Question

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

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

Answer 1

The correct option is B Z\{0}

$[x+1]=0$ if $0\le x+1<1$ i.e., $-1\le x<0$

Thus, domain of $f=\frac{R}{[-1,0)}$

We have, $\sin \left(\frac{\pi }{[x+1]}\right)$ continuous at all points of $\frac{R}{[-1,0)}$ and $\left[x\right]$ continuous on $\frac{R}{Z},$ where $Z$ denotes the set of integers.

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

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

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

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 $\frac{Z}{\left\{0\right\}}.$