iii) was required to reach the final answer.