| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| 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 finding intersection points by equating equations, then computing area using the polar area formula with two integrals. While it involves Further Maths content (making it inherently harder than single maths), the techniques are routine for F2 students: solve 2a sin 2θ = a, set up ½∫r² dθ for each curve over appropriate intervals, and simplify to match the given form. The algebraic manipulation is moderate but straightforward. |
| Spec | 4.09a Polar coordinates: convert to/from cartesian4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a = 2a\sin 2\theta \Rightarrow \sin 2\theta = \frac{1}{2} \Rightarrow 2\theta = \ldots\) | M1 | \(C_1 = C_2\) and attempt to solve for \(2\theta\) |
| \(\sin 2\theta = \frac{1}{2} \Rightarrow 2\theta = \frac{\pi}{6}, \frac{5\pi}{6}\) | A1 | \(2\theta = \frac{\pi}{6}\) or \(\frac{5\pi}{6}\) or both. Decimals allowed (min 3 sf) |
| \(\left(a, \frac{\pi}{12}\right), \left(a, \frac{5\pi}{12}\right)\) | A1 | Both points. Can be written \(r=a\), \(\theta = \frac{\pi}{12}, \frac{5\pi}{12}\). Decimals allowed (min 3 sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times a^2 \times \frac{\pi}{3}\) oe | B1 | Correct expression for the sector |
| \(\frac{1}{2}\int r^2 d\theta = \frac{1}{2}\int(2a\sin 2\theta)^2 d\theta\) | M1 | Use of correct formula. Limits not needed (ignore any shown) |
| \(\cos 4\theta = 1 - 2\sin^2 2\theta \Rightarrow \sin^2 2\theta = \frac{1}{2}(1-\cos 4\theta)\) | M1 | Uses \(\sin^2 2\theta = \frac{\pm 1 \pm \cos 4\theta}{2}\) |
| \(\int(1-\cos 4\theta)d\theta = \theta - \frac{1}{4}\sin 4\theta\) | A1 | Correct integration. Limits not needed (ignore any shown) |
| \(I = a^2\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^{\frac{\pi}{12}} = a^2\left\{\left(\frac{\pi}{12} - \frac{1}{4}\sin\frac{4\pi}{12}\right) - (0)\right\}\) | ddM1 | Attempt to find one or both regions either side of sector. Uses limits \(0, \frac{\pi}{12}\) and/or \(\frac{5\pi}{12}, \frac{\pi}{2}\). Both previous method marks must have been scored |
| \(R = 2I + \frac{a^2\pi}{6} = 2a^2\left(\frac{\pi}{12} - \frac{\sqrt{3}}{8}\right) + \frac{a^2\pi}{6}\) | M1 | Correct strategy for complete area (sector \(+ 2I\)). All areas must be positive |
| \(R = \frac{1}{12}a^2(4\pi - 3\sqrt{3})\) | A1 | If decimals seen anywhere (either in rt 3 or in the limits) this mark is lost |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a = 2a\sin 2\theta \Rightarrow \sin 2\theta = \frac{1}{2} \Rightarrow 2\theta = \ldots$ | M1 | $C_1 = C_2$ and attempt to solve for $2\theta$ |
| $\sin 2\theta = \frac{1}{2} \Rightarrow 2\theta = \frac{\pi}{6}, \frac{5\pi}{6}$ | A1 | $2\theta = \frac{\pi}{6}$ or $\frac{5\pi}{6}$ or both. Decimals allowed (min 3 sf) |
| $\left(a, \frac{\pi}{12}\right), \left(a, \frac{5\pi}{12}\right)$ | A1 | Both points. Can be written $r=a$, $\theta = \frac{\pi}{12}, \frac{5\pi}{12}$. Decimals allowed (min 3 sf) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times a^2 \times \frac{\pi}{3}$ oe | B1 | Correct expression for the sector |
| $\frac{1}{2}\int r^2 d\theta = \frac{1}{2}\int(2a\sin 2\theta)^2 d\theta$ | M1 | Use of correct formula. Limits not needed (ignore any shown) |
| $\cos 4\theta = 1 - 2\sin^2 2\theta \Rightarrow \sin^2 2\theta = \frac{1}{2}(1-\cos 4\theta)$ | M1 | Uses $\sin^2 2\theta = \frac{\pm 1 \pm \cos 4\theta}{2}$ |
| $\int(1-\cos 4\theta)d\theta = \theta - \frac{1}{4}\sin 4\theta$ | A1 | Correct integration. Limits not needed (ignore any shown) |
| $I = a^2\left[\theta - \frac{1}{4}\sin 4\theta\right]_0^{\frac{\pi}{12}} = a^2\left\{\left(\frac{\pi}{12} - \frac{1}{4}\sin\frac{4\pi}{12}\right) - (0)\right\}$ | ddM1 | Attempt to find one or both regions either side of sector. Uses limits $0, \frac{\pi}{12}$ and/or $\frac{5\pi}{12}, \frac{\pi}{2}$. Both previous method marks must have been scored |
| $R = 2I + \frac{a^2\pi}{6} = 2a^2\left(\frac{\pi}{12} - \frac{\sqrt{3}}{8}\right) + \frac{a^2\pi}{6}$ | M1 | Correct strategy for complete area (sector $+ 2I$). All areas must be positive |
| $R = \frac{1}{12}a^2(4\pi - 3\sqrt{3})$ | A1 | If decimals seen anywhere (either in rt 3 or in the limits) this mark is lost |
**Total: 10**9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{77d00a35-e947-41ef-8d80-5a573702ed39-14_643_1274_251_342}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the curve $C _ { 1 }$ with polar equation $r = 2 a \sin 2 \theta , 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$, and the circle $C _ { 2 }$ with polar equation $r = a , 0 \leqslant \theta \leqslant 2 \pi$, where $a$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, the polar coordinates of the points where the curve $C _ { 1 }$ meets the circle $C _ { 2 }$
The regions enclosed by the curve $C _ { 1 }$ and the circle $C _ { 2 }$ overlap and the common region $R$ is shaded in Figure 1.
\item Find the area of the shaded region $R$, giving your answer in the form $\frac { 1 } { 12 } a ^ { 2 } ( p \pi + q \sqrt { 3 } )$, where $p$ and $q$ are integers to be found.
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2014 Q9 [10]}}