Given A=sin^2x+cos^4x,then for all real x, min≤A≤max

Question

Ritesh Khatri · Answer

We Know that Sin^2x Also Cos^2x >= Cos^4x Add sin^2x both sides Cos^2x + sin^2x >= Cos^4x + sin^2x 1 >= Cos^4x + sin^2x Therefore max of A is 1 Also both terms Cos^4x + sin^2x = (1 – sin^2x)^2 + sin^2x = 1 + sin^4x – 2sin^2x + sin^2x =Sin^4x – sinn^2x + 1 = (sin^2x – 1/2)^2 + 3/4 now Min =3/4 and Max = 1

Rishi Sharma · Answer

Dear Student, Please find the solution to your problem in the attached image. Thanks and Regards

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Given A=sin^2x+cos^4x,then for all real x, min≤A≤max

Given A=sin^2x+cos^4x,then for all real x, min≤A≤max

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