A differentiable function f Is satifying the relation ` f(x+y)= f(x) + f(y) + 2xy( x+y) -(1)/(3)AA x, y in R and lim_( h to 0) ( 3f(h)-1)/(6h)=(2)/(3)`. Then the value of [f(2)]is ( where [x] represents the greatest integer function )_________.

Answer

To solve the problem step by step, we start with the given functional equation and the limit condition.

### Step 1: Analyze the functional equation
We are given the functional equation:
\[ f(x+y) = f(x) + f(y) + 2xy(x+y) - \frac{1}{3} \]
for all \( x, y \in \mathbb{R} \).

### Step 2: Substitute \( x = 0 \) and \( y = 0 \)
Substituting \( x = 0 \) and \( y = 0 \) into the functional equation:
\[ f(0 + 0) = f(0) + f(0) + 2 \cdot 0 \cdot 0 \cdot (0 + 0) - \frac{1}{3} \]
This simplifies to:
\[ f(0) = 2f(0) - \frac{1}{3} \]
Rearranging gives:
\[ f(0) - 2f(0) = -\frac{1}{3} \]
\[ -f(0) = -\frac{1}{3} \]
Thus, we find:
\[ f(0) = \frac{1}{3} \]

### Step 3: Analyze the limit condition
We are also given:
\[ \lim_{h \to 0} \frac{3f(h) - 1}{6h} = \frac{2}{3} \]
Multiplying both sides by \( 6h \) gives:
\[ 3f(h) - 1 = \frac{2}{3} \cdot 6h \]
\[ 3f(h) - 1 = 4h \]
Thus:
\[ 3f(h) = 4h + 1 \]
Dividing by 3:
\[ f(h) = \frac{4h + 1}{3} \]

### Step 4: Find \( f'(0) \)
Taking the limit as \( h \to 0 \):
\[ f(0) = \frac{1}{3} \]
Now, we can find \( f'(0) \):
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{\frac{4h + 1}{3} - \frac{1}{3}}{h} \]
This simplifies to:
\[ = \lim_{h \to 0} \frac{\frac{4h}{3}}{h} = \frac{4}{3} \]

### Step 5: General form of \( f(x) \)
Now we can assume a polynomial form for \( f(x) \). Given the structure of the functional equation, we can propose:
\[ f(x) = ax^3 + bx + c \]
Using \( f(0) = \frac{1}{3} \), we have \( c = \frac{1}{3} \).

### Step 6: Differentiate and match coefficients
Differentiating:
\[ f'(x) = 3ax^2 + b \]
Setting \( f'(0) = \frac{4}{3} \):
\[ b = \frac{4}{3} \]

### Step 7: Substitute back into the functional equation
Now substitute \( f(x) = ax^3 + \frac{4}{3}x + \frac{1}{3} \) back into the functional equation to find \( a \).

### Step 8: Calculate \( f(2) \)
After determining \( a \) (which can be found through matching coefficients), we can compute:
\[ f(2) = a(2^3) + \frac{4}{3}(2) + \frac{1}{3} \]
This gives:
\[ f(2) = 8a + \frac{8}{3} + \frac{1}{3} = 8a + 3 \]

### Step 9: Find the greatest integer function
Finally, we need to find \( \lfloor f(2) \rfloor \).