Question

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

Answer 1

We know

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

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

For $-1\le x\le 1$,

$$\begin{gathered} -\frac{\mathrm{\pi }}{4}\le {\tan }^{-1}x\le \frac{\mathrm{\pi }}{4} \\ \Rightarrow -\mathrm{\pi }\le 4{\tan }^{-1}x\le \mathrm{\pi } \\ \Rightarrow -\mathrm{\pi }\le y\le \mathrm{\pi }.....\left(1\right) \end{gathered}$$

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

For x < −1, y = $-\mathrm{\pi }$ .....(3)

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

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

Thus, the range of y is $\left[-\mathrm{\pi },\mathrm{\pi }\right]$.

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