Question

Let \( f(x) \) be a real differentiable function such that \( f(0) = 1 \) and \( f(x + y) = f(x)f'(y) + f'(x)f(y) \) for all \( x, y \in \mathbb{R} \). Then \( \sum_{n=1}^{100} \log_e f(n) \) is equal to :

• A. 2384
• B. 2525
• C. 5220
• D. 2406

Hint:

When solving functional equations, try substituting specific values for \( x \) and \( y \) to simplify the problem.

Answer 1

The Correct Option is B

We are given the functional equation: \[ f(x + y) = f(x)f'(y) + f'(x)f(y) \] First, substitute \( x = y = 0 \) into the equation: \[ f(0) = f(0)f'(0) + f'(0)f(0) \] Since \( f(0) = 1 \), we get: \[ 1 = 2f'(0) \] \[ f'(0) = \frac{1}{2} \] Next, substitute \( y = 0 \) into the original equation: \[ f(x) = f(x)f'(0) + f'(x)f(0) \] \[ f(x) = \frac{1}{2}f(x) + f'(x) \] \[ f'(x) = \frac{1}{2}f(x) \] Thus, solving the differential equation \( f'(x) = \frac{1}{2}f(x) \) yields: \[ f(x) = e^{x/2} \] Now, we compute the sum: \[ \sum_{n=1}^{100} \log f(n) = \sum_{n=1}^{100} \log e^{n/2} = \sum_{n=1}^{100} \frac{n}{2} \] The sum of the first 100 integers is \( \sum_{n=1}^{100} n = 5050 \). Thus, the required sum is: \[ \frac{1}{2} \times 5050 = 2525 \] Thus, the answer is \( \boxed{2525} \).

Answer 2

Put $y=0$ in the functional equation: $$f(x)=f(x)f'(0)+f'(x)f(0).$$ Using $f(0)=1$ this gives $$f(x)=f(x)f'(0)+f'(x)\quad\Rightarrow\quad f'(x)=(1-f'(0))f(x).$$

So $f$ satisfies the linear ODE $f'(x)=c\,f(x)$ with constant $c=1-f'(0)$. Hence $$f(x)=Ae^{cx}.$$ Using $f(0)=1$ gives $A=1$, so $f(x)=e^{cx}$.

Now compute $f'(0)=c e^{0}=c$. But by definition $c=1-f'(0)=1-c$, so $2c=1\Rightarrow c=\tfrac{1}{2}$. Therefore $$\boxed{f(x)=e^{x/2}}.$$

Thus $\ln f(n)=\dfrac{n}{2}$ and $$\sum_{n=1}^{100}\ln f(n)=\sum_{n=1}^{100}\frac{n}{2}=\frac{1}{2}\cdot\frac{100\cdot101}{2}=\boxed{2525}.$$

Answer

2525 (Option 2)