| Exam Board | OCR MEI |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch rational with reciprocal terms |
| Difficulty | Standard +0.3 This is a straightforward curve sketching question requiring standard calculus techniques: finding stationary points via differentiation, using second derivative test, identifying asymptotes, and determining concavity. All steps are routine A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07f Convexity/concavity: points of inflection1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{dy}{dx} = 1 - \dfrac{1}{(x-2)^2}\) | M1 (1.1a) | Attempt to differentiate with one term correct |
| A1 (1.1) | Correct derivative | |
| \(1 - \dfrac{1}{(x-2)^2} = 0\) at stationary points | M1 (1.1a) | |
| \(x - 2 = \pm 1\) so \(x = 1,\ 3\) | A1 (2.2a) | Both values of \(x\) |
| \((1,\ -5)\ (3,\ -1)\) | A1 (1.1) | Both values of \(y\) - ft their \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = \frac{2}{(x-2)^3}\) | M1 | AO 1.1a |
| \(x=3\): \(\frac{d^2y}{dx^2} > 0\), so minimum (2) | A1 | AO 2.4 |
| \(x=1\): \(\frac{d^2y}{dx^2} < 0\), so maximum (-2) | A1 [3] | AO 2.4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = 2\) | B1 [1] | AO 1.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x > 2\) | A1 [1] | AO 2.2a |
**Question 4(i):**
$\dfrac{dy}{dx} = 1 - \dfrac{1}{(x-2)^2}$ | M1 (1.1a) | Attempt to differentiate with one term correct
| A1 (1.1) | Correct derivative
$1 - \dfrac{1}{(x-2)^2} = 0$ at stationary points | M1 (1.1a) |
$x - 2 = \pm 1$ so $x = 1,\ 3$ | A1 (2.2a) | Both values of $x$
$(1,\ -5)\ (3,\ -1)$ | A1 (1.1) | Both values of $y$ - ft their $x$
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2} = \frac{2}{(x-2)^3}$ | M1 | AO 1.1a |
| $x=3$: $\frac{d^2y}{dx^2} > 0$, so minimum (2) | A1 | AO 2.4 |
| $x=1$: $\frac{d^2y}{dx^2} < 0$, so maximum (-2) | A1 [3] | AO 2.4 | OR: Allow consideration of gradient either side of stationary point for M1; correct gradients above and below each tp A1; correct convincing conclusions (possibly with sketches) A1 |
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 2$ | B1 [1] | AO 1.2 |
## Question 4(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x > 2$ | A1 [1] | AO 2.2a | FT their (iii) if region is to right of their $x$ value |4 In this question you must show detailed reasoning.\\
A curve has equation $y = x - 5 + \frac { 1 } { x - 2 }$. The curve is shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{31bc8bde-8d37-4e97-94e2-e3e73aab55e9-5_723_844_424_612}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Determine the coordinates of the stationary points on the curve.\\
(ii) Determine the nature of each stationary point.\\
(iii) Write down the equation of the vertical asymptote.\\
(iv) Deduce the set of values of $x$ for which the curve is concave upwards.
\hfill \mbox{\textit{OCR MEI Paper 3 2018 Q4 [10]}}