Let `f(x)=|[1+sin^(2)x,cos^(2)x,sin2x],[sin^(2)x,1+cos^(2)x,sin2x],[sin^(2)x,cos^(2)x,1+sin2x]|,x in[(pi)/(6),pi/(3)]]`.If `alpha` and `beta` respectively are the maximum and the minimum values of `f,` then

Options

A. `beta^(2)+2sqrt(alpha)=(19)/(4)`
B. `alpha^(2)+beta^(2)=(9)/(2)`
C. `alpha^(2)-beta^(2)=4sqrt(3)`
D. `beta^(2)-2sqrt(alpha)=(19)/(4)`

Answer

To solve the given problem, we need to find the maximum and minimum values of the function \( f(x) \) defined by the determinant of a 3x3 matrix. The matrix is given as:

\[ \begin{bmatrix} 1 + \sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{bmatrix} \]

where \( x \in \left[\frac{\pi}{6}, \frac{\pi}{3}\right] \).

### Step-by-Step Solution:

1. **Matrix Simplification using Row Operations:**

Let's perform the row operations \( R1 \rightarrow R1 - R2 \) and \( R2 \rightarrow R2 - R3 \) to simplify the determinant.

\[ \begin{vmatrix} 1 + \sin^2 x & \cos^2 x & \sin 2x \\ \sin^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix} \]

After applying \( R1 \rightarrow R1 - R2 \):

\[ \begin{vmatrix} 1 & -1 & 0 \\ \sin^2 x & 1 + \cos^2 x & \sin 2x \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix} \]

After applying \( R2 \rightarrow R2 - R3 \):

\[ \begin{vmatrix} 1 & -1 & 0 \\ 0 & 1 & 0 \\ \sin^2 x & \cos^2 x & 1 + \sin 2x \end{vmatrix} \]

2. **Determinant Calculation:**

The determinant of the simplified matrix is:

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

Simplifying the determinant:

\[ f(x) = 1 \cdot (1 \cdot (1 + \sin 2x) - 0 \cdot \cos^2 x) - (-1) \cdot (0 \cdot (1 + \sin 2x) - 0 \cdot \sin^2 x) + 0 \]

\[ f(x) = 1 + \sin 2x + \cos^2 x + \sin^2 x \]

Since \(\cos^2 x + \sin^2 x = 1\):

\[ f(x) = 1 + \sin 2x + 1 = 2 + \sin 2x \]

3. **Finding Maximum and Minimum Values:**

The function \( f(x) = 2 + \sin 2x \) is dependent on \(\sin 2x\). The range of \(\sin 2x\) is \([-1, 1]\).

- The maximum value of \( \sin 2x \) is 1.
- The minimum value of \( \sin 2x \) is -1.

Therefore:

- Maximum value of \( f(x) \) (denoted as \(\alpha\)) is \( 2 + 1 = 3 \).
- Minimum value of \( f(x) \) (denoted as \(\beta\)) is \( 2 - 1 = 1 \).

### Final Answer:

\[ \alpha = 3, \quad \beta = 1 \]