Question

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

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

Answer 1

The correct option is D

$f’\left(x\right)$exists for all $x$

Explanation for the correct option:

Finding the value of given function:

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

As we know,

$\pi [x-\pi ]$ $=n\mathrm{\pi }$ and $\tan \left(n\mathrm{\pi }\right)=0$ $\left(\because \left[x-\pi \right]=integer\right)$

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

Now, $[1+{\left[x\right]}^2]$ is not equal to $0$

$f\left(x\right)=0$ , for all values of $x$.

$\Rightarrow$ $f\text{'}\left(x\right)=0$

$\Rightarrow$ $f\text{'}\text{'}\left(x\right)=0$

$\therefore$$f\text{'}\left(x\right)=0$, for all values of $\mathit{x}$ and exists for all $\mathit{x}$

Hence, option (D) is correct.