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JEE Main202225 Jun 2022Morning ShiftMathematicsFunctionsActual

Let f : N → R be a function such that f x + y = 2 f x f y for natural numbers x and y . If f 1 = 2 , then the value of α for which ∑ k = 1 10 f α + k = 512 3 2 20 - 1 holds, is

Options

  1. A3
  2. B4
  3. C5
  4. D6

Correct answer

B. 4

Step-by-step solution

Given f x + y = 2 f x · f y & f 1 = 2 Now putting x = 1 & y = 1 in f x + y = 2 f x · f y we get f 1 + 1 = 2 f 1 · f 1 = 2 × 2 2 = 2 3 So f 2 = 2 3 , Similarly f 3 = 2 5 ,   f 4 = 2 7 . . . . . Now ∑ k = 1 10 f α + k = ∑ k = 1 10 2 f α · f k = 512 3 2 20 - 1 ⇒    2 f α ∑ k = 1 10 f k = 512 3 2 20 - 1 ⇒    2 f α f 1 + f 2 ⋯ f 10 = 512 3 2 20 - 1 ⇒    2 f α 2 + 2 3 + 2 5 … = 512 3 2 20 -

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