Question

Which of the following is/are the solution(s) of the inequality $\frac{1}{\left(x\right|-3}\ge \frac{1}{2}$?

• A. $\left(-\infty ,-5\right)$
• B. $[-5,-3)$
• C. $\left(-3,3\right)$
• D. $(3,5]$
• E. $\left(5,\infty \right)$

Answer 1

Answer: (B), (D)

$(3,5]$

Given the linear inequality,
$\frac{1}{\left(x\right|-3}\ge \frac{1}{2}\Rightarrow \frac{1}{\left(x\right|-3}-\frac{1}{2}\ge 0\Rightarrow \frac{2-\left(x\right|+3}{\left(x\right|-3}\ge 0\Rightarrow \frac{5-\left(x\right|}{\left(x\right|-3}\ge 0\Rightarrow \frac{\left(x\right|-5}{\left(x\right|-3}\le 0\mathrm{So},\mathrm{if}\left(x\right|-5=0\Rightarrow x=-5\mathrm{or}5\mathrm{or},\mathrm{if}\left(x\right|-3=0\Rightarrow x=-3\mathrm{or}3\therefore \mathrm{The}\mathrm{critical}\mathrm{points}\mathrm{are}-5,-3,3\mathrm{and}5.$


$\therefore x\in [-5,-3)\cup (3,5]$