June 2025 Paper 2 Q10
10 A curve \(C\) has equation
\[y = x^{k}\ln x \quad \text{for } x \gt 0\]where \(k\) is a positive integer.
(a) Show that\[\frac{\mathrm{d}y}{\mathrm{d}x} = x^{k-1}\left[A + k\ln x\right]\]where \(A\) is a constant to be found. [4 marks]
(b) Hence show that the \(y\)-coordinate of the stationary point of \(C\) can be written as \(-\dfrac{1}{k\mathrm{e}}\)
Fully justify your answer.
[5 marks](c) Given that the stationary point of \(C\) has coordinates \(\left(\dfrac{1}{\mathrm{e}}, -\dfrac{1}{\mathrm{e}}\right)\) state the value of \(k\) [1 mark]
(d) Prove that \(C\) does not have a point of inflection. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Uses the product rule with one term correct Condone vu’ – uv’ or v’u – u’v | M1 | 3.1a |
| Obtains \(kx^{k-1}\ln x + x^{k} \times \dfrac{1}{x}\) | A1 | 1.1b |
| Simplifies \(x^{k} \times \dfrac{1}{x} = x^{k-1}\) | A1 | 1.1b |
| Completes reasoned argument to show \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^{k-1}\left[1 + k\ln x\right]\) Argument must include \(x^{k} \times \dfrac{1}{x}\) or \(\dfrac{x^{k}}{x}\) before \(x^{k-1}\) seen | R1 | 2.1 |
| (4) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}x} = kx^{k-1}\ln x + x^{k} \times \frac{1}{x}\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = kx^{k-1}\ln x + x^{k-1}\]\[\frac{\mathrm{d}y}{\mathrm{d}x} = x^{k-1}\left[1 + k\ln x\right]\]| Scheme | Marks | AO |
|---|---|---|
| Equates the given expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to zero. PI \(1 + k\ln x = 0\) and Solves to obtain a non-zero value for \(x\) | M1 | 1.1a |
| Gives a correct reason for rejecting \(x = 0\) | E1 | 2.4 |
| Deduces \(x = \mathrm{e}^{-\frac{1}{k}}\) Accept \(x = \mathrm{e}^{-\frac{A}{k}}\) | A1 | 2.2a |
| Substitutes their value for \(x\) into equation of \(C\). | M1 | 1.1a |
| Completes reasoned argument to show the given result. Must see \(\left(\mathrm{e}^{-\frac{1}{k}}\right)^{k}\) simplified to \(\mathrm{e}^{-1}\) or \(\dfrac{1}{\mathrm{e}}\) and \(\ln \mathrm{e}^{-\frac{1}{k}}\) simplified to \(-\dfrac{1}{k}\) before the final answer AG | R1 | 2.1 |
| (5) |
Typical solution
\[x^{k-1}\left[1 + k\ln x\right] = 0\]\[\Rightarrow 1 + k\ln x = 0 \text{ or } x^{k-1} = 0\]\[\ln x = -\frac{1}{k}\]\[x = \mathrm{e}^{-\frac{1}{k}}\]Since \(x \gt 0,\ x^{k-1} \neq 0\)
\[y = \left(\mathrm{e}^{-\frac{1}{k}}\right)^{k} \times \ln \mathrm{e}^{-\frac{1}{k}}\]\[y = \frac{1}{\mathrm{e}} \times \left(-\frac{1}{k}\right)\]\[= -\frac{1}{k\mathrm{e}}\]| Scheme | Marks | AO |
|---|---|---|
| States 1 | B1 | 1.1b |
| (1) |
Typical solution
1
| Scheme | Marks | AO |
|---|---|---|
| Differentiates \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = x^{k-1}\left[1 + k\ln x\right]\) where their \(k \gt 0\) accept \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) in terms of \(A\) and/or \(k\) | M1 | 3.1a |
| Obtains \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = \dfrac{1}{x}\) | A1 | 1.1a |
| Completes reasoned argument by comparing \(\dfrac{1}{x}\) with zero and concludes that there is no point of inflection. | R1 | 2.1 |
| (3) | ||
| (13 marks) |
Typical solution
When \(k = 1\)
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 1 + \ln x\]If there is a point of inflection \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0\)
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{1}{x} \neq 0\]Therefore, \(C\) has no point of inflection.