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The function $f(x)=\frac{\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}}{2+[x]^{2}},$ where $[x]$ denotes the greatest integer $\leq x$, is
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continuous for all values of $x$
Given, $f(x)=\frac{\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}}{2+[x]^{2}}$
Since. $\left[x-\frac{\pi}{2}\right]$ is an integer for all $x$, therefore $\pi\left[x-\frac{\pi}{2}\right]$ is an integral multiple of $\pi$ for all $x$. Hence, $\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}=0$ for all $x$
Also, $2+[x]^{2} \neq 0$ for all $x$
Hence, $f(x)=0$ for all $x$ Hence, $f(x)$ is continuous and derivable for all $x$.
Since. $\left[x-\frac{\pi}{2}\right]$ is an integer for all $x$, therefore $\pi\left[x-\frac{\pi}{2}\right]$ is an integral multiple of $\pi$ for all $x$. Hence, $\tan \left\{\pi\left[x-\frac{\pi}{2}\right]\right\}=0$ for all $x$
Also, $2+[x]^{2} \neq 0$ for all $x$
Hence, $f(x)=0$ for all $x$ Hence, $f(x)$ is continuous and derivable for all $x$.
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