Let the function satisfy the equation $f(x + y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$, where $f(0)
eq 0$. If $f(5) = 3$ and $f'(0) = 2$, then $f'(5)$ is:

Question

Let the function satisfy the equation $f(x + y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$, where $f(0) 
eq 0$. If $f(5) = 3$ and $f'(0) = 2$, then $f'(5)$ is:

cdquestions Admin · Accepted Answer

The functional equation $f(x + y) = f(x)f(y)$ implies $f(0) = 1$, as: $$ f(x + 0) = f(x)f(0) \implies f(0) = 1. $$ Assume $f(x) = e^{kx}$ (a standard solution for such functional equations). Differentiating both sides of the functional equation: $$ f'(x + y) = f'(x)f(y) + f(x)f'(y). $$ At $y = 0$, we get: $$ f'(x) = f'(x)f(0) + f(x)f'(0). $$ Given $f'(0) = 2$ and $f(0) = 1$, we find: $$ f'(x) = 2f(x). $$ Using the condition $f(5) = 3$, and $f'(x) = 2f(x)$, we calculate: $$ f'(5) = 2f(5) = 2 \cdot 3 = 6. $$

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Let the function satisfy the equation $f(x + y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$, where $f(0) \neq 0$. If $f(5) = 3$ and $f'(0) = 2$, then $f'(5)$ is:

The Correct Option is C

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