If `f(x+y) = f(x) + f(y) + |x|y+xy^(2),AA x, y in R and f'(0) = 0`, then

Options

A. f need not be differentiable at every non-zero x
B. f is differentiable for all `x in R`
C. f is twice differentiable at x = 0
D. None of the above

Answer

To solve the problem given, we need to analyze the functional equation and the condition provided.

### Step-by-Step Solution:

1. **Understanding the Functional Equation**:
We start with the equation:
\[ f(x+y) = f(x) + f(y) + |x|y + xy^2 \]
for all \( x, y \in \mathbb{R} \).

2. **Substituting \( x = 0 \)**:
Let's substitute \( x = 0 \) into the equation:
\[ f(0+y) = f(0) + f(y) + |0|y + 0 \cdot y^2 \]
This simplifies to:
\[ f(y) = f(0) + f(y) \]
From this, we can conclude that:
\[ f(0) = 0 \]

3. **Finding the Derivative**:
We know that \( f'(0) = 0 \). To find \( f'(x) \), we can use the definition of the derivative:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Now, substituting \( y = h \) in our original functional equation:
\[ f(x+h) = f(x) + f(h) + |x|h + xh^2 \]
Rearranging gives:
\[ f(x+h) - f(x) = f(h) + |x|h + xh^2 \]

4. **Taking the Limit**:
Now, we divide both sides by \( h \):
\[ \frac{f(x+h) - f(x)}{h} = \frac{f(h)}{h} + |x| + xh \]
Taking the limit as \( h \to 0 \):
\[ f'(x) = \lim_{h \to 0} \left( \frac{f(h)}{h} + |x| + xh \right) \]
Since \( f'(0) = 0 \), we have:
\[ f'(0) = \lim_{h \to 0} \frac{f(h)}{h} = 0 \]
Thus, we can conclude that:
\[ f'(x) = |x| \quad \text{for all } x \]

5. **Integrating to Find \( f(x) \)**:
To find \( f(x) \), we integrate \( f'(x) \):
\[ f(x) = \int |x| \, dx \]
This gives us two cases:
- For \( x \geq 0 \): \( f(x) = \frac{x^2}{2} + C \)
- For \( x < 0 \): \( f(x) = -\frac{x^2}{2} + C \)

6. **Using the Condition \( f(0) = 0 \)**:
Since \( f(0) = 0 \), we can find \( C \):
\[ f(0) = C = 0 \]
Thus, we have:
\[ f(x) = \frac{x^2}{2} \quad \text{for } x \geq 0 \]
\[ f(x) = -\frac{x^2}{2} \quad \text{for } x < 0 \]

### Final Form of \( f(x) \):
Combining both cases, we can write:
\[ f(x) = \frac{x^2}{2} \text{sgn}(x) \quad \text{for all } x \in \mathbb{R} \]

### Conclusion:
From our analysis, we can conclude that:
- \( f(x) \) is differentiable for all \( x \in \mathbb{R} \).
- \( f(x) \) is not twice differentiable at \( x = 0 \).