Question

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

• A. $R-\left\{1\right]\}$
• B. $R^{+}\cup \left\{1\right]\}$
• C. $[0,1]$
• D. $Noneofthese$

Answer 1

The correct option is C $[0,1]$

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

$\Rightarrow y=\frac{x^2}{1+x^2}$ Range

$\Rightarrow (y-1)n^2+y=0$ $\to$ This equation must have real roots

So, $D\ge 0$

$-4(y-1)\left(y\right)\ge 0$

$\Rightarrow y\in [0,1]$.