| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.3 Part (i) is a standard product rule application with chain rule (routine C3 differentiation). Part (ii) requires setting the derivative to zero and algebraic manipulation to reach the given answer, which is straightforward once the derivative is found. This is a typical textbook exercise testing core differentiation skills with minimal problem-solving demand, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = e^{-x} \sin 2x\) | M1, B1, A1 [3] | Product rule: \(u \times \text{their } v' + v \times \text{their } u'\); Any correct expression but mark final answer |
| \(\frac{dy}{dx} = e^{-x} \cdot 2\cos 2x + (-e^{-x})\sin 2x\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 0\) when \(2\cos 2x - \sin 2x = 0\) | M1, M1, A1 [3] | ft their \(\frac{dy}{dx}\) but must eliminate \(e^{-x}\); derivative must have 2 terms; \(\sin 2x / \cos 2x = \tan 2x\) used [or \(\tan^{-1}\)]; substituting \(\frac{1}{2}\arctan 2\) into their deriv M0 (unless \(\cos 2x = 1/\sqrt{5}\) and \(\sin 2x = 2/\sqrt{5}\) found) must show previous step |
| \(\Rightarrow 2 = \tan 2x\) | ||
| \(\Rightarrow 2x = \arctan 2\) | ||
| \(\Rightarrow x = \frac{1}{2}\arctan 2\) * |
### (i)
$y = e^{-x} \sin 2x$ | M1, B1, A1 [3] | Product rule: $u \times \text{their } v' + v \times \text{their } u'$; Any correct expression but mark final answer
$\frac{dy}{dx} = e^{-x} \cdot 2\cos 2x + (-e^{-x})\sin 2x$ | |
### (ii)
$\frac{dy}{dx} = 0$ when $2\cos 2x - \sin 2x = 0$ | M1, M1, A1 [3] | ft their $\frac{dy}{dx}$ but must eliminate $e^{-x}$; derivative must have 2 terms; $\sin 2x / \cos 2x = \tan 2x$ used [or $\tan^{-1}$]; substituting $\frac{1}{2}\arctan 2$ into their deriv M0 (unless $\cos 2x = 1/\sqrt{5}$ and $\sin 2x = 2/\sqrt{5}$ found) must show previous step
$\Rightarrow 2 = \tan 2x$ | |
$\Rightarrow 2x = \arctan 2$ | |
$\Rightarrow x = \frac{1}{2}\arctan 2$ * | |
---\begin{enumerate}[label=(\roman*)]
\item Given that $y = e^{-x} \sin 2x$, find $\frac{dy}{dx}$. [3]
\item Hence show that the curve $y = e^{-x} \sin 2x$ has a stationary point when $x = \frac{1}{2} \arctan 2$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2013 Q1 [6]}}