Which of the following intervals is a possibl | vedclass

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(C) The function $f(x) = \log_{\{x\}}[x] + \log_{[x]}\{x\}$ is defined under the following conditions:$1$. For $\log_{\{x\}}[x]$ to be defined:Base $\

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$(2, 3)$

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(C) The function $f(x) = \log_{\{x\}}[x] + \log_{[x]}\{x\}$ is defined under the following conditions:$1$. For $\log_{\{x\}}[x]$ to be defined:Base $\{x\} > 0$ and $\{x\} 
eq 1$. Since $\{x\} = x - [x]$,$\{x\} \in [0, 1)$. Thus,$\{x\} \in (0, 1)$.Argument $[x] > 0$,which implies $x \geq 1$. Since $\{x\} 
eq 0$,$x$ cannot be an integer.$2$. For $\log_{[x]}\{x\}$ to be defined:Base $[x] > 0$ and $[x] 
eq 1$. This implies $[x] \geq 2$,so $x \geq 2$.Argument $\{x\} > 0$,which implies $x$ is not an integer.Combining these conditions,we require $x \geq 2$ and $x 
otin \mathbb{Z}$.Among the given options,the interval $(2, 3)$ satisfies $x > 2$ and $x$ is not an integer.Therefore,the correct option is $(c)$.

$A$. $f(x)=\frac{\|x+2\|}{x+2}, x \neq-2$	$1$. $[\frac{1}{3}, 1]$
$B$. $g(x)=\|[x]\|, x \in R$	$2$. $Z$
$C$. $h(x)=\|x-[x]\|, x \in R$	$3$. $W$
$D$. $f(x)=\frac{1}{2-\sin 3x}, x \in R$	$4$. $[0, 1)$
	$5$. $\{-1, 1\}$

Which of the following intervals is a possible domain of the function $f(x) = \log_{\{x\}}[x] + \log_{[x]}\{x\}$,where $[x]$ is the greatest integer not exceeding $x$ and $\{x\} = x - [x]$?

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