Given `A=sin^2theta+cos^4theta,` then for all real `theta,` (a)`1lt=Alt=2` (b) `3/4lt=Alt=1` (c)`(13)/(16)lt=Alt=1` (d) `3/4lt=Alt=(13)/(16)`

Explanation

`A = sin^2theta+cos^4theta`

`=>A = 1-cos^2theta+cos^4theta`

`=>A= 1-cos^2theta(1-cos^2theta)`

`=>A= 1-cos^2theta(sin^2theta)`

`=>A= 1-1/4(2sinthetacostheta)^2`

`=>A= 1-1/4sin^2 2theta`

Now, we know, `0 le sin^2 2theta le 1`

`=>0 le 1/4sin^2 2theta le 1/4`

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