Suppose f and g are differentiabel functions such that `xg (f(x))f'(g (x))g '(x) =f(g(x))g '(f(x)) f'(x) AA x in R and f` is positive `AA n in R.` Also `int _( 0)^(x) f (g(t )) dt =1/2 (1-e^(-2x))AA x in R, g (f(0))=1 and h (x) = (g(f (x)))/(f (g(x)))AA x in R.`
The graph of `y =h (x)` is symmetric with respect to line:

Options

A. `x=-1`
B. `x=0`
C. `x=1`
D. `x=2`

Answer

To solve the problem step by step, we will analyze the given information and derive the necessary results.

### Step 1: Understand the given equation
We are given the equation:
\[ xg(f(x))f'(g(x))g'(x) = f(g(x))g'(f(x))f'(x) \]
This equation holds for all \( x \in \mathbb{R} \).

### Step 2: Analyze the integral
We also have the integral:
\[ \int_0^x f(g(t)) \, dt = \frac{1}{2}(1 - e^{-2x}) \]
for all \( x \in \mathbb{R} \).

### Step 3: Differentiate the integral
Differentiating both sides with respect to \( x \):
\[ f(g(x)) = \frac{d}{dx} \left( \frac{1}{2}(1 - e^{-2x}) \right) = \frac{1}{2}(2e^{-2x}) = e^{-2x} \]
Thus, we have:
\[ f(g(x)) = e^{-2x} \]

### Step 4: Take the logarithm
Taking the logarithm of both sides:
\[ \log(f(g(x))) = -2x \]

### Step 5: Substitute into the original equation
Substituting \( f(g(x)) \) into the original equation:
\[ xg(f(x))f'(g(x))g'(x) = e^{-2x}g'(f(x))f'(x) \]

### Step 6: Simplify the equation
Rearranging gives:
\[ xg(f(x))f'(g(x))g'(x) = e^{-2x}g'(f(x))f'(x) \]

### Step 7: Analyze the function \( h(x) \)
We have:
\[ h(x) = \frac{g(f(x))}{f(g(x))} \]
Using the results from previous steps, we can express \( g(f(x)) \) and \( f(g(x)) \).

### Step 8: Substitute known values
From our earlier results:
1. \( g(f(x)) = e^{-x^2} \)
2. \( f(g(x)) = e^{-2x} \)

Thus:
\[ h(x) = \frac{e^{-x^2}}{e^{-2x}} = e^{2x - x^2} \]

### Step 9: Determine symmetry
To find the symmetry of \( h(x) \), we need to analyze the function \( h(x) = e^{2x - x^2} \).

### Step 10: Find the vertex of the parabola
The function \( 2x - x^2 \) is a downward-opening parabola. The vertex occurs at:
\[ x = -\frac{b}{2a} = -\frac{2}{-2} = 1 \]
Thus, the graph of \( h(x) \) is symmetric about the line \( x = 1 \).

### Conclusion
The graph of \( y = h(x) \) is symmetric with respect to the line:
\[ \boxed{x = 1} \]