If `f(x+y)=f(x) xx f(y)` for all `x,y in R` and` f(5)=2, f'(0)=3, then f'(5)=`

Options

A. 3
B. 1
C. `-6`
D. 6

Answer

To solve the problem, we start with the functional equation given:

\[ f(x+y) = f(x) \cdot f(y) \]

for all \( x, y \in \mathbb{R} \). We also know that \( f(5) = 2 \) and \( f'(0) = 3 \). We need to find \( f'(5) \).

### Step 1: Finding \( f(0) \)

To find \( f(0) \), we can set \( x = 0 \) and \( y = 0 \) in the functional equation:

\[ f(0+0) = f(0) \cdot f(0) \]

This simplifies to:

\[ f(0) = f(0)^2 \]

Rearranging gives:

\[ f(0)^2 - f(0) = 0 \]

Factoring out \( f(0) \):

\[ f(0)(f(0) - 1) = 0 \]

This gives us two possible solutions:

1. \( f(0) = 0 \)
2. \( f(0) = 1 \)

### Step 2: Verifying \( f(0) = 1 \)

If \( f(0) = 0 \), then substituting \( y = 5 \) in the original equation gives:

\[ f(5) = f(0) \cdot f(5) = 0 \cdot 2 = 0 \]

This contradicts the fact that \( f(5) = 2 \). Therefore, we conclude that:

\[ f(0) = 1 \]

### Step 3: Differentiate the functional equation

Next, we differentiate the functional equation 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:

\[ f'(x+y) = f(x) \cdot f'(y) \]

Now, we set \( x = 0 \) and \( y = 5 \):

\[ f'(0+5) = f(0) \cdot f'(5) \]

This simplifies to:

\[ f'(5) = f(0) \cdot f'(5) \]

Substituting \( f(0) = 1 \):

\[ f'(5) = 1 \cdot f'(5) = f'(5) \]

This does not help us directly, so we need to express \( f'(5) \) in terms of \( f'(0) \).

### Step 4: Use the differentiation result

From our earlier differentiation, we have:

\[ f'(x+y) = f(x) \cdot f'(y) \]

Setting \( x = 5 \) and \( y = 0 \):

\[ f'(5+0) = f(5) \cdot f'(0) \]

This gives:

\[ f'(5) = f(5) \cdot f'(0) \]

Substituting the known values \( f(5) = 2 \) and \( f'(0) = 3 \):

\[ f'(5) = 2 \cdot 3 = 6 \]

### Final Answer

Thus, the value of \( f'(5) \) is:

\[ \boxed{6} \]