JEE Main 2022MathematicsDefinite IntegrationHardMCQ

JEE Main 2022 — Definite Integration Question with Solution

JEE Main 2022 (28 Jun Shift 2)

Question

Let f:R→R be a differentiable function such that fπ4=2,fπ2=0 and f'π2=1 and let gx=∫xπ4f'tsect+tantsectftdt for x∈π4,π2. Then limx→π2-gx is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B3

Step-by-step explanation

Given,

gx=∫xπ4f'tsect+tantsectftdt

⇒gx=∫xπ4dft·sect⇒gx=ftsectxπ4

gx=fπ4secπ4-fx·secx

gx=2-fxsecx=2-fxcosx

Now taking limit both side, we get

limx→π2-gx=2-limx→π2-fxcosx

Using L'Hospital Rule

=2-limx→π2-f'x-sinx

=2+f'π2sinπ2=2+11=3

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 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.