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JEE Advanced Mathematics Definite Integration 2020 JEE Advanced 2020 (Paper 2)

JEE Advanced Mathematics Question (2020) — Solution

Question

Let  f : ℝ → ℝ be a differentiable function such that its derivative f ' is continuous and  f π = - 6 . If  F : 0 , π → ℝ is defined by  F x = ∫ 0 x f t d t ,  and if ∫ 0 π f ' x + F x cos   x   d x = 2 , then the value of f 0 is_______

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

I = ∫ 0 π f ' x · cos   x + F x · cos   x d x = 2 = ∫ 0 π f ' x · cos   x · d x + ∫ 0 π F x cos   x · d x = 2 = cos x · f ( x ) 0 π - ∫ 0 π - sin x · f x d x + ∫ 0 π F x · cos   x · d x = 2 ⇒    cos π . f π - cos 0 . f 0 + ∫ 0 π sin x · f x d x + ∫ 0 π F x · cos   x d x = 2 ⇒    - 1 · - 6 - f 0 + sin x . F ( x ) 0 π - ∫ 0 π cos x · F x d x + ∫ 0 π F x · cos x · d x = 2 ⇒    6 - f 0 + sin π · F π - sin 0 . F 0 = 2 ⇒    f 0 = 4 .

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Related: Mathematics — Definite Integration · All PYQ Banks