If `f(x) = |{:( cos (2x) ,, cos ( 2x ) ,, sin ( 2x) ), ( - cos x,, cosx ,, - sin x ), ( sinx,, sin x,, cos x ):}|`, then

Options

A. f'(x) =0 at exactly three points in `(-pi,pi)`
B. f(x) attains its maximum at x=0
C. f(x) attains its minimum at x=0
D. f'(x) =0 at more than three points in `(-pi,pi)`

Explanation

To solve the problem, we need to evaluate the determinant given by the function \( f(x) \) and analyze its properties, particularly focusing on its critical points and whether it attains maximum or minimum values.

### Step 1: Write down the determinant
The function is defined as:
\[ f(x) = \begin{vmatrix} \cos(2x) & \cos(2x) & \sin(2x) \\ -\cos(x) & \cos(x) & -\sin(x) \\ \sin(x) & \sin(x) & \cos(x) \end{vmatrix} \]

### Step 2: Simplify the determinant
We can simplify the determinant using the property of determinants. We will perform the operation \( C_2 \leftarrow C_2 - C_1 \):
\[ f(x) = \begin{vmatrix} \cos(2x) & 0 & \sin(2x) \\ -\cos(x) & 2\cos(x) & -\sin(x) \\ \sin(x) & 0 & \cos(x) \end{vmatrix} \]

### Step 3: Expand the determinant
Now we can expand the determinant along the second column:
\[ f(x) = 2\cos(x) \cdot \begin{vmatrix} \cos(2x) & \sin(2x) \\ \sin(x) & \cos(x) \end{vmatrix} \]
Calculating the 2x2 determinant:
\[ \begin{vmatrix} \cos(2x) & \sin(2x) \\ \sin(x) & \cos(x) \end{vmatrix} = \cos(2x) \cos(x) - \sin(2x) \sin(x) = \cos(2x + x) = \cos(3x) \]
Thus,
\[ f(x) = 2\cos(x) \cos(3x) \]

### Step 4: Find the derivative
Now we need to find the derivative \( f'(x) \):
Using the product rule:
\[ f'(x) = 2\left(-\sin(x) \cos(3x) + \cos(x)(-3\sin(3x))\right) \]
This simplifies to:
\[ f'(x) = -2\sin(x)\cos(3x) - 6\cos(x)\sin(3x) \]

### Step 5: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):
\[ -2\sin(x)\cos(3x) - 6\cos(x)\sin(3x) = 0 \]
Factoring out common terms:
\[ -2\left(\sin(x)\cos(3x) + 3\cos(x)\sin(3x)\right) = 0 \]
This gives us two cases:
1. \( \sin(x) = 0 \)
2. \( \sin(3x + x) = 0 \)

### Step 6: Solve for critical points
1. From \( \sin(x) = 0 \):
- \( x = n\pi \) for \( n \in \mathbb{Z} \)
- In the interval \( (-\pi, \pi) \), we have \( x = 0 \).

2. From \( \sin(4x) = 0 \):
- \( 4x = n\pi \) implies \( x = \frac{n\pi}{4} \)
- In the interval \( (-\pi, \pi) \), we have \( x = -\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{2}, -\frac{\pi}{2} \).

### Step 7: Analyze maximum and minimum
To determine whether \( x = 0 \) is a maximum or minimum, we compute the second derivative:
\[ f''(x) = -2\left(\cos(x)\cos(3x) - 3\sin(x)\sin(3x)\right) \]
Evaluating at \( x = 0 \):
\[ f''(0) = -2\left(1 \cdot 1 - 0\right) = -2 < 0 \]
Since \( f''(0) < 0 \), \( x = 0 \) is a point of local maximum.

### Conclusion
Thus, the function \( f(x) \) attains its maximum at \( x = 0 \) and \( f'(x) = 0 \) at exactly three points in the interval \( (-\pi, \pi) \).