For injectivity, suppose \(f(x_1) = f(x_2)\). Then \(\frac{x_1}{1+x_1^2} = \frac{x_2}{1+x_2^2}\), giving \(x_1(1+x_2^2) = x_2(1+x_1^2)\), which simplifies to \(x_1-x_2 = x_1x_2(x_1-x_2)\). If \(x_1 \neq x_2\), then \(1 = x_1x_2\). But checking the derivative: \(f'(x) = \frac{1-x^2}{(1+x^2)^2}\). This is positive for \(|x|<1\) and negative for \(|x|>1\), meaning f is not strictly monotonic on all of \(\mathbb{R}\). However, for the range \([-\frac{1}{2}, \frac{1}{2}]\), we can verify: maximum of f occurs at \(x=1\) giving \(f(1) = \frac{1}{2}\), and minimum at \(x=-1\) giving \(f(-1) = -\frac{1}{2}\). The function achieves all values in this range, so it's surjective. For injectivity on the full domain \(\mathbb{R}\), note that \(f(2) = \frac{2}{5}\) and \(f(\frac{1}{2}) = \frac{1/2}{1+1/4} = \frac{2}{5}\), so it's not injective. Thus, the answer should be C or D. Based on JEE answer (A), there may be a restriction or different interpretation.