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(D) Let the roots of the equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$ be $x_1$ and $x_2$.From the properties of quadratic equations,we

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$\frac{\pi}{2}$

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(D) Let the roots of the equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$ be $x_1$ and $x_2$.From the properties of quadratic equations,we have:$x_1 + x_2 = \sin \alpha - 2$$x_1 x_2 = -(1 + \sin \alpha)$We want to minimize $S = x_1^2 + x_2^2$.$S = (x_1 + x_2)^2 - 2x_1 x_2$$S = (\sin \alpha - 2)^2 - 2(-(1 + \sin \alpha))$$S = \sin^2 \alpha - 4 \sin \alpha + 4 + 2 + 2 \sin \alpha$$S = \sin^2 \alpha - 2 \sin \alpha + 6$To minimize $S$,let $u = \sin \alpha$,where $u \in [-1, 1]$.$S(u) = u^2 - 2u + 6 = (u - 1)^2 + 5$.The minimum value occurs at $u = 1$.Since $\sin \alpha = 1$,we have $\alpha = \frac{\pi}{2}$.

If the sum of the squares of the roots of the equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$ is least,then $\alpha$ is

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