Question

The range of the function $f\left(x\right)=\frac{x}{1+x^2}$ is

• A. $-\infty ,\infty$
• B. $[-1,1]$
• C. $[-\frac{1}{2},\frac{1}{2}]$
• D. $[-\sqrt{2},\sqrt{2}]$

Answer 1

The correct option is C $[-\frac{1}{2},\frac{1}{2}]$

Let $y=\frac{x}{1+x^2}$
$\Rightarrow x^{2}y-x+y=0$
For $x$ to be real, $1-4y^2\ge 0$
$\Rightarrow (1-2y)(1+2y)\ge 0$
$\Rightarrow (\frac{1}{2}-y)(\frac{1}{2}+y)\ge 0$
$\Rightarrow -\frac{1}{2}\le y\le \frac{1}{2}$
$\therefore y=f\left(x\right)ϵ[-\frac{1}{2},\frac{1}{2}]$