JEE Main 2022MathematicsContinuity and DifferentiabilityHardMCQ

JEE Main 2022 — Continuity and Differentiability Question with Solution

JEE Main 2022 (26 Jun Shift 2)

Question

Let fx=min1,1+xsinx,0≤x≤2π. If m is the number of points, where f is not differentiable and n is the number of points, where f is not continuous, then the ordered pair m,n is equal to

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Show full solutionCorrect option: B
Correct answer
B1,0

Step-by-step explanation

Given,

fx=min1,1+xsinxx∈0,2π

fx=1,  0≤x≤π1+xsinx,π≤x≤2π

Now, at x=π,limx→π-fx=1=limx→π+fx

∴  fx is continuous in 0,2π

So, number of points of discontinuity is zero or n=0

Now, at x=π

L.H.D =limh→0fπ-h-fπ-h=0

R.H.D =limh→0fπ+h-fπh=limh→01-π+hsinh-1h

=-π

∴fx is not differentiable at x=π so, m=1

∴m,n=1,0

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About this question

This is a previous-year question from JEE Main 2022, 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.