The function f(x)=frac{tan {pi[x-frac{pi}{2}] | vedclass

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(A) Given,$f(x)=\frac{	an \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$.Since $[x-\frac{\pi}{2}]$ is an integer for all $x$,let $[x-\frac{\pi}{2}] = k$,where

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continuous for all values of $x$

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(A) Given,$f(x)=\frac{	an \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$.Since $[x-\frac{\pi}{2}]$ is an integer for all $x$,let $[x-\frac{\pi}{2}] = k$,where $k \in \mathbb{Z}$.Then the numerator becomes $	an(\pi k)$,which is equal to $0$ for all integers $k$.Since the denominator $2+[x]^{2}$ is always $\geq 2$ and never zero,the function simplifies to $f(x) = \frac{0}{2+[x]^{2}} = 0$ for all $x \in \mathbb{R}$.$A$ constant function $f(x) = 0$ is continuous and differentiable for all values of $x$.Therefore,the function is continuous for all values of $x$.

The function $f(x)=\frac{\tan \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$,where $[x]$ denotes the greatest integer $\leq x$,is

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