Question

If $f\left(x\right)=\left[x\right]\sin \left(\frac{\pi }{[x+1]}\right),$ where $[.]$ denotes the greatest integer function, then the points of discontinuity of $f$ in the domain are

• A. $Z$
• B. $Z-\left\{0\right\}$
• C. $R-[1,0)$
• D. None of these

Answer 1

Answer: (B)

$Z-\left\{0\right\}$

$[x+1]=0$ if $0\le x+1<1\Rightarrow -1\le x<0$

Thus domain of $f=R-[1,0)$

We have, $\sin \left(\frac{\pi }{[x+1]}\right)$ continuous at all points of $R-[1,0)$

and $\left[x\right]$ continuous on $R-Z$, where $Z$ denotes the set of integers.

Thus, the points where $f$ can possibly be discontinuous are $...,-3,-2,-1,0,1,2,\mathrm{3...}$

For $0\le x<1,\left[x\right]=0$ and $\sin \left(\frac{\pi }{[x+1]}\right)$ is defined.

$\therefore f\left(x\right)=0$ and $0\le x<1$

Also, $f$ is not defined on $[-1,0)$, so the continuity of $f$ at $0$ means continuity of $f$ from right at $0$.

Since $f$ is continuous from right at $0$, so $f$ is continuous at $0$.

Hence the set of points of discontinuity of $f$ is $Z-\left\{0\right\}$.