JEE Main 2019MathematicsDefinite IntegrationMediumMCQ

JEE Main 2019 — Definite Integration Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

If ∫0xftdt=x2+∫x1t2ftdt, then f'12 is

Choose an option

Show full solutionCorrect option: B
Correct answer
B2425

Step-by-step explanation

Differentiating both the sides w.r.t. x

fx=2x-x2fx
⇒fx=2x1+x2
⇒f'x=1+x2×2-2x×2x1+x22
=2-2x21+x22

Hence, f'12=2425

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Definite Integration chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, 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.