| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.8 Part (a) is a standard quotient rule application to find stationary points. Part (b) requires second derivative test. Part (c) is the challenging element: students must manipulate the inequality ln(e)/e > ln(a)/a algebraically to derive e^a > a^e, which requires non-routine algebraic insight and isn't a typical textbook exercise. The multi-step proof structure and the need to connect calculus results to an exponential inequality elevates this above average difficulty. |
| Spec | 1.06d Natural logarithm: ln(x) function and properties1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{\frac{1}{x} - \ln x}{x^2}\) o.e. | M1 | M1 for attempt to use quotient rule (allow one error) |
| A1 | ||
| \(\frac{1-\ln x}{x^2} = 0 \Rightarrow \ln x = 1 \Rightarrow x = e\) | E1 | Convincing completion (AG); subbing \(x=e\) gets M0 |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{1}{x^2} - \frac{\ln x}{x^2}\) | ||
| \(\frac{d^2y}{dx^2} = -\frac{2}{x^3} - \frac{x^2 \cdot \frac{1}{x} - 2x\ln x}{x^4}\) | M1 | Attempt to differentiate again (allow one error); OR M1 subst values either side of \(e\) into derivative |
| \(= -\frac{2}{x^3} - \frac{x - 2x\ln x}{x^4}\) or \(\frac{x^2\left(-\frac{1}{x}\right)-(1-\ln x)2x}{x^4}\) | A1 | Correct second derivative; A1 correct conclusion about sign of gradient |
| \(= \frac{-3+2\ln x}{x^3}\) | ||
| When \(x=e\), \(\frac{d^2y}{dx^2} = -\frac{1}{e^3} < 0\) hence maximum | E1 | Correct conclusion from correct working; \((-0.05)\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\ln e}{e} > \frac{\ln a}{a} \Rightarrow \frac{1}{e} > \frac{\ln a}{a}\) or \(a\ln e > e\ln a\) | M1 | |
| \(e^{\frac{a}{e}} > a\) hence \(e^a > a^e\) or \(\ln e^a > \ln a^e\) | A1 | Convincing completion (AG) |
| \(e^a > a^e\) | ||
| [2] |
## Question 12(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{\frac{1}{x} - \ln x}{x^2}$ o.e. | M1 | M1 for attempt to use quotient rule (allow one error) |
| | A1 | |
| $\frac{1-\ln x}{x^2} = 0 \Rightarrow \ln x = 1 \Rightarrow x = e$ | E1 | Convincing completion (AG); subbing $x=e$ gets M0 |
| **[3]** | | |
---
## Question 12(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{1}{x^2} - \frac{\ln x}{x^2}$ | | |
| $\frac{d^2y}{dx^2} = -\frac{2}{x^3} - \frac{x^2 \cdot \frac{1}{x} - 2x\ln x}{x^4}$ | M1 | Attempt to differentiate again (allow one error); OR M1 subst values either side of $e$ into derivative |
| $= -\frac{2}{x^3} - \frac{x - 2x\ln x}{x^4}$ or $\frac{x^2\left(-\frac{1}{x}\right)-(1-\ln x)2x}{x^4}$ | A1 | Correct second derivative; A1 correct conclusion about sign of gradient |
| $= \frac{-3+2\ln x}{x^3}$ | | |
| When $x=e$, $\frac{d^2y}{dx^2} = -\frac{1}{e^3} < 0$ hence maximum | E1 | Correct conclusion from correct working; $(-0.05)$ |
| **[3]** | | |
---
## Question 12(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\ln e}{e} > \frac{\ln a}{a} \Rightarrow \frac{1}{e} > \frac{\ln a}{a}$ or $a\ln e > e\ln a$ | M1 | |
| $e^{\frac{a}{e}} > a$ hence $e^a > a^e$ or $\ln e^a > \ln a^e$ | A1 | Convincing completion (AG) |
| $e^a > a^e$ | | |
| **[2]** | | |
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\begin{enumerate}[label=(\alph*)]
\item Show that the only stationary point on the curve $\mathrm { y } = \frac { \ln \mathrm { x } } { \mathrm { x } }$ occurs where $x = \mathrm { e }$, as given in line 45.
\item Show that the stationary point is a maximum.
\item It follows from part (b) that, for any positive number $a$ with $a \neq \mathrm { e }$,\\
$\frac { \ln \mathrm { e } } { \mathrm { e } } > \frac { \ln a } { a }$.\\
Use this fact to show that $\mathrm { e } ^ { a } > a ^ { \mathrm { e } }$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 3 2020 Q12 [8]}}