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. $(-\infty ,-\frac{2}{3}]\cup [4,\infty )$

Answer 1

The correct option is C $(-\infty ,-\frac{1}{3})\cup [2,\infty )$

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

$\Rightarrow \frac{x-2}{3x+1}\ge 0$

$\Rightarrow x-2\ge 0$ and $3x+1>0$ or

$x-2\le 0$ and $3x+1<0$

i.e. $x\ge 2$ and $x>-\frac{1}{3}$

or $x\le 2$ and $x<-\frac{1}{3}\Rightarrow x\in (-\infty ,-\frac{1}{3})\cup [2,\infty )$