On the same axes, sketch the curves with equations $y = \left| \frac{ax + 1}{1 - ax} \right|$ and $y = \frac{1}{a}x + 1$, where $a > 1$, giving the equations of the asymptotes and the coordinates of the points where the curves meet the axes. Solve the inequality $\left| \frac{ax + 1}{1 - ax} \right| > \frac{1}{a}x + 1$, giving your answers in terms of $a$.

1. Find the vertical and horizontal asymptotes of the rational curve.

For the curve $y = \left| \frac{ax + 1}{1 - ax} \right|$:
The denominator cannot be zero, giving the vertical asymptote:
$1 - ax = 0 \implies x = \frac{1}{a}$
To find the horizontal asymptote, consider the limit as $x \to \pm\infty$:
$\frac{ax + 1}{1 - ax} = \frac{a + \frac{1}{x}}{\frac{1}{x} - a} \to \frac{a}{-a} = -1$
Taking the modulus, the horizontal asymptote is:
$y = |-1| = 1$

2. Determine the axial intercepts of the curve.

To find the $y$-intercept, substitute $x = 0$:
$y = \left| \frac{a(0) + 1}{1 - a(0)} \right| = 1 \implies (0, 1)$
To find the $x$-intercept, substitute $y = 0$:
$\frac{ax + 1}{1 - ax} = 0 \implies ax + 1 = 0 \implies x = -\frac{1}{a}$
The curve meets the axes at $(0, 1)$ and $\left(-\frac{1}{a}, 0\right)$.

3. Determine the axial intercepts of the line and establish relative positions for sketching.

For the line $y = \frac{1}{a}x + 1$:
When $x = 0$, $y = 1 \implies (0, 1)$.
When $y = 0$:
$\frac{1}{a}x + 1 = 0 \implies x = -a \implies (-a, 0)$
Since $a > 1$, we can deduce that $a > \frac{1}{a}$. Multiplying by $-1$ flips the inequality, giving $-a < -\frac{1}{a}$.
This ensures the line's $x$-intercept is further to the left than the curve's $x$-intercept on the sketch.

1. Set up the equations to find the points of intersection between the modulus curve and the line.

To find the intersections, we solve $\left| \frac{ax + 1}{1 - ax} \right| = \frac{1}{a}x + 1 = \frac{x + a}{a}$.
This gives two possible cases when removing the absolute value:
$\frac{ax + 1}{1 - ax} = \frac{x + a}{a} \quad \text{or} \quad \frac{ax + 1}{1 - ax} = -\left(\frac{x + a}{a}\right)$

2. Solve the positive case and identify any extraneous roots.

For the first case:
$a(ax + 1) = (1 - ax)(x + a)$
$a^2x + a = x + a - ax^2 - a^2x$
$ax^2 + 2a^2x - x = 0$
$x(ax + 2a^2 - 1) = 0$
$x = 0 \quad \text{or} \quad x = \frac{1 - 2a^2}{a}$
Since $a > 1$, for $x = \frac{1 - 2a^2}{a}$, the right-hand side of our original equation evaluates to $\frac{1 - 2a^2 + a^2}{a^2} = \frac{1 - a^2}{a^2} < 0$.
Since an absolute value cannot equal a negative number, $x = \frac{1 - 2a^2}{a}$ is an extraneous root and is rejected. Thus, $x = 0$ is the only valid intersection from this case.

3. Solve the negative case to find the remaining valid intersection points.

For the second case:
$a(ax + 1) = -(1 - ax)(x + a)$
$a^2x + a = ax^2 + a^2x - x - a$
$ax^2 - x - 2a = 0$
Using the quadratic formula:
$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(a)(-2a)}}{2a} = \frac{1 \pm \sqrt{1 + 8a^2}}{2a}$
Both of these roots satisfy $\frac{x + a}{a} \ge 0$ (since $a > 1$) and are therefore valid intersection points.

4. Use the graph sketched in part (i) to determine the intervals where the curve strictly exceeds the line.

From the sketch in part (i), the curve $y = \left| \frac{ax + 1}{1 - ax} \right|$ has a vertical asymptote at $x = \frac{1}{a}$.
The inequality requires the modulus curve to be strictly above the line $y = \frac{1}{a}x + 1$.
Based on the $x$-coordinates of the intersections ($x = \frac{1 - \sqrt{1 + 8a^2}}{2a}$, $x = 0$, and $x = \frac{1 + \sqrt{1 + 8a^2}}{2a}$), and comparing the heights of the graphed functions, the curve is above the line in three disjoint intervals:

1. To the left of the first intersection.
2. Between $x=0$ and the vertical asymptote.
3. Between the vertical asymptote and the third intersection.

Hence, the required intervals are:
$x < \frac{1 - \sqrt{1 + 8a^2}}{2a} \quad \text{or} \quad 0 < x < \frac{1}{a} \quad \text{or} \quad \frac{1}{a} < x < \frac{1 + \sqrt{1 + 8a^2}}{2a}$