JEE Main 2023MathematicsDefinite IntegrationEasyMCQ

JEE Main 2023 — Definite Integration Question with Solution

JEE Main 2023 (11 Apr Shift 2)

Question

Let the function f:0,2→ℝ be defined as fx=eminx2,x-x,x∈[0,1)ex-logex,x∈1,2, where t denotes the greatest integer less than or equal to t. Then the value of the integral ∫02xfxdx is

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Show full solutionCorrect option: D
Correct answer
D2e-12
 

Step-by-step explanation

Given,

fx=eminx2,x-x,x∈[0,1)ex-logex,x∈1,2
⇒fx=ex2x∈[0,1)e,x∈1,2

As x-lnx∈[1,2) for x∈[1,2], sox-lnx=1 and x-x=x, so x∈0,1, x2<x

Now solving the integral we get,∫02xfx=∫01x·ex2dx+∫12x·e dx

Now in first integral, let x2=t⇒2xdx=dt we get,

⇒∫02xfx=12∫01et dt+e·x2212

⇒∫02xfx=e-12+3e2

⇒∫02xfx=2e-12

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About this question

This is a previous-year question from JEE Main 2023, covering the Definite 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.