The function $\text{f}$ is defined by
$\text{f} : x \mapsto \frac{2x+k}{x-2}, x \in \mathbb{R}, x \ne 2,$
where $k$ is a positive constant.

1. Find the equations of the vertical and horizontal asymptotes.

For $y = \frac{2x+k}{x-2}$:
The denominator is zero when $x - 2 = 0$, so the vertical asymptote is $x = 2$.
By polynomial long division or taking the limit as $x \to \pm\infty$, $y = \frac{2x+k}{x-2} = 2 + \frac{k+4}{x-2}$.
As $x \to \pm\infty$, $y \to 2$, so the horizontal asymptote is $y = 2$.

2. Find the coordinates of the intercepts on the $x$ and $y$ axes.

$y$-intercept: Let $x = 0$, then $y = \frac{k}{-2} = -\frac{k}{2}$. The coordinates are $\left(0, -\frac{k}{2}\right)$.
$x$-intercept: Let $y = 0$, then $2x + k = 0 \implies x = -\frac{k}{2}$. The coordinates are $\left(-\frac{k}{2}, 0\right)$.

3. Sketch the graph showing asymptotes and intercepts.

Draw the asymptotes $x = 2$ and $y = 2$ as dashed lines.
Since $k > 0$, $-\frac{k}{2}$ is negative. The graph passes through the negative parts of both the $x$ and $y$ axes.
The graph is a hyperbola with two branches:

• One branch passes through $\left(-\frac{k}{2}, 0\right)$ and $\left(0, -\frac{k}{2}\right)$ in the region where $x < 2$ and $y < 2$.
• The other branch lies entirely in the upper right quadrant relative to the asymptotes, where $x > 2$ and $y > 2$.

1. Express $y = \text{f}(x)$ in the form $y = a + \frac{b}{x-c}$ by algebraic manipulation.

$y = \frac{2x+k}{x-2}$
$y = \frac{2(x-2)+4+k}{x-2}$
$y = \frac{k+4}{x-2} + 2$

2. Identify the scaling transformation from $y = \frac{1}{x}$.

Comparing $y = \frac{1}{x}$ and $y = \frac{k+4}{x}$, the first transformation is a scaling (or stretch) parallel to the $y$-axis by a scale factor of $k+4$.

3. Identify the translation transformation.

Comparing $y = \frac{k+4}{x}$ and $y = \frac{k+4}{x-2} + 2$, the second transformation is a translation of $2$ units in the positive $x$-direction and $2$ units in the positive $y$-direction. This can be represented by a translation by the vector $\begin{pmatrix} 2 \\ 2 \end{pmatrix}$.

1. Set $y = \text{f}(x)$ and rearrange to make $x$ the subject.

Let $y = \frac{2x+k}{x-2}$.
$y(x-2) = 2x+k$
$xy - 2y = 2x+k$
$xy - 2x = 2y+k$
$x(y - 2) = 2y+k$
$x = \frac{2y+k}{y-2}$

2. Express $\text{f}^{-1}$ in the mapping notation, stating its domain.

The inverse function is formed by replacing $y$ with $x$. Since the denominator cannot be zero, $x \neq 2$.

In mapping notation:
$\text{f}^{-1} : x \mapsto \frac{2x+k}{x-2}, x \in \mathbb{R}, x \ne 2$

3. Determine the range of $\text{f}^{-1}$.

The range of $\text{f}^{-1}$ is exactly the domain of $\text{f}$.
The domain of $\text{f}$ is given as $x \in \mathbb{R}, x \ne 2$.
Therefore, the range of $\text{f}^{-1}$ is all real numbers except $2$.

Range of $\text{f}^{-1} = \mathbb{R} \setminus \{2\}$ (or $\text{f}^{-1}(x) \ne 2$)

1. Find $\text{f}^2(x)$ by first determining $\text{f}^{-1}(x)$ (as required in part (iii)) and using the 'Hence' method.

Let $y = \frac{2x+k}{x-2}$.
$y(x-2) = 2x+k$
$xy - 2y = 2x+k$
$x(y-2) = 2y+k$
$x = \frac{2y+k}{y-2}$
So $\text{f}^{-1}(x) = \frac{2x+k}{x-2} = \text{f}(x)$.
Since $\text{f}$ is its own inverse, $\text{f}^2(x) = \text{f}(\text{f}(x)) = \text{f}(\text{f}^{-1}(x)) = x$.

2. Use the result for $\text{f}^2(x)$ to simplify the expression for $\text{f}^{2017}(x)$.

Since $\text{f}^2(x) = x$, applying $\text{f}$ an even number of times yields $x$.
$\text{f}^{2017}(x) = \text{f}\left(\text{f}^{2016}(x)\right) = \text{f}\left(\left(\text{f}^2\right)^{1008}(x)\right) = \text{f}(x)$

3. Evaluate $\text{f}^{2017}\left(\frac{1}{2}\right)$ by substituting into $\text{f}(x)$.

$\text{f}^{2017}\left(\frac{1}{2}\right) = \text{f}\left(\frac{1}{2}\right)$
$= \frac{2\left(\frac{1}{2}\right) + k}{\frac{1}{2} - 2}$
$= \frac{1 + k}{-\frac{3}{2}}$
$= -\frac{2}{3}(1+k)$

1. Find the range of $\text{g}$.

Given the domain $x > 3$, we have $x - 3 > 0$, which means $\sqrt{x-3} > 0$.
Therefore, $\text{g}(x) = a + \sqrt{x-3} > a$.
The range of $\text{g}$ is $\text{R}_\text{g} = (a, \infty)$.

2. Use the existence condition for composite function $\text{fg}$ to find an inequality for $a$.

For $\text{fg}$ to exist, the range of $\text{g}$ must be a subset of the domain of $\text{f}$, so $\text{R}_\text{g} \subseteq \text{D}_\text{f}$.
We are given $\text{D}_\text{f} = \mathbb{R} \setminus \{2\}$.
Thus, the value $2$ cannot be in $(a, \infty)$, which implies that $2 \le a$.
Therefore, $a \ge 2$.

3. Explain why the composite function $\text{gf}$ does not exist.

For $\text{gf}$ to exist, we need $\text{R}_\text{f} \subseteq \text{D}_\text{g}$.
From earlier parts (or by finding the horizontal asymptote), $\text{f}(x) = \frac{2x+k}{x-2} = 2 + \frac{k+4}{x-2}$. Since $k>0$, $\text{R}_\text{f} = \mathbb{R} \setminus \{2\}$.
We are given $\text{D}_\text{g} = (3, \infty)$.
Clearly, $\mathbb{R} \setminus \{2\}$ is not a subset of $(3, \infty)$ (for example, $0 \in \text{R}_\text{f}$ but $0 \notin \text{D}_\text{g}$).
Since $\text{R}_\text{f} \not\subseteq \text{D}_\text{g}$, the composite function $\text{gf}$ does not exist.