Question

For real number y, let $\left[y\right]$ denote the greatest integer less than or equal to y. Then $f\left(x\right)=\frac{\tan \left(\pi [x-\pi ]\right)}{1+{\left[x\right]}^2}$ is

• A. discontinuous at some x
• B. continuous at all x, but the derivative $f^{\prime }\left(x\right)$ does not exist for some x
• C. $f^{\prime }\left(x\right)$ exists for all x but the second derivative $f^{\prime \prime }\left(x\right)$ does not exist
• D. $f^{\prime }\left(x\right)$ exists for all x

Answer 1

The correct option is D $f^{\prime }\left(x\right)$ exists for all x

Here $f\left(x\right)=\frac{\tan \pi \left[(x-\pi )\right]}{1+{\left[x\right]}^2}$
Since, we know $\pi \left[(x-\pi )\right]=n\pi$ and $\tan n\pi =0$
$\because 1+{\left[x\right]}^2\ne 0\therefore f\left(x\right)=0$ for all x
Thus $f\left(x\right)$ is constant.
$\therefore f^{\prime }\left(x\right)f^{\prime \prime }\left(x\right)...$ all exists for every x, their value being 0.
$\Rightarrow f^{\prime }\left(x\right)$ exists for all x