Let m and M be respectively the minimum and maximum values of | \beginmatrix \cos ^2x & 1 + \sin ^2x & \sin 2x 1 + \cos ^2x & \sin ^2x & \sin 2x \cos ^2x & \sin ^2x & 1 + \sin 2x \endmatrix | Then the ordered pair (m, M) is equal to :

Matrices and Determinants PYQ — JEE Main 2020 · 6 Sep · Shift 1 · Q37 with Solution

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Correct answer: A

\cos^2 x & 1+\sin^2 x & \sin 2x\\ 1+\cos^2 x & \sin^2 x & \sin 2x\\ \cos^2 x & \sin^2 x & 1+\sin 2x \end{vmatrix}.$$ We need the minimum and maximum values of $D$. 2. Use row operations that do not change the determinant: - $R_1 \to R_1-R_3$ - $R_2 \to R_2-R_3$ Then $$D=\begin{vmatrix} 0 & 1 & -1\\ 1 & 0 & -1\\ \cos^2 x & \sin^2 x & 1+\sin 2x \end{vmatrix}.$$ 3. Expand along the first row: $$D=0\cdot C_{11}+1\cdot C_{12}+(-1)\cdot C_{13}.$$ Now, $$C_{12}=(-1)^{1+2}\begin{vmatrix}1 & -1\\ \cos^2 x & 1+\sin 2x\end{vmatrix} =-\big((1)(1+\sin 2x)-(-1)\cos^2 x\big) =-(1+\sin 2x+\cos^2 x),$$ and $$C_{13}=(-1)^{1+3}\begin{vmatrix}1 & 0\\ \cos^2 x & \sin^2 x\end{vmatrix} =\sin^2 x.$$ Hence $$D=-(1+\sin 2x+\cos^2 x)-\sin^2 x.$$ Since $$\cos^2 x+\sin^2 x=1,$$ we get $$D=-(1+\sin 2x+1)=-(2+\sin 2x).$$ 4. Since $$-1\le \sin 2x\le 1,$$ therefore $$2+\sin 2x\in [1,3].$$ So $$D=-(2+\sin 2x)\in [-3,-1].$$ Thus, $$m=-3,\quad M=-1.$$ 5. Therefore the ordered pair is $$\boxed{(-3,-1)}.$$ This corresponds to option A.

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Correct answer: A

\cos^2 x & 1+\sin^2 x & \sin 2x\\ 1+\cos^2 x & \sin^2 x & \sin 2x\\ \cos^2 x & \sin^2 x & 1+\sin 2x \end{vmatrix}.$$ We need the minimum and maximum values of $D$. 2. Use row operations that do not change the determinant: - $R_1 \to R_1-R_3$ - $R_2 \to R_2-R_3$ Then $$D=\begin{vmatrix} 0 & 1 & -1\\ 1 & 0 & -1\\ \cos^2 x & \sin^2 x & 1+\sin 2x \end{vmatrix}.$$ 3. Expand along the first row: $$D=0\cdot C_{11}+1\cdot C_{12}+(-1)\cdot C_{13}.$$ Now, $$C_{12}=(-1)^{1+2}\begin{vmatrix}1 & -1\\ \cos^2 x & 1+\sin 2x\end{vmatrix} =-\big((1)(1+\sin 2x)-(-1)\cos^2 x\big) =-(1+\sin 2x+\cos^2 x),$$ and $$C_{13}=(-1)^{1+3}\begin{vmatrix}1 & 0\\ \cos^2 x & \sin^2 x\end{vmatrix} =\sin^2 x.$$ Hence $$D=-(1+\sin 2x+\cos^2 x)-\sin^2 x.$$ Since $$\cos^2 x+\sin^2 x=1,$$ we get $$D=-(1+\sin 2x+1)=-(2+\sin 2x).$$ 4. Since $$-1\le \sin 2x\le 1,$$ therefore $$2+\sin 2x\in [1,3].$$ So $$D=-(2+\sin 2x)\in [-3,-1].$$ Thus, $$m=-3,\quad M=-1.$$ 5. Therefore the ordered pair is $$\boxed{(-3,-1)}.$$ This corresponds to option A.