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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

```
3 years ago

```							We Know thatSin^2x Also Cos^2x >= Cos^4xAdd sin^2x both sidesCos^2x + sin^2x >= Cos^4x + sin^2x 1 >= Cos^4x + sin^2xTherefore max of A is 1Also 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  + 3/4now Min =3/4 and Max = 1
```
3 years ago
```							Dear Student,Please find the solution to your problem in the attached image.Thanks and Regards ```
6 months ago
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