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

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(C) Given $A = \sin^2 x + \cos^4 x$.Let $t = \sin^2 x$,then $0 \le t \le 1$. Since $\cos^2 x = 1 - \sin^2 x = 1 - t$,we have $\cos^4 x = (1 - t)^2$.Su

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

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(C) Given $A = \sin^2 x + \cos^4 x$.Let $t = \sin^2 x$,then $0 \le t \le 1$. Since $\cos^2 x = 1 - \sin^2 x = 1 - t$,we have $\cos^4 x = (1 - t)^2$.Substituting these into the expression for $A$:$A = t + (1 - t)^2 = t + 1 - 2t + t^2 = t^2 - t + 1$.This is a quadratic expression in $t$ where $t \in [0, 1]$.The vertex of the parabola $f(t) = t^2 - t + 1$ is at $t = -(-1)/(2 	imes 1) = 1/2$.Since $1/2$ is within the interval $[0, 1]$,the minimum value is $f(1/2) = (1/2)^2 - (1/2) + 1 = 1/4 - 1/2 + 1 = 3/4$.The maximum value occurs at the boundaries $t=0$ or $t=1$:$f(0) = 0^2 - 0 + 1 = 1$.$f(1) = 1^2 - 1 + 1 = 1$.Thus,the range of $A$ is $\frac{3}{4} \le A \le 1$.

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

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