Suppose that `F(alpha)` denotes the area of the region bounded by `x = 0, x=2, y^2=4x` and `y=|alphax-1|+|alphax-2|+alphax`, where `alpha in {0,1}`. Then the value of `F(alpha)+(8sqrt(2))/3` when `alpha = 0` is (A) 4 (B) 5 (C) 6 (D) 9

Options

A. 4
B. 5
C. 6
D. 9

Explanation

To solve the problem, we need to find the area \( F(\alpha) \) when \( \alpha = 0 \) and then compute \( F(\alpha) + \frac{8\sqrt{2}}{3} \).

### Step 1: Define the functions
When \( \alpha = 0 \), the equation for \( y \) simplifies to:
\[ y = |0 \cdot x - 1| + |0 \cdot x - 2| + 0 \cdot x = | -1 | + | -2 | = 1 + 2 = 3 \]
So, the line \( y = 3 \) is our upper boundary.

### Step 2: Identify the lower boundary
The lower boundary is given by the parabola \( y^2 = 4x \). We can express this as:
\[ y = \pm 2\sqrt{x} \]
Since we are interested in the area above the parabola and below the line \( y = 3 \), we will use the positive branch \( y = 2\sqrt{x} \).

### Step 3: Set up the integral for the area
The area \( F(0) \) is given by the integral of the upper function minus the lower function from \( x = 0 \) to \( x = 2 \):
\[ F(0) = \int_{0}^{2} (3 - 2\sqrt{x}) \, dx \]

### Step 4: Calculate the integral
Now, we compute the integral:
\[ F(0) = \int_{0}^{2} 3 \, dx - \int_{0}^{2} 2\sqrt{x} \, dx \]

Calculating the first integral:
\[ \int_{0}^{2} 3 \, dx = 3x \bigg|_{0}^{2} = 3(2) - 3(0) = 6 \]

Calculating the second integral:
\[ \int_{0}^{2} 2\sqrt{x} \, dx = 2 \cdot \frac{2}{3} x^{3/2} \bigg|_{0}^{2} = \frac{4}{3} (2^{3/2}) - 0 = \frac{4}{3} (2\sqrt{2}) = \frac{8\sqrt{2}}{3} \]

### Step 5: Combine the results
Now substituting back into the area calculation:
\[ F(0) = 6 - \frac{8\sqrt{2}}{3} \]

### Step 6: Compute \( F(0) + \frac{8\sqrt{2}}{3} \)
Now we need to compute:
\[ F(0) + \frac{8\sqrt{2}}{3} = \left( 6 - \frac{8\sqrt{2}}{3} \right) + \frac{8\sqrt{2}}{3} = 6 \]

### Final Answer
Thus, the value of \( F(0) + \frac{8\sqrt{2}}{3} \) is:
\[ \boxed{6} \]