JEE Main 2019MathematicsContinuity and DifferentiabilityEasyMCQ

JEE Main 2019 — Continuity and Differentiability Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

If  fx=sin(p+1)x+sinxx,x<0q,x=0x+x2-xx3/2,x>0 is continuous at x=0 , then the ordered pair (p, q) is equal to:

Choose an option

Show full solutionCorrect option: D
Correct answer
D-32,12

Step-by-step explanation

∵   fx=sin⁡p+1x +sinxxqx2+x-xx3/2  :x<0:x=0:x>0 

∵f is continuous at x=0

L.H.L=

limh→0 sinp+1-h+sin(-h)-h=limh→0sinp+1h+sinhh

=limh→0sinp+1hhp+1p+1+limh→0sinhh=p+1+1

=p+2

R.H.L= limh→0 h2+h- hh32× h2+h+hh2+h+h=limh→0h2+h-hh32 h12h+1+1

limh→011+h+1=12

⇒L.H.L.=R.H.L.=f0⇒p+2=12=q
=p=-32     &  q=12 

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About this question

This is a previous-year question from JEE Main 2019, 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.