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

$\begin{array}{1 1}(a)\;\text{discontinuous at some x}\\(b)\;\text{continuous at all x,but the derivative f'(x) does not exist for some x}\\(c)\;\text{f'(x) exists for all x,but the second derivative f'(x) does not exist for some x}\\(d)\;\text{f'(x) exists for all x}\end{array}$

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By differentiating $[x-\pi]$ is an integer whatever be the value of x.And so $\pi[x-\pi]$ is an integral multiple of $\pi$ consequently.
$\tan(\pi[x-\pi])=0\forall x$
And since $1+[x]^2\neq 0$ for any x we conclude that
Thus $f(x)$ is constant function and so it is continuous and differentiable any no of times,that is $f'(x),f''(x),f'''(x)$ all exist for every x,their value being 0 at every point x.
Hence (d) is the correct answer.
answered Dec 31, 2013 by sreemathi.v

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