If $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}-1$, then on the interval $[0, \pi]$ where [.] represents…

If $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}-1$, then on the interval $[0, \pi]$ where [.] represents greatest integer function
  1. $\tan [\mathrm{f}(\mathrm{x})]$ is continuous but $\frac{1}{\mathrm{f}(\mathrm{x})}$ is not continuous.
  2. $\tan [\mathrm{f}(\mathrm{x})]$ and $\frac{1}{\mathrm{f}(\mathrm{x})}$ are both continuous.
  3. $\tan [\mathrm{f}(\mathrm{x})]$ and $\frac{1}{\mathrm{f}(\mathrm{x})}$ are both discontinuous.
  4. $\tan [\mathrm{f}(\mathrm{x})]$ is discontinuous and $\frac{1}{\mathrm{f}(\mathrm{x})}$ is continuous.

Solution

$\tan [\mathrm{f}(\mathrm{x})]=\tan \left[\frac{\mathrm{x}}{2}-1\right]=\left\{\begin{array}{lll}\tan (-1), & \text { if } & 0 \leq \mathrm{x}<2 \\ \tan (0)=0, & \text { if } & 2 \leq \mathrm{x} \leq \pi\end{array}\right.$ which is discontinuous at $\mathrm{x}=2$ $\frac{1}{f(x)}=\frac{1}{\frac{x}{2}-1}$ which is discontinuous at $x=2$

Asked in: MHT CET 2022 (05 Aug Shift 1)

Practice more Continuity and Differentiability questions on Aicharya