if $f\text{'}\left(x\right)=\left|x\right|-\left\{x\right\}$, where $\left\{x\right\}$ denotes the fractional part of $x$ , then $f\left(x\right)$ is decreasing in 

$f\text{'}\left(x\right)=\left|x\right|-\left\{x\right\}=\left|x\right|-(x-\left[x\right])=\left|x\right|-x+\left[x\right]$

for $$\begin{gathered} x\in (-1/2, 0) \\ f\text{'}x=-x-x-1=-2x-1 \end{gathered}$$

Also, for $-\frac{1}{2}<x<0$ or $0<-2x<1$ or $-1<-2x-1<0$

or $f\text{'}\left(x\right)<0,f \left(x\right)$decreases in $(-1/2, 0)$

Similarly, we can check for other given options say for

$x\in (1/2, 2)$

$f\text{'}x=\left\{\begin{array}{l}(-x)-x-1, -\frac{1}{2}<x<0\\ x-x+0, 0\le x<1\\ x-x+1, 1\le x<2\\ ⋮\end{array}\right.$

Here, $f \left(x\right)$decreases only in$(-1/2, 0)$; otherwise $f \left(x\right)$ in other
intervals is constant.