Question:
Let $f: R \rightarrow R$ be defined by $f(x)=\frac{x}{1+x^{2}}, x \in R$. Then
the range of $f$ is :
Correct Option: 1
Solution:
$f(x)=\frac{x}{1+x^{2}}, x \in R$
Let, $y=\frac{x}{1+x^{2}}$
$\Rightarrow \quad y x^{2}-x+y=0 \quad \Rightarrow \quad x=\frac{1 \pm \sqrt{1-4 y^{2}}}{2}$
$\Rightarrow \quad 1-4 y^{2} \geq 0$
$\Rightarrow \quad 1 \geq 4 y^{2}$
$\Rightarrow \quad|y| \leq \frac{1}{2}$
$\Rightarrow \quad-\frac{1}{2} \leq y \leq \frac{1}{2}$
$\Rightarrow$ The range of $f$ is $\left[-\frac{1}{2}, \frac{1}{2}\right]$.
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