Let `f` and `g` be differentiable functions such that: `xg(f(x))f\'(g(x))g\'(x)=f(g(x))g\'(f(x))f\'(x) AA x in R` Also, `f(x)gt0` and `g(x)gt0 AA x in R` `int_0^xf(g(t))dt=1-e^(-2x)/2, AA x in R` and `g(f(0))=1, h(x)=g(f(x))/f(g(x)) AA x in R` Now answer the question: `f(g(0))+g(f(0))=` (A) `1` (B) `2` (C) `3` (D) `4`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to analyze the given information step by step.

### Step 1: Differentiate the integral equation
We start with the equation given in the problem:
\[ \int_0^x f(g(t)) dt = 1 - \frac{e^{-2x}}{2} \]
To find \( f(g(x)) \), we differentiate both sides with respect to \( x \) using the Leibniz rule:
\[ \frac{d}{dx} \left( \int_0^x f(g(t)) dt \right) = f(g(x)) \]
The right-hand side becomes:
\[ \frac{d}{dx} \left( 1 - \frac{e^{-2x}}{2} \right) = 0 + \frac{1}{2} \cdot 2 e^{-2x} = e^{-2x} \]
Thus, we have:
\[ f(g(x)) = e^{-2x} \]

### Step 2: Evaluate \( f(g(0)) \)
Next, we substitute \( x = 0 \) into the equation we derived:
\[ f(g(0)) = e^{-2 \cdot 0} = e^0 = 1 \]

### Step 3: Use the given information about \( g(f(0)) \)
We are also given that:
\[ g(f(0)) = 1 \]

### Step 4: Calculate \( f(g(0)) + g(f(0)) \)
Now, we can find the value of \( f(g(0)) + g(f(0)) \):
\[ f(g(0)) + g(f(0)) = 1 + 1 = 2 \]

### Conclusion
Thus, the final answer is:
\[ f(g(0)) + g(f(0)) = 2 \]