JEE 2025 — Mathematics PYQ

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Question

JEE 2025 — Mathematics PYQ

JEE | Mathematics | 2025

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 ∈ R. Then $\sum_{n = 1}^{100} lo g_{e} f (n)$ is equal to:

Choose the correct answer:

Explanation

Given, $f (x + y) = f (x) \cdot f^{'} (y) + f^{'} (x) \cdot f (y)$

$Put y = 0$

$f (x) = f (x) \cdot f^{'} (0) + f^{'} (x) f (0)$

$(∵ f (0) = 1)$

$f (x) = f (x) \cdot f^{'} (0) + f^{'} (x) ... ($ i)

$Put x = y = 0$

$f (0) = f^{'} (0) + f^{'} (0) \Rightarrow f^{'} (0) = \frac{1}{2}$

Hence, equation (i) becomes

$f (x) = \frac{f ( x )}{2} + f^{'} (x)$

$f^{'} (x) = \frac{f ( x )}{2}$

$\frac{f ^{'} ( x )}{f ( x )} = \frac{1}{2}$

Integration both of sides

$ln$ $f (x) = \frac{x}{2} + C$

$Put x = 0 \Rightarrow C = 0$

$\Rightarrow ln f (x) = \frac{x}{2}$

$f (x) = e^{\frac{x}{2}}$
$∴ f (n) = e^{\frac{n}{2}}$
$lo g_{e} f (n) = \frac{n}{2}$
$∴ n = 1 \sum 100 lo g_{e} f (n) = \frac{1}{2} n = 1 \sum 100 n$
$= \frac{1}{2} (\frac{100}{2} (100 + 1))$
$= 25 \times 101$
$= 2525$

Choose the correct answer:

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