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

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$\frac{\pi}{2}$

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(D) Given the quadratic equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$.Let the roots be $x_1$ and $x_2$.From the properties of roots,$x_1 + x_2 = \sin \alpha - 2$ and $x_1 x_2 = -(1 + \sin \alpha)$.The sum of the squares of the roots is given by $S = x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2$.Substituting the values: $S = (\sin \alpha - 2)^2 - 2(-(1 + \sin \alpha)) = \sin^2 \alpha - 4 \sin \alpha + 4 + 2 + 2 \sin \alpha = \sin^2 \alpha - 2 \sin \alpha + 6$.To find the minimum value,we can rewrite this as $S = (\sin \alpha - 1)^2 + 5$.Since the range of $\sin \alpha$ is $[-1, 1]$,the expression $( \sin \alpha - 1)^2$ is minimum when $\sin \alpha = 1$.Thus,$\sin \alpha = 1$ implies $\alpha = \frac{\pi}{2}$.

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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