A function f is defined by `f(x) = int_0^pi cos t cos(x-t)dt,0 <= x <= 2 pi` then which of the following.hold(s) good?

Options

A. `(pi)/4`
B. `(pi)/2`
C. `(-pi)/2`
D. `(-pi)/4`

Explanation

To solve the problem, we need to evaluate the function defined by the integral:

\[ f(x) = \int_0^\pi \cos t \cos(x - t) \, dt \]

for \( 0 \leq x \leq 2\pi \).

### Step 1: Use the product-to-sum identities
We can use the product-to-sum identities to simplify the integral. The identity states:

\[ \cos A \cos B = \frac{1}{2} (\cos(A + B) + \cos(A - B)) \]

Applying this to our function:

\[ \cos t \cos(x - t) = \frac{1}{2} \left( \cos(t + (x - t)) + \cos(t - (x - t)) \right) = \frac{1}{2} \left( \cos x + \cos(2t - x) \right) \]

### Step 2: Rewrite the integral
Now, substituting this back into the integral:

\[ f(x) = \int_0^\pi \cos t \cos(x - t) \, dt = \frac{1}{2} \int_0^\pi \left( \cos x + \cos(2t - x) \right) dt \]

### Step 3: Evaluate the integral
We can separate the integral:

\[ f(x) = \frac{1}{2} \left( \int_0^\pi \cos x \, dt + \int_0^\pi \cos(2t - x) \, dt \right) \]

The first integral is straightforward:

\[ \int_0^\pi \cos x \, dt = \cos x \cdot \pi \]

For the second integral, we can use a substitution. Let \( u = 2t - x \), then \( du = 2 dt \) or \( dt = \frac{du}{2} \). When \( t = 0 \), \( u = -x \) and when \( t = \pi \), \( u = 2\pi - x \):

\[ \int_0^\pi \cos(2t - x) \, dt = \frac{1}{2} \int_{-x}^{2\pi - x} \cos u \, du \]

Evaluating this integral gives:

\[ \frac{1}{2} [\sin u]_{-x}^{2\pi - x} = \frac{1}{2} (\sin(2\pi - x) - \sin(-x)) = \frac{1}{2} (\sin x + \sin x) = \sin x \]

### Step 4: Combine results
Now substituting back into our expression for \( f(x) \):

\[ f(x) = \frac{1}{2} \left( \pi \cos x + \sin x \right) \]

### Step 5: Analyze the function
Now we can analyze \( f(x) \):

1. The function \( f(x) \) is periodic with period \( 2\pi \).
2. To find the minimum value, we can differentiate \( f(x) \) and set it to zero.

### Step 6: Find critical points
Setting the derivative \( f'(x) = 0 \):

\[ f'(x) = -\frac{\pi}{2} \sin x + \frac{1}{2} \cos x = 0 \]

This gives:

\[ \pi \sin x = \cos x \quad \Rightarrow \quad \tan x = \frac{1}{\pi} \]

### Step 7: Evaluate minimum value
The minimum value of \( f(x) \) occurs at the critical points found above. Evaluating \( f(x) \) at these points will give us the minimum value.

### Conclusion
Thus, we find that the function has a minimum value of \( -\frac{\pi}{2} \) at some point in the interval \( [0, 2\pi] \).

### Final Answer
The correct option is that the minimum value of \( f(x) \) is \( -\frac{\pi}{2} \).

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