Question

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 $f\left(x\right)=y$
$\Rightarrow \frac{x}{1+x^2}=y\Rightarrow x^{2}y-x+y=0$
$x=\frac{1\pm \sqrt{1-4y^2}}{2y}$
Now $\frac{1\pm \sqrt{1-4y^2}}{2y}$ is a real number if
$1-4y^2\ge 0$ and $y\ne 0$
$\Rightarrow 4y^2-1\le 0$ $y\ne 0$
$\Rightarrow (y^2-\frac{1}{4})\le 0$
$y\ne 0$
$\Rightarrow (y-\frac{1}{2})(y+\frac{1}{2})\le 0$
$\Rightarrow -\frac{1}{2}\le y\le \frac{1}{2}$ and $y\ne 0$
$\Rightarrow y\in [-\frac{1}{2},0)\cup (0,\frac{1}{2}]$
for $x=0,y=0$ $\therefore$ range is $[-\frac{1}{2},\frac{1}{2}]$