| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Year | 2020 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Second derivative test justification |
| Difficulty | Moderate -0.8 This is a straightforward application of standard differentiation techniques with fractional powers, followed by routine stationary point analysis. The question requires only mechanical application of power rule, solving a simple equation, and using second derivative test—all standard AS-level procedures with no problem-solving insight required. |
| 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 |
|---|---|---|
| \(Ax^{-\frac{1}{2}}\) | M1 (1.1) | |
| \((+/-)\, Bx^{\frac{1}{2}}\) | M1 (1.1) | |
| \(\frac{12}{\sqrt{x}} - 12\sqrt{x}\) | A1 (1.1) [3] | allow equivalent in index form, but coefficients must be 12 and \(-12\); accept mixed, e.g. \(12x^{-0.5} - 12\sqrt{x}\) |
| Answer | Marks |
|---|---|
| their \(\frac{dy}{dx} = 0\) seen | M1 (3.1a) |
| \(x = 1\) | A1 (1.1) |
| \(y = 32\) | A1 (1.1) [3] |
| Answer | Marks | Guidance |
|---|---|---|
| their 1 substituted in their \(\frac{d^2y}{dx^2}\) | M1 (3.1a) | NB \(-6x^{-\frac{3}{2}} - 6x^{-\frac{1}{2}}\); alternatively, substitution of their \(1 \pm \delta\) in their \(\frac{dy}{dx}\) (either) |
| \(= -12\) so \((1, 32)\) is a (local) maximum | A1 (3.2a) [2] | allow explanation that for \(x > 0\) both terms in 2nd derivative are negative so must be a local maximum; correct values obtained for \(\frac{dy}{dx}\) and correct conclusion made |
## Question 9(a):
$Ax^{-\frac{1}{2}}$ | M1 (1.1) |
$(+/-)\, Bx^{\frac{1}{2}}$ | M1 (1.1) |
$\frac{12}{\sqrt{x}} - 12\sqrt{x}$ | A1 (1.1) **[3]** | allow equivalent in index form, but coefficients must be 12 and $-12$; accept mixed, e.g. $12x^{-0.5} - 12\sqrt{x}$
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## Question 9(b):
their $\frac{dy}{dx} = 0$ seen | M1 (3.1a) |
$x = 1$ | A1 (1.1) |
$y = 32$ | A1 (1.1) **[3]** |
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## Question 9(c):
their 1 substituted in their $\frac{d^2y}{dx^2}$ | M1 (3.1a) | NB $-6x^{-\frac{3}{2}} - 6x^{-\frac{1}{2}}$; alternatively, substitution of their $1 \pm \delta$ in their $\frac{dy}{dx}$ (either)
$= -12$ so $(1, 32)$ is a (local) maximum | A1 (3.2a) **[2]** | allow explanation that for $x > 0$ both terms in 2nd derivative are negative so must be a local maximum; correct values obtained for $\frac{dy}{dx}$ and correct conclusion made
---9 The equation of a curve is $y = 24 \sqrt { x } - 8 x ^ { \frac { 3 } { 2 } } + 16$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { dy } } { \mathrm { dx } }$.
\item Find the coordinates of the turning point.
\item Determine the nature of the turning point.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 2 2020 Q9 [8]}}