If $$x = \frac{2\sqrt{24}}{\sqrt{3}+\sqrt{2}}$$, then the value of $$\frac{x+\sqrt{8}}{x-\sqrt{8}}+\frac{x+\sqrt{12}}{x-\sqrt{12}}$$ is

Given : $$x = \frac{2\sqrt{24}}{\sqrt{3}+\sqrt{2}}$$

=> $$x = \frac{2\sqrt{24}}{\sqrt{3}+\sqrt{2}}\times(\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2})$$

=> $$x=\frac{2\sqrt{24}(\sqrt3-\sqrt2)}{3-2}$$

=> $$x=2\sqrt{72}-2\sqrt{48}$$

=> $$x=6\sqrt8-4\sqrt{12}$$ ---------------(i)

To find : $$\frac{x+\sqrt{8}}{x-\sqrt{8}}+\frac{x+\sqrt{12}}{x-\sqrt{12}}$$

= $$\frac{6\sqrt8-4\sqrt{12}+\sqrt{8}}{6\sqrt8-4\sqrt{12}-\sqrt{8}} + \frac{6\sqrt8-4\sqrt{12}+\sqrt{12}}{6\sqrt8-4\sqrt{12}-\sqrt{12}}$$     [Using (i)]

= $$\frac{7\sqrt{8}-4\sqrt{12}}{5\sqrt{8}-4\sqrt{12}} + \frac{6\sqrt8-3\sqrt{12}}{6\sqrt8-5\sqrt{12}}$$

= $$\frac{(336-35\sqrt{96}-24\sqrt{96}+240)+(240-15\sqrt{96}-24\sqrt{96}+144)}{240-25\sqrt{96}-24\sqrt{96}+240}$$

= $$\frac{960-98\sqrt{96}}{480-49\sqrt{96}}$$

= $$\frac{2(480-49\sqrt6)}{480-49\sqrt6}=2$$

=> Ans - (B)