JEE Main 2022MathematicsIndefinite IntegrationMediumMCQ

JEE Main 2022 — Indefinite Integration Question with Solution

JEE Main 2022 (27 Jun Shift 1)

Question

∫x2+1exx+12dx=fxex+C, where C is a constant, then d3fdx3 at x=1 is equal to

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Show full solutionCorrect option: A
Correct answer
A34

Step-by-step explanation

 ∫x2+1exdxx+12=fxex+C

∫exx2-1x+12+2x+12dx=fxex+C

∫exx-1x+1+2x+12dx=fxex+C

We know that ∫exfx+f'x=exfx+c

Here fx=x-1x+1 & f'x=2x+12

So ∫exx-1x+1+2(x+1)2dx=fxex+C

⇒exx-1x+1+C=exfx+C

On comparing both sides we get fx=x-1x+1

So f'x=2x+12 & f"x=-4x+13

f'''x=12x+14 ,

Now f'''1=121+14=1216=34

 

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About this question

This is a previous-year question from JEE Main 2022, covering the Indefinite Integration 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.