| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area enclosed by polar curve |
| Difficulty | Standard +0.8 This is a Further Maths polar coordinates question requiring (i) finding when r=0 to determine tangent directions at the pole, and (ii) setting up and evaluating the polar area integral ½∫r²dθ with correct limits for one loop of a rose curve. While the integration itself is standard, identifying the correct limits and handling the four-petaled rose curve r=2cos(2θ) requires solid understanding of polar curves beyond standard A-level, 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 |
|---|---|---|
| (i) \(r = 0 \Rightarrow \cos 2\theta = 0 \Rightarrow \theta = \pm\frac{1}{4}\pi, \pm\frac{3}{4}\pi\) | M1 | For equating \(r\) to zero and solving for \(\theta\) |
| \(-\) | A1 | For any two correct values |
| \(-\) | A1 | For all four correct values and no others |
| \(-\) | 3 | |
| (ii) Area is \(\frac{1}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 4\cos^2 2\theta \, d\theta\) | M1 | For us of correct formula \(\frac{1}{2}\int r^2 \, d\theta\) |
| \(-\) | B1√ | For correct limits from (i) |
| i.e. \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 + \cos 4\theta \, d\theta = \left[\theta + \frac{1}{4}\sin 4\theta\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{1}{2}\pi\) | M1 | For using double-angle formula |
| \(-\) | A1 | For \(\theta + \frac{1}{4}\sin 4\theta\) correct |
| \(-\) | A1 | For correct (exact) answer |
| \(-\) | 5 |
(i) $r = 0 \Rightarrow \cos 2\theta = 0 \Rightarrow \theta = \pm\frac{1}{4}\pi, \pm\frac{3}{4}\pi$ | M1 | For equating $r$ to zero and solving for $\theta$ |
$-$ | A1 | For any two correct values |
$-$ | A1 | For all four correct values and no others |
$-$ | **3** | |
(ii) Area is $\frac{1}{2}\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 4\cos^2 2\theta \, d\theta$ | M1 | For us of correct formula $\frac{1}{2}\int r^2 \, d\theta$ |
$-$ | B1√ | For correct limits from (i) |
i.e. $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 1 + \cos 4\theta \, d\theta = \left[\theta + \frac{1}{4}\sin 4\theta\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \frac{1}{2}\pi$ | M1 | For using double-angle formula |
$-$ | A1 | For $\theta + \frac{1}{4}\sin 4\theta$ correct |
$-$ | A1 | For correct (exact) answer |
$-$ | **5** | |
---4 The equation of a curve, in polar coordinates, is
$$r = 2 \cos 2 \theta \quad ( - \pi < \theta \leqslant \pi ) .$$
(i) Find the values of $\theta$ which give the directions of the tangents at the pole.
One loop of the curve is shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{e4e1c424-8dd5-4d18-9950-e902de0301b0-3_362_720_653_708}\\
(ii) Find the exact value of the area of the region enclosed by the loop.
\hfill \mbox{\textit{OCR FP2 Q4 [8]}}