Question

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

• A. $(0,1)$
• B. $[0,1)$
• C. $R$
• D. $[0,1]$

Answer 1

The correct option is B $[0,1)$

$\text{Let}y=\frac{x^2}{1+x^2}$
$\Rightarrow x^2-y-y{x}^2=0$
$\Rightarrow (1-y)x^2=y$
$\Rightarrow x=\sqrt{\frac{y}{1-y}}$
For $x$ to be defined,
$\frac{y}{1-y}\ge 0\text{and}y\ne 1$
$\Rightarrow y\in [0,1)$

Alternate:
$\text{As}0\le x^2<1+x^2,$
$0\le \frac{x^2}{1+x^2}<\frac{1+x^2}{1+x^2}$
$\Rightarrow 0\le f\left(x\right)<1$