Question

If $x<2$, then $\frac{1}{x}$ lies in the interval

• A. $(-\infty ,\frac{1}{2})$
• B. $(-\infty ,0)\cup (\frac{1}{2},\infty )$
• C. $(-\infty ,0)\cup (0,\frac{1}{2})$
• D. $(\frac{1}{2},\infty )$

Answer 1

The correct option is B $(-\infty ,0)\cup (\frac{1}{2},\infty )$

$x<2\Rightarrow x\in (-\infty ,2)$
$\to 1/x$ is not defined for $x=0$
$\therefore$ whenever $0$ lies in the interval while taking the reciprocal, we break it into two parts such that $0$ can be excluded.
$\therefore \frac{1}{x}$ only defined for $(-\infty ,2)-\left\{0\right\}=(-\infty ,0)\cup (0,2)$
$\to x\in (-\infty ,0)\Rightarrow \frac{1}{x}\in (-\infty ,0)$
$\to x\in (0,2)\Rightarrow \frac{1}{x}\in (\frac{1}{2},\infty )$
$\Rightarrow \frac{1}{x}\in (-\infty ,0)\cup (\frac{1}{2},\infty )$