Let $E_1 = \left\{ x \in \mathbb{R} : x \neq 1 \text{ and } \frac{x}{x-1} > 0 \right\}$ and $E_2 = \left\{ x \in E_1 : \sin^{-1}\left( \log_e\left( \frac{x}{x-1} \right) \right) \text{ is a real number} \right\}$. (Here, the inverse trigonometric function $\sin^{-1} x$ assumes values in $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$.) Let $f : E_1 \to \mathbb{R}$ be the function defined by $f(x) = \log_e\left( \frac{x}{x-1} \right)$ and $g : E_2 \to \mathbb{R}$ be the function defined by $g(x) = \sin^{-1}\left( \log_e\left( \frac{x}{x-1} \right) \right)$. | List-I | List-II | | --- | --- | | P. The range of f is | 1. $\left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right)$ | | Q. The range of g contains | 2. $(0, 1)$ | | R. The domain of f contains | 3. $\left[ -\frac{1}{2}, \frac{1}{2} \right]$ | | S. The domain of g is | 4. $(-\infty, 0) \cup (0, \infty)$ | | | 5. $\left( -\infty, \frac{e}{e-1} \right]$ | | | 6. $(-\infty, 0) \cup \left( \frac{1}{2}, \frac{e}{e-1} \right]$ |

Let E₁ = { x ∈ R : x ≠ 1 and x/(x-1) > 0 } and E₂ = { x ∈ E₁ : sin⁻¹( logₑ( x/(x-1) ) ) is a real number }. (Here, the inverse trigonometric function sin⁻¹ x assumes values in [ -π/2, π/2 ].) Let f : E₁ → R be the function defined by f(x) = logₑ( x/(x-1) ) and g : E₂ → R be the function defined by g(x) = sin⁻¹( logₑ( /

$E_1 = \left\{ x : x \neq 1,\ \dfrac{x}{x-1} > 0 \right\}$

$E_2 = \left\{ x \in E_1 : \sin^{-1}\!\left( \ln\dfrac{x}{x-1} \right) \in \mathbb{R} \right\}$

$f(x) = \ln\dfrac{x}{x-1},\qquad g(x) = \sin^{-1}\!\left( \ln\dfrac{x}{x-1} \right)$

$\text{Match: } P(\text{range of }f),\ Q(\text{range of }g),\ R(\text{domain of }f),\ S(\text{domain of }g)$

$\dfrac{x}{x-1} > 0 \iff x < 0 \text{ or } x > 1$

$E_1 = (-\infty, 0) \cup (1, \infty)$

$t = \dfrac{x}{x-1}:\quad x<0 \Rightarrow t\in(0,1);\quad x>1 \Rightarrow t\in(1,\infty)$

$\text{Range}(f) = (-\infty,0)\cup(0,\infty) = \mathbb{R}\setminus\{0\}$

$\Rightarrow\ P \to 4$

$\text{Domain}(f) = E_1 = (-\infty,0)\cup(1,\infty)$

$\left(-\infty,\tfrac{1}{1-e}\right]\cup\left[\tfrac{e}{e-1},\infty\right) \subset E_1$

$\Rightarrow\ R \to 1$

$-1 \le \ln t \le 1 \iff \tfrac{1}{e} \le t \le e$

$x = \dfrac{t}{t-1} \Rightarrow \text{Domain}(g) = \left(-\infty,\tfrac{1}{1-e}\right]\cup\left[\tfrac{e}{e-1},\infty\right)$

$\Rightarrow\ S \to 1$

$\ln t \in [-1,1]\setminus\{0\} \Rightarrow \text{Range}(g) = \left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$

$(0,1) \subset \text{Range}(g),\qquad \left[-\tfrac12,\tfrac12\right] \not\subset \text{Range}(g)$

$\Rightarrow\ Q \to 2$

$P \to 4,\quad Q \to 2,\quad R \to 1,\quad S \to 1$

$\boxed{\text{option } (1)}$