Let $$f(x) = \begin{vmatrix} 1+\sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1+\cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix}$$, $$x \in [\frac{\pi}{6}, \frac{\pi}{3}]$$. If $$\alpha$$ and $$\beta$$ respectively are the maximum and the minimum values of $$f$$, then

We need to find the maximum $$\alpha$$ and minimum $$\beta$$ of the determinant function on $$[\pi/6, \pi/3]$$.

Simplify the determinant.

$$ f(x) = \begin{vmatrix} 1+\sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1+\cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1+\sin 2x \end{vmatrix} $$

Apply the column operation $$C_1 \to C_1 + C_2$$ (this does not change the determinant value):

Since $$\sin^2 x + \cos^2 x = 1$$, the first column entries become:

- Row 1: $$(1 + \sin^2 x) + \cos^2 x = 1 + 1 = 2$$

- Row 2: $$\sin^2 x + (1 + \cos^2 x) = 1 + 1 = 2$$

- Row 3: $$\sin^2 x + \cos^2 x = 1$$

$$ f(x) = \begin{vmatrix} 2 & \cos^2 x & \sin 2x \\ 2 & 1+\cos^2 x & \sin 2x \\ 1 & \cos^2 x & 1+\sin 2x \end{vmatrix} $$

Apply $$R_1 \to R_1 - R_2$$:

$$ f(x) = \begin{vmatrix} 0 & -1 & 0 \\ 2 & 1+\cos^2 x & \sin 2x \\ 1 & \cos^2 x & 1+\sin 2x \end{vmatrix} $$

Expand along Row 1.

Expanding along $$R_1$$ (which has two zeros):

$$ f(x) = 0 - (-1) \cdot \begin{vmatrix} 2 & \sin 2x \\ 1 & 1+\sin 2x \end{vmatrix} + 0 $$

$$ = 1 \cdot [2(1 + \sin 2x) - \sin 2x \cdot 1] $$

$$ = 2 + 2\sin 2x - \sin 2x = 2 + \sin 2x $$

Find the maximum and minimum on $$[\pi/6, \pi/3]$$.

$$f(x) = 2 + \sin 2x$$

On $$x \in [\pi/6, \pi/3]$$, we have $$2x \in [\pi/3, 2\pi/3]$$.

The function $$\sin(2x)$$ on $$[\pi/3, 2\pi/3]$$:

- At $$2x = \pi/3$$: $$\sin(\pi/3) = \frac{\sqrt{3}}{2}$$

- At $$2x = \pi/2$$: $$\sin(\pi/2) = 1$$ (maximum)

- At $$2x = 2\pi/3$$: $$\sin(2\pi/3) = \frac{\sqrt{3}}{2}$$

Therefore:

- Maximum of $$\sin 2x$$ = 1 (at $$x = \pi/4$$), giving $$\alpha = 2 + 1 = 3$$

- Minimum of $$\sin 2x$$ = $$\frac{\sqrt{3}}{2}$$ (at endpoints), giving $$\beta = 2 + \frac{\sqrt{3}}{2}$$

Verify Option 1: $$\beta^2 - 2\sqrt{\alpha} = \frac{19}{4}$$.

$$ \beta^2 = \left(2 + \frac{\sqrt{3}}{2}\right)^2 = 4 + 2 \cdot 2 \cdot \frac{\sqrt{3}}{2} + \frac{3}{4} = 4 + 2\sqrt{3} + \frac{3}{4} = \frac{19}{4} + 2\sqrt{3} $$

$$ 2\sqrt{\alpha} = 2\sqrt{3} $$

$$ \beta^2 - 2\sqrt{\alpha} = \frac{19}{4} + 2\sqrt{3} - 2\sqrt{3} = \frac{19}{4} \quad \checkmark $$

The correct answer is Option 1: $$\beta^2 - 2\sqrt{\alpha} = \dfrac{19}{4}$$.