Question

Let $f:\left(-1,1\right)\to B$ be a function defined by $f\left(x\right)={\sin }^{-1}\left[\frac{2x}{\left(1+x^2\right)}\right]$ then $f$ is both one-one and onto then $B$is in the interval

• A. $\left(-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$
• B. $\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$
• C. $\left[-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right]$
• D. $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Answer 1

The correct option is B

$\left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)$

Explanation for the correct option:

Finding the interval:

Consider the given function,

$f\left(x\right)={\sin }^{-1}\left[\frac{2x}{1+x^2}\right]$

Let us assume that,

$x=\tan \theta ,x\in \left(-1,1\right)$

$\theta \in \left(-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$

Substitute the above value in above function

$f\left(x\right)={\sin }^{-1}\left[\frac{2\tan \theta }{1+{\tan }^2\theta }\right]$

we know that,

$\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$

Then,

$$\begin{gathered} f\left(x\right)={\sin }^{-1}\left(\sin 2\theta \right) \\ f\left(x\right)=2\theta \end{gathered}$$

Then, $\theta \in \left(-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right)$

Since,

$$\begin{gathered} f\left(x\right)\in 2\left(-\frac{\mathrm{\pi }}{4},\frac{\mathrm{\pi }}{4}\right) \\ f\left(x\right)\in \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$

Hence, the correct answer is Option (B).