| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.3 This is a straightforward application of the product rule to find stationary points, followed by a simple second derivative test. The exponential function makes the algebra clean, and sketching y = axe^(-x) is a standard curve. Slightly easier than average due to the routine nature of all steps. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.06a Exponential function: a^x and e^x graphs and properties1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = axe^{-x} \Rightarrow \frac{dy}{dx} = ae^{-x} - axe^{-x} = ae^{-x}(1-x)\) | M1, A1 | Product |
| \(\frac{dy}{dx} = 0 \Rightarrow x = 1\) only at \(\left(1, \frac{a}{e}\right)\) | M1, A1 | \(= 0\) |
| \(\frac{d^2y}{dx^2} = -ae^{-x}(1-x) - ae^{-x}\): When \(x=1\), \(\frac{d^2y}{dx^2} < 0 \Rightarrow\) Maximum | B1 | or any equivalent argument |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sketch of curve | B1 | For curve |
| Stationary point marked at \(\left(1, \frac{a}{e}\right)\) | B1 | For stationary point |
| Total: 2 |
## Question 4:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = axe^{-x} \Rightarrow \frac{dy}{dx} = ae^{-x} - axe^{-x} = ae^{-x}(1-x)$ | M1, A1 | Product |
| $\frac{dy}{dx} = 0 \Rightarrow x = 1$ only at $\left(1, \frac{a}{e}\right)$ | M1, A1 | $= 0$ |
| $\frac{d^2y}{dx^2} = -ae^{-x}(1-x) - ae^{-x}$: When $x=1$, $\frac{d^2y}{dx^2} < 0 \Rightarrow$ Maximum | B1 | or any equivalent argument |
| **Total: 5** | | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sketch of curve | B1 | For curve |
| Stationary point marked at $\left(1, \frac{a}{e}\right)$ | B1 | For stationary point |
| **Total: 2** | | |
---4 (i) Show that $y = a x e ^ { - x }$ for $a > 0$ has only one stationary point for all values of $x$. Determine whether this stationary value is a maximum or minimum point.\\
(ii) Sketch the curve.
\hfill \mbox{\textit{OCR MEI C3 Q4 [7]}}