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JEE Main202226 Jul 2022Morning ShiftMathematicsDefinite IntegrationActual

If a = lim n → ∞ ∑ k = 1 n 2 n n 2 + k 2 and f x = 1 - cos x 1 + cos x , x ∈ 0 , 1 , then:

Options

  1. A2 2 f a 2 = f ' a 2
  2. Bf a 2 f ' a 2 = 2
  3. C2 f a 2 = f ' a 2
  4. Df a 2 = 2 f ' a 2

Correct answer

C. 2 f a 2 = f ' a 2

Step-by-step solution

Given, a = lim n → ∞ ∑ k = 1 n 2 n n 2 + k 2 ⇒ a = 1 n ∑ k = 1 n 2 1 + k n 2 = ∫ 0 1 2 1 + x 2 d x ⇒ a = 2 tan - 1 x 0 1 ⇒ a = 2 π 4 - 0 = π 2 Now, f x = 1 - cos x 1 + cos x ⇒ f x = tan x 2 ; x ∈ 0 , 1 So, f a 2 = f π 4 ⇒ f π 4 = 2 - 1 And f ' π 4 = 1 2 sec 2 π 8 = 2 2 + 1 So, f ' π 4 = 2 f π 4

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