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 (a) -1 (b) 0 (c) 1 (d) none of these

Options

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

Explanation

To solve the problem, we need to find the value of \( f(0) + f'(0) \) given the equation:

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

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

First, we substitute \( x = 0 \) into the equation to find \( f(0) \):

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

This simplifies to:

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

### Step 2: Differentiate the equation

Next, we differentiate both sides of the 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 have:

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

This simplifies to:

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

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

Now, we substitute \( x = 0 \) into the differentiated equation to find \( f'(0) \):

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

This simplifies to:

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

### Step 4: Combine \( f(0) \) and \( f'(0) \)

Now we can find \( f(0) + f'(0) \):

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

### Final Answer

Thus, the value of \( f(0) + f'(0) \) is:

\[ \boxed{0} \]