Mathematics - Functions Question with Solution | TestHub

The denominator is $2\left[\frac{x+21\pi }{\pi }\right]-41=2\left[\frac{x}{\pi }+21\right]-41$.

Therefore, $f(x)=\frac{x(\sin x+\tan x)}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}$.

This implies $f(-x)=\frac{-x\{\sin (-x)+\tan (-x)\}}{\left[-\frac{x}{\pi }\right]+\frac{1}{2}}=\frac{x(\sin x+\tan x)}{-1-\left[\frac{x}{\pi }\right]+\frac{1}{2}}$ (if $x\ne n\pi$).

So, $f(-x)=-\frac{x(\sin x+\tan x)}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}=-f(x)$ (if $x=n\pi$, then $f(x)=0$).