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 :

We have $$f : \mathbb{R} - \{0, 1\} \to \mathbb{R}$$ satisfying $$f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$$. Generate a system of equations by substituting related values.

Put $$x = 2$$: $$f(2) + f\left(\frac{1}{1-2}\right) = 3 \implies f(2) + f(-1) = 3 \quad \cdots (1)$$

Put $$x = -1$$ (note $$\frac{1}{1-(-1)} = \frac{1}{2}$$): $$f(-1) + f\left(\frac{1}{2}\right) = 0 \quad \cdots (2)$$

Put $$x = \frac{1}{2}$$ (note $$\frac{1}{1-\frac{1}{2}} = 2$$): $$f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2} \quad \cdots (3)$$

Solve the system. Adding all three equations:

$$2\left(f(2) + f(-1) + f\left(\frac{1}{2}\right)\right) = \frac{9}{2}$$

$$f(2) + f(-1) + f\left(\frac{1}{2}\right) = \frac{9}{4}$$

From equation (2): $$f(-1) + f\left(\frac{1}{2}\right) = 0$$. Therefore, $$f(2) = \frac{9}{4} - 0 = \frac{9}{4}$$.

The correct answer is Option B: $$\frac{9}{4}$$.