A differentiable function satisfies `f(x) = int_(0)^(x) (f(t) cot t - cos(t - x))dt`.
Which of the following hold(s) good?

Options

A. f(x) has a minimum value 1 - e
B. f(x) has a maximum value 1 - `e^(-1)`
C. `f''((pi)/(2))=e`
D. `f'(0) = 1`

Explanation

To solve the given problem, we start with the equation:

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

### Step 1: Rewrite the Integral
We can rewrite the integral by separating the two terms inside the integral:

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

### Step 2: Evaluate the Second Integral
The second integral can be evaluated as follows:

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

Using the property of cosine, we have:

\[ \int_0^x \cos(t - x) \, dt = \sin(t - x) \bigg|_0^x = \sin(0) - \sin(-x) = 0 + \sin x = \sin x \]

### Step 3: Substitute Back into the Equation
Now substituting back into our equation, we have:

\[ f(x) = \int_0^x f(t) \cot t \, dt - \sin x \]

### Step 4: Differentiate Both Sides
Next, we differentiate both sides with respect to \(x\):

\[ f'(x) = f(x) \cot x + \int_0^x f(t) \frac{d}{dx}(\cot t) dt - \cos x \]

Since \(\frac{d}{dx}(\cot t) = 0\) (as \(t\) is a constant with respect to \(x\)), we simplify to:

\[ f'(x) = f(x) \cot x - \cos x \]

### Step 5: Rearranging the Equation
Rearranging gives us:

\[ f'(x) + \cos x = f(x) \cot x \]

### Step 6: Solve the Differential Equation
This is a first-order linear differential equation. We can rewrite it as:

\[ \frac{dy}{dx} + \cos x = y \cot x \]

Where \(y = f(x)\). To solve this, we can use an integrating factor. The integrating factor \(I(x)\) is given by:

\[ I(x) = e^{\int \cot x \, dx} = e^{\ln(\sin x)} = \sin x \]

### Step 7: Multiply through by the Integrating Factor
Multiplying the entire equation by \(\sin x\):

\[ \sin x f'(x) + \sin x \cos x = f(x) \sin x \cot x \]

This simplifies to:

\[ \frac{d}{dx}(\sin x f(x)) = \sin x \cos x \]

### Step 8: Integrate Both Sides
Now we integrate both sides:

\[ \int d(\sin x f(x)) = \int \sin x \cos x \, dx \]

The right side can be integrated using the identity \(\sin x \cos x = \frac{1}{2} \sin(2x)\):

\[ \sin x f(x) = \frac{1}{2} \sin(2x) + C \]

### Step 9: Solve for \(f(x)\)
Thus, we have:

\[ f(x) = \frac{\frac{1}{2} \sin(2x) + C}{\sin x} \]

### Step 10: Apply Initial Conditions
Assuming \(f(0) = 0\) (as derived from the original equation), we can find \(C\):

\[ f(0) = \frac{\frac{1}{2} \sin(0) + C}{\sin(0)} \implies C = 0 \]

### Final Solution
Thus, the function simplifies to:

\[ f(x) = \frac{1}{2} \frac{\sin(2x)}{\sin x} \]

### Checking Options
Now we can check the options provided in the problem statement based on our derived function.