Step-by-step Solution

1. Define the Region R and its Area

The region R is defined by the inequalities $x^3 \le y \le x$ for $0 \le x \le 1$. This is the area enclosed between the curves $y=x$ and $y=x^3$ from $x=0$ to $x=1$.

In the interval $(0, 1)$, for any value of $x$, we have $x > x^3$. For example, if $x=1/2$, $x=1/2$ and $x^3=1/8$. Thus, the curve $y=x$ is above the curve $y=x^3$.

The total area of the region R, let's call it $A_{total}$, is calculated by the integral: $A_{total} = \int_{0}^{1} (\text{upper curve} - \text{lower curve}) \,dx$ $A_{total} = \int_{0}^{1} (x - x^3) \,dx$

2. Calculate the Total Area

We evaluate the integral: $A_{total} = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{1}$ $A_{total} = \left( \frac{1^2}{2} - \frac{1^4}{4} \right) - \left( \frac{0^2}{2} - \frac{0^4}{4} \right)$ $A_{total} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

3. Set up the Condition for the Dividing Line

The line $x = \alpha$ divides the area of region R into two equal parts. This means the area of the region from $x=0$ to $x=\alpha$ is half of the total area.

Area from $x=0$ to $x=\alpha$ is given by: $A_1 = \int_{0}^{\alpha} (x - x^3) \,dx$ The condition is: $A_1 = \frac{1}{2} A_{total} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$

4. Formulate and Solve the Equation for $\alpha$

First, we evaluate the integral for $A_1$: $A_1 = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_{0}^{\alpha} = \frac{\alpha^2}{2} - \frac{\alpha^4}{4}$ Now, we set this equal to $1/8$: $\frac{\alpha^2}{2} - \frac{\alpha^4}{4} = \frac{1}{8}$ To eliminate the denominators, we multiply the entire equation by 8: $8 \left( \frac{\alpha^2}{2} \right) - 8 \left( \frac{\alpha^4}{4} \right) = 8 \left( \frac{1}{8} \right)$ $4\alpha^2 - 2\alpha^4 = 1$ Rearranging the terms, we get: $2\alpha^4 - 4\alpha^2 + 1 = 0$ This equation matches Option A. Thus, Option A is correct.

5. Determine the Value and Range of $\alpha$

To find the value of $\alpha$, we can solve the equation $2\alpha^4 - 4\alpha^2 + 1 = 0$. Let $u = \alpha^2$. The equation becomes a quadratic in $u$: $2u^2 - 4u + 1 = 0$ Using the quadratic formula, $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $u = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)} = \frac{4 \pm \sqrt{16 - 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4} = \frac{2 \pm \sqrt{2}}{2}$ So, we have two possible values for $\alpha^2$: $\alpha^2 = \frac{2 + \sqrt{2}}{2} \quad \text{or} \quad \alpha^2 = \frac{2 - \sqrt{2}}{2}$ Since the line $x=\alpha$ divides the region for $x \in [0, 1]$, it must be that $0 < \alpha < 1$. This implies $0 < \alpha^2 < 1$.

Let's analyze the two possible values for $\alpha^2$:

• $\alpha^2 = \frac{2 + \sqrt{2}}{2} \approx \frac{2 + 1.414}{2} = 1.707 > 1$. This solution is rejected.
• $\alpha^2 = \frac{2 - \sqrt{2}}{2} \approx \frac{2 - 1.414}{2} = 0.293$. This value is between 0 and 1, so it is the correct one.

Now we need to check the range of $\alpha$. We have $\alpha^2 = \frac{2 - \sqrt{2}}{2}$. We are asked to compare $\alpha$ with $1/2$. This is equivalent to comparing $\alpha^2$ with $(1/2)^2 = 1/4$.

Let's check if $\alpha^2 > 1/4$: $\frac{2 - \sqrt{2}}{2} > \frac{1}{4}$ Multiply by 4: $2(2 - \sqrt{2}) > 1$ $4 - 2\sqrt{2} > 1$ $3 > 2\sqrt{2}$ Squaring both sides (which is valid as both are positive): $9 > (2\sqrt{2})^2$ $9 > 8$ This is true. Therefore, $\alpha^2 > (1/2)^2$. Since $\alpha > 0$, we have $\alpha > 1/2$.

Combining with the condition that $\alpha < 1$, we get $1/2 < \alpha < 1$. This matches Option C. Thus, Option C is also correct.

6. Evaluate Other Options

• Option B: $\alpha^4 + 4\alpha^2 - 1 = 0$ is incorrect as it is different from the derived equation $2\alpha^4 - 4\alpha^2 + 1 = 0$.
• Option D: $0 < \alpha \le 1/2$ is incorrect because we proved that $\alpha > 1/2$.

Conclusion

The correct options are A and C.