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Let m and n be respectively the min and max value of \begin{vmatrix} \cos^{2}x &1+\sin^2x &\sin2x \\1+\cos^{2}x & \sin^2x &\sin2x \\ \cos^2x&\sin^2x & 1+\sin2x \end{vmatrix} then ordered pair (m,n) = ?
Option: 1 (-4,-1) 
Option: 2 (1,3)   
Option: 3 (-3,3)   
Option: 4 (-3,-1)

Answers (1)

best_answer

\begin{array}{l} R_{2} \rightarrow R_{2}-R_{1} \\ R_{3} \rightarrow R_{3}-R_{1} \end{array}

\left|\begin{array}{ccc} \cos ^{2} x & 1+\sin ^{2} x & \sin 2 x \\ 1 & -1 & 0 \\ 0 & -1 & 1 \end{array}\right|

\cos ^{2} x(-1)-\left(1+\sin ^{2} x\right)[1]+\sin 2 x(-1) \\ \\-\cos ^{2} x-1-\sin ^{2} x-\sin 2 x

-\sin 2 x-2

Minimum = -3 and Maximum = -1

Correct Answer: Option D

Posted by

himanshu.meshram

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