If x = a(>0) divides the area bounded by X–axis, part of the curve $\mathrm{y}=1+\frac{8}{{\mathrm{x}}^2}$ and the ordinates x=2, x=4 into equal parts then a =

$\text{ Here, }{\int }_2^a \left(1+\frac{8}{x^2}\right)dx={\int }_a^4 \left(1+\frac{8}{x^2}\right)dx$

$\Rightarrow {\left(x-\frac{8}{x}\right)}_2^a={\left(x-\frac{8}{x^2}\right)}_a^4$

$\Rightarrow \left(a-\frac{8}{a}\right)-(2-4)=(4-2)-\left(a-\frac{8}{a}\right)$

$\Rightarrow \mathrm{a}-\frac{8}{\mathrm{a}}+2=2-\mathrm{a}+\frac{8}{\mathrm{a}}$

$\Rightarrow 2\mathrm{a}-\frac{16}{\mathrm{a}}=0\text{ or }2\left({\mathrm{a}}^2-8\right)=0$

$\Rightarrow \mathrm{a}=\pm 2\sqrt{2}\text{, neglecting }(-\mathrm{ve})$

$a=2\sqrt{2}$