Let E 1 = { x : x 1 and x x - 1 0 } and E 2 = { x E 1 : sin - 1 ( log e ( x x - 1 ) ) is a real number } . (Here, the inverse trigonometric function sin - 1 x assumes values in [ - 2 , 2 ] ). Let f : E 1 be the function defined by f ( x

Question

Let $E_{1} = {x \in ℝ : x \neq 1 and \frac{x}{x - 1} > 0}$ and

$E_{2} = {x \in E_{1} : \sin^{- 1} (\log_{e} (\frac{x}{x - 1})) is a real number} .$

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

Let $f : E_{1} \to ℝ$ be the function defined by $f (x) = \log_{e} (\frac{x}{x - 1})$ and $g : E_{2} \to ℝ$ be the function defined by $g (x) = \sin^{- 1} (\log_{e} (\frac{x}{x - 1})) .$ [2018]

	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 $g$ contains	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:

Smart Achievers · Accepted Answer

(1)

For $E_{1}$ , $\frac{x}{x - 1} > 0$ and $x \neq 1$ $\Rightarrow x \in (- \infty, 0) \cup (1, \infty)$

For $E_{2}$ , $- 1 \leq \log_{e} (\frac{x}{x - 1}) \leq 1$ $\Rightarrow \frac{1}{e} \leq \frac{x}{x - 1} \leq e$

$\Rightarrow \frac{1}{e} \leq 1 + \frac{1}{x - 1} \leq e$ $\Rightarrow \frac{1}{e} - 1 \leq \frac{1}{x - 1} \leq e - 1$

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

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

For $E_{1}$ , $\frac{x}{x - 1} \in (0, \infty) - {1}$

$\Rightarrow \log_{e} (\frac{x}{x - 1}) \in (- \infty, \infty) - {0}$

$\Rightarrow f (x) \in (- \infty, 0) \cup (0, \infty)$

$g (x) = \sin^{- 1} (\log_{e} (\frac{x}{x - 1})) \in [- \frac{π}{2}, \frac{π}{2}] - {0}$

Solution

1 P → 4; Q → 2; R → 1; S → 1

2 P → 3; Q → 3; R → 6; S → 5

3 P → 4; Q → 2; R → 1; S → 6

4 P → 4; Q → 3; R → 6; S → 5

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