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If $\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}1+\sin ^{2} \mathrm{x} & \cos ^{2} \mathrm{x} & 4 \sin 2 \mathrm{x} \\ \sin ^{2} \mathrm{x} & 1+\cos ^{2} \mathrm{x} & 4 \sin 2 \mathrm{x} \\ \sin ^{2} \mathrm{x} & \cos ^{2} \mathrm{x} & 1+4 \sin 2 \mathrm{x}\end{array}\right|$
What is the maximum value of $\mathrm{f}(\mathrm{x})$?
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What is the maximum value of $\mathrm{f}(\mathrm{x})$?
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Verified Answer
The correct answer is:
6
$\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}1+\sin ^{2} \mathrm{x} & \cos ^{2} x & 4 \sin 2 \mathrm{x} \\ \sin ^{2} \mathrm{x} & 1+\cos ^{2} \mathrm{x} & 4 \sin 2 \mathrm{x} \\ \sin ^{2} \mathrm{x} & \cos ^{2} \mathrm{x} & 1+4 \sin 2 \mathrm{x}\end{array}\right|$
Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}$
$=\left|\begin{array}{ccc}2 & \cos ^{2} x & 4 \sin 2 x \\ 2 & 1+\cos ^{2} x & 4 \sin 2 x \\ 1 & \cos ^{2} x & 1+4 \sin 2 x\end{array}\right|$
(Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\left.\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\right)$
$=\left|\begin{array}{ccc}2 & \cos ^{2} x & 4 \sin 2 x \\ 0 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$
$\mathrm{f}(\mathrm{x})=2+4 \sin 2 \mathrm{x}$
$\therefore \quad-1 \leq \sin 2 x \leq 1$, maximum value of $\sin 2 x=1$
Thus, maximum value of $f(x)=2+4=6$
Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}$
$=\left|\begin{array}{ccc}2 & \cos ^{2} x & 4 \sin 2 x \\ 2 & 1+\cos ^{2} x & 4 \sin 2 x \\ 1 & \cos ^{2} x & 1+4 \sin 2 x\end{array}\right|$
(Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\left.\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\right)$
$=\left|\begin{array}{ccc}2 & \cos ^{2} x & 4 \sin 2 x \\ 0 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$
$\mathrm{f}(\mathrm{x})=2+4 \sin 2 \mathrm{x}$
$\therefore \quad-1 \leq \sin 2 x \leq 1$, maximum value of $\sin 2 x=1$
Thus, maximum value of $f(x)=2+4=6$
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