Question

The solution of $\frac{6x}{4x-1}<\frac{1}{2}$ is

• A. $x<\frac{-1}{8}$
• B. $\frac{-1}{8}<x<\frac{1}{4}$
• C. $x<\frac{-1}{8}$ and $x>\frac{1}{4}$
• D. $x>\frac{1}{8}$

Answer 1

Answer: (B)

Given inequality is,
$\frac{6x}{4x-1}<\frac{1}{2}\Rightarrow \frac{12x}{4x-1}<1$
$\Rightarrow \frac{12x}{4x-1}-1<0\Rightarrow \frac{8x+1}{4x-1}<0$ ...(i)
For above inequality (i), we have two cases either $(I)8x+1>0,4x-1<0$
or $(II)8x+1<0,4x-1>0$
From (I), we get $x>\frac{-1}{8},x<\frac{1}{4}$
$\therefore$ $x\in \left(-\frac{1}{8},\frac{1}{4}\right)$ ....(ii)
From (II), $x<\frac{-1}{8},x>\frac{1}{4}$ , both of which can not be true
$\therefore$ (ii) represent the solution set of given inequality.