Question

If $f:(-1,1)\to B$is a function defined by $f\left(x\right)=ta{n}^{-1}\left[\frac{2x}{(1-x^2)}\right]$ then $f$ is both one-one and onto when $B$ is the interval

• A. $\left[\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$
• B. $\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$
• C. $\left[0,\frac{\mathrm{\pi }}{2}\right]$
• D. $\left[0,-\frac{\mathrm{\pi }}{2}\right]$

Answer 1

The correct option is B

$\left[-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right]$

Explanation for the correct option:

Step 1. Evaluating the given conditions:

Given that $f:(-1,1)\to B$such that

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

Now, Put $x=\tan y$we get

$f(\tan y)={\tan }^{-1}\left[\frac{2\tan y}{1–{\tan }^{2}y}\right]$

$\begin{array}{rcl}f(\tan y)& =& {\tan }^{-1}\left[\frac{2\tan y}{1–{\tan }^{2}y}\right]\\ & =& {\tan }^{-1}(\tan 2y)\\ & =& 2y\\ & =& 2{\tan }^{-1}x\\ f\left(x\right)& =& 2{\tan }^{-1}x\end{array}$ $\left[\mathbf{\because }\mathbf{tan}\mathbf{}\mathbf{2}\mathit{\theta }\mathbf{=}\mathbf{}\frac{\mathbf{2}\mathbf{}\mathbf{tan}\mathbf{}\mathbf{\theta }}{\mathbf{1}\mathbf{–}{\mathbf{tan}}^{\mathbf{2}}\mathbf{\theta }}\right]$

Step 2. Finding the value of interval $B$:

Since ${\tan }^{-1}x$ is strictly increasing function in $(-1,1)$

If $\begin{array}{rcl}x& =& -1,\end{array}$

$\begin{array}{rcl}f\left(x\right)& =& 2{\tan }^{-1}(-1)\\ & =& 2\times (-\frac{\pi }{4})\\ & =& -\frac{\pi }{2}\end{array}$

If $x=1,$

$\begin{array}{rcl}f\left(x\right)& =& 2{\tan }^{-1}\left(1\right)\\ & =& 2\times \left(\frac{\pi }{4}\right)\\ & =& \frac{\pi }{2}\end{array}$

The function is Onto.

$B=$ range

$B=\mathbf{(}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{,}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{)}$

Hence, Option ‘B’ is Correct.