Question

Let $E_1=\left\{x\in \mathbb{R}:x\ne 1and\frac{x}{x-1}>0\right\}$ and $E_2=\left\{x\in E_1:{\sin }^{-1}\left({\log }_e\left(\frac{x}{x-1}\right)\right)isarealnumber\right\}$.
(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 $f(x)={\log }_e(\frac{x}{x-1})$
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 f 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:

• A. P $\to$ 4; Q $\to$ 2; R $\to$ 1; S $\to$ 1
• B. P $\to$ 3; Q $\to$ 3; R $\to$ 6; S $\to$ 5
• C. P $\to$ 4; Q $\to$ 2; R $\to$ 1; S $\to$ 6
• D. P $\to$ 4; Q $\to$ 3; R $\to$ 6; S $\to$ 5

Answer 1

Answer: (A)

$E_1:\frac{x}{x-1}>0$
$\Rightarrow E_1:x\in \left(-\infty ,0\right)\cup \left(1,\infty \right)$
$E_2:-1\le ln\left(\frac{x}{x+1}\right)\le 1$
$\frac{1}{e}\le \frac{x}{x+1}\le e$
Now $\frac{x}{x-1}-\frac{1}{e}\ge 0$
$\Rightarrow \frac{\left(e-1\right)x+1}{e\left(x-1\right)}\ge 0$
Now P $\to$ 4, Q $\to$ 2, R $\to$ 1 , S $\to$ 1
Hence A is correct