Question

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

• A. $[-\frac{1}{2},\frac{1}{2}]$
• B. $[0,1]$
• C. $[0,\frac{1}{2}]$
• D. $[0,\infty )$

Answer 1

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

$f\left(x\right)=\frac{x}{1+x^2}$

$f^{\prime }\left(x\right)=\frac{(1+x^2)\times 1-2x\times x}{(1+x^2{)}^2}$

$=\frac{1-x^2}{(1+x^2{)}^2}$

$=\frac{(1+x)(1-x)}{(1+x^2{)}^2}$

$\Rightarrow f^{\prime }\left(x\right)=0$

At $x=\pm 1$
So maxima will be at $1$ & minima will be at $-1$
$f\left(1\right)=\frac{1}{2}$
$f(-1)=\frac{-1}{2}$
Range of $f\left(x\right)$ is $[\frac{-1}{2},\frac{1}{2}]$