The function f: R → [- (1)/(2), (1)/(2)] defined as f(x) = x 1 + x...

Let, $y = \dfrac{x}{1 + {x}^{2}}$

$⇒ y + y{x}^{2} - x = 0$

$⇒ {x}^{2}y - x + y = 0$

$⇒ x = \dfrac{+ 1 \pm \sqrt{1 - 4{y}^{2}}}{2y}$

$⇒ 1 - 4{y}^{2} \ge 0$

$⇒ 4{y}^{2} \le 1$

$⇒$ So, $y \in \left[- \dfrac{1}{2}, \dfrac{1}{2}\right]$

Here, Co-Domain=Range

Hence, the function is surjective.

Now, for injective function

$⇒ \dfrac{{x}_{1}}{1 + {x}_{1}^{2}} = \dfrac{{x}_{2}}{1 + {x}_{2}^{2}}$

$⇒$${x}_{1} + {x}_{2}{x}_{2}^{2} - {x}_{2} - {x}_{2}{x}_{1}^{2} = 0$

$⇒$$\left({x}_{1} - {x}_{2}\right) + {x}_{1}{x}_{2}\left({x}_{2} - {x}_{1}\right) = 0$

$⇒$$\left({x}_{2} - {x}_{1}\right)\left({x}_{1}{x}_{2} - 1\right) = 0$

$⇒$${x}_{2} = {x}_{1}$ or ${x}_{1} = \dfrac{1}{{x}_{2}}$

Hence, the function is not injective.