# if

Question:

If $y=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ for all $x$, then $y$ lies in the interval _________________.

Solution:

We know

$2 \tan ^{-1} x= \begin{cases}\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), & -1 \leq x \leq 1 \\ \pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), & x>1 \\ -\pi-\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right), & x<-1\end{cases}$

$\therefore y=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)= \begin{cases}4 \tan ^{-1} x, & -1 \leq x \leq 1 \\ \pi, & x>1 \\ -\pi, & x<-1\end{cases}$

For $-1 \leq x \leq 1$

$-\frac{\pi}{4} \leq \tan ^{-1} x \leq \frac{\pi}{4}$

$\Rightarrow-\pi \leq 4 \tan ^{-1} x \leq \pi$

$\Rightarrow-\pi \leq y \leq \pi$                      ....(1)

For $x>1, y=\pi$                                             ...(2)

For $x<-1, y=-\pi$                                          ...(3)

From (1), (2) and (3), we get

$y \in[-\pi, \pi]$, for all $x \in \mathrm{R}$

Thus, the range of $y$ is $[-\pi, \pi]$.

If $y=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ for all $x$, then $y$ lies in the interval $[-\pi, \pi]$