ML=(-2, -3) and LN=(-6, -1). Find (i) MN, (ii) |ML|. (b) The diagram shows vectors PT = a, PS = b and PS is divided into 3 equal parts at Q and R. U is the midpoint of TR. Express, in simplest form, in terms of a and/or b. (i) TS (ii) TR (c) Using the information in (b), show that PT = 2QU.

$$\overrightarrow{MN} = \overrightarrow{ML} + \overrightarrow{LN}$$ $$\overrightarrow{MN} = \begin{pmatrix} -2 \\ 3 \end{pmatrix} + \begin{pmatrix} -6 \\ -1 \end{pmatrix}$$

$$\overrightarrow{MN} = \begin{pmatrix} -2 + (-6) \\ 3 + (-1) \end{pmatrix}$$ $$\overrightarrow{MN} = \begin{pmatrix} -8 \\ 2 \end{pmatrix}$$

The vector $\overrightarrow{MN}$ is $\boxed{\begin{pmatrix} -8 \\ 2 \end{pmatrix}}$.

Question 5 (a) (ii): Find $|\overrightarrow{ML}|$
Given: $\overrightarrow{ML} = \begin{pmatrix} -2 \\ 3 \end{pmatrix}$.

$$|\overrightarrow{ML}| = \sqrt{4 + 9}$$ $$|\overrightarrow{ML}| = \sqrt{13}$$

The magnitude of $\overrightarrow{ML}$ is $\boxed{\sqrt{13} \text{ units}}$.

Question 5 (b) (i): Express $\overrightarrow{TS}$ in terms of $a$ and/or $b$
Given: $\overrightarrow{PT} = a$, $\overrightarrow{PS} = b$.

Since $\overrightarrow{PT} = a$, then $\overrightarrow{TP} = -a$.

$$\overrightarrow{TS} = \overrightarrow{TP} + \overrightarrow{PS}$$ $$\overrightarrow{TS} = -a + b$$

The vector $\overrightarrow{TS}$ is $\boxed{b - a}$.

Question 5 (b) (ii): Express $\overrightarrow{TR}$ in terms of $a$ and/or $b$
Given: $\overrightarrow{PT} = a$, $\overrightarrow{PS} = b$. $PS$ is divided into 3 equal parts at $Q$ and $R$.

We know $\overrightarrow{TP} = -a$.
Since $PS$ is divided into 3 equal parts at $Q$ and $R$, $\overrightarrow{PR} = \frac{2}{3}\overrightarrow{PS}$.

$$\overrightarrow{PR} = \frac{2}{3}b$$

$$\overrightarrow{TR} = -a + \frac{2}{3}b$$

The vector $\overrightarrow{TR}$ is $\boxed{\frac{2}{3}b - a}$.

Question 5 (c): Using the information in (b), show that $\overrightarrow{PT} = 2\overrightarrow{QU}$
Given: $\overrightarrow{PT} = a$, $\overrightarrow{PS} = b$. $PS$ is divided into 3 equal parts at $Q$ and $R$. $U$ is the midpoint of $TR$.

$$\overrightarrow{QU} = \overrightarrow{QR} + \overrightarrow{RU}$$

From part (b), $\overrightarrow{QR} = \frac{1}{3}\overrightarrow{PS} = \frac{1}{3}b$.
Since $U$ is the midpoint of $TR$, $\overrightarrow{RU} = \frac{1}{2}\overrightarrow{RT}$.
From part (b)(ii), $\overrightarrow{TR} = \frac{2}{3}b - a$. So, $\overrightarrow{RT} = -\overrightarrow{TR} = -(\frac{2}{3}b - a) = a - \frac{2}{3}b$.

$$\overrightarrow{RU} = \frac{1}{2}\left(a - \frac{2}{3}b\right) = \frac{1}{2}a - \frac{1}{3}b$$

$$\overrightarrow{QU} = \frac{1}{3}b + \left(\frac{1}{2}a - \frac{1}{3}b\right)$$ $$\overrightarrow{QU} = \frac{1}{2}a$$

$$2\overrightarrow{QU} = 2\left(\frac{1}{2}a\right) = a$$

We are given $\overrightarrow{PT} = a$.
Therefore, $\overrightarrow{PT} = 2\overrightarrow{QU}$. (Shown)

Question 6 (a) (i): Find $g(-6)-3$
Given: $g(x) = x^2 + 1$.

$$g(-6) = (-6)^2 + 1$$ $$g(-6) = 36 + 1$$ $$g(-6) = 37$$

$$g(-6) - 3 = 37 - 3$$ $$g(-6) - 3 = 34$$

The value is $\boxed{34}$.

Question 6 (a) (ii): Find $h^{-1}g(x)$, in its simplest form.
Given: $h(x) = 2x - 5$ and $g(x) = x^2 + 1$.

Let $y = h(x)$.

$$y = 2x - 5$$

Swap $x$ and $y$:

$$x = 2y - 5$$

Solve for $y$:

$$x + 5 = 2y$$ $$y = \frac{x + 5}{2}$$

So, $h^{-1}(x) = \frac{x + 5}{2}$.

$$h^{-1}g(x) = h^{-1}(x^2 + 1)$$ $$h^{-1}g(x) = \frac{(x^2 + 1) + 5}{2}$$ $$h^{-1}g(x) = \frac{x^2 + 6}{2}$$

The expression for $h^{-1}g(x)$ is $\boxed{\frac{x^2 + 6}{2}}$.

Question 6 (b): $2g(x) - h(x)$ can be expressed in the form $px^2 + qx + 7$. Find the value of $p$ and $q$.
Given: $h(x) = 2x - 5$ and $g(x) = x^2 + 1$.

$$2g(x) - h(x) = 2x^2 + 2 - 2x + 5$$ $$2g(x) - h(x) = 2x^2 - 2x + 7$$

By comparing $2x^2 - 2x + 7$ with $px^2 + qx + 7$:
$p = 2$
$q = -2$
The values are $\boxed{p = 2}$ and $\boxed{q = -2}$.

Question 6 (c): Find the value of $x$ for which $xh(x) - 2g(x) = -12$
Given: $h(x) = 2x - 5$ and $g(x) = x^2 + 1$.

$$2x^2 - 5x - (2x^2 + 2) = -12$$ $$2x^2 - 5x - 2x^2 - 2 = -12$$ $$-5x - 2 = -12$$

$$-5x = -12 + 2$$ $$-5x = -10$$ $$x = \frac{-10}{-5}$$ $$x = 2$$

The value of $x$ is $\boxed{2}$.