FP2 June 2009 Q4
4.

Figure 1 shows a sketch of the curve with polar equation \[r = a + 3\cos\theta, \quad a > 0, \quad 0 \leqslant \theta < 2\pi\]
The area enclosed by the curve is \(\dfrac{107}{2}\pi\).
Find the value of \(a\). (8)
| Scheme | Marks |
|---|---|
| \(A = \dfrac{1}{2}\displaystyle\int_0^{2\pi}(a + 3\cos\theta)^2\,\mathrm{d}\theta\) Applies \(\frac{1}{2}\int_0^{2\pi} r^2(\mathrm{d}\theta)\) with correct limits. Ignore \(\mathrm{d}\theta\). | B1 |
| \((a + 3\cos\theta)^2 = a^2 + 6a\cos\theta + 9\cos^2\theta\) | |
| \(= \underline{a^2 + 6a\cos\theta + 9\left(\dfrac{1 + \cos 2\theta}{2}\right)}\) \(\cos^2\theta = \frac{\pm 1 \pm \cos 2\theta}{2}\) Correct underlined expression. | M1 A1 |
| \(A = \dfrac{1}{2}\displaystyle\int_0^{2\pi}\left(a^2 + 6a\cos\theta + \frac{9}{2} + \frac{9}{2}\cos 2\theta\right)\mathrm{d}\theta\) | |
| \(= \left(\dfrac{1}{2}\right)\left[a^2\theta + 6a\sin\theta + \dfrac{9}{2}\theta + \dfrac{9}{4}\sin 2\theta\right]_0^{2\pi}\) Integrated expression with at least 3 out of 4 terms of the form \(\pm A\theta \pm B\sin\theta \pm C\theta \pm D\sin 2\theta\). Ignore the \(\frac{1}{2}\). Ignore limits. \(a^2\theta + 6a\sin\theta +\) correct ft integration. Ignore the \(\frac{1}{2}\). Ignore limits. | M1* A1 ft |
| \(= \dfrac{1}{2}\left[(2\pi a^2 + 0 + 9\pi + 0) - (0)\right]\) | |
| \(= \pi a^2 + \dfrac{9\pi}{2}\) \(\pi a^2 + \frac{9\pi}{2}\) | A1 |
| Hence, \(\pi a^2 + \dfrac{9\pi}{2} = \dfrac{107}{2}\pi\) Integrated expression equal to \(\frac{107}{2}\pi\). | dM1* |
| \(a^2 + \dfrac{9}{2} = \dfrac{107}{2}\) \(a^2 = 49\) | |
| As \(a > 0\), \(a = 7\) \(a = 7\) | A1 cso |
| (8 marks) |
Notes
Some candidates may achieve \(a = 7\) from incorrect working. Such candidates will not get full marks