If the ordinate x = a divides the area bounded by the curve y = 1 + $\frac{8}{x^2}$ and the ordinates x = 2, x = 4 into two equal parts, then a =

$2\sqrt{2}$

Area = $\int\limits_2^4\left(1+\frac{8}{x^2}\right) d x=\left[x-\frac{8}{x}\right]_2^4=4$

Also $A_1=\int\limits_2^a y d x=\frac{1}{2} A=2 \Rightarrow\left[x-\frac{8}{x}\right]_2^a=2$

$\Rightarrow (a-2)-8\left(\frac{1}{a}-\frac{1}{2}\right)=2 \Rightarrow a-\frac{8}{a}=0$

$\Rightarrow a=2 \sqrt{2}$

Hence (2) is the correct answer.