Question

The complete set of values of $x$ that satisfies the inequality $\sqrt{{\log }_2\left(\frac{2x-3}{x-1}\right)}<1$ is

• A. $x\in [2,\infty )$
• B. $x\in [\infty ,4)$
• C. $x\in [-\infty ,\infty )$
• D. $x\in [4,\infty )$

Answer 1

The correct option is A $x\in [2,\infty )$

For, $\sqrt{{\log }_2\left(\frac{2x-3}{x-1}\right)}<1$
$\log$ is defined, when
$\frac{2x-3}{x-1}>0\Rightarrow (2x-3)(x-1)>0$
$\Rightarrow x\in (-\infty ,1)\cup (\frac{3}{2},\infty )$ ...(1)
Now,
$\sqrt{{\log }_2\left(\frac{2x-3}{x-1}\right)}<1\Rightarrow {\log }_2\left(\frac{2x-3}{x-1}\right)<1$
$\Rightarrow \left(\frac{2x-3}{x-1}\right)<2^1$
For, $x-1>0\Rightarrow x\in (1,\infty )$ ...(2)

$\therefore -2x-3<2(x-1)$
$\Rightarrow -3<-1$

From (1) and (2)

$x\in [2,\infty ]$