If `f(x) = \begin{vmatrix} cos2x & cos2x & sin2x \\ -cosx & cosx & -sinx\\ sinx & sinx & cosx \end{vmatrix}` then

Options

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

Answer

To solve the given problem step-by-step, we need to evaluate the determinant and then analyze the function \( f(x) \).

### Step 1: Evaluate the Determinant

Given:
\[ f(x) = \begin{vmatrix} \cos 2x & \cos 2x & \sin 2x \\ -\cos x & \cos x & -\sin x \\ \sin x & \sin x & \cos x \end{vmatrix} \]

To evaluate the determinant, we can use the rule of Sarrus or cofactor expansion. Here, we will use cofactor expansion along the first row.

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

Calculating the first determinant:
\[ \begin{vmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{vmatrix} = \cos^2 x + \sin^2 x = 1 \]

Calculating the second determinant:
\[ \begin{vmatrix} -\cos x & -\sin x \\ \sin x & \cos x \end{vmatrix} = -\cos x \cdot \cos x - (-\sin x)(\sin x) = -\cos^2 x + \sin^2 x \]

Calculating the third determinant:
\[ \begin{vmatrix} -\cos x & \cos x \\ \sin x & \sin x \end{vmatrix} = -\cos x \cdot \sin x - \sin x \cdot \cos x = -2 \sin x \cos x = -\sin 2x \]

Now substituting back into the expression for \( f(x) \):
\[ f(x) = \cos 2x (1 - (-\cos^2 x + \sin^2 x)) + \sin 2x (-\sin 2x) \]
\[ = \cos 2x (1 + \cos^2 x - \sin^2 x) - \sin^2 2x \]
Using the identity \( \cos^2 x - \sin^2 x = \cos 2x \):
\[ = \cos 2x (1 + \cos 2x) - \sin^2 2x \]
\[ = \cos 2x + \cos^2 2x - \sin^2 2x \]
\[ = \cos 2x + \cos 4x \]

### Final Expression for \( f(x) \)
Thus, we have:
\[ f(x) = \cos 2x + \cos 4x \]

### Step 2: Find the First Derivative \( f'(x) \)

To find the critical points, we will differentiate \( f(x) \):
\[ f'(x) = -2 \sin 2x - 4 \sin 4x \]

### Step 3: Find Critical Points

Setting \( f'(x) = 0 \):
\[ -2 \sin 2x - 4 \sin 4x = 0 \]
\[ \sin 2x + 2 \sin 4x = 0 \]

### Step 4: Analyze the Second Derivative \( f''(x) \)

Now, we differentiate \( f'(x) \) to find \( f''(x) \):
\[ f''(x) = -4 \cos 2x - 16 \cos 4x \]

### Step 5: Determine Maxima or Minima

Evaluate \( f''(0) \):
\[ f''(0) = -4 \cdot 1 - 16 \cdot 1 = -20 \]
Since \( f''(0) < 0 \), there is a local maximum at \( x = 0 \).

### Step 6: Check for Other Critical Points

From \( \sin 2x + 2 \sin 4x = 0 \), we can find other critical points in the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

### Conclusion

The critical points are \( x = 0, -\frac{\pi}{2}, \frac{\pi}{2} \).