| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 12 |
| 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 C1 differentiation question requiring routine application of power rule, evaluation at given points, and interpretation of sign of derivative. All steps are standard textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.07b Gradient as rate of change: dy/dx notation1.07d Second derivatives: d^2y/dx^2 notation1.07n Stationary points: find maxima, minima using derivatives1.07o Increasing/decreasing: functions using sign of dy/dx |
| Answer | Marks | Guidance |
|---|---|---|
| [Modulus graph sketch] | M1 | Modulus graph |
| Correct shape including cusp at \((\pi, 0)\), ignore any part of graph beyond \(0 \leq x \leq 2\pi\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(k = 1\) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| [Two branch curve sketch] | M1 | Two branch curve, general shape correct |
| Min at \((\alpha, 1)\), Max at \((\beta, 1)\) with \(\alpha\) roughly halfway between \(0\) and \(\pi\), and \(\beta\) roughly halfway between \(\pi\) and \(2\pi\), curve asymptotic to \(x = 0\), \(x = \pi\) and \(x = 2\pi\) | A1 | 2 |
# Question 3(a)(i) [XMCA2]:
[Modulus graph sketch] | M1 | Modulus graph
Correct shape including cusp at $(\pi, 0)$, ignore any part of graph beyond $0 \leq x \leq 2\pi$ | A1 | **2** |
---
# Question 3(a)(ii) [XMCA2]:
$k = 1$ | B1 | **1** |
---
# Question 3(b) [XMCA2]:
[Two branch curve sketch] | M1 | Two branch curve, general shape correct
Min at $(\alpha, 1)$, Max at $(\beta, 1)$ with $\alpha$ roughly halfway between $0$ and $\pi$, and $\beta$ roughly halfway between $\pi$ and $2\pi$, curve asymptotic to $x = 0$, $x = \pi$ and $x = 2\pi$ | A1 | **2** |
---3 The depth of water, $y$ metres, in a tank after time $t$ hours is given by
$$y = \frac { 1 } { 8 } t ^ { 4 } - 2 t ^ { 2 } + 4 t , \quad 0 \leqslant t \leqslant 4$$
\begin{enumerate}[label=(\alph*)]
\item Find:
\begin{enumerate}[label=(\roman*)]
\item $\frac { \mathrm { d } y } { \mathrm {~d} t }$;
\item $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} t ^ { 2 } }$.
\end{enumerate}\item Verify that $y$ has a stationary value when $t = 2$ and determine whether it is a maximum value or a minimum value.
\item \begin{enumerate}[label=(\roman*)]
\item Find the rate of change of the depth of water, in metres per hour, when $t = 1$.
\item Hence determine, with a reason, whether the depth of water is increasing or decreasing when $t = 1$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2010 Q3 [12]}}