| Exam Board | Edexcel |
|---|---|
| Module | F2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 8 |
| 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 verification of intersection and area calculation using the polar area formula. Part (a) is routine substitution, and part (b) involves setting up and evaluating ∫½r²dθ for two curves with standard trigonometric integrals. The algebra is straightforward and the answer form guides the simplification. Slightly above average difficulty due to being Further Maths content, but a textbook-style exercise. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| VIIIV SIHI NI JIIIM ION OC | VIIIV SIHI NI JIHM I I ON OC | VI4V SIHI NI JIIYM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | 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 (coordinate brackets not needed) |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equate \(r\)s: \(\sqrt{3}\sin\theta = 1 + \cos\theta\) and verify \(\theta = \frac{\pi}{3}\), or solve using \(t = \tan\frac{\theta}{2}\), or: \(\frac{\sqrt{3}}{2}\sin\theta - \frac{1}{2}\cos\theta = \frac{1}{2}\), \(\sin\!\left(\theta - \frac{\pi}{6}\right) = \frac{1}{2}\), \(\theta = \frac{\pi}{3}\) | M1 | Squaring allowed as \(\theta\) is known to be between \(0\) and \(\pi\) |
| Use \(\theta = \frac{\pi}{3}\) in either equation to obtain \(r = \frac{3}{2}\) | A1 | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2}\int(\sqrt{3}\sin\theta)^2\,d\theta\), \(\frac{1}{2}\int(1+\cos\theta)^2\,d\theta\) | M1 | Correct formula used on at least one curve (\(\frac{1}{2}\) may appear later); integrals may be separate or added/subtracted |
| \(= \frac{1}{2}\int 3\sin^2\theta\,d\theta\), \(\frac{1}{2}\int(1 + 2\cos\theta + \cos^2\theta)\,d\theta\) | ||
| \(= \left(\frac{1}{2}\right)\int\frac{3}{2}(1-\cos 2\theta)\,d\theta\), \(\left(\frac{1}{2}\right)\int\left(1 + 2\cos\theta + \frac{1}{2}(1+\cos 2\theta)\right)d\theta\) | 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 |
| \(= \frac{3}{4}\!\left[\theta - \frac{1}{2}\sin 2\theta\right]_0^{(\pi/3)}\), \(\frac{1}{2}\!\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_{(\pi/3)}^{(\pi)}\) | 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 |
| (6) |
## Question 7(a):
| Working/Answer | Mark | 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 (coordinate brackets not needed) |
| | **(2)** | |
**Alternative:**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equate $r$s: $\sqrt{3}\sin\theta = 1 + \cos\theta$ and verify $\theta = \frac{\pi}{3}$, or solve using $t = \tan\frac{\theta}{2}$, or: $\frac{\sqrt{3}}{2}\sin\theta - \frac{1}{2}\cos\theta = \frac{1}{2}$, $\sin\!\left(\theta - \frac{\pi}{6}\right) = \frac{1}{2}$, $\theta = \frac{\pi}{3}$ | M1 | Squaring allowed as $\theta$ is known to be between $0$ and $\pi$ |
| Use $\theta = \frac{\pi}{3}$ in either equation to obtain $r = \frac{3}{2}$ | A1 | |
| | **(2)** | |
## Question 7(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\int(\sqrt{3}\sin\theta)^2\,d\theta$, $\frac{1}{2}\int(1+\cos\theta)^2\,d\theta$ | M1 | Correct formula used on at least one curve ($\frac{1}{2}$ may appear later); integrals may be separate or added/subtracted |
| $= \frac{1}{2}\int 3\sin^2\theta\,d\theta$, $\frac{1}{2}\int(1 + 2\cos\theta + \cos^2\theta)\,d\theta$ | | |
| $= \left(\frac{1}{2}\right)\int\frac{3}{2}(1-\cos 2\theta)\,d\theta$, $\left(\frac{1}{2}\right)\int\left(1 + 2\cos\theta + \frac{1}{2}(1+\cos 2\theta)\right)d\theta$ | 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 |
| $= \frac{3}{4}\!\left[\theta - \frac{1}{2}\sin 2\theta\right]_0^{(\pi/3)}$, $\frac{1}{2}\!\left[\frac{3}{2}\theta + 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_{(\pi/3)}^{(\pi)}$ | 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 |
| | **(6)** | |
---7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b197811e-1df5-4937-b0d8-f98f82412c76-24_480_926_217_511}
\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.\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIIIV SIHI NI JIIIM ION OC & VIIIV SIHI NI JIHM I I ON OC & VI4V SIHI NI JIIYM IONOO \\
\hline
\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel F2 2018 Q7 [8]}}