If f(x) be continuous function for all real values of x and satisfies, $x^2+\{f(x)-2\} x+2 \sqrt{3}-3-\sqrt{3} f(x)=0, \forall x \in R$. Then find the value of $f(\sqrt{3})$.
Answer & explanation
Correct answer: option 3
From the given equation, we have
$f(x)=\frac{x^2-2 x+2 \sqrt{3}-3}{\sqrt{3}-x}, x \neq \sqrt{3}$
As f(x) is continuous for all $x \in R$, we must have
$f(\sqrt{3})=\lim\limits_{x \rightarrow \sqrt{3}} f(x)$
$=\lim\limits_{x \rightarrow \sqrt{3}} \frac{x^2-2 x+2 \sqrt{3}-3}{\sqrt{3}-x}$
$=\lim\limits_{x \rightarrow \sqrt{3}} \frac{(2-\sqrt{3}-x)(\sqrt{3}-x)}{(\sqrt{3}-x)}$
$= 2(1-\sqrt{3})$
Hence, $f(\sqrt{3})=2(1-\sqrt{3})$