Let f x = sin 2 x - 2 + cos 2 x cos 2 x 2 + sin 2 x cos 2 x cos 2 x sin 2 x cos 2 x 1 + cos 2 x ,   x…

Let fx=sin2x-2+cos2xcos2x2+sin2xcos2xcos2xsin2xcos2x1+cos2x, x0,π. Then the maximum value of  fx is equal to

Solution

Given,

fx=sin2x-2+cos2xcos2x2+sin2xcos2xcos2xsin2xcos2x1+cos2x

=-2-2020-1sin2xcos2x1+cos2xR1R1-R2& R2R2-R3

=-2cos2x+22+2cos2x+sin2x

=4+4cos2x-2cos2x-sin2x

fx=4+2cos2x

We know, -1cos2x1

So, fxmax=4+2=6

Asked in: JEE Main 2021 (27 Jul Shift 1)

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