Question

Range of the function $y =\sin^{-1} \left(\frac{x^{2}}{1+x^{2}}\right)$ is

• A. $\left(0, \frac{\pi}{2}\right)$
• B. $\left[0, \frac{\pi}{2}\right)$
• C. $\left(0, \frac{\pi}{2}\right]$
• D. $\left[0, \frac{\pi}{2}\right]$

Answer 1

Answer: (B)

$\left[0, \frac{\pi}{2}\right)$

Answer 2

We have the function $y=\sin ^{-1}\left(\frac{x^{2}}{1+x^{2}}\right)$ For $y$ to be defined $\left|\frac{x^{2}}{1+x^{2}}\right|<1$ which is true for all $x \in R$ Now, $y=\sin ^{-1}\left(\frac{x^{2}}{1+x^{2}}\right)$ $\Rightarrow \frac{x^{2}}{1+x^{2}}=\sin y$ $\Rightarrow x=\sqrt{\frac{\sin y}{1-\sin y}}$ For the existance of $x$ $\sin y \geq 0$ and $1-\sin y>0$ $\Rightarrow 0 \leq \sin y<1$ $\Rightarrow 0 \leq y