f(x)=∣sin2x1+sin2xsin2x1+cos2xcos2xcos2xcos2xcos2xsin2x∣,x∈R
C1→C1+C2
=∣2211+cos2xcos2xcos2xcos2xcos2xsin2x∣
R1→R1−R2
=∣0211cos2xcos2x0cos2xsin2x∣
Expanding w.r.t. R1
=−(2sin2x−cos2x)
So, f(x)=cos2x−2sin2x
f(x)max=1+4=5
JEE Main 2021 — Mathematics Algebra
The maximum value of f(x)=∣sin2x1+sin2xsin2x1+cos2xcos2xcos2xcos2xcos2xsin2x∣,x∈R is
Held on 16 Mar 2021 · Verified 6 Jul 2026.
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