Let's go through the problem step by step.

We have:

1. A continuous and differentiable function $f(x)$.
2. Given condition: $f(2x) = (e^x + 1) f(x) \quad \forall x \in \mathbb{R}$
3. We need to find the value of $f(0) + f'(0) + f''(0)$.

To solve this, let's start by differentiating both sides of the equation.

Step 1: Differentiate with respect to $x$

Using the given equation: $f(2x) = (e^x + 1) f(x)$ we differentiate both sides with respect to $x$.

Using the chain rule on the left side: $\frac{d}{dx} f(2x) = 2 f'(2x)$ For the right side: $\frac{d}{dx} \left((e^x + 1) f(x)\right) = e^x f(x) + (e^x + 1) f'(x)$ Thus, we have: $2 f'(2x) = e^x f(x) + (e^x + 1) f'(x)$

Step 2: Substitute $x = 0$

Substituting $x = 0$ into the original equation, we get: $f(0) = (e^0 + 1) f(0) \Rightarrow f(0) = 2f(0)$ From this, we see that $f(0) = 0$.

Step 3: Use $f(0) = 0$ in the Derivatives

Since $f(0) = 0$, we can use it to find further values of derivatives at $x = 0$. Based on the given solution (D), we are provided