If $$x = \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}$$, then the value of $$\frac{x+\sqrt{2}}{x-\sqrt{2}} + \frac{x+\sqrt{3}}{x-\sqrt{3}}$$ is

Given : $$x = \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}$$

=> $$x = \frac{2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\times(\frac{\sqrt3-\sqrt2}{\sqrt3-\sqrt2})$$

=> $$x=\frac{2\sqrt6(\sqrt3-\sqrt2)}{3-2}$$

=> $$x=2\sqrt{18}-2\sqrt{12}$$

=> $$x=6\sqrt2-4\sqrt3$$ ---------------(i)

To find : $$\frac{x+\sqrt{2}}{x-\sqrt{2}} + \frac{x+\sqrt{3}}{x-\sqrt{3}}$$

= $$\frac{6\sqrt2-4\sqrt3+\sqrt{2}}{6\sqrt2-4\sqrt3-\sqrt{2}} + \frac{6\sqrt2-4\sqrt3+\sqrt{3}}{6\sqrt2-4\sqrt3-\sqrt{3}}$$     [Using (i)]

= $$\frac{7\sqrt{2}-4\sqrt3}{5\sqrt{2}-4\sqrt3} + \frac{6\sqrt2-3\sqrt{3}}{6\sqrt2-5\sqrt{3}}$$

= $$\frac{(84-35\sqrt6-24\sqrt6+60)+(60-15\sqrt6-24\sqrt6+36)}{60-25\sqrt6-24\sqrt6+60}$$

= $$\frac{240-98\sqrt6}{120-49\sqrt6}$$

= $$\frac{2(120-49\sqrt6)}{120-49\sqrt6}=2$$

=> Ans - (D)