Question

The solution set of $\frac{1}{4}+\frac{1}{x}<\frac{1}{2}+\frac{1}{3}$, when $xϵR,$ is:

• A. $\frac{1}{x}<\frac{7}{12}$
• B. $x\ge \frac{12}{7}$
• C. $x<\frac{14}{7}$
• D. $\frac{1}{x}>\frac{3}{2}$

Answer 1

The correct option is A $\frac{1}{x}<\frac{7}{12}$

$\frac{1}{4}+\frac{1}{x}<\frac{5}{6}$ $\therefore \frac{1}{x}<\frac{5}{6}-\frac{1}{4}$
$\therefore \frac{1}{x}<\frac{7}{12}$