Sets, Relations and Functions

Let ${E}_{1}=\left\{x\in \mathrm{\mathbb{R}}:x\ne 1\mathrm{and}\frac{x}{x-1}0\right\}$ and ${E}_{2}=\left\{x\in {E}_{1}:{\mathrm{sin}}^{-1}\left({\mathrm{log}}_{e}\left(\frac{x}{x-1}\right)\right)\mathrm{is}\mathrm{a}\mathrm{real}\mathrm{number}\right\}.$

$\left(\mathrm{Here},\mathrm{the}\mathrm{inverse}\mathrm{trigonometric}\mathrm{function}{\mathrm{sin}}^{-1}x\mathrm{assumes}\mathrm{values}\mathrm{in}\left[-\frac{\mathrm{\pi}}{2},\frac{\mathrm{\pi}}{2}\right].\right)$

Let

and

The correct option is:

$\left(\mathrm{Here},\mathrm{the}\mathrm{inverse}\mathrm{trigonometric}\mathrm{function}{\mathrm{sin}}^{-1}x\mathrm{assumes}\mathrm{values}\mathrm{in}\left[-\frac{\mathrm{\pi}}{2},\frac{\mathrm{\pi}}{2}\right].\right)$

Let

*f*:*E*_{1}→ ℝ be the function defined by $f\left(x\right)={\mathrm{log}}_{e}\left(\frac{x}{x-1}\right)$and

*g*:*E*_{2}→ ℝ be the function defined by $g\left(x\right)={\mathrm{sin}}^{-1}\left({\mathrm{log}}_{e}\left(\frac{x}{x-1}\right)\right).$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 g 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. | (–∞, 0) ∪ (0, ∞) |

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:

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JEE Advanced 2018