Question

The set of real values of $x$ satisfying
${\log }_{0.2}\left(\frac{x+2}{x}\right)\le 1$ is

• A. $(-\infty ,-\frac{5}{2}]\cup (0,\infty )$
• B. $[\frac{5}{2},\infty )$
• C. $(-\infty ,-2)\cup (0,\infty )$
• D. None of these

Answer 1

The correct option is A $(-\infty ,-\frac{5}{2}]\cup (0,\infty )$

${\log }_{0.2}\left(\frac{x+2}{x}\right)$ is defined, if
$\frac{x+2}{x}>0\Rightarrow x\in (-\infty ,-2)\cup (0,\infty )$
Now,
${\log }_{0.2}\left(\frac{x+2}{x}\right)\le 1$
$\Rightarrow \frac{x+2}{x}\ge (0.2{)}^1$
$\Rightarrow \frac{x+2}{x}\ge \frac{1}{5}$
$\Rightarrow \frac{x+2}{x}-\frac{1}{5}\ge 0$
$\Rightarrow \frac{4x+10}{5x}\ge 0\Rightarrow \frac{2x+5}{x}\ge 0\Rightarrow x\in (-\infty ,-\frac{5}{2}]\cup (0,\infty )$

Hence, the solution is
$x\in (-\infty ,-\frac{5}{2}]\cup (0,\infty )$