Question

The following system of inequalities
$\frac{x}{2x + 1} \ge \frac{1}{4}$, $\frac{6x}{4x - 1} < \frac{1}{2}$ have

• A. infinitely many solution
• B. unique solution
• C. no solution
• D. None of these

Answer 1

Answer: (C)

From the first inequality, we have $\frac{x}{2x + 1} - \frac{1}{4} \ge 0$
$\Rightarrow \quad \frac{2x - 1}{2x + 1} \ge 0$
$\Rightarrow \quad$ ($2x - 1 \ge 0$ and $2x + 1 > 0$)
or $\quad$($2x - 1 \le 0$ and $2x + 1 < 0$) $\quad$ [Since $2x + 1 \ne 0$]
$\Rightarrow \quad$ ($x \ge \frac{1}{2}$ and $x > - \frac{1}{2}$) or ($x \le \frac{1}{2}$ and $x < -\frac{1}{2}$)
$\Rightarrow \quad x \ge \frac{1}{2}$ or $x < -\frac{1}{2}\quad\Rightarrow x\in \left(-\infty,\,-\frac{1}{2}\right)\cup\left[\frac{1}{2},\,\infty\right]...\left(1\right)$
From the second inequality, we have $\frac{6x}{4x - 1} - \frac{1}{2} < 0$
$\Rightarrow \quad \frac{8x + 1}{4x - 1}< 0$
$\Rightarrow \quad$ ($8x + 1 < 0$ and $4x - 1 > 0$)
or $\quad$ ($8x + 1 > 0$ and $4x - 1 < 0$)
$\Rightarrow \quad$ ($x < -\frac{1}{8}$ and $x > \frac{1}{4}$, it is not possible)
or $\quad$($x > - \frac{1}{8}$ and $x < \frac{1}{4}$)
$\Rightarrow \quad x \in \left(-\frac{1}{8},\, \frac{1}{4}\right)\quad...\left(2\right)$
Note that the common solution of $(1)$ and $(2)$ is null set. Hence, the given system of inequalities has no solution.