Check whether the function $f(x)=\frac{x(\sin x+\tan x)}{[\frac{x+π}{π}]-\frac{1}{2}}$, where [.] denotes greatest integer function is even or odd function.
Answer & explanation
Correct answer: option 2
We have
$f(x)= \frac{x(\sin x+\tan x)}{[\frac{x+π}{π}]-\frac{1}{2}}=\frac{x(\sin x+\tan x)}{[\frac{x}{π}]+1-\frac{1}{2}}=\frac{x(\sin x+\tan x)}{[\frac{x}{π}]+0.5}$
$⇒f(-x)=\frac{-x(\sin (-x)+\tan (-x))}{[-\frac{x}{π}]+0.5}$
$=\left\{\begin{matrix}\frac{x(\sin x+\tan x)}{-1[-\frac{x}{π}]+0.5},&x≠nπ,\,n∈Z\\0,&x=nπ,\,n∈Z\end{matrix}\right.$
$=\left\{\begin{matrix}-\frac{x(\sin x+\tan x)}{[\frac{x}{π}]+0.5},&x≠nπ\\0,&x=nπ\end{matrix}\right.$
Thus f(-x)=-f(x)
Hence, f(x) is an odd function.