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

Correct Options:

1. 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.
2. 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.
3. 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.

Step-by-Step Explanation:

Step 1: Analyze the bounds of f(x) on [0, 1]

The given function is:

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

Taking the first derivative to find critical points:

$f^{\prime }\left(x\right)=x^2-2x+\frac{5}{9}=(x-\frac{1}{3})(x-\frac{5}{3})$

Within the interval [0, 1], $f^{\prime }\left(x\right)=0$ at $x=\frac{1}{3}$. Evaluating the function at boundary points and the critical point gives:

$f\left(0\right)=\frac{17}{36}$

$f\left(1\right)=\frac{1}{3}-1+\frac{5}{9}+\frac{17}{36}=\frac{13}{36}$

$f\left(\frac{1}{3}\right)=\frac{1}{81}-\frac{1}{9}+\frac{5}{27}+\frac{17}{36}=\frac{181}{324}$

Thus, for all $x\in [0,1]$, the range of f(x) satisfies:

$\frac{1}{4}<\frac{13}{36}\le f\left(x\right)\le \frac{181}{324}<\frac{2}{3}$

Step 2: Calculate the Total Area of Regions R and G

The total area of the red region R (under y = f(x)) inside the square region S = [0, 1] × [0, 1] is:

$A_R={\int }_0^{1}f\left(x\right)dx={[\frac{x^4}{12}-\frac{x^3}{3}+\frac{5x^2}{18}+\frac{17x}{36}]}_0^1=\frac{1}{12}-\frac{1}{3}+\frac{5}{18}+\frac{17}{36}=\frac{1}{2}$

Since the total area of square S is 1, the area of the green region G is:

$A_G=1-A_{\mathrm{Rpopular}}=1-\frac{1}{2}=\frac{1}{2}$

Step 3: Define Area Relations for Horizontal Line L_h

Let for any height h ∈ [0, 1]:
- R_above(h): Area of red region above L_h
- R_below(h): Area of red region below L_h
- G_above(h): Area of green region above L_h
- G_below(h): Area of green region below L_h

By definition:

$R_{\text{above}}\left(h\right)+R_{\text{below}}\left(h\right)=\frac{1}{2}$

$G_{\text{above}}\left(h\right)+G_{\text{below}}\left(h\right)=\frac{1}{2}$

$R_{\text{below}}\left(h\right)+G_{\text{below}}\left(h\right)=h$

Step 4: Verification of Options

Verification of Statement 2: R_above(h) = R_below(h)

$R_{\text{above}}\left(h\right)=R_{\text{below}}\left(h\right)\Rightarrow \frac{1}{2}-R_{\text{below}}\left(h\right)=R_{\text{below}}\left(h\right)\Rightarrow R_{\text{below}}\left(h\right)=\frac{1}{4}$

Since $f\left(x\right)>\frac{1}{4}$ for all x ∈ [0, 1], at $h=\frac{1}{4}$, the entire region below y = 1/4 lies strictly within the red region R. Therefore, $R_{\text{below}}\left(\frac{1}{4}\right)=\frac{1}{4}$.

Hence, $h=\frac{1}{4}\in [\frac{1}{4},\frac{2}{3}]$ satisfies the condition, making Statement 2 TRUE.

Verification of Statement 3: G_above(h) = R_below(h)

Using the area relations:

$G_{\text{above}}\left(h\right)=(1-h)-R_{\text{above}}\left(h\right)=1-h-(\frac{1}{2}-R_{\text{below}}(h\left)\right)=\frac{1}{2}-h+R_{\text{below}}\left(h\right)$

Equating G_above(h) to R_below(h):

$\frac{1}{2}-h+R_{\text{below}}\left(h\right)=R_{\text{below}}\left(h\right)\Rightarrow h=\frac{1}{2}$

Since $h=\frac{1}{2}\in [\frac{1}{4},\frac{2}{3}]$, Statement 3 is TRUE.

Verification of Statement 4: R_above(h) = G_below(h)

Substitute R_above(h) = 1/2 - R_below(h) and G_below(h) = h - R_below(h):

Since $h=\frac{1}{2}\in [\frac{1}{4},\frac{2}{3}]$, Statement 4 is TRUE.

Thus, statements 2, 3, and 4 are all correct.