Last updated at May 29, 2018 by Teachoo

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Misc 16 Solve sin-1(1 – x) – 2sin-1 x = π/2 , then x is equal to (A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2 sin-1 (1 – x) – 2sin-1 x = π/2 – 2sin-1 x = π/2 – sin-1 (1 – x) − 2sin-1 x = cos-1 (1 – x) − 2sin-1 x = cos-1 (1 – x) …(1) Let sin-1 x = a, Hence our equation becomes − 2a = cos-1 (1 – x) cos (−2a) = 1 – x cos (2a)= (1 – x) 1 – 2 sin2 a = 1 – x We assumed that sin-1 x = a 1 – 2 [sin(sin-1 x)]2 = 1 – x 1 – 2x2 = 1 – x 1 – 2x2 – 1 + x = 0 1 – 1 – 2x2 + x = 0 – 2x2 + x = 0 0 = 2x2 – x 2x2 – x = 0 x (2x – 1) = 0 x = 0 and 2x – 1 = 0 x = 0 and x = 1/2 But x = 𝟏/𝟐 does not satisfy the equation Taking equation sin-1(1 – x) – 2sin-1 x = π/2 Putting x = 1/2 in L.H.S sin-1(1− 1/2) – 2 sin-1 (1/2) = sin-1(1/2) – 2 sin-1 (1/2) = 𝜋/6 – 2 × 𝜋/6 = (𝜋 − 2𝜋)/6 = (− 𝜋)/6 ≠ 𝜋/2 Hence x = 1/2 not possible ∴ x = 0 is the only solution Option C is correct Answer

Chapter 2 Class 12 Inverse Trigonometric Functions

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.