Question

If $f\left(x\right)=\frac{x}{2}-1$, then on the interval $[0,\pi ]$ which one of the following is correct?

• A. $\tan \left[f\right(x\left)\right]$, where $[.]$ is the greatest integer function, and $\frac{1}{f\left(x\right)}$ are both continuous
• B. $\tan \left[f\right(x\left)\right]$, where $[.]$ is the greatest integer function, and $f^{-1}\left(x\right)$ are both continuous
• C. $\tan \left[f\right(x\left)\right]$, where $[.]$ is the greatest integer function, and $\frac{1}{f\left(x\right)}$ are both discontinuous
• D. $\tan \left[f\right(x\left)\right]$, where $[.]$ is the greatest integer function, is discontinuous but $\frac{1}{f\left(x\right)}$ is continuous

Answer 1

Answer: (C)

$\tan \left[f\right(x\left)\right]$, where $[.]$ is the greatest integer function, and $\frac{1}{f\left(x\right)}$ are both discontinuous

Given that

$f\left(x\right)=\frac{x}{2}-1$

$\therefore \frac{1}{f\left(x\right)}=\frac{2}{x-2}$

The denominator of $\frac{1}{f\left(x\right)}$ vanishes at x = 2. This function has a discontinuity in the interval $[0,\pi ]$ since

${lim}_{x\to 2}\frac{2}{x-2}$

does not exist at x = 2.

Hence, $\frac{1}{f\left(x\right)}$ is discontinuous on the interval $[0,\pi ]$.

Now, consider greatest integer function applied to f(x):

$\left[f\right(x\left)\right]=[\frac{x}{2}-1]$

Note: The greatest integer function applied to x returns the largest integer less than or equal to x.

At x = 2, the one-sided limits are given as:

${lim}_{x\to 2^{+}}[\frac{x}{2}-1]=0$

and

${lim}_{x\to 2^{-}}[\frac{x}{2}-1]=-1$

The two limits are not equal, as such there is a discontinuity for

$\left[f\right(x\left)\right]=[\frac{x}{2}-1]$

at x = 2.

Hence, the function $\tan \left[f\right(x\left)\right]$ also has a discontinuity at x = 2.

Therefore, $\tan \left[f\right(x\left)\right]$ and $\frac{1}{f\left(x\right)}$ both are discontinuous on the interval $[0,\pi ]$.

The answer: option C.