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JEE Advanced Mathematics Area Under Curves 2023 JEE Advanced 2023 (Paper 1)

JEE Advanced Mathematics Question (2023) — Solution

Question

Let f : 0 ,   1 → 0 ,   1 be the function defined by f x = x 3 3 - x 2 + 5 9 x + 17 36 . Consider the square region S = 0 ,   1 × 0 ,   1 . Let G = x ,   y ∈ S : y > f x be called the green region and  R = x ,   y ∈ S : y < f x be called the red region. Let L h = x ,   h ∈ S : x ∈ 0 ,   1 be the horizontal line drawn at a height h ∈ 0 ,   1 . Then which of the following statements is(are) true?

Options

  1. A. There exists an h ∈ 1 4 ,   2 3  such that the area of the green region above the line L h  equals the area of the green region below the line  L h
  2. B. There exists an  h ∈ 1 4 ,   2 3  such that the area of the red region above the line L h equals the area of the red region below the line L h
  3. C. There exists an  h ∈ 1 4 ,   2 3  such that the area of the green region above the line L h equals the area of the red region below the line  L h
  4. D. There exists an  h ∈ 1 4 ,   2 3  such that the area of the red region above the line L h equals the area of the green region below the line  L h

Answer

D. There exists an  h ∈ 1 4 ,   2 3  such that the area of the red region above the line L h equals the area of the green region below the line  L h

Step-by-step solution

Given, Function  f x = x 3 3 - x 2 + 5 9 x + 17 36 Square region S = 0 ,   1 × 0 ,   1 Green region given by  G = x ,   y ∈ S : y > f x Red region is given by  R = x ,   y ∈ S : y < f x And  L h = x ,   h ∈ S : x ∈ 0 ,   1 be the horizontal line drawn at a height h ∈ 0 ,   1 Now plotting the diagram of the above function we get, Now differentiating the function, f x = x 3 3 - x 2 + 5 9 x + 17 36 ⇒ f ' x = x 2 - 2 x + 5 9 For maxima/minima,  f ' x = 0 ⇒ x = 1 3 Now finding the area of red region we get, A R = ∫ 0 1 f x   d x = 1 2 So, area of the green region will be, A G = 1 2 , as total area is  1  sq.unit Now solving option  A  we get, 1 - h = h - 1 2 ⇒ h = 3 4 ,   3 4 > 2 3 So, option A is incorrect For option  B   h = 1 2 - h ⇒ h = 1 4 So, option B is correct as  h ∈ 1 4 , 2 3 Now solving option  C  we get, ∫ 0 1 f x   d x = 1 2 , ∫ 0 1 1 2 d x = 1 2 ⇒ ∫ 0 1 f x - 1 2 d x = 0 ⇒ h = 1 2 So, option C is correct. D   ∵  Option C is correct ⇒ option D is also correct.

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