If f and g are two functions having derivative of order three for all x satisfying `f(x)g(x)=C` (constant) and `(f''')/(f')-"A" (f'')/f -(g''')/(g')+(3g")/g=0`. Then A is equal to

Answer

To solve the problem, we need to analyze the given conditions and derive the value of \( A \).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We have two functions \( f(x) \) and \( g(x) \) such that:
\[ f(x)g(x) = C \]
where \( C \) is a constant.

2. **Differentiating the Product**:
We differentiate both sides of the equation \( f(x)g(x) = C \) with respect to \( x \):
\[ f'(x)g(x) + f(x)g'(x) = 0 \]
This implies:
\[ f'(x)g(x) = -f(x)g'(x) \tag{1} \]

3. **Differentiating Again**:
We differentiate equation (1) again:
\[ f''(x)g(x) + f'(x)g'(x) + f'(x)g'(x) + f(x)g''(x) = 0 \]
Simplifying this gives:
\[ f''(x)g(x) + 2f'(x)g'(x) + f(x)g''(x) = 0 \tag{2} \]

4. **Differentiating a Third Time**:
We differentiate equation (2) again:
\[ f'''(x)g(x) + 3f''(x)g'(x) + 3f'(x)g''(x) + f(x)g'''(x) = 0 \]
Rearranging gives:
\[ f'''(x)g(x) + 3f''(x)g'(x) + 3f'(x)g''(x) + f(x)g'''(x) = 0 \tag{3} \]

5. **Dividing by \( f(x)g(x) \)**:
Now, we divide equation (3) by \( f(x)g(x) \):
\[ \frac{f'''(x)}{f(x)} + 3\frac{f''(x)}{f(x)}\frac{g'(x)}{g(x)} + 3\frac{f'(x)}{f(x)}\frac{g''(x)}{g(x)} + \frac{g'''(x)}{g(x)} = 0 \]

6. **Substituting the Given Equation**:
We are given:
\[ \frac{f'''(x)}{f'(x)} - A\frac{f''(x)}{f(x)} - \frac{g'''(x)}{g'(x)} + 3\frac{g''(x)}{g(x)} = 0 \]
We can compare the terms from our derived equation with this equation.

7. **Comparing Coefficients**:
By comparing the coefficients from both equations, we find:
\[ A = 3 \]

### Conclusion:
Thus, the value of \( A \) is:
\[ \boxed{3} \]