Question

The solution set of $\frac{2x}{x^2-9}\le \frac{1}{x+2}$ is

• A. $(-\infty ,-2)\cup (3,\infty )$
• B. $(-\infty ,-3)\cup (-2,3)$
• C. $(-3,0)\cup (3,\infty )$
• D. $(-3,-2)\cup (3,\infty )$

Answer 1

The correct option is B $(-\infty ,-3)\cup (-2,3)$

$\frac{2x}{x^2-9}\le \frac{1}{x+2}\Rightarrow \frac{2x}{x^2-9}-\frac{1}{x+2}\le 0\Rightarrow \frac{x^2+4x+9}{(x^2-9)(x+2)}\le 0\Rightarrow \frac{(x+2{)}^2+5}{(x+3)(x-3)(x+2)}\le 0\Rightarrow \frac{1}{(x+3)(x-3)(x+2)}\le 0(\because (x+2{)}^2+5>0,\forall x\in R)$


$\therefore x\in (-\infty ,-3)\cup (-2,3)$