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If $f(\mathrm{x})=\frac{\mathrm{x}}{2}-1$, then on the interval $[0, \pi]$ which one of the following is correct?
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The correct answer is:
$\tan [f(\mathrm{x})]$, where $[\cdot]$ is the greatest integer function, and $\frac{1}{f(\mathrm{x})}$ are both discontinuous.
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}-1, \quad[0, \pi] \\ & \tan \cdot \mathrm{f}(\mathrm{x})=\tan \left(\frac{\mathrm{x}}{2}-1\right) \end{aligned}$
$\frac{1}{\mathrm{f}(\mathrm{x})}=\frac{1}{\frac{\mathrm{x}}{2}-1}$ is discontinuous at $\mathrm{x}=2$
tan. $\mathrm{f}(\mathrm{x})$ is discontinuous for $\mathrm{x}=2$ in $[0, \pi]$
$\frac{1}{\mathrm{f}(\mathrm{x})}=\frac{1}{\frac{\mathrm{x}}{2}-1}$ is discontinuous at $\mathrm{x}=2$
tan. $\mathrm{f}(\mathrm{x})$ is discontinuous for $\mathrm{x}=2$ in $[0, \pi]$
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