JEE Advanced 2018 · Paper 2 · Mathematics Q15

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MathematicsList match+3 / -1Question 51 of 54

Let $E_1 = \left\{x \in \mathbb{R} : x \ne 1 \text{ and } \frac{x}{x-1} > 0\right\}$ and $E_2 = \left\{x \in E_1 : \sin^{-1}\left(\log_e\left(\frac{x}{x-1}\right)\right) \text{ is a real number}\right\}$ .

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

Let $f: E_1 \to \mathbb{R}$ be the function defined by $f(x) = \log_e\left(\frac{x}{x-1}\right)$ and $g: E_2 \to \mathbb{R}$ be the function defined by $g(x) = \sin^{-1}\left(\log_e\left(\frac{x}{x-1}\right)\right)$ .

The correct option is:

LIST-I

P
The range of $f$ is
Q
The range of $g$ contains
R
The domain of $f$ contains
S
The domain of $g$ is

LIST-II

1
$\left(-\infty, \frac{1}{1-e}\right] \cup \left[\frac{e}{e-1}, \infty\right)$
2
$(0, 1)$
3
$\left[-\frac{1}{2}, \frac{1}{2}\right]$
4
$(-\infty, 0) \cup (0, \infty)$
5
$\left(-\infty, \frac{e}{e-1}\right]$
6
$(-\infty, 0) \cup \left(\frac{1}{2}, \frac{e}{e-1}\right]$

P → 4; Q → 2; R → 1; S → 1

P → 3; Q → 3; R → 6; S → 5

P → 4; Q → 2; R → 1; S → 6

P → 4; Q → 3; R → 6; S → 5

Answer and worked solution

Official answer: A

Domain $E_1$ . $\dfrac x{x-1}>0$ when $x<0$ or $x>1$ . Let $t=\dfrac x{x-1}$ . Then $t\in(0,1)$ for $x<0$ and $t\in(1,\infty)$ for $x>1$ .

P. The range of $f=\ln t$ is $(-\infty,0)\cup(0,\infty)$ : 4.
S. The domain of $g$ needs $-1\le\ln t\le1$ , i.e. $\dfrac1e\le t\le e$ .
- For $x<0$ : $\dfrac x{x-1}\ge\dfrac1e\iff x\le\dfrac1{1-e}$ .
- For $x>1$ : $\dfrac x{x-1}\le e\iff x\ge\dfrac e{e-1}$ .
So the domain is $\left(-\infty,\dfrac1{1-e}\right]\cup\left[\dfrac e{e-1},\infty\right)$ : 1.
R. The domain of $f$ is $E_1$ , which contains the set in 1: 1.
Q. The range of $g$ is $\sin^{-1}$ of $[-1,0)\cup(0,1]$ , i.e. $[-\sin^{-1}1,0)\cup(0,\sin^{-1}1]$ . This contains $(0,1)$ : 2.

P → 4; Q → 2; R → 1; S → 1.

(A)

Topics:Domain and range Inverse trigonometric functions Inequalities and modulus

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JEE Advanced 2018 · Paper 2, Mathematics question 15