Question

The real values of $x$ satisfying ${\log }_{0.5}\left(\frac{x+1}{x+2}\right)\le 1$

• A. $(-\infty ,-2)\cup [0,\infty )$
• B. $(-\infty ,-2]\cup [0,\infty )$
• C. $(-\infty ,-2)\cup (-1,\infty )$
• D. $R-[-2,-1]$

Answer 1

The correct option is A $(-\infty ,-2)\cup [0,\infty )$

$\frac{x+1}{x+2}>0\Rightarrow x\notin [-2,-1]\cdots \left(1\right)$
Now ${\log }_{0.5}\left(\frac{x+1}{x+2}\right)\le 1$
$\Rightarrow \frac{x+1}{x+2}\ge \frac{1}{2}\Rightarrow \frac{x+1}{x+2}-\frac{1}{2}\ge 0\Rightarrow \frac{x}{x+2}\ge 0\Rightarrow x\notin [-2,0)\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$ $x\in (-\infty ,-2)\cup [0,\infty )$