JEE Main 2023 (Online) 13th April Evening ShiftQuestion 1
The range of $f(x)=4\sin^{-1}\left(\frac{x^{2}}{x^{2}+1}\right)$ is
Inverse Trigonometric Functions
Solution
Let $u=\frac{x^2}{x^2+1}=1-\frac1{x^2+1}$; for real $x$, $u\in[0,1)$. $\sin^{-1}u\in[0,\frac\pi2)$, so $f=4\sin^{-1}u\in[0,2\pi)$. Option (D).