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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

Grade:11

2 Answers

Ritesh Khatri
76 Points
6 years ago
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
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find the solution to your problem in the attached image.

Thanks and Regards
645-366_2.PNG

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