Verified Solution • 2025–26 Session moderate

QJEEAdv18Ma929c: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Ma929c Solution: Let and E 2 = { x ∈ E 1 : sin − 1 ( log e ( x x − 1 ) ) i s a r …

Let {E_1} = \left\{ {x \in R:x \ne 1\,and\,{x \over {x - 1}} > 0} \right\} and


E2={x∈E1:sin−1(loge(xx−1))isarealnumber}

(Here, the inverse trigonometric function sin⁡−1{\sin ^{ - 1}} x assumes values in [−π2,π2]\left[ { - {\pi \over 2},{\pi \over 2}} \right] .).

Let f : E1 →\to R be the function defined by f(x) = log⁡e(xx−1){{{\log }_e}\left( {{x \over {x - 1}}} \right)} and g : E2 →\to R be the function defined by g(x) = sin⁡−1(log⁡e(xx−1)){\sin ^{ - 1}}\left( {{{\log }_e}\left( {{x \over {x - 1}}} \right)} \right) .
LIST-I LIST-II
P. The range of ff is 1. (−∞,11−e]∪[ee−1,∞)\left( -\infty, \frac{1}{1-e} \right] \cup \left[ \frac{e}{e-1}, \infty \right)
Q. The range of gg contains 2. (0,1)(0, 1)
R. The domain of ff contains 3. [−12,12]\left[ -\frac{1}{2}, \frac{1}{2} \right]
S. The domain of gg is 4. (−∞,0)∪(0,∞)(-\infty, 0) \cup (0, \infty)
5. (−∞,ee−1)\left( -\infty, \frac{e}{e-1} \right)
6. (−∞,0)∪(12,ee−1](-\infty, 0) \cup \left( \frac{1}{2}, \frac{e}{e-1} \right]
The correct option is :

Multiple Choice Options

  • A P →\to 4; Q →\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

Step-by-Step Verified Solution

We have,

{E_1} = \left\{ {x \in R:x \ne 1\,and\,{x \over {x - 1}} > 0} \right\}

∴\therefore {E_1} = {x \over {x - 1}} > 0

JEE Advanced 2018 Paper 2 Offline Mathematics - Functions Question 22 English Explanation 1

E1=x∈(−∞,0)∪(1,∞){E_1} = x \in ( - \infty ,0) \cup (1,\infty )

and E2={x∈E1:sin−1(loge(xx−1))isarealnumber}

E2=−1≤log⁡exx−1≤1{E_2} = - 1 \le {\log _e}{x \over {x - 1}} \le 1

⇒e−1≤xx−1≤e\Rightarrow {e^{ - 1}} \le {x \over {x - 1}} \le e

Now, xx−1≥e−1⇒xx−1−1e≥0{x \over {x - 1}} \ge {e^{ - 1}} \Rightarrow {x \over {x - 1}} - {1 \over e} \ge 0

⇒ex−x+1e(x−1)≥0⇒x(e−1)+1(x−1)e≥0\Rightarrow {{ex - x + 1} \over {e(x - 1)}} \ge 0 \Rightarrow {{x(e - 1) + 1} \over {(x - 1)e}} \ge 0

JEE Advanced 2018 Paper 2 Offline Mathematics - Functions Question 22 English Explanation 2

⇒x∈(−∞, 11−e)∪(1,∞)\Rightarrow x \in \left( { - \infty ,\,{1 \over {1 - e}}} \right) \cup (1,\infty )

Also, xx−1≤e{x \over {x - 1}} \le e

⇒(e−1)x−ex−1≥0\Rightarrow {{(e - 1)x - e} \over {x - 1}} \ge 0

JEE Advanced 2018 Paper 2 Offline Mathematics - Functions Question 22 English Explanation 3

⇒x∈(−∞, 1)∪[ee−1,∞)\Rightarrow x \in ( - \infty ,\,1) \cup \left[ {{e \over {e - 1}},\infty } \right)

So, E2=(−∞, 11−e]∪[ee−1, ∞){E_2} = \left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)

∴\therefore The domain of f and g are

(−∞, 11−e]∪[ee−1, ∞)\left( { - \infty ,\,{1 \over {1 - e}}} \right] \cup \left[ {{e \over {e - 1}},\,\infty } \right)

and Range of xx−1{x \over {x - 1}} is R+ −- {1}

⇒\Rightarrow Range of f is R −- {0} or ( −- ∞\infty , 0) ∪\cup (0, ∞\infty )

Range of g is [−π2,π2]−{0}\left[ { - {\pi \over 2},{\pi \over 2}} \right] - \{ 0\} or [−π2,0)∪(0,π2]\left[ { - {\pi \over 2},0} \right) \cup \left( {0,{\pi \over 2}} \right]

Now, P →\to 4, Q →\to 2, R →\to 1, S →\to 1

Hence, option (a) is correct answer.

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