Question

Let $f\left(x\right)=\frac{\sin \left\{\pi [x-\pi ]\right\}}{1+{\left[x\right]}^2}$ where $\left[x\right]$ stands for 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 $f^{\prime \prime }\left(x\right),f^{\prime \prime \prime }\left(x\right),...$ do not exist
• D. differentiable for all $x$

Answer 1

The correct option is D differentiable for all $x$

${\left[x\right]}^2+1\ne 0.$

Also $[x-\pi ]$ is always an integer .

$\therefore p[x-\pi ]$ is of the form $k\pi$ $($kis any integer$)$ and hence $f\left(x\right)$

is identically the zero function for all $x.$ So it is a constant function.