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JEE Advanced 2018 Paper 2

Section: Mathematics

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Question : 51 of 54

Marks: +1, -0

E_1=\{X \in \mathbb{R}: x \neq 1 \text{ and } \frac{x}{x-1} > 0\}

E_2=\{X \in R_1: \sin^{-1}\left(\text{log}_e\left(\frac{x}{x-1}\right)\right) \text{ is a real number}\}

.(Here the inverse trigonometric function function

\sin^{-1}

assume values in

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

f: E_1 \rightarrow \mathbb{R}

be the function defined by

f(x)=\text{log}_e\left(\frac{x}{x-1}\right)

g: E_2 \rightarrow \mathbb{R}

be the function defined by

g(x)=\sin^{-1}\left(\text{log}\left(\frac{x}{x-1}\right)\right)

List - I	List - II
P. The range of f is	1. $\left(-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty\right)$
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. $(-\infty, 0] \cup (0, \infty)$
	5. $(-\infty, \frac{e}{e-1}]$
	6. $(-\infty, 0) \cup (\frac{1}{2}, \frac{e}{e-1}]$

The correct options is:

	P	Q	R	S
A)	4	2	1	1
b)	3	3	6	5
c)	4	2	1	6
D)	4	3	6	5

[JEE Adv 2018 P2]

A
B
C
D

Solution:

E_1: \frac{x}{x-1} > 0

\Rightarrow E_1: x \in (-\infty, 0) \cup (1, \infty)

E_2: -1 \leq \text{ln}\left(\frac{x}{x+1}\right) \leq 1

\frac{1}{e} \leq \frac{x}{x-1} \leq e

Now

\frac{x}{x-1} = -\frac{1}{e} \geq 0

\Rightarrow \frac{(e-1)x+1}{e(x+1)} \geq 0

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

also

\frac{x}{x-1} - e \leq 0

\frac{(e-1)x-e}{x-1} \geq 0

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

So

E_2: \left(-\infty, \frac{1}{1-e}\right) \cup \left[\frac{e}{e-1}, \infty\right]

as Range of

\frac{x}{x-1}

is

R^{+} - \{1\}

⇒ Range of f is

R - \{0\}

or

(-\infty, 0) \cup (0, \infty)

Range of g is

\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \setminus \{0\}

or

[-\frac{\pi}{2}, 0) \cup \left(0, \frac{\pi}{2}\right)

Now P→4,Q→2,R→1,S→1

Hence a is correct

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