If f(x) = frac{x}{2} - 1,then on the interval | vedclass

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(C) Given $f(x) = \frac{x}{2} - 1$. On the interval $[0, \pi]$,we analyze the functions at $x = 2$. For $	an[f(x)]$: As $x 	o 2^-$,$[f(x)] = [\frac{

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$	an [f(x)]$ and $\frac{1}{f(x)}$ are both discontinuous.

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(C) Given $f(x) = \frac{x}{2} - 1$. On the interval $[0, \pi]$,we analyze the functions at $x = 2$. For $	an[f(x)]$: As $x 	o 2^-$,$[f(x)] = [\frac{x}{2} - 1] = -1$,so $	an[f(x)] 	o 	an(-1)$. As $x 	o 2^+$,$[f(x)] = [\frac{x}{2} - 1] = 0$,so $	an[f(x)] 	o 	an(0) = 0$. Since $	an(-1) 
eq 0$,$	an[f(x)]$ is discontinuous at $x = 2$. For $\frac{1}{f(x)}$: $f(x) = \frac{x}{2} - 1$. At $x = 2$,$f(2) = 0$. Thus,$\frac{1}{f(x)}$ is undefined at $x = 2$,making it discontinuous at $x = 2$. Therefore,both functions are discontinuous on the interval $[0, \pi]$.

If $f(x) = \frac{x}{2} - 1$,then on the interval $[0, \pi]$,where $[.]$ represents the greatest integer function,which of the following is true?

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