Question

The solution set of the inequality ${\log }_3\frac{|x^2-4x|+3}{x^2+|x-5|}\ge 0$ is

• A. $x\le \frac{-2}{3},\frac{1}{2}\le x\le 2$
• B. $x\le \frac{2}{3},\frac{1}{2}\le x\le 2$
• C. $x\le \frac{-2}{3},\frac{2}{3}\le x\le 2$
• D. $x\le \frac{-1}{2},\frac{2}{3}\le x\le 2$

Answer 1

The correct option is A $x\le \frac{-2}{3},\frac{1}{2}\le x\le 2$

${\log }_3\frac{|x^2-4x|+3}{x^2+|x-5|}\ge 0$

Case I: When $x\ge 5$

${\log }_3\frac{x^2-4x+3}{x^2+x-5}\ge 0$

$\Rightarrow \frac{x^2-4x+3}{x^2+x-5}\ge 1$

$\Rightarrow x^2-4x+3\ge x^2+x-5$

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

Case II: When $4\le x<5$

${\log }_3\frac{x^2-4x+3}{x^2-(x-5)}\ge 0$

$\Rightarrow \frac{x^2-4x+3}{x^2-(x-5)}\ge 1$

$\Rightarrow x^2-4x+3\ge x^2-x+5$

$\Rightarrow x\le -\frac{2}{3}$

Case III: When $x<4$

${\log }_3\frac{{-(x}^2-4x)+3}{x^2-(x-5)}\ge 0$

$\Rightarrow -x^{+}4x+3\ge x^2-x+5$

$\Rightarrow 2x^2-5x+2\le 0$

$\Rightarrow (x-2)(2x-1)\le 0$

$\Rightarrow x\in [\frac{1}{2},2]$

Hence, from all the cases iit follows that

$x\le -\frac{2}{3},\frac{1}{2}\le x\le 2$