| Exam Board | CAIE |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Area between two polar curves |
| Difficulty | Challenging +1.2 This is a multi-part polar coordinates question requiring finding intersections, sketching curves, and computing an area between curves. While it involves several steps and the area calculation requires careful setup of integration limits, the techniques are standard for Further Maths: solving trigonometric equations, recognizing standard polar curves (circle and cardioid), and applying the polar area formula. The final part guides students to a specific answer, reducing problem-solving demand. Moderately above average difficulty due to the multi-step nature and polar integration, but well within expected FM1 scope. |
| 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 |
|---|---|---|
| (i) \(1 + \sin\theta = 3\sin\theta \Rightarrow \sin\theta = \frac{1}{2}\) | M1 | |
| \(\left(\frac{3}{2}, \frac{\pi}{6}\right)\) and \(\left(\frac{3}{2}, \frac{5\pi}{6}\right)\) | A1 | (both) |
| (ii) | B1 | circle |
| B1 | cardioid behaviour at origin | |
| B1 | cardioid closed and symmetry | [3] |
| (iii) Subtract integrands | M1 | |
| \(2 \times \frac{1}{2}\int_{\pi/6}^{5\pi/6}(3 - 4\cos 2\theta - 2\sin\theta)d\theta\) | M1 | |
| \(= [3\theta - 2\sin 2\theta + 2\cos\theta]_{\pi/6}^{5\pi/6}\) | M1A1 | |
| \(= \pi\) | A1 |
| Answer | Marks |
|---|---|
| \(2 \times \frac{1}{2}\int_{\pi/6}^{5\pi/6} 9\sin^2\theta d\theta = \frac{9}{2}\left[\theta - \frac{1}{2}\sin 2\theta\right]_{\pi/6}^{5\pi/6}\) | M1 |
| \(= \frac{9}{2}\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2 \times \frac{1}{2}\int_{\pi/6}^{5\pi/6}[1 + 2\sin\theta + \frac{1}{2}(1-\cos 2\theta)]d\theta\) | M1 | |
| \(= \left[\frac{3\theta}{2} - 2\cos\theta - \frac{1}{4}\sin 2\theta\right]_{\pi/6}^{5\pi/6}\) | M1 | |
| \(= \left(\frac{\pi}{2} + \frac{9\sqrt{3}}{8}\right)\) | A1 | if not earned earlier |
| \(\text{Subtraction}\) | M1 | |
| \(\text{Required area} = \pi\) | A1 |
**(i)** $1 + \sin\theta = 3\sin\theta \Rightarrow \sin\theta = \frac{1}{2}$ | M1
$\left(\frac{3}{2}, \frac{\pi}{6}\right)$ and $\left(\frac{3}{2}, \frac{5\pi}{6}\right)$ | A1 | (both) | [2]
**(ii)** | B1 | circle
| | B1 | cardioid behaviour at origin
| | B1 | cardioid closed and symmetry | [3]
**(iii)** Subtract integrands | M1
$2 \times \frac{1}{2}\int_{\pi/6}^{5\pi/6}(3 - 4\cos 2\theta - 2\sin\theta)d\theta$ | M1
$= [3\theta - 2\sin 2\theta + 2\cos\theta]_{\pi/6}^{5\pi/6}$ | M1A1
$= \pi$ | A1 | | [5]
**Alternative:**
**Area inside $C_1$:**
$2 \times \frac{1}{2}\int_{\pi/6}^{5\pi/6} 9\sin^2\theta d\theta = \frac{9}{2}\left[\theta - \frac{1}{2}\sin 2\theta\right]_{\pi/6}^{5\pi/6}$ | M1
$= \frac{9}{2}\left(\frac{\pi}{3} + \frac{\sqrt{3}}{4}\right)$ | A1
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## Question 8 (continued)
**Area inside $C_2$:**
$2 \times \frac{1}{2}\int_{\pi/6}^{5\pi/6}[1 + 2\sin\theta + \frac{1}{2}(1-\cos 2\theta)]d\theta$ | M1
$= \left[\frac{3\theta}{2} - 2\cos\theta - \frac{1}{4}\sin 2\theta\right]_{\pi/6}^{5\pi/6}$ | M1
$= \left(\frac{\pi}{2} + \frac{9\sqrt{3}}{8}\right)$ | A1 | if not earned earlier
$\text{Subtraction}$ | M1
$\text{Required area} = \pi$ | A1 | | [5]
---8 The curves $C _ { 1 }$ and $C _ { 2 }$ have polar equations given by
$$\begin{array} { l l r }
C _ { 1 } : & r = 3 \sin \theta , & 0 \leqslant \theta < \pi , \\
C _ { 2 } : & r = 1 + \sin \theta , & - \pi < \theta \leqslant \pi .
\end{array}$$
(i) Find the polar coordinates of the points, other than the pole, where $C _ { 1 }$ and $C _ { 2 }$ meet.\\
(ii) In a single diagram, draw sketch graphs of $C _ { 1 }$ and $C _ { 2 }$.\\
(iii) Show that the area of the region which is inside $C _ { 1 }$ but outside $C _ { 2 }$ is $\pi$.
\hfill \mbox{\textit{CAIE FP1 2010 Q8 [10]}}