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, $\displaystyle\lim _{x \rightarrow \sqrt{3}} f(x)=f(\sqrt{3})$
where $f(x)=\frac{x^{2}-2 x+2 \sqrt{3}-3}{\sqrt{3}-x}, x \neq \sqrt{3}$
$\displaystyle\lim _{x \rightarrow \sqrt{3}} f(x)=\displaystyle\lim _{x \rightarrow \sqrt{3}} \frac{x^{2}-2 x+2 \sqrt{3}-3}{\sqrt{3}-x}$
$=\displaystyle\lim _{x \rightarrow \sqrt{3}} \frac{(2-\sqrt{3}-x)(\sqrt{3}-x)}{(\sqrt{3}-x)}=2(1-\sqrt{3})$
$\Rightarrow \quad f(\sqrt{3})=2(1-\sqrt{3})$