Consider a region $$R = \{(x, y) \in R^2 : x^2 \le y \le 2x\}$$. If a line $$y = \alpha$$ divides the area of region $$R$$ into two equal parts, then which of the following is true?

We have to cut the region $$R=\{(x,y)\in\mathbb R^{2}\;:\;x^{2}\le y\le 2x\}$$ into two equal areas by the horizontal line $$y=\alpha$$. In order to do so, we first compute the total area of $$R$$.

Along any vertical line $$x=\text{constant}$$ with $$0\le x\le 2$$, the ordinate of the upper curve is $$y=2x$$ and that of the lower curve is $$y=x^{2}$$. Hence the vertical thickness at that $$x$$ equals $$2x-x^{2}$$. Using the elementary area formula

$$\text{Area}=\int_{x=a}^{b}(\text{upper }y-\text{lower }y)\,dx,$$

we obtain

$$ \begin{aligned} A_{\text{total}}&=\int_{0}^{2}(2x-x^{2})\,dx =\left[x^{2}-\frac{x^{3}}{3}\right]_{0}^{2} \\ &=\left(4-\frac{8}{3}\right)-0 =\frac{12}{3}-\frac{8}{3} =\frac{4}{3}. \end{aligned} $$

Now the desired line $$y=\alpha$$ must divide this area into two equal parts, so the area lying below $$y=\alpha$$ must be

$$\frac{1}{2}A_{\text{total}}=\frac{1}{2}\cdot\frac{4}{3}=\frac{2}{3}.$$

It is more convenient to describe the region slice-by-slice in the horizontal direction. For a fixed height $$y$$ (with $$0\le y\le 4$$), the two bounding curves give the $$x$$-limits by solving $$x^{2}=y\quad\text{and}\quad 2x=y.$$

Thus we find

$$x=\sqrt{y}\qquad\text{and}\qquad x=\frac{y}{2},$$

and since $$\frac{y}{2}\le\sqrt{y}$$ for $$0\le y\le4$$, the horizontal length of the strip at that height is

$$\sqrt{y}-\frac{y}{2}.$$

Using the horizontal-strip area formula

$$\text{Area}=\int_{y=c}^{d}(\text{right }x-\text{left }x)\,dy,$$

the area from $$y=0$$ up to $$y=\alpha$$ becomes

$$ \begin{aligned} A(\alpha)&=\int_{0}^{\alpha}\left(\sqrt{y}-\frac{y}{2}\right)dy \\ &=\left[\frac{2}{3}y^{3/2}-\frac{y^{2}}{4}\right]_{0}^{\alpha} \\ &=\frac{2}{3}\alpha^{3/2}-\frac{\alpha^{2}}{4}. \end{aligned} $$

Setting this equal to the required half-area $$\dfrac{2}{3}$$, we write

$$\frac{2}{3}\alpha^{3/2}-\frac{\alpha^{2}}{4}=\frac{2}{3}.$$

To clear the denominators, we multiply every term by $$12$$ (the least common multiple of 3 and 4):

$$ 12\left(\frac{2}{3}\alpha^{3/2}\right)\;-\; 12\left(\frac{\alpha^{2}}{4}\right)\;=\; 12\left(\frac{2}{3}\right). $$

That gives

$$8\alpha^{3/2}-3\alpha^{2}=8.$$

Re-arranging all terms to one side yields the polynomial equation

$$3\alpha^{2}-8\alpha^{3/2}+8=0.$$

This is precisely the statement in Option B. None of the other options matches this relation.

Hence, the correct answer is Option 2.