Question

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

• A. $(-\infty -2)\cup (3,\infty )$
• B. $(-\infty ,-3)\cup (-2,3)$
• C. $(-3,0]\cup (3,\infty )$
• D. none of these

Answer 1

Answer: (B)

We have $x^2-9\ne 0$ and $x+2\ne 0$ and
$\frac{2x}{x^2-9}-\frac{1}{x+2}\le 0\Rightarrow \frac{2x^2+4x-x^2+9}{\left(x+2\right)\left(x^2-9\right)}\le 0$
$\Rightarrow \frac{x^2+4x+9}{\left(x+2\right)\left(x^2-9\right)}\le 0\Rightarrow \left(x+2\right)\left(x+3\right)\left(x-3\right)<0$
$\left(\because x^2+4x+9>0\forall x\in R\right)$
From the wavy curve shown, we have
$x\in (-\infty ,-2)\cup (-2,3)$