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3 and 4 .Determinants and Matrices
hard
माना $m$ तथा $M \left|\begin{array}{ccc}\cos ^{2} x & 1+\sin ^{2} x & \sin 2 x \\ 1+\cos ^{2} x & \sin ^{2} x & \sin 2 x \\ \cos ^{2} x & \sin ^{2} x & 1+\sin 2 x \end{array}\right|$ के, क्रमशः न्यूनतम तथा अधिकतम मान हैं, तो क्रमित युग्म $( m , M )$ बराबर है
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
