Question

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

• A. $(-\infty ,-\frac{5}{2}]\cup (0,+\infty )$
• B. $[\frac{5}{2}+\infty )$
• C. $(-\infty ,0)\cup (\frac{8}{5},+\infty )$
• D. none of these

Answer 1

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

${\log }_{0.2}\frac{x+2}{x}\le 1$
For above expression to be defined, $\frac{x+2}{x}>0$
$\Rightarrow x>0$ ...(1)
or $x<-2$ ...(2)
Now,
${\log }_{0.2}\frac{x+2}{x}\le 1$
$\frac{x+2}{x}\ge (.2{)}^1=\frac{1}{5},$ since base of log is less than 1.
$\frac{x+2}{x}-\frac{1}{5}\ge 0$
$\frac{5x+10-x}{5x}\ge 0$
$\frac{4x+10}{x}\ge 0$
$\frac{2x+5}{x}\ge 0$
$\Rightarrow x\in (-\infty ,-\frac{5}{2}]\cup (0,\infty )$ .......(3)
From (1), (2) & (3) we get,
$x\in (-\infty ,-\frac{5}{2}]\cup (0,+\infty )$
Ans: A