Question

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

• A. $D\left(f\right)=R,R\left(f\right)=R-[-1,1]$
• B. $D\left(f\right)=R,R\left(f\right)=R-(-1,1)$
• C. $D=R,R\left(f\right)=[-1,1]$
• D. $D=R,R\left(f\right)=(-1,1)$

Answer 1

The correct option is C $D=R,R\left(f\right)=[-1,1]$

$y=\frac{2x}{1+x^2}$
Clearly, $f\left(x\right)$ is defined for all real values of $x$
Hence, domain of $f$ is R.
Range:
$y=\frac{2x}{1+x^2}$
$y+x^{2}y=2x$
$x^{2}y-2x+y=0$
Since, $x$ is real , so $D\ge 0$
$\Rightarrow 4-4y^2\ge 0$
$\Rightarrow y^2-1\le 0$
$\Rightarrow y\in [-1,1]$
Hence, range is $[-1,1]$