1. Find $E_1$ (domain of $f$)

We need $x\neq 1 \quad \text{and} \quad \frac{x}{x-1}>0.$

Now, the fraction is positive when numerator and denominator have the same sign.

• If $x>0$ and $x-1>0$, then $x>1$.
• If $x<0$ and $x-1<0$, then $x<0$.

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

So the domain of $f$ is $(-\infty,0)\cup(1,\infty),$ which is clearly contained in $(-\infty,0)\cup(0,\infty).$ Thus, $R \to 4.$

2. Find the range of $f(x)=\ln\left(\frac{x}{x-1}\right)$

Let $y=\frac{x}{x-1}=1+\frac{1}{x-1}.$ We analyze $y$ on the domain $(-\infty,0)\cup(1,\infty)$.

• For $x<0$: $\frac{x}{x-1}\in(0,1).$ Indeed, as $x\to -\infty$, $\frac{x}{x-1}\to 1^{-}$, and as $x\to 0^{-}$, $\frac{x}{x-1}\to 0^{+}$.

• For $x>1$: $\frac{x}{x-1}>1.$ As $x\to 1^{+}$, $\frac{x}{x-1}\to +\infty$, and as $x\to +\infty$, it tends to $1^{+}$.

Therefore, $\frac{x}{x-1}\in (0,1)\cup(1,\infty).$ Applying $\ln$, $f(x)=\ln\left(\frac{x}{x-1}\right)\in (-\infty,0)\cup(0,\infty).$ So the range of $f$ is $(-\infty,0)\cup(0,\infty),$ which matches item 4. Thus, $P \to 4.$

3. Find $E_2$ (domain of $g$)

For $g(x)=\sin^{-1}\left(\ln\left(\frac{x}{x-1}\right)\right)$ to be real, we need $-1\le \ln\left(\frac{x}{x-1}\right)\le 1.$ Exponentiating, $e^{-1}\le \frac{x}{x-1}\le e.$

Now solve this on the domain $E_1=(-\infty,0)\cup(1,\infty)$.

Case 1: $x<0$

Here $x-1<0$.

(i) Solve $\dfrac{x}{x-1}\ge e^{-1}$

Since $x-1<0$, inequality reverses on multiplying: $x\le e^{-1}(x-1).$ So, $ex\le x-1$ $x(e-1)\le -1$ $x\le -\frac{1}{e-1}=\frac{1}{1-e}.$

(ii) Solve $\dfrac{x}{x-1}\le e$

Again $x-1<0$, so inequality reverses: $x\ge e(x-1)$ $x\ge ex-e$ $x(1-e)\ge -e$ $x\le \frac{e}{e-1}.$ This is automatically true for all $x<0$.

Thus for $x<0$, $x\le \frac{1}{1-e}.$ So we get $(-\infty,\tfrac{1}{1-e}].$

Case 2: $x>1$

Here $x-1>0$.

(i) Solve $\dfrac{x}{x-1}\ge e^{-1}$

Multiply directly: $x\ge e^{-1}(x-1).$ This is automatically true for all $x>1$.

(ii) Solve $\dfrac{x}{x-1}\le e$

Multiply directly: $x\le e(x-1)$ $x\le ex-e$ $e\le x(e-1)$ $x\ge \frac{e}{e-1}.$

Thus for $x>1$, we get $\left[\frac{e}{e-1},\infty\right).$

Hence, $E_2=(-\infty,\tfrac{1}{1-e}]\cup\left[\tfrac{e}{e-1},\infty\right).$ This matches item 1. Thus, $S \to 1.$

4. Find the range of $g$

Since $g(x)=\sin^{-1}(f(x)),$ and for $x\in E_2$, we have $f(x)\in[-1,1]\setminus\{0\}?$ Let us check whether $0$ occurs.

If $f(x)=0,$ then $\ln\left(\frac{x}{x-1}\right)=0 \implies \frac{x}{x-1}=1,$ which gives $x=x-1,$ impossible.

So actually $f(E_2)=[-1,0)\cup(0,1].$ Applying $\sin^{-1}$, $g(E_2)=\left[-\frac{\pi}{2},0\right)\cup\left(0,\frac{\pi}{2}\right].$

Now among the listed sets, this range certainly contains $(0,1)$, because $(0,1)\subset \left(0,\frac{\pi}{2}\right].$ But it does not contain $\left[-\frac12,\frac12\right]$, since $0$ is not in the range.

Hence, $Q \to 2.$

5. Final matching

We have found:

• $P \to 4$
• $Q \to 2$
• $R \to 4$ (domain of $f$ contains item 4)
• $S \to 1$

Now compare with options:

• Option A: $P\to 4,\; Q\to 2,\; R\to 1,\; S\to 1$
• Option B: wrong
• Option C: wrong
• Option D: wrong

There is an inconsistency in the options regarding $R$.

The correct statement for $R$ should be $R\to 4$, because $\text{Dom}(f)=(-\infty,0)\cup(1,\infty)\subset (-\infty,0)\cup(0,\infty),$ and item 1 is not contained in the domain of $f$ since it includes numbers in $(0,1]$.

So the mathematically correct matching is: $P\to 4,\quad Q\to 2,\quad R\to 4,\quad S\to 1.$ Since no option matches this, the stored answer A appears to be incorrect, likely due to a misprint in LIST-II or the options.