## Question 10:

**Part 1 (Showing initial results):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $z^n = \cos n\theta + i\sin n\theta$ and $z^{-n} = \cos n\theta - i\sin n\theta$ | M1A1 | AG |
| $\Rightarrow z^n + \frac{1}{z^n} = 2\cos n\theta$ and $z^n - \frac{1}{z^n} = 2i\sin n\theta$ | | [2] |

**Part 2 (Showing $\sin^4\theta\cos^2\theta$ result):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\left(z-\frac{1}{z}\right)^4\left(z+\frac{1}{z}\right)^2 = \left(z-\frac{1}{z}\right)^2\left(z^2-\frac{1}{z^2}\right)^2$ | M1 | Or 1st bracket expanded |
| $= \left(z^2-2+\frac{1}{z^2}\right)\left(z^4-2+\frac{1}{z^4}\right)$ | M1 | 2nd bracket expanded |
| $= \left(z^6+\frac{1}{z^6}\right)-2\left(z^4+\frac{1}{z^4}\right)-\left(z^2+\frac{1}{z^2}\right)+4$ | M1A1 | Expanded and grouped |
| $\Rightarrow 64\sin^4\theta\cos^2\theta = 2\cos 6\theta - 4\cos 4\theta - 2\cos 2\theta + 4$ | M1A1 | |
| $\Rightarrow \sin^4\theta\cos^2\theta = \frac{1}{32}(\cos 6\theta - 2\cos 4\theta - \cos 2\theta + 2)$ | A1 | AG, uses initial result [7] |

**Part 3 (Integration):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int_0^{\frac{1}{4}\pi}\sin^4\theta\cos^2\theta\, d\theta = \frac{1}{32}\left[\frac{1}{6}\sin 6\theta - \frac{1}{2}\sin 4\theta - \frac{1}{2}\sin 2\theta + 2\theta\right]_0^{\frac{1}{4}\pi}$ | M1A1 | |
| $= \frac{1}{32}\left(-\frac{1}{6}-0-\frac{1}{2}+\frac{\pi}{2}\right) = \left(\frac{3\pi-4}{192}\right)$ or $0.0283$ (3sf) | A1 | [3] |

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## Question 11E:

**Part 1 (Finding p and q):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -3 & -1\\ 3 & -4 & -2\end{vmatrix} = \begin{pmatrix}2\\-1\\5\end{pmatrix}$ | M1A1 | |
| Obtains three equations from $\overrightarrow{OP} + \overrightarrow{PQ} = \overrightarrow{OQ}$: $t-3s+2k=4$, $-3t+4s-k=-2$, $-t+2s+5k=4$ | M1A1 | |
| Solves: $s=-1$, $t=-1$, $k=1$ | M1A1 | |
| $\mathbf{p} = 3\mathbf{i}+\mathbf{j}+2\mathbf{k}$ and $\mathbf{q} = \mathbf{i}+2\mathbf{j}-3\mathbf{k}$ | A1 | Both required [7] |

**Part 2 (Equation of plane $\Pi$ and line $l_3$):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Direction of $l_3$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\3&-4&-2\\0&0&1\end{vmatrix} = \begin{pmatrix}-4\\-3\\0\end{pmatrix} \sim \begin{pmatrix}4\\3\\0\end{pmatrix}$ | B1 | |
| Normal to $\Pi$: $\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\22&19&-5\\2&-1&5\end{vmatrix} = \begin{pmatrix}90\\-120\\-60\end{pmatrix} = \begin{pmatrix}3\\-4\\-2\end{pmatrix}$ | M1 | |
| Cartesian equation of $\Pi$: $3x-4y-2z=1$ | A1 | Since $P$ lies in $\Pi$ |
| Vector equation of $l_3$: $\mathbf{r} = \begin{pmatrix}-1\\-1\\0\end{pmatrix}+\lambda\begin{pmatrix}4\\3\\0\end{pmatrix}$ | B1 | OE, since $z=0$ [4] |

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## Question 11O:

**Preliminary (AG):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\dot{x}^2+\dot{y}^2 = 36t^2+1-18t^2+81t^4 = 1+18t^2+81t^4 = (1+9t^2)^2$ | M1A1 | AG [2] |

**Part (i):**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $s = \int_0^{\frac{1}{\sqrt{3}}}(1+9t^2)\,dt = \left[t+3t^3\right]_0^{\frac{1}{\sqrt{3}}} = \frac{2}{\sqrt{3}}$ | M1A1 | [2] |

**Part (ii) — Surface area:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $S = 2\pi\int_0^{\frac{1}{\sqrt{3}}}t(1-3t^2)(1+9t^2)\,dt = 2\pi\int_0^{\frac{1}{\sqrt{3}}}(t+6t^3-27t^5)\,dt$ | *M1 | |
| $= 2\pi\left[\frac{1}{2}t^2+\frac{3}{2}t^4-\frac{9}{2}t^6\right]_0^{\frac{1}{\sqrt{3}}} = \frac{\pi}{3}$ | DM1A1 | [3] |

**Part (ii) — Polar/Cartesian conversion:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| Substitute $t=\frac{y}{x}$ to obtain $x = 1-3\frac{y^2}{x^2}$ | M1 | |
| Use $x=r\cos\theta$, $y=r\sin\theta \Rightarrow r\cos\theta = 1-3\tan^2\theta \Rightarrow r=\sec\theta(1-3\tan^2\theta)$ | M1A1 | |
| Domain: $0 \leq \theta \leq \frac{\pi}{6}$ | B1 | [4] |

**Part (ii) — Area in polar:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $A = \frac{1}{2}\int_0^{\frac{\pi}{6}}\sec^2\theta(1-6\tan^2\theta+9\tan^4\theta)\,d\theta$ | M1 | |
| $= \frac{1}{2}\left[\tan\theta - 2\tan^3\theta+\frac{9}{5}\tan^5\theta\right]_0^{\frac{\pi}{6}} = \frac{4}{45}\sqrt{3}$ | A1A1 | [3] |