JEE Main 2020MathematicsContinuity and DifferentiabilityMediumMCQ

JEE Main 2020 — Continuity and Differentiability Question with Solution

JEE Main 2020 (09 Jan Shift 1)

Question

If fx=sina+2x+sinxx;x<0b;x=0x+3x21/3-x1/3x1/3;x>0 is continuous at x=0 , then a+2b is equal to:

Choose an option

Show full solutionCorrect option: C
Correct answer
C0

Step-by-step explanation

f0-=limx→0-sinaxcos2xx+sin2xcosaxx+sinxx

=limx→0-acos2x+2cosax+1

=a+3

Also, f0=b
f0+=limx→0x13x131+3x13-1x , this is 00form

Apply LH Rule 

⇒f0+=1

∴ a=-2

b=1

∴a+2b=0

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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.