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(A) Let the determinant be $\Delta$. Applying row operations $R_{1} \rightarrow R_{1} - R_{2}$ and $R_{2} \rightarrow R_{2} - R_{3}$:$\Delta = \left|\

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$(-3, -1)$

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(A) Let the determinant be $\Delta$. Applying row operations $R_{1} ightarrow R_{1} - R_{2}$ and $R_{2} ightarrow R_{2} - R_{3}$:$\Delta = \left|\begin{array}{ccc} -1 & 1 & 0 \ 1 & 0 & -1 \ \cos ^{2} x & \sin ^{2} x & 1+\sin 2 x \end{array}ight|$Expanding along the first row:$\Delta = -1(0 - (-\sin ^{2} x)) - 1(1 + \sin 2 x + \cos ^{2} x) + 0$$\Delta = -\sin ^{2} x - 1 - \sin 2 x - \cos ^{2} x$Since $\sin ^{2} x + \cos ^{2} x = 1$,we have:$\Delta = -1 - 1 - \sin 2 x = -2 - \sin 2 x$We know that $-1 \leq \sin 2 x \leq 1$. For the minimum value $m$,$\sin 2 x = 1$,so $m = -2 - 1 = -3$.For the maximum value $M$,$\sin 2 x = -1$,so $M = -2 - (-1) = -1$.Thus,$(m, M) = (-3, -1)$.

Let $m$ and $M$ be respectively the minimum and maximum values of $\left|\begin{array}{ccc}\cos ^{2} x & 1+\sin ^{2} x & \sin 2 x \\ 1+\cos ^{2} x & \sin ^{2} x & \sin 2 x \\ \cos ^{2} x & \sin ^{2} x & 1+\sin 2 x\end{array}\right|$. Then the ordered pair $(m, M)$ is equal to

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