Question

Let $E_1=\{xϵR:x\ne 1and\frac{x}{x-1}>0\}$
and $E_2=\{xϵE_1:{\sin }^{-1}\left({\log }_e\left(\frac{x}{x-1}\right)\right)\text{is a real number}\}$.
(Here, the inverse trigonometric function ${\sin }^{-1}x$ assumes values in $[-\frac{\pi }{2},\frac{\pi }{2}]$)
Let $f:E_1\to R$ be the function define by $f\left(x\right)={\log }_e\left(\frac{x}{x-1}\right)$ and $g:E_2\to R$ be the function defined by $g\left(x\right)={\sin }^{-1}\left({\log }_e\left(\frac{x}{x-1}\right)\right)$.

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 $g$ contins	$2.(0,1)$
R. The domain of $f$ contains	$3.[-\frac{1}{2},\frac{1}{2}]$
S. The domain of $g$ is	$4.(-\infty ,0)\cup (0,\infty )$
	$5.(-\infty ,\frac{e}{e-1}]$
	$6.(-\infty ,0)\cup (\frac{1}{2},\frac{e}{e-1}]$

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

The correct option is A $P\to 4;Q\to 2;R\to 1;S\to 1$

$E_1:\frac{x}{x-1}>0$
$\Rightarrow E_1:xϵ(-\infty ,0)\cup (1,\infty )$

$E_2:-1\le ln\left(\frac{x}{x-1}\right)\le 1$
$\frac{1}{e}\le \frac{x}{x-1}\le e$

Now $\frac{x}{x-1}-\frac{1}{e}\ge 0$
$\Rightarrow \frac{(e-1)x+1}{e(x-1)}\ge 0$

$\Rightarrow xϵ(-\infty ,\frac{1}{1-e}]\cup (1,\infty )$
also $\frac{x}{x-1}-e\le 0$

$\frac{(e-1)x-e}{x-1}\ge 0$

$\Rightarrow xϵ(-\infty ,1)\cup [\frac{e}{e-1},\infty ]$

So, $E_2:(-\infty ,\frac{1}{1-e}]\cup [\frac{e}{e-1},\infty ]$

as Range of $\frac{x}{x-1}$ is $R^{+}-\left\{1\right\}$
$\Rightarrow$ Range of $f$ is $R-\left\{0\right\}$ or $(-\infty ,0)\cup (0,\infty )$

Range of $g$ is $[\frac{-\pi }{2},\frac{\pi }{2}]\cup \left\{0\right\}$ or $[-\frac{\pi }{2},0)\cup (0,\frac{\pi }{2}]$

Now $P\to 4;Q\to 2;R\to 1;S\to 1$
Hence $A$ is correct.