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 start with the given equation:

\[ \frac{f(x+y) - f(x)}{2} = \frac{f(y) - 1}{2} + xy \]

for all \( x, y \in \mathbb{R} \).

### Step 1: Simplify the Equation
Multiply both sides by 2 to eliminate the fraction:

\[ f(x+y) - f(x) = f(y) - 1 + 2xy \]

### Step 2: Set \( x = 0 \) and \( y = 0 \)
Substituting \( x = 0 \) and \( y = 0 \):

\[ f(0+0) - f(0) = f(0) - 1 + 2(0)(0) \]

This simplifies to:

\[ f(0) - f(0) = f(0) - 1 \]

Thus, we have:

\[ 0 = f(0) - 1 \implies f(0) = 1 \]

### Step 3: Differentiate the Function
We know \( f(x) \) is differentiable, and we can find \( f'(x) \). We rewrite the previous equation by substituting \( y = h \):

\[ f(x+h) - f(x) = f(h) - 1 + 2xh \]

Now, divide by \( h \):

\[ \frac{f(x+h) - f(x)}{h} = \frac{f(h) - 1}{h} + 2x \]

Taking the limit as \( h \to 0 \):

\[ f'(x) = \lim_{h \to 0} \left( \frac{f(h) - 1}{h} + 2x \right) \]

### Step 4: Evaluate \( f'(0) \)
Since \( f'(0) = 1 \) is given:

\[ f'(0) = \lim_{h \to 0} \frac{f(h) - 1}{h} + 0 = 1 \]

This implies:

\[ \lim_{h \to 0} \frac{f(h) - 1}{h} = 1 \]

### Step 5: Find \( f'(x) \)
From the previous steps, we can express \( f'(x) \):

\[ f'(x) = 2x + 1 \]

### Step 6: Integrate to Find \( f(x) \)
Integrating \( f'(x) \):

\[ f(x) = \int (2x + 1) \, dx = x^2 + x + C \]

### Step 7: Determine the Constant \( C \)
Using \( f(0) = 1 \):

\[ f(0) = 0^2 + 0 + C = 1 \implies C = 1 \]

Thus, we have:

\[ f(x) = x^2 + x + 1 \]

### Step 8: Determine the Domain of \( \log(f(x)) \)
The domain of \( \log(f(x)) \) requires \( f(x) > 0 \):

\[ f(x) = x^2 + x + 1 \]

### Step 9: Analyze \( f(x) \)
The quadratic \( x^2 + x + 1 \) has a discriminant:

\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]

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 \( x \in \mathbb{R} \).

### Conclusion
Thus, the domain of \( \log(f(x)) \) is:

\[ \text{Domain of } \log(f(x)) = \mathbb{R} \]