Explanation

(c)
Let $y = \frac{x}{1 + x^{2}}$
$\Rightarrow x^{2}y - x + y = 0$
For $x$ to be real
$\ \ \ \ 1 - 4y^{2} \geq 0$
$\Rightarrow \ \ (1 - 2y)(1 + 2y) \geq 0$
$\Rightarrow \ \left( \frac{1}{2} - y \right)\left( \frac{1}{2} + y \right) \geq 0$
$\Rightarrow \ - \frac{1}{2} \leq y \leq \frac{1}{2}$
$\therefore\ \ \ y = f(x) \in \left\lbrack - \frac{1}{2},\ \frac{1}{2} \right\rbrack$