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)

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 $x$ $\sin y\ge 0$ and $1-\sin y>0$
$\Rightarrow 0\le \sin y<1$
$\Rightarrow 0\le y<\frac{\pi }{2}$
Thus, range of the given function is $\left[0,\frac{\pi }{2}\right)$.