October 2020 Paper 3 Q12
12
The questions in this section refer to the article on the Insert. You should read the article before attempting the questions.
The relevant parts of the article “Which is bigger?” are reproduced below; the line numbers are those printed on the Insert.
Line 43
Using a similar method, it can be shown that \(\mathrm{e}^a > a^{\mathrm{e}}\) for any positive number \(a \neq \mathrm{e}\).Lines 44–45
An alternative method for showing that \(\mathrm{e}^a > a^{\mathrm{e}}\) for any positive number \(a\) is to show that the only stationary point on the curve \(y = \dfrac{\ln x}{x}\) (a maximum) occurs where \(x = \mathrm{e}\).
\(\dfrac{\ln\mathrm{e}}{\mathrm{e}} > \dfrac{\ln a}{a}\).
Use this fact to show that \(\mathrm{e}^a > a^{\mathrm{e}}\). [2]
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\frac{x}{x} - \ln x}{x^2}\) oe | M1 A1 | 1.1a 1.1 |
| \(\dfrac{1 - \ln x}{x^2} = 0 \Rightarrow \ln x = 1 \Rightarrow x = \mathrm{e}\) | E1 | 2.2a |
| [3] |
Notes
M1: M1 for attempt to use quotient rule (allow one error)
E1: Convincing completion (AG)
Subbing \(x = \mathrm{e}\) gets M0
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{x^2} - \dfrac{\ln x}{x^2}\) | ||
| \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\dfrac{2}{x^3} - \dfrac{\;\dfrac{x^2}{x} - 2x\ln x\;}{x^4}\) | M1 | 1.1a |
| \(\quad = -\dfrac{2}{x^3} - \dfrac{x - 2x\ln x}{x^4}\) or \(\dfrac{x^2\left(-\frac{1}{x}\right) - (1 - \ln x)2x}{x^4}\) | A1 | 1.1 |
| \(\quad = \dfrac{-3 + 2\ln x}{x^3}\) | ||
| When \(x = \mathrm{e}\), \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\dfrac{1}{\mathrm{e}^3} < 0\) hence maximum | E1 | 2.1 |
| [3] |
Notes
M1: Attempt to differentiate again (allow one error)
A1: Correct second derivative
E1: Correct conclusion from correct working
(−0.05)
OR M1 subst (must be seen) values either side of e into derivative
A1 correct conclusion about sign of gradient
E1 correct conclusion from correct working regarding maximum
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\ln\mathrm{e}}{\mathrm{e}} > \dfrac{\ln a}{a} \Rightarrow \dfrac{1}{\mathrm{e}} > \dfrac{\ln a}{a}\) or \(a\ln\mathrm{e} > \mathrm{e}\ln a\) | M1 | 3.1a |
| \(\mathrm{e}^{\frac{a}{\mathrm{e}}} > a\) hence \(\mathrm{e}^a > a^{\mathrm{e}}\) or \(\ln\mathrm{e}^a > \ln a^{\mathrm{e}}\) \(\mathrm{e}^a > a^{\mathrm{e}}\) | A1 | 2.4 |
| [2] |
Notes
A1: Convincing completion (AG)