| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a standard C3 differentiation question covering routine techniques: product rule for dy/dx and d²y/dx², finding stationary points, curve sketching, and integration. Part (v) requires solving a transcendental equation which adds slight challenge, but overall this is a typical multi-part question testing well-practiced methods with no novel insight required—slightly easier than the average A-level question. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks |
|---|---|
| \(\Rightarrow x = -2\) | B1 B1 |
| so \((-2,0)\) and \((0,2)\) | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \frac{dy}{dx} = -e^{-x}(x+1) = 0 \Rightarrow x = -1\) | M1 A1 | Product rule |
| SP is \((-1, e)\) | M1 A1 | \(= 0\) |
| 4 | ||
| (iii) \(\Rightarrow \frac{d^2y}{dx^2} = xe^{-x}\) | M1 A1 | |
| At \((-1, e)\) this is negative, so SP is a maximum. | A1 | |
| 3 | ||
| (iv) Sketch of curve showing maximum at \((-1, e)\), crossing axes at \((-2, 0)\) and \((0, 2)\), approaching \(y = 0\) as \(x \to \infty\) | B1 | |
| 1 | ||
| (v) At \((0,2)\) gradient is \(-1\) so gradient of normal is \(1\) | B1 | |
| Normal is \(y = x+2\). | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow x = -2\) (or \(0\)) | A1 | |
| New intersection point is \((-2, 0)\). | ||
| 3 | ||
| (vi) Required area is \(\int_1^3 (x+2)e^{-x} \, dx\) | B1 | |
| \(= \left[-e^{-x}(x+2)\right]_1^3 + \int_1^3 e^{-x} \, dx\) | M1 A1 | |
| \(= \left[-e^{-x}(x+2)\right]_1^3 + \left[-e^{-x}\right]_1^3\) | A1 | |
| \(= \frac{-6}{e^3} + \frac{4}{e}\) | A1 | or equivalent |
| 5 |
**(i)** $0 = (x+2)e^{-x}$
$\Rightarrow x = -2$ | B1 B1 |
so $(-2,0)$ and $(0,2)$ | |
| 2 |
**(ii)** $y = (x+2)e^{-x}$
$\Rightarrow \frac{dy}{dx} = -e^{-x}(x+1) = 0 \Rightarrow x = -1$ | M1 A1 | Product rule
SP is $(-1, e)$ | M1 A1 | $= 0$
| 4 |
**(iii)** $\Rightarrow \frac{d^2y}{dx^2} = xe^{-x}$ | M1 A1 |
At $(-1, e)$ this is negative, so SP is a maximum. | A1 |
| 3 |
**(iv)** Sketch of curve showing maximum at $(-1, e)$, crossing axes at $(-2, 0)$ and $(0, 2)$, approaching $y = 0$ as $x \to \infty$ | B1 |
| 1 |
**(v)** At $(0,2)$ gradient is $-1$ so gradient of normal is $1$ | B1 |
Normal is $y = x+2$. | M1 |
$y = x+2$, $y = (x+2)e^{-x}$
$\Rightarrow 0 = (x+2)(1-e^{-x})$
$\Rightarrow x = -2$ (or $0$) | A1 |
New intersection point is $(-2, 0)$. | |
| 3 |
**(vi)** Required area is $\int_1^3 (x+2)e^{-x} \, dx$ | B1 |
$= \left[-e^{-x}(x+2)\right]_1^3 + \int_1^3 e^{-x} \, dx$ | M1 A1 |
$= \left[-e^{-x}(x+2)\right]_1^3 + \left[-e^{-x}\right]_1^3$ | A1 |
$= \frac{-6}{e^3} + \frac{4}{e}$ | A1 | or equivalent
| 5 |8 A curve has equation $y = ( x + 2 ) \mathrm { e } ^ { - x }$.\\
(i) Find the coordinates of the points where the curve cuts the axes.\\
(ii) Find the coordinates of the stationary point, S , on the curve.\\
(iii) By evaluating $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at S , determine whether the stationary point is a maximum or a minimum.\\
(iv) Sketch the curve in the domain $- 3 < x < 3$.\\
(v) Find where the normal to the curve at the point $( 0,2 )$ cuts the curve again.\\
(vi) Find the area of the region bounded by the curve, the $x$-axis and the lines $x = 1$ and $x = 3$.
\hfill \mbox{\textit{OCR MEI C3 Q8 [18]}}