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Question: If \(f:\left( -1,1 \right)\to B\) is a function defined by \(f\left( x \right)={{\tan }^{-1}}\left[ ...

If f:(−1,1)→Bf:\left( -1,1 \right)\to B is a function defined by f(x)=tan⁡−1[2x(1−x2)]f\left( x \right)={{\tan }^{-1}}\left[ \dfrac{2x}{\left( 1-{{x}^{2}} \right)} \right] , then ff is both one-one and onto when BB is the interval:

  1. [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]
  2. [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]
  3. [0,π2]\left[ 0,\dfrac{\pi }{2} \right]
  4. [0,π2]\left[ 0,\dfrac{\pi }{2} \right]
Explanation

Solution

Here in this question we have been given that f:(−1,1)→Bf:\left( -1,1 \right)\to B is a function defined by f(x)=tan⁡−1[2x(1−x2)]f\left( x \right)={{\tan }^{-1}}\left[ \dfrac{2x}{\left( 1-{{x}^{2}} \right)} \right] and ff is both one-one and onto. We have been asked to find the interval BB. Let us assume x=tan⁡θx=\tan \theta and simplify f(x)f\left( x \right) then evaluate the interval.

Complete step-by-step solution:
Now considering from the question, we have been given that f:(−1,1)→Bf:\left( -1,1 \right)\to B is a function defined by f(x)=tan⁡−1[2x(1−x2)]f\left( x \right)={{\tan }^{-1}}\left[ \dfrac{2x}{\left( 1-{{x}^{2}} \right)} \right] andff is both one-one and onto.
We have been asked to find the interval BB.
Let us assume x=tan⁡θx=\tan \theta .
By using this we will have tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } .
Therefore we can say that f(x)=2θf\left( x \right)=2\theta . Hence we can simply say that f(x)=2tan⁡−1xf\left( x \right)=2{{\tan }^{-1}}x .
Now from the definition of f(x)f\left( x \right) we know that it is one-one and onto and −1<x<1-1 < x < 1 .
Therefore
2tan⁡−1(−1)≤f(x)≤2tan⁡−1(1) ⇒−π2≤f(x)≤π2 \begin{aligned} & 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{aligned} .
Therefore we can conclude that when it is given that f:(−1,1)→Bf:\left( -1,1 \right)\to B is a function defined by f(x)=tan⁡−1[2x(1−x2)]f\left( x \right)={{\tan }^{-1}}\left[ \dfrac{2x}{\left( 1-{{x}^{2}} \right)} \right] and ff is both one-one and onto then the interval BB is [−π2,π2]\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.