JEE Main 2020MathematicsDefinite IntegrationMediumMCQ

JEE Main 2020 — Definite Integration Question with Solution

JEE Main 2020 (06 Sep Shift 1)

Question

limx→1∫0(x-1)2tcost2dt(x-1)sin(x-1)

Choose an option

Show full solutionCorrect option: D
Correct answer
Dis equal to 0.

Step-by-step explanation

Applying L'Hospital's Rule, for numerator we have to apply Newton Leibnitz integral rule

limx→12(x-1)×(x-1)2cosx-14-0(x-1)cos(x-1)+sin(x-1)  00

Divide x-1

=limx→12(x-1)2cos(x-1)4cos(x-1)+sin(x-1)(x-1)

=01+1

=0

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About this question

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