Question

Let $f\left(x\right)=\frac{x}{1+x^2}$. Then range of $f$ is

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

Answer 1

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

$f\left(x\right)=\frac{x}{1+x^2}$
Denominator is $1+x^2$.
$1+x^2\ne 0$ for any real value of $x$.
Hence, domain of $f$ is $R$.

Now, let $y=f\left(x\right)$
$\therefore y=\frac{x}{1+x^2}\Rightarrow y{x}^2-x+y=0\Rightarrow x=\frac{1\pm \sqrt{1-4y^2}}{2y}$
For $x$ to be defined,
$1-4y^2\ge 0,y\ne 0$
$\Rightarrow y\in [-\frac{1}{2},\frac{1}{2}]-\left\{0\right\}$
But for $x=0,y=0$
$\therefore R\left(f\right)=[-\frac{1}{2},\frac{1}{2}]$