JEE Main 2022MathematicsContinuity and DifferentiabilityMediumMCQ

JEE Main 2022 — Continuity and Differentiability Question with Solution

JEE Main 2022 (27 Jul Shift 2)

Question

Let fx=2+x-x-1+x+1, x∈R.

Consider

S1:f'-32+f'-12+f'12+f'32=2

S2: ∫-22 fxdx=12

Then,

Choose an option

Show full solutionCorrect option: D
Correct answer
Donly S2 is correct

Step-by-step explanation

Given fx=2+x-x-1+x+1, x∈R ⇒fx=-x,x<-1x+2,-1≤x<03x+2,0≤x<1x+4,x≥1

⇒f'x=-1,x<-11,-1<x<03,0<x<11,x>1

∴  f'-32+f'-12+f'12+f'32

=-1+1+3+1=4

i.e. S1 is incorrect.

Now  ∫-22fxdx=∫-2-1fxdx+∫-10fxdx+∫01fxdx+∫12fxdx

=-x22-2-1+x+222-10+3x+22601+x+42212

-12+2+2-12+256-46+18-252

=32+32+72+112=12

i.e. S2 is correct.

 

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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.