Let f: [0, 1] → [0, 1] be the function defined by f(x)=x33−x2+59x+1736. 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 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?

f(x)=x33−x2+59x+1736 f′(x)=3x23−2x+59=x2−2x+59=(x−1)2−49f′(x)=0⇒x=13,53A) ∫01 (1−h)dx=∫01 (h−f(x))dx if h lies above y = f(x)1+12=2h⇒h=34⇒A is incorrect. В) ∫01 hdx=∫01 (f(x)−h)dx⇒2h=12⇒h=14 if h lies below the curve y=f(x)⇒B is correctC&D) Area of green region = ∫01 (1−f(x))dx=1−12=12Area of red region = ∫01 (1−f(x))dx=1−12=1212−A2=12−A2+A3( Option C)⇒A2=A1+A3( Option D)By using intermediate value theorem ⇒Option C and D are correcty=f(x) is a curve, x = h is a line which divides the curve such that the area between the line and curve are equal.

Let f: [0, 1] → [0, 1] be the function defined by f(x)=x33−x2+59x+1736. 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 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?

f(x)=x33−x2+59x+1736 f′(x)=3x23−2x+59=x2−2x+59=(x−1)2−49f′(x)=0⇒x=13,53A) ∫01 (1−h)dx=∫01 (h−f(x))dx if h lies above y = f(x)1+12=2h⇒h=34⇒A is incorrect. В) ∫01 hdx=∫01 (f(x)−h)dx⇒2h=12⇒h=14 if h lies below the curve y=f(x)⇒B is correctC&D) Area of green region = ∫01 (1−f(x))dx=1−12=12Area of red region = ∫01 (1−f(x))dx=1−12=1212−A2=12−A2+A3( Option C)⇒A2=A1+A3( Option D)By using intermediate value theorem ⇒Option C and D are correcty=f(x) is a curve, x = h is a line which divides the curve such that the area between the line and curve are equal.

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Let f: [0, 1] → [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] × [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 ∈[0, 1] . Then which of the following statements is(are) true?

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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

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

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

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

answer is B, C, D.

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Detailed Solution

$f (x) = \frac{x^{3}}{3} - x^{2} + \frac{5}{9} x + \frac{17}{36} \begin{array}{l} f^{'} (x) = \frac{3 x^{2}}{3} - 2 x + \frac{5}{9} = x^{2} - 2 x + \frac{5}{9} = (x - 1)^{2} - \frac{4}{9} \\ f^{'} (x) = 0 \Rightarrow x = \frac{1}{3}, \frac{5}{3} \end{array}$
Question Image
$A) \int_{0}^{1} (1 - h) dx = \int_{0}^{1} (h - f (x)) dx if h lies above y = f (x)$
$1 + \frac{1}{2} = 2 h \Rightarrow h = \frac{3}{4} \Rightarrow A is incorrect.$

$В) \int_{0}^{1} hdx = \int_{0}^{1} (f (x) - h) dx$
$\Rightarrow 2 h = \frac{1}{2} \Rightarrow h = \frac{1}{4}$ if h lies below the curve $y = f (x) \Rightarrow B$ is correct
C&D) Area of green region = $\int_{0}^{1} (1 - f (x)) dx = 1 - \frac{1}{2} = \frac{1}{2}$
Area of red region = $\int_{0}^{1} (1 - f (x)) dx = 1 - \frac{1}{2} = \frac{1}{2}$
Question Image
$\begin{array}{l} \frac{1}{2} - A_{2} = \frac{1}{2} - (A_{2} + A_{3}) (Option C) \\ \Rightarrow A_{2} = A_{1} + A_{3} (Option D) \end{array}$
By using intermediate value theorem $\Rightarrow$ Option C and D are correct
$y = f (x)$ is a curve, x = h is a line which divides the curve such that the area between the line and curve are equal.