Question:

The number of points of discontinuity of the function \[ f(x)=[x]+|x-2|, \qquad -3<x<3 \] is

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For expressions involving the greatest integer function, \[ [x], \] first locate all integers in the given interval. The modulus term \[ |x-a| \] is continuous and does not affect the jump discontinuities of \[ [x]. \]
Updated On: Jul 9, 2026
  • \(5\)
  • \(3\)
  • \(4\)
  • \(2\) \bigskip
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The Correct Option is A

Solution and Explanation

Concept: \[ [x] \] (the greatest integer function) is discontinuous at every integer, whereas \[ |x-2| \] is continuous for all real \(x\). Therefore, discontinuities of \[ f(x)=[x]+|x-2| \] can occur only at integer points.

Step 1:
List all integers in the interval \((-3,3)\). The integers lying in \[ -3<x<3 \] are \[ -2,\,-1,\,0,\,1,\,2. \]

Step 2:
Check discontinuity at these points. Since \[ |x-2| \] is continuous everywhere, adding it does not remove the jump discontinuity of \[ [x]. \] At every integer \(n\), \[ \lim_{x\to n^-}[x]=n-1, \] \[ \lim_{x\to n^+}[x]=n. \] Hence the jump is \[ 1. \] Therefore \(f(x)\) is discontinuous at each of \[ -2,\,-1,\,0,\,1,\,2. \]

Step 3:
Count the discontinuity points. Number of discontinuity points \[ =5. \]

Step 4:
Write the final answer. \[ \boxed{5} \]
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