If the line sx $= \alpha$ divides the area of region $R = \{(x,y)$ in $R^{2} : x^{3} \le y \le x, 0 \le x \le 1\}$ into two equal parts, then $(1) 0 < \alpha \le \frac{1}{2} (2) \frac{1}{2} < \alpha < 1 (3) 2 \alpha^{4} - 4 \alpha^{2} + 1 = 0 (4) \alpha^{4} + 4 \alpha^{2} - 1 = 0$

$R = \{(x,y) : x^3 \le y \le x,\ 0 \le x \le 1\}$

$\text{vertical line } sx = \alpha \text{ splits the area in half}$

$\text{To find: the correct statement about } \alpha$

$x \ge x^3 \text{ on } [0,1]: \ \text{upper } y=x, \ \text{lower } y=x^3$

$A = \int_0^1 (x - x^3)\,dx = \dfrac12 - \dfrac14 = \dfrac14$

$\text{left part} = \int_0^{\alpha}(x - x^3)\,dx = \dfrac{\alpha^2}{2} - \dfrac{\alpha^4}{4} = \dfrac18$

$4\alpha^2 - 2\alpha^4 = 1 \ \Rightarrow \ 2\alpha^4 - 4\alpha^2 + 1 = 0 \ \Rightarrow (3)\ \text{is true}$

$2u^2 - 4u + 1 = 0 \Rightarrow u = \alpha^2 = 1 \pm \dfrac{\sqrt2}{2}$

$\alpha \in (0,1) \Rightarrow \alpha^2 = 1 - \dfrac{\sqrt2}{2} \Rightarrow \alpha \approx 0.541 \in \left(\dfrac12, 1\right)$

$\boxed{\text{Answer: statements (2) and (3)}}$