# Question 7:

## Part (a)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Area} = \frac{1}{2}\int_{-\pi/2}^{\pi/2} r^2 \, d\theta$ | M1 | Correct formula with correct limits |
| $= \frac{1}{2}\int_{-\pi/2}^{\pi/2} (1+\cos 2\theta)^2 \, d\theta$ | | |
| $= \frac{1}{2}\int_{-\pi/2}^{\pi/2} (1 + 2\cos 2\theta + \cos^2 2\theta) \, d\theta$ | M1 | Expanding and using $\cos^2 2\theta = \frac{1}{2}(1+\cos 4\theta)$ |
| $= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \left(\frac{3}{2} + 2\cos 2\theta + \frac{1}{2}\cos 4\theta\right) d\theta$ | A1 | Correct integrand |
| $= \frac{1}{2}\left[\frac{3\theta}{2} + \sin 2\theta + \frac{\sin 4\theta}{8}\right]_{-\pi/2}^{\pi/2}$ | A1 | Correct integration |
| $= \frac{3\pi}{4}$ | A1 | Correct answer |

## Part (b)(i)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Set $1+\cos 2\theta = 1+\sin\theta$ | M1 | Equating the two curves |
| $\cos 2\theta = \sin\theta$ | | |
| $1 - 2\sin^2\theta = \sin\theta$ | M1 | Using $\cos 2\theta = 1-2\sin^2\theta$ |
| $2\sin^2\theta + \sin\theta - 1 = 0$ | | |
| $(2\sin\theta - 1)(\sin\theta + 1) = 0$ | M1 | Factorising |
| $\sin\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{6}$ | A1 | $\theta = \frac{\pi}{6}$, $r = 1 + \sin\frac{\pi}{6} = \frac{3}{2}$ |
| $A = \left(\frac{3}{2}, \frac{\pi}{6}\right)$ | A1 | Both coordinates correct |

## Part (b)(ii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| Line through $A$ parallel to initial line has equation $r\sin\theta = \frac{3}{2}\sin\frac{\pi}{6} = \frac{3}{4}$ | M1 | Correct equation of line |
| So $r\sin\theta = \frac{3}{4}$ | A1 | |
| On $C_1$: $r = 1+\cos 2\theta$ so $(1+\cos 2\theta)\sin\theta = \frac{3}{4}$ | M1 | Substituting into $C_1$ |
| $(1+1-2\sin^2\theta)\sin\theta = \frac{3}{4}$ | M1 | Using double angle formula |
| $2\sin\theta - 2\sin^3\theta = \frac{3}{4}$ | | |
| $8\sin^3\theta - 8\sin\theta + 3 = 0$ | M1 | Forming cubic |
| $(2\sin\theta - 1)(4\sin^2\theta + 2\sin\theta - 3) = 0$ | M1 | Factorising, knowing $\sin\theta = \frac{1}{2}$ is a factor |
| $\sin\theta = \frac{-2\pm\sqrt{4+48}}{8} = \frac{-1\pm\sqrt{13}}{4}$ | A1 | Correct solution for $B$ |
| $r = \frac{3}{4} \div \sin\theta = \frac{3}{4} \cdot \frac{4}{\sqrt{13}-1} = \frac{3}{\sqrt{13}-1} = \frac{3(\sqrt{13}+1)}{12} = \frac{1}{4}(\sqrt{13}+1)$ | A1 | Completion of proof (shown) |

## Part (b)(iii)

| Answer/Working | Marks | Guidance |
|---|---|---|
| $AB^2 = OA^2 + OB^2 - 2\cdot OA\cdot OB\cos(\angle AOB)$ or use Cartesian coordinates | M1 | Correct method |
| $OA = \frac{3}{2}$, $OB = \frac{1}{4}(\sqrt{13}+1)$ | | |
| Difference in $\theta$ values found | M1 | Correct calculation |
| $AB \approx 1.14$ | A1 | Correct answer to 3 s.f. |

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