Let $$f : R - \{0, 1\} \to R$$ be a function such that $$f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$$. Then $$f(2)$$ is equal to:

$$f : \mathbb{R} - \{0, 1\} \to \mathbb{R}$$ such that $$f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$$. Find $$f(2)$$.

Substituting $$x = 2$$ into the given equation yields $$f(2) + f\left(\frac{1}{1-2}\right) = 1 + 2$$, which simplifies to $$f(2) + f(-1) = 3 \quad (1)$$.

Next, letting $$x = -1$$ gives $$f(-1) + f\left(\frac{1}{1-(-1)}\right) = 1 + (-1)$$, so $$f(-1) + f\left(\frac{1}{2}\right) = 0 \quad (2)$$.

Finally, putting $$x = \frac{1}{2}$$ leads to $$f\left(\frac{1}{2}\right) + f\left(\frac{1}{1-\frac{1}{2}}\right) = 1 + \frac{1}{2}$$, hence $$f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2} \quad (3)$$.

From (1) we have $$f(-1) = 3 - f(2)$$. Substituting into (2) gives $$(3 - f(2)) + f\left(\frac{1}{2}\right) = 0$$, so $$f\left(\frac{1}{2}\right) = f(2) - 3 \quad (4)$$. Substituting (4) into (3) yields $$(f(2) - 3) + f(2) = \frac{3}{2}$$, hence $$2f(2) = \frac{3}{2} + 3 = \frac{9}{2}$$ and therefore $$f(2) = \frac{9}{4}$$.

Answer: Option B $$\left(\frac{9}{4}\right)$$