JEE Main 2021MathematicsContinuity and DifferentiabilityEasyMCQ

JEE Main 2021 — Continuity and Differentiability Question with Solution

JEE Main 2021 (27 Jul Shift 1)

Question

Let f:-π4,π4→R be defined as, fx=1+sinx3asinx,  -π4<x<0b,  x=0ecot4x/cot2x,  0<x<π4

If f is continuous at x=0 then the value of 6a+b2 is equal to:

Choose an option

Show full solutionCorrect option: C
Correct answer
C1+e

Step-by-step explanation

If fx is continuous at point x=a, then limx→0fx=limx→0+fx=limx→0-fx

From the question, we can say that limx→0fx=b

And limx→0+ecot4xcot2x=limx→0+etan2xtan4x=limx→0+etan2x2×2xtan4x4x=e12=b

And limx→0-1+sinx3asinx=elimx→0-sinx3asinx=e3a

From the above we can say that

e3a=e12

⇒a=16⇒6a=1

So, 6a+b2=1+e

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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.