| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area between two polar curves |
| Difficulty | Standard +0.8 This is a Further Maths polar coordinates question requiring students to find the area between two curves. Part (a) is routine verification, but part (b) requires setting up the correct integral with two different curves over different angular ranges, recognizing where the curves intersect, and performing non-trivial integration involving trigonometric identities. The algebraic manipulation to reach the specific form is also demanding. This is moderately challenging for Further Maths students. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\theta = \frac{\pi}{3} \Rightarrow r = \sqrt{3}\sin\!\left(\frac{\pi}{3}\right) = \frac{3}{2}\) | M1 | Attempt to verify coordinates in at least one of the polar equations |
| \(\theta = \frac{\pi}{3} \Rightarrow r = 1 + \cos\!\left(\frac{\pi}{3}\right) = \frac{3}{2}\) | A1 | Coordinates verified in both curves |
| Answer | Marks | Guidance |
|---|---|---|
| Equate \(r\)s: \(\sqrt{3}\sin\theta = 1 + \cos\theta\) and verify \(\theta = \frac{\pi}{3}\) is a solution; or solve using \(t = \tan\frac{\theta}{2}\); or \(\frac{\sqrt{3}}{2}\sin\theta - \frac{1}{2}\cos\theta = \frac{1}{2} \Rightarrow \sin\!\left(\theta - \frac{\pi}{6}\right) = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}\) | M1 | — |
| Use \(\theta = \frac{\pi}{3}\) in either equation to obtain \(r = \frac{3}{2}\) | A1 | — |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}\int(\sqrt{3}\sin\theta)^2\,d\theta,\quad \frac{1}{2}\int(1+\cos\theta)^2\,d\theta\) | M1 | Correct formula used on at least one curve (\(\frac{1}{2}\) may appear later); integrals may be separate or added/subtracted |
| \(= \frac{1}{2}\int 3\sin^2\theta\,d\theta,\quad \frac{1}{2}\int(1 + 2\cos\theta + \cos^2\theta)\,d\theta\) | — | — |
| \(= \left(\frac{1}{2}\right)\int\frac{3}{2}(1-\cos 2\theta)\,d\theta,\quad \left(\frac{1}{2}\right)\int\!\left(1 + 2\cos\theta + \frac{1}{2}(1+\cos 2\theta)\right)d\theta\) | M1 | Attempt to use \(\sin^2\theta\) or \(\cos^2\theta = \pm\frac{1}{2} \pm \frac{1}{2}\cos 2\theta\) on either integral; not dependent, \(\frac{1}{2}\) may be missing |
| \(= \frac{3}{4}\!\left[\theta - \frac{1}{2}\sin 2\theta\right]_0^{\pi/3},\quad \frac{1}{2}\!\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_{\pi/3}^{\pi}\) | A1, A1 | Correct integration (ignore limits); A1A1 or A1A0 |
| \(R = \frac{3}{4}\!\left[\frac{\pi}{3} - \frac{\sqrt{3}}{4} - 0\right] + \frac{1}{2}\!\left[\frac{3\pi}{2} - \left(\frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8}\right)\right]\) | ddM1 | Correct use of limits for both integrals; integrals must be added; dep on both previous M marks |
| \(= \frac{3}{4}(\pi - \sqrt{3})\) | A1 | Cao; no equivalents allowed |
## Question 7:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\theta = \frac{\pi}{3} \Rightarrow r = \sqrt{3}\sin\!\left(\frac{\pi}{3}\right) = \frac{3}{2}$ | M1 | Attempt to verify coordinates in at least one of the polar equations |
| $\theta = \frac{\pi}{3} \Rightarrow r = 1 + \cos\!\left(\frac{\pi}{3}\right) = \frac{3}{2}$ | A1 | Coordinates verified in both curves |
**Alternative:**
| Equate $r$s: $\sqrt{3}\sin\theta = 1 + \cos\theta$ and verify $\theta = \frac{\pi}{3}$ is a solution; or solve using $t = \tan\frac{\theta}{2}$; or $\frac{\sqrt{3}}{2}\sin\theta - \frac{1}{2}\cos\theta = \frac{1}{2} \Rightarrow \sin\!\left(\theta - \frac{\pi}{6}\right) = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3}$ | M1 | — |
| Use $\theta = \frac{\pi}{3}$ in either equation to obtain $r = \frac{3}{2}$ | A1 | — |
**Total: (2)**
---
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}\int(\sqrt{3}\sin\theta)^2\,d\theta,\quad \frac{1}{2}\int(1+\cos\theta)^2\,d\theta$ | M1 | Correct formula used on at least one curve ($\frac{1}{2}$ may appear later); integrals may be separate or added/subtracted |
| $= \frac{1}{2}\int 3\sin^2\theta\,d\theta,\quad \frac{1}{2}\int(1 + 2\cos\theta + \cos^2\theta)\,d\theta$ | — | — |
| $= \left(\frac{1}{2}\right)\int\frac{3}{2}(1-\cos 2\theta)\,d\theta,\quad \left(\frac{1}{2}\right)\int\!\left(1 + 2\cos\theta + \frac{1}{2}(1+\cos 2\theta)\right)d\theta$ | M1 | Attempt to use $\sin^2\theta$ or $\cos^2\theta = \pm\frac{1}{2} \pm \frac{1}{2}\cos 2\theta$ on either integral; not dependent, $\frac{1}{2}$ may be missing |
| $= \frac{3}{4}\!\left[\theta - \frac{1}{2}\sin 2\theta\right]_0^{\pi/3},\quad \frac{1}{2}\!\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_{\pi/3}^{\pi}$ | A1, A1 | Correct integration (ignore limits); A1A1 or A1A0 |
| $R = \frac{3}{4}\!\left[\frac{\pi}{3} - \frac{\sqrt{3}}{4} - 0\right] + \frac{1}{2}\!\left[\frac{3\pi}{2} - \left(\frac{\pi}{2} + \sqrt{3} + \frac{\sqrt{3}}{8}\right)\right]$ | ddM1 | Correct use of limits for both integrals; integrals must be added; dep on both previous M marks |
| $= \frac{3}{4}(\pi - \sqrt{3})$ | A1 | Cao; no equivalents allowed |
**Total: (6)**
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ee9a9df3-f7a4-41d0-bf8b-e44340c401d6-13_458_933_251_504}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the two curves given by the polar equations
$$\begin{array} { l l }
r = \sqrt { 3 } \sin \theta , & 0 \leqslant \theta \leqslant \pi \\
r = 1 + \cos \theta , & 0 \leqslant \theta \leqslant \pi
\end{array}$$
\begin{enumerate}[label=(\alph*)]
\item Verify that the curves intersect at the point $P$ with polar coordinates $\left( \frac { 3 } { 2 } , \frac { \pi } { 3 } \right)$.
The region $R$, bounded by the two curves, is shown shaded in Figure 1.
\item Use calculus to find the exact area of $R$, giving your answer in the form $a ( \pi - \sqrt { 3 } )$, where $a$ is a constant to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2015 Q7 [8]}}