The function f is defined as follows: $f(x)=x+\frac{1}{x-a},\qquad a<x\le b$ where a is a positive constant.

1. Differentiate $f(x)$ to investigate its monotonic properties.

$f'(x) = \frac{d}{dx}\left(x + (x-a)^{-1}\right) = 1 - \frac{1}{(x-a)^2}$

2. Find the stationary point and analyse the sign of $f'(x)$.

Set $f'(x) = 0$:
$1 - \frac{1}{(x-a)^2} = 0 \implies (x-a)^2 = 1$
Since the domain is restricted to $x > a$, we have $x - a > 0$.
$x - a = 1 \implies x = a + 1$
For $a < x < a + 1$, $(x-a)^2 < 1 \implies f'(x) < 0$ (so $f$ is strictly decreasing).
For $x > a + 1$, $(x-a)^2 > 1 \implies f'(x) > 0$ (so $f$ is strictly increasing).

3. Apply the condition for the existence of an inverse function to deduce the restriction on $b$.

For $f^{-1}$ to exist, $f$ must be a one-to-one function, meaning it must be strictly monotonic over its entire defined domain $a < x \le b$.
Since $f(x)$ changes from decreasing to increasing at $x = a + 1$, the upper limit $b$ of the domain must not exceed this minimum turning point. Otherwise, the function will fail the horizontal line test.
Therefore, $b \le a + 1$.

1. Set up an equation $y = f(x)$ and rearrange it into a quadratic equation in $x$.

Given $a=1, b=2$, the function is $f(x) = x + \frac{1}{x-1}$ for $1 < x \le 2$.
Let $y = x + \frac{1}{x-1}$.
Multiplying throughout by $(x-1)$:
$y(x-1) = x(x-1) + 1$
$xy - y = x^2 - x + 1$
Rearranging to form a quadratic equation in $x$:
$x^2 - (y+1)x + (y+1) = 0$

2. Solve for $x$ in terms of $y$ using the quadratic formula.

Applying the quadratic formula:
$x = \frac{(y+1) \pm \sqrt{(-(y+1))^2 - 4(1)(y+1)}}{2}$
$x = \frac{y+1 \pm \sqrt{y^2 + 2y + 1 - 4y - 4}}{2}$
$x = \frac{y+1 \pm \sqrt{y^2 - 2y - 3}}{2}$

3. Choose the correct root based on the domain restriction of $f$.

Since the domain of $f$ is $1 < x \le 2$, we must have $x \le 2$.
Substituting our expression for $x$:
$\frac{y+1 \pm \sqrt{y^2 - 2y - 3}}{2} \le 2$
$\pm \sqrt{y^2 - 2y - 3} \le 3 - y$
As we will see in the next step, the range of $f$ is $y \ge 3$, meaning $3-y \le 0$. A positive square root cannot be less than or equal to a negative or zero value (except when exactly $0$). Hence, we must reject the positive root.
Taking the negative root:
$x = \frac{y+1 - \sqrt{y^2 - 2y - 3}}{2}$
Therefore, replacing $y$ with $x$,
$f^{-1}(x) = \frac{x+1 - \sqrt{x^2 - 2x - 3}}{2}$

4. Find the range of $f(x)$ to state the domain of $f^{-1}(x)$.

The domain of $f^{-1}$ is the range of $f$ over $1 < x \le 2$.
Differentiating $f(x)$ to test for monotonicity:
$f'(x) = 1 - \frac{1}{(x-1)^2}$
For $1 < x \le 2$, we have $0 < x-1 \le 1$. Squaring this gives $(x-1)^2 \le 1$.
Thus, $\frac{1}{(x-1)^2} \ge 1$, which implies $f'(x) \le 0$.
The function $f$ is strictly decreasing on this interval.
The minimum value occurs at $x = 2$:
$f(2) = 2 + \frac{1}{2-1} = 3$
As $x \to 1^+$, $\frac{1}{x-1} \to \infty$, so $f(x) \to \infty$.
Thus, the range of $f$ is $[3, \infty)$.
Therefore, the domain of $f^{-1}$ is $x \ge 3$ (or $[3, \infty)$).