JEE Main 2021MathematicsContinuity and DifferentiabilityEasyMCQ

JEE Main 2021 — Continuity and Differentiability Question with Solution

JEE Main 2021 (18 Mar Shift 2)

Question

Let f : R→R be a function defined as

fx=sin(a+1)x+sin2x2x,if x<0b,if x=0x+bx3-xbx5/2,if x>0

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

Choose an option

Show full solutionCorrect option: D
Correct answer
D-32

Step-by-step explanation

f(x) is continuous at x=0

limx→0+fx=f0=limx→0-fx   ...1

f(0)=b   ...2

limx→0-fx=limx→0-sin(a+1)x2x+sin2x2x

=a+12+1   ...3

limx→0+fx=limx→0+x+bx3-xbx5/2

=limx→0+x+bx3-xbx5/2x+bx3+x

=limx→0+xx1+bx2+1=12   ...4

Use (2),(3) & (4) in (1)

12=b=a+12+1

⇒b=12,a=-2
a+b=-32

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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.