Question

If $f\left(x\right)$ is a continuous function for all real values of $x$ and satisfies $x^2+xf\left(x\right)-2x+2\sqrt{3}-3-\sqrt{3}f\left(x\right)=0,\forall x\in R$, then the value of $f\left(\sqrt{3}\right)$ is

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

Answer 1

The correct option is B $2(1-\sqrt{3})$

Given: $x^2+xf\left(x\right)-2x+2\sqrt{3}-3-\sqrt{3}f\left(x\right)=0$
As $f\left(x\right)$ is continuous for all $x\in R.$
Thus, $\underset{x\to \sqrt{3}}{lim}f\left(x\right)=f\left(\sqrt{3}\right)$
Where
$f\left(x\right)=\frac{x^2-2x+2\sqrt{3}-3}{\sqrt{3}-x},x\ne \sqrt{3}$
Now,
$\underset{x\to \sqrt{3}}{lim}f\left(x\right)=\underset{x\to \sqrt{3}}{lim}\frac{x^2-2x+2\sqrt{3}-3}{\sqrt{3}-x}$
Using L-Hospital's rule
$=\underset{x\to \sqrt{3}}{lim}\frac{2x-2}{-1}=2(1-\sqrt{3})\therefore f\left(\sqrt{3}\right)=2(1-\sqrt{3})$