Let $$f : A \to B$$ be a function defined as $$f(x) = \frac{x-1}{x-2}$$, where $$A = R - \{2\}$$ and $$B = R - \{1\}$$. Then f is:

We are given the function $$f : A \to B$$ with the definition $$f(x)=\dfrac{x-1}{x-2}$$, where the domain is $$A=\mathbb R-\{2\}$$ and the codomain is $$B=\mathbb R-\{1\}$$. Our task is to decide whether the function is invertible and, if it is, to compute its inverse explicitly.

First, we check injectivity. We take two arbitrary points $$x_1, x_2\in A$$ and assume $$f(x_1)=f(x_2)$$. That equality means

$$\dfrac{x_1-1}{x_1-2}=\dfrac{x_2-1}{x_2-2}.$$

Because both denominators are non-zero (each $$x_i\neq2$$), we cross-multiply:

$$ (x_1-1)(x_2-2)=(x_2-1)(x_1-2). $$

We expand each side completely:

$$ x_1x_2-2x_1-x_2+2 = x_1x_2-2x_2-x_1+2. $$

Now we cancel the common term $$x_1x_2$$ on both sides and gather the remaining terms. Subtract the right side from the left side:

$$ (-2x_1 - x_2 + 2) - (-2x_2 - x_1 + 2)=0. $$

Simplifying the expression inside the brackets, we get

$$ -2x_1 - x_2 + 2 + 2x_2 + x_1 - 2 = 0. $$

The constants $$+2$$ and $$-2$$ cancel, leaving

$$ -2x_1 - x_2 + 2x_2 + x_1 = 0. $$

Combining like terms, we have

$$ (-2x_1 + x_1) + (-x_2 + 2x_2) = 0, $$

which is

$$ -x_1 + x_2 = 0. $$

Thus $$x_2 = x_1$$. Because the only way $$f(x_1) = f(x_2)$$ occurs is when the two inputs are identical, the function is injective.

Next, we verify surjectivity onto $$B$$. Let an arbitrary $$y \in B$$ be given. We must find an $$x \in A$$ satisfying $$f(x)=y$$, that is,

$$ y = \dfrac{x-1}{x-2}. $$

We now solve this equation for $$x$$. First we write the relation in the equivalent cross-multiplied form:

$$ y(x-2)=x-1. $$

Expanding the left side gives

$$ yx-2y = x-1. $$

We want to gather all terms containing $$x$$ on one side, so we subtract $$x$$ from both sides:

$$ yx - x - 2y = -1. $$

We factor out $$x$$ from the two terms containing it, using the distributive law $$ax+bx=(a+b)x$$:

$$ x(y-1) - 2y = -1. $$

Now we add $$2y$$ to both sides, yielding

$$ x(y-1) = 2y - 1. $$

Because every $$y \in B$$ satisfies $$y \neq 1$$, the coefficient $$(y-1)$$ is never zero, so we can safely divide by it. Doing so gives us the explicit solution

$$ x = \dfrac{2y-1}{y-1}. $$

The obtained expression is defined for every $$y \neq 1$$, and that is precisely the set $$B$$. We also observe that the value $$x=\dfrac{2y-1}{y-1}$$ can never equal $$2$$, for if it did we would have

$$ 2 = \dfrac{2y-1}{y-1}. $$

Cross-multiplying that hypothetical equality yields $$2y-2=2y-1,$$ which simplifies to $$-2=-1,$$ an impossibility. Hence $$x \neq 2$$, so $$x \in A$$. Therefore, for every $$y\in B$$ we have located an $$x\in A$$ satisfying $$f(x)=y$$, and surjectivity is proved.

Having shown both injectivity and surjectivity, we conclude that $$f$$ is bijective, i.e., invertible. The algebraic manipulation above already produced the inverse correspondence. We merely rewrite it in the standard inverse notation. By definition of an inverse, if $$y=f(x)$$, then $$x=f^{-1}(y)$$. The equation we derived was

$$ x = \dfrac{2y-1}{y-1}. $$

So the inverse function is

$$ f^{-1}(y)=\dfrac{2y-1}{y-1}, \qquad y \in B = \mathbb R-\{1\}. $$

This formula matches exactly what is written in Option D. No other option provides the correct inverse expression, and we have rigorously demonstrated both the bijectivity and the precise inverse.

Hence, the correct answer is Option D.