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

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(D) Given $A = \cos^2 	heta + \sin^4 	heta$.Since $\sin^2 	heta \le 1$,we have $A = \cos^2 	heta + \sin^2 	heta \cdot \sin^2 	heta \le \cos^2

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$3/4 \le A \le 1$

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(D) Given $A = \cos^2 	heta + \sin^4 	heta$.Since $\sin^2 	heta \le 1$,we have $A = \cos^2 	heta + \sin^2 	heta \cdot \sin^2 	heta \le \cos^2 	heta + \sin^2 	heta = 1$.Thus,$A \le 1$.Now,express $A$ in terms of $\sin^2 	heta$:$A = (1 - \sin^2 	heta) + \sin^4 	heta = \sin^4 	heta - \sin^2 	heta + 1$.Let $x = \sin^2 	heta$,where $0 \le x \le 1$. Then $A = x^2 - x + 1$.Completing the square: $A = (x - 1/2)^2 + 3/4$.Since $(x - 1/2)^2 \ge 0$,the minimum value of $A$ is $3/4$ when $x = 1/2$ (i.e.,$\sin^2 	heta = 1/2$).Therefore,$3/4 \le A \le 1$.

If $A = \cos^2 \theta + \sin^4 \theta$,then for all values of $\theta$:

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