Let S = { sin^2 2theta : (sin^4 theta + cos^4 | vedclass

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(A) Let $u = \sin^2 2	heta$. We know $\sin^4 	heta + \cos^4 	heta = 1 - \frac{1}{2}\sin^2 2	heta = 1 - \frac{u}{2}$ and $\sin^6 	heta + \cos^6

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$4$

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(A) Let $u = \sin^2 2	heta$. We know $\sin^4 	heta + \cos^4 	heta = 1 - \frac{1}{2}\sin^2 2	heta = 1 - \frac{u}{2}$ and $\sin^6 	heta + \cos^6 	heta = 1 - \frac{3}{4}\sin^2 2	heta = 1 - \frac{3u}{4}$.For the quadratic equation to have real roots,the discriminant $D \ge 0$.$D = (\sin 2	heta)^2 - 4(1 - \frac{u}{2})(1 - \frac{3u}{4}) \ge 0$.$u - 4(1 - \frac{3u}{4} - \frac{u}{2} + \frac{3u^2}{8}) \ge 0$.$u - 4 + 3u + 2u - \frac{3u^2}{2} \ge 0$.$-\frac{3}{2}u^2 + 6u - 4 \ge 0 \implies 3u^2 - 12u + 8 \le 0$.The roots of $3u^2 - 12u + 8 = 0$ are $u = \frac{12 \pm \sqrt{144 - 96}}{6} = \frac{12 \pm 4\sqrt{3}}{6} = 2 \pm \frac{2}{\sqrt{3}}$.Since $u = \sin^2 2	heta \in [0, 1]$,the range for $u$ is $[0, 2 - \frac{2}{\sqrt{3}}]$.Thus,$\alpha = 0$ and $\beta = 2 - \frac{2}{\sqrt{3}}$.Calculating $3((\alpha - 2)^2 + (\beta - 1)^2) = 3((0 - 2)^2 + (2 - \frac{2}{\sqrt{3}} - 1)^2) = 3(4 + (1 - \frac{2}{\sqrt{3}})^2) = 3(4 + 1 - \frac{4}{\sqrt{3}} + \frac{4}{3}) = 3(5 + \frac{4}{3} - \frac{4}{\sqrt{3}}) = 15 + 4 - 4\sqrt{3} = 19 - 4\sqrt{3}$.

Let $S = \{ \sin^2 2\theta : (\sin^4 \theta + \cos^4 \theta)x^2 + (\sin 2\theta)x + (\sin^6 \theta + \cos^6 \theta) = 0 \text{ has real roots} \}$. If $\alpha$ and $\beta$ are the smallest and largest elements of the set $S$,respectively,then $3((\alpha - 2)^2 + (\beta - 1)^2)$ equals:

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