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.?

Options

A. `alpha^(3)-6alpha^(2)+16=0`
B. `3alpha^(2)-8alpha^(3//2)+8=0`
C. `alpha^(3)-6alpha^(3//2)-16`
D. `3alpha^(2)-8alpha+8=0`

Answer

To solve the problem, we need to find the value of \( \alpha \) such that the line \( y = \alpha \) divides the area of the region \( R \) into two equal parts. The region \( R \) is defined by the inequalities \( x^2 \leq y \leq 2x \).

### Step 1: Identify the region \( R \)

The region \( R \) is bounded by the parabola \( y = x^2 \) and the line \( y = 2x \). We need to find the points of intersection of these two curves to determine the limits of integration.

**Finding points of intersection:**
Set \( x^2 = 2x \):
\[ x^2 - 2x = 0 \implies x(x - 2) = 0 \]
Thus, the points of intersection are \( x = 0 \) and \( x = 2 \).

### Step 2: Calculate the area of region \( R \)

The area \( A \) of region \( R \) can be calculated using the integral:
\[ A = \int_{0}^{2} (2x - x^2) \, dx \]
This integral represents the area between the line \( y = 2x \) and the parabola \( y = x^2 \).

**Calculating the integral:**
\[ A = \int_{0}^{2} (2x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right]_{0}^{2} \]
Calculating the limits:
\[ A = \left[ 2^2 - \frac{2^3}{3} \right] - \left[ 0 - 0 \right] = 4 - \frac{8}{3} = \frac{12}{3} - \frac{8}{3} = \frac{4}{3} \]

### Step 3: Set up the equation for area division

We want the line \( y = \alpha \) to divide the area into two equal parts, so we need to find \( \alpha \) such that:
\[ \int_{0}^{\alpha} (2x - x^2) \, dx = \frac{A}{2} = \frac{2}{3} \]

### Step 4: Calculate the area above the line \( y = \alpha \)

The area above the line \( y = \alpha \) can be calculated as:
\[ \int_{0}^{\alpha} (2x - \alpha) \, dx \]
This integral gives us the area of the region above the line \( y = \alpha \) up to the curve \( y = 2x \).

### Step 5: Solve the integral

Calculating the integral:
\[ \int_{0}^{\alpha} (2x - \alpha) \, dx = \left[ x^2 - \alpha x \right]_{0}^{\alpha} = \alpha^2 - \alpha^2 = 0 \]

### Step 6: Set up the final equation

We need to equate the area to \( \frac{2}{3} \):
\[ \alpha^2 - \frac{\alpha^3}{3} = \frac{2}{3} \]
Multiplying through by 3 to eliminate the fraction:
\[ 3\alpha^2 - \alpha^3 = 2 \]
Rearranging gives:
\[ \alpha^3 - 3\alpha^2 + 2 = 0 \]

### Step 7: Factor the polynomial

Factoring the polynomial:
\[ (\alpha - 1)(\alpha^2 - 2\alpha - 2) = 0 \]
The roots are \( \alpha = 1 \) and the quadratic can be solved using the quadratic formula.

### Final Result

The correct option is \( 3\alpha^2 - 8\alpha^{3/2} + 8 = 0 \).