Range and Domain Analysis of Logarithmic and Inverse Trigonometric Composite Functions

To determine the correct match between LIST-I and LIST-II, let us analyze each function step-by-step.

---

1. Domain of $f$ ($E_1$) and Range of $f$ (P & R)

The function $f: E_1 \to \mathbb{R}$ is defined by: $f(x) = \log_e \left( \frac{x}{x-1} \right)$

For $f(x)$ to be defined, the argument of the logarithm must be strictly positive: $\frac{x}{x-1} > 0$

Using the critical points $x = 0$ and $x = 1$:

• For $x > 1$, $\frac{x}{x-1} > 0$.
• For $0 < x < 1$, $\frac{x}{x-1} < 0$.
• For $x < 0$, $\frac{x}{x-1} > 0$.

Thus, the domain of $f$ is: $E_1 = (-\infty, 0) \cup (1, \infty)$

Analysis of P (Range of $f$):

Let $y = \frac{x}{x-1} = 1 + \frac{1}{x-1}$.

• When $x \in (-\infty, 0)$, we have $x - 1 \in (-\infty, -1)$, so $\frac{1}{x-1} \in (-1, 0)$, which implies $y \in (0, 1)$.
• When $x \in (1, \infty)$, we have $x - 1 \in (0, \infty)$, so $\frac{1}{x-1} \in (0, \infty)$, which implies $y \in (1, \infty)$.

Thus, $y \in (0, 1) \cup (1, \infty)$. Since $\log_e(y)$ maps $(0, 1) \to (-\infty, 0)$ and $(1, \infty) \to (0, \infty)$, the range of $f$ is: $\text{Range}(f) = (-\infty, 0) \cup (0, \infty)$ This matches 4 in LIST-II. So, $\mathbf{P \rightarrow 4}$.

Analysis of R (The domain of $f$ contains):

Consider item 1 in LIST-II: $\left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right)$.

• Since $e \approx 2.718$, $\frac{1}{1-e} \approx -0.582 < 0$. Thus, $\left( -\infty, \frac{1}{1-e} \right] \subset (-\infty, 0)$.
• Since $\frac{e}{e-1} \approx 1.582 > 1$. Thus, $\left[ \frac{e}{e-1}, \infty \right) \subset (1, \infty)$.

Therefore: $\left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right) \subset E_1$ This means the domain of $f$ contains item 1. So, $\mathbf{R \rightarrow 1}$.

---

2. Domain of $g$ ($E_2$) (S)

The function $g: E_2 \to \mathbb{R}$ is defined by: $g(x) = \sin^{-1} \left( \log_e \left( \frac{x}{x-1} \right) \right)$

For $g(x)$ to be defined as a real number, we require $x \in E_1$ and: $-1 \le \log_e \left( \frac{x}{x-1} \right) \le 1$

Taking the exponential on all sides (since $e > 1$): $\frac{1}{e} \le \frac{x}{x-1} \le e$

1. Solving $\frac{x}{x-1} \le e$: $\frac{x}{x-1} - e \le 0 \implies \frac{(1-e)x + e}{x-1} \le 0 \implies \frac{(e-1)x - e}{x-1} \ge 0$ Since $\frac{e}{e-1} > 1$, this inequality yields: $x \in (-\infty, 1) \cup \left[ \frac{e}{e-1}, \infty \right)$

2. Solving $\frac{x}{x-1} \ge \frac{1}{e}$: $\frac{x}{x-1} - \frac{1}{e} \ge 0 \implies \frac{(e-1)x + 1}{e(x-1)} \ge 0$ Since $\frac{1}{1-e} < 0$, this inequality yields: $x \in \left( -\infty, \frac{1}{1-e} \right] \cup (1, \infty)$

Taking the intersection of both solutions and restricting to $x \in E_1$: $E_2 = \left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right)$

Thus, the domain of $g$ is item 1 in LIST-II. So, $\mathbf{S \rightarrow 1}$.

---

3. Range of $g$ (Q)

For $x \in E_2$, the argument $t = \log_e \left( \frac{x}{x-1} \right)$ spans $[-1, 0) \cup (0, 1]$. Since $\sin^{-1}(t)$ is continuous and strictly increasing, its range is: $\text{Range}(g) = \left[ -\frac{\pi}{2}, 0 \right) \cup \left( 0, \frac{\pi}{2} \right]$

Since $\frac{\pi}{2} \approx 1.57 > 1$, the interval $(0, 1)$ is entirely contained inside $\left( 0, \frac{\pi}{2} \right] \subset \text{Range}(g)$. Hence, the range of $g$ contains $(0, 1)$, which is item 2 in LIST-II. So, $\mathbf{Q \rightarrow 2}$.

---

Conclusion

Matching the results:

• $\text{P} \rightarrow 4$
• $\text{Q} \rightarrow 2$
• $\text{R} \rightarrow 1$
• $\text{S} \rightarrow 1$

This matches option (A).