| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| 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) requires simple substitution, (b) is algebraic expansion, (c) uses the standard formula with straightforward integration of exponential terms, and (d) is basic graph transformation. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.06a Exponential function: a^x and e^x graphs and properties1.06c Logarithm definition: log_a(x) as inverse of a^x4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = -8\) | B1 | |
| \(e^{2x} - 9 = 0\) | M1 | |
| \(e^{2x} = 9\) | ||
| \(2x = \ln 9\) | ||
| \(x = \ln 3\) | A1 | AG Condone verification |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((e^{2x}-9)^2 = e^{4x} - 18e^{2x} + 81\) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V = \pi\int y^2\,dx\) | B1 | |
| \(= (\pi)\int e^{4x} - 18e^{2x} + 81\,dx\) | M1 | |
| \(= (\pi)\left[\frac{e^{4x}}{4} - 9e^{2x} + 81x\right]_0^{\ln 3}\) | M1 A1 | \(1^{st}\) or \(2^{nd}\) term correct; All correct |
| \(= (\pi)\left[\left(\frac{e^{\ln 81}}{4} - 9e^{\ln 9} + 81\ln 3\right) - \left(\frac{1}{4}-9\right)\right]\) | m1 | Attempt at limits with \(\ln 3\) |
| \(= \pi[81\ln 3 - 52]\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Modulus graph sketch (V-shape), vertex at \((-a, 8)\), crossing at \((b, \ln 3)\), \(-a\) on \(y\)-axis | M1 | Modulus graph |
| All correct | A1F |
## Question 5:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = -8$ | B1 | |
| $e^{2x} - 9 = 0$ | M1 | |
| $e^{2x} = 9$ | | |
| $2x = \ln 9$ | | |
| $x = \ln 3$ | A1 | AG Condone verification |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(e^{2x}-9)^2 = e^{4x} - 18e^{2x} + 81$ | B1 | AG |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = \pi\int y^2\,dx$ | B1 | |
| $= (\pi)\int e^{4x} - 18e^{2x} + 81\,dx$ | M1 | |
| $= (\pi)\left[\frac{e^{4x}}{4} - 9e^{2x} + 81x\right]_0^{\ln 3}$ | M1 A1 | $1^{st}$ or $2^{nd}$ term correct; All correct |
| $= (\pi)\left[\left(\frac{e^{\ln 81}}{4} - 9e^{\ln 9} + 81\ln 3\right) - \left(\frac{1}{4}-9\right)\right]$ | m1 | Attempt at limits with $\ln 3$ |
| $= \pi[81\ln 3 - 52]$ | A1 | |
### Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Modulus graph sketch (V-shape), vertex at $(-a, 8)$, crossing at $(b, \ln 3)$, $-a$ on $y$-axis | M1 | Modulus graph |
| All correct | A1F | |
---5 The diagram shows part of the graph of $y = \mathrm { e } ^ { 2 x } - 9$. The graph cuts the coordinate axes at $( 0 , a )$ and $( b , 0 )$.\\
\includegraphics[max width=\textwidth, alt={}, center]{908f530c-076d-47b1-90dd-38dbfe44f898-03_826_924_477_541}
\begin{enumerate}[label=(\alph*)]
\item State the value of $a$, and show that $b = \ln 3$.
\item Show that $y ^ { 2 } = \mathrm { e } ^ { 4 x } - 18 \mathrm { e } ^ { 2 x } + 81$.
\item The shaded region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis. Find the volume of the solid formed, giving your answer in the form $\pi ( p \ln 3 + q )$, where $p$ and $q$ are integers.
\item Sketch the curve with equation $y = \left| \mathrm { e } ^ { 2 x } - 9 \right|$ for $x \geqslant 0$.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2006 Q5 [12]}}