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. $(-\infty ,-2)\cup [0,\infty )$
• D. none of these

Answer 1

Answer: (A)

The inequality is $lo{g}_{0.2}\frac{x+2}{x}\le 1$
The L.H.S is valid if $\frac{x+2}{x}>0$
$\Rightarrow x(x+2)>0\Rightarrow x<-2orx>0$
Solving the inequality, we get (note that base < 1).
$\frac{x+2}{x}\ge 0.2=\frac{1}{5}$
$\Rightarrow \frac{x+2}{x}-\frac{1}{5}\ge 0\Rightarrow \frac{4x+10}{5x}\ge 0$
$x\left(2x+5\right)\ge 0\Rightarrow x\le -\frac{5}{2}$ or $x\ge 0$
Taking the intersection, we get
$x\le -\frac{5}{2}$ or $x>0$
$\Rightarrow x\in \left(-\infty ,-\frac{5}{2}\right]\cup (0,\infty )$