Let `f :R ^(+) to R` be a differentiable function with `f (1)=3` and satisfying :
`int _(1) ^(xy) f(t) dt =y int_(1) ^(x) f (t) dt +x int_(1) ^(y) f (t) dt AA x, y in R^(+),` then `f (e) =`

Options

A. 3
B. 4
C. 1
D. None of these

Answer

To solve the problem, we need to analyze the given integral equation and find the value of \( f(e) \).

### Step 1: Write down the given equation
We start with the equation:
\[ \int_{1}^{xy} f(t) \, dt = y \int_{1}^{x} f(t) \, dt + x \int_{1}^{y} f(t) \, dt \]
for all \( x, y \in \mathbb{R}^+ \).

### Step 2: Differentiate with respect to \( x \)
We differentiate both sides of the equation with respect to \( x \), treating \( y \) as a constant:
\[ \frac{d}{dx} \left( \int_{1}^{xy} f(t) \, dt \right) = \frac{d}{dx} \left( y \int_{1}^{x} f(t) \, dt + x \int_{1}^{y} f(t) \, dt \right) \]
Using the Fundamental Theorem of Calculus and the chain rule on the left side, we get:
\[ f(xy) \cdot y = y f(x) + \int_{1}^{y} f(t) \, dt \]

### Step 3: Substitute \( x = 1 \)
Now, we set \( x = 1 \):
\[ f(y) \cdot y = y f(1) + \int_{1}^{y} f(t) \, dt \]
Since \( f(1) = 3 \), we have:
\[ f(y) \cdot y = 3y + \int_{1}^{y} f(t) \, dt \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ y f(y) - 3y = \int_{1}^{y} f(t) \, dt \]
This simplifies to:
\[ y (f(y) - 3) = \int_{1}^{y} f(t) \, dt \]

### Step 5: Differentiate with respect to \( y \)
Next, we differentiate both sides with respect to \( y \):
\[ f(y) + y f'(y) - 3 = f(y) \]
This simplifies to:
\[ y f'(y) = 3 \]

### Step 6: Solve for \( f'(y) \)
From the equation \( y f'(y) = 3 \), we can solve for \( f'(y) \):
\[ f'(y) = \frac{3}{y} \]

### Step 7: Integrate to find \( f(y) \)
Integrating \( f'(y) \):
\[ f(y) = 3 \ln y + C \]
where \( C \) is a constant.

### Step 8: Use the condition \( f(1) = 3 \)
Substituting \( y = 1 \):
\[ f(1) = 3 \ln(1) + C = 3 \cdot 0 + C = C \]
Thus, \( C = 3 \), and we have:
\[ f(y) = 3 \ln y + 3 \]

### Step 9: Find \( f(e) \)
Now we can find \( f(e) \):
\[ f(e) = 3 \ln(e) + 3 = 3 \cdot 1 + 3 = 6 \]

### Final Answer
Thus, the value of \( f(e) \) is:
\[ \boxed{6} \]