JEE Main 2022MathematicsDefinite IntegrationHardMCQ

JEE Main 2022 — Definite Integration Question with Solution

JEE Main 2022 (29 Jun Shift 2)

Question

Let f be a real valued continuous function on 0, 1 and fx=x+∫01x-tftdt. Then which of the following points x, y lies on the curve y=fx?

Choose an option

Show full solutionCorrect option: D
Correct answer
D6, 8

Step-by-step explanation

Given,

fx=x1+∫01ftdt-∫01tftdt

fx=Ax-B ...i

A=1+∫01ftdt=1+∫01At-Bdt

⇒A=21-B ...ii

Also B=∫01tftdt=∫01At2-Btdt

A=92B ...iii

From ii, iii

A=1813, B=413

So, fx=1813x-413

 ⇒f6=18×6-413=8

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About this question

This is a previous-year question from JEE Main 2022, 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.