JEE Main202430 Jan 2024Evening ShiftMathematicsContinuity and DifferentiabilityActual

Let f : R - 0 → R be a function satisfying f x y = f ( x ) f ( y ) for all x , y , f ( y ) ≠ 0 . If f ' ( 1 ) = 2024 , then

Options

Ax f ' x - 2024 f x = 0
Bx f ' x + 2024 f x = 0
Cx ' ( x ) + f ( x ) = 2024
Dx f ' ( x ) - 2023 f ( x ) = 0

Correct answer

A. x f ' x - 2024 f x = 0

Step-by-step solution

Given: f x y = f x f y And we know that, if f x y = f x f y then f x = x n . So, differentiating the function we get, f ' x = n x n - 1 and given f ' 1 = 2024 . Hence, on comparing we get, n = 2024 ⇒ x f ' x = x × 2024 x 2023 ⇒ x f ' x = 2024 x 2024 ⇒ x f ' x - 2024 x 2024 = 0 ⇒ x f ' x - 2024 f x = 0

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