Question

The number of points of discontinuity of $f\left(x\right)=\{\begin{array}{cc}\left[cos\pi x\right],& 0\le x\le 1\\ |2x-3|[x-2],& 1<x\le 2\end{array}$ is/are where $[.]$ denotes the greatest integral function

• A. Two
• B. Three
• C. Four
• D. Zero

Answer 1

Answer: (B)

Three

Consider $x\in [0,1]$
From the graph given, it is clear that $\left[cos\pi x\right]$ is discontinuous at $x=0,1/2$ (1)
Now, consider $x\in (1,2].$
$f\left(x\right)=[x-2]|2x-3|$
For $x\in (1,2),[x-2]=-1$, and for $x=2,[x-2]=0$.
Also, $|2x-3|=0$ or $x=3/2$.
Therefore, $x=3/2$ and 2 may be the points at which f(x) is discontinuous.
$f\left(x\right)=\{\begin{array}{cc}1,& x=0\\ x,& 0<x\le \frac{1}{2}\\ -1,& \frac{1}{2}<x\le 1\\ -(3-2x),& 1<x\le 3/2\\ -(2x-3),& 3/2<x\le 2\\ 0,& x=2\end{array}$
Thus, f(x) is continuous when $x\in [0,2]-\{0,1/2,2\}$.

option B is correct