Question

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

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

$f\left(x\right)=\frac{\tan \left(\pi (x-\pi )\right)}{1+{\left[x\right]}^2}$

$[x-\pi ]$ will always give an integer

We know that

$\tan \left(n\pi \right)=0,nϵI$

$\Rightarrow f\left(x\right)=\frac{0}{1+{\left[x\right]}^2}=0$

$\Rightarrow f\left(x\right)$ is a continuous function

$f^{\prime }\left(x\right)=0$

$\Rightarrow f^{\prime }\left(x\right)$ is defined for all x.