For injectivity: if \(f(x_1) = f(x_2)\), then \(\frac{2x_1}{x_1-1} = \frac{2x_2}{x_2-1}\), giving \(2x_1(x_2-1) = 2x_2(x_1-1)\), so \(x_1x_2 - x_1 = x_2x_1 - x_2\), thus \(x_1 = x_2\). So f is injective. For surjectivity to \(\mathbb{R}\): given \(y \in \mathbb{R}\), solve \(\frac{2x}{x-1} = y\), getting \(2x = yx - y\), so \(x(y-2) = -y\), giving \(x = \frac{y}{2-y}\) (provided \(y \neq 2\)). We need to check if this x is in domain A (not a positive integer). For most y, this x won't be a positive integer, but there are some values of y that map to positive integers, which should be excluded from the range. So f is not surjective to all of \(\mathbb{R}\). The answer is D.