Let `f(x)=|(2cosx,1,0),(1,2cosx,1),(0,1,2cosx)|` then

Options

A. `f((pi)/3)=1`
B. `f'((pi)/3)=-sqrt(3)`
C. `f((pi)/2)=-1`
D. None of these

Answer

To solve the problem, we need to find the determinant of the matrix given by the function \( f(x) = \left| \begin{array}{ccc} 2\cos x & 1 & 0 \\ 1 & 2\cos x & 1 \\ 0 & 1 & 2\cos x \end{array} \right| \).

### Step 1: 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:

\[ A = \begin{pmatrix} 2\cos x & 1 & 0 \\ 1 & 2\cos x & 1 \\ 0 & 1 & 2\cos x \end{pmatrix} \]

Assigning the elements:

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

Now substituting into the determinant formula:

\[ \text{det}(A) = 2\cos x \left( (2\cos x)(2\cos x) - (1)(1) \right) - 1 \left( (1)(2\cos x) - (1)(0) \right) + 0 \]

Calculating the terms:

1. \( (2\cos x)(2\cos x) - 1 = 4\cos^2 x - 1 \)
2. \( 1(2\cos x) - 0 = 2\cos x \)

Putting it all together:

\[ \text{det}(A) = 2\cos x (4\cos^2 x - 1) - 2\cos x \]

This simplifies to:

\[ \text{det}(A) = 8\cos^3 x - 2\cos x - 2\cos x = 8\cos^3 x - 4\cos x \]

Thus, we have:

\[ f(x) = 8\cos^3 x - 4\cos x \]

### Step 2: Evaluate \( f\left(\frac{\pi}{2}\right) \)
Now we evaluate \( f\left(\frac{\pi}{2}\right) \):

\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

Substituting into \( f(x) \):

\[ f\left(\frac{\pi}{2}\right) = 8(0)^3 - 4(0) = 0 \]

### Step 3: Evaluate \( f\left(\frac{\pi}{3}\right) \)
Next, we evaluate \( f\left(\frac{\pi}{3}\right) \):

\[ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Substituting into \( f(x) \):

\[ f\left(\frac{\pi}{3}\right) = 8\left(\frac{1}{2}\right)^3 - 4\left(\frac{1}{2}\right) = 8\left(\frac{1}{8}\right) - 2 = 1 - 2 = -1 \]

### Step 4: Find the Derivative \( f'(x) \)
To find \( f'(x) \), we differentiate \( f(x) = 8\cos^3 x - 4\cos x \):

Using the chain rule:

\[ f'(x) = 24\cos^2 x (-\sin x) + 4\sin x = -24\cos^2 x \sin x + 4\sin x \]

Factoring out \( \sin x \):

\[ f'(x) = \sin x (-24\cos^2 x + 4) \]

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

\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \quad \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Substituting into \( f'(x) \):

\[ f'\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \left(-24\left(\frac{1}{2}\right)^2 + 4\right) = \frac{\sqrt{3}}{2} \left(-24 \cdot \frac{1}{4} + 4\right) \]

This simplifies to:

\[ = \frac{\sqrt{3}}{2} \left(-6 + 4\right) = \frac{\sqrt{3}}{2} (-2) = -\sqrt{3} \]

### Conclusion
Thus, we have:

- \( f\left(\frac{\pi}{2}\right) = 0 \)
- \( f\left(\frac{\pi}{3}\right) = -1 \)
- \( f'\left(\frac{\pi}{3}\right) = -\sqrt{3} \)