Question

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

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

Answer 1

Answer: (D)

$2(1-\sqrt{3})$

According to given question,

$(x^2+x(f\left(x\right)-2)+2\sqrt{3}-3-\sqrt{3}f(x\left)\right)=0$
$\Rightarrow f\left(x\right)[\sqrt{3}-x]=x^2-2x+2\sqrt{3}-3$
$\Rightarrow f\left(x\right)=\frac{x^2-2x+2\sqrt{3}-3}{\sqrt{3}-x}$............(1)
It is given that, function is continuous, Hence limit will be equal to the value at $x=\sqrt{3}$ ,
Therefore,
$Li{m}_{x\to \sqrt{3}}f\left(x\right)=f\left(\sqrt{3}\right)$...........(2)
Solving limit using L.hospital's rule (i.e. derivative approach)
=> $Li{m}_{x\to \sqrt{3}}f\left(x\right)=\frac{2x-2}{-1}$..................From(1)

=> $f\left(\sqrt{3}\right)=2-2\sqrt{3}$................................From(2)

$=2(1-\sqrt{3})$

Hence, the answer is option $D$