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Question: Answered & Verified by Expert
If $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}-1$, then on the interval $[0, \pi]$ where [.] represents greatest integer function
MathematicsContinuity and DifferentiabilityMHT CETMHT CET 2022 (05 Aug Shift 1)
Options:
  • A $\tan [\mathrm{f}(\mathrm{x})]$ is continuous but $\frac{1}{\mathrm{f}(\mathrm{x})}$ is not continuous.
  • B $\tan [\mathrm{f}(\mathrm{x})]$ and $\frac{1}{\mathrm{f}(\mathrm{x})}$ are both continuous.
  • C $\tan [\mathrm{f}(\mathrm{x})]$ and $\frac{1}{\mathrm{f}(\mathrm{x})}$ are both discontinuous.
  • D $\tan [\mathrm{f}(\mathrm{x})]$ is discontinuous and $\frac{1}{\mathrm{f}(\mathrm{x})}$ is continuous.
Solution:
1295 Upvotes Verified Answer
The correct answer is: $\tan [\mathrm{f}(\mathrm{x})]$ and $\frac{1}{\mathrm{f}(\mathrm{x})}$ are both discontinuous.
$\tan [\mathrm{f}(\mathrm{x})]=\tan \left[\frac{\mathrm{x}}{2}-1\right]=\left\{\begin{array}{lll}\tan (-1), & \text { if } & 0 \leq \mathrm{x}<2 \\ \tan (0)=0, & \text { if } & 2 \leq \mathrm{x} \leq \pi\end{array}\right.$
which is discontinuous at $\mathrm{x}=2$
$\frac{1}{f(x)}=\frac{1}{\frac{x}{2}-1}$ which is discontinuous at $x=2$

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