Let f(x) =`|{:(cos x ,1,0 ),(1,2cosx,1),(0,1,2cosx):}|` then

Options

A. `f""((pi)/(3))=-1`
B. `f""'((pi)/(3))=sqrt (3)`
C. `int_(0)^(pi)f(x)dx=0`
D. `int_(-pi)^(pi)f(x)dx=0`

Answer

To solve the problem, we need to evaluate the determinant given by the function \( f(x) = \begin{vmatrix} \cos x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{vmatrix} \).

### Step 1: Write the Determinant
We start with the determinant:

\[ f(x) = \begin{vmatrix} \cos x & 1 & 0 \\ 1 & 2 \cos x & 1 \\ 0 & 1 & 2 \cos x \end{vmatrix} \]

### Step 2: Calculate the Determinant
We can calculate the determinant using the formula for a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix, we have:

- \( a = \cos x \), \( b = 1 \), \( c = 0 \)
- \( d = 1 \), \( e = 2 \cos x \), \( f = 1 \)
- \( g = 0 \), \( h = 1 \), \( i = 2 \cos x \)

Substituting these values into the determinant formula, we get:

\[ f(x) = \cos x \left( (2 \cos x)(2 \cos x) - (1)(1) \right) - 1 \left( (1)(2 \cos x) - (1)(0) \right) + 0 \]

### Step 3: Simplify the Expression
Now, let's simplify the expression:

\[ = \cos x \left( 4 \cos^2 x - 1 \right) - (2 \cos x) \]

\[ = 4 \cos^3 x - \cos x - 2 \cos x \]

\[ = 4 \cos^3 x - 3 \cos x \]

### Step 4: Recognize the Trigonometric Identity
We can recognize that:

\[ 4 \cos^3 x - 3 \cos x = \cos(3x) \]

Thus, we have:

\[ f(x) = \cos(3x) \]

### Step 5: Evaluate \( f\left(\frac{\pi}{3}\right) \)
Now we will evaluate \( f\left(\frac{\pi}{3}\right) \):

\[ f\left(\frac{\pi}{3}\right) = \cos\left(3 \cdot \frac{\pi}{3}\right) = \cos(\pi) = -1 \]

### Step 6: Find the Derivative \( f'(x) \)
Next, we find the derivative \( f'(x) \):

\[ f'(x) = -3 \sin(3x) \]

### Step 7: Evaluate \( f'\left(\frac{\pi}{3}\right) \)
Now we evaluate \( f'\left(\frac{\pi}{3}\right) \):

\[ f'\left(\frac{\pi}{3}\right) = -3 \sin\left(3 \cdot \frac{\pi}{3}\right) = -3 \sin(\pi) = 0 \]

### Step 8: Compute the Integral \( \int_0^\pi f(x) \, dx \)
We compute the integral:

\[ \int_0^\pi f(x) \, dx = \int_0^\pi \cos(3x) \, dx \]

The integral of \( \cos(3x) \) is:

\[ \frac{1}{3} \sin(3x) \Big|_0^\pi = \frac{1}{3} \left( \sin(3\pi) - \sin(0) \right) = \frac{1}{3} (0 - 0) = 0 \]

### Step 9: Compute the Integral \( \int_{-\pi}^\pi f(x) \, dx \)
Using the property of integrals:

\[ \int_{-\pi}^\pi f(x) \, dx = 2 \int_0^\pi f(x) \, dx = 2 \cdot 0 = 0 \]

### Final Result
Thus, we have:

- \( f\left(\frac{\pi}{3}\right) = -1 \)
- \( f'\left(\frac{\pi}{3}\right) = 0 \)
- \( \int_0^\pi f(x) \, dx = 0 \)
- \( \int_{-\pi}^\pi f(x) \, dx = 0 \)