Let $$f : \mathbb{R} \to \mathbb{R}$$ be defined as $$f(x) = x - 1$$ and $$g : R \to \{1, -1\} \to \mathbb{R}$$ be defined as $$g(x) = \frac{x^2}{x^2 - 1}$$. Then the function $$fog$$ is:

We need to determine whether $$f \circ g$$ is one-one and/or onto, where $$f : \mathbb{R} \to \mathbb{R}$$ is defined as $$f(x) = x - 1$$ and $$g : \mathbb{R} \setminus \{1, -1\} \to \mathbb{R}$$ is defined as $$g(x) = \frac{x^2}{x^2 - 1}$$.

First, we find the composite function. The domain of $$f \circ g$$ is the domain of $$g$$, which is $$\mathbb{R} \setminus \{-1, 1\}$$, and the codomain is $$\mathbb{R}$$. Substituting gives $$f(g(x)) = g(x) - 1 = \frac{x^2}{x^2 - 1} - 1 = \frac{x^2 - (x^2 - 1)}{x^2 - 1} = \frac{1}{x^2 - 1}.$$

Next, we check if $$f \circ g$$ is one-one (injective) by testing whether $$f(g(x_1)) = f(g(x_2))$$ implies $$x_1 = x_2$$. Since $$f(g(x)) = \frac{1}{x^2 - 1}$$ depends only on $$x^2$$, we have $$f(g(x)) = f(g(-x))$$ for all $$x$$ in the domain. For example, take $$x = 2$$ and $$x = -2$$: $$f(g(2)) = \frac{1}{4 - 1} = \frac{1}{3}, \quad f(g(-2)) = \frac{1}{4 - 1} = \frac{1}{3}.$$ Since $$2 \neq -2$$ but $$f(g(2)) = f(g(-2))$$, the function is not one-one.

Now, we check if $$f \circ g$$ is onto (surjective). We need to determine whether every real number $$y$$ is in the range of $$h(x) = \frac{1}{x^2 - 1}$$. Setting $$y = \frac{1}{x^2 - 1}$$ gives $$x^2 - 1 = \frac{1}{y}, \quad x^2 = 1 + \frac{1}{y}.$$ For a real solution to exist, we need $$y \neq 0$$ and $$1 + \frac{1}{y} \geq 0$$, i.e., $$\frac{y + 1}{y} \geq 0.$$ Therefore, such values of $$y$$ satisfy $$y > 0$$ or $$y \leq -1$$ (and additionally $$x \neq \pm 1$$, which means $$x^2 \neq 1$$, i.e., $$\frac{1}{y} \neq 0$$, which is always true).

Case-by-case analysis of values NOT in the range:

Case 1: $$y = 0$$. Since $$\frac{1}{x^2 - 1} = 0$$ has no solution, $$y = 0$$ is not in the range.

Case 2: $$-1 < y < 0$$. Here $$\frac{y + 1}{y} < 0$$, so $$x^2 < 0$$, which gives no real solution. These values are not in the range. For instance, $$y = -\frac{1}{2}$$ gives $$x^2 = 1 - 2 = -1 < 0$$, which has no real solution.

Since not every real number is achieved (the interval $$(-1, 0]$$ is not in the range), the function is not onto.

Therefore, $$f \circ g$$ is neither one-one nor onto.

The answer is Option D: Neither one-one nor onto.