| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| 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 C2 question testing routine application of differentiation techniques: substitution to verify a stationary point, finding the second derivative using standard rules (including differentiating x^{-2}), applying the second derivative test, and integrating to find the curve equation. All steps are mechanical with no problem-solving or insight required, making it easier than average. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07i Differentiate x^n: for rational n and sums1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x=2\), \(\frac{dy}{dx} = 12 - 1 - 11 = 0\) | B1 (1 mark) | AG Must see intermediate evaluations |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{4}{x^2} = 4x^{-2}\), so \(\frac{dy}{dx} = 3x^2 - 4x^{-2} - 11\) | B1 | \(\frac{4}{x^2} = 4x^{-2}\) seen; PI by \(\pm 8x^{-3}\) term in answer |
| \(\frac{d^2y}{dx^2} = 6x + 8x^{-3}\) | M1, A1 | Correct powers of \(x\) obtained from differentiating first two terms |
| When \(x=2\), \(\frac{d^2y}{dx^2} = 12 + \frac{8}{8} = 13\) | A1 (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Since \(\frac{d^2y}{dx^2} > 0\), \(P\) is a minimum point | E1F (1 mark) | Must see reference to sign of \(y''(2)\) and correct conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int\left(3x^2 - \frac{4}{x^2} - 11\right)dx = x^3 + 4x^{-1} - 11x \ (+c)\) | M1, A1 | Attempt to integrate with at least two of three terms correct |
| When \(x=2\), \(y=1 \Rightarrow 1 = 8 + 2 - 22 + c\) | M1 | Substituting \(x=2\), \(y=1\) to find \(c\) |
| \(y = x^3 + 4x^{-1} - 11x + 13\) | A1 (4 marks) | ACF |
## Question 6:
### Part (a)(i):
| When $x=2$, $\frac{dy}{dx} = 12 - 1 - 11 = 0$ | B1 (1 mark) | AG Must see intermediate evaluations |
### Part (a)(ii):
| $\frac{4}{x^2} = 4x^{-2}$, so $\frac{dy}{dx} = 3x^2 - 4x^{-2} - 11$ | B1 | $\frac{4}{x^2} = 4x^{-2}$ seen; PI by $\pm 8x^{-3}$ term in answer |
| $\frac{d^2y}{dx^2} = 6x + 8x^{-3}$ | M1, A1 | Correct powers of $x$ obtained from differentiating first two terms |
| When $x=2$, $\frac{d^2y}{dx^2} = 12 + \frac{8}{8} = 13$ | A1 (4 marks) | |
### Part (a)(iii):
| Since $\frac{d^2y}{dx^2} > 0$, $P$ is a minimum point | E1F (1 mark) | Must see reference to sign of $y''(2)$ and correct conclusion |
### Part (b):
| $\int\left(3x^2 - \frac{4}{x^2} - 11\right)dx = x^3 + 4x^{-1} - 11x \ (+c)$ | M1, A1 | Attempt to integrate with at least two of three terms correct |
| When $x=2$, $y=1 \Rightarrow 1 = 8 + 2 - 22 + c$ | M1 | Substituting $x=2$, $y=1$ to find $c$ |
| $y = x^3 + 4x^{-1} - 11x + 13$ | A1 (4 marks) | ACF |
---6 At the point $( x , y )$, where $x > 0$, the gradient of a curve is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = 3 x ^ { 2 } - \frac { 4 } { x ^ { 2 } } - 11$$
The point $P ( 2,1 )$ lies on the curve.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Verify that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$ when $x = 2$.\\
(l mark)
\item Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ when $x = 2$.
\item Hence state whether $P$ is a maximum point or a minimum point, giving a reason for your answer.
\end{enumerate}\item Find the equation of the curve.
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2012 Q6 [10]}}