Question

Consider the function $f\left(x\right)={\cos }^{-1}\left(\right[2^x\left]\right)+{\sin }^{-1}\left(\right[2^x-1\left]\right)$ (where $[.]$ denotes greatest integer function), then

• A. Domain of $f\left(x\right)$ is $x\in (-\infty ,0]$
• B. Range of $f\left(x\right)$ is singleton set
• C. $f\left(x\right)$ is an even function
• D. $f\left(x\right)$ is an odd function

Answer 1

The correct option is B Range of $f\left(x\right)$ is singleton set

Given : $f\left(x\right)={\cos }^{-1}\left(\right[2^x\left]\right)+{\sin }^{-1}\left(\right[2^x-1\left]\right)$
For $f\left(x\right)$ to be defined
$-1\le \left[2^x\right]\le 1\text{and}-1\le [2^x-1]\le 1-1\le \left[2^x\right]\le 1\text{and}-1\le \left[2^x\right]-1\le 1\Rightarrow -1\le \left[2^x\right]\le 1\text{and}0\le \left[2^x\right]\le 2\Rightarrow \left[2^x\right]=0,1\Rightarrow 2^x\in (0,2)(\because 2^x\ne 0)\Rightarrow x\in (-\infty ,1)$

When $\left[2^x\right]=0,$ then
$f\left(x\right)={\cos }^{-1}\left(0\right)+{\sin }^{-1}(-1)\Rightarrow f\left(x\right)=\frac{\pi }{2}-\frac{\pi }{2}=0$

When $\left[2^x\right]=1,$ then
$f\left(x\right)={\cos }^{-1}\left(1\right)+{\sin }^{-1}\left(0\right)\Rightarrow f\left(x\right)=0+0=0$

$\therefore$ Range of $f\left(x\right)$ is $\left\{0\right\}$
Hence, $f\left(x\right)$ is neither even nor odd function.