Question

If $f\left(x\right)={\sin }^{-1}\left(\ln \left[x\right]\right)+\ln \left({\sin }^{-1}\left[x\right]\right)$ (where $[.]$ denotes the greatest integer function), then

• A. range of $f\left(x\right)$ is $\left\{\log \frac{\pi }{2}\right\}.$
• B. domain of $f\left(x\right)$ is $[1,2).$
• C. domain of $f\left(x\right)$ is $(0,2).$
• D. range of $f\left(x\right)$ is $\{0,\log \frac{\pi }{2}\}.$

Answer 1

Answer: (A), (B)

domain of $f\left(x\right)$ is $[1,2).$

Given, $f\left(x\right)={\sin }^{-1}\left(\ln \left[x\right]\right)+\ln \left({\sin }^{-1}\left[x\right]\right)$

For given function to exist
$-1\le \ln \left[x\right]\le 1$
For $\log$ function to defined, $x\ge 1$
$\Rightarrow 0\le \ln \left[x\right]\le 1$
$\Rightarrow 1\le \left[x\right]\le e$
$\Rightarrow x\in [1,3)\dots \left(1\right)$

and $0<{\sin }^{-1}\left[x\right]\le \frac{\pi }{2}$
$\Rightarrow 0<\left[x\right]\le 1$
$\Rightarrow x\in [1,2)\dots \left(2\right)$

Thus, domain is $\left(1\right)\cap \left(2\right)$ i.e. $[1,2)$

and range $={\sin }^{-1}\left(\ln 1\right)+\ln \left({\sin }^{-1}1\right)$
$={\sin }^{-1}0+\log \frac{\pi }{2}$
$=\log \frac{\pi }{2}$