If `f(x)=1+1/x int_1^x f(t)dt`, then the value of `f(e^(-1))` is (a) `1` (b) `0` (c) `-1` (d) none of these

Options

A. `1`
B. `0`
C. `-1`
D. none of these

Explanation

To solve the problem, we need to find the value of \( f(e^{-1}) \) given the function defined as:

\[ f(x) = 1 + \frac{1}{x} \int_1^x f(t) \, dt \]

### Step-by-Step Solution:

1. **Multiply both sides by \( x \)**:
\[ x f(x) = x + \int_1^x f(t) \, dt \]

2. **Differentiate both sides with respect to \( x \)**:
Using the product rule on the left side and Leibniz's rule on the right side:
\[ \frac{d}{dx}(x f(x)) = f(x) + x f'(x) \]
\[ \frac{d}{dx}\left(x + \int_1^x f(t) \, dt\right) = 1 + f(x) \]

Therefore, we have:
\[ f(x) + x f'(x) = 1 + f(x) \]

3. **Simplify the equation**:
By cancelling \( f(x) \) from both sides, we get:
\[ x f'(x) = 1 \]

4. **Solve for \( f'(x) \)**:
\[ f'(x) = \frac{1}{x} \]

5. **Integrate to find \( f(x) \)**:
\[ f(x) = \int \frac{1}{x} \, dx = \log x + C \]

6. **Determine the constant \( C \)**:
Substitute \( x = 1 \) into the original function to find \( f(1) \):
\[ f(1) = 1 + \frac{1}{1} \int_1^1 f(t) \, dt \]
The integral from 1 to 1 is 0:
\[ f(1) = 1 \]
Now substituting into \( f(x) = \log x + C \):
\[ 1 = \log(1) + C \implies C = 1 \]

7. **Final expression for \( f(x) \)**:
\[ f(x) = \log x + 1 \]

8. **Find \( f(e^{-1}) \)**:
Substitute \( x = e^{-1} \):
\[ f(e^{-1}) = \log(e^{-1}) + 1 = -1 + 1 = 0 \]

### Conclusion:
The value of \( f(e^{-1}) \) is \( 0 \).