| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Verify stationary point location |
| Difficulty | Standard +0.3 This is a straightforward product rule application followed by routine algebraic manipulation. Part (i) requires applying the product rule to e^(-x) and sin(2x), which is standard C3 material. Part (ii) involves setting the derivative to zero and verifying a given answer rather than finding it independently, making it easier than if students had to solve the equation themselves. The arctan manipulation is slightly above routine but the 'show that' format significantly reduces difficulty. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = e^{-x}\sin 2x\) | ||
| \(\frac{dy}{dx} = e^{-x} \cdot 2\cos 2x + (-e^{-x})\sin 2x\) | M1 | Product rule: \(u \times\) their \(v' + v \times\) their \(u'\) |
| \(\frac{d}{dx}(\sin 2x) = 2\cos 2x\) | B1 | |
| Any correct expression | A1 | But mark final answer |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 0\) when \(2\cos 2x - \sin 2x = 0\) | M1 | Fit their \(\frac{dy}{dx}\) but must eliminate \(e^{-x}\); derivative must have 2 terms |
| \(2 = \tan 2x\) | M1 | \(\sin 2x / \cos 2x = \tan 2x\) used; substituting \(\frac{1}{2}\arctan 2\) into their deriv M0 (unless \(\cos 2x = 1/\sqrt{5}\) and \(\sin 2x = 2/\sqrt{5}\) found) |
| \(2x = \arctan 2\) | ||
| \(x = \frac{1}{2}\arctan 2\) * | A1 | NB AG; must show previous step |
| [3] |
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = e^{-x}\sin 2x$ | | |
| $\frac{dy}{dx} = e^{-x} \cdot 2\cos 2x + (-e^{-x})\sin 2x$ | M1 | Product rule: $u \times$ their $v' + v \times$ their $u'$ |
| $\frac{d}{dx}(\sin 2x) = 2\cos 2x$ | B1 | |
| Any correct expression | A1 | But mark final answer |
| **[3]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 0$ when $2\cos 2x - \sin 2x = 0$ | M1 | Fit their $\frac{dy}{dx}$ but must eliminate $e^{-x}$; derivative must have 2 terms |
| $2 = \tan 2x$ | M1 | $\sin 2x / \cos 2x = \tan 2x$ used; substituting $\frac{1}{2}\arctan 2$ into their deriv M0 (unless $\cos 2x = 1/\sqrt{5}$ and $\sin 2x = 2/\sqrt{5}$ found) |
| $2x = \arctan 2$ | | |
| $x = \frac{1}{2}\arctan 2$ * | A1 | **NB AG**; must show previous step |
| **[3]** | | |
---3 (i) Given that $y = \mathrm { e } ^ { - x } \sin 2 x$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(ii) Hence show that the curve $y = \mathrm { e } ^ { - x } \sin 2 x$ has a stationary point when $x = \frac { 1 } { 2 } \arctan 2$.
\hfill \mbox{\textit{OCR MEI C3 Q3 [6]}}