Question

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

Answer 1

${\log }_{0.2}\frac{x+2}{x}\le 1$
For $\log$ to be defined,
$\frac{x+2}{x}>0\cdots \left(i\right)$
Now,
${\log }_{0.2}\left(\frac{x+2}{x}\right)\le {\log }_{0.2}0.2$
$\Rightarrow \frac{x+2}{x}\ge 0.2$
$\Rightarrow \frac{x+2}{x}\ge \frac{1}{5},\text{Satisfying equation (i)}$
$\Rightarrow \frac{x+2}{x}-\frac{1}{5}\ge 0$
$\Rightarrow \frac{5x+10-x}{5x}\ge 0$
$\Rightarrow \frac{4x+10}{5x}\ge 0\Rightarrow \frac{2x+5}{5x}\ge 0$
Here, Critical points $=0,-\frac{5}{2}$
Now, draw wavy curve along this critical points


$\therefore x\in (-\infty ,-\frac{5}{2}]\cup (0,\infty )$.
Hence, The correct answer is option (a)