Let $f(\mathrm{x})$ be a real differentiable function such that $f(0)=1$ and $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f^{\prime}(\mathrm{y})+f^{\prime}(\mathrm{x}) f(\mathrm{y})$ for all $\mathrm{x}, \mathrm{y} \in \mathbf{R}$. Then $\sum_{\mathrm{n}=1}^{100} \log _{\mathrm{e}} f(\mathrm{n})$ is equal to :

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Differential Equations

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} lo g_{e} f (n)

is equal to :

Answer: (B)

Set

x = y = 0

f (x + y) = f (x) f^{'} (y) + f^{'} (x) f (y)

f (0) = 2 f (0) f^{'} (0) ⟹ 1 = 2 f^{'} (0) ⟹ f^{'} (0) = \frac{1}{2} .

Now set

y = 0

f (x) = f (x) f^{'} (0) + f^{'} (x) f (0) = \frac{1}{2} f (x) + f^{'} (x) .

f^{'} (x) = \frac{1}{2} f (x) ⟹ \frac{f ^{'} ( x )}{f ( x )} = \frac{1}{2} ⟹ ln f (x) = \frac{x}{2} + C .

Since

f (0) = 1

ln f (0) = 0 = C

, so

ln f (x) = \frac{x}{2}, i.e. ln f (n) = \frac{n}{2} .

n = 1 \sum 100 ln f (n) = \frac{1}{2} n = 1 \sum 100 n = \frac{1}{2} \cdot \frac{100 \cdot 101}{2} = \frac{5050}{2} = 2525.

JEE Main · 2025 · 22 Jan 2025 · Shift 1 · Single correct option

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