• # question_answer If $f'(x)=\,|x|\,-\{x\},$ where {x} denotes the fractional part function of $x,$ then $f(x)$ is decreasing in A)  $\left( \frac{-1}{2},\,0 \right)$                       B)  $\left( \frac{-1}{2},\,2 \right)$ C)  $\left( \frac{-1}{2},\,-2 \right]$                      D)  $\left( \frac{1}{2},\,\infty \right)$

$\because$    $f(x)=|x|-\{x\}$ $\because$     $f(x)$ is decreasing. $\therefore$    $f'(x)<0$ $\Rightarrow$            $|x|-\{x\}<0$ $\Rightarrow$            $|x|\,<\{x\}$ From figure, $x\in \,\left( -\frac{1}{2},\,\,0 \right)$ From graph, it is clear that, $f(x)$ has local maxima at $x=1$.