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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)
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The correct answer is:
$\mathrm{x}=0,1$
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 $\mathrm{x}=0,1$.
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 $\mathrm{x}=0,1$.
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