Given,
f(x)=∣sin2x2+sin2xsin2x−2+cos2xcos2xcos2xcos2xcos2x1+cos2x∣
=∣−22sin2x−20cos2x0−11+cos2x∣(R1→R1−R2R2→R2−R3)
=−2(cos2x)+2(2+2cos2x+sin2x)
=4+4cos2x−2(cos2x−sin2x)
⇒f(x)=4+2cos2x
We know, −1≤cos2x≤1
So, f(x)max=4+2=6
Let f(x)=∣sin2x2+sin2xsin2x−2+cos2xcos2xcos2xcos2xcos2x1+cos2x∣,x∈[0,π]. Then the maximum value of f(x) is equal to
Held on 27 Jul 2021 · Verified 6 Jul 2026.
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