Q. The set of admissible values of $x$ such that $\frac{2x + 3}{2x - 9} < 0$ is
A) $\left( - \infty,\ - \frac{3}{2} \right) \cup \left( \frac{9}{2},\ \infty \right)$
B) $( - \infty,\ 0) \cup \left( \frac{9}{2},\ \infty \right)$
C) $\left( - \frac{3}{2},\ 0 \right)$
D) $\left( - \frac{3}{2},\ \frac{9}{2} \right)$
Explanation
(d)
Given, $\frac{2x + 3}{2x - 9} < 0$
$\Rightarrow 2x + 3 < 0$ and $2x - 9 > 0$
Or $2x + 3 > 0$ and $2x - 9 < 0$ and $x \neq \frac{9}{2}$
$\Rightarrow \ \ x < - \frac{3}{2}$ and $x > \frac{9}{2}$ or $x > - \frac{3}{2}$ and $x < \frac{9}{2}$ and $x \neq \frac{9}{2}$
$\Rightarrow x \in \left( - \frac{3}{2},\ \frac{9}{2} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.