JEE Main 2020MathematicsContinuity and DifferentiabilityHardNumerical

JEE Main 2020 — Continuity and Differentiability Question with Solution

JEE Main 2020 (05 Sep Shift 1)

Question

Let f(x)=x·x2, for -10<x<10, where t denotes the greatest integer function. Then the number of points of discontinuity of fx is equal to

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Show full solutionCorrect answer: 8
Correct answer
8

Step-by-step explanation

We know that the greatest integer function is discontinuous at integral points.

-5<x2<5

⇒  x2=-5,-4,-3,-2,-1,0,1,2,3,4, are the integer values.

At x=0, f0+=f0=f0-=0, which means that function is continuous at x=0.

∵x≠-10⇒x2≠-5

Hence, the function is discontinuous at points -4,-3,-2,-1,1,2,3,4

The number of values is 8 .

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About this question

This is a previous-year question from JEE Main 2020, covering the Continuity and Differentiability chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.