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If the line $x=\alpha$ divides the area of region R={(x,y) $\in$ R² : x³≤ y≤ x, o≤x≤1 } into two equal parts, then

$2\alpha^{4}-4\alpha^{2}+1=0$

$\alpha^{4}+4\alpha^{2}-1=0$

$\frac{1}{2}<\alpha<1$

$0<\alpha\leq\frac{1}{2}$

Option A,C

$\int_{0}^{1} (x-x^{3})dx=2\int_{0}^{\alpha} (x-x^{3})dx$

$\frac{1}{4}=2(\frac{\alpha^{2}}{2}-\frac{\alpha^{4}}{4})$

$2\alpha^{4}-4\alpha^{2}+1=0$

$\alpha^{2}=\frac{4-\sqrt{16-8}}{4}(\because\alpha\in(0,1))$

$\alpha^{2}=1-\frac{1}{\sqrt{2}}$

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