If maximum and minimum values of the determinant `|{:(1 + cos^(2)x , sin^(2) x, cos 2x),(cos^(2) x , 1 + sin^(2)x, cos 2x),(cos^(2) x , sin^(2) x , 1 + cos 2 x):}|` are `alpha and beta ` then

Options

A. `alpha^(2) + beta^(101) = 10 `
B. `alpha^(3) - beta^(99) = 26 `
C. `2 alpha^(2) - 18 beta^(11) = 0`
D. `alpha^(3) + 2 beta^(2) = 0 `

Answer

To solve the determinant

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

we will perform a series of operations to simplify it.

### Step 1: Simplifying the First Column

We will perform the operation \( C_1 \leftarrow C_1 + C_2 \):

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

Since \( \cos^2 x + \sin^2 x = 1 \), we can simplify:

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

### Step 2: Subtracting Rows

Next, we perform the operation \( R_2 \leftarrow R_2 - R_1 \):

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

### Step 3: Expanding the Determinant

Now we can expand the determinant along the first row:

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

Calculating the 2x2 determinant:

\[ D = 2 \left( \sin^2 x (1 + \cos 2x) - \sin^2 x \cos 2x \right) \]

This simplifies to:

\[ D = 2 \sin^2 x \]

### Step 4: Finding Maximum and Minimum Values

The maximum value of \( \sin^2 x \) is 1 and the minimum value is 0. Therefore:

- Maximum value \( \alpha = 2 \cdot 1 = 2 \)
- Minimum value \( \beta = 2 \cdot 0 = 0 \)

### Final Result

Thus, the maximum and minimum values of the determinant are \( \alpha = 2 \) and \( \beta = 0 \).