Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

For $x=0,f\left(x\right)=1$
For $0<x\le \frac{1}{2},f\left(x\right)=0$
For $\frac{1}{2}<x\le 1,f\left(x\right)=-1$

For $x\in (1,2)$,
$[x-2]=-1$ and $|x-1|=x-1$

For $x=2,[x-2]=0$

$\therefore f\left(x\right)=\{\begin{array}{cc}1,& x=0\\ 0,& 0<x\le \frac{1}{2}\\ -1,& \frac{1}{2}<x\le 1\\ -(x-1),& 1<x<2\\ 0,& x=2\end{array}$

Clearly, $f$ is discontinuous at four points i.e., $\{0,\frac{1}{2},1,2\}$ in the interval $[0,2]$