Question

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

• A. discontinuous at some x
• B. continuous at all x, but the derivative f'(x)does not exist for some x
• C. f'(x) exists for all x, but the derivative f''(x) does not exist for some x
• D. f'(x) exists for all x

Answer 1

The correct option is D f'(x) exists for all x

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