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

We have $$f(x) = 2x - 1$$ and $$g(x) = \frac{x - \frac{1}{2}}{x - 1}$$, with domain of $$g$$ being $$\mathbb{R} - \{1\}$$.

The composition is $$f(g(x)) = 2 \cdot \frac{x - \frac{1}{2}}{x - 1} - 1 = \frac{2x - 1}{x - 1} - 1 = \frac{2x - 1 - (x - 1)}{x - 1} = \frac{x}{x - 1}$$.

The domain of $$f(g(x))$$ is $$\mathbb{R} - \{1\}$$.

To check one-one: suppose $$\frac{x_1}{x_1 - 1} = \frac{x_2}{x_2 - 1}$$. Then $$x_1(x_2 - 1) = x_2(x_1 - 1)$$, giving $$x_1 x_2 - x_1 = x_1 x_2 - x_2$$, so $$x_1 = x_2$$. The function is one-one.

To check onto: let $$\frac{x}{x - 1} = y$$. Then $$x = y(x - 1) = yx - y$$, so $$x(1 - y) = -y$$, giving $$x = \frac{y}{y - 1}$$. This requires $$y \neq 1$$. So the range is $$\mathbb{R} - \{1\}$$, which is not all of $$\mathbb{R}$$.

Therefore $$f(g(x))$$ is one-one but not onto.

Hence, the correct answer is Option B.