Suppose f and g are differentiabel functions such that `xg (f(x))f'(g (x))g '(x) =(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 follow the reasoning outlined in the video transcript while providing a more structured approach.

### Step 1: Understanding the Given Information
We are given two differentiable functions \( f \) and \( g \) satisfying the equation:
\[ x g(f(x)) f'(g(x)) g'(x) = g(x) g'(f(x)) f'(x) \quad \forall x \in \mathbb{R} \]
and the integral condition:
\[ \int_0^x f(g(t)) dt = \frac{1}{2} (1 - e^{-2x}) \quad \forall x \in \mathbb{R} \]
Additionally, we know that \( g(f(0)) = 1 \) and \( h(x) = \frac{g(f(x))}{f(g(x))} \).

### Step 2: Rearranging the Given Equation
We start by rearranging the given equation:
\[ x g(f(x)) f'(g(x)) g'(x) = g(x) g'(f(x)) f'(x) \]
Dividing both sides by \( g(f(x)) f(g(x)) \) (assuming they are non-zero):
\[ \frac{x f'(g(x)) g'(x)}{f(g(x))} = \frac{g'(f(x)) f'(x)}{g(x)} \]

### Step 3: Integrating Both Sides
Next, we multiply both sides by \( dx \) and integrate:
\[ \int \frac{x f'(g(x)) g'(x)}{f(g(x))} dx = \int \frac{g'(f(x)) f'(x)}{g(x)} dx \]
Using integration by parts or recognizing the structure, we can express this in terms of logarithmic functions.

### Step 4: Using the Integral Condition
From the integral condition, we differentiate both sides:
\[ f(g(x)) = e^{-2x} \]
This gives us a relationship between \( f \) and \( g \).

### Step 5: Finding \( g(f(x)) \)
Using the relationship we derived, we can express \( g(f(x)) \):
\[ g(f(x)) = e^{-x^2} \]

### Step 6: Finding \( h(x) \)
Now we can substitute these into the function \( h(x) \):
\[ h(x) = \frac{g(f(x))}{f(g(x))} = \frac{e^{-x^2}}{e^{-2x}} = e^{-x^2 + 2x} \]

### Step 7: Analyzing the Symmetry of \( h(x) \)
To find the symmetry of \( h(x) \), we analyze the function:
\[ h(x) = e^{-x^2 + 2x} \]
This is a quadratic function in the exponent. The vertex of the parabola \( -x^2 + 2x \) 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 \( h(x) \) is symmetric with respect to the line:
\[ \boxed{x = 1} \]