For a real numbers y, ley [y] denotes the greatest integer less than (or) equal to y let $f (x) = \frac{\tan (π [x - π])}{1 + [x]^{2}}$

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f(x) Is discontinuous at some x

f(x) Is continuous at all x, but the derivative f¹(x) does not exist for some x

f¹(x) exists for all x

none

answer is C.

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Detailed Solution

Now $π [x - π]$ is an integral multiple of $πtan (π [x - π]) = 0 \forall xf (x) = 0 \forall x$

$f^{1} (x) = 0 \forall x$

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