Question

Range of $f\left(x\right)={\sin }^{-1}x+{\tan }^{-1}x+{\cos }^{-1}x$ is

• A. $[0,\pi ]$
• B. $[\frac{\pi }{4},\frac{3\pi }{4}]$
• C. $[-\pi ,2\pi ]$
• D. None of these

Answer 1

Answer: (B)

$[\frac{\pi }{4},\frac{3\pi }{4}]$

Given $f\left(x\right)={\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)+{\tan }^{-1}\left(x\right)$

First we will calculate domain of function $f\left(x\right)$

function ${\sin }^{-1}\left(x\right)$ is defined in $x\in [-1,1]$

Similarly function ${\cos }^{-1}\left(x\right)$ is defined in $x\in [-1,1]$

and function ${\tan }^{-1}\left(x\right)$ is defined in $x\in (-\infty ,+\infty )$

So $f\left(x\right)={\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)+{\tan }^{-1}\left(x\right)$ is defined in $x\in [-1,1]$

So $f\left(x\right)={\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)+{\tan }^{-1}\left(x\right)$ is defined in

$x\in [-1,1]$

So $f\left(x\right)=\frac{\pi }{2}+{\tan }^{-1}\left(x\right)$

Now $f\text{'}\left(x\right)=\frac{1}{1+x^2}>0\forall x\in [-1,1]$

So $f\left(x\right)$ is Strictly Increasing function.

So $f(-1)=\frac{\pi }{2}+{\tan }^{-1}(-1)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$

and $f(+1)=\frac{\pi }{2}+{\tan }^{-1}\left(1\right)=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$

So $f\left(x\right)\in [\frac{\pi }{4},\frac{3\pi }{4}]$