JEE Main 2022MathematicsIndefinite IntegrationHardMCQ

JEE Main 2022 — Indefinite Integration Question with Solution

JEE Main 2022 (25 Jun Shift 1)

Question

Let g:0,∞→R be a differentiable function such that ∫xcosx-sinxex+1+gxex+1-xexex+12dx=xgxex+1+C, for all x>0, where C is an arbitrary constant. Then

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Show full solutionCorrect option: B
Correct answer
Bg-g' is increasing in 0,π2

Step-by-step explanation

∫xex+1cosx-sinxdx+∫gxex+1-xexex+12dx

=xex+1sinx+cosx-∫ex+1-xexex+12sinx+cosxdx+∫gxex+1-xexex+12dx

By comparison, we get, gx=sinx+cosx
⇒ gx=2sinx+π4

Since x∈0,π4 so, x+π4∈π4,π2

So gx is increasing in 0,π4

g'x=cosx-sinx

i.e. gx-g'x=2sinx is an increasing function in 0,π2.

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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.