JEE Main 2022MathematicsLimitsMediumMCQ

JEE Main 2022 — Limits Question with Solution

JEE Main 2022 (25 Jun Shift 2)

Question

limx→π2tan2x2sin2x+3sinx+412-sin2x+6sinx+212 is equal to

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Show full solutionCorrect option: A
Correct answer
A112

Step-by-step explanation

Given, limx→π2tan2x2sin2x+3sinx+4-sin2x+6sinx+2

=limx→π2tan2x2sin2x+3sinx+42-sin2x+6sinx+222sin2x+3sinx+4+sin2x+6sinx+2

=limx→π2tan2xsin2x-3sinx+29+9

=limx→π2tan2xsinx-1sinx-26

=16limx→π2tan2x1-sinx

=16limx→π2sin2x1-sinx1-sinx1+sinx=112

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About this question

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