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

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Question

(A) Given $p = \sin^2 	heta + \cos^4 	heta$. Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we have: $p = \sin^2 	heta + (1 - \sin^2 	heta

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $p = \sin^2 	heta + \cos^4 	heta$. Using the identity $\cos^2 	heta = 1 - \sin^2 	heta$,we have: $p = \sin^2 	heta + (1 - \sin^2 	heta)^2$ $p = \sin^2 	heta + 1 + \sin^4 	heta - 2 \sin^2 	heta$ $p = \sin^4 	heta - \sin^2 	heta + 1$. Let $x = \sin^2 	heta$. Since $	heta$ is real,$0 \le x \le 1$. Then $p = f(x) = x^2 - x + 1$. This is a parabola opening upwards with vertex at $x = -b/(2a) = -(-1)/(2 	imes 1) = 1/2$. Since $1/2$ lies 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 $[0, 1]$. $f(0) = 0^2 - 0 + 1 = 1$. $f(1) = 1^2 - 1 + 1 = 1$. Thus,the range of $p$ is $\frac{3}{4} \le p \le 1$.

If $p = \sin^2 \theta + \cos^4 \theta$,then for all real values of $\theta$,which of the following is true?

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