To solve the given problem, we need to analyze the function defined by the integral:
\[
f(x) = \int_0^\pi \cos t \cos(x - t) \, dt
\]
### Step 1: Simplifying the Integral
Using the product-to-sum identities, we can simplify the integral:
\[
\cos t \cos(x - t) = \frac{1}{2} \left( \cos(t + (x - t)) + \cos(t - (x - t)) \right) = \frac{1}{2} \left( \cos(x) + \cos(2t - x) \right)
\]
Thus, we can rewrite \( f(x) \) as:
\[
f(x) = \frac{1}{2} \int_0^\pi \left( \cos(x) + \cos(2t - x) \right) dt
\]
### Step 2: Evaluating the Integral
Now, we can separate the integral:
\[
f(x) = \frac{1}{2} \left( \cos(x) \int_0^\pi dt + \int_0^\pi \cos(2t - x) dt \right)
\]
Calculating the first integral:
\[
\int_0^\pi dt = \pi
\]
Now for the second integral, we can use the substitution \( u = 2t - x \), which gives \( du = 2 dt \) or \( dt = \frac{du}{2} \). Changing the limits accordingly:
- When \( t = 0 \), \( u = -x \)
- When \( t = \pi \), \( u = 2\pi - x \)
Thus,
\[
\int_0^\pi \cos(2t - x) dt = \frac{1}{2} \int_{-x}^{2\pi - x} \cos(u) du
\]
Evaluating this integral:
\[
\int \cos(u) du = \sin(u)
\]
So,
\[
\int_{-x}^{2\pi - x} \cos(u) du = \sin(2\pi - x) - \sin(-x) = \sin(x) + \sin(x) = 2\sin(x)
\]
Thus, we have:
\[
\int_0^\pi \cos(2t - x) dt = \frac{1}{2} \cdot 2\sin(x) = \sin(x)
\]
### Step 3: Final Expression for \( f(x) \)
Putting it all together, we get:
\[
f(x) = \frac{1}{2} \left( \pi \cos(x) + \sin(x) \right)
\]
### Step 4: Analyzing the Function
Now we need to analyze the properties of \( f(x) \):
1. **Continuity and Differentiability**:
- The function \( f(x) \) is composed of continuous functions (cosine and sine), thus it is continuous on \([0, 2\pi]\) and differentiable on \((0, 2\pi)\).
2. **Finding Critical Points**:
- To find critical points, we differentiate \( f(x) \):
\[
f'(x) = \frac{1}{2} \left( -\pi \sin(x) + \cos(x) \right)
\]
Setting \( f'(x) = 0 \):
\[
-\pi \sin(x) + \cos(x) = 0 \implies \cos(x) = \pi \sin(x) \implies \tan(x) = \frac{1}{\pi}
\]
There exists at least one \( c \in (0, 2\pi) \) such that \( f'(c) = 0 \).
3. **Maximum and Minimum Values**:
- The maximum value of \( f(x) \) occurs when \( \cos(x) = 1 \) (i.e., \( x = 0 \)):
\[
f(0) = \frac{1}{2}(\pi \cdot 1 + 0) = \frac{\pi}{2}
\]
- The minimum value occurs when \( \cos(x) = -1 \) (i.e., \( x = \pi \)):
\[
f(\pi) = \frac{1}{2}(-\pi + 0) = -\frac{\pi}{2}
\]
### Conclusion
Based on the analysis, we can conclude:
- (A) False: \( f(x) \) is continuous and differentiable in \( (0, 2\pi) \).
- (B) True: There exists at least one \( c \in (0, 2\pi) \) such that \( f'(c) = 0 \).
- (C) True: Maximum value of \( f \) is \( \frac{\pi}{2} \).
- (D) True: Minimum value of \( f \) is \( -\frac{\pi}{2} \).
