If the maximum and minimum values of the determinant
`|(1+sin^(2)x,cos^(2)x,sin2x),(sin^(2)x,1+cos^(2)x,sin2x),(sin^(2)x,cos^(2)x,1+sin2x)|` are `alpha` and `beta` respectively, then `alpha+2beta` is equal to

Answer

To solve the determinant

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

we will simplify it step by step.

### Step 1: Simplify the determinant

We can first observe that:

\[ 1 + \sin^2 x = 1 + \sin^2 x, \quad 1 + \cos^2 x = 1 + \cos^2 x, \quad 1 + \sin 2x = 1 + \sin 2x \]

Now we can apply the column operation \(C_1 \to C_1 + C_2\):

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

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

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

### Step 2: Row operations

Now, we will perform the row operation \(R_2 \to R_2 - R_1\) and \(R_3 \to R_3 - \frac{1}{2} R_1\):

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

This simplifies to:

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

### Step 3: Calculate the determinant

Now we can calculate the determinant using the first row:

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

This gives:

\[ D = 2 \left(1 \cdot \left(\frac{1}{2} + \frac{1}{2} \sin 2x\right) - 0\right) = 2 \left(\frac{1}{2} + \frac{1}{2} \sin 2x\right) = 1 + \sin 2x \]

### Step 4: Find maximum and minimum values

The maximum value of \(\sin 2x\) is \(1\) and the minimum value is \(-1\):

- Maximum value of \(D\) (denoted as \(\alpha\)):
\[ \alpha = 1 + 1 = 2 \]
- Minimum value of \(D\) (denoted as \(\beta\)):
\[ \beta = 1 - 1 = 0 \]

### Step 5: Calculate \(\alpha + 2\beta\)

Now we can calculate \(\alpha + 2\beta\):

\[ \alpha + 2\beta = 2 + 2 \cdot 0 = 2 \]

### Final Result

Thus, the value of \(\alpha + 2\beta\) is:

\[ \boxed{2} \]