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)equals to

Options

A. -1
B. 2
C. 1
D. 0

Answer

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: Find \( f(0) \)

First, we will evaluate \( f(0) \):

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

This simplifies to:

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

### Step 2: Differentiate \( f(x) \)

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 get:

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

This simplifies to:

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

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

Now, we substitute \( x = 0 \):

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

This simplifies to:

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

### Step 4: Differentiate \( f'(x) \) to find \( f''(x) \)

Next, we differentiate \( f'(x) \) to find \( f''(x) \):

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

Using the previously derived expression for \( f'(x) \):

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

This gives us:

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

### Step 5: Evaluate \( f''(0) \)

Substituting \( x = 0 \):

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

This simplifies to:

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

### Step 6: Calculate \( f'(0) + f''(0) \)

Finally, we find:

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

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

\[ \boxed{0} \]