Let $E_1=${$x\in \mathrm{ℝ}: x\ne 1$ and $\frac{x}{x-1}>0$} and $E_2=${$x\in E_1: {\sin }^{-1}\left({\log }_e\left(\frac{x}{x-1}\right)\right)$ is a real number}

(Here the inverse trigonometric function ${\sin }^{-1}\left(x\right)$ assumes values in $\left[-\frac{\mathrm{\pi }}{2}. \frac{\mathrm{\pi }}{2}\right]$.)
Let $f:E_1\to \mathrm{ℝ}$ be the function defined by $f\left(x\right)={\log }_e\left(\frac{x}{x-1}\right)$
And $g:E_2\to \mathbb{R}$ be the function defined by $g\left(x\right)={\sin }^{-1}\left({\log }_e\left(\frac{x}{x-1}\right)\right)$ .

LIST-I	LIST-II
A. The range of $f$ is	P. $\left(-\infty ,\frac{1}{1-e}\right]\cup \left[\frac{e}{e-1}, \infty \right)$
B. The range of $g$ contains	Q. (0, 1)
C. The domain of $f$ contains	R. $\left[-\frac{1}{2},\frac{1}{2}\right]$
D. The domain of $g$ is	S. $\left(-\infty ,0\right)\cup \left(0, \infty \right)$
	T. $\left(-\infty ,\frac{e}{e-1}\right]$
	U. $\left(-\infty ,0\right)\cup \left(\frac{1}{2},\frac{e}{e-1}\right]$

The correct option is:

• A

  a-p;b-s;c-r;d-q;
• B

  a-u;b-p;c-t;d-r;
• C

  a-q;b-r;c-p;d-u;
• Correct answer
  D

  a-s;b-q;c-p;d-p;

Solution

$E_1:\frac{x}{x-1}>0$

$\Rightarrow$ $E_1: x\in \left(-\infty ,0\right)\cup \left(1,\infty \right)$
$E_2:-1\le \ln \left(\frac{x}{x+1}\right)\le 1$
$\frac{1}{e}\le \frac{x}{x-1}\le e$
Now $\frac{x}{x-1}-\frac{1}{e}\ge 0$
$\Rightarrow$ $\frac{\left(e-1\right)x+1}{e\left(x-1\right)}\ge 0$

$\Rightarrow$ $x ϵ \left(-\infty ,\frac{1}{1-e}\right]\cup \left(1,\infty \right)$
Also $\frac{x}{x-1}-e\le 0$
$\frac{\left(e-1\right)x-e}{x-1}\ge 0$

$\Rightarrow$ $x\in \left(-\mathrm{\infty },\frac{1}{1-e}\right]\cup \left[\frac{e}{e-1}, \infty \right]$
As Range of $\frac{x}{x-1}$ is $R^{+}-\left\{1\right\}$
$\Rightarrow$ Range of $f$ is $R-\left\{0\right\}$ or $\left(-\mathrm{\infty }, 0\right)\cup \left(0,\mathrm{ }\mathrm{\infty }\right)$
Range of $g$ is $\left[–\frac{\pi }{2},\frac{\pi }{2}\right]$ $-\left\{0\right\}$ or $\left[–\frac{\pi }{2}, 0\right)\cup \left(0,\frac{\pi }{2}\right]$
Now $P\to 4, Q\to 2, R\to 1, S\to 1$
Hence $A$ is correct.

Hence, the correct answer is option: 'D'.