Let f(x) = | cases 1 + ^ 2 x & ^ 2 x & 2x ^ 2 x & 1 + ^ 2 x & 2x ^...

Given:

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

Applying ${C}_{1} \to {C}_{1} + {C}_{2} + {C}_{3}$

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

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

Applying ${R}_{2} \to {R}_{2} - {R}_{1}$ and ${R}_{3} \to {R}_{3} - {R}_{1}$

f\left(x\right) = \left(2 + \sin 2x\right)\left|\begin{cases}1 & \cos^{2}x & \sin 2x \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}\right|

$⇒ f\left(x\right) = \left(2 + \sin 2x\right)\left(1\right) = 2 + \sin 2x$

Now, for

$x \in \left[\dfrac{\pi}{6}, \dfrac{\pi}{3}\right]$

$⇒ 2x \in \left[\dfrac{\pi}{3}, \dfrac{2\pi}{3}\right]$

$⇒ \sin 2x \in \left[\dfrac{\sqrt{3}}{2}, 1\right]$

$⇒ 2 + \sin 2x \in \left[2 + \dfrac{\sqrt{3}}{2}, 3\right]$

Hence,

$\beta = 2 + \dfrac{\sqrt{3}}{2}$

$\alpha = 3$

So,

${\beta}^{2} - 2\alpha = 4 + \dfrac{3}{4} + 2\sqrt{3} - 2\sqrt{3} = \dfrac{19}{4}$