| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | January |
| 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 C1 stationary points question requiring routine differentiation of a polynomial, solving dy/dx = 0, and applying the second derivative test. All steps are standard textbook procedures with no problem-solving insight needed, making it easier than average but not trivial due to the cubic polynomial arithmetic. |
| 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 |
|---|---|---|
| \(\frac{dy}{dx} = 18 + 6x - 12x^2\) | M1, A1, A1 | one of these terms correct; another term correct; all correct (no + c etc) (penalise + c once only in question) |
| Answer | Marks | Guidance |
|---|---|---|
| \(18 + 6x - 12x^2 = 0\) | M1 | putting their \(\frac{dy}{dx} = 0\), PI by attempt to solve or factorise |
| \(6(3 - 2x)(x + 1) = 0\) | m1 | attempt at factors of their quadratic or use of quadratic equation formula |
| \(x = -1, x = \frac{3}{2}\) OE | A1 | must see both values unless \(x = -1\) is verified separately |
| If M1 not scored, award SC B1 for verifying that \(x = -1\) leads to \(\frac{dy}{dx} = 0\) and a further SC B2 for finding \(x = \frac{3}{2}\) as other value |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d^2y}{dx^2} = 6 - 24x\) | B1√ | FT their \(\frac{dy}{dx}\) but \(\frac{d^2y}{dx^2}\) must be correct if 3 marks earned in part (a) |
| When \(x = -1\), \(\frac{d^2y}{dx^2} = 6 - (24 \times -1)\) | M1 | Sub \(x = -1\) into their \(\frac{d^2y}{dx^2}\) |
| \(\frac{d^2y}{dx^2} = 30\) | A1cso | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Minimum point | E1√ | must have a value in (c)(i); FT "maximum" if their value of \(\frac{d^2y}{dx^2} < 0\) |
**1(a)**
$\frac{dy}{dx} = 18 + 6x - 12x^2$ | M1, A1, A1 | one of these terms correct; another term correct; all correct (no + c etc) (penalise + c once only in question)
**1(b)**
$18 + 6x - 12x^2 = 0$ | M1 | putting their $\frac{dy}{dx} = 0$, PI by attempt to solve or factorise
$6(3 - 2x)(x + 1) = 0$ | m1 | attempt at factors of their quadratic or use of quadratic equation formula
$x = -1, x = \frac{3}{2}$ OE | A1 | must see both values unless $x = -1$ is verified separately
| | | If M1 not scored, award SC B1 for verifying that $x = -1$ leads to $\frac{dy}{dx} = 0$ and a further SC B2 for finding $x = \frac{3}{2}$ as other value
**1(c)(i)**
$\frac{d^2y}{dx^2} = 6 - 24x$ | B1√ | FT their $\frac{dy}{dx}$ but $\frac{d^2y}{dx^2}$ must be correct if 3 marks earned in part (a)
When $x = -1$, $\frac{d^2y}{dx^2} = 6 - (24 \times -1)$ | M1 | Sub $x = -1$ into their $\frac{d^2y}{dx^2}$
$\frac{d^2y}{dx^2} = 30$ | A1cso | 3
**1(c)(ii)**
Minimum point | E1√ | must have a value in (c)(i); FT "maximum" if their value of $\frac{d^2y}{dx^2} < 0$ | 1
---1 The curve with equation $y = 13 + 18 x + 3 x ^ { 2 } - 4 x ^ { 3 }$ passes through the point $P$ where $x = - 1$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Show that the point $P$ is a stationary point of the curve and find the other value of $x$ where the curve has a stationary point.
\item \begin{enumerate}[label=(\roman*)]
\item Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at the point $P$.
\item Hence, or otherwise, determine whether $P$ is a maximum point or a minimum point.\\
(l mark)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2011 Q1 [10]}}