3 and 4 .Determinants and Matrices
hard

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

 

A

$(-3,-1)$

B

$(-4,-1)$

C

$(1,3)$

D

$(-3,3)$

(JEE MAIN-2020)

Solution

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

$R_{1} \rightarrow R_{1}-R_{2}, R_{2} \rightarrow R_{2}-R_{3}$

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

$=-1\left(\sin ^{2} x\right)-1\left(1+\sin 2 x+\cos ^{2} x\right)$

$=-\sin 2 x-2$

$m=-3, M=-1$

Standard 12
Mathematics

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