Question

Let $f\left(x\right)=\frac{\sin x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}$, where $\left[x\right]$ denotes greatest integer function, then $f\left(x\right)$ is

• A. odd function
• B. even function
• C. neither odd or even
• D. both odd and even function

Answer 1

The correct option is B even function

$f\left(0\right)=\frac{\sin 0}{\left[\frac{0}{\pi }\right]+\frac{1}{2}}=0$
Now putting $x\to -x$
$f(-x)=\frac{\sin (-x)}{\left[\frac{-x}{\pi }\right]+\frac{1}{2}}$
Now we know that, when $\frac{x}{\pi }$ is not an integer
$\left[\frac{-x}{\pi }\right]+\left[\frac{x}{\pi }\right]=-1$
Using this we can write,
$f(-x)=-\frac{\sin \left(x\right)}{-\left[\frac{x}{\pi }\right]-1+\frac{1}{2}}\Rightarrow f(-x)=\frac{\sin x}{\left[\frac{x}{\pi }\right]+\frac{1}{2}}=f\left(x\right)$
When $\frac{x}{\pi }$ is an integer,
$\frac{x}{\pi }=n\Rightarrow x=n\pi$
So the function will be,
$f\left(n\pi \right)=\frac{\sin \left(n\pi \right)}{n+\frac{1}{2}}=0$
So, the given function is even