Question

Let f :[0,1] → [0,1] be the function defined by \(\frac{x^3}{3}-x^2+\frac{5}{9}x+\frac{17}{36}.\) Consider the square region S = [0,1] x [0,1]. Let G = {(x,y) ∈ S: y > f(s)} be called the green region and R = {(x,y) ∈ S: y < f(s)} 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?

• A. There exists an h ∈ [\(\frac{1}{4},\frac{2}{3}\)] 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 ∈ [\(\frac{1}{4},\frac{2}{3}\)] 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 ∈ [\(\frac{1}{4},\frac{2}{3}\)] 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 ∈ [\(\frac{1}{4},\frac{2}{3}\)] 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 1

The Correct Option is B, C, D

To solve this problem, we need to understand the regions defined by the function \( f(x) = \frac{x^3}{3}-x^2+\frac{5}{9}x+\frac{17}{36} \) on the domain \([0,1]\) and how these regions interact with the horizontal line \(L_h\). We have three types of regions within the square \(S = [0,1] \times [0,1]\):

• The green region \(G = \{(x,y) \in S: y > f(x)\}\). 
• The red region \(R = \{(x,y) \in S: y < f(x)\}\).
• The line \(L_h = \{(x,h) \in S: x \in [0,1]\}\).

We need to find \(h \in \left(\frac{1}{4},\frac{2}{3}\right)\) such that different areas of these regions above and below \(L_h\) meet certain criteria. We have the following observations:

1. The red region above \(L_h\) is bounded by the lines \(y = f(x)\) and \(y = h\).
2. The green region below \(L_h\) is bounded by the lines \(y = h\) and \(y = f(x)\).
3. We are looking for points where the areas satisfy certain conditions above and below the line \(L_h\).

The total area over the unit square is 1. Function \(f(x)\) is continuous and differentiable over the interval \([0,1]\), and its integral describes the curve within this square. Analyzing by symmetry and the intermediate value theorem, we can describe the potential solutions:

1. The area of the red region above the line \(L_h\) equals the area of the red region below the line \(L_h\).

When \(h\) is adjusted, the red regions above and below this line vary but given that \(f(x)\) is continuous, there exists such an \(h\) meeting this requirement.

2. The area of the green region above the line \(L_h\) equals the area of the red region below the line \(L_h\).

3. The area of the red region above the line \(L_h\) equals the area of the green region below the line \(L_h\).

For these conditions, there are again intervals of continuity and intermediate value which assure the equality of these areas. By adjusting \(h\) within the specified range, corresponding equalities can be achieved.

Given the nature of \(f(x)\), these scenarios are possible with proper transference of areas across the threshold of \(h\), justifying the following statements as true.

Answer 2

\(f(x)=\frac{x^3}{3}-x^2+\frac{5}{9}x+\frac{17}{36},f'(x)=x^2-2x+\frac{5}{9}\)
For maxima or minima, \(f'(x)=0\Rightarrow x=\frac{1}{3}\)
\(A_R = \int_{0}^{1}f(x)dx=\frac{1}{2}\Rightarrow A_G=\frac{1}{2}\)
Now, checking each options :
(A) 1 - h = h - \(\frac{1}{2}\)
⇒ \(h=\frac{3}{4},\frac{3}{4}\gt\frac{2}{3}\)    So, the option (A) is incorrect
(B) h = \(\frac{1}{2}-h\)
⇒ h =  \(\frac{1}{4}\)      So, the option (B) is correct
(C) \(\int^1_0f(x)dx=\frac{1}{2},\int_0^1\frac{1}{2}dx=\frac{1}{2}\)
⇒ \(\int_0^1(f(x)-\frac{1}{2})dx=0\)
⇒ \(h=\frac{1}{2}\)    So, the option (B) is correct.
(D) As the option (C) is correct, the option (D) is also correct.
So, the correct options are (B), (C) and (D).