Question

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

• A. $[-\frac{1}{2},0)\cup (0,\frac{1}{2}]$
• B. $[-1,0)\cup (0,1]$
• C. $\left|y\right|\ge 1$
• D. $[-\frac{1}{2},\frac{1}{2}]$

Answer 1

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

$f\left(x\right)=\frac{x}{1+x^2}=y\left(say\right)$
$\Rightarrow x^{2}y-x+y=0$
$\Rightarrow D\ge 0$ for $xϵR$
$\Rightarrow 1-4y^2\ge 0\Rightarrow 4y^2\le 1$
$\Rightarrow y^2\le \frac{1}{4}\Rightarrow \left|y\right|\le \frac{1}{2}$
$\therefore y\in [\frac{-1}{2},\frac{1}{2}]$