1. Use the functional equation

Given $f(x+y)=f(x)f(y) \quad \forall x,y\in\mathbb R.$ Also, $f$ is twice differentiable, hence continuous.

A continuous solution of $f(x+y)=f(x)f(y)$ is of exponential form. Let us derive it using the given derivative information.

First, put $y=0$: $f(x)=f(x)f(0).$ So either $f\equiv 0$ or $f(0)=1$.

But $f'(0)=4a$ and $a>0$, so $f'(0)\neq 0$, hence $f$ is not identically zero. Therefore, $f(0)=1.$

Now differentiate $f(x+y)=f(x)f(y)$ with respect to $y$: $f'(x+y)=f(x)f'(y).$ Set $y=0$: $f'(x)=f(x)f'(0).$ Given $f'(0)=4a$, we get $f'(x)=4a\,f(x).$ Thus, $\frac{f'(x)}{f(x)}=4a,$ which gives $f(x)=Ce^{4ax}.$ Since $f(0)=1$, we get $C=1$. Hence $f(x)=e^{4ax}.$

2. Use the differential equation to find $a$

Given $f''(x)-3af'(x)-f(x)=0.$

For $f(x)=e^{4ax},$ we have $f'(x)=4ae^{4ax}, \qquad f''(x)=16a^2 e^{4ax}.$ Substitute into the differential equation: $16a^2 e^{4ax}-3a(4a e^{4ax})-e^{4ax}=0.$ So $\left(16a^2-12a^2-1\right)e^{4ax}=0,$ $(4a^2-1)e^{4ax}=0.$ Since $e^{4ax}\neq 0$, $4a^2-1=0 \implies a^2=\frac14.$ Given $a>0$, $a=\frac12.$

Therefore, $f(x)=e^{4ax}=e^{2x}.$

3. Find $f(ax)$

Since $a=\frac12$, $f(ax)=f\left(\frac{x}{2}\right)=e^{2\cdot x/2}=e^x.$

4. Compute the required area

The region is $R=\{(x,y)\mid 0\le y\le f(ax),\ 0\le x\le 2\}.$ So area is $\text{Area} = \int_0^2 f(ax)\,dx = \int_0^2 e^x\,dx.$ Thus, $\text{Area} = \left[e^x\right]_0^2 = e^2-1.$

5. Check options

• A: $e^2-1$ ✅
• B: $e^4+1$ ❌
• C: $e^2+1$ ❌
• D: $e^4-1$ ❌

Hence the correct answer is A.