JEE Main 2020MathematicsIndefinite IntegrationHardMCQ

JEE Main 2020 — Indefinite Integration Question with Solution

JEE Main 2020 (05 Sep Shift 1)

Question

If ∫e2x+2ex-e-x-1eex+e-xdx=g(x)eex+e-x+c, where c is a constant of integration, then g(0) is

Choose an option

Show full solutionCorrect option: D
Correct answer
D2

Step-by-step explanation

I=∫e2x+2ex-e-x-1eex+e-xdx

⇒I=∫e2x+ex-1eex+e-xdx+∫ex-e-xeex+e-xdx

⇒I=∫ex+1-e-xeex+e-x+xdx+eex+e-x+c

ex+e-x+x=u

ex-e-x+1dx=du

I=eex+e-x+x+eex+e-x+c=eex+e-xex+1+c

then g(x)=ex+1

g(0)=2

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About this question

This is a previous-year question from JEE Main 2020, 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.