Let `f :R to R` be a continous and differentiable function such that `f (x+y ) =f (x). F(y)AA x, y, f(x) ne 0 and f (0 )=1 and f '(0) =2.`
Let `f (xy) =g (x). G (y) AA x, y and g'(1)=2.g (1) ne =0`
Identify the correct option:

Options

A. `f (2) =e ^(4)`
B. `f (2) =2e ^(2)`
C. `f(1) lt 4`
D. `f (3) gt 729`

Answer

To solve the problem step by step, we will analyze the given functional equation and the conditions provided.

### Step 1: Analyze the Functional Equation
We are given that \( f(x+y) = f(x) \cdot f(y) \) for all \( x, y \in \mathbb{R} \) and \( f(0) = 1 \). This is a well-known functional equation that suggests that \( f(x) \) could be an exponential function.

### Step 2: Differentiate the Functional Equation
To find the form of \( f(x) \), we differentiate both sides with respect to \( y \):
\[ \frac{d}{dy} f(x+y) = \frac{d}{dy} (f(x) \cdot f(y)) \]
Using the chain rule on the left side, we get:
\[ f'(x+y) = f(x) \cdot f'(y) \]
Now, set \( y = 0 \):
\[ f'(x+0) = f(x) \cdot f'(0) \]
This simplifies to:
\[ f'(x) = f(x) \cdot f'(0) \]
Given that \( f'(0) = 2 \), we can write:
\[ f'(x) = 2 f(x) \]

### Step 3: Solve the Differential Equation
This is a separable differential equation. We can separate variables:
\[ \frac{f'(x)}{f(x)} = 2 \]
Integrating both sides gives:
\[ \ln |f(x)| = 2x + C \]
Exponentiating both sides, we find:
\[ f(x) = e^{2x + C} = e^C e^{2x} \]
Let \( k = e^C \), so:
\[ f(x) = k e^{2x} \]

### Step 4: Determine the Constant \( k \)
Using the condition \( f(0) = 1 \):
\[ f(0) = k e^{2 \cdot 0} = k = 1 \]
Thus, we have:
\[ f(x) = e^{2x} \]

### Step 5: Calculate \( f(2) \) and \( f(1) \)
Now we can calculate \( f(2) \):
\[ f(2) = e^{2 \cdot 2} = e^4 \]
And for \( f(1) \):
\[ f(1) = e^{2 \cdot 1} = e^2 \]

### Step 6: Analyze the Function \( g(x) \)
We are given that \( f(xy) = g(x) g(y) \). Since \( f(x) = e^{2x} \), we have:
\[ f(xy) = e^{2xy} \]
Assuming \( g(x) = e^{h(x)} \), we can write:
\[ e^{2xy} = e^{h(x)} e^{h(y)} \implies h(x+y) = h(x) + h(y) \]
This implies that \( h(x) \) is a linear function, say \( h(x) = ax \). Thus:
\[ g(x) = e^{ax} \]

### Step 7: Find \( g'(1) \)
The derivative \( g'(x) \) is given by:
\[ g'(x) = a e^{ax} \]
At \( x = 1 \):
\[ g'(1) = a e^a \]
We know \( g'(1) = 2 g(1) \) and \( g(1) = e^a \), so:
\[ g'(1) = 2 e^a \]
Thus:
\[ a e^a = 2 e^a \implies a = 2 \]
So, \( g(x) = e^{2x} \).

### Step 8: Verify the Conditions
Now we check if \( g(1) \neq 0 \) and \( g'(1) = 2 g(1) \):
\[ g(1) = e^2 \neq 0 \]
And:
\[ g'(1) = 2 e^2 = 2 g(1) \]
Both conditions are satisfied.

### Conclusion
Now we can summarize the results:
- \( f(2) = e^4 \)
- \( f(1) = e^2 \)
- \( f(3) = e^6 \)

### Final Options
1. \( g(2) = e^4 \) is correct.
2. \( f(2) = e^4 \) is correct.
3. \( f(1) < 4 \) is incorrect since \( e^2 \approx 7.39 > 4 \).
4. \( f(3) < 729 \) is incorrect since \( e^6 \approx 403.43 < 729 \).

### Correct Option
The correct option is \( f(2) = e^4 \).