Question

If $\frac{5}{x-2}<5,x\ne 2$, then the interval(s) in which $x$ can be lie is/are

• A. $(2,\infty )$
• B. $(3,\infty )$
• C. $(-\infty ,2)$
• D. $R-[2,3]$

Answer 1

Answer: (B), (C), (D)

$R-[2,3]$

Given
$\frac{5}{x-2}<5,x\ne 2$
$\Rightarrow \frac{1}{x-2}<1\Rightarrow -\infty <\frac{1}{x-2}<1$
$\because 0$ is lying in the interval. So, we have to break the interval into two parts such that $0$ can be neglected
$\Rightarrow -\infty <\frac{1}{x-2}<0\text{or}0<\frac{1}{x-2}<1\Rightarrow -\infty <x-2<0\text{or}1<x-2<\infty \Rightarrow -\infty <x<2\text{or}3<x<\infty \therefore x\in (-\infty ,2)\cup (3,\infty )$
​​​​​​​or $x\in R-[2,3]$