Let $E_1=\{x\in R:x\ne 1$ and $\frac{x}{x-1}>0\}$and $E_2=\{x\in E_1:{\sin }^{-1}\left({\left(\log \right)}_e\left(\frac{x}{x-1}\right)\right)$ is a real number}.Here, the inverse trigonometric function ${\sin }^{-1}x$assumes values in $\left[\frac{\pi }{2},\frac{\pi }{2}\right]$Let $f:E_1\to R$be the function defined by $f\left(x\right)={\left(\log \right)}_e\left(\frac{x}{x-1}\right)$and $g:E_2\to R$be the function defined by $g\left(x\right)={\sin }^{-1}\left({\left(\log \right)}_e\left(\frac{x}{x-1}\right)\right)$LIST-I LIST-IIP. 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. $\left(0,1\right)$R. The domain of $f$contains 3. $\left[\frac{1}{2},\frac{1}{2}\right]$S. The domain of $g$is 4. $\left(-\infty ,0\right)\cup \left(0,\infty \right)$5. $\left(-\infty ,\frac{e}{e-1}\right]$6. $\left(-\infty ,0\right)\cup \left(\frac{1}{2},\frac{e}{e-1}\right]$The correct option is:$P\to 4;\to 2;R\to 1;S\to 1$(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) $P\to 4;Q\to 3;R\to 6;S\to 5$

Views: 759 students