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JEE Main 20231 FebruaryMorning Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main 2023 · 1 February · Morning Shift · Mathematics · Inverse Trigonometric Functions

Let S be the set of all solutions of the equation ^ - 1 (2x) - 2 ^...

Let SS be the set of all solutions of the equation cos⁡−1(2x)−2cos⁡−1(1−x2)=π,x∈[−12,12]\cos^{- 1}\left(2x\right) - 2\cos^{- 1}\left(\sqrt{1 - {x}^{2}}\right) = \pi, x \in \left[- \dfrac{1}{2}, \dfrac{1}{2}\right]. Then ∑x∈S2sin⁡−1(x2−1)\mathop{∑}_{x \in S}2\sin^{- 1}\left({x}^{2} - 1\right) is equal to

Options

  1. A00
  2. B−2π3\dfrac{- 2\pi}{3}
  3. Cπ−sin⁡−1(34)\pi - \sin^{- 1}\left(\dfrac{\sqrt{3}}{4}\right)
  4. Dπ−2sin⁡−1(34)\pi - 2\sin^{- 1}\left(\dfrac{\sqrt{3}}{4}\right)

Correct answer

A. 00

Step-by-step solution

Given equation is:

cos⁡−1(2x)−2cos⁡−1(1−x2)=π,x∈[−12,12]\cos^{- 1}\left(2x\right) - 2\cos^{- 1}\left(\sqrt{1 - {x}^{2}}\right) = \pi, x \in \left[- \dfrac{1}{2}, \dfrac{1}{2}\right]

It is possible if

cos⁡−1(2x)=π\cos^{- 1}\left(2x\right) = \pi ....(1)

and 2cos⁡−1(1−x2)=02\cos^{- 1}\left(\sqrt{1 - {x}^{2}}\right) = 0 ....(2)

From equation (1), x=−12x = - \dfrac{1}{2} which is not satisfying equation (2).

So, no such xx exists.

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