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JEE Main 202229 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
29 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2022 · 29 June · Morning Shift · Mathematics · Functions

The domain of the function ^ - 1 ( 2 ^ - 1 ( 1 4 x ^ 2 - 1 ) π ) is

The domain of the function cos⁡−1\cos^{- 1}\left(\dfrac{2\sin^{- 1}\left(\dfrac{1}{4{x}^{2} - 1}\right)}{\pi}\right) is

Options

  1. A(−∞,−12]∪[12,∞)∪{0}\left(- \infty , - \dfrac{1}{\sqrt{2}}\right] \cup \left[\dfrac{1}{\sqrt{2}}, \infty\right) \cup \left\{0\right\}
  2. B(−∞,−12]∪[12,∞)\left(- \infty , - \dfrac{1}{\sqrt{2}}\right] \cup \left[\dfrac{1}{\sqrt{2}}, \infty\right)
  3. C(−∞,−12)∪(12,∞)∪{0}\left(- \infty , - \dfrac{1}{\sqrt{2}}\right) \cup \left(\dfrac{1}{2}, \infty\right) \cup \left\{0\right\}
  4. DR−{−12,12}R - \left\{- \dfrac{1}{2}, \dfrac{1}{2}\right\}

Correct answer

A. (−∞,−12]∪[12,∞)∪{0}\left(- \infty , - \dfrac{1}{\sqrt{2}}\right] \cup \left[\dfrac{1}{\sqrt{2}}, \infty\right) \cup \left\{0\right\}

Step-by-step solution

f(x)=cos⁡−1(2sin⁡−1(14x2−1)π)f\left(x\right) = \cos^{- 1}\left(\dfrac{2\sin^{- 1}\left(\dfrac{1}{4{x}^{2} - 1}\right)}{\pi}\right)

Case I: −1≤14x2−1≤1- 1 \le \dfrac{1}{4{x}^{2} - 1} \le 1

⇒4x2−1≥1⇒ 4{x}^{2} - 1 \ge 1 or 4x2−1≤−14{x}^{2} - 1 \le - 1 and 4x2−1≠04{x}^{2} - 1 \ne 0

∴x2≥12∴ {x}^{2} \ge \dfrac{1}{2} or x2≤0⇒x=0{x}^{2} \le 0 ⇒ x = 0 and x≠±12x \ne \pm \dfrac{1}{2}

∴x∈(−∞,−12]∪[12,∞)∪{0}∴ x \in \left(- \infty , - \dfrac{1}{\sqrt{2}}\right] \cup \left[\dfrac{1}{\sqrt{2}}, \infty\right) \cup \left\{0\right\}

Case II: −1≤2sin⁡−1(14x2−1)π≤1- 1 \le \dfrac{2\sin^{- 1}\left(\dfrac{1}{4{x}^{2} - 1}\right)}{\pi} \le 1

⇒−π2≤sin⁡−1(14x2−1)≤π2⇒ - \dfrac{\pi}{2} \le \sin^{- 1}\left(\dfrac{1}{4{x}^{2} - 1}\right) \le \dfrac{\pi}{2} which is always true.

Hence, x∈(−∞,−12]∪[12,∞)∪{0}x \in \left(- \infty , - \dfrac{1}{\sqrt{2}}\right] \cup \left[\dfrac{1}{\sqrt{2}}, \infty\right) \cup \left\{0\right\}

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