Question

Find whether the given function is even or odd function, where $f\left(x\right)=\frac{x(\sin x+\tan x)}{\left[\frac{x+\pi }{\pi }\right]-\frac{1}{2}}$, where $x\ne n\pi$, where $\left[\right]$ denotes the greatest integer function.

• A. $f\left(x\right)$ is an odd function
• B. $f\left(x\right)$ is an even function
• C. $f\left(x\right)$ is neither even nor odd function
• D. $f\left(x\right)$ is both even and odd function

Answer 1

The correct option is A $f\left(x\right)$ is an odd function

$f\left(x\right)=\frac{x(\sin x+\tan x)}{\left[\frac{x+\pi }{\pi }\right]-\frac{1}{2}}=\frac{x(\sin x+\tan x)}{\left[\frac{x}{\pi }\right]+1-\frac{1}{2}}$
$f\left(x\right)=\frac{x(\sin x+\tan x)}{\left[\frac{x}{\pi }\right]+0.5}$
$\Rightarrow f(-x)=\frac{-x(\sin (-x)+\tan (-x))}{[-\frac{x}{\pi }]+0.5}$
$\Rightarrow f(-x)=\frac{x(\sin x+\tan x)}{-1-\left[\frac{x}{\pi }\right]+0.5}$
Hence, $f(-x)=-\left(\frac{x(\sin x+\tan x)}{\left[\frac{x}{\pi }\right]+0.5}\right)$
$\Rightarrow f(-x)=-f\left(x\right)$
Hence, $f\left(x\right)$ is an odd function (if $x\ne n\pi )$.