Let E1={x∈R:x≠1 and xx−1>0}and E2={x∈E1:sin−1(loge(xx−1))is a real number}.(Here, the inverse trigonometric function sin−1x assumes values in[−π2,π2].)Let f:E1→R be the function defined by f(x)=loge(xx−1)and g:E2→R be the function defined byg(x)=sin−1(loge(xx−1)). List I List IIP.The range of f is 1.(−∞,11−e]∪[ee−1,∞) Q.The range of g contains 2.(0,1)R.The domain of f contains 3.[−12,12]S.The domain of g is 4.(−∞,0)∪(0,∞) 5.(−∞,ee−1] 6.(−∞,0)∪(12,ee−1]The correct option is:

Let E1={x∈R:x≠1 and xx−1>0}and E2={x∈E1:sin−1(loge(xx−1))i

1.	Let E1={x∈R:x≠1 and xx−1>0}and E2={x∈E1:sin−1(loge(xx−1))is a real number}.(Here, the inverse trigonometric function sin−1x assumes values in[−π2,π2].)Let f:E1→R be the function defined by f(x)=loge(xx−1)and g:E2→R be the function defined byg(x)=sin−1(loge(xx−1)). List I List IIP.The range of f is 1.(−∞,11−e]∪[ee−1,∞) Q.The range of g contains 2.(0,1)R.The domain of f contains 3.[−12,12]S.The domain of g is 4.(−∞,0)∪(0,∞) 5.(−∞,ee−1] 6.(−∞,0)∪(12,ee−1]The correct option is:
Answer» Let $E_{1} = {x \in R : x \neq 1 and \frac{x}{x - 1} > 0}$ and $E_{2} = {x \in E_{1} : {sin}^{- 1} ({log}_{e} (\frac{x}{x - 1})) is a real number} .$ $($ Here, the inverse trigonometric function ${sin}^{- 1} x$ assumes values in $[- \frac{π}{2}, \frac{π}{2}]$ . $)$ Let $f : E_{1} \to R$ be the function defined by $f (x) = {log}_{e} (\frac{x}{x - 1})$ and $g : E_{2} \to R$ be the function defined by $g (x) = {sin}^{- 1} ({log}_{e} (\frac{x}{x - 1})) .$ $\begin{matrix} List I & List II P . & The range of f is & 1. & (- \infty, \frac{1}{1 - e}] \cup [\frac{e}{e - 1}, \infty) Q . & The range of g contains & 2. & (0, 1) R . & The domain of f contains & 3. & [- \frac{1}{2}, \frac{1}{2}] S . & The domain of g is & 4. & (- \infty, 0) \cup (0, \infty) 5. & (- \infty, \frac{e}{e - 1}] 6. & (- \infty, 0) \cup (\frac{1}{2}, \frac{e}{e - 1}] \end{matrix}$ The correct option is: