## Question 10:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$ | B1 | |
| $(c+is)^5 = c^5 + 5c^4s\,i - 10c^3s^2 - 10ic^2s^3i + 5cs^4 + s^5i$ | M1A1 | |
| $\tan 5\theta = \dfrac{5c^4s - 10c^3s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4}$ | M1 | |
| Divide numerator and denominator by $c^5$: $\tan 5\theta = \dfrac{5\tan\theta - 10\tan^3\theta + \tan^5\theta}{1 - 10\tan^2\theta + 5\tan^4\theta}$ | A1 | AG, [5] |
| $\tan 5\theta = 0 \Rightarrow \theta = \frac{1}{5}\pi, \frac{2}{5}\pi, \frac{3}{5}\pi, \frac{4}{5}\pi, \pi$ | B1 | |
| $t^5 - 10t^3 + 5t = 0$ has roots $\tan\!\left(\frac{1}{5}\pi\right), \tan\!\left(\frac{2}{5}\pi\right), \tan\!\left(\frac{3}{5}\pi\right), \tan\!\left(\frac{4}{5}\pi\right), \tan\pi$ | | |
| $\Rightarrow t^4 - 10t^2 + 5 = 0$ has roots $\tan\!\left(\frac{1}{5}\pi\right), \tan\!\left(\frac{2}{5}\pi\right), \tan\!\left(\frac{3}{5}\pi\right), \tan\!\left(\frac{4}{5}\pi\right)$ | B1 | |
| $\Rightarrow \left(t^2 - \tan^2\!\left(\frac{1}{5}\pi\right)\right)\!\left(t^2 - \tan^2\!\left(\frac{2}{5}\pi\right)\right) = 0$ | | |
| since $\tan\!\left(\frac{1}{5}\pi\right) = -\tan\!\left(\frac{4}{5}\pi\right)$ and $\tan\!\left(\frac{2}{5}\pi\right) = -\tan\!\left(\frac{3}{5}\pi\right)$ | M1 | |
| $\Rightarrow x^2 - 10x + 5 = 0$ has roots $\tan^2\!\left(\frac{1}{5}\pi\right)$ and $\tan^2\!\left(\frac{2}{5}\pi\right)$ | A1 | AG, [4] |
| $\sec^2\alpha = 1 + \tan^2\alpha$ | M1 | |
| $y = 1 + x \Rightarrow x = y-1 \Rightarrow (y-1)^2 - 10(y-1) + 5 = 0$ | M1 | |
| $\Rightarrow y^2 - 12y + 16 = 0$ | A1 | [3], Total 12 |

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## Question 11E:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 0 \\ -1 & 0 & 4 \end{vmatrix} = \begin{pmatrix}8\\4\\2\end{pmatrix} \sim \begin{pmatrix}4\\2\\1\end{pmatrix}$ | M1A1 | |
| $\dfrac{\begin{pmatrix}1\\0\\0\end{pmatrix}\cdot\begin{pmatrix}4\\2\\1\end{pmatrix}}{\sqrt{4^2+2^2+1^2}} = \dfrac{4}{\sqrt{21}}$ | M1A1 | AG, [4] |
| $\mathbf{p} = \dfrac{3}{\sqrt{21}}\left(\dfrac{4\mathbf{i}+2\mathbf{j}+\mathbf{k}}{\sqrt{21}}\right) = \dfrac{1}{7}(4\mathbf{i}+2\mathbf{j}+\mathbf{k})$ | B1 | |
| Line $AP$: $\mathbf{r} = \begin{pmatrix}1\\0\\0\end{pmatrix} + t\begin{pmatrix}-3\\2\\1\end{pmatrix}$ | M1A1 | |
| For $Q$: $1-3t=0 \Rightarrow t=\frac{1}{3} \Rightarrow \mathbf{q} = \frac{1}{3}\begin{pmatrix}0\\2\\1\end{pmatrix}$ | M1A1 | [5] |
| $\overrightarrow{AB} = \begin{pmatrix}-1\\2\\0\end{pmatrix}$, $\overrightarrow{BQ} = \frac{1}{3}\begin{pmatrix}0\\-4\\1\end{pmatrix}$ | B1 | |
| $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\-1&2&0\\0&-4&1\end{vmatrix} = \begin{pmatrix}2\\1\\4\end{pmatrix}$ | M1A1 | |
| $\cos^{-1}\dfrac{\begin{pmatrix}4\\2\\1\end{pmatrix}\cdot\begin{pmatrix}2\\1\\4\end{pmatrix}}{\sqrt{21}\cdot\sqrt{21}} = \cos^{-1}\dfrac{8+2+4}{21} = \cos^{-1}\dfrac{14}{21} = \cos^{-1}\dfrac{2}{3}$ | M1A1 | AG, [5], Total 14 |

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## Question 11O:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Closed curve through pole with correct orientation | B1 | |
| Completely correct | B1 | [2] |
| $2 \times \frac{1}{2}a^2\int_{\frac{1}{2}\pi}^{\pi}(1-2\cos\theta+\cos^2\theta)d\theta = a^2\int_{\frac{1}{2}\pi}^{\pi}\left(\frac{3}{2}-2\cos\theta+\frac{1}{2}\cos 2\theta\right)d\theta$ | M1M1 | |
| $= a^2\left[\frac{3}{2}\theta - 2\sin\theta + \frac{1}{4}\sin 2\theta\right]_{\frac{1}{2}\pi}^{\pi}$ | M1A1 | |
| $= a^2\left(\frac{3}{4}\pi + 2\right)$ | A1 | [5] |
| $\left(\dfrac{ds}{d\theta}\right)^2 = a^2(1-2\cos\theta+\cos^2\theta+\sin^2\theta)$ | B1 | |
| $= 2a^2(1-\cos\theta) = 2a^2\cdot 2\sin^2\frac{1}{2}\theta = 4a^2\sin^2\frac{1}{2}\theta$ | M1A1 | AG |
| $s = 2\times\int_{\frac{1}{2}\pi}^{\pi} 2a\sin\frac{1}{2}\theta\,d\theta$ | M1 | |
| $= 4a\left[-2\cos\frac{1}{2}\theta\right]_{\frac{1}{2}\pi}^{\pi}$ | A1 | |
| $= 4\sqrt{2}\,a$ | M1A1 | [7], Total 14 |