| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2011 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with exponential functions |
| Difficulty | Standard +0.3 This is a standard C3 volumes of revolution question with routine exponential manipulation. Part (a) is basic logarithm work, parts (b)(i-iii) involve straightforward curve analysis (substitution, solving e^{-2x}=4, differentiation), and part (iv) requires expanding (4e^{-2x}-e^{-4x})^2 and integrating exponentials—all textbook techniques with no novel insight required. Slightly easier than average due to the structured scaffolding. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(-2x = \ln 4\) | M1 | Taking logarithm |
| \(x = -\frac{\ln 4}{2}\) or \(x = -\ln 2\) | A1 | Exact value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y\)-coordinate of \(A\) is \(3\) | B1 | \(4(1)-1=3\) at \(x=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(4e^{-2x} - e^{-4x} = 0 \Rightarrow e^{-4x}(4e^{2x}-1)=0\) | M1 | Setting \(y=0\), factoring |
| \(e^{2x} = \frac{1}{4}\), so \(2x = -\ln 4\) | A1 | Correct equation |
| \(x = -\ln 2\) | A1 | Exact answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = -8e^{-2x} + 4e^{-4x} = 0\) | M1 | Differentiating and setting to zero |
| \(4e^{-4x}(e^{2x}-2)=0 \Rightarrow e^{2x} = 2\) | A1 | Correct equation |
| \(x = \frac{\ln 2}{2}\) | A1 | Exact answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(V = \pi\int_0^{\ln 2}(4e^{-2x}-e^{-4x})^2\,dx\) | M1 | Correct volume formula with \(y^2\) |
| \(= \pi\int_0^{\ln 2}(16e^{-4x} - 8e^{-6x} + e^{-8x})\,dx\) | A1 | Correct expansion |
| \(= \pi\left[-4e^{-4x}+\frac{4}{3}e^{-6x}-\frac{1}{8}e^{-8x}\right]_0^{\ln 2}\) | A1 | Correct integration |
| At \(x=\ln 2\): \(-\frac{4}{16}+\frac{4}{3\cdot 64}-\frac{1}{8\cdot 256} = -\frac{1}{4}+\frac{1}{48}-\frac{1}{2048}\) | M1 | Substituting upper limit |
| At \(x=0\): \(-4+\frac{4}{3}-\frac{1}{8}\) | A1 | Substituting lower limit |
| \(V = \frac{p}{q}\pi\) where final answer \(= \frac{81}{16}\pi\) | A1A1 | Correct \(p\) and \(q\) integers |
# Question 8:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $-2x = \ln 4$ | M1 | Taking logarithm |
| $x = -\frac{\ln 4}{2}$ or $x = -\ln 2$ | A1 | Exact value |
## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y$-coordinate of $A$ is $3$ | B1 | $4(1)-1=3$ at $x=0$ |
## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $4e^{-2x} - e^{-4x} = 0 \Rightarrow e^{-4x}(4e^{2x}-1)=0$ | M1 | Setting $y=0$, factoring |
| $e^{2x} = \frac{1}{4}$, so $2x = -\ln 4$ | A1 | Correct equation |
| $x = -\ln 2$ | A1 | Exact answer |
## Part (b)(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = -8e^{-2x} + 4e^{-4x} = 0$ | M1 | Differentiating and setting to zero |
| $4e^{-4x}(e^{2x}-2)=0 \Rightarrow e^{2x} = 2$ | A1 | Correct equation |
| $x = \frac{\ln 2}{2}$ | A1 | Exact answer |
## Part (b)(iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $V = \pi\int_0^{\ln 2}(4e^{-2x}-e^{-4x})^2\,dx$ | M1 | Correct volume formula with $y^2$ |
| $= \pi\int_0^{\ln 2}(16e^{-4x} - 8e^{-6x} + e^{-8x})\,dx$ | A1 | Correct expansion |
| $= \pi\left[-4e^{-4x}+\frac{4}{3}e^{-6x}-\frac{1}{8}e^{-8x}\right]_0^{\ln 2}$ | A1 | Correct integration |
| At $x=\ln 2$: $-\frac{4}{16}+\frac{4}{3\cdot 64}-\frac{1}{8\cdot 256} = -\frac{1}{4}+\frac{1}{48}-\frac{1}{2048}$ | M1 | Substituting upper limit |
| At $x=0$: $-4+\frac{4}{3}-\frac{1}{8}$ | A1 | Substituting lower limit |
| $V = \frac{p}{q}\pi$ where final answer $= \frac{81}{16}\pi$ | A1A1 | Correct $p$ and $q$ integers |8
\begin{enumerate}[label=(\alph*)]
\item Given that $\mathrm { e } ^ { - 2 x } = 4$, find the exact value of $x$.
\item The diagram shows the curve $y = 4 \mathrm { e } ^ { - 2 x } - \mathrm { e } ^ { - 4 x }$.\\
\includegraphics[max width=\textwidth, alt={}, center]{6761e676-48ae-47e9-9617-153342cdf5c4-9_490_1185_463_440}
The curve crosses the $y$-axis at the point $A$, the $x$-axis at the point $B$, and has a stationary point at $M$.
\begin{enumerate}[label=(\roman*)]
\item State the $y$-coordinate of $A$.
\item Find the $x$-coordinate of $B$, giving your answer in an exact form.
\item Find the $x$-coordinate of the stationary point, $M$, giving your answer in an exact form.
\item The shaded region $R$ is bounded by the curve $y = 4 \mathrm { e } ^ { - 2 x } - \mathrm { e } ^ { - 4 x }$, the lines $x = 0$ and $x = \ln 2$ and the $x$-axis.
Find the volume of the solid generated when the region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis, giving your answer in the form $\frac { p } { q } \pi$, where $p$ and $q$ are integers.\\
(7 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2011 Q8 [16]}}