6 The curves \(C _ { 1 }\) and \(C _ { 2 }\) have polar equations $$\begin{array} { l l } C _ { 1 } : & r = a \\ C _ { 2 } : & r = 2 a \cos 2 \theta , \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 4 } \pi \end{array}$$ where \(a\) is a positive constant. Sketch \(C _ { 1 }\) and \(C _ { 2 }\) on the same diagram. The curves \(C _ { 1 }\) and \(C _ { 2 }\) intersect at the point with polar coordinates ( \(a , \beta\) ). State the value of \(\beta\). Show that the area of the region bounded by the initial line, the arc of \(C _ { 1 }\) from \(\theta = 0\) to \(\theta = \beta\), and the arc of \(C _ { 2 }\) from \(\theta = \beta\) to \(\theta = \frac { 1 } { 4 } \pi\) is $$a ^ { 2 } \left( \frac { 1 } { 6 } \pi - \frac { 1 } { 8 } \sqrt { } 3 \right)$$

# Question 6:

| Working/Answer | Marks | Guidance |
|---|---|---|
| Sketch of $C_1$ (relevant part only). Sketch of $C_2$ (generous on tangency features). | B1, B2 | Sketches each curve on same diagram |
| $\beta = \frac{\pi}{6}$ | B1 | States value of $\beta$ |
| $\frac{1}{12}\pi a^2 + \frac{1}{2}\int_{\pi/6}^{\pi/4} 4a^2\cos^2 2\theta\,d\theta$ | B1M1 | Adds $\frac{1}{12}$ of area of circle to sector of $C_2$ from $\theta = \frac{\pi}{6}$ to $\theta = \frac{\pi}{4}$ |
| $= \frac{1}{12}\pi a^2 + a^2\int_{\pi/6}^{\pi/4}(\cos 4\theta + 1)\,d\theta$ | — | Uses double angle formula |
| $= \frac{1}{12}\pi a^2 + a^2\left[\frac{\sin 4\theta}{4} + \theta\right]_{\pi/6}^{\pi/4}$ | M1 | Integrates |
| $= a^2\left(\frac{\pi}{6} - \frac{\sqrt{3}}{8}\right)$ (AG) | A1 | Obtains printed result |

6 The curves $C _ { 1 }$ and $C _ { 2 }$ have polar equations

$$\begin{array} { l l } 
C _ { 1 } : & r = a \\
C _ { 2 } : & r = 2 a \cos 2 \theta , \text { for } 0 \leqslant \theta \leqslant \frac { 1 } { 4 } \pi
\end{array}$$

where $a$ is a positive constant. Sketch $C _ { 1 }$ and $C _ { 2 }$ on the same diagram.

The curves $C _ { 1 }$ and $C _ { 2 }$ intersect at the point with polar coordinates ( $a , \beta$ ). State the value of $\beta$.

Show that the area of the region bounded by the initial line, the arc of $C _ { 1 }$ from $\theta = 0$ to $\theta = \beta$, and the arc of $C _ { 2 }$ from $\theta = \beta$ to $\theta = \frac { 1 } { 4 } \pi$ is

$$a ^ { 2 } \left( \frac { 1 } { 6 } \pi - \frac { 1 } { 8 } \sqrt { } 3 \right)$$

\hfill \mbox{\textit{CAIE FP1 2011 Q6 [8]}}