JEE Advanced 2023 — Paper 1 — MATH-I solution

JEE Advanced2023Paper 1MATH-I

Let $f : [0, 1] \to [0, 1]$ be the function defined by $f (x) = \frac{x ^{3}}{3} - x^{2} + \frac{5 x}{9} + \frac{17}{36}$ . 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?

A
There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the green region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$
B
There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the red region above the line $L_{h}$ equals the area of the red region below the line $L_{h}$
C
There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the green region above the line $L_{h}$ equals the area of the red region below the line $L_{h}$
D
There exists an $h \in [\frac{1}{4}, \frac{2}{3}]$ such that the area of the red region above the line $L_{h}$ equals the area of the green region below the line $L_{h}$

Solution

Compute $f^{'} (x) = x^{2} - 2 x + \frac{5}{9}$ , which vanishes at $x = \frac{1}{3}$ in $[0, 1]$ . Evaluate:

$f (0) = \frac{17}{36} \approx 0.472$ , $f (1) = \frac{13}{36} \approx 0.361$ , $f (\frac{1}{3}) = \frac{181}{324} \approx 0.558$ .

The integral $\int_{0}^{1} f (x) d x = \frac{1}{2}$ , so $A_{R} = \frac{1}{2}$ (area of red region) and $A_{G} = 1 - \frac{1}{2} = \frac{1}{2}$ (area of green region).

(A) Equal green areas above and below $L_{h}$ requires $h = \frac{3}{4}$ , which lies outside $[\frac{1}{4}, \frac{2}{3}]$ . False.

(B) Equal red areas above and below $L_{h}$ is achievable at $h = \frac{1}{4}$ , which lies in the interval. True.

(C) Define $ϕ (h) =$ (green area above $L_{h}$ ) $-$ (red area below $L_{h}$ ). At $h = \frac{181}{324}$ : green above $< \frac{1}{2}$ and red below $= \frac{1}{2}$ , so $ϕ < 0$ . At $h = \frac{13}{36}$ : green above $= \frac{1}{2}$ and red below $< \frac{1}{2}$ , so $ϕ > 0$ . By continuity (IVT), some $h \in (\frac{13}{36}, \frac{181}{324}) \subset [\frac{1}{4}, \frac{2}{3}]$ gives equality. True.

(D) Since the four sub-areas (green above + green below + red above + red below) sum to $1$ , the equation in (D) is the complementary statement of (C); whenever (C) holds, (D) holds for the same $h$ . True.

$∴$ The correct options are (B), (C) and (D).

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