Let f: [0, 1] → [0, 1] be the function defined by f(x) = x ^ 3 3 -...

Given,

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

Square region $S = \left[0, 1\right] \times \left[0, 1\right]$

Green region given by $G = \left\{\left(x, y\right) \in S: y > f\left(x\right)\right\}$

Red region is given by $R = \left\{\left(x, y\right) \in S: y < f\left(x\right)\right\}$

And ${L}_{h} = \left\{\left(x, h\right) \in S: x \in \left[0, 1\right]\right\}$ be the horizontal line drawn at a height $h \in \left[0, 1\right]$

Now plotting the diagram of the above function we get,

Now differentiating the function,

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

$⇒ {f}^{'}\left(x\right) = {x}^{2} - 2x + \dfrac{5}{9}$

For maxima/minima, ${f}^{'}\left(x\right) = 0 ⇒ x = \dfrac{1}{3}$

Now finding the area of red region we get,

${A}_{R} = {∫}_{0}^{1}f\left(x\right)dx = \dfrac{1}{2}$

So, area of the green region will be,

${A}_{G} = \dfrac{1}{2}$, as total area is $1$ sq.unit

Now solving option \left(A\right)$we get,</p><p>$1 - h = h - \dfrac{1}{2} ⇒ h = \dfrac{3}{4}, \dfrac{3}{4} > \dfrac{2}{3}$</p><p>So, option$\left(A\right)$is incorrect</p><p>For option$\left(B\right)$h = \dfrac{1}{2} - h ⇒ h = \dfrac{1}{4}$

So, option $\left(B\right)$ is correct as $h \in$\left[\dfrac{1}{4}, \dfrac{2}{3}\right]$</p><p>Now solving option \left(C\right)$ we get,

${∫}_{0}^{1}f$\left(x\right)dx = \dfrac{1}{2}, {∫}_{0}^{1}\dfrac{1}{2}dx = \dfrac{1}{2} ⇒ {∫}_{0}^{1}\left(f\left(x\right) - \dfrac{1}{2}\right)dx = 0$</p><p>$⇒ h = \dfrac{1}{2}$</p><p>So, option \left(C\right)$ is correct.

$\left(D\right)$∵$Option \left(C\right)$ is correct $⇒$ option $\left(D\right)$ is also correct.