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

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(C) Given $A = \sin^2 x + \cos^4 x$.We can write this as $A = \sin^2 x + (\cos^2 x)^2$.Since $\sin^2 x = 1 - \cos^2 x$,let $t = \cos^2 x$,where $0 \le

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

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(C) Given $A = \sin^2 x + \cos^4 x$.We can write this as $A = \sin^2 x + (\cos^2 x)^2$.Since $\sin^2 x = 1 - \cos^2 x$,let $t = \cos^2 x$,where $0 \le t \le 1$.Then $A = (1 - t) + t^2 = t^2 - t + 1$.This is a quadratic expression in $t$. The vertex of the parabola $f(t) = t^2 - t + 1$ occurs at $t = -b/(2a) = -(-1)/(2 	imes 1) = 1/2$.Since $1/2$ is in 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 endpoints of the interval $[0, 1]$.At $t = 0$,$f(0) = 0^2 - 0 + 1 = 1$.At $t = 1$,$f(1) = 1^2 - 1 + 1 = 1$.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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