Let E _ 1 = x ∈ ℝ: x ≠ 1 and (x)/(x - 1) > 0 and E _ 2 = x ∈ E _ 1...

Let ${E}_{1} =${$x \in \mathbb{R}: x \ne 1 and$\dfrac{x}{x - 1} > 0} and{E}_{2} =${x \in {E}_{1}: \sin^{- 1}\left(\log_{e}\left(\dfrac{x}{x - 1}\right)\right) is a real number}$

(Here the inverse trigonometric function $\sin^{- 1}\left(x\right)$ assumes values in \left[- \dfrac{\pi}{2} . \dfrac{\pi}{2}\right]$.)<br />Let$f: {E}_{1} \to \mathbb{R}$be the function defined by$f\left(x\right) = \log_{e} \left(\dfrac{x}{x - 1}\right)$<br />And$g: {E}_{2} \to \mathbb{R}$be the function defined by$g\left(x\right) = \sin^{- 1} \left(\log_{e} \left(\dfrac{x}{x - 1}\right)\right)$.</p><table border="1"> <tbody> <tr> <td>LIST-I</td> <td>LIST-II</td> </tr> <tr> <td>A. The range of$f$is</td> <td>P. \left(- \infty, \dfrac{1}{1 - e}\right] \cup \left[\dfrac{e}{e - 1}, \infty\right)$ B. The range of $g$ contains Q. (0, 1) C. The domain of $f$ contains R. \left[- \dfrac{1}{2}, \dfrac{1}{2}\right]$</td> </tr> <tr> <td>D. The domain of$g$is</td> <td>S. \left(- \infty, 0\right) \cup \left(0, \infty\right)$ T. \left(- \infty, \dfrac{e}{e - 1}\right]$</td> </tr> <tr> <td> </td> <td>U.$( - \infty, 0 ) \cup ( \dfrac{1}{2} , \dfrac{e}{e - 1} ]$ The correct option is: