Question

The domain of $f\left(x\right)=\sqrt{{\log }_x\left\{x\right\}}$ is
$\left(\right\{x\}$ represents the fractional part of $x)$

• A. $R$
• B. $[0,1]$
• C. $(0,1)$
• D. $(0,1]$

Answer 1

The correct option is C $(0,1)$

For $f$ to be defined,
${\log }_x\left\{x\right\}\ge 0$
and $x\ne 1,\left\{x\right\}>0\cdots \left(1\right)$

We know that ${\log }_{a}x$ is strictly decreasing for $0<a<1$ and strictly increasing for $a>1$
So, if $0<x<1$,
then ${\log }_x\left\{x\right\}\ge 0\Rightarrow \left\{x\right\}\le x^0$
i.e., $\left\{x\right\}\le 1$
But $0\le \left\{x\right\}<1\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,
$\left\{x\right\}\in (0,1)\Rightarrow 0<x<1$

If $x>1$,
then $\left\{x\right\}\ge 1$, which is not possible.

Hence, $D\left(f\right)=(0,1)$