Let `(f(x+y)-f(x))/(2)=(f(y)-1)/(2)+xy`, for all `x,yinR,f(x)` is differentiable and `f'(0)=1.` Domain of `log(f(x)),` is

Options

A. `R^(+)`
B. `R-{0}`
C. `R`
D. `R^(-)`

Answer

To solve the problem, we need to analyze the given equation and find the function \( f(x) \) such that we can determine the domain of \( \log(f(x)) \).

### Step-by-Step Solution:

1. **Start with the Given Equation:**
\[ \frac{f(x+y) - f(x)}{2} = \frac{f(y) - 1}{2} + xy \]
Multiply both sides by 2 to eliminate the fraction:
\[ f(x+y) - f(x) = f(y) - 1 + 2xy \]

2. **Substitute \( x = 0 \) and \( y = 0 \):**
This will help us find \( f(0) \):
\[ f(0+0) - f(0) = f(0) - 1 + 2(0)(0) \]
Simplifying gives:
\[ 0 = f(0) - 1 \implies f(0) = 1 \]

3. **Differentiate the Given Equation:**
We know \( f'(0) = 1 \). To find \( f'(x) \), we can differentiate both sides of the equation with respect to \( y \):
\[ \frac{d}{dy}\left(f(x+y) - f(x)\right) = \frac{d}{dy}\left(f(y) - 1 + 2xy\right) \]
This gives:
\[ f'(x+y) = f'(y) + 2x \]

4. **Set \( y = 0 \) in the Derivative:**
This will help us find \( f'(x) \):
\[ f'(x+0) = f'(0) + 2x \implies f'(x) = 1 + 2x \]

5. **Integrate \( f'(x) \):**
To find \( f(x) \), integrate \( f'(x) \):
\[ f(x) = \int (1 + 2x) \, dx = x + x^2 + C \]

6. **Determine the Constant \( C \):**
Use \( f(0) = 1 \):
\[ f(0) = 0 + 0 + C \implies C = 1 \]
Thus, we have:
\[ f(x) = x^2 + x + 1 \]

7. **Find the Domain of \( \log(f(x)) \):**
The logarithm function is defined for positive arguments. Therefore, we need to ensure:
\[ f(x) > 0 \]
Analyze \( f(x) = x^2 + x + 1 \):
- The discriminant of the quadratic \( x^2 + x + 1 \) is:
\[ D = b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3 \]
Since the discriminant is negative, the quadratic has no real roots and opens upwards (as the coefficient of \( x^2 \) is positive). Therefore, \( f(x) > 0 \) for all \( x \in \mathbb{R} \).

8. **Conclusion:**
Since \( f(x) > 0 \) for all \( x \in \mathbb{R} \), the domain of \( \log(f(x)) \) is:
\[ \text{Domain of } \log(f(x)) = \mathbb{R} \]