JEE Main 2021MathematicsDeterminantsEasyNumerical

JEE Main 2021 — Determinants Question with Solution

JEE Main 2021 (27 Jul Shift 1)

Question

Let fx=sin2x-2+cos2xcos2x2+sin2xcos2xcos2xsin2xcos2x1+cos2x, x∈0,π. Then the maximum value of  fx is equal to

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Show full solutionCorrect answer: 6
Correct answer
6

Step-by-step explanation

Given,

fx=sin2x-2+cos2xcos2x2+sin2xcos2xcos2xsin2xcos2x1+cos2x

=-2-2020-1sin2xcos2x1+cos2xR1→R1-R2& R2→R2-R3

=-2cos2x+22+2cos2x+sin2x

=4+4cos2x-2cos2x-sin2x

⇒fx=4+2cos2x

We know, -1≤cos2x≤1

So, fxmax=4+2=6

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About this question

This is a previous-year question from JEE Main 2021, covering the Determinants 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.