Question

If $f(x+y) = f(x) \cdot f(y)$, $f(3) = 3$, $f'(0) = 11$, then $f'(3)$ is equal to

• A. $11 \cdot e^{33}$
• B. $33$
• C. $11$
• D. $\log 33$

Hint:

The functional equation $f(x+y) = f(x)f(y)$ implies $f(x) = e^{kx}$.

Answer 1

The Correct Option is B

Step 1: Basic Principle
The functional equation $f(x+y) = f(x)f(y)$ with $f$ differentiable implies $f(x) = e^{kx}$.
Step 2: Solution Procedure:
From $f(x+y) = f(x)f(y)$, differentiate with respect to $x$: $f'(x+y) = f'(x)f(y)$. Put $x=0$: $f'(y) = f'(0)f(y) = 11 f(y)$. So $f'(y) = 11 f(y)$.
Solving: $f(y) = Ce^{11y}$. Given $f(3) = 3$, so $3 = Ce^{33} \Rightarrow C = 3e^{-33}$. Thus $f(y) = 3e^{11(y-3)}$.
Then $f'(y) = 33e^{11(y-3)}$, so $f'(3) = 33$.
Step 3: Required Answer:
$f'(3) = 33$.