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MHT CET · Maths · Functions

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)

  1. A \(\mathrm{x}=-1,0,1,2\)
  2. B \(\mathrm{x}=-1,0,1\)
  3. C \(\mathrm{x}=0,1\)
  4. D \(\mathrm{x}=2\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\mathrm{x}=0,1\)

Step-by-step Solution

Detailed explanation

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 \(\mathrm{x}=0,1\).