| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area between two polar curves |
| Difficulty | Challenging +1.3 This is a Further Pure 2 polar coordinates question requiring finding intersection points and computing areas. Part (i) involves solving a trigonometric equation (standard technique). Part (ii) requires setting up and evaluating polar area integrals with appropriate limits, using the formula A = ½∫r²dθ. While the integration requires careful setup and trigonometric manipulation, these are well-practiced FP2 techniques. The 9 total marks and multi-step nature place it above average difficulty, but it follows standard polar area methodology without requiring exceptional insight. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\cos^2\alpha = 2\sin 2\alpha = 4\sin\alpha\cos\alpha\) | M1 | For equation in \(\cos\alpha\) and \(\sin\alpha\) (only – ie dealing with sin2α leading to AG(\(\theta\) may be used instead of \(\alpha\))) |
| \(\Rightarrow \tan\alpha = \frac{1}{2}\) | A1 | SR Allow verification only if exact |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= \frac{1}{2}\int_0^\alpha r_2^2 d\theta + \frac{1}{2}\int_\alpha^{\pi} r_1^2 d\theta\) | M1 | For both integrals added with limits soi Allow \(\theta\) for \(\alpha\), and reversal of r² terms |
| \(= \frac{1}{2}\int_0^\alpha 2\sin 2\theta\, d\theta + \frac{1}{2}\int_\alpha^\pi 1 + \cos 2\theta\, d\theta\) | M1 | For using \(2\cos^2\theta = 1 + \cos 2\theta\) in 2nd integral |
| \(= \left[-\frac{1}{2}\cos 2\theta\right]_0^\alpha + \left[\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta\right]_\alpha^\pi\) | M1 | For k cos2θ as first integrated term |
| \(= \left(-\frac{1}{2}\cos 2\alpha + \frac{1}{2}\right) + \left(\frac{1}{4}\pi - \frac{\alpha}{4} - \frac{1}{4}\sin 2\alpha \cos\alpha\right)\) | A1 | For correct first area |
| \(= \left(-\frac{1}{2}(1-2\sin^2\alpha) + \frac{1}{2}\right) + \left(\frac{1}{4}\pi - \frac{\alpha}{2} - \frac{1}{2}\sin\alpha\cos\alpha\right)\) | A1 | For correct second area |
| \(= \frac{1}{5} + \frac{1}{4}\pi - \frac{1}{2}\alpha - \frac{1}{2} \cdot \frac{2}{\sqrt{5}}\cdot\frac{\sqrt{5}}{}\) | M1 | For using Pythagoras to find sinα or cosα; OR t formula for cos2α or sin2α |
| \(= \frac{1}{4}\pi - \frac{1}{2}\alpha\) | A1 | For simplification to AG |
| [7] |
### (i)
$2\cos^2\alpha = 2\sin 2\alpha = 4\sin\alpha\cos\alpha$ | M1 | For equation in $\cos\alpha$ and $\sin\alpha$ (only – ie dealing with sin2α leading to AG($\theta$ may be used instead of $\alpha$))
$\Rightarrow \tan\alpha = \frac{1}{2}$ | A1 | SR Allow verification only if exact
| [2] |
### (ii)
Area $= \frac{1}{2}\int_0^\alpha r_2^2 d\theta + \frac{1}{2}\int_\alpha^{\pi} r_1^2 d\theta$ | M1 | For both integrals added with limits soi Allow $\theta$ for $\alpha$, and reversal of r² terms
$= \frac{1}{2}\int_0^\alpha 2\sin 2\theta\, d\theta + \frac{1}{2}\int_\alpha^\pi 1 + \cos 2\theta\, d\theta$ | M1 | For using $2\cos^2\theta = 1 + \cos 2\theta$ in 2nd integral
$= \left[-\frac{1}{2}\cos 2\theta\right]_0^\alpha + \left[\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta\right]_\alpha^\pi$ | M1 | For k cos2θ as first integrated term
$= \left(-\frac{1}{2}\cos 2\alpha + \frac{1}{2}\right) + \left(\frac{1}{4}\pi - \frac{\alpha}{4} - \frac{1}{4}\sin 2\alpha \cos\alpha\right)$ | A1 | For correct first area
$= \left(-\frac{1}{2}(1-2\sin^2\alpha) + \frac{1}{2}\right) + \left(\frac{1}{4}\pi - \frac{\alpha}{2} - \frac{1}{2}\sin\alpha\cos\alpha\right)$ | A1 | For correct second area
$= \frac{1}{5} + \frac{1}{4}\pi - \frac{1}{2}\alpha - \frac{1}{2} \cdot \frac{2}{\sqrt{5}}\cdot\frac{\sqrt{5}}{}$ | M1 | For using Pythagoras to find sinα or cosα; OR t formula for cos2α or sin2α
$= \frac{1}{4}\pi - \frac{1}{2}\alpha$ | A1 | For simplification to AG
| [7] |
---\includegraphics{figure_8}
The diagram shows two curves, $C_1$ and $C_2$, which intersect at the pole $O$ and at the point $P$. The polar equation of $C_1$ is $r = \sqrt{2}\cos\theta$ and the polar equation of $C_2$ is $r = \sqrt{2}\sin 2\theta$. For both curves, $0 \leq \theta \leq \frac{1}{2}\pi$. The value of $\theta$ at $P$ is $\alpha$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\tan\alpha = \frac{1}{2}$. [2]
\item Show that the area of the region common to $C_1$ and $C_2$, shaded in the diagram, is $\frac{1}{4}\pi - \frac{1}{2}\alpha$. [7]
\end{enumerate}
\hfill \mbox{\textit{OCR FP2 2012 Q8 [9]}}