Question

Let $E_{1} = \left\{x \in \mathbb R : x \ne1 and \frac{x}{x-1} > 0\right\}$ and $ E_{2} =\left\{x \in E_{1} : \sin^{-1}\left(\log_{e} \left(\frac{x}{x-1}\right)\right) is \, a \, real \, number\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) \le1 $
$ \frac{1}{e} \le \frac{x}{x+1} \le e $
Now $ \frac{x}{x-1} - \frac{1}{e} \ge0 $
$ \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