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JEE Main202329 Jan 2023Evening ShiftMathematicsFunctionsActual

Consider a function f : ℕ → ℝ , satisfying f 1 + 2 f 2 + 3 f 3 + … + x f x = x x + 1 f x ; x ≥ 2 with f 1 = 1 . Then 1 f 2022 + 1 f 2028 is equal to

Options

  1. A8200
  2. B8000
  3. C8400
  4. D8100

Correct answer

D. 8100

Step-by-step solution

Given: f 1 + 2 f 2 + 3 f 3 + … + x f x = x x + 1 f x ⇒ f 1 + 2 f 2 + 3 f 3 + … + x - 1 f x - 1 = x x - 1 f x - 1 Now when x = 2 , then f 1 + 2 f 2 = 6 f 2 ⇒ 1 = 4 f 2 ⇒ f 2 = 1 4 When x = 3 , f 1 + 2 f 2 = 9 f 3 ⇒ f 3 = 1 6 When x = 4 , f 1 + 2 f 2 + 3 f 3 = 16 f 4 ⇒ f 4 = 1 8 So, f x = 1 2 x Hence, 1 f 2022 + 1 f 2028 = 4044 + 4056 = 8100

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