JEE Advanced 2018 — Paper 2 — Question 42

Domain of $f$ is $E_1 = \{x \in \mathbb{R} : x \neq 1 \text{ and } \frac{x}{x-1} > 0\}$. Solving $\frac{x}{x-1} > 0$ gives $x \in (-\infty, 0) \cup (1, \infty)$. Range of $f$: as $x$ varies over $(-\infty, 0)$, $\frac{x}{x-1} \in (0,1)$ so $f(x) \in (-\infty, 0)$; as $x$ varies over $(1, \infty)$, $\frac{x}{x-1} \in (1, \infty)$ so $f(x) \in (0, \infty)$. Hence $\text{Range}(f) = (-\infty, 0) \cup (0, \infty)$. $\Rightarrow \mathbf{P} \to 4$. For $g$, require $f(x) \in [-1,1]$. Solving $-1 \leq \ln\left(\frac{x}{x-1}\right) \leq 1$ gives $\frac{1}{e} \leq \frac{x}{x-1} \leq e$. Solving these inequalities with $x \in E_1$ yields $x \in \left(-\infty, \frac{1}{1-e}\right] \cup \left[\frac{e}{e-1}, \infty\right)$, which is $E_2$. $\Rightarrow \mathbf{S} \to 1$. On $E_2$, $f(x) \in [-1,0) \cup (0,1]$, so $g(x) = \sin^{-1}(f(x)) \in [-\pi/2, 0) \cup (0, \pi/2]$. This range contains $(0,1)$. $\Rightarrow \mathbf{Q} \to 2$. Domain of $f$ is $(-\infty, 0) \cup (1, \infty)$, which contains $\left(-\infty, \frac{1}{1-e}\right] \cup \left[\frac{e}{e-1}, \infty\right)$.

$\Rightarrow \mathbf{R} \to 1$. Thus the correct mapping is $\mathbf{P} \to 4, \mathbf{Q} \to 2, \mathbf{R} \to 1, \mathbf{S} \to 1$, which matches option A.