| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area between two polar curves |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates question requiring intersection points and area calculation. Part (a) involves solving 1 = 2 - 2sin(θ), which is straightforward. Part (b) requires setting up the area integral ½∫(r₂² - r₁²)dθ with correct limits from part (a), then integrating (2-2sin(θ))² - 1. While this involves more steps than a typical C3/C4 question and requires FM-specific polar area techniques, it follows a standard template without requiring novel insight or particularly complex manipulation. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| \(2 - 2\sin\theta = 1 \rightarrow \sin\theta = 0.5\ \therefore \theta = \frac{\pi}{6}\ \left(\text{or } \frac{5\pi}{6}\right)\) | M1 A1 | |
| Points are \(\left(1, \frac{\pi}{6}\right)\) and \(\left(1, \frac{5\pi}{6}\right)\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Uses Area \(= \frac{1}{2}\left[\int(2-2\sin\theta)^2\,d\theta\right]\) | M1 | |
| \(= \frac{1}{2}\left[\int\left(4 - 8\sin\theta + (2-2\cos 2\theta)\right)d\theta\right]\) | M1 | |
| \(= \frac{1}{2}\left[(6\theta + 8\cos\theta - \sin 2\theta)\right]\) | A1 | |
| Uses limits \(\frac{\pi}{2}\) and \(\frac{\pi}{6}\) to give \(\pi - \frac{7\sqrt{3}}{4}\) or \(2\pi - \frac{7\sqrt{3}}{2}\) | M1 A1 | |
| Finds area of sector of circle \(\frac{2\pi}{3}\) | B1 | |
| So required area is \(\frac{8\pi}{3} - \frac{7\sqrt{3}}{2}\) | M1 A1 | (8 marks) |
## Question 9:
### Part (a):
| $2 - 2\sin\theta = 1 \rightarrow \sin\theta = 0.5\ \therefore \theta = \frac{\pi}{6}\ \left(\text{or } \frac{5\pi}{6}\right)$ | M1 A1 | |
| Points are $\left(1, \frac{\pi}{6}\right)$ and $\left(1, \frac{5\pi}{6}\right)$ | A1 | (3 marks) |
### Part (b):
| Uses Area $= \frac{1}{2}\left[\int(2-2\sin\theta)^2\,d\theta\right]$ | M1 | |
| $= \frac{1}{2}\left[\int\left(4 - 8\sin\theta + (2-2\cos 2\theta)\right)d\theta\right]$ | M1 | |
| $= \frac{1}{2}\left[(6\theta + 8\cos\theta - \sin 2\theta)\right]$ | A1 | |
| Uses limits $\frac{\pi}{2}$ and $\frac{\pi}{6}$ to give $\pi - \frac{7\sqrt{3}}{4}$ or $2\pi - \frac{7\sqrt{3}}{2}$ | M1 A1 | |
| Finds area of sector of circle $\frac{2\pi}{3}$ | B1 | |
| So required area is $\frac{8\pi}{3} - \frac{7\sqrt{3}}{2}$ | M1 A1 | (8 marks) |
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To extract mark scheme content, please share the pages that contain the actual question markings (typically the interior pages showing question numbers, model answers, mark allocations, and examiner notes).9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1f8a7998-613b-449b-9758-9bf105c56a8f-9_370_820_316_626}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the curves given by the polar equations
$$r = 1 \text { and } r = 2 - 2 \sin \theta$$
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the points where the curves intersect.
The region $S$, between the curves, for which $r < 1$ and for which $r < 2 - 2 \sin \theta$, is shown shaded in Figure 1.
\item Find, by integration, the area of the shaded region $S$, giving your answer in the form $a \pi + b \sqrt { } 3$, where $a$ and $b$ are rational numbers.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2013 Q9 [11]}}