# Question 10:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\sin\theta\cos\!\left(\theta - \tfrac{1}{4}\pi\right) = \sin\!\left(2\theta - \tfrac{1}{4}\pi\right) + \sin\!\left(\tfrac{1}{4}\pi\right)$ | M1 | |
| $= \cos\!\left(\tfrac{1}{2}\pi - 2\theta + \tfrac{1}{4}\pi\right) + \tfrac{1}{\sqrt{2}}$ | M1A1 | |
| $= \cos\!\left(2\theta - \tfrac{3}{4}\pi\right) + \tfrac{1}{\sqrt{2}}$ (AG) | | 3 marks |
| Closed loop through origin, in correct position | B1 | |
| For line of symmetry $2\theta - \tfrac{3}{4}\pi = 0 \Rightarrow \theta = \tfrac{3}{8}\pi$ | B1B1 | 3 marks |
| $A = \tfrac{1}{2}\int_0^{\frac{3}{4}\pi}\left\{\cos^2\!\left(2\theta - \tfrac{3}{4}\pi\right) + \sqrt{2}\cos\!\left(2\theta - \tfrac{3}{4}\pi\right) + \tfrac{1}{2}\right\}d\theta$ | M1A1 | |
| $= \tfrac{1}{2}\int_0^{\frac{3}{4}\pi}\left\{\tfrac{1}{2}\cos\!\left(4\theta - \tfrac{3}{2}\pi\right) + \sqrt{2}\cos\!\left(2\theta - \tfrac{3}{4}\pi\right) + 1\right\}d\theta$ | dM1A1 | |
| $= \left[\tfrac{1}{16}\sin\!\left(4\theta - \tfrac{3}{2}\pi\right) + \tfrac{1}{2\sqrt{2}}\sin\!\left(2\theta - \tfrac{3}{4}\pi\right) + \tfrac{\theta}{2}\right]_0^{\frac{3\pi}{4}}$ | dM1 | |
| $= \left[-\tfrac{1}{16} + \tfrac{1}{4} + \tfrac{3}{8}\pi\right] - \left[\tfrac{1}{16} - \tfrac{1}{4}\right]$ | | |
| $= \tfrac{3}{8}(\pi + 1)$ (AG) | A1 | 6 marks, **[12]** |
| | | N.B. Method marks dependent in final part. If $\frac{1}{2}$ factor missing throughout – award M's (Max 3). If $2 \times \frac{1}{2}\int_0^{\frac{3}{8}\pi} r^2\,d\theta$, penultimate line is $= \left[\frac{3}{8}\pi\right] - \left[\frac{1}{8} - \frac{1}{2}\right]$ |

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# Question 11E:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{PQ} = \begin{pmatrix}-3+\mu-3\lambda \\ -6\mu-2\lambda \\ 12-2\mu+\lambda\end{pmatrix}$ | M1A1 | |
| $\begin{vmatrix} i & j & k \\ 3 & 2 & -1 \\ 1 & -6 & -2 \end{vmatrix} = \begin{pmatrix}-10\\5\\-20\end{pmatrix} \sim \begin{pmatrix}2\\-1\\4\end{pmatrix}$ | M1 | Direction of common perpendicular (or uses scalar product of $\mathbf{PQ}$ with direction vector of each line) |
| $-3 + \mu - 3\lambda = 12\mu + 4\lambda$ | A1 | |
| $-24\mu - 8\lambda = -12 + 2\mu - \lambda$ | A1 | |
| $\mu = 1,\ \lambda = -2$ | M1A1 | |
| $\mathbf{p} = \begin{pmatrix}4+3\lambda\\7+2\lambda\\-1-\lambda\end{pmatrix} = \begin{pmatrix}-2\\3\\1\end{pmatrix}$, $\mathbf{q} = \begin{pmatrix}1+\mu\\7-6\mu\\11-2\mu\end{pmatrix} = \begin{pmatrix}2\\1\\9\end{pmatrix}$ | A1 | 8 marks |
| $\mathbf{AB} \times \mathbf{PQ} = \begin{vmatrix} i & j & k \\ -1 & 0 & 4 \\ 2 & -1 & 4 \end{vmatrix} = \begin{pmatrix}4\\12\\1\end{pmatrix}$ | M1A1 | |
| $\mathbf{PA}\begin{pmatrix}6\\4\\-2\end{pmatrix}$, $\mathbf{QA}\begin{pmatrix}2\\6\\-10\end{pmatrix}$, $\mathbf{PB}\begin{pmatrix}3\\4\\10\end{pmatrix}$, $\mathbf{QB}\begin{pmatrix}-1\\6\\2\end{pmatrix}$ | B1 | |
| $\dfrac{\begin{pmatrix}6\\4\\-2\end{pmatrix} \cdot \begin{pmatrix}4\\12\\1\end{pmatrix}}{\sqrt{16+144+1}} = \dfrac{24+48-2}{\sqrt{161}} = \dfrac{70}{\sqrt{161}} = 5.52$ | M1A1 A1 | 6 marks, **[14]** |
| Alternative for last 4 marks: Plane through e.g. $P$ in direction $\mathbf{PA}$: $4x+12y-29=0$. Then use distance of point from line formula to get $\dfrac{70}{\sqrt{161}}$ | | Award M1A1, then M1A1 |

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# Question 11O:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Shows position of 3 cube roots on Argand diagram | B1 | 1 mark |
| $1 = \cos 2k\pi + i\sin 2k\pi$; $k = 0, 1, 2$ | | |
| $\omega = \cos\dfrac{2\pi}{3} + i\sin\dfrac{2\pi}{3}$ and $\omega^2 = \cos\dfrac{4\pi}{3} + i\sin\dfrac{4\pi}{3}$ | M1A1 | |
| $\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$, $\omega^2 = -\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}$ | A1 | 3 marks |
| S.C. Award B1 for cube roots without/incorrect working. $(z-1)(z^2+z+1)=0$ then squaring one to get the other, scores M1A1A1 | | |
| $(6 - \omega^3) - 3\omega(9\omega^2 - 2\omega^2) + 2\omega^2(3\omega - 4\omega)$ | M1A1 | |
| $= 5 - 21 - 2 = -18$ | B1 | 3 marks |
| $z = 4\sqrt{3}\!\left(-\dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i\right) - 4\!\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i\right)$ | M1A1 | |
| $= -2\sqrt{3} - 6i - 2\sqrt{3} + 2i = -4(\sqrt{3}+i)$ | A1 | |
| $= -8\!\left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}i\right)$ | | |
| $= -8\!\left(\cos\dfrac{1}{6}\pi + i\sin\dfrac{1}{6}\pi\right) = 8\!\left(\cos\dfrac{7}{6}\pi + i\sin\dfrac{7}{6}\pi\right)$ | M1A1 | 5 marks |
| Cube roots are: $2\!\left(\cos\dfrac{7}{18}\pi + i\sin\dfrac{7}{18}\pi\right)$ | B1 | |
| $2\!\left(\cos\dfrac{19}{18}\pi + i\sin\dfrac{19}{18}\pi\right)$, $2\!\left(\cos\dfrac{31}{18}\pi + i\sin\dfrac{31}{18}\pi\right)$ | B1 | 2 marks, **[14]** |