1.

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)`

Answer» `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`
`=>3/4 le 1- 1/4sin^2 2theta le 1`
`=>3/4 le A le 1`
So, option - `(b)` is the correct option.


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