Question

The set of admissible values of$x$ such that$\frac{2x+3}{2x-9}<0$ is

• A. $(-\infty ,–\frac{3}{2})\cup (\frac{9}{2},ɑ)$
• B. $(-\infty ,0)\cup (\frac{9}{2},ɑ)$
• C. $(-\frac{3}{2},0)$
• D. $(0,\frac{9}{2})$
• E. $\left(-\frac{3}{2},\frac{9}{2}\right)$

Answer 1

The correct option is E

$\left(-\frac{3}{2},\frac{9}{2}\right)$

The explanation for the correct answer.

Compute the required set:

$\frac{2x+3}{2x-9}<0$

$2x+3<0$ and $2x–9>0$

or $2x+3>0$ and $2x–9<0$ and $x\ne \frac{9}{2}$

$\Rightarrow x<–\frac{3}{2}$ and $x>\frac{9}{2}$

or $x>–\frac{3}{2}$ and $x<\frac{9}{2}$ and $x\ne \frac{9}{2}$

$\Rightarrow x\in \left(-\frac{3}{2},\frac{9}{2}\right)$

Hence, option $E$ is the correct answer.