Let `f(x)` be a differentiable function on `x in R` such that `f(x+y)=f(x). F(y)" for all, "x,y`. If `f(0) ne 0, f(5)=12 and f'(0)=16`, then `f'(5)` is equal to

Options

A. 190
B. 186
C. 196
D. 192

Answer

To solve the problem step by step, we start with the given functional equation and the conditions provided.

### Step 1: Understand the functional equation
We are given that:
\[ f(x+y) = f(x) \cdot f(y) \]
for all \( x, y \in \mathbb{R} \).

### Step 2: Find \( f(0) \)
Let’s substitute \( y = 0 \):
\[ f(x + 0) = f(x) \cdot f(0) \]
This simplifies to:
\[ f(x) = f(x) \cdot f(0) \]
Since \( f(0) \neq 0 \), we can divide both sides by \( f(x) \):
\[ 1 = f(0) \]
Thus, we have:
\[ f(0) = 1 \]

### Step 3: Use the given value \( f(5) = 12 \)
Now, we know:
\[ f(5) = 12 \]

### Step 4: Differentiate the functional equation
Next, we differentiate the functional equation with respect to \( x \):
\[ \frac{d}{dx}[f(x+y)] = \frac{d}{dx}[f(x) \cdot f(y)] \]
Using the chain rule on the left side and the product rule on the right side, we get:
\[ f'(x+y) = f'(x) \cdot f(y) \]

### Step 5: Substitute \( y = 0 \) in the differentiated equation
Let’s substitute \( y = 0 \):
\[ f'(x + 0) = f'(x) \cdot f(0) \]
This simplifies to:
\[ f'(x) = f'(x) \cdot 1 \]
Thus, we have:
\[ f'(x) = f'(x) \]

### Step 6: Use \( y = 5 \)
Now, let’s substitute \( y = 5 \) in the differentiated equation:
\[ f'(x + 5) = f'(x) \cdot f(5) \]
This gives us:
\[ f'(x + 5) = f'(x) \cdot 12 \]

### Step 7: Find \( f'(5) \)
Now, let’s set \( x = 0 \):
\[ f'(0 + 5) = f'(0) \cdot 12 \]
This simplifies to:
\[ f'(5) = f'(0) \cdot 12 \]
We know \( f'(0) = 16 \):
\[ f'(5) = 16 \cdot 12 \]
Calculating this gives:
\[ f'(5) = 192 \]

### Conclusion
Thus, the value of \( f'(5) \) is:
\[ \boxed{192} \]