Question

The number of solutions of the equation $\sin \left(\frac{\pi x}{2\sqrt{3}}\right)=x^2-2\sqrt{3}x+4$

• A. forms an empty set
• B. is only one
• C. is only two
• D. is greater than $2$

Answer 1

The correct option is B is only one

$\sin \left(\frac{\pi x}{2\sqrt{3}}\right)=x^2-2\sqrt{3}x+4$
$\because -1\le \sin \left(\frac{\pi x}{2\sqrt{3}}\right)\le 1$
$\Rightarrow -1\le x^2-2\sqrt{3}x+4\le 1$
$\Rightarrow x^2-2\sqrt{3}x+4\le 1$
and $x^2-2\sqrt{3}x+4\ge -1$

$\Rightarrow (x-\sqrt{3}{)}^2\le 0$
and $(x-\sqrt{3}{)}^2+2\ge 0$
This is possible only when $x-\sqrt{3}=0$
$\therefore x=\sqrt{3}$