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(D) Given $f(	heta) = \left|\begin{array}{ccc} -\sin^2 	heta & -1-\sin^2 	heta & 1 \ -\cos^2 	heta & -1-\cos^2 	heta & 1 \ 12 & 10 & -2 \end{ar

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$(-4, 0)$

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(D) Given $f(	heta) = \left|\begin{array}{ccc} -\sin^2 	heta & -1-\sin^2 	heta & 1 \ -\cos^2 	heta & -1-\cos^2 	heta & 1 \ 12 & 10 & -2 \end{array}ight|$.Applying the column operation $C_2 ightarrow C_2 - C_1$:$f(	heta) = \left|\begin{array}{ccc} -\sin^2 	heta & -1 & 1 \ -\cos^2 	heta & -1 & 1 \ 12 & -2 & -2 \end{array}ight|$.Applying $C_3 ightarrow C_3 + C_2$:$f(	heta) = \left|\begin{array}{ccc} -\sin^2 	heta & -1 & 0 \ -\cos^2 	heta & -1 & 0 \ 12 & -2 & -4 \end{array}ight|$.Expanding along $C_3$:$f(	heta) = -4 [(-\sin^2 	heta)(-1) - (-1)(-\cos^2 	heta)] = -4 [\sin^2 	heta - \cos^2 	heta] = -4 [-\cos 2	heta] = 4 \cos 2	heta$.Given $	heta \in \left[\frac{\pi}{4}, \frac{\pi}{2}ight]$,then $2	heta \in \left[\frac{\pi}{2}, \piight]$.For $2	heta \in [\frac{\pi}{2}, \pi]$,$\cos 2	heta$ ranges from $-1$ to $0$.Thus,$f(	heta) = 4 \cos 2	heta$ ranges from $4(-1) = -4$ to $4(0) = 0$.Therefore,$m = -4$ and $M = 0$,so the ordered pair $(m, M) = (-4, 0)$.

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