Question

For the given expression $\frac{\sqrt{2x-1}}{x-2}<1,x\in$ :

• A. $[\frac{1}{2},2)\cup (5,\infty )$
• B. $[\frac{1}{2},1)\cup (5,\infty )$
• C. $[0,2)\cup (5,\infty )$
• D. $(-\infty ,2)\cup (5,\infty )$

Answer 1

The correct option is A $[\frac{1}{2},2)\cup (5,\infty )$

Given expression $\frac{\sqrt{2x-1}}{x-2}<1$ will be valid
if $2x-1\ge 0,x\ne 2$
$x\ge \frac{1}{2},x\ne 2$

Case $1:$ When $x>2$,
then $\frac{\sqrt{2x-1}}{x-2}>0$
Hence $\frac{2x-1}{(x-2{)}^2}<1^2$
$\Rightarrow 2x-1<x^2+4-4x\Rightarrow x^2-6x+5>0\Rightarrow (x-5)(x-1)>0$
$\therefore x\in (5,\infty )\cdots \left(1\right)$

Case $2:$ When $\frac{1}{2}\le x<2$,
then $\frac{\sqrt{2x-1}}{x-2}<0$
Hence $\frac{\sqrt{2x-1}}{x-2}<1$
$\therefore x\in [\frac{1}{2},2)\cdots \left(2\right)$

Hence from $\left(1\right)$ and $\left(2\right)$
$x\in [\frac{1}{2},2)\cup (5,\infty )$