If $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$, for $\mathrm{x} \in(-1,2)$, then $\mathrm{f}$ is discontinuous at…
If $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$, for $\mathrm{x} \in(-1,2)$, then $\mathrm{f}$ is discontinuous at (where $[\mathrm{x}]$ represents floor function)
$\mathrm{x}=-1,0,1,2$
$\mathrm{x}=-1,0,1$
$\mathrm{x}=0,1$
$\mathrm{x}=2$
Solution
We have $\mathrm{f}(\mathrm{x})=[\mathrm{x}]$
Let $[\mathrm{x}]=\mathrm{K}$, an integer.
$\therefore \lim _{x \rightarrow K^{+}} f(x)=K \text { and } \lim _{x \rightarrow K^{-}} f(x)=K-1$
Thus given function is not continuous at all integral values in its domain.
$\because \mathrm{f}$ is discontinuous at $\mathrm{x}=0,1$.