Question

The domain of the function $f\left(x\right)=si{n}^{-1}(x+[x\left]\right).,$where

Answer 1

We have,

$\begin{array}{c}f\left(x\right)={\sin }^{-1}\left(x+\left[x\right]\right)\\ -1\le x+\left[x\right]\le 1\\ -1\le x+x-\left\{x\right\}\le 1\left(\because x=\left[x\right]+\left\{x\right\}\right)\\ \Rightarrow -1\le 2x-\left\{x\right\}\le 1\\ \Rightarrow -1\le \left\{x\right\}\le 2x\le 1+\left\{x\right\}\\ \Rightarrow \frac{-1+\left\{x\right\}}{2}\le x\le \frac{1+\left\{x\right\}}{2}\because \left\{x=\left\{x\right\}<1\right\}\\ \therefore \frac{-1}{2}\le x\le 1\end{array}$

Hence, this is the answer.