| Exam Board | AQA |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| 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 straightforward multi-part calculus question covering standard C2 techniques: differentiating powers of x (including negative powers), finding stationary points by setting dy/dx = 0, finding a normal line, and integration with definite integrals. All parts follow routine procedures with no novel problem-solving required, though the multiple steps and marks make it slightly more substantial than the most basic exercises. |
| 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 derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) | \(y = x + 1 + 4x^{-2} \Rightarrow \frac{dy}{dx} = 1 - 8x^{-3}\) | M1, A2,1,0 |
| (ii) | \(1 - 8x^{-3} = 0\) | M1 |
| \(x^3 = 8\) | m1 | Using \(x^{-k} = \frac{1}{x^k}\) to reach \(x^a = b\), \(a>0\) or correct use of logs. |
| \(x = 2\); When \(x = 2\), \(y = 4\) | A1, A1ft | |
| Total: 4 | ||
| (iii) | At \((1, 6)\), \(\frac{dy}{dx} = 1 - 8 = -7\) | M1 |
| Gradient of normal \(= \frac{1}{7}\) | M1 | Use of or stating \(m \times m' = -1\) |
| Equation of normal is \(y - 6 = m[x - 1]\) | M1 | \(m\) numerical |
| \(y - 6 = \frac{1}{7}(x - 1)\); \(7y = x + 41\) | A1ft | OE ft on c's answer for (a)(i) provided at least A1 given in (a)(i) and previous 3M marks awarded |
| Total: 4 | ||
| (b)(i) | \(\int x(1 + \frac{4}{x^2})dx = ........ = \frac{x^2}{2} + x - 4x^{-1} + \{c\}\) | M1, A2,1,0 |
| (ii) | \(\{\text{Area}\} = \int_1^4 x + 1 + \frac{4}{x^2}dx =\) | M1 |
| \([\frac{x^2}{2} + x - \frac{4}{x}]_1^4 = (8 + 4 - 1) - (\frac{1}{2} + 1 - 4)\) | M1 | Dealing correctly with limits; F(4)−F(1) (must have integrated) |
| \(= 13.5\) | A1 | |
| Total: 16 |
(a)(i) | $y = x + 1 + 4x^{-2} \Rightarrow \frac{dy}{dx} = 1 - 8x^{-3}$ | M1, A2,1,0 | Power $p \rightarrow p-1$ (A1 if $1 + ax^n$ with $a = -8$ or $n = -3$) |
(ii) | $1 - 8x^{-3} = 0$ | M1 | Puts c's $\frac{dy}{dx} = 0$ |
| $x^3 = 8$ | m1 | Using $x^{-k} = \frac{1}{x^k}$ to reach $x^a = b$, $a>0$ or correct use of logs. |
| $x = 2$; When $x = 2$, $y = 4$ | A1, A1ft | |
| | **Total: 4** | |
(iii) | At $(1, 6)$, $\frac{dy}{dx} = 1 - 8 = -7$ | M1 | Attempt to find $y'(1)$ |
| Gradient of normal $= \frac{1}{7}$ | M1 | Use of or stating $m \times m' = -1$ |
| Equation of normal is $y - 6 = m[x - 1]$ | M1 | $m$ numerical |
| $y - 6 = \frac{1}{7}(x - 1)$; $7y = x + 41$ | A1ft | OE ft on c's answer for (a)(i) provided at least A1 given in (a)(i) and previous 3M marks awarded |
| | **Total: 4** | |
(b)(i) | $\int x(1 + \frac{4}{x^2})dx = ........ = \frac{x^2}{2} + x - 4x^{-1} + \{c\}$ | M1, A2,1,0 | One of three terms correct. For A2 need all three terms as printed or better (A1 if 2 of 3 terms correct) |
(ii) | $\{\text{Area}\} = \int_1^4 x + 1 + \frac{4}{x^2}dx =$ | M1 | |
| $[\frac{x^2}{2} + x - \frac{4}{x}]_1^4 = (8 + 4 - 1) - (\frac{1}{2} + 1 - 4)$ | M1 | Dealing correctly with limits; F(4)−F(1) (must have integrated) |
| $= 13.5$ | A1 | |
| | **Total: 16** | |
---6 A curve $C$ is defined for $x > 0$ by the equation $y = x + 1 + \frac { 4 } { x ^ { 2 } }$ and is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{c16d94a6-52f2-4bf3-acee-0b227ae55a1a-4_545_784_420_628}
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = x + 1 + \frac { 4 } { x ^ { 2 } }$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item The curve $C$ has a minimum point $M$. Find the coordinates of $M$.
\item Find an equation of the normal to $C$ at the point ( 1,6 ).
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Find $\int \left( x + 1 + \frac { 4 } { x ^ { 2 } } \right) \mathrm { d } x$.
\item Hence find the area of the region bounded by the curve $C$, the lines $x = 1$ and $x = 4$ and the $x$-axis.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C2 2007 Q6 [16]}}