| Exam Board | AQA |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2008 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area of region with line boundary |
| Difficulty | Challenging +1.2 This is a standard Further Maths polar coordinates area question requiring the formula ½∫r²dθ and integration of (1+tan θ)². Part (a) is routine integration with a given answer to verify. Part (b) requires finding the triangle area and subtracting, which is straightforward geometry. The integration involves standard techniques (expanding, using sec²θ) but is more involved than typical A-level Core questions, placing it moderately above average difficulty. |
| Spec | 4.09b Sketch polar curves: r = f(theta)4.09c Area enclosed: by polar curve |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\text{Area} = \frac{1}{2}\int (1+\tan\theta)^2\,d\theta\) | M1 | |
| \(\ldots = \frac{1}{2}\int (1+2\tan\theta + \tan^2\theta)\,d\theta\) | B1 | |
| \(= \frac{1}{2}\int (\sec^2\theta + 2\tan\theta)\,d\theta\) | M1 | |
| \(= \frac{1}{2}[\tan\theta + 2\ln(\sec\theta)]_0^{\pi/3}\) | A1/, B1/ | |
| \(= \frac{1}{2}[(\sqrt{3} + 2\ln 2)-0] = \frac{\sqrt{3}}{2} + \ln 2\) | A1 | Total: 6 |
| (b) \(OP = 1\); \(OQ = 1 + \tan\frac{\pi}{3}\) | B1 | |
| Shaded area \(= \text{'answer (a)'} - \frac{1}{2}OP \times OQ \times \sin(\frac{\pi}{3})\) | M1 | |
| \(= \frac{\sqrt{3}}{2} + \ln 2 - \frac{\sqrt{3}}{4}(1+\sqrt{3})\) | A1 | Total: 3 |
| \(= \frac{\sqrt{3}}{4} + \ln 2 - \frac{3}{4}\) |
**(a)** $\text{Area} = \frac{1}{2}\int (1+\tan\theta)^2\,d\theta$ | M1 | | Use of $\frac{1}{2}\int r^2\,d\theta$
$\ldots = \frac{1}{2}\int (1+2\tan\theta + \tan^2\theta)\,d\theta$ | B1 | | Correct expansion of $(1+\tan\theta)^2$
$= \frac{1}{2}\int (\sec^2\theta + 2\tan\theta)\,d\theta$ | M1 | | $1+\tan^2\theta = \sec^2\theta$ used
$= \frac{1}{2}[\tan\theta + 2\ln(\sec\theta)]_0^{\pi/3}$ | A1/, B1/ | | Integrating $\sec^2\theta$ correctly; Integrating $\tan\theta$ correctly
$= \frac{1}{2}[(\sqrt{3} + 2\ln 2)-0] = \frac{\sqrt{3}}{2} + \ln 2$ | A1 | Total: 6 | Completion. AG CSO be convinced
**(b)** $OP = 1$; $OQ = 1 + \tan\frac{\pi}{3}$ | B1 | | Both needed. Accept 2.73 for OQ
Shaded area $= \text{'answer (a)'} - \frac{1}{2}OP \times OQ \times \sin(\frac{\pi}{3})$ | M1 | |
$= \frac{\sqrt{3}}{2} + \ln 2 - \frac{\sqrt{3}}{4}(1+\sqrt{3})$ | A1 | Total: 3 | ACF. Condone 0.376... if exact 'value' for area of triangle seen
$= \frac{\sqrt{3}}{4} + \ln 2 - \frac{3}{4}$ | | |2 The diagram shows a sketch of part of the curve $C$ whose polar equation is $r = 1 + \tan \theta$. The point $O$ is the pole.\\
\includegraphics[max width=\textwidth, alt={}, center]{0c177d90-02ae-4e91-bc9d-d0c7051799b8-3_561_629_406_772}
The points $P$ and $Q$ on the curve are given by $\theta = 0$ and $\theta = \frac { \pi } { 3 }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that the area of the region bounded by the curve $C$ and the lines $O P$ and $O Q$ is
$$\frac { 1 } { 2 } \sqrt { 3 } + \ln 2$$
(6 marks)
\item Hence find the area of the shaded region bounded by the line $P Q$ and the arc $P Q$ of $C$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP3 2008 Q2 [9]}}