Let f:0, 1→0, 1 be the function defined by fx=x33-x2+59x+1736. Consider the square region S=0, 1×0, 1. Let G=x, y∈S:y>fx be called the green region and R=x, y∈S:y<fx be called the red region. Let Lh=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?

(A) There exists anh∈14, 23 such that the area of the green region above the line Lh equals the area of the green region below the line Lh

(C) There exists an h∈14, 23 such that the area of the green region above the line Lh equals the area of the red region below the line Lh

JEE Advanced · Mathematics · 28. Area Under Curves

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

Verified Solution

Answer & Solution

Correct Answer

(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 $L_{h}$

Step-by-step Solution

Detailed explanation

Given, Function

f (x) = \frac{x^{3}}{3} - x^{2} + \frac{5}{9} x + \frac{17}{36}

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Let $\vec{w}=\hat{i}+\hat{j}-2 \hat{k}$, and $\overrightarrow{\mathrm{u}}$ and $\overrightarrow{\mathrm{v}}$ be two vectors such that $\vec{u} \times \vec{v}=\vec{w}$ and $\vec{v} \times \vec{w}=\vec{u}$. Let $\alpha, \beta, \gamma$ and $t$ be real numbers such that $\vec{u}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k},-t \alpha+\beta+\gamma=0, \alpha-t \beta+\gamma=0$, and $\alpha+\beta-t \gamma=0$.
Match each entry in List-I to the correct entry in List-II and choose the correct option.

	LIST - I		LIST - II
(P)	$\|\vec{v}\|^2$ is equal to	(1)	0
(Q)	If $\alpha=\sqrt{3}$,then $\gamma^2$ is equal to	(2)	1
(R)	If $\alpha=\sqrt{3}$,then $(\beta+\gamma)^2$ is equal to	(3)	2
(S)	If $\alpha=\sqrt{2}$,then $t+3$ is equal to	(4)	3
		(5)	5

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	LIST - I		LIST - II
(P)	\(\|\vec{v}\|^2\) is equal to	(1)	0
(Q)	If \(\alpha=\sqrt{3}\),then \(\gamma^2\) is equal to	(2)	1
(R)	If \(\alpha=\sqrt{3}\),then \((\beta+\gamma)^2\) is equal to	(3)	2
(S)	If \(\alpha=\sqrt{2}\),then \(t+3\) is equal to	(4)	3
		(5)	5