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

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

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$\frac{3}{4} \le A \le 1$

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(B) Given $A = \sin^2 	heta + \cos^4 	heta$.Since $\cos^2 	heta \le 1$,we have $\cos^4 	heta \le \cos^2 	heta$.Thus,$A = \sin^2 	heta + \cos^4 	heta \le \sin^2 	heta + \cos^2 	heta = 1$. So,$A \le 1$.Now,express $A$ in terms of $\cos^2 	heta$:$A = (1 - \cos^2 	heta) + \cos^4 	heta = \cos^4 	heta - \cos^2 	heta + 1$.Let $x = \cos^2 	heta$,where $0 \le x \le 1$.Then $A = x^2 - x + 1 = (x - \frac{1}{2})^2 + \frac{3}{4}$.Since $(x - \frac{1}{2})^2 \ge 0$,the minimum value is $\frac{3}{4}$ (at $x = \frac{1}{2}$).Since $x \in [0, 1]$,the maximum value occurs at the boundaries $x=0$ or $x=1$,which is $1$.Therefore,$\frac{3}{4} \le A \le 1$.

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

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