Question

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

• A. $(-\frac{1}{2},0)$
• B. $(-\frac{1}{2},2)$
• C. $(-\frac{1}{2},3)$
• D. $(\frac{1}{2},\infty )$

Answer 1

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

$f^{\prime }\left(x\right)=|x|-\left\{x\right\}\Rightarrow f^{\prime }\left(x\right)=|x|-(x-[x\left]\right)\Rightarrow f^{\prime }\left(x\right)=|x|-x+\left[x\right]\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{c}-2x+\left[x\right],x<0\\ \left[x\right],x\ge 0\end{array}$

Now, we know $\left[x\right]\ge 0$ for $x\ge 0$, so
$f^{\prime }\left(x\right)<0$ is only possible when $x<0$, then
$-2x+\left[x\right]<0\Rightarrow \left[x\right]<2x\Rightarrow x\in (-\frac{1}{2},0)$


Alternate solution:
$f^{\prime }\left(x\right)=|x|-\left\{x\right\}$
For the function to be strictly decreasing,
$f^{\prime }\left(x\right)<0\Rightarrow |x|-\left\{x\right\}<0\Rightarrow |x|<\left\{x\right\}$
We know, $\left\{x\right\}\in (0,1)$, so the inequality can holds for $x\in (-1,1)$

Now, when $x\in (-1,0)$
$-x<1+x\Rightarrow x>-\frac{1}{2}\Rightarrow x\in (-\frac{1}{2},0)$
When $x\in [0,1)$
$x<x$
Which is not possible.
Hence, $x\in (-\frac{1}{2},0)$