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 $x, y \in R$. Then $\sum_{n=1}^{100} \log_e f(n)$ is equal to:
- (1) 2384
- (2) 2525
- (3) 5220
- (4) 2406
Solution:
Given $f(x+y) = f(x)f'(y) + f'(x)f(y)$.
Put $x=0, y=0$:
$$ f(0) = f(0)f'(0) + f'(0)f(0) $$
$$ 1 = 1 \cdot f'(0) + f'(0) \cdot 1 \Rightarrow 2f'(0) = 1 \Rightarrow f'(0) = \frac{1}{2} $$
Now put $y=0$ in the original equation:
$$ f(x) = f(x)f'(0) + f'(x)f(0) $$
$$ f(x) = \frac{1}{2}f(x) + f'(x) $$
$$ f'(x) = \frac{1}{2}f(x) \Rightarrow \frac{f'(x)}{f(x)} = \frac{1}{2} $$
Integrating both sides with respect to $x$:
$$ \int \frac{f'(x)}{f(x)} dx = \int \frac{1}{2} dx $$
$$ \log_e f(x) = \frac{x}{2} + C $$
Using $f(0)=1 \Rightarrow \log_e(1) = 0 + C \Rightarrow C = 0$.
So, $\log_e f(n) = \frac{n}{2}$.
We need the sum:
$$ S = \sum_{n=1}^{100} \log_e f(n) = \sum_{n=1}^{100} \frac{n}{2} = \frac{1}{2} \sum_{n=1}^{100} n $$
$$ S = \frac{1}{2} \times \frac{100(101)}{2} = \frac{5050}{2} = 2525 $$
Ans. (2)