| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find stationary points coordinates |
| Difficulty | Moderate -0.3 This is a standard C2 stationary points question requiring routine differentiation of a power function, setting derivative to zero, and finding normal equations. While multi-part with several marks, each step follows textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = 6 - 3x^{\frac{1}{2}}\) | B1 | For either 6 or \(6x^0\) |
| M1 | \(Ax^{\frac{3}{2}-1}\), \(A \neq 0\) OE | |
| \(6 - 3x^{\frac{1}{2}}\) or \(6 - 3\sqrt{x}\) with no '+c' | A1 (Total: 3) | If unsimplified here, A1 can be awarded retrospectively if correct simplified expression seen in (b)(i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6 - 3x^{\frac{1}{2}} = 0\) | M1 | Equating c's \(\frac{dy}{dx}\) to 0, PI by correct ft; rearrangement of c's \(dy/dx=0\) |
| \(x^{\frac{1}{2}} = 2 \Rightarrow x = 2^2\) | m1 | \(x^{\frac{1}{2}} = k\) \((k>0)\), to \(x = k^2\). PI by correct value of \(x\) if no error seen |
| \(M(4, 8)\) | A1 (Total: 3) | SC If M0 award B1 for \((4, 8)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of normal at \(M\) is \(x = 4\) | B1F (Total: 1) | Ft on \(x =\) c's \(x_M\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| When \(x = \frac{9}{4}\), \(\frac{dy}{dx} = 6 - 3 \times \frac{3}{2} = \frac{3}{2}\) | M1 | Attempt to find \(\frac{dy}{dx}\) when \(x = \frac{9}{4}\) |
| Gradient of normal at \(P = -\frac{2}{3}\) | m1 | \(m \times m' = -1\) used |
| \(y - \frac{27}{4} = -\frac{2}{3}\left(x - \frac{9}{4}\right)\) | A1 | ACF e.g. \(y = -\frac{2}{3}x + \frac{33}{4}\) |
| \(12y - 81 = -8x + 18 \Rightarrow 8x + 12y = 99\) | A1 (Total: 4) | Coefficients and constant must be positive integers, but accept different order e.g. \(12y + 8x = 99\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8(4) + 12y = 99\) | M1 | Solving c's answer (b)(ii), must be in form \(x =\) positive const, with c's answer (c)(i). PI by correct earlier work and correct coordinates for \(R\) |
| \(R\left(4, \frac{67}{12}\right)\) | A1 (Total: 2) | Accept 5.58 or better as equivalent to \(\frac{67}{12}\) |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 6 - 3x^{\frac{1}{2}}$ | B1 | For either 6 or $6x^0$ |
| | M1 | $Ax^{\frac{3}{2}-1}$, $A \neq 0$ OE |
| $6 - 3x^{\frac{1}{2}}$ or $6 - 3\sqrt{x}$ with no '+c' | A1 (Total: 3) | If unsimplified here, A1 can be awarded retrospectively if correct simplified expression seen in (b)(i) |
### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6 - 3x^{\frac{1}{2}} = 0$ | M1 | Equating c's $\frac{dy}{dx}$ to 0, PI by correct ft; rearrangement of c's $dy/dx=0$ |
| $x^{\frac{1}{2}} = 2 \Rightarrow x = 2^2$ | m1 | $x^{\frac{1}{2}} = k$ $(k>0)$, to $x = k^2$. PI by correct value of $x$ if no error seen |
| $M(4, 8)$ | A1 (Total: 3) | SC If M0 award B1 for $(4, 8)$ |
### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of normal at $M$ is $x = 4$ | B1F (Total: 1) | Ft on $x =$ c's $x_M$ |
### Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| When $x = \frac{9}{4}$, $\frac{dy}{dx} = 6 - 3 \times \frac{3}{2} = \frac{3}{2}$ | M1 | Attempt to find $\frac{dy}{dx}$ when $x = \frac{9}{4}$ |
| Gradient of normal at $P = -\frac{2}{3}$ | m1 | $m \times m' = -1$ used |
| $y - \frac{27}{4} = -\frac{2}{3}\left(x - \frac{9}{4}\right)$ | A1 | ACF e.g. $y = -\frac{2}{3}x + \frac{33}{4}$ |
| $12y - 81 = -8x + 18 \Rightarrow 8x + 12y = 99$ | A1 (Total: 4) | Coefficients and constant must be positive integers, but accept different order e.g. $12y + 8x = 99$ |
### Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8(4) + 12y = 99$ | M1 | Solving c's answer (b)(ii), must be in form $x =$ positive const, with c's answer (c)(i). PI by correct earlier work and correct coordinates for $R$ |
| $R\left(4, \frac{67}{12}\right)$ | A1 (Total: 2) | Accept 5.58 or better as equivalent to $\frac{67}{12}$ |
---5 The diagram shows part of a curve with a maximum point $M$.\\
\includegraphics[max width=\textwidth, alt={}, center]{258f0400-6e3b-406c-9f86-acc9fff4e094-4_480_645_354_694}
The curve is defined for $x \geqslant 0$ by the equation
$$y = 6 x - 2 x ^ { \frac { 3 } { 2 } }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(3 marks)
\item \begin{enumerate}[label=(\roman*)]
\item Hence find the coordinates of the maximum point $M$.
\item Write down the equation of the normal to the curve at $M$.
\end{enumerate}\item The point $P \left( \frac { 9 } { 4 } , \frac { 27 } { 4 } \right)$ lies on the curve.
\begin{enumerate}[label=(\roman*)]
\item Find an equation of the normal to the curve at the point $P$, giving your answer in the form $a x + b y = c$, where $a , b$ and $c$ are positive integers.
\item The normals to the curve at the points $M$ and $P$ intersect at the point $R$. Find the coordinates of $R$.\\
$6 \quad$ A curve $C$, defined for $0 \leqslant x \leqslant 2 \pi$ by the equation $y = \sin x$, where $x$ is in radians, is sketched below. The region bounded by the curve $C$, the $x$-axis from 0 to 2 and the line $x = 2$ is shaded.\\
\includegraphics[max width=\textwidth, alt={}, center]{258f0400-6e3b-406c-9f86-acc9fff4e094-5_441_789_466_612}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q5 [13]}}