Question

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

• A. $x\in (-\infty ,-3)$
• B. $(-2,3)$
• C. $x\in (-10,-5)$
• D. $(5,10)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $x\in (-\infty ,-3)$
B $(-2,3)$
C $x\in (-10,-5)$

We have $x^2-9\ne 0$ and $x+2\ne 0$ and
$\frac{2x}{x^2-9}-\frac{1}{x+2}\le 0\Rightarrow \frac{2x^2+4x-x^2+9}{(x+2)(x^2-9)}\le 0$
$\Rightarrow \frac{x^2+4x+9}{(x+2)(x^2-9)}\le 0$

$\Rightarrow \frac{x^2+4x+4+5}{(x+2)(x^2-9)}\le 0$

$\Rightarrow \frac{(x+2{)}^2+5}{(x+2)(x^2-9)}\le 0$

$\Rightarrow (x+2)(x+3)(x-3)<0$ ...... $(\because (x+2{)}^2+5>0\forall x\in R)$
From the wavy curve shown, we have
$x\in (-\infty ,-3)\cup (-2,3)$