Question

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

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

Answer 1

The correct option is A $1$

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