Le `f : R to R and g : R to R ` be functions satisfying `f(x+y) = f(x) +f(y) +f(x)f(y) and f(x) = xg (x)` for all `x , y in R . If lim_(x to 0) g(x) = 1` , then which of the following statements is/are TRUE ?

Options

A. f is differentiable at every `x in R `
B. If `g(0) = 1 ` , then g is differentiable at every `x in R `
C. The derivative f(1) is equal to 1
D. The derivative f'(0) is equal to 1

Answer

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

### Step 1: Analyze the functional equation
We are given the functional equation:
\[ f(x+y) = f(x) + f(y) + f(x)f(y) \]
This resembles the form of an exponential function.

### Step 2: Substitute \( x = 0 \) and \( y = 0 \)
Let's substitute \( x = 0 \) and \( y = 0 \) into the functional equation:
\[ f(0 + 0) = f(0) + f(0) + f(0)f(0) \]
This simplifies to:
\[ f(0) = 2f(0) + f(0)^2 \]
Rearranging gives:
\[ f(0)^2 + f(0) = 0 \]
Factoring out \( f(0) \):
\[ f(0)(f(0) + 1) = 0 \]
Thus, \( f(0) = 0 \) or \( f(0) = -1 \).

### Step 3: Analyze the second functional equation
We also have:
\[ f(x) = x g(x) \]
Substituting \( x = 0 \):
\[ f(0) = 0 \cdot g(0) = 0 \]
Thus, we conclude that \( f(0) = 0 \).

### Step 4: Differentiate the functional equation
We can differentiate the functional equation with respect to \( y \):
\[ \frac{d}{dy}[f(x+y)] = \frac{d}{dy}[f(x) + f(y) + f(x)f(y)] \]
Using the chain rule on the left side:
\[ f'(x+y) = f'(y) + f(x)f'(y) \]
Setting \( y = 0 \):
\[ f'(x) = f'(0) + f(x)f'(0) \]
Let \( f'(0) = c \):
\[ f'(x) = c(1 + f(x)) \]

### Step 5: Evaluate \( f'(0) \)
From the earlier analysis, we found \( f(0) = 0 \):
\[ f'(0) = c(1 + f(0)) = c(1 + 0) = c \]
Thus, \( f'(0) = c \).

### Step 6: Find \( g(0) \)
We know:
\[ g(0) = \lim_{x \to 0} g(x) = 1 \]

### Step 7: Evaluate \( f'(1) \)
Using the expression \( f'(x) = c(1 + f(x)) \):
\[ f'(1) = c(1 + f(1)) \]
We need to find \( f(1) \). Using the functional equation:
\[ f(1) = 1g(1) \]
We do not have the explicit value of \( g(1) \) yet.

### Step 8: Check differentiability of \( g(x) \)
Since \( g(x) = \frac{f(x)}{x} \), we can differentiate it:
\[ g'(x) = \frac{f'(x)x - f(x)}{x^2} \]
At \( x = 0 \):
\[ g'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h^2} = \lim_{h \to 0} \frac{h g(h)}{h^2} = \lim_{h \to 0} \frac{g(h)}{h} \]
Since \( g(0) = 1 \), we find \( g'(0) \) exists.

### Conclusion
From our analysis:
- \( f \) is differentiable everywhere.
- \( g(0) = 1 \) implies \( g \) is differentiable everywhere.
- \( f'(0) = 1 \) and \( f'(1) \) needs further evaluation.

Thus, the true statements are:
- \( f \) is differentiable everywhere (True).
- \( g(0) = 1 \) implies \( g \) is differentiable everywhere (True).
- \( f'(0) = 1 \) (True).

### Final Answer
The true statements are:
- A: True
- B: True
- D: True