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Match the Columns
Let \( E_{1}=\left\{x \in R: x \neq 1\right. \) and \( \left.\frac{x}{x-1}0\right\} \) and \( E_{2}=\left\{x \in E_{1}: \sin ^{-1}\left(\log _{e}\left(\frac{x}{x-1}\right)\right)\right. \) is a real number \( \} \)
\( \mathrm{P} \)
W
(Here, the inverse trigonometric function \( \sin ^{-1} x \) assumes values in \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right\} \). Let \( f: E_{1} \rightarrow R \) be the function defined by \( f(x)=\log _{e}\left(\frac{x}{x-1}\right) \) and \( g: E_{2} \rightarrow R \) be the function defined by \( g(x)=\sin ^{-1}\left(\log _{e}\left(\frac{x}{x-1}\right)\right) \).
(2018 Adv.)
\begin{tabular}{lll}
\hline \multicolumn{1}{c}{ List I } & \multicolumn{2}{c}{ List II } \\
\hline The range of \( f \) is & 1. & \( \left(-\infty, \frac{1}{1-e}\right] \cup\left[\frac{e}{e-1}, \infty\right) \) \\
\hline The range of \( g \) contains & 2. & \( (0,1) \) \\
\hline \( \begin{array}{l}\text { The domain of } \\
\text { contains }\end{array} \) & 3. & {\( \left[-\frac{1}{2}, \frac{1}{2}\right] \)} \\
\hline The domain of \( g \) is & 4. & \( (-\infty, 0) \cup(0, \infty) \) \\
\hline & 5. \( \left(-\infty, \frac{e}{e-1}\right] \) \\
\hline
\end{tabular}
The correct option is
(a) \( \mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 1 \)
(b) \( \mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 6 ; \mathrm{S} \rightarrow 5 \)
(c) \( \mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 6 \)
(d) \( \mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 6 ; \mathrm{S} \rightarrow 5 \)
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