If the ordinate x = a divides the area bounded by the curve `y=1+(8)/(x^(2))` and the ordinates `x=2, x=4` into two equal parts, then a is equal to

Options

A. `2sqrt3`
B. `2sqrt2`
C. 3
D. none of these

Answer

To solve the problem, we need to find the value of \( a \) such that the area bounded by the curve \( y = 1 + \frac{8}{x^2} \) and the ordinates \( x = 2 \) and \( x = 4 \) is divided into two equal parts by the line \( x = a \).

### Step-by-step Solution:

1. **Find the total area between the curve and the x-axis from \( x = 2 \) to \( x = 4 \)**:
\[ \text{Area} = \int_{2}^{4} \left( 1 + \frac{8}{x^2} \right) \, dx \]

2. **Calculate the integral**:
\[ \int \left( 1 + \frac{8}{x^2} \right) \, dx = x - \frac{8}{x} + C \]
Now, evaluate the definite integral from \( 2 \) to \( 4 \):
\[ \text{Area} = \left[ x - \frac{8}{x} \right]_{2}^{4} = \left( 4 - \frac{8}{4} \right) - \left( 2 - \frac{8}{2} \right) \]
Simplifying this:
\[ = \left( 4 - 2 \right) - \left( 2 - 4 \right) = 2 + 2 = 4 \]

3. **Since the area is 4, the area from \( x = 2 \) to \( x = a \) must be half of this**:
\[ \text{Area from } 2 \text{ to } a = \frac{4}{2} = 2 \]
Thus, we set up the equation:
\[ \int_{2}^{a} \left( 1 + \frac{8}{x^2} \right) \, dx = 2 \]

4. **Calculate the integral from \( 2 \) to \( a \)**:
\[ \int_{2}^{a} \left( 1 + \frac{8}{x^2} \right) \, dx = \left[ x - \frac{8}{x} \right]_{2}^{a} = \left( a - \frac{8}{a} \right) - \left( 2 - 4 \right) \]
This simplifies to:
\[ a - \frac{8}{a} + 2 \]

5. **Set the equation equal to 2**:
\[ a - \frac{8}{a} + 2 = 2 \]
Simplifying gives:
\[ a - \frac{8}{a} = 0 \]
Multiplying through by \( a \) (assuming \( a \neq 0 \)):
\[ a^2 - 8 = 0 \]

6. **Solve for \( a \)**:
\[ a^2 = 8 \implies a = \sqrt{8} = 2\sqrt{2} \]

### Final Answer:
The value of \( a \) is \( 2\sqrt{2} \).