Question

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

• A. $[0,1]$
• B. $[0,1)$
• C. $[-1,1]$
• D. $[1,\infty )$

Answer 1

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

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

Now, $x^2\in [0,\infty ]\therefore \frac{1}{x^2+1}\in (0,1]$

$\therefore f\left(x\right)=\sqrt{1-(0,1]}$

$\in [0,1)$