Question

Range of the expression $f\left(x\right)=\frac{x^2-1}{x^2+1},x\in R$ is

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

Answer 1

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

Let
$y=\frac{x^2-1}{x^2+1}\Rightarrow (y-1)x^2+(y+1)=0$
$\Rightarrow y\ne 1$ because coefficient of $x^2$ can not be $0.$
$x\in R\Rightarrow \Delta \ge 0\Rightarrow -4(y-1)(y+1)\ge 0\Rightarrow (y^2-1)\le 0$
$\Rightarrow y\in [-1,1]$
$\therefore \text{Range of}y=[-1,1)$