Q. The set of all $x$ satisfying the inequality $\frac{4x - 1}{3x + 1} \geq 1$ is
A) $\left( - \infty,\ - \frac{1}{3} \right) \cup \left\lbrack \frac{1}{4},\ \infty \right\rbrack$
B) $\left( - \infty,\ - \frac{2}{3} \right) \cup \left\lbrack \frac{5}{4},\ \infty \right\rbrack$
C) $\left( - \infty,\ - \frac{1}{3} \right) \cup \lbrack 2,\ \infty)$
D) $\left( - \infty,\ - \frac{2}{3} \right)\lbrack 4,\ \infty)$
Explanation
(c)
Given, $\frac{4x - 1}{3x + 1} - 1 \geq 0$
$\Rightarrow \ \ \ \frac{x - 2}{3x + 1} \geq 0$
$\Rightarrow \ \ x - 2 \geq 0$ and $3x + 1 > 0$
Or $x - 2 \leq 0$ and $3x + 1 < 0$
$\Rightarrow \ \ x \geq 2$ and $x < - \frac{1}{3}$
Or $x \leq 2$ and $x > - \frac{1}{3}$
$\Rightarrow \ \ x \in \left( - \infty,\ - \frac{1}{3} \right) \cup \lbrack 2,\ \infty)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.