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

Let, $y=\frac{x}{1+{x}^{2}}$y=1+x2x​

$$\Rightarrow y+y{x}^{2}-x=0$$⇒y+yx2−x=0

$$\Rightarrow {x}^{2}y-x+y=0$$⇒x2y−x+y=0

$$\Rightarrow x=\frac{+1\pm \sqrt{1-4{y}^{2}}}{2y}$$⇒x=2y+1±1−4y2​​

$$\Rightarrow 1-4{y}^{2} \geq 0$$⇒1−4y2≥0

$$\Rightarrow 4{y}^{2} \leq 1$$⇒4y2≤1

$\Rightarrow$⇒ So, $y\in [-\frac{1}{2},\frac{1}{2}]$y∈[−21​,21​]

Here, Co-Domain$=$=Range

Hence, the function is surjective.

Now, for injective function

$$\Rightarrow \frac{{x}_{1}}{1+{x}_{1}^{2}}=\frac{{x}_{2}}{1+{x}_{2}^{2}}$$⇒1+x12​x1​​=1+x22​x2​​

$\Rightarrow$⇒${x}_{1}+{x}_{2}{x}_{2}^{2}-{x}_{2}-{x}_{2}{x}_{1}^{2}=0$x1​+x2​x22​−x2​−x2​x12​=0

$\Rightarrow$⇒$({x}_{1}-{x}_{2})+{x}_{1}{x}_{2}({x}_{2}-{x}_{1})=0$(x1​−x2​)+x1​x2​(x2​−x1​)=0

$\Rightarrow$⇒$({x}_{2}-{x}_{1}) ({x}_{1}{x}_{2}-1)=0$(x2​−x1​)(x1​x2​−1)=0

$\Rightarrow$⇒${x}_{2}={x}_{1}$x2​=x1​ or ${x}_{1}=\frac{1}{{x}_{2}}$x1​=x2​1​

Hence, the function is not injective.