Question
For a real number $$y,$$ let $$\left[ y \right]$$ denotes the greatest integer less
than or equal to $$y:$$ Then the function $$f\left( x \right) = \frac{{\tan \left( {\pi \left[ {x - \pi } \right]} \right)}}{{1 + {{\left[ x \right]}^2}}}$$ is-
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 :
$$f'\left( x \right)$$ exists for all $$x$$
Solution :
$$f\left( x \right) = \frac{{\tan \left( {\pi \left[ {x - \pi } \right]} \right)}}{{1 + {{\left[ x \right]}^2}}}$$
By def. $$\left[ {x - \pi } \right]$$ is an integer whatever be the value of $$x.$$
And so $$\pi \left[ {x - \pi } \right]$$ is an integral multiple of $$\pi .$$
Consequently $$\tan \left( {\pi \left[ {x - \pi } \right]} \right) = 0,\,\,\forall \,x.$$
And since $$1 + {\left[ x \right]^2} \ne 0$$ for any $$x,$$ we conclude that $$f\left( x \right) = 0.$$
Thus $$f\left( x \right)$$ is constant function and so, it is continuous and differentiable any number of times, that is $$f'\left( x \right),\,f''\left( x \right),\,f'''\left( x \right),.....$$ all exist for every $$x,$$ their value being 0 at every pt. $$x.$$
Hence, out of all the alternatives only (D) is correct.