| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2011 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find stationary points and nature |
| Difficulty | Moderate -0.3 This is a comprehensive multi-part question covering standard C2 techniques (differentiation, tangent equations, stationary points, integration, transformations). While it has many parts (7 marks total based on typical AQA marking), each individual step is routine and algorithmic. The differentiation uses basic power rule, finding stationary points is standard equating derivative to zero, and the integration is straightforward. The final transformation part requires minimal insight. Slightly easier than average due to being purely procedural with no problem-solving required. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07i Differentiate x^n: for rational n and sums1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x + 3 + 8x^{-4}\) | B1 | Rewriting in index form |
| \(\frac{dy}{dx} = 1 - 32x^{-5}\) | M1 A1 | M1 for attempt to differentiate; A1 correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(x=1\): \(y = 1+3+8 = 12\) | B1 | Correct \(y\)-value |
| \(\frac{dy}{dx} = 1 - 32 = -31\) | M1 | Substituting \(x=1\) into derivative |
| \(y - 12 = -31(x-1)\), i.e. \(y = -31x + 43\) | A1 | Correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1 - 32x^{-5} = 0\) | M1 | Setting \(\frac{dy}{dx}=0\) |
| \(x^5 = 32\), \(x = 2\) | A1 | Correct \(x\)-value |
| \(y = 2 + 3 + \frac{8}{16} = 5.5\) | M1 A1 | Substituting back; \(M = (2, 5.5)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\left(x+3+8x^{-4}\right)dx\) | M1 | Attempt to integrate each term |
| \(= \frac{x^2}{2} + 3x - \frac{8}{3}x^{-3}\) | A1 A1 | A1 for first two terms, A1 for \(-\frac{8}{3}x^{-3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\left[\frac{x^2}{2}+3x-\frac{8}{3x^3}\right]_1^2\) | M1 | Applying limits |
| \(= \left(2+6-\frac{1}{3}\right)-\left(\frac{1}{2}+3-\frac{8}{3}\right) = \frac{23}{3} - \frac{11}{6} = \frac{35}{6}\) | A1 | Correct area |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(k = -5.5\) (or \(-\frac{11}{2}\)) | B1 | The minimum value of \(y\) is \(5.5\), so translate down by \(5.5\) |
# Question 7:
## Part (a) – Find $\frac{dy}{dx}$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x + 3 + 8x^{-4}$ | B1 | Rewriting in index form |
| $\frac{dy}{dx} = 1 - 32x^{-5}$ | M1 A1 | M1 for attempt to differentiate; A1 correct answer |
## Part (b) – Equation of tangent at $x=1$
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $x=1$: $y = 1+3+8 = 12$ | B1 | Correct $y$-value |
| $\frac{dy}{dx} = 1 - 32 = -31$ | M1 | Substituting $x=1$ into derivative |
| $y - 12 = -31(x-1)$, i.e. $y = -31x + 43$ | A1 | Correct equation |
## Part (c) – Minimum point M
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1 - 32x^{-5} = 0$ | M1 | Setting $\frac{dy}{dx}=0$ |
| $x^5 = 32$, $x = 2$ | A1 | Correct $x$-value |
| $y = 2 + 3 + \frac{8}{16} = 5.5$ | M1 A1 | Substituting back; $M = (2, 5.5)$ |
## Part (d)(i) – Find integral
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\left(x+3+8x^{-4}\right)dx$ | M1 | Attempt to integrate each term |
| $= \frac{x^2}{2} + 3x - \frac{8}{3}x^{-3}$ | A1 A1 | A1 for first two terms, A1 for $-\frac{8}{3}x^{-3}$ |
## Part (d)(ii) – Area between $x=1$ and $x=2$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[\frac{x^2}{2}+3x-\frac{8}{3x^3}\right]_1^2$ | M1 | Applying limits |
| $= \left(2+6-\frac{1}{3}\right)-\left(\frac{1}{2}+3-\frac{8}{3}\right) = \frac{23}{3} - \frac{11}{6} = \frac{35}{6}$ | A1 | Correct area |
## Part (e) – Value of $k$
| Answer/Working | Mark | Guidance |
|---|---|---|
| $k = -5.5$ (or $-\frac{11}{2}$) | B1 | The minimum value of $y$ is $5.5$, so translate down by $5.5$ |
---\begin{enumerate}[label=(\alph*)]
\item Given that $y = x + 3 + \frac { 8 } { x ^ { 4 } }$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Find an equation of the tangent at the point on the curve $C$ where $x = 1$.
\item The curve $C$ has a minimum point $M$. Find the coordinates of $M$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\int \left( x + 3 + \frac { 8 } { x ^ { 4 } } \right) \mathrm { d } x$.
\item Hence find the area of the region bounded by the curve $C$, the $x$-axis and the lines $x = 1$ and $x = 2$.
\end{enumerate}\item The curve $C$ is translated by $\left[ \begin{array} { l } 0 \\ k \end{array} \right]$ to give the curve $y = \mathrm { f } ( x )$. Given that the $x$-axis is a tangent to the curve $y = \mathrm { f } ( x )$, state the value of the constant $k$.\\
(1 mark)
\end{enumerate}
\hfill \mbox{\textit{AQA C2 2011 Q7 [16]}}