Question

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

Answer 1

$\begin{array}{c}Here\\ x^2\ge 0\\ x^2+1>0\end{array}$

Since, the denominator is always greater than the numerator for all values other than 0 and at

$\begin{array}{c}0,f\left(x\right)=0\\ \therefore Range=\left[0,1\right)\end{array}$