Properties and Differentiability of Functional Equations

To solve the given problem, we analyze the functional relations and limit conditions provided:

1. $f(x + y) = f(x) + f(y) + f(x)f(y)$ for all $x, y \in \mathbb{R}$
2. $f(x) = x g(x)$ for all $x \in \mathbb{R}$
3. $\lim_{x \to 0} g(x) = 1$

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Step 1: Value of $f(0)$ and $f'(0)$

From the relation $f(x) = x g(x)$, substituting $x = 0$: $f(0) = 0 \cdot g(0) = 0$

Now, using the definition of derivative at $x = 0$: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h g(h) - 0}{h} = \lim_{h \to 0} g(h)$

Given that $\lim_{x \to 0} g(x) = 1$, we get: $f'(0) = 1$

Thus, Statement (D) is TRUE.

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Step 2: Differentiability of $f(x)$ for all $x \in \mathbb{R}$

For any arbitrary $x \in \mathbb{R}$, we evaluate the derivative $f'(x)$ using the first principle of differentiation: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Using the given functional equation $f(x + h) = f(x) + f(h) + f(x)f(h)$, we can rewrite the numerator as: $f(x + h) - f(x) = f(h)(1 + f(x))$

Substituting this into the limit expression gives: $f'(x) = \lim_{h \to 0} \frac{f(h)(1 + f(x))}{h} = (1 + f(x)) \lim_{h \to 0} \frac{f(h)}{h} = (1 + f(x)) \cdot f'(0)$

Since $f'(0) = 1$, we have: $f'(x) = 1 + f(x)$

Since $f(x)$ is continuous at $x = 0$ (because $\lim_{h \to 0} f(h) = \lim_{h \to 0} h g(h) = 0 = f(0)$), it follows that $f(x)$ is continuous and differentiable for all $x \in \mathbb{R}$.

Thus, Statement (A) is TRUE.

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Step 3: Determining $f(x)$ and calculating $f'(1)$

From the differential equation derived above: $f'(x) = 1 + f(x) \implies \frac{f'(x)}{1 + f(x)} = 1$

Integrating both sides with respect to $x$: $\ln |1 + f(x)| = x + C$

Using the initial condition $f(0) = 0$: $\ln |1 + 0| = 0 + C \implies C = 0$

Therefore: $1 + f(x) = e^x \implies f(x) = e^x - 1$

Taking the derivative of $f(x)$: $f'(x) = e^x$

Evaluating at $x = 1$: $f'(1) = e^1 = e \neq 1$

Thus, Statement (C) is FALSE.

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Step 4: Differentiability of $g(x)$ given $g(0) = 1$

Given $f(x) = x g(x)$, if $g(0) = 1$, we define $g(x)$ as:

$$g(x) = \begin{cases} \frac{e^x - 1}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$$

For any $x \neq 0$, $g(x)$ is a quotient of differentiable functions where the denominator is non-zero, making $g(x)$ differentiable for all $x \neq 0$.

At $x = 0$, we test for differentiability using the limit definition: $g'(0) = \lim_{x \to 0} \frac{g(x) - g(0)}{x - 0} = \lim_{x \to 0} \frac{\frac{e^x - 1}{x} - 1}{x} = \lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$

Using the Taylor series expansion of $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$: $g'(0) = \lim_{x \to 0} \frac{\left(1 + x + \frac{x^2}{2} + O(x^3)\right) - 1 - x}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2} + O(x^3)}{x^2} = \frac{1}{2}$

Since $g'(0)$ exists and is finite, $g(x)$ is differentiable at $x = 0$ as well. Hence, $g(x)$ is differentiable at every $x \in \mathbb{R}$.

Thus, Statement (B) is TRUE.

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Conclusion

The correct options are (A), (B), and (D).