Question

Let $E_{1}=\left\{x \in \mathbb{R}: x \neq 1\right.$ and $\left.\frac{x}{x-1}>0\right\}$
and $E_{2}=\left\{x \in E_{1}: \sin ^{-1}\left(\log _{e}\left(\frac{x}{x-1}\right)\right)\right.$ is a real number $\}$.
(Here, the inverse trigonometric function $\sin ^{-1} x$ assumes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.)
Let $f: E_{1} \rightarrow \mathbb{R}$ be the function defined by $f(x)=\log _{e}\left(\frac{x}{x-1}\right)$
and $g: E_{2} \rightarrow \mathbb{R}$ 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. | $(-\infty, 0) \cup\left(\frac{1}{2}, \frac{e}{e-1}\right]$ |

The correct option is:

Accepted Answer

(A) $\mathbf{P} \rightarrow 4 ; \mathbf{Q} \rightarrow 2 ; \mathbf{R} \rightarrow 1 ; \mathbf{S} \rightarrow \mathbf{1}$

Answer

(B) $\mathbf{P} \rightarrow \mathbf{3 ;} \mathbf{Q} \rightarrow \mathbf{3 ; R} \rightarrow \mathbf{6 ; S} \rightarrow \mathbf{5}$

Answer

(C) $\mathbf{P} \rightarrow \mathbf{4 ; ~ Q} \rightarrow \mathbf{2 ;} \mathbf{R} \rightarrow \mathbf{1} ; \mathbf{S} \rightarrow \mathbf{6}$

Answer

(D) $\mathbf{P} \rightarrow \mathbf{4 ; ~ Q} \rightarrow \mathbf{3 ; ~ R} \rightarrow \mathbf{6 ; ~ S} \rightarrow \mathbf{5}$

JEE Advanced · 2018 · 20 May 2018 · Paper 2 · Sets Relations and Functions

Solution

More JEE Advanced questions on Sets Relations and Functions

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.	$(- \infty, 0) \cup (\frac{1}{2}, \frac{e}{e - 1}]$