If the maximum area bounded by `y^(2)=4x` and the line `y=mx(AA m in [1, 3])` is k square units, then the smallest prime number greater than 3k is

Options

A. 3
B. 5
C. 7
D. 11

Answer

To solve the problem, we need to find the maximum area bounded by the curve \( y^2 = 4x \) and the line \( y = mx \) for \( m \) in the interval \([1, 3]\). We will then determine the smallest prime number greater than \( 3k \), where \( k \) is the maximum area.

### Step-by-Step Solution:

1. **Identify the curves and the area to be calculated**:
The curve is given by \( y^2 = 4x \), which can be rewritten as \( y = 2\sqrt{x} \) (considering the positive branch since we are dealing with area). The line is given by \( y = mx \).

2. **Find the points of intersection**:
To find the points where the line intersects the curve, we set \( 2\sqrt{x} = mx \).
\[ 2\sqrt{x} = mx \implies 2 = mx^{1/2} \implies x^{1/2} = \frac{2}{m} \implies x = \left(\frac{2}{m}\right)^2 = \frac{4}{m^2} \]
The corresponding \( y \) value is:
\[ y = mx = m\left(\frac{4}{m^2}\right) = \frac{4}{m} \]

3. **Set up the area integral**:
The area \( A \) bounded by the curve and the line from \( x = 0 \) to \( x = \frac{4}{m^2} \) is given by:
\[ A = \int_0^{\frac{4}{m^2}} (2\sqrt{x} - mx) \, dx \]

4. **Calculate the integral**:
We compute the integral:
\[ A = \int_0^{\frac{4}{m^2}} 2\sqrt{x} \, dx - \int_0^{\frac{4}{m^2}} mx \, dx \]
The first integral:
\[ \int 2\sqrt{x} \, dx = \frac{2}{\frac{3}{2}} x^{\frac{3}{2}} = \frac{4}{3} x^{\frac{3}{2}} \]
Evaluating from \( 0 \) to \( \frac{4}{m^2} \):
\[ \frac{4}{3} \left(\frac{4}{m^2}\right)^{\frac{3}{2}} = \frac{4}{3} \cdot \frac{8}{m^3} = \frac{32}{3m^3} \]

The second integral:
\[ \int mx \, dx = \frac{m}{2} x^2 \]
Evaluating from \( 0 \) to \( \frac{4}{m^2} \):
\[ \frac{m}{2} \left(\frac{4}{m^2}\right)^2 = \frac{m}{2} \cdot \frac{16}{m^4} = \frac{8}{m^3} \]

5. **Combine the results**:
Thus, the area \( A \) becomes:
\[ A = \frac{32}{3m^3} - \frac{8}{m^3} = \frac{32 - 24}{3m^3} = \frac{8}{3m^3} \]

6. **Maximize the area**:
To maximize \( A \) with respect to \( m \), we note that \( A \) is maximized when \( m \) is minimized (since \( m^3 \) is in the denominator). The minimum value of \( m \) in the interval \([1, 3]\) is \( m = 1 \).
\[ A_{\text{max}} = \frac{8}{3 \cdot 1^3} = \frac{8}{3} \]

7. **Determine \( k \) and calculate \( 3k \)**:
Here, \( k = \frac{8}{3} \). Therefore:
\[ 3k = 3 \cdot \frac{8}{3} = 8 \]

8. **Find the smallest prime number greater than \( 3k \)**:
The smallest prime number greater than \( 8 \) is \( 11 \).

### Final Answer:
The smallest prime number greater than \( 3k \) is \( \boxed{11} \).