Suppose `f(x) and g(x)` are two continuous functions defined for `0<=x<=1`.Given, `f(x)=int_0^1 e^(x+1) .f(t) dt and g(x)=int_0^1 e^(x+1) *g(t) dt+x` The value of `f( 1)` equals

Options

A. 0
B. 1
C. `e^(-1)`
D. e

Answer

To solve the problem, we need to find the value of \( f(1) \) given the integral equation for \( f(x) \):

\[ f(x) = \int_0^1 e^{x+1} f(t) \, dt \]

### Step-by-step Solution:

1. **Understanding the Integral Equation**:
The equation states that \( f(x) \) is defined as an integral involving \( f(t) \). The term \( e^{x+1} \) is a function of \( x \) that will factor out of the integral.

2. **Factor Out \( e^{x+1} \)**:
We can rewrite the equation:
\[ f(x) = e^{x+1} \int_0^1 f(t) \, dt \]
Let \( C = \int_0^1 f(t) \, dt \), then we have:
\[ f(x) = C e^{x+1} \]

3. **Finding \( C \)**:
Substitute \( f(x) \) back into the integral to find \( C \):
\[ C = \int_0^1 f(t) \, dt = \int_0^1 C e^{t+1} \, dt \]
This simplifies to:
\[ C = C \int_0^1 e^{t+1} \, dt \]

4. **Evaluating the Integral**:
Calculate the integral:
\[ \int_0^1 e^{t+1} \, dt = e \int_0^1 e^t \, dt = e \left[ e^t \right]_0^1 = e (e - 1) = e^2 - e \]
Thus, we have:
\[ C = C (e^2 - e) \]

5. **Solving for \( C \)**:
Rearranging gives:
\[ C (1 - (e^2 - e)) = 0 \]
This implies either \( C = 0 \) or \( 1 - (e^2 - e) = 0 \). Since \( C = 0 \) leads to a trivial solution, we solve:
\[ 1 = e^2 - e \implies e^2 - e - 1 = 0 \]

6. **Using the Quadratic Formula**:
The roots of the equation \( e^2 - e - 1 = 0 \) can be found using the quadratic formula:
\[ e = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2} \]
Since we are looking for a positive value, we take:
\[ C = \frac{1 + \sqrt{5}}{2} \]

7. **Finding \( f(1) \)**:
Now substituting back to find \( f(1) \):
\[ f(1) = C e^{1+1} = C e^2 = \frac{1 + \sqrt{5}}{2} e^2 \]

### Final Result:
The value of \( f(1) \) is:
\[ f(1) = e^2 \cdot \frac{1 + \sqrt{5}}{2} \]