# Question 11:

| Answer/Working | Mark | Notes |
|---|---|---|
| $\overrightarrow{AB} = \begin{pmatrix}6\\4\\-8\end{pmatrix}$, $\overrightarrow{CD} = \begin{pmatrix}5\\-2\\-4\end{pmatrix}$ | B1 | |
| Common perpendicular is $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\3 & 2 & -4\\5 & -2 & -4\end{vmatrix} = \begin{pmatrix}-16\\-8\\-16\end{pmatrix} \sim \begin{pmatrix}2\\1\\2\end{pmatrix}$ | M1A1 | |
| $\overrightarrow{AD} = \begin{pmatrix}1\\-4\\9\end{pmatrix} \Rightarrow$ shortest distance $= \frac{\begin{vmatrix}\begin{pmatrix}1\\-4\\9\end{pmatrix}\cdot\begin{pmatrix}2\\1\\2\end{pmatrix}\end{vmatrix}}{\sqrt{4+1+4}} = \frac{16}{3}$ or 5.33 | M1A1 [5] | |
| Normal to $\Pi_1$ is $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & -4 & 9\\3 & 2 & -4\end{vmatrix} = \begin{pmatrix}-2\\31\\14\end{pmatrix}$ | M1A1 | |
| Normal to $\Pi_2$ is $\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & -4 & 9\\5 & -2 & -4\end{vmatrix} = \begin{pmatrix}34\\49\\18\end{pmatrix}$ | A1 | |
| $\cos\theta = \frac{-2\times34 + 31\times49 + 14\times18}{\sqrt{2^2+31^2+14^2}\sqrt{34^2+49^2+18^2}}$ | M1A1 | |
| $\Rightarrow \theta = 36.7°$ | A1 [6] | CAO |

---

# Question 12E:

## Part (i):
| Answer/Working | Mark | Notes |
|---|---|---|
| $\dot{x} = 2t$, $\dot{y} = -\frac{1}{2}(2-t)^{-\frac{1}{2}} \Rightarrow \frac{dy}{dx} = \frac{-1}{4t(2-t)^{\frac{1}{2}}}$ | M1A1 | |
| $\frac{d^2y}{dx^2} = \left\{4t(2-t)^{\frac{1}{2}}\right\}^{-2}\left\{4(2-t)^{\frac{1}{2}} - \frac{4t}{2(2-t)^{\frac{1}{2}}}\right\}\times\frac{1}{2t}$ (OE) | M1A1 | |
| $= \frac{1}{16t^3(2-t)^{\frac{3}{2}}}(4-3t)$ (any correct simplified form) | A1 [5] | |

## Part (ii):
| Answer/Working | Mark | Notes |
|---|---|---|
| $\int_0^4 y\,dx = \int_0^2 2t(2-t)^{\frac{1}{2}}\,dt$ | | |
| $= 2\left\{\left[-\frac{2}{3}t(2-t)^{\frac{3}{2}}\right]_0^2 + \int_0^2\frac{2}{3}(2-t)^{\frac{3}{2}}\,dt\right\}$ | M1A1 | LNR |
| $= 2\left\{0 + \left[-\frac{4}{15}(2-t)^{\frac{5}{2}}\right]_0^2\right\} = \frac{32}{15}\sqrt{2}$ | M1A1 | LR |
| $MV = \frac{\int_0^4 y\,dx}{4-0} = \frac{1}{4}\times\frac{32}{15}\sqrt{2} = \frac{8}{15}\sqrt{2}$ or 0.754 | M1A1 [6] | |

## Part (iii):
| Answer/Working | Mark | Notes |
|---|---|---|
| $\frac{1}{2}\int_0^4 y^2\,dx = \frac{1}{2}\int_0^2 2t(2-t)\,dt = \left[t^2 - \frac{1}{3}t^3\right]_0^2 = \frac{4}{3}$ or 1.33 | M1A1 | |
| $\bar{y} = \frac{\frac{1}{2}\int_0^4 y^2\,dx}{\int_0^4 y\,dx} = \frac{4}{3}\times\frac{15}{32\sqrt{2}} = \frac{5}{8\sqrt{2}} = \frac{5}{16}\sqrt{2}$ or 0.442 | A1 [3] | |

---

# Question 12O:

## Part (i):
| Answer/Working | Mark | Notes |
|---|---|---|
| $d = -2$ | B1 [1] | |

## Part (ii):
| Answer/Working | Mark | Notes |
|---|---|---|
| $x=0 \Rightarrow y = -2 = \frac{c}{d} \Rightarrow c = 4$ | M1A1 | |
| $y = ax + (b+2a) + \frac{4+4a+2b}{x-2}$ | M1A1 | |
| Oblique asymptote $y = x+1 \Rightarrow a=1$, $b+2a = 1 \Rightarrow b = -1$ | A1A1 [6] | |

## Part (iii):
| Answer/Working | Mark | Notes |
|---|---|---|
| $y = \frac{x^2-x+4}{x-2} \Rightarrow x^2-(y+1)x+2y+4=0$ | M1 | |
| For real $x$: $B^2-4AC > 0 \Rightarrow (y+1)^2 - 4(2y+4) > 0$ | M1 | |
| $\Rightarrow y^2 - 6y - 15 > 0$ | A1 | |
| $\Rightarrow (y-3)^2 > 24$ | M1A1 | |
| $\Rightarrow y-3 < -2\sqrt{6}$ or $y-3 > -2\sqrt{6}$ | M1 | for thumbnail sketch |
| $\Rightarrow y < 3-2\sqrt{6}$ or $y > 3+2\sqrt{6}$ | A1 [7] | AG; Note: if done by differentiation, max and min must be demonstrated for full marks |