| Exam Board | OCR MEI |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2005 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Volume with exponential functions |
| Difficulty | Standard +0.3 This is a straightforward volumes of revolution question requiring the standard formula V = π∫y² dx with y = √(1 + e^(-2x)). The integration is direct since y² = 1 + e^(-2x), which integrates easily to x - (1/2)e^(-2x). While it involves an exponential function, no substitution or novel technique is needed, making it slightly easier than average for a C4 question. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\phi}{1} = \frac{1}{\phi - 1}\) | M1 | |
| \(\Rightarrow \phi^2 - \phi = 1 \Rightarrow \phi^2 - \phi - 1 = 0\) | ||
| Using the quadratic formula gives \(\phi = \frac{1 \pm \sqrt{5}}{2}\) | E1 | Or complete verification B2 |
$\frac{\phi}{1} = \frac{1}{\phi - 1}$ | M1 |
$\Rightarrow \phi^2 - \phi = 1 \Rightarrow \phi^2 - \phi - 1 = 0$ | |
Using the quadratic formula gives $\phi = \frac{1 \pm \sqrt{5}}{2}$ | E1 | Or complete verification B2 |4 Fig. 4 shows a sketch of the region enclosed by the curve $\sqrt { 1 + \mathrm { e } ^ { - 2 x } }$, the $x$-axis, the $y$-axis and the line $x = 1$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7a1123f8-53cd-4b24-bec6-8c3bccc22653-3_517_755_1576_649}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
Find the volume of the solid generated when this region is rotated through $360 ^ { \circ }$ about the $x$-axis. Give your answer in an exact form.
\hfill \mbox{\textit{OCR MEI C4 2005 Q4 [5]}}