JEE Main 2021MathematicsContinuity and DifferentiabilityHardMCQ

JEE Main 2021 — Continuity and Differentiability Question with Solution

JEE Main 2021 (16 Mar Shift 2)

Question

Let α∈R be such that the function fx=cos-11-x2sin-11-xx-x3,x≠0α,x=0 is continuous at x=0, where x=x-x,x is the greatest integer less than or equal to x. Then :

Choose an option

Show full solutionCorrect option: C
Correct answer
Cno such α exists

Step-by-step explanation

limx→0+fx=f0=Limx→0-x

=limx→0+cos-11-x2·sin-11-xx1-x1+x

=limx→0+cos-11-x2x·1·1·π2

Let 1-x2=cosθ

=π2limθ→0+θ1-cosθ

=π2limθ→0+θ2sinθ2=π2

Now, limx→0-cos-11-1+x2sin-1-x1+x-1+x3

limx→0-π2-sin-1x1+x2+x-x

limx→0-π21·2·sin-1xx=π4

⇒RHL≠LHL

Function can't be continuous

⇒ No value of α exist

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Continuity and Differentiability chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2021, 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.