If A = sin^2 theta + cos^4 theta,then for all | vedclass

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(B) We have,$A = \sin^2 	heta + \cos^4 	heta$ $= (1 - \cos^2 	heta) + \cos^4 	heta$ $= \cos^4 	heta - \cos^2 	heta + 1$ To find the range,let $x

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$[\frac{3}{4}, 1]$

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(B) We have,$A = \sin^2 	heta + \cos^4 	heta$ $= (1 - \cos^2 	heta) + \cos^4 	heta$ $= \cos^4 	heta - \cos^2 	heta + 1$ To find the range,let $x = \cos^2 	heta$,where $x \in [0, 1]$. Then $A = x^2 - x + 1$. This is a quadratic in $x$ with vertex at $x = -b/(2a) = -(-1)/(2 	imes 1) = 1/2$. Since $1/2 \in [0, 1]$,the minimum value is at $x = 1/2$: $A_{min} = (1/2)^2 - (1/2) + 1 = 1/4 - 1/2 + 1 = 3/4$. The maximum value occurs at the boundaries $x=0$ or $x=1$: At $x = 0$,$A = 0^2 - 0 + 1 = 1$. At $x = 1$,$A = 1^2 - 1 + 1 = 1$. Thus,the range of $A$ is $[3/4, 1]$.

If $A = \sin^2 \theta + \cos^4 \theta$,then for all values of $\theta$,$A$ lies in the interval

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