**16.** 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: - (1) 2384 - (2) 2525 - (3) 5220 - (4) 2406

1. Spotting a Functional Equation that Hints at a Differential Equation

The given law
$f(x+y)=f(x)f'(y)+f'(x)f(y)$
looks very similar to how the derivative of a product behaves: $\bigl[ f(x)f(y) \bigr]' = f'(x)f(y)+f(x)f'(y)$ This suggests that maybe $f(x+y)$ behaves like the derivative of a product evaluated at $(x,y)$.

2. Use a Clever Substitution (Set $y=0$)

Plugging $y=0$ (and knowing $f(0)=1$):

$f(x) = f(x)f'(0) + f'(x)f(0)$ $\Downarrow$ $f(x) - f'(0)f(x) = f'(x)$ $\Downarrow$ $f'(x) = \bigl[1-f'(0)\bigr]f(x)$

This is a first-order linear ODE of the form $f'(x)=k\,f(x)$ with $k=1-f'(0)$.

3. Solve the ODE

General solution:

$f(x)=C e^{k x}$ Use $f(0)=1$ to get $C=1$:

$f(x)=e^{kx}$

4. Self-Consistency to Find $k$

Compute $f'(x)=k e^{kx}$ so $f'(0)=k$.
But $k$ was defined as $1-f'(0)$; substitute:

$k = 1-k \;\;\Longrightarrow\;\; 2k = 1 \Longrightarrow k = \frac12$

Thus the only function satisfying all conditions is

$f(x)=e^{x/2}$

5. Convert the Requested Sum into Something Simple

We want

$\sum_{n=1}^{100}\ln f(n) = \sum_{n=1}^{100} \ln\bigl(e^{n/2}\bigr) = \sum_{n=1}^{100} \frac{n}{2}$

6. Perform the Arithmetic

The sum of the first $100$ natural numbers is

$\sum_{n=1}^{100} n = \frac{100\times101}{2} = 5050$ Therefore,

$\text{Required sum} = \frac12\times5050 = 2525$

That matches option (2).

Think of "f" as a Magic Growing Number

1. Starting Point: At $x = 0$, the magic number is $1$.

2. Rule for Growing: Whenever you give it two numbers $x$ and $y$, it grows in a very special way:

   $f(x+y)$ equals this–times–that plus that–times–this (it mixes the value of $f$ with the speed $f'$ at each point).

3. What We Need: Add up all the natural‐log ($\log_e$) values of $f(1), f(2), \dots, f(100)$.

4. Sneaky Shortcut: After a little detective work, we discover the only magic rule that fits is

   $f(x) = e^{x/2}$

5. Logs Make Life Easy: $\log_e\bigl(e^{x/2}\bigr) = x/2$. So the problem turns into adding $1/2+2/2+3/2+\dots+100/2$.

6. Quick Sum:

   $\text{Sum} = \frac12\,(1+2+\dots+100) = \frac12\times5050 = 2525$

So the answer is 2525 (option 2).

Step 1 – Identify the Function

Given

$f(x+y)=f(x)f'(y)+f'(x)f(y)$

Plug $y=0$ (using $f(0)=1$):

$f(x)=f(x)f'(0)+f'(x)\cdot1$ $\Rightarrow f'(x)=\bigl[1-f'(0)\bigr]f(x)$ Let $k=1-f'(0)$. Then

$f'(x)=k f(x)$

Solve the differential equation:

$f(x)=Ce^{kx}$ Apply $f(0)=1\;\Rightarrow\;C=1$, so

$f(x)=e^{kx}$

Compute $f'(0)=k$ and recall $k=1-f'(0)$:

$k=1-k\;\Rightarrow\;2k=1\;\Rightarrow\;k=\frac12$

Thus

$f(x)=e^{x/2}$

Step 2 – Evaluate the Required Sum

$\ln f(n)=\ln\bigl(e^{n/2}\bigr)=\frac{n}{2}$

Hence

$\sum_{n=1}^{100}\ln f(n)=\frac12\sum_{n=1}^{100}n$

The well-known sum:

$\sum_{n=1}^{100}n=\frac{100\times101}{2}=5050$

Therefore

$\text{Answer}=\frac12\times5050=2525$

Answer: 2525 (option 2)