# Question 8:

## Part (a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $2\sin\theta = 1.5 - \sin\theta \Rightarrow \theta = \ldots$ or $\sin\theta = \frac{r}{2} \Rightarrow r = 1.5 - r \Rightarrow r = \ldots$ | M1 | Equate and attempt to solve for $\theta$, or eliminate $\sin\theta$ and solve for $r$ |
| $P\!\left(1, \frac{\pi}{6}\right)$ | A1 | Correct coordinates; allow wrong way round; allow awrt $0.524$ for $\frac{\pi}{6}$ |
| $Q\!\left(1, \frac{5\pi}{6}\right)$ | A1 | Correct coordinates; allow wrong way round; allow awrt $2.62$ for $\frac{5\pi}{6}$ |

# Question (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left(\frac{1}{2}\right)\int(1.5-\sin\theta)^2\,d\theta$ or $\left(\frac{1}{2}\right)\int(2\sin\theta)^2\,d\theta$ | M1 | Correct area formula for polar curves |
| Attempts to use $\int(\sin\theta)^2\,d\theta$ or $\int(1.5-\sin\theta)^2\,d\theta$ | M1 | |
| $(1.5-\sin\theta)^2 = 2.25 - 3\sin\theta + \sin^2\theta = 2.25 - 3\sin\theta + \frac{(1-\cos 2\theta)}{2}$ | M1 | Expands (allow poor squaring e.g. $(1.5-\sin\theta)^2 = 2.25+\sin^2\theta$) **and** attempts to use $\sin^2\theta = \pm\frac{1}{2}\pm\frac{\cos 2\theta}{2}$ |
| $\frac{1}{2}\int(1.5-\sin\theta)^2\,d\theta = \frac{1}{2}\left[\frac{11}{4}\theta + 3\cos\theta - \frac{1}{4}\sin 2\theta\right]$ | M1A1 | M1: Attempt to integrate reaching form $\alpha\theta + \beta\cos\theta + \gamma\sin 2\theta$; A1: Correct integration (with or without the $\frac{1}{2}$) |
| $\frac{1}{2}\Big[\Big]_{\frac{\pi}{6}}^{\frac{5\pi}{6}} = \frac{1}{2}\left\{\left(\frac{11}{4}\cdot\frac{5\pi}{6}+3\cos\frac{5\pi}{6}-\frac{1}{4}\sin 2\cdot\frac{5\pi}{6}\right)-\left(\frac{11}{4}\cdot\frac{\pi}{6}+3\cos\frac{\pi}{6}-\frac{1}{4}\sin 2\cdot\frac{\pi}{6}\right)\right\}$ | M1 | Key step: must use $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ (or twice limits of $\frac{\pi}{6}$ and $\frac{\pi}{2}$) |
| $\frac{1}{2}\int(2\sin\theta)^2\,d\theta = \int(1-\cos 2\theta)\,d\theta = \left[\theta - \frac{1}{2}\sin 2\theta\right]_0^{\frac{\pi}{6}} = \left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right)(-0)$ | M1 | Uses limits 0 and $\frac{\pi}{6}$ for at least one segment; if integrating must obtain $p\theta + q\sin 2\theta$ with correct limits. NB can be done as $\frac{1}{2}(1)^2\left(\frac{\pi}{3}\right)-\frac{1}{2}(1)^2\sin\left(\frac{\pi}{3}\right)$ |
| $\frac{11}{12}\pi - \frac{11\sqrt{3}}{8} + 2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right) = \frac{5}{4}\pi - \frac{15}{8}\sqrt{3}$ | ddM1A1 | ddM1: Adds two areas to give numerical value for shaded area; dependent on previous 2 M marks; must be completely correct strategy attempting: $\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}(1.5-\sin\theta)^2\,d\theta$ or $2\times\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(1.5-\sin\theta)^2\,d\theta$ **+** $2\times\frac{1}{2}\int_0^{\frac{\pi}{6}}(2\sin\theta)^2\,d\theta$ or $\left(\frac{1}{2}\int_0^{\frac{\pi}{6}}(2\sin\theta)^2\,d\theta+\frac{1}{2}\int_{\frac{5\pi}{6}}^{\pi}(2\sin\theta)^2\,d\theta\right)$; A1: Correct answer (allow equivalent fractions) |

**Total: 8 marks for part (b), 11 marks total**

**Note:** Attempts using $\frac{1}{2}\int(C_1-C_2)^2\,d\theta$ e.g. $\frac{1}{2}\int(2\sin\theta-(1.5-\sin\theta))^2\,d\theta$ will probably score maximum of first 3 marks only: M1 for correct form, M1 for expanding and using $\sin^2\theta$, M1 for integrating to form $\alpha\theta+\beta\cos\theta+\gamma\sin 2\theta$ | M1 | |