If $$x=\frac{4}{2\sqrt{3}+3\sqrt{2}}$$,then find the value of  $$(x+\frac{1}{x})$$

Given : $$x=\frac{4}{2\sqrt{3}+3\sqrt{2}}$$

Rationalizing the denominator, we get :

=> $$x=\frac{4}{2\sqrt{3}+3\sqrt{2}}\times\frac{(2\sqrt3-3\sqrt2)}{(2\sqrt3-3\sqrt2)}$$

=> $$x=\frac{4(2\sqrt3-3\sqrt2)}{(12-18)}=\frac{2(3\sqrt2-2\sqrt3)}{3}$$ ---------(i)

Now, $$\frac{1}{x}=\frac{3}{2(3\sqrt2-2\sqrt3)}\times\frac{(3\sqrt2+2\sqrt3)}{(3\sqrt2+2\sqrt3)}$$

=> $$\frac{1}{x}=\frac{3(3\sqrt2+2\sqrt3)}{2(18-12)}$$

=> $$\frac{1}{x}=\frac{3\sqrt2+2\sqrt3}{4}$$ -------------(ii)

Adding equations (i) and (ii), we get :

=> $$x+\frac{1}{x}=(\frac{2(3\sqrt2-2\sqrt3)}{3})+(\frac{3\sqrt2+2\sqrt3}{4})$$

= $$\frac{8(3\sqrt2-2\sqrt3)+3(3\sqrt2+2\sqrt3)}{12}$$

= $$\frac{24\sqrt2-16\sqrt3+9\sqrt2+6\sqrt3}{12}$$

= $$\frac{33\sqrt2-10\sqrt3}{12}$$

= $$\frac{-(10\sqrt3-33\sqrt2)}{12}$$

=> Ans - (C)