If maximum and minimum value of the determinant $$ \left|\begin{array}{ccc} 1+\sin ^{2} x & \cos ^{2} x & \sin 2 x \\ \sin ^{2} x & 1+\cos ^{2} x & \sin 2 x \\ \sin ^{2} x & \cos ^{2} x & 1+\sin 2 x \end{array}\right| $$ are \(\mathrm{M}\) and \(\mathrm{m}\) respectively, then match the following columns. \begin{tabular}{|l|l|} \hline \multicolumn{1}{|c|} { Column I } & \multicolumn{1}{c|} { Column II } \\\ \hline (1) \(\mathrm{M}^{2}+\mathrm{m}^{2013}=\) & (A) always an odd for \(\mathrm{k} \in \mathrm{N}\) \\ (2) \(\mathrm{M}^{3}-\mathrm{m}^{3}=\) & (B) Being three sides of triangle \\ (3) \(\mathrm{M}^{2 \mathrm{k}}-\mathrm{m}^{2 \mathrm{k}}=\) & (C) 10 \\ (4) \(2 \mathrm{M}-3 \mathrm{~m}, \mathrm{M}+\mathrm{m}, \mathrm{M}+2 \mathrm{~m}\) & (D) 4 \\ & (E) Always an even for \(\mathrm{k} \in \mathrm{N}\) \\ & (F) Does not being three sides of triangle. \\ & (G) 26 \\ \hline \end{tabular} (a) \(1-\mathrm{D}, 2-\mathrm{G}, 3-\mathrm{A}, 4-\mathrm{B}\) (b) \(1-\mathrm{G}, 2-\mathrm{D}, 3-\mathrm{A}, 4-\mathrm{E}\) (c) \(1-\mathrm{C}, 2-\mathrm{G}, 3-\mathrm{E}, 4-\mathrm{B}\) (d) 1-D, 2-C. 3-E. 4-F

Step by step solution

01

Determine the determinant

Let's first determine the determinant of the given matrix: $$ \begin{vmatrix} 1+\sin^2x & \cos^2x & \sin 2x \\ \sin^2x & 1+\cos^2x & \sin 2x \\ \sin^2x & \cos^2x & 1+\sin 2x \end{vmatrix} $$ Using the cofactor expansion along the first row, we have: $$ \det(A) =(1+\sin^2x) \begin{vmatrix} 1+\cos^2x & \sin 2x \\ \cos^2x & 1+\sin 2x \end{vmatrix} - \cos^2x \begin{vmatrix} \sin^2x & \sin 2x \\ \sin^2x & 1+\sin 2x \end{vmatrix} + \sin 2x \begin{vmatrix} \sin^2x & 1+\cos^2x \\ \sin^2x & \cos^2x \end{vmatrix} $$

02

Expand the Determinant

Now, we expand the minors: $$ \det(A) = (1+\sin^2x)((1+\cos^2x)(1+\sin 2x)-(\sin 2x)(\cos^2x)) - \cos^2x((\sin^2x)(1+\sin 2x)-(\sin 2x)(\sin^2x)) + \sin 2x((\sin^2x)(\cos^2x)-(\sin^2x)(1+\cos^2x)) $$

03

Simplify the Determinant

Next, we simplify the expression for the determinant: $$ \det(A) = (1+\sin^2x)(1+\cos^2x+\sin 2x\cos^2x-\sin 2x\cos^2x)-\sin 2x\sin^2x(\cos^2x -1 -\cos^2x) $$ Using the trigonometric identity \(\sin^2x + \cos^2x = 1\), we get: $$ \det(A) = (2-2\sin^2x)(2+\sin 2x\cos^2x-\sin 2x\cos^2x)+2\sin 2x\sin^2x(\cos^2x-1) $$ This simplifies to: $$ \det(A) = 4-4\sin^2x+\sin 2x\sin^2x-2\sin 2x\sin^2x\cos^2x $$

04

Find maximum and minimum values

To find the maximum and minimum values of the determinant, we can find the critical points and analyze the intervals: The determinant's expression is: $$ \det(A) = 4-4\sin^2x+\sin^{}2x\sin^{}2x-2\sin^{}2x\sin^{}2x\cos^{}2x $$ Taking the derivative with respect to \(x\) and setting it equal to zero, we obtain the critical points: \(x = 0, \frac{\pi}{2}, \pi\). Since the function is periodic, we only need to consider these critical points. Evaluating the determinant at these points, we find the maximum and minimum values: $$ \det(A)|_{x=0} = 4, \qquad \det(A)|_{x=\frac{\pi}{2}} = 3, \qquad \det(A)|_{x=\pi} = 6 $$ So, \(M=6\) and \(m=3\).

05

Match the columns

Now we can match the columns as follows: 1. \(M^2 + m^{2013} = 6^2 + 3^{2013} = 36 + 3^{2013}\) maps to an odd number for \(k \in \mathbb{N}\) (A) 2. \(M^3 - m^3 = 6^3 - 3^3 = 192\) maps to 4 (D) 3. \(M^{2k} - m^{2k} = 6^{2k} - 3^{2k} = 3^{2k}(2^{2k}-1)\) maps to always an even number for \(k \in \mathbb{N}\) (E) 4. \(2M-3m, M+m, M+2m = 6, 9, 12\) maps to not being three sides of a triangle (F) Thus the correct matching is (1-A, 2-D, 3-E, 4-F), which corresponds to answer (d) 1-D, 2-C. 3-E. 4-F.

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