| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Classify nature of stationary points |
| Difficulty | Standard +0.3 This is a straightforward stationary points question requiring differentiation of a simple function (polynomial plus reciprocal term), solving a cubic equation that factors nicely, and using the second derivative test. While it requires multiple steps, all techniques are standard A-level procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = -2x + 32x^{-3} = 0\) | M1 | Attempt to differentiate and equate to zero |
| \(x^4 = 16\), so \(x = \pm 2\) | A1 | Both \(x\)-values and no others |
| When \(x = \pm 2\), \(y = 7\); points are \((-2, 7)\) and \((2, 7)\) | A1 | FT their \(x\)-coordinates. Do not FT \(x = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{d^2y}{dx^2} = -2 - 96x^{-4}\) | M1 | Attempts to find second derivative. FT their \(\frac{dy}{dx}\) |
| When \(x = 2\): \(\frac{d^2y}{dx^2} = -2 - \frac{96}{16} < 0\); When \(x = -2\): \(\frac{d^2y}{dx^2} = -2 - \frac{96}{16} < 0\) | Or convincing statement that \(\frac{d^2y}{dx^2} < 0\) for any \(x[\neq 0]\) because \(x^{-4}\) is always positive | |
| So both points are maximum points | E1 | Complete argument required from correct second derivative and \(x \neq 0\). Also allow for one point established and argument from symmetry |
| Alternative: \(\frac{dy}{dx} > 0\) for \(0 < x < 2\) and \(\frac{dy}{dx} < 0\) for \(x > 2\); and \(\frac{dy}{dx} > 0\) for \(x < -2\) and \(\frac{dy}{dx} < 0\) for \(-2 < x < 0\) | M1 | Evaluating gradient for suitable values of \(x\) on either side of each turning point. Also allow \(y\)-coordinates in these ranges |
| So both points are maximum points | E1 | Complete argument from correct first derivative and \(x \neq 0\). Also allow for one point established and argument from symmetry |
## Question 5:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = -2x + 32x^{-3} = 0$ | M1 | Attempt to differentiate and equate to zero |
| $x^4 = 16$, so $x = \pm 2$ | A1 | Both $x$-values and no others |
| When $x = \pm 2$, $y = 7$; points are $(-2, 7)$ and $(2, 7)$ | A1 | FT their $x$-coordinates. Do not FT $x = 0$ |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d^2y}{dx^2} = -2 - 96x^{-4}$ | M1 | Attempts to find second derivative. FT their $\frac{dy}{dx}$ |
| When $x = 2$: $\frac{d^2y}{dx^2} = -2 - \frac{96}{16} < 0$; When $x = -2$: $\frac{d^2y}{dx^2} = -2 - \frac{96}{16} < 0$ | | Or convincing statement that $\frac{d^2y}{dx^2} < 0$ for any $x[\neq 0]$ because $x^{-4}$ is always positive |
| So both points are maximum points | E1 | Complete argument required from correct second derivative and $x \neq 0$. Also allow for one point established and argument from symmetry |
| **Alternative:** $\frac{dy}{dx} > 0$ for $0 < x < 2$ and $\frac{dy}{dx} < 0$ for $x > 2$; and $\frac{dy}{dx} > 0$ for $x < -2$ and $\frac{dy}{dx} < 0$ for $-2 < x < 0$ | M1 | Evaluating gradient for suitable values of $x$ on either side of each turning point. Also allow $y$-coordinates in these ranges |
| So both points are maximum points | E1 | Complete argument from correct first derivative and $x \neq 0$. Also allow for one point established and argument from symmetry |
---5 In this question you must show detailed reasoning.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the two stationary points on the graph of $y = 15 - x ^ { 2 } - \frac { 16 } { x ^ { 2 } }$.
\item Show that both these stationary points are maximum points.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2023 Q5 [5]}}