Area Division by Vertical Line for Cubic Region

To determine the correct options, we first calculate the total area of the region $R$, bounded by $y = x^3$ and $y = x$ for $0 \le x \le 1$.

Step 1: Calculate the Total Area of Region $R$

The total area $A$ is given by the definite integral: $A = \int_{0}^{1} (x - x^3) \, dx$

Evaluating the integral: $A = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

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Step 2: Establish the Equation for $\alpha$

The line $x = \alpha$ divides the region $R$ into two equal parts, so the area from $x = 0$ to $x = \alpha$ must be equal to half of the total area $A$: $\int_{0}^{\alpha} (x - x^3) \, dx = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$

Evaluating the left-hand side integral: $\left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{\alpha} = \frac{\alpha^2}{2} - \frac{\alpha^4}{4}$

Equating this to $\frac{1}{8}$: $\frac{\alpha^2}{2} - \frac{\alpha^4}{4} = \frac{1}{8}$

Multiply the entire equation by $8$: $4\alpha^2 - 2\alpha^4 = 1 \implies 2\alpha^4 - 4\alpha^2 + 1 = 0$

This verifies option (C).

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Step 3: Solve for $\alpha$ and Determine its Range

Let $t = \alpha^2$. Substituting $t$ into the equation gives a quadratic equation in $t$: $2t^2 - 4t + 1 = 0$

Solving for $t$ using the quadratic formula: $t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)} = \frac{4 \pm \sqrt{8}}{4} = 1 \pm \frac{1}{\sqrt{2}}$

Since $0 < \alpha < 1$, we must have $0 < t < 1$. Thus, we choose the smaller root: $\alpha^2 = 1 - \frac{1}{\sqrt{2}}$

Now, let's compare $\alpha^2$ with $\frac{1}{4}$: $1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} = 1 - \frac{\sqrt{2}}{2} \approx 1 - 0.7071 = 0.2929$

Since $0.2929 > 0.25 = \frac{1}{4}$: $\alpha^2 > \frac{1}{4} \implies \alpha > \frac{1}{2}$

Similarly, since $1 - \frac{1}{\sqrt{2}} < 1$: $\alpha^2 < 1 \implies \alpha < 1$

Combining these bounds gives: $\frac{1}{2} < \alpha < 1$

This verifies option (B).

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Conclusion:

The correct statements are (B) and (C).