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 will use the properties of the function \( f(x) \) given in the question.

### Step 1: Understanding the Functional Equation
We are given that:
\[ f(x+y) = f(x) \cdot f(y) \]
for all \( x, y \in \mathbb{R} \). This is a well-known functional equation that suggests that \( f(x) \) could be of the form \( f(x) = a^x \) for some constant \( a \).

### Step 2: Finding \( f(0) \)
Substituting \( x = 0 \) and \( y = 0 \) into the functional equation:
\[ f(0 + 0) = f(0) \cdot f(0) \implies f(0) = f(0)^2 \]
This implies that \( f(0) \) can either be \( 0 \) or \( 1 \). However, since we are given that \( f(0) \neq 0 \), we conclude:
\[ f(0) = 1 \]

### Step 3: Finding the Derivative
We need to find \( f'(5) \). The derivative of \( f(x) \) can be expressed using the limit definition:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Substituting \( x = 5 \):
\[ f'(5) = \lim_{h \to 0} \frac{f(5+h) - f(5)}{h} \]

### Step 4: Using the Functional Equation
Using the functional equation, we can express \( f(5+h) \):
\[ f(5+h) = f(5) \cdot f(h) \]
Thus,
\[ f'(5) = \lim_{h \to 0} \frac{f(5) \cdot f(h) - f(5)}{h} = \lim_{h \to 0} \frac{f(5)(f(h) - 1)}{h} \]

### Step 5: Evaluating the Limit
We know that \( f(0) = 1 \) and we can use the derivative at \( 0 \):
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{f(h) - 1}{h} \]
Given that \( f'(0) = 16 \), we can substitute this into our expression for \( f'(5) \):
\[ f'(5) = f(5) \cdot f'(0) \]
Substituting \( f(5) = 12 \) and \( f'(0) = 16 \):
\[ f'(5) = 12 \cdot 16 = 192 \]

### Final Answer
Thus, \( f'(5) \) is equal to:
\[ \boxed{192} \]