Question

Let $g\left(x\right)=1+x-\left[x\right],$
$f\left(x\right)=\{\begin{array}{cc}-1,& x<0\\ 0,& x=0\\ 1,& x>0\end{array}.$
Then for all $x$, $f\left(g\left(x\right)\right)$ is
(where $[.]$ is greatest integer function)

• A. $x$
• B. $1$
• C. $f\left(x\right)$
• D. $g\left(x\right)$

Answer 1

The correct option is B $1$

$g\left(x\right)=1+x-\left[x\right]$
$g\left(x\right)=1+\left\{x\right\}$
$\therefore g\left(x\right)>0\forall x\in R$
$\therefore f\left(g\left(x\right)\right)=1$