If `f(x)=cosx-int_0^x(x-t)f(t)dt ,t h e nf^(prime)(x)+f(x)` is equal to
a) `-cosx`
(b) `-sinx`
c)`int_0^x(x-t)f(t)dt`
(d) 0

Options

A. `-cosx`
B. `-sinx`
C. `int_(0)^(x)(x-t)f(t)dt`
D. `0`

Explanation

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

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

### Step 1: Differentiate \( f(x) \)

We start by differentiating \( f(x) \) with respect to \( x \):

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

Using the Leibniz rule for differentiation under the integral sign, we have:

\[ f'(x) = -\sin x - \left[ (x - t) f(t) \bigg|_{t=0}^{t=x} + \int_0^x \frac{\partial}{\partial x}((x - t) f(t)) dt \right] \]

Evaluating the boundary term:

\[ = -\sin x - \left[ (x - x)f(x) - (x - 0)f(0) \right] + \int_0^x f(t) dt \]

This simplifies to:

\[ f'(x) = -\sin x + \int_0^x f(t) dt \]

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

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

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

Again applying the Leibniz rule:

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

### Step 3: Combine \( f''(x) \) and \( f(x) \)

Now we combine \( f''(x) \) and \( f(x) \):

\[ f''(x) + f(x) = (-\cos x + f(x)) + f(x) = -\cos x + 2f(x) \]

However, we need to simplify this further. Since we have \( f(x) = \cos x - \int_0^x (x - t) f(t) dt \), we can substitute \( f(x) \) back into the equation.

### Step 4: Final Result

From our previous calculations, we have:

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

Thus, the final answer is:

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

### Conclusion

The correct option is:

**(a) \(-\cos x\)**