Question

Let $f\left(x\right)=\frac{\text{sin}\left(\pi [x-\pi ]\right)}{1+{\left[x\right]}^2},$ where $[.]$ denotes the greatest integer function. Then $f\left(x\right)$ is

• A. discontinuous at integral points
• B. continuous everywhere but not differentiable
• C. differentiable once but $f^{\prime \prime }\left(x\right),f^{\prime \prime \prime }\left(x\right)\cdots$ doesn't exists
• D. differentiable for all$x$

Answer 1

The correct option is D differentiable for all$x$

Given : $f\left(x\right)=\frac{\sin \left(\pi [x-\pi ]\right)}{1+{\left[x\right]}^2}$
And $[x-\pi ]$is always an integer, so
$f\left(x\right)=\frac{\sin \left(k\pi \right)}{1+{\left[x\right]}^2},k\in I\Rightarrow f\left(x\right)=0(\because 1+[x{]}^2\ne 0)$