JEE Main 2025 Jan 22 Shift 1, Mathematics Q16: Methods of Solving First Order, First Degree Differential Equation

JEE Main2025Jan 22, Shift 1Mathematics

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 $n = 1 \sum 100 lo g_{e} f (n)$ is equal to :

A
2384
B
2525
C
5220
D
2406

Solution

Set $x = y = 0$ : $f (0) = 2 f (0) f^{'} (0) \Rightarrow 1 = 2 f^{'} (0) \Rightarrow f^{'} (0) = \frac{1}{2}$ .

Set $y = 0$ : $f (x) = f (x) f^{'} (0) + f^{'} (x) f (0) = \frac{f ( x )}{2} + f^{'} (x)$ , so $f^{'} (x) = \frac{f ( x )}{2}$ .

This gives $f (x) = f (0) e^{x /2} = e^{x /2}$ , hence $ln f (n) = \frac{n}{2}$ .

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

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JEE Main 2025 Jan 22 Shift 1, Mathematics Q16: Methods of Solving First Order, First Degree Differential Equation

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