Question

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:

• A. 3
• B. 0
• C. 6
• D. -6

Answer 1

The Correct Option is C

1. Understand the problem:

Given the functional equation \( f(x+y) = f(x)f(y) \) for all \( x, y \in \mathbb{R} \), with \( f(0) \neq 0 \), \( f(5) = 3 \), and \( f'(0) = 2 \), we need to find \( f'(5) \).

2. Solve the functional equation:

The equation \( f(x+y) = f(x)f(y) \) is satisfied by exponential functions of the form:

\[ f(x) = e^{kx} \]

Given \( f(0) = e^{k \cdot 0} = 1 \neq 0 \), which is consistent.

3. Determine the constant \( k \):

From \( f(5) = 3 \):

\[ e^{5k} = 3 \implies k = \frac{\ln 3}{5} \]

From \( f'(0) = 2 \):

\[ f'(x) = k e^{kx} \implies f'(0) = k e^{0} = k = 2 \]

Thus, \( k = 2 \).

4. Find \( f'(5) \):

Using \( k = 2 \):

\[ f'(5) = 2 e^{2 \cdot 5} = 2 e^{10} \]

However, this contradicts the given \( f(5) = 3 \). Let's re-evaluate:

The general solution is \( f(x) = f(0) e^{kx} \). Given \( f(0) = 1 \), \( f(5) = e^{5k} = 3 \), and \( f'(0) = k = 2 \), we have:

\[ f'(5) = k e^{5k} = 2 \cdot 3 = 6 \]

Correct Answer: (C) 6

Answer 2

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