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3 and 4 .Determinants and Matrices
medium
The maximum value of
$f(x)=\left|\begin{array}{ccc} \sin ^{2} x & 1+\cos ^{2} x & \cos 2 x \\ 1+\sin ^{2} x & \cos ^{2} x & \cos 2 x \\ \sin ^{2} x & \cos ^{2} x & \sin 2 x \end{array}\right|, x \in R \text { is }$
A
$\sqrt{7}$
B
$\frac{3}{4}$
C
$\sqrt{5}$
D
$5$
(JEE MAIN-2021)
Solution
$C _{1}+ C _{2} \rightarrow C _{1}$
$\left|\begin{array}{ccc}2 & 1+\cos ^{2} x & \cos 2 x \\ 2 & \cos ^{2} x & \cos 2 x \\ 1 & \cos ^{2} x & \sin 2 x\end{array}\right|$
$R _{1}- R _{2} \rightarrow R _{1}$
$\left|\begin{array}{ccc}0 & 1 & 0 \\ 2 & \cos ^{2} x & \cos 2 x \\ 1 & \cos ^{2} x & \sin 2 x\end{array}\right|$
Open w.r.t. $R _{1}$
$-(2 \sin 2 x-\cos 2 x)$
$\cos 2 x-2 \sin 2 x=f(x)$
$\left. f ( x )\right|_{\max }=\sqrt{1+4}=\sqrt{5}$
Standard 12
Mathematics
