Question

If $f(x)$ be continuous function for all real values of $x$ and satisfies; $x^2+\{f(x)-2\}x+2\sqrt{3}-3-\sqrt{3}\cdot f(x)=0$, $\forall x\in R$. Then the value of $f(\sqrt{3})$ is

• A. $2(1+\sqrt{3})$
• B. $(\sqrt{3}-1)$
• C. $\sqrt{3}-1$
• D. $2(1-\sqrt{3})$

Answer 1

Answer: (D)

As $f(x)$ is continuous for all $x\in R$.
Thus, $\underset{x\to \sqrt{3}}{\lim }f(x)=f(\sqrt{3})$
where $f(x)=\frac{x^2-2x+2\sqrt{3}-3}{\sqrt{3}-x},x\ne \sqrt{3}$
$\underset{x\to \sqrt{3}}{\lim }f(x)=\underset{x\to \sqrt{3}}{\lim }\frac{x^2-2x+2\sqrt{3}-3}{\sqrt{3}-x}$
$=\underset{x\to \sqrt{3}}{\lim }\frac{(2-\sqrt{3}-x)(\sqrt{3}-x)}{(\sqrt{3}-x)}=2(1-\sqrt{3})$
$\Rightarrow f(\sqrt{3})=2(1-\sqrt{3})$