Question

Let $f\left(x\right)=\frac{x(\sin x+\tan x)}{\left[\frac{x+\pi }{\pi }\right]-\frac{1}{2}}$, then which among the following option is incorrect
(where [.] denotes the greatest integer function)

• A. $f\left(x\right)$ is an odd function when $x\ne n\pi$
• B. $f\left(x\right)$ is an even function when $x=n\pi$
• C. $f\left(x\right)$ is an odd function when $x=n\pi$
• D. $f\left(x\right)$ is an even function when $x\ne n\pi$

Answer 1

The correct option is D $f\left(x\right)$ is an even function when $x\ne n\pi$

$f\left(x\right)=\frac{x(\sin x+\tan x)}{\left[\frac{x+\pi }{\pi }\right]-\frac{1}{2}}\Rightarrow f\left(x\right)=\frac{x(\sin x+\tan x)}{\left[\frac{x}{\pi }\right]+1-\frac{1}{2}}\Rightarrow f\left(x\right)=\frac{x(\sin x+\tan x)}{\left[\frac{x}{\pi }\right]+0.5}\Rightarrow f\left(x\right)=0\text{if}x\in n\pi ,n\in Z$
$\Rightarrow f\left(x\right)$ is both even and odd function if $x=n\pi$

We know,
$x\ne n\pi \Rightarrow [-\frac{x}{\pi }]=-1-\left[\frac{x}{\pi }\right]$

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