Step 1: Determine the function f(x)

1. The given functional equation is f(x + y) = f(x) + f(y) + f(x)f(y) for all $x, y \in R$.
2. We can rewrite this equation by adding 1 to both sides: $1 + f(x + y) = 1 + f(x) + f(y) + f(x)f(y)$ $1 + f(x + y) = (1 + f(x))(1 + f(y))$
3. Let's define a new function h(x) = 1 + f(x). Substituting this into the equation, we get: $h(x + y) = h(x)h(y)$
4. This is a standard functional equation. If h is a continuous function, its solutions are of the form $h(x) = a^x$ for some constant a > 0, or h(x) = 0 for all x.
   • If h(x) = 0, then f(x) = -1. In this case, g(x) = f(x)/x = -1/x. The limit $\lim_{x \to 0} g(x)$ does not exist, which contradicts the given condition. So, h(x) cannot be identically zero.
   • Thus, we must have $h(x) = a^x$ for some a > 0.
5. This implies $1 + f(x) = a^x$, so $f(x) = a^x - 1$.
6. We are given f(x) = xg(x) and $\lim_{x \to 0} g(x) = 1$. From this, we can find the value of a: $\lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{f(x)}{x} = 1$ $\lim_{x \to 0} \frac{a^x - 1}{x} = 1$
7. We know the standard limit $\lim_{t \to 0} \frac{a^t - 1}{t} = \ln(a)$.
8. Therefore, $\ln(a) = 1$, which means a = e.
9. So, the function is $f(x) = e^x - 1$.

Step 2: Evaluate each statement

A: f is differentiable at every x ∈ R

• We found $f(x) = e^x - 1$.
• The derivative is $f'(x) = \frac{d}{dx}(e^x - 1) = e^x$.
• The function $e^x$ is defined and finite for all $x \in R$.
• Therefore, f(x) is differentiable at every $x \in R$.
• Statement A is TRUE.

D: The derivative f'(0) is equal to 1

• We have $f'(x) = e^x$.
• At x = 0, the derivative is $f'(0) = e^0 = 1$.
• Statement D is TRUE.

C: The derivative f'(1) is equal to 1

• We have $f'(x) = e^x$.
• At x = 1, the derivative is $f'(1) = e^1 = e$.
• Since $e \approx 2.718$, $f'(1) \neq 1$.
• Statement C is FALSE.

B: If g(0) = 1, then g is differentiable at every x ∈ R

• From f(x) = xg(x), we have $g(x) = \frac{f(x)}{x} = \frac{e^x - 1}{x}$ for $x \neq 0$.
• Given g(0) = 1, the function g(x) is defined as: $g(x) = \begin{cases} \frac{e^x - 1}{x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases}$
• Note that $\lim_{x \to 0} \frac{e^x - 1}{x} = 1 = g(0)$, so g(x) is continuous at x=0.
• For $x \neq 0$, we find the derivative using the quotient rule: $g'(x) = \frac{(e^x)(x) - (e^x - 1)(1)}{x^2} = \frac{xe^x - e^x + 1}{x^2}$ This exists for all $x \neq 0$.
• To check differentiability at x = 0, we use the limit definition of the derivative: $g'(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h} = \lim_{h \to 0} \frac{\frac{e^h - 1}{h} - 1}{h}$ $g'(0) = \lim_{h \to 0} \frac{e^h - 1 - h}{h^2}$
• This is an indeterminate form 0/0. We can apply L'Hopital's Rule: $g'(0) = \lim_{h \to 0} \frac{\frac{d}{dh}(e^h - 1 - h)}{\frac{d}{dh}(h^2)} = \lim_{h \to 0} \frac{e^h - 1}{2h}$
• This is still 0/0. Applying L'Hopital's Rule again: $g'(0) = \lim_{h \to 0} \frac{\frac{d}{dh}(e^h - 1)}{\frac{d}{dh}(2h)} = \lim_{h \to 0} \frac{e^h}{2} = \frac{e^0}{2} = \frac{1}{2}$
• Since g'(0) exists (and is equal to 1/2), g(x) is differentiable at x=0.
• Since g(x) is differentiable for all $x \neq 0$ and also at x=0, it is differentiable at every $x \in R$.
• Statement B is TRUE.

Conclusion

The true statements are A, B, and D.