| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Moderate -0.3 This is a straightforward implicit differentiation question with standard techniques. Part (i) requires routine application of implicit differentiation rules, and part (ii) involves setting dy/dx = 0 and solving simultaneous equations. While it requires multiple steps and careful algebra, it follows a predictable pattern with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x + 4y\frac{dy}{dx} = 4\) | M1, A1, A1 [3] | Rearranging for \(y\) and differentiating explicitly is M0; Ignore superfluous \(\frac{dy}{dx} = \ldots\) unless used subsequently; correct equation; o.e., but mark final answer |
| \(\Rightarrow \frac{dy}{dx} = \frac{4 - 2x}{4y}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 0 \Rightarrow x = 2\) | B1dep, B1B1 [3] | dep correct derivative; \(\sqrt{2}, -\sqrt{2}\); can isw, penalise inexact answers of \(\pm 1.41\) or better once only; \(-1\) for extra solutions found from using \(y = 0\) |
| \(\Rightarrow 4 + 2y^2 = 8 \Rightarrow y^2 = 2, y = \sqrt{2}\) or \(-\sqrt{2}\) |
### (i)
$2x + 4y\frac{dy}{dx} = 4$ | M1, A1, A1 [3] | Rearranging for $y$ and differentiating explicitly is M0; Ignore superfluous $\frac{dy}{dx} = \ldots$ unless used subsequently; correct equation; o.e., but mark final answer
$\Rightarrow \frac{dy}{dx} = \frac{4 - 2x}{4y}$ | |
### (ii)
$\frac{dy}{dx} = 0 \Rightarrow x = 2$ | B1dep, B1B1 [3] | dep correct derivative; $\sqrt{2}, -\sqrt{2}$; can isw, penalise inexact answers of $\pm 1.41$ or better once only; $-1$ for extra solutions found from using $y = 0$
$\Rightarrow 4 + 2y^2 = 8 \Rightarrow y^2 = 2, y = \sqrt{2}$ or $-\sqrt{2}$ | |
---A curve has equation $x^2 + 2y^2 = 4x$.
\begin{enumerate}[label=(\roman*)]
\item By differentiating implicitly, find $\frac{dy}{dx}$ in terms of $x$ and $y$. [3]
\item Hence find the exact coordinates of the stationary points of the curve. [You need not determine their nature.] [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2013 Q2 [6]}}