Question

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

$(-\infty , - 3) \cup (-2 , 3)$

Answer 2

We have $x^2 - 9 \neq 0$ and $x + 2 \neq 0$ and $\frac{2x}{x^{2}-9}-\frac{1}{x+2}\le0 \Rightarrow \frac{2x^{2}+4x-x^{2}+9}{\left(x+2\right)\left(x^{2}-9\right)}\le0$ $\Rightarrow \frac{x^{2}+4x+9}{\left(x+2\right)\left(x^{2}-9\right)} \le0\Rightarrow \left(x+2\right)\left(x+3\right)\left(x-3\right)<0$ $\left( \because x^{2}+4x + 9>0 \, \forall x \, \in R\right)$ From the wavy curve shown, we have $x \in ( -\infty , - 2) \cup ( - 2 , 3)$