The maximum value of $f\left(x\right)=\left|\begin{array}{ccc}{\sin }^{2}x& 1+{\cos }^{2}x& \cos 2x\\ 1+{\sin }^{2}x& {\cos }^{2}x& \cos 2x\\ {\sin }^{2}x& {\cos }^{2}x& \sin 2x\end{array}\right|$$,x\in R$ is:

The correct option is B

$\sqrt{5}$

Explanation for the correct option:

Solve for the maximum value of $f\left(x\right)=\left|\begin{array}{ccc}{\sin }^{2}x& 1+{\cos }^{2}x& \cos 2x\\ 1+{\sin }^{2}x& {\cos }^{2}x& \cos 2x\\ {\sin }^{2}x& {\cos }^{2}x& \sin 2x\end{array}\right|$

$\begin{array}{rcl}{C}_1& \to & C_1+C_2\\ & =& \left|\begin{array}{ccc}2& 1+{\cos }^{2}x& \cos 2x\\ 2& {\cos }^{2}x& \cos 2x\\ 1& {\cos }^{2}x& \sin 2x\end{array}\right|\\ R_1& \to & R_1-R_2\\ & =& \left|\begin{array}{ccc}0& 1& 0\\ 2& {\cos }^{2}x& \cos 2x\\ 1& {\cos }^{2}x& \sin 2x\end{array}\right|\\ & =& (-1)\left[2\sin 2x-\cos 2x\right]\\ f\left(x\right)& =& \cos 2x-2\sin 2x\end{array}$

The maximum value of $a\cos x+b\sin x$ is $\sqrt{a^2+b^2}$

Therefore the Maximum value of $f\left(x\right)=\sqrt{1^2+{\left(-2\right)}^2}$

$=\sqrt{5}$

Hence the maximum value of $f\left(x\right)=\left|\begin{array}{ccc}{\sin }^{2}x& 1+{\cos }^{2}x& \cos 2x\\ 1+{\sin }^{2}x& {\cos }^{2}x& \cos 2x\\ {\sin }^{2}x& {\cos }^{2}x& \sin 2x\end{array}\right|$is $\sqrt{5}$.

Hence, the correct answer is option (B)