Question

Solve $\left|\frac{x^2-5x+4}{x^2-4}\right|\le 1$

Answer 1

$\left|\frac{x^2-5x+4}{x^2-4}\right|=\left|\frac{(x-4)(x-1)}{(x-2)(x+2)}\right|\le 1$

Case (1) $\therefore$ if $x\in (-\infty ,-2)\cup [1,2)\cup [4,\infty ]$

then

$\frac{(x-4)(x-1)}{(x-2)(x+2)}\le 1$

$\Rightarrow x^2-5x+4\le x^2-4$

$8\le 5x$

$\frac{8}{5}\le x$

$\therefore x\in [\frac{8}{5},2)\cup [4,\infty ).......\left(1\right)$

Case (2) Now, if $x\in (-2,1]\cup (2,4]$

$\therefore -\left(\frac{x^2-5x+4}{x^2-4}\right)\le 1$

$x^2-5x+4\ge -x^2+4$

$2x^2-5x\ge 0$

$\left(x\right)(2x-5)\ge 0$

$x\in (-\infty ,0]\cup [\frac{5}{2},\infty )$

$\therefore x\in (-2,0]\cup (\frac{5}{2},4]$........ (2)$

$\therefore$ final answer $\Rightarrow \left(1\right)\cup \left(2\right)$

$x\in (-2,0]\cup [\frac{8}{5},2)\cup (\frac{5}{2},\infty )$