Let `y=f(x)` satisfies the equation
`f(x) = (e^(-x)+e^(x))cosx-2x+int_(0)^(x)(x-t)f^(')(t)dt`
The value of `f(0)+f^(')(0)` equal

Options

A. `-1`
B. 0
C. 1
D. 1

Explanation

To solve the problem, we need to find the values of \( f(0) \) and \( f'(0) \) from the given equation:

\[ f(x) = (e^{-x} + e^{x}) \cos x - 2x + \int_{0}^{x} (x - t) f'(t) \, dt \]

### Step 1: Calculate \( f(0) \)

Substituting \( x = 0 \) into the equation:

\[ f(0) = (e^{0} + e^{0}) \cos(0) - 2(0) + \int_{0}^{0} (0 - t) f'(t) \, dt \]

Calculating each term:

- \( e^{0} = 1 \)
- \( \cos(0) = 1 \)
- The integral from \( 0 \) to \( 0 \) is \( 0 \).

Thus,

\[ f(0) = (1 + 1) \cdot 1 - 0 + 0 = 2 \]

### Step 2: Calculate \( f'(x) \)

To find \( f'(x) \), we differentiate both sides of the original equation with respect to \( x \):

\[ f'(x) = \frac{d}{dx} \left( (e^{-x} + e^{x}) \cos x - 2x + \int_{0}^{x} (x - t) f'(t) \, dt \right) \]

Using the product rule and the Fundamental Theorem of Calculus, we differentiate each term:

1. For \( (e^{-x} + e^{x}) \cos x \):
\[ \frac{d}{dx} \left( (e^{-x} + e^{x}) \cos x \right) = (e^{-x} + e^{x}) (-\sin x) + (e^{-x} - e^{x}) \cos x \]

2. For \( -2x \):
\[ \frac{d}{dx} (-2x) = -2 \]

3. For the integral:
\[ \frac{d}{dx} \left( \int_{0}^{x} (x - t) f'(t) \, dt \right) = (x - x) f'(x) + \int_{0}^{x} f'(t) \, dt = 0 + (x - t) f'(x) \bigg|_{t=x} = 0 \]

Combining these results, we have:

\[ f'(x) = (e^{-x} + e^{x})(-\sin x) + (e^{-x} - e^{x}) \cos x - 2 \]

### Step 3: Calculate \( f'(0) \)

Now substituting \( x = 0 \):

\[ f'(0) = (e^{0} + e^{0})(-\sin(0)) + (e^{0} - e^{0}) \cos(0) - 2 \]

Calculating each term:

- \( e^{0} = 1 \)
- \( \sin(0) = 0 \)
- \( \cos(0) = 1 \)

Thus,

\[ f'(0) = (1 + 1)(0) + (1 - 1)(1) - 2 = 0 + 0 - 2 = -2 \]

### Step 4: Calculate \( f(0) + f'(0) \)

Now we can find the final result:

\[ f(0) + f'(0) = 2 - 2 = 0 \]

### Final Answer

The value of \( f(0) + f'(0) \) is \( \boxed{0} \).

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