Question

The greatest integral value of $x$ which can satisfy the inequality $\frac{x^2-6x+3}{x^2-4x+3}<0$ is

Answer 1

$\frac{x^2-6x+3}{x^2-4x+3}<0$
$\Rightarrow \frac{x^2-6x+3}{(x-3)(x-1)}<0$
Numerator
$x^2-6x+3=0D=36-12=24>0$
The roots of the equation $x^2-6x+3=0$ are
$\Rightarrow x=\frac{6\pm \sqrt{24}}{2}\Rightarrow x=3\pm \sqrt{6}$

The inequality becomes
$\frac{x^2-6x+3}{(x-3)(x-1)}<0\Rightarrow \frac{(x-(3-\sqrt{6}\left)\right)(x-(3+\sqrt{6}\left)\right)}{(x-1)(x-3)}<0$

Critical point are $3-\sqrt{6},1,3,3+\sqrt{6}$


$\therefore x\in (3-\sqrt{6},1)\cup (3,3+\sqrt{6})$

Hence, the greatest integral value which satisfies the inequality is $5$.