Question

If $f:\left( -1,1 \right)\to B$ is a function defined by $f\left( x \right)={{\tan }^{-1}}\left[ \dfrac{2x}{\left( 1-{{x}^{2}} \right)} \right]$ , then $f$ is both one-one and onto when $B$ is the interval:
1) $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$
2) $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$
3) $\left[ 0,\dfrac{\pi }{2} \right]$
4) $\left[ 0,\dfrac{\pi }{2} \right]$

Answer 1

Hint: Here in this question we have been given that $f:\left( -1,1 \right)\to B$ is a function defined by $f\left( x \right)={{\tan }^{-1}}\left[ \dfrac{2x}{\left( 1-{{x}^{2}} \right)} \right]$ and $f$ is both one-one and onto. We have been asked to find the interval $B$. Let us assume $x=\tan \theta $ and simplify $f\left( x \right)$ then evaluate the interval.

Complete step-by-step solution:
Now considering from the question, we have been given that $f:\left( -1,1 \right)\to B$ is a function defined by $f\left( x \right)={{\tan }^{-1}}\left[ \dfrac{2x}{\left( 1-{{x}^{2}} \right)} \right]$ and$f$ is both one-one and onto.
We have been asked to find the interval $B$.
Let us assume $x=\tan \theta $.
By using this we will have $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ .
Therefore we can say that $f\left( x \right)=2\theta $ . Hence we can simply say that $f\left( x \right)=2{{\tan }^{-1}}x$ .
Now from the definition of $f\left( x \right)$ we know that it is one-one and onto and $-1 < x < 1$ .
Therefore
$\begin{align}
  & 2{{\tan }^{-1}}\left( -1 \right) \le f\left( x \right) \le 2{{\tan }^{-1}}\left( 1 \right) \\
 & \Rightarrow \dfrac{-\pi }{2} \le f\left( x \right) \le \dfrac{\pi }{2} \\
\end{align}$ .
Therefore we can conclude that when it is given that $f:\left( -1,1 \right)\to B$ is a function defined by $f\left( x \right)={{\tan }^{-1}}\left[ \dfrac{2x}{\left( 1-{{x}^{2}} \right)} \right]$ and $f$ is both one-one and onto then the interval $B$ is $\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]$ .
Hence we will mark the option “2” as correct.

Note: During the process of answering questions of this type we should be sure with the concepts that we are going to apply in between the steps. This is a very simple question but the instant we look at it, it seems to be a complex one because the domain has many values and we can’t verify each and every value of it.