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. **Calculate the total area between the curve and the ordinates:**
We need to find the area \( A \) between \( x = 2 \) and \( x = 4 \):
\[ A = \int_{2}^{4} \left(1 + \frac{8}{x^2}\right) \, dx \]

2. **Integrate the function:**
The integral can be split:
\[ A = \int_{2}^{4} 1 \, dx + \int_{2}^{4} \frac{8}{x^2} \, dx \]
The first integral is:
\[ \int_{2}^{4} 1 \, dx = [x]_{2}^{4} = 4 - 2 = 2 \]
The second integral is:
\[ \int_{2}^{4} \frac{8}{x^2} \, dx = 8 \left[-\frac{1}{x}\right]_{2}^{4} = 8 \left(-\frac{1}{4} + \frac{1}{2}\right) = 8 \left(-\frac{1}{4} + \frac{2}{4}\right) = 8 \cdot \frac{1}{4} = 2 \]

3. **Total area:**
Therefore, the total area \( A \) is:
\[ A = 2 + 2 = 4 \]

4. **Area divided by \( x = a \):**
We want to find \( a \) such that the area from \( x = 2 \) to \( x = a \) is equal to half of the total area:
\[ \int_{2}^{a} \left(1 + \frac{8}{x^2}\right) \, dx = 2 \]

5. **Set up the integral for the area from \( x = 2 \) to \( x = a \):**
\[ \int_{2}^{a} 1 \, dx + \int_{2}^{a} \frac{8}{x^2} \, dx = 2 \]
This simplifies to:
\[ [x]_{2}^{a} + 8 \left[-\frac{1}{x}\right]_{2}^{a} = 2 \]
Which gives:
\[ (a - 2) + 8 \left(-\frac{1}{a} + \frac{1}{2}\right) = 2 \]

6. **Solve the equation:**
Expanding the equation:
\[ a - 2 - \frac{8}{a} + 4 = 2 \]
Simplifying:
\[ a - \frac{8}{a} + 2 = 2 \]
Thus:
\[ a - \frac{8}{a} = 0 \]
Multiplying through by \( a \) (assuming \( a \neq 0 \)):
\[ a^2 - 8 = 0 \]
Therefore:
\[ a^2 = 8 \quad \Rightarrow \quad a = \sqrt{8} = 2\sqrt{2} \]

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