| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2011 |
| Session | June |
| 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 area question requiring students to find where r=5/2, set up the integral ½∫r²dθ, and subtract the rectangular region. While it involves multiple steps and careful geometric reasoning about the boundary, the techniques are routine for FP2 students who have practiced polar coordinates. The integration itself is straightforward (cos²θ), making this moderately above average difficulty but not requiring novel insight. |
| Spec | 4.09c Area enclosed: by polar curve |
| Answer | Marks |
|---|---|
| \(2 + \cos\theta = \frac{5}{2} \Rightarrow \theta = \frac{\pi}{3}\) | B1 |
| \(\frac{1}{2}\int(2+\cos\theta)^2 d\theta = \frac{1}{2}\int(4 + 4\cos\theta + \cos^2\theta) d\theta\) | M1 |
| \(= \frac{1}{2}\left[4\theta + 4\sin\theta + \frac{\sin 2\theta}{4} + \frac{\theta}{2}\right]\) | M1 A1 |
| Substituting limits: \(\left(\frac{1}{2}\left[\frac{9\pi}{6} + 4\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{8} - \frac{1}{2}\left(\frac{3\pi}{2} + \frac{17\sqrt{3}}{8}\right)\right]\right)\) | M1 |
| Area of triangle \(= \frac{1}{2}(r\cos\theta)(r\sin\theta) = \frac{1}{2} \times \frac{5}{4} \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{32}\) | M1 A1 |
| Area of \(R = \frac{3\pi}{4} + \frac{17\sqrt{3}}{16} - \frac{25\sqrt{3}}{32} = \frac{3\pi}{4} + \frac{9\sqrt{3}}{32}\) | M1 A1 |
| (9) 9 | 1st M1 for use of \(\frac{1}{2}\int r^2 d\theta\) and correct attempt to expand. 2nd M1 for use of double angle formula - \(\sin 2\theta\) required in square brackets. 3rd M1 for substituting their limits. 4th M1 for use of \(\frac{1}{2}\) base \(\times\) height. 5th M1 area of sector - area of triangle. Please note there are no follow through marks on accuracy. |
| $2 + \cos\theta = \frac{5}{2} \Rightarrow \theta = \frac{\pi}{3}$ | B1 | |
| $\frac{1}{2}\int(2+\cos\theta)^2 d\theta = \frac{1}{2}\int(4 + 4\cos\theta + \cos^2\theta) d\theta$ | M1 | |
| $= \frac{1}{2}\left[4\theta + 4\sin\theta + \frac{\sin 2\theta}{4} + \frac{\theta}{2}\right]$ | M1 A1 | |
| Substituting limits: $\left(\frac{1}{2}\left[\frac{9\pi}{6} + 4\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{8} - \frac{1}{2}\left(\frac{3\pi}{2} + \frac{17\sqrt{3}}{8}\right)\right]\right)$ | M1 | |
| Area of triangle $= \frac{1}{2}(r\cos\theta)(r\sin\theta) = \frac{1}{2} \times \frac{5}{4} \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{32}$ | M1 A1 | |
| Area of $R = \frac{3\pi}{4} + \frac{17\sqrt{3}}{16} - \frac{25\sqrt{3}}{32} = \frac{3\pi}{4} + \frac{9\sqrt{3}}{32}$ | M1 A1 | |
| | (9) 9 | 1st M1 for use of $\frac{1}{2}\int r^2 d\theta$ and correct attempt to expand. 2nd M1 for use of double angle formula - $\sin 2\theta$ required in square brackets. 3rd M1 for substituting their limits. 4th M1 for use of $\frac{1}{2}$ base $\times$ height. 5th M1 area of sector - area of triangle. Please note there are no follow through marks on accuracy. |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{893efbc9-8321-469f-bd5e-89f9d5827737-09_650_937_269_482}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The curve $C$ shown in Figure 1 has polar equation
$$r = 2 + \cos \theta , \quad 0 \leqslant \theta \leqslant \frac { \pi } { 2 }$$
At the point $A$ on $C$, the value of $r$ is $\frac { 5 } { 2 }$.\\
The point $N$ lies on the initial line and $A N$ is perpendicular to the initial line.\\
The finite region $R$, shown shaded in Figure 1, is bounded by the curve $C$, the initial line and the line $A N$.
Find the exact area of the shaded region $R$.
\hfill \mbox{\textit{Edexcel FP2 2011 Q6 [9]}}