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NTA Abhyas JEE Main2020MathematicsInverse Trigonometric FunctionsPractice

The value(s) of x satisfying the equation sin - 1 ⁡ 1 - x - 2 sin - 1 ⁡ x = π 2 is/are

Options

  1. A0
  2. B1 2
  3. C0 , 1 2
  4. D- 1 2

Correct answer

A. 0

Step-by-step solution

We have, sin - 1 ⁡ 1 - x - 2 sin - 1 ⁡ x = π 2 ⇒ sin - 1 ⁡ 1 - x = π 2 + 2 sin - 1 ⁡ x ⇒ 1 - x = sin ⁡ π 2 + 2 sin - 1 ⁡ x ⇒ 1 - x = cos ⁡ 2 sin - 1 ⁡ x ⇒ 1 - x = cos ⁡ cos - 1 ⁡ 1 - 2 x 2 ∵ 2 sin - 1 ⁡ x = cos - 1 ⁡ 1 - 2 x 2 ⇒ 1 - x = 1 - 2 x 2 ⇒ x = 2 x 2 ⇒ x 2 x - 1 = 0 ⇒ x = 0 , 1 2 For, x = 1 2 , we have LHS = sin - 1 ⁡ 1 - x - 2 sin - 1 ⁡ x = sin - 1 ⁡ 1 2 - 2 sin - 1 ⁡ 1 2 = - sin - 1 ⁡ 1 2 = - π 6 ≠ R . H . S . So, x = 1 2 is not a root of the given equation. Clearly, x = 0 satisfies the equation Here, x =

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