If f(x) is a continuous function for all real values of x satisfying ${\mathrm{x}}^2+\left(\mathrm{f}\right(\mathrm{x})-2)\mathrm{x}+2\sqrt{3}-3-\sqrt{3}\mathrm{f}\left(\mathrm{x}\right)=0$ then the value of $\mathrm{f}\left(\sqrt{3}\right)$ can be  

As f (x) is continuous for all $\mathrm{x}\in \mathrm{R}$ Thus ${\lim }_{\mathrm{x}\to \sqrt{3}} \mathrm{f}\left(\mathrm{x}\right)=\mathrm{f}\left(\sqrt{3}\right)$

$\begin{array}{ll}{\mathrm{x}}^2+\left(\mathrm{f}\right(\mathrm{x})-2)\mathrm{x}+2\sqrt{3}-3-\sqrt{3}\mathrm{f}\left(\mathrm{x}\right)=0& \\ \Rightarrow x^2-2x+2\sqrt{3}-3=f\left(x\right)\left(\sqrt{3}-x\right)& \\ f\left(x\right)=\frac{x^2-2x+2\sqrt{3}-3}{\sqrt{3}-x}& \\ \therefore \underset{x\to \sqrt{3}}{\lim } f\left(x\right)=\underset{x\to \sqrt{3}}{\lim } \frac{x^2-2x+2\sqrt{3}-3}{\sqrt{3}-x}& \\ \underset{x\to \sqrt{3}}{\lim } \frac{(x-\sqrt{3})(x+\sqrt{3}-2)}{(\sqrt{3}-x)}=2(1-\sqrt{3})& \end{array}$

Thus $\mathrm{f}\left(\sqrt{3}\right)=2(1-\sqrt{3})$