Let `f:(-oo,oo)vec[0,oo)` be a continuous function such that `f(x+y)=f(x)+f(y)+f(x)f(y),AAx in Rdot` Also `f'(0)=1.` Then `[f(2)]` equal `([dot]` represents the greatest integer function`)` `5` b. `6` c. `7` d. `8`

Options

A. 5
B. 6
C. 7
D. 8

Explanation

To solve the problem step by step, we need to analyze the given functional equation and the derivative condition.

### Step 1: Understanding the Functional Equation
We are given the functional equation:
\[ f(x+y) = f(x) + f(y) + f(x)f(y) \]
for all \( x, y \in \mathbb{R} \).

### Step 2: Transforming the Equation
To simplify the equation, we can rewrite it as:
\[ 1 + f(x+y) = 1 + f(x) + f(y) + f(x)f(y) \]
This can be rearranged to:
\[ 1 + f(x+y) = (1 + f(x))(1 + f(y)) \]

### Step 3: Defining a New Function
Let us define a new function:
\[ g(x) = 1 + f(x) \]
Then the equation becomes:
\[ g(x+y) = g(x)g(y) \]
This indicates that \( g(x) \) is a multiplicative function.

### Step 4: Finding the Form of \( g(x) \)
The general solution for a continuous multiplicative function is:
\[ g(x) = e^{kx} \]
for some constant \( k \). Thus, we have:
\[ 1 + f(x) = e^{kx} \]
which leads to:
\[ f(x) = e^{kx} - 1 \]

### Step 5: Using the Derivative Condition
We are given \( f'(0) = 1 \). To find \( f'(x) \), we differentiate \( f(x) \):
\[ f'(x) = ke^{kx} \]
Now, evaluating at \( x = 0 \):
\[ f'(0) = ke^{k \cdot 0} = k \]
Setting this equal to 1 gives us:
\[ k = 1 \]

### Step 6: Finding the Function \( f(x) \)
Substituting \( k = 1 \) back into the equation for \( f(x) \):
\[ f(x) = e^x - 1 \]

### Step 7: Calculating \( f(2) \)
Now we can find \( f(2) \):
\[ f(2) = e^2 - 1 \]

### Step 8: Evaluating \( e^2 \)
Using the approximation \( e \approx 2.718 \):
\[ e^2 \approx (2.718)^2 \approx 7.389 \]
Thus,
\[ f(2) \approx 7.389 - 1 = 6.389 \]

### Step 9: Applying the Greatest Integer Function
Now we apply the greatest integer function:
\[ [f(2)] = [6.389] = 6 \]

### Final Answer
Thus, the answer is:
\[ \boxed{6} \]