Let m and M be respectively the minimum and maximum values of ( cos^2x1+ sin^2x sin2x1+ cos^2x sin^2x sin2x cos^2x sin^2x1+ sin2x). Then the ordered pair (m, M) is equal to:

Question

Accepted Answer

Correct answer: B. (b) JEE Main 2020 Let f (x )=( cos^2x1+ sin^2x sin2x1+ cos^2x sin^2x sin2x cos^2x sin^2x1+ sin2x) After expanding it we get f (x )=-2- sin2x f^ ' (x )=-2 cos2x=0 ⇒ cos2x=0 2x= 2 , 3 2 x= 4 , 3 4 f^ ' (x )=4 sin2x So, f ' ' ( 4 )=4>0 (minima) f ( 3 4 )=-4<0 (maxima) So m=f_ min=f ( 4 )=-2-1=-3 And M=f_ max= ( 3 4 )=-2+1=-1 So (m, M) =(-3 ,-1)

$Let m and M be respectively the minimum and maximum values of (cos^{2} x 1 + sin^{2} x sin 2 x 1 + cos^{2} x sin^{2} x sin 2 x cos^{2} x sin^{2} x 1 + sin 2 x) . Then the ordered pair (m, M) is equal to:$

Solution

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