What is the maximum value of $F\left( x \right) = \begin{vmatrix}{{{\sin }^2}x}&{1 + {{\cos }^2}x}&{\cos 2x}\\{1 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x}\\{{{\sin }^2}x}&{{{\cos }^2}x}&{\sin 2x}\end{vmatrix}$, $x \in \mathbb{R}$?
A. $\sqrt 7 $
B. $\sqrt 5 $
C. $5$
D. $\dfrac{3}{4}$

Hint: Expand the determinant. The value of the determinant will be a function of $\sin x$ and $\cos x$. Using the fact that the value of $\sin x$ and $\cos x$ lies between $\left( { - 1} \right)$ and $\left( 1 \right)$, find the maximum value of the function $F\left( x \right)$.

Formula Used:
${\sin ^2}x + {\cos ^2}x = 1$
The maximum value of the expression $A\sin \theta + B\cos \theta $ is $\sqrt {{A^2} + {B^2}} $

Complete step by step solution:
The given function is $F\left( x \right) = \begin{vmatrix}{{{\sin }^2}x}&{1 + {{\cos }^2}x}&{\cos 2x}\\{1 + {{\sin }^2}x}&{{{\cos }^2}x}&{\cos 2x}\\{{{\sin }^2}x}&{{{\cos }^2}x}&{\sin 2x}\end{vmatrix}$, $x \in \mathbb{R}$
Expanding the determinant with respect to the first row, we get
$F\left( x \right) = {\sin ^2}x\left( {{{\cos }^2}x\sin 2x - \cos 2x{{\cos }^2}x} \right) - \left( {1 + {{\cos }^2}x} \right)\left\{ {\left( {1 + {{\sin }^2}x} \right)\sin 2x - \cos 2x{{\sin }^2}x} \right\} + \cos 2x\left\{ {\left( {1 + {{\sin }^2}x} \right){{\cos }^2}x - {{\cos }^2}x{{\sin }^2}x} \right\}$
$ = {\sin ^2}x{\cos ^2}x\left( {\sin 2x - \cos 2x} \right) - \left( {1 + {{\cos }^2}x} \right)\left( {\sin 2x + {{\sin }^2}x\sin 2x - \cos 2x{{\sin }^2}x} \right) + \cos 2x\left( {{{\cos }^2}x + {{\sin }^2}x{{\cos }^2}x - {{\cos }^2}x{{\sin }^2}x} \right)$
$ = {\sin ^2}x{\cos ^2}x\sin 2x - {\sin ^2}x{\cos ^2}x\cos 2x - \sin 2x - {\sin ^2}x\sin 2x + \cos 2x{\sin ^2}x - {\cos ^2}x\sin 2x - {\cos ^2}x{\sin ^2}x\sin 2x + {\cos ^2}x\cos 2x{\sin ^2}x + \cos 2x{\cos ^2}x$
$ = - \sin 2x - {\sin ^2}x\sin 2x + \cos 2x{\sin ^2}x - {\cos ^2}x\sin 2x + \cos 2x{\cos ^2}x$
$ = - \sin 2x\left( {1 + {{\sin }^2}x + {{\cos }^2}x} \right) + \cos 2x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)$
Use the identity ${\sin ^2}x + {\cos ^2}x = 1$
$F\left( x \right) = - 2\sin 2x + \cos 2x$
The maximum value of the expression $A\sin \theta + B\cos \theta $ is $\sqrt {{A^2} + {B^2}} $
Here $A = - 2$, $B = 1$ and $\theta = 2x$
So, maximum value of $F\left( x \right)$ is $\sqrt {{{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt {4 + 1} = \sqrt 5 $

Option ‘B’ is correct

Note: The sign convention for a $3 \times 3$ matrix is $\begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix}$. The maximum value of the expression $A\sin \theta + B\cos \theta $ is $\sqrt {{A^2} + {B^2}} $ and the minimum value of the expression $A\sin \theta + B\cos \theta $ is $ - \sqrt {{A^2} + {B^2}} $.