Question

Let $E_1=\{x\in R:x\ne 1\text{and}\frac{x}{x-1}>0\}$
and $E_2=\{x\in E_1:{\sin }^{-1}({\log }_e(\frac{x}{x-1}\left))\text{is a real number}\right\}.$
$($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 defined by $f\left(x\right)={\log }_e\left(\frac{x}{x-1}\right)$ and $g:E_2\to R$ be the function defined by$g\left(x\right)={\sin }^{-1}({\log }_e\left(\frac{x}{x-1}\right)).$
$$\begin{array}{cccc}& ListA& & ListB\\ P.& \text{The range of}f\text{is}& 1.& (-\infty ,\frac{1}{1-e}]\cup [\frac{e}{e-1},\infty )\\ Q.& \text{The range of}g\text{contains}& 2.& (0,1)\\ R.& \text{The domain of}f\text{contains}& 3.& [-\frac{1}{2},\frac{1}{2}]\\ S.& \text{The domain of}g\text{is}& 4.& (-\infty ,0)\cup (0,\infty )\\ & & 5.& (-\infty ,\frac{e}{e-1}]\\ & & 6.& (-\infty ,0)\cup (\frac{1}{2},\frac{e}{e-1}]\end{array}$$
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 $P\to 4;Q\to 2;R\to 1;S\to 1$

$E_1=\{x\in R:x\ne 1\text{and}\frac{x}{x-1}>0\}$
$E_1:\frac{x}{x-1}>0$
$\Rightarrow x\in (-\infty ,0)\cup (1,\infty )$

$E_2=\{x\in E_1:{\sin }^{-1}({\log }_e(\frac{x}{x-1}\left))\text{is a real number}\right\}.$
$E_2:-1\le \ln \left(\frac{x}{x-1}\right)\le 1\frac{1}{e}\le \frac{x}{x-1}\le e$
Now, $\frac{x}{x-1}-\frac{1}{e}\ge 0\Rightarrow \frac{x(e-1)+1}{e(x-1)}\ge 0$


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


$x\in (-\infty ,1]\cup [\frac{e}{e-1},\infty )$
So, intersection:$\Rightarrow x\in (-\infty ,\frac{1}{1-e}]\cup [\frac{e}{e-1},\infty )$
Hence, the domain of $f,g$ is contained in $(-\infty ,0)\cup (1,\infty )$
and Range of $f$ is $R-\left\{0\right\}\text{or,}(-\infty ,0)\cup (0,\infty )$
Range of $g$ is $[-\frac{\pi }{2},\frac{\pi }{2}]-\left\{0\right\}\text{or,}[-\frac{\pi }{2},0)\cup (0,\frac{\pi }{2}]$