| Exam Board | AQA |
|---|---|
| Module | Further Paper 2 (Further Paper 2) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.8 This is a substantial Further Maths polar coordinates question requiring: (a) converting polar line to Cartesian form to identify perpendicularity, (b) solving simultaneous polar equations and justifying which solutions are valid, and (c) computing area between a curve and line using polar integration with careful attention to bounds. The multi-step nature, need for geometric insight about which region to integrate, and technical integration make this significantly harder than average, though the individual techniques are standard for Further Maths students. |
| 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 |
| \(r = \frac{7}{4}\sec\theta\), so \(r\cos\theta = \frac{7}{4}\), and \(x = r\cos\theta\), therefore \(x = \frac{7}{4}\) | B1 | Obtains \(x = \frac{7}{4}\) |
| \(x = \frac{7}{4}\) is perpendicular to the \(x\)-axis, therefore \(L\) is perpendicular to the initial line | E1 | Must explain that a vertical line is perpendicular to the initial line (AO 2.1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{7}{4}\sec\theta = 3 + \cos\theta\) | M1 | Obtains equation in \(\theta\) or \(r\) (AO 1.1a) |
| \(\cos^2\theta + 3\cos\theta - \frac{7}{4} = 0\), so \(4\cos^2\theta + 12\cos\theta - 7 = 0\) | M1 | Rearranges and solves for \(\cos\theta\) or \(\sec\theta\) or \(r\) (AO 3.1a) |
| \(\cos\theta = -\frac{7}{2}\) rejected as \(-1 \leq \cos\theta \leq 1\) | E1 | Rejects with reason the impossible value (AO 2.2a) |
| \(\cos\theta = \frac{1}{2}\) | A1 | Obtains correct value of \(\cos\theta\) or \(r\) (AO 1.1b) |
| Points are \(\left(\frac{7}{2}, \frac{\pi}{3}\right)\) and \(\left(\frac{7}{2}, -\frac{\pi}{3}\right)\) | A1 | Obtains correct polar coordinates (AO 1.1b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Identifies required region (as shown in diagram with points \(P\left(\frac{7}{2},\frac{\pi}{3}\right)\) and \(Q\left(\frac{7}{2},-\frac{\pi}{3}\right)\)) | M1 | AO 3.1a |
| Area of triangle \(OPQ = \frac{1}{2} \times \frac{7\sqrt{3}}{2} \times \frac{7}{4} = \frac{49\sqrt{3}}{16}\), where \(PQ = \frac{7\sqrt{3}}{2}\) | B1 | Obtains correct area of triangle (AO 1.1b) |
| Area of sector \(OPQ = 2\left(\int_0^{\frac{\pi}{3}} \frac{1}{2}r^2\, d\theta\right)\) | M1 | Obtains \(k\int(3+\cos\theta)^2\,d\theta\) with or without limits (AO 3.1a) |
| \(= \int_0^{\frac{\pi}{3}}(3+\cos\theta)^2\,d\theta = \int_0^{\frac{\pi}{3}}(9 + 6\cos\theta + \cos^2\theta)\,d\theta\) | M1 | Uses \(\cos^2\theta = \frac{1}{2}\cos 2\theta + \frac{1}{2}\) (AO 3.1a) |
| \(= \int_0^{\frac{\pi}{3}}\left(9 + 6\cos\theta + \frac{1}{2}\cos 2\theta + \frac{1}{2}\right)d\theta = \left[\frac{19\theta}{2} + 6\sin\theta + \frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{3}}\) | A1F | Integrates three-part expression correctly (AO 1.1b) |
| \(= \frac{19\pi}{6} + 3\sqrt{3} + \frac{\sqrt{3}}{8} = \frac{19\pi}{6} + \frac{25\sqrt{3}}{8}\) | A1 | Obtains correct area of sector or half sector (AO 1.1b) |
| Required area \(= \frac{19\pi}{6} + \frac{25\sqrt{3}}{8} - \frac{49\sqrt{3}}{16} = \frac{19\pi}{6} + \frac{\sqrt{3}}{16}\) | A1 | Obtains exact correct answer (AO 1.1b) |
# Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = \frac{7}{4}\sec\theta$, so $r\cos\theta = \frac{7}{4}$, and $x = r\cos\theta$, therefore $x = \frac{7}{4}$ | B1 | Obtains $x = \frac{7}{4}$ |
| $x = \frac{7}{4}$ is perpendicular to the $x$-axis, therefore $L$ is perpendicular to the initial line | E1 | Must explain that a vertical line is perpendicular to the initial line (AO 2.1) |
# Question 9(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{7}{4}\sec\theta = 3 + \cos\theta$ | M1 | Obtains equation in $\theta$ or $r$ (AO 1.1a) |
| $\cos^2\theta + 3\cos\theta - \frac{7}{4} = 0$, so $4\cos^2\theta + 12\cos\theta - 7 = 0$ | M1 | Rearranges and solves for $\cos\theta$ or $\sec\theta$ or $r$ (AO 3.1a) |
| $\cos\theta = -\frac{7}{2}$ rejected as $-1 \leq \cos\theta \leq 1$ | E1 | Rejects with reason the impossible value (AO 2.2a) |
| $\cos\theta = \frac{1}{2}$ | A1 | Obtains correct value of $\cos\theta$ or $r$ (AO 1.1b) |
| Points are $\left(\frac{7}{2}, \frac{\pi}{3}\right)$ and $\left(\frac{7}{2}, -\frac{\pi}{3}\right)$ | A1 | Obtains correct polar coordinates (AO 1.1b) |
# Question 9(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Identifies required region (as shown in diagram with points $P\left(\frac{7}{2},\frac{\pi}{3}\right)$ and $Q\left(\frac{7}{2},-\frac{\pi}{3}\right)$) | M1 | AO 3.1a |
| Area of triangle $OPQ = \frac{1}{2} \times \frac{7\sqrt{3}}{2} \times \frac{7}{4} = \frac{49\sqrt{3}}{16}$, where $PQ = \frac{7\sqrt{3}}{2}$ | B1 | Obtains correct area of triangle (AO 1.1b) |
| Area of sector $OPQ = 2\left(\int_0^{\frac{\pi}{3}} \frac{1}{2}r^2\, d\theta\right)$ | M1 | Obtains $k\int(3+\cos\theta)^2\,d\theta$ with or without limits (AO 3.1a) |
| $= \int_0^{\frac{\pi}{3}}(3+\cos\theta)^2\,d\theta = \int_0^{\frac{\pi}{3}}(9 + 6\cos\theta + \cos^2\theta)\,d\theta$ | M1 | Uses $\cos^2\theta = \frac{1}{2}\cos 2\theta + \frac{1}{2}$ (AO 3.1a) |
| $= \int_0^{\frac{\pi}{3}}\left(9 + 6\cos\theta + \frac{1}{2}\cos 2\theta + \frac{1}{2}\right)d\theta = \left[\frac{19\theta}{2} + 6\sin\theta + \frac{1}{4}\sin 2\theta\right]_0^{\frac{\pi}{3}}$ | A1F | Integrates three-part expression correctly (AO 1.1b) |
| $= \frac{19\pi}{6} + 3\sqrt{3} + \frac{\sqrt{3}}{8} = \frac{19\pi}{6} + \frac{25\sqrt{3}}{8}$ | A1 | Obtains correct area of sector or half sector (AO 1.1b) |
| Required area $= \frac{19\pi}{6} + \frac{25\sqrt{3}}{8} - \frac{49\sqrt{3}}{16} = \frac{19\pi}{6} + \frac{\sqrt{3}}{16}$ | A1 | Obtains exact correct answer (AO 1.1b) |9
\begin{enumerate}[label=(\alph*)]
\item The line $L$ has polar equation
$$r = \frac { 7 } { 4 } \sec \theta \quad \left( - \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 } \right)$$
Show that $L$ is perpendicular to the initial line.\\
9
\item The curve $C$ has polar equation
$$r = 3 + \cos \theta \quad ( - \pi < \theta \leq \pi )$$
Find the polar coordinates of the points of intersection of $L$ and $C$ Fully justify your answer.\\
9
\item The region $R$ is the set of points such that\\
and
$$r > \frac { 7 } { 4 } \sec \theta$$
Find the exact area of $R$
$$r < 3 + \cos \theta$$
Find the exact area of $R$\\[0pt]
[7 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 2 2021 Q9 [14]}}