JEE Main 2021MathematicsContinuity and DifferentiabilityHardNumerical

JEE Main 2021 — Continuity and Differentiability Question with Solution

JEE Main 2021 (25 Jul Shift 2)

Question

Consider the function fx=Pxsinx-2, x≠2, and fx=7, x=2where Px is a polynomial such that P"x is always a constant and P3=9. If fx is continuous at x=2, then P5 is equal to __________.

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Show full solutionCorrect answer: 39
Correct answer
39

Step-by-step explanation

We have, fx=Pxsinx-2,7,x≠2x=2

P"x= Constant

⇒Px is a 2 degree polynomial

Let Px=x-2ax+b

fx is continuous at x=2

f2+=f2-

Now, limx→2+Pxsinx-2=7

⇒limx→2+x-2ax+bsinx-2=7

⇒2a+b=7

⇒P3=3-23a+b=9

⇒3a+b=9

⇒a=2,b=3

Hence, P5=5-22.5+3

=3.13

=39.

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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.