| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Determine constant from stationary point condition |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard differentiation techniques: finding constants using stationary point conditions, computing the second derivative, and applying the second derivative test. All steps are routine with no problem-solving insight required, making it easier than average but not trivial due to the algebraic manipulation involved. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = p - 16x^{-3}\) | M1 | Uses negative powers to attempt \(\frac{dy}{dx}\); a term in \(x^{-3}\) needed |
| \(p - 16 \times 2^{-3} = 0\) | M1 | Equates derivative to zero and attempts to solve using \(x = 2\) |
| \(p = 2\) | A1 | |
| When \(x = 2\), \(y = 7\) so \(7 = 2p + \frac{8}{2^2} + q\) | M1 | Uses given coordinates to link \(p\) and \(q\) in the Cartesian equation |
| \(2p + q = 5\), so \(q = 1\) | A1 | FT their \(p\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{d^2y}{dx^2} = 48x^{-4} = \left[\frac{48}{x^4}\right]\) | B1 | Allow even from wrong values of \(p\) and \(q\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \((2, 7)\): \(\frac{d^2y}{dx^2} = 48 \times 2^{-4} = [3] > 0\) | M1 | Substitutes \(x = 2\) into their (b); also allow arguing \(48x^{-4}\) is always positive |
| So the stationary point is a minimum | A1 | Clear statement using positivity of second derivative; FT their second derivative |
## Question 13:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = p - 16x^{-3}$ | M1 | Uses negative powers to attempt $\frac{dy}{dx}$; a term in $x^{-3}$ needed |
| $p - 16 \times 2^{-3} = 0$ | M1 | Equates derivative to zero and attempts to solve using $x = 2$ |
| $p = 2$ | A1 | |
| When $x = 2$, $y = 7$ so $7 = 2p + \frac{8}{2^2} + q$ | M1 | Uses given coordinates to link $p$ and $q$ in the Cartesian equation |
| $2p + q = 5$, so $q = 1$ | A1 | FT their $p$ |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{d^2y}{dx^2} = 48x^{-4} = \left[\frac{48}{x^4}\right]$ | B1 | Allow even from wrong values of $p$ and $q$ |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $(2, 7)$: $\frac{d^2y}{dx^2} = 48 \times 2^{-4} = [3] > 0$ | M1 | Substitutes $x = 2$ into their (b); also allow arguing $48x^{-4}$ is always positive |
| So the stationary point is a minimum | A1 | Clear statement using positivity of second derivative; FT their second derivative |
---13 The curve with equation $\mathrm { y } = \mathrm { px } + \frac { 8 } { \mathrm { x } ^ { 2 } } + \mathrm { q }$, where $p$ and $q$ are constants, has a stationary point at $( 2,7 )$.
\begin{enumerate}[label=(\alph*)]
\item Determine the values of $p$ and $q$.
\item Find $\frac { d ^ { 2 } y } { d x ^ { 2 } }$.
\item Hence determine the nature of the stationary point at (2, 7).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2024 Q13 [8]}}