| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Determine nature of stationary points |
| Difficulty | Moderate -0.3 This is a straightforward C2 differentiation question requiring standard techniques: differentiating a simple expression, finding stationary points by solving dy/dx=0, using the second derivative test, and integrating to find the curve equation. All steps are routine applications of core calculus rules with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07e Second derivative: as rate of change of gradient1.07n Stationary points: find maxima, minima using derivatives1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = -\frac{4}{x^3} - \frac{1}{4}\) | M1 A1 A1 | M1 for differentiating \(\frac{2}{x^2}\) or \(-\frac{x}{4}\); A1 for each correct term |
| Answer | Marks | Guidance |
|---|---|---|
| At stationary point \(\frac{dy}{dx} = 0\), so \(\frac{2}{x^2} - \frac{x}{4} = 0\) | M1 | Setting derivative to zero |
| \(x^3 = 8\), so \(x = 2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| At \(x = 2\): \(\frac{d^2y}{dx^2} = -\frac{4}{8} - \frac{1}{4} = -\frac{1}{2} - \frac{1}{4} = -\frac{3}{4} < 0\), therefore maximum | A1 | Must reference negative value and conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = \int \left(\frac{2}{x^2} - \frac{x}{4}\right)dx\) | M1 | Attempt to integrate |
| \(y = -\frac{2}{x} - \frac{x^2}{8} + c\) | A1 A1 | A1 for each correct term |
| Using \(\left(2, \frac{5}{2}\right)\): \(\frac{5}{2} = -1 - \frac{1}{2} + c\), so \(c = 4\) | M1 A1 | Substituting point to find \(c\) |
| \(y = -\frac{2}{x} - \frac{x^2}{8} + 4\) |
## Question 4:
**Part (a):** Find $\frac{d^2y}{dx^2}$
| $\frac{d^2y}{dx^2} = -\frac{4}{x^3} - \frac{1}{4}$ | M1 A1 A1 | M1 for differentiating $\frac{2}{x^2}$ or $-\frac{x}{4}$; A1 for each correct term |
**Part (b)(i):** Find the $x$-coordinate of $M$
| At stationary point $\frac{dy}{dx} = 0$, so $\frac{2}{x^2} - \frac{x}{4} = 0$ | M1 | Setting derivative to zero |
| $x^3 = 8$, so $x = 2$ | A1 | |
**Part (b)(ii):** Show $M$ is a maximum
| At $x = 2$: $\frac{d^2y}{dx^2} = -\frac{4}{8} - \frac{1}{4} = -\frac{1}{2} - \frac{1}{4} = -\frac{3}{4} < 0$, therefore maximum | A1 | Must reference negative value and conclusion |
**Part (b)(iii):** Find the equation of the curve
| $y = \int \left(\frac{2}{x^2} - \frac{x}{4}\right)dx$ | M1 | Attempt to integrate |
| $y = -\frac{2}{x} - \frac{x^2}{8} + c$ | A1 A1 | A1 for each correct term |
| Using $\left(2, \frac{5}{2}\right)$: $\frac{5}{2} = -1 - \frac{1}{2} + c$, so $c = 4$ | M1 A1 | Substituting point to find $c$ |
| $y = -\frac{2}{x} - \frac{x^2}{8} + 4$ | | |
---4 A curve is defined for $x > 0$. The gradient of the curve at the point $( x , y )$ is given by
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 } { x ^ { 2 } } - \frac { x } { 4 }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$.
\item The curve has a stationary point $M$ whose $y$-coordinate is $\frac { 5 } { 2 }$.
\begin{enumerate}[label=(\roman*)]
\item Find the $x$-coordinate of $M$.
\item Use your answers to parts (a) and (b)(i) to show that $M$ is a maximum point.
\item Find the equation of the curve.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2015 Q4 [10]}}