JEE Advanced 2023 Paper 1 Q03 Mathematics Integration & Differential Equations Area Under Curves Hard

JEE Advanced 2023 Paper 1 · Q03 · Area Under Curves

Let $f: [0,1] \to [0,1]$ be the function defined by $f(x) = \dfrac{x^3}{3} - x^2 + \dfrac{5}{9}x + \dfrac{17}{36}$. Consider the square region $S = [0,1] \times [0,1]$. Let $G = \{(x,y) \in S : y > f(x)\}$ be called the green region and $R = \{(x,y) \in S : y < f(x)\}$ be called the red region. Let $L_h = \{(x, h) \in S : x \in [0,1]\}$ be the horizontal line drawn at a height $h \in [0,1]$. Then which of the following statements is(are) true?

  1. A. There exists an $h \in [1/4, 2/3]$ such that the area of the green region above $L_h$ equals the area of the green region below $L_h$
  2. B. There exists an $h \in [1/4, 2/3]$ such that the area of the red region above $L_h$ equals the area of the red region below $L_h$
  3. C. There exists an $h \in [1/4, 2/3]$ such that the area of the green region above $L_h$ equals the area of the red region below $L_h$
  4. D. There exists an $h \in [1/4, 2/3]$ such that the area of the red region above $L_h$ equals the area of the green region below $L_h$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, C, D

Solution

Direct computation gives $\int_0^1 f(x)\,dx = \dfrac{1}{2}$, so $\text{Area}(G) = \text{Area}(R) = \dfrac{1}{2}$. Note $f(1/4) = 13/36$ and $f(2/3) = 181/324$ both lie in $(1/4, 2/3)$ and $f$ is monotone on this range. Option (A) requires $1-h = 1/4$, giving $h = 3/4 \notin [1/4, 2/3]$ — INCORRECT. Option (B) gives $h = 1/4 \in [1/4, 2/3]$ — CORRECT. For (C) and (D), define $A(h) = R_h^{\text{above}} - G_h^{\text{below}}$; at the endpoints $A$ changes sign, so by the Intermediate Value Theorem an $h$ exists in $(1/4, 2/3)$ — both (C) and (D) are CORRECT.

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