Question

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

• A. $(-\frac{1}{2},0)$
• B. $(-\frac{1}{2},2)$
• C. $(1,2)$
• D. $(\frac{1}{2},\infty )$

Answer 1

The correct option is A $(-\frac{1}{2},0)$

$f^{\prime }\left(x\right)=|x|-\left\{x\right\}=|x|-(x-[x\left]\right)=|x|-x+\left[x\right]$
For $x\in (\frac{-1}{2},0)$,
$f^{\prime }\left(x\right)=-x-x-1=-2x-1$
As,
$-\frac{1}{2}<x<0$
or $0<-2x<1$
or $-1<-2x-1<0$
or $f^{\prime }\left(x\right)<0$,

$f\left(x\right)$ decreases in $(-\frac{1}{2},0)$
Similarly we can check for other given options say for $x\in (\frac{-1}{2},2)$,
$f^{\prime }\left(x\right)=\{\begin{array}{cc}(-x)-x-1,& -\frac{1}{2}<x<0\\ x-x+0,& 0\le x<1\\ x-x+1,& 1\le x<2\\ .& \\ .& \\ .& \end{array}$

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