| Exam Board | OCR |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - given gradient condition |
| Difficulty | Standard +0.8 This question requires chain rule differentiation, quotient rule for the second derivative, finding a maximum by setting the second derivative to zero, then verifying a tangent passes through a specific point. The multi-step nature, optimization component, and algebraic manipulation of logarithmic expressions elevate this above a standard C3 differentiation question. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = \frac{2\ln x}{x}\) | M1, A1 | For relevant attempt at the chain rule; For correct result, in any form |
| (ii) \(\frac{d^2y}{dx^2} = \frac{x(2/x) - 2\ln x}{x^2} = \frac{2 - 2\ln x}{x^2}\) | M1, A1 | For relevant attempt at quotient rule; For correct simplified answer |
| (iii) For maximum gradient, \(2 - 2\ln x = 0 \Rightarrow x = e\) Hence \(P\) is \((e, 1)\) The gradient at \(P\) is \(\frac{2}{e}\) Tangent at \(P\) is \(y - 1 = \frac{2}{e}(x - e)\) Hence, when \(x = 0, y = -1\) as required | M1, A1, A1✓, A1✓, M1, A1 | For equating second derivative to zero; For correct value \(e\); For stating or using the y-coordinate; For stating or using the gradient at \(P\); For forming the equation of the tangent; For correct verification of (0, −1) |
**(i)** $\frac{dy}{dx} = \frac{2\ln x}{x}$ | M1, A1 | For relevant attempt at the chain rule; For correct result, in any form
**(ii)** $\frac{d^2y}{dx^2} = \frac{x(2/x) - 2\ln x}{x^2} = \frac{2 - 2\ln x}{x^2}$ | M1, A1 | For relevant attempt at quotient rule; For correct simplified answer
**(iii)** For maximum gradient, $2 - 2\ln x = 0 \Rightarrow x = e$ Hence $P$ is $(e, 1)$ The gradient at $P$ is $\frac{2}{e}$ Tangent at $P$ is $y - 1 = \frac{2}{e}(x - e)$ Hence, when $x = 0, y = -1$ as required | M1, A1, A1✓, A1✓, M1, A1 | For equating second derivative to zero; For correct value $e$; For stating or using the y-coordinate; For stating or using the gradient at $P$; For forming the equation of the tangent; For correct verification of (0, −1)
**Total for Question 8: 10 marks**
---8\\
\includegraphics[max width=\textwidth, alt={}, center]{b6b6e55a-a5ba-466c-ac9f-b5ef5bca7a3c-4_476_608_287_756}
The diagram shows the curve $y = ( \ln x ) ^ { 2 }$.\\
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.\\
(ii) The point $P$ on the curve is the point at which the gradient takes its maximum value. Show that the tangent at $P$ passes through the point $( 0 , - 1 )$.
\hfill \mbox{\textit{OCR C3 Q8 [10]}}