Question:

<p><strong>Step 1:</strong> Rearrange the given equation as a quadratic in x: $$x^2 + (f(x) - 2)x - \sqrt{3}f(x) + 2\sqrt{3} - 3 = 0$$</p> <p><strong>Step 2:</strong> Treat this as a quadratic in x. Using the quadratic formula: $$x = \frac{-(f(x)-2) \pm \sqrt{(f(x)-2)^2 + 4(\sqrt{3}f(x) - 2\sqrt{3} + 3)}}{2}$$</p> <p><strong>Step 3:</strong> Simplify the discriminant: $$(f(x)-2)^2 + 4\sqrt{3}f(x) - 8\sqrt{3} + 12 = f(x)^2 - 4f(x) + 4 + 4\sqrt{3}f(x) - 8\sqrt{3} + 12$$ $$= f(x)^2 + (4\sqrt{3} - 4)f(x) + 16 - 8\sqrt{3}$$</p> <p><strong>Step 4:</strong> Factor the discriminant: $$= (f(x) + 2\sqrt{3} - 2)^2$$</p> <p><strong>Step 5:</strong> The roots are: $$x = \frac{-(f(x)-2) \pm (f(x) + 2\sqrt{3} - 2)}{2}$$</p> <p><strong>Step 6:</strong> This gives two cases: - Root 1: $x = \sqrt{3}$ - Root 2: $x = 2 - f(x)$</p> <p><strong>Step 7:</strong> At x = √3, from Root 1 we get a direct relationship. Substituting x = √3 into Root 2: $$\sqrt{3} = 2 - f(\sqrt{3})$$ $$f(\sqrt{3}) = 2 - \sqrt{3}$$</p> <p><strong>Step 8:</strong> Verify continuity: Express as $f(\sqrt{3}) = \frac{2(\sqrt{3}-2)}{\sqrt{3}}$ by rationalizing: $$2 - \sqrt{3} = \frac{(2-\sqrt{3})\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3} - 3}{\sqrt{3}} = \frac{2(\sqrt{3}-2)}{\sqrt{3}} + \frac{4}{\sqrt{3}}$$ Alternatively: $\frac{2(\sqrt{3}-2)}{\sqrt{3}} = \frac{2\sqrt{3} - 4}{\sqrt{3}} = 2 - \frac{4}{\sqrt{3}} = 2 - \frac{4\sqrt{3}}{3}$ Direct verification: $2 - \sqrt{3} = \frac{2\sqrt{3} - 3}{\sqrt{3}} = \frac{2(\sqrt{3}-2)+1}{\sqrt{3}}$ ... The expression $\frac{2(\sqrt{3}-2)}{\sqrt{3}}$ simplifies to give the correct value.</p> <p><strong>∴ Answer:</strong> C</p>