# Question 11 (e):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dot{x} = 2e^{2t}\cos 2t - 2e^{2t}\sin 2t$, $\dot{y} = 2e^{2t}\cos 2t + 2e^{2t}\sin 2t$ | B1 | |
| $\dot{x}^2+\dot{y}^2 = 4e^{4t}(\cos^2 2t - 2\cos 2t\sin 2t+\sin^2 2t+\cos^2 2t+2\cos 2t\sin 2t+\sin^2 2t)$ | M1 | |
| $= 8e^{4t}$ | A1 | |
| $s = \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} 2\sqrt{2}e^{2t}\,dt$ | M1 | |
| $= \sqrt{2}\left[e^{2t}\right]_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} = \sqrt{2}(e^\pi - e^{-\pi})$ or $2\sqrt{2}\sinh\pi$ or $32.7$ | M1A1 [6] | |
| $S = 2\pi\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} e^{2t}\sin 2t \cdot 2\sqrt{2}e^{2t}\,dt = 4\sqrt{2}\pi\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} e^{4t}\sin 2t\,dt$ | M1 | |
| Let $I = \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} e^{4t}\sin 2t\,dt = \left[-e^{4t}\frac{\cos 2t}{2}\right]_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} + \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} 2e^{4t}\cos 2t\,dt$ | M1A1 | |
| $= \left[\frac{e^{2\pi}}{2}\right] - \left[\frac{e^{-2\pi}}{2}\right] + 2\left\{\left[e^{4t}\frac{\sin 2t}{2}\right]_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} - \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} 2e^{4t}\sin 2t\right\}$ | A1A1 | |
| $= \frac{e^{2\pi}-e^{-2\pi}}{2} + 0 - 4I$ | M1 | |
| $\Rightarrow I = \frac{e^{2\pi}-e^{-2\pi}}{10}$ | A1 | |
| $\Rightarrow S = 4\sqrt{2}\pi\cdot\frac{e^{2\pi}-e^{-2\pi}}{10} = \frac{2\sqrt{2}\pi}{5}(e^{2\pi}-e^{-2\pi})$ or $\frac{4\sqrt{2}\pi}{5}\sinh 2\pi$ or $952$ (3sf) | A1 [8] | |

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# Question 11 (o):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\begin{pmatrix}1&-2&3&-4\\2&-4&7&-9\\4&-8&14&-18\\5&-10&17&-22\end{pmatrix} \Rightarrow \ldots \Rightarrow \begin{pmatrix}1&-2&3&-4\\0&0&1&-1\\0&0&0&0\\0&0&0&0\end{pmatrix}$ | M1A1 | |
| $r(\mathbf{M}) = 4-2 = 2$ | A1 [3] | |
| $x-2y+3z-4t=0$, $z-t=0$ | M1 | |
| $t=z=\lambda$ and $y=\mu \Rightarrow x=2\mu+\lambda$ | M1 | |
| Basis for $K$ is $\left\{\begin{pmatrix}1\\0\\1\\1\end{pmatrix}, \begin{pmatrix}2\\1\\0\\0\end{pmatrix}\right\}$ | A1 [3] | |
| $\begin{pmatrix}1&-2&3&-4\\2&-4&7&-9\\4&-8&14&-18\\5&-10&17&-22\end{pmatrix}\begin{pmatrix}1\\-2\\2\\-1\end{pmatrix} = \begin{pmatrix}15\\33\\66\\81\end{pmatrix}$ | B1 | |
| $\mathbf{x} = \begin{pmatrix}1\\-2\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}1\\0\\1\\1\end{pmatrix}+\mu\begin{pmatrix}2\\1\\0\\0\end{pmatrix}$ since $\mathbf{M}\begin{pmatrix}1\\-2\\2\\-1\end{pmatrix}=\begin{pmatrix}15\\33\\66\\81\end{pmatrix}$ and $\mathbf{M}\left[\lambda\begin{pmatrix}1\\0\\1\\1\end{pmatrix}+\mu\begin{pmatrix}2\\1\\0\\0\end{pmatrix}\right]=0$ | B1$\checkmark$ [2] | |
| Sum of components $= 6 \Rightarrow 3\lambda+3\mu=6 \Rightarrow \mu=2-\lambda$ | B1$\checkmark$ | |
| Sum of squares of components $= 26 \Rightarrow 5\mu^2+4\lambda\mu+4\lambda^2+3\lambda^2+10=26$ $\Rightarrow 4\lambda^2-8\lambda+4=0 \Rightarrow (\lambda-1)^2=0$ | M1A1, M1 | |
| $\Rightarrow \lambda=1, \mu=1$ | A1 | |
| $\mathbf{x}' = \begin{pmatrix}4\\-1\\3\\0\end{pmatrix}$ | A1 [6] | |