Question

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

• A. domain of $f\left(x\right)=(-\ln 2,\ln 2)$
• B. range of $f\left(x\right)=\left\{\pi \right\}$
• C. $f\left(x\right)$ is discontinuous at $x=0$
• D. $f\left(x\right)={\cos }^{-1}x$ has only one solution

Answer 1

Answer: (A), (C)

$f\left(x\right)$ is discontinuous at $x=0$

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

For $f\left(x\right)$ to be defined,
$-1\le \left[e^x\right]\le 1$ and $-1\le \left[e^{-x}\right]\le 1$
​​​​​​As $e^x,e^{-x}$ are always positive, so
$0<e^x<2$ and $0<e^{-x}<2$
$\Rightarrow -\infty <x<\ln 2$ and $-\ln 2<x<\infty$
$\therefore x\in (-\ln 2,\ln 2)$


If $x\in (-\ln 2,0)$
$e^x\in (\frac{1}{2},1),e^{-x}\in (1,2)\Rightarrow \left[e^x\right]=0,\left[e^{-x}\right]=1\Rightarrow f\left(x\right)={\sin }^{-1}0+{\sin }^{-1}1=\frac{\pi }{2}$

If $x\in (0,\ln 2),$
$e^x\in (1,2),e^{-x}\in (\frac{1}{2},1)\Rightarrow \left[e^x\right]=1,\left[e^{-x}\right]=0\Rightarrow f\left(x\right)={\sin }^{-1}1+{\sin }^{-1}0=\frac{\pi }{2}$

If $x=0,$
$f\left(x\right)={\sin }^{-1}1+{\sin }^{-1}1=\pi$

Therefore, $f\left(x\right)=\{\begin{array}{c}\pi ,x=0\\ \frac{\pi }{2},x\in (-\ln 2,0)\cup (0,\ln 2)\end{array}$

So, the range $=\{\pi ,\frac{\pi }{2}\}$
Also, $f\left(x\right)$ is discontinuous at $x=0$
We know that ${\cos }^{-1}x=\{\begin{array}{c}\pi ,x=-1\\ \frac{\pi }{2},x=0\end{array}$
Therefore, $f\left(x\right)={\cos }^{-1}x$ has no solution.