Let f : ℝ → ℝ be a twice-differentiable function such that   f(x + y) = f(x) f(y) for all x, y ∈ ℝ. If f′(0) = 4a and f satisfies   f″(x) − 3a f′(x) − f(x) = 0, a > 0, then the area of the region   R = { (x, y) | 0 ≤ y ≤ f(ax), 0 ≤ x ≤ 2 } is: (1) e² − 1  (2) e⁴ + 1  (3) e⁴ − 1  (4) e² + 1

1. Recognising the Functional Equation

The identity
$f(x+y)=f(x)f(y)$
with $f:\mathbb R\to\mathbb R$ twice-differentiable is the well-known Cauchy Exponential functional equation. Under differentiability, its non-trivial solutions are of the form
$f(x)=e^{kx}$
for some constant $k$.

2. Checking $f(0)$

Set $x=y=0$:
$f(0)=f(0)f(0)\;\;\Rightarrow\;\;f(0)=0 \text{ or } 1.$
If $f(0)=0$, the functional equation forces $f\equiv0$, contradicting $f'(0)=4a\,(>0)$. Hence
$f(0)=1.$

3. First Derivative Condition

For $f(x)=e^{kx}$ we get
$f'(x)=k e^{kx},\qquad f'(0)=k.$
Given $f'(0)=4a$, we immediately have
$k=4a.$

4. Satisfying the Second-Order ODE

Insert $f(x)=e^{kx}$ into
$f''(x)-3a f'(x)-f(x)=0.$
Because $f''(x)=k^{2}e^{kx}$ and $f'(x)=k e^{kx}$, the exponential factor $e^{kx}$ never vanishes, so
$k^{2}-3a k-1=0.$
Substitute $k=4a$ into the quadratic:
$16a^{2}-12a^{2}-1=0 \;\;\Longrightarrow\;\;4a^{2}=1 \;\;\Longrightarrow\;\;a=\frac12\;(a>0).$
Hence

5. Function Inside the Area

We need $f(ax)$ with $a=\frac12$:

6. Computing the Area

The required region is

Its area is the definite integral

Thus the correct option is (1) $e^{2}-1$.

Imagine a Magic Balloon

1. Balloon Rule (Functional Equation)
   If you blow the balloon for $x$ seconds and then for $y$ more seconds, its size becomes the same as blowing it for $x+y$ seconds in one go.
   Mathematically, that is $f(x+y)=f(x)f(y)$.

2. Extra Clues

   • The way the balloon’s size changes at the very start is $4a$ (that’s $f'(0)=4a$).
   • The size also obeys a special bending rule while it grows:
     $f''(x)-3a f'(x)-f(x)=0$

3. Goal
   We draw the curve $y=f(ax)$ from $x=0$ to $x=2$ and colour everything under it (down to $y=0$).
   We want the area of that coloured part.

4. What Happens
   a. The only shape that always follows the rule $f(x+y)=f(x)f(y)$ and is smooth is an exponential curve $f(x)=e^{kx}$.
   b. Using the clues, we find $k=2$ and $a=\frac12$.
   c. So $f(ax)=e^{x}$.
   d. The coloured area is just the area under $y=e^{x}$ from $0$ to $2$, which is $e^{2}-1$.

Answer: $e^{2}-1$.

Step 1 – Assume Exponential Form

Because $f$ is twice differentiable and obeys
$f(x+y)=f(x)f(y),$
write
$f(x)=e^{kx}\quad (k\in\mathbb R).$

Step 2 – Determine $k$ Using $f'(0)$

For $f(x)=e^{kx}$,
$f'(x)=k e^{kx}\quad\Longrightarrow\quad f'(0)=k.$
Given $f'(0)=4a$, we get
k=4a.\tag{1}

Step 3 – Impose the Second-Order ODE

Plug $f(x)=e^{kx}$ into
$f''(x)-3a f'(x)-f(x)=0.$
We have
$f''(x)=k^{2}e^{kx},\;f'(x)=k e^{kx}\;\Longrightarrow\;k^{2}e^{kx}-3a k e^{kx}-e^{kx}=0.$
Divide by $e^{kx}$ (never zero):
k^{2}-3a k-1=0.\tag{2}

Step 4 – Solve for $a$ and $k$

Insert (1) into (2):
$\bigl(4a\bigr)^{2}-3a\bigl(4a\bigr)-1=0 \Longrightarrow 16a^{2}-12a^{2}-1=0 \Longrightarrow 4a^{2}=1 \Longrightarrow a=\frac12 \;(a>0).$
Hence
$k=4a=2.$
So
f(x)=e^{2x}.\tag{3}

Step 5 – Function Inside the Region

We need $f(ax)$: with $a=\tfrac12$,
f(ax)=e^{2\left(\frac12 x\right)}=e^{x}.\tag{4}

Step 6 – Area Under the Curve

The region is bounded by $y=e^{x}$, the $x$-axis, and the verticals $x=0$ and $x=2$. Hence
\text{Area}=\int_{0}^{2} e^{x}\,dx = \left[ e^{x} \right]_{0}^{2}=e^{2}-1.\tag{5}

Final Answer

$\boxed{e^{2}-1}$
Option (1).