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)  \[\left( \frac{1}{2},\,\infty  \right)\]

Correct Answer: A

Solution :

 \[\because \]    \[f(x)=|x|-\{x\}\] \[\because \]     \[f(x)\] 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 \[x=1\].