Question

Which of the following is/are true regarding the range of $f\left(x\right)={\tan }^{-1}\left[\frac{2}{\pi }(2{\tan }^{-1}x-{\sin }^{-1}x+{\cot }^{-1}x-{\cos }^{-1}x)\right]$

• A. Range contains only one integer
• B. Range contains more than $2$ integers
• C. Range contains the element $0$
• D. Range contains the element $1$ and $2$

Answer 1

Answer: (A), (C)

Range contains the element $0$

${\tan }^{-1}\left[\frac{2}{\pi }(2{\tan }^{-1}x-{\sin }^{-1}x+{\cot }^{-1}x-{\cos }^{-1}x)\right]$

$={\tan }^{-1}\left[\frac{2}{\pi }({\tan }^{-1}x+{\tan }^{-1}x+{\cot }^{-1}x-({\sin }^{-1}x+{\cos }^{-1}x\left)\right)\right]$
$[\because {\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}\text{and}{\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}]$
$={\tan }^{-1}\left(\frac{2}{\pi }{\tan }^{-1}x\right)$
As we have the range of ${\tan }^{-1}x$ is $(-\frac{\pi }{4},\frac{\pi }{4})$ since the domain of the given function is $x\in (-1,1)$
$-\frac{\pi }{4}<{\tan }^{-1}x<\frac{\pi }{4}$

$\Rightarrow -\frac{1}{2}<\frac{2}{\pi }{\tan }^{-1}x<\frac{1}{2}$
$\Rightarrow -{\tan }^{-1}\frac{1}{2}<{\tan }^{-1}\left(\frac{2}{\pi }{\tan }^{-1}x\right)<{\tan }^{-1}\frac{1}{2}$
We know ${\tan }^{-1}$ is an increasing function. Also ${\tan }^{-1}\sqrt{3}$ is approximately equal to 1.
And ${\tan }^{-1}\sqrt{3}>{\tan }^{-1}\frac{1}{2}$
Clearly, we have $1>{\tan }^{-1}\frac{1}{2}$
$\therefore$ We have ${\tan }^{-1}\frac{1}{2}<1$
So, the range $f\left(x\right)$ has only $1$ integer value and it is $0$.