Question

The set of all values of ${}^{\prime }{x}^{\prime }$ satisfying the inequation $\left(\frac{1}{|x|-3}\right)\le \frac{1}{2}$ is

• A. $(-\infty ,-5)\cup (-3,3)\cup [5,\infty )$
• B. $(-\infty ,-5]\cup [-3,3]\cup [5,\infty )$
• C. $(-\infty ,-5)\cup (-3,3)\cup (5,\infty )$
• D. None of the above

Answer 1

Answer: (A)

$(-\infty ,-5)\cup (-3,3)\cup [5,\infty )$

$\frac{1}{\left|x\right|-3}\le \frac{1}{2}$

If $x>0$

$\frac{1}{x-3}\le \frac{1}{2}$

$\frac{1}{x-3}-\frac{1}{2}\le 0$

$\frac{2-x+3}{2(x-3)}\le 0$

$\frac{2-x+3}{2(x-3)}\le 0$

$\frac{5-x}{2(x-3)}\ge 0$

From $shape\left(i\right)$ $x\in (-\infty ,3)\cup (5,\infty ).⟶\left(1\right)$

If $x<0$

$\frac{1}{-x-3}\le \frac{1}{2}$

$\frac{1}{x+3}\ge -\frac{1}{2}$

$\frac{1}{x+3}+\frac{1}{2}\ge 0$

$\frac{2+x+3}{2(x+3)}\ge 0$

$\frac{5+x}{2(x+3)}\ge 0$

From $shape\left(ii\right)$ $x\in (-\infty ,-5)\cup (-3,\infty ).⟶\left(2\right)$

From $\left(1\right)$ & $\left(2\right),$

$x\in (-\infty ,-5)\cup (-3,3)\cup [5,\infty ).$