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 \( y = mx \) where \( m \) is in the interval \([1, 3]\).

### Step-by-Step Solution:

1. **Identify the Points of Intersection**:
The curve \( y^2 = 4x \) can be rewritten as \( x = \frac{y^2}{4} \). The line \( y = mx \) can be rewritten as \( x = \frac{y}{m} \). To find the points of intersection, we set these equations equal to each other:
\[ \frac{y^2}{4} = \frac{y}{m} \]
Rearranging gives:
\[ y^2 - \frac{4y}{m} = 0 \]
Factoring out \( y \):
\[ y \left( y - \frac{4}{m} \right) = 0 \]
Thus, the points of intersection are \( y = 0 \) and \( y = \frac{4}{m} \).

2. **Calculate the Corresponding x-Coordinates**:
For \( y = \frac{4}{m} \):
\[ x = \frac{y^2}{4} = \frac{\left( \frac{4}{m} \right)^2}{4} = \frac{16}{4m^2} = \frac{4}{m^2} \]
Therefore, the points of intersection are \( (0, 0) \) and \( \left( \frac{4}{m^2}, \frac{4}{m} \right) \).

3. **Set Up the Area Integral**:
The area \( A \) bounded by the curve and the line can be calculated using the integral:
\[ A = \int_0^{\frac{4}{m^2}} \left( \sqrt{4x} - mx \right) \, dx \]
Here, \( \sqrt{4x} = 2\sqrt{x} \).

4. **Evaluate the Integral**:
\[ A = \int_0^{\frac{4}{m^2}} \left( 2\sqrt{x} - mx \right) \, dx \]
Calculate the integral:
- The integral of \( 2\sqrt{x} \) is \( \frac{4}{3} x^{3/2} \).
- The integral of \( mx \) is \( \frac{mx^2}{2} \).

Thus,
\[ A = \left[ \frac{4}{3} x^{3/2} - \frac{mx^2}{2} \right]_0^{\frac{4}{m^2}} \]
Evaluating at the upper limit:
\[ A = \frac{4}{3} \left( \frac{4}{m^2} \right)^{3/2} - \frac{m}{2} \left( \frac{4}{m^2} \right)^2 \]
Simplifying:
\[ A = \frac{4}{3} \cdot \frac{64}{m^3} - \frac{m}{2} \cdot \frac{16}{m^4} = \frac{256}{3m^3} - \frac{8}{m^3} = \frac{256 - 24}{3m^3} = \frac{232}{3m^3} \]

5. **Maximize the Area**:
To find the maximum area, we need to minimize \( m^3 \) since \( A \) is inversely proportional to \( m^3 \). The minimum value of \( m \) in the interval \([1, 3]\) is \( m = 1 \):
\[ A_{max} = \frac{232}{3 \cdot 1^3} = \frac{232}{3} \]

6. **Calculate \( k \)**:
Here, \( k = \frac{232}{3} \).

7. **Find \( 3k \)**:
\[ 3k = 3 \cdot \frac{232}{3} = 232 \]

8. **Find the Smallest Prime Number Greater than \( 3k \)**:
The smallest prime number greater than 232 is 233.

### Final Answer:
The smallest prime number greater than \( 3k \) is **233**.