Sketch the graphs of $y = 1 + \frac{a-2}{x-a}$ and $y = -\frac{1}{a}x + \frac{2}{a}$ on a single diagram, where $a$ is a positive constant and $1 < a < 2$, showing all asymptotes and axial intercepts clearly.

1. Identify the $x$-coordinates of the intersections and the vertical asymptote.

From the equations given for the graphs, substituting $x=0$ gives $y = 1 + \frac{a-2}{-a} = \frac{2}{a}$ for the curve and $y = \frac{2}{a}$ for the line.
Substituting $y=0$ gives $0 = 1 + \frac{a-2}{x-a} \implies x = 2$ for the curve and $0 = -\frac{1}{a}x + \frac{2}{a} \implies x = 2$ for the line.
Thus, the graphs intersect at the points $(0, \frac{2}{a})$ and $(2, 0)$, so the intersection $x$-coordinates are $x=0$ and $x=2$.

The vertical asymptote of the curve $y = 1 + \frac{a-2}{x-a}$ is $x = a$.
Since $1 < a < 2$, the points lie in the order $x=0$, $x=a$, and $x=2$.

2. Solve the inequality by determining where the curve lies above the line.

The inequality $1 + \frac{a-2}{x-a} > -\frac{1}{a}x + \frac{2}{a}$ asks for the range of values of $x$ where the curve $C$ is strictly above the line $L$.

Based on the branches of the hyperbola (since $a-2 < 0$) and the line $L$ with negative gradient $-\frac{1}{a}$:

• For $x < a$ (left branch): The curve passes through $(0, \frac{2}{a})$ and tends to $+\infty$ as $x \to a^-$. It is above the line for $0 < x < a$.
• For $x > a$ (right branch): The curve passes through $(2, 0)$ and tends to $1$ as $x \to \infty$. It is above the line for $x > 2$.

Hence, the required solution is $0 < x < a \text{ or } x > 2$.

1. Identify the substitution linking (ii) to (i)

Replace $x$ by $\dfrac{1}{x}$ in the inequality of part (i):
$1+\frac{a-2}{\frac1x-a}=1+\frac{(a-2)x}{1-ax}=1+\frac{ax-2x}{1-ax},$
which is exactly the left-hand side of (ii). The right-hand side $-\frac1a x+\frac2a$ becomes $-\frac{1}{ax}+\frac{2}{a}$. So (ii) is just (i) with $x$ replaced by $\dfrac1x$.

2. Transfer the solution of (i) using the substitution

From part (i) the inequality holds for $0<x<a$ or $x>2$. Replacing $x$ by $\dfrac1x$, the inequality in (ii) holds when
$0<\frac1x<a\qquad\text{or}\qquad \frac1x>2.$

3. Solve the transformed inequalities and combine

For $0<\dfrac1x<a$: $\dfrac1x>0\Rightarrow x>0$, and $\dfrac1x<a\Rightarrow x>\dfrac1a$ (since $x>0$). This gives $x>\dfrac1a$.

For $\dfrac1x>2$: this requires $x>0$ and $x<\dfrac12$, i.e. $0<x<\dfrac12$.

Since $1<a<2$ gives $\dfrac12<\dfrac1a<1$, the two ranges are separate. Hence
$0<x<\frac12\quad\text{or}\quad x>\frac1a.$