Let f(x) = | cases ^ 2 x & - 2 + ^ 2 x & 2x 2 + ^ 2 x & ^ 2 x & 2x...

Given,

f\left(x\right) = \left|\begin{cases}\sin^{2}x & - 2 + \cos^{2}x & \cos 2x \\ 2 + \sin^{2}x & \cos^{2}x & \cos 2x \\ \sin^{2}x & \cos^{2}x & 1 + \cos 2x\end{cases}\right|

= \left|\begin{cases}- 2 & - 2 & 0 \\ 2 & 0 & - 1 \\ \sin^{2}x & \cos^{2}x & 1 + \cos 2x\end{cases}\right|\left($\begin{array}{ll}{R}_{1} \to {R}_{1} - {R}_{2} \\ \text{ and } {R}_{2} \to {R}_{2} - {R}_{3}\end{array}$\right)

$= - 2\left(\cos^{2}x\right) + 2\left(2 + 2\cos 2x + \sin^{2}x\right)$

$= 4 + 4\cos 2x - 2\left(\cos^{2}x - \sin^{2}x\right)$

$⇒ f\left(x\right) = 4 + 2\cos 2x$

We know, $- 1 \le \cos 2x \le 1$

So, f{\left(x\right)}_\max = 4 + 2 = 6