Question

The set of all $x$satisfying the inequality$\frac{4x-1}{3x+1}\ge 1$ is

• A. $(-\infty ,–\frac{1}{3})\cup (\frac{1}{4},\infty )$
• B. $(-\infty ,–\frac{2}{3})\cup (\frac{5}{4},\infty )$
• C. $(-\infty ,–\frac{1}{3})\cup [2,\infty )$
• D. $(-\infty ,–\frac{2}{3})\cup [4,\infty )$
• E. $(-\infty ,–\frac{1}{3})\cup (\frac{1}{2},\infty )$

Answer 1

The correct option is C

$(-\infty ,–\frac{1}{3})\cup [2,\infty )$

Explanation for the correct option.

Given, $\frac{4x-1}{3x+1}\ge 1$

$\frac{4x-1}{3x+1}-1\ge 0$

$\frac{x–2}{3x+1}\ge 0$

$x–2\ge 0$ and $3x+1>0$ or $x–2\le 0$ and $3x+1<0$

$x\ge 2$and $x>–\frac{1}{2}$ or $x\le 2$and $x<–\frac{1}{3}$

$\Rightarrow x\in \left(-\infty ,–\frac{1}{3}\right)\cup [2,\infty )$

Hence, option C is the correct option.