Question

Consider
$f\left(x\right)={\sin }^{-1}\left(\sec \left({\tan }^{-1}x\right)\right)+{\cos }^{-1}\left(\text{cosec}\left({\cot }^{-1}x\right)\right)$
Statement$-1:$ Domain of $f\left(x\right)$ is a singleton set.

Reason
Statement$-2:$ Range of the function $f\left(x\right)$ is a singleton set.

• A. Statement$-1$ is true, Statement$-2$ is true and Statement$-2$ is correct explantion for Statement$-1.$
• B. Statement$-1$ is true, Statement$-2$ is true and Statement$-2$ is NOT the correct explanation for Statement$-1.$
• C. Statement$-1$ is true, Statement$-2$ is false.
• D. Statement$-1$ is false, Statement$-2$ is true.

Answer 1

The correct option is B Statement$-1$ is true, Statement$-2$ is true and Statement$-2$ is NOT the correct explanation for Statement$-1.$

$f\left(x\right)={\sin }^{-1}\left(\sec \left({\tan }^{-1}\right(x\left)\right)\right)+{\cos }^{-1}\left(\text{cosec}\left({\cot }^{-1}x\right)\right)\Rightarrow f\left(x\right)={\sin }^{-1}\left(\sqrt{1+{\tan }^2\left({\tan }^{-1}\right(x\left)\right)}\right)+{\cos }^{-1}\left(\sqrt{1+{\cot }^2{\cot }^{-1}\left(x\right)}\right)\Rightarrow f\left(x\right)={\sin }^{-1}\left(\sqrt{x^2+1}\right)+{\cos }^{-1}\left(\sqrt{x^2+1}\right)$
Hence,
$-1\le \sqrt{x^2+1}\le 1\Rightarrow x^2+1\le 1\Rightarrow x^2\le 0\therefore x=0$

Hence, domain is a singleton set and
$f\left(x\right)=\frac{\pi }{2}$
Therefore, the range is also a singleton set.

Both statements are correct but second statement is not the correct explanation.