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Question 139 of 177

Q.If sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1-x) - 2\sin^{-1}x = \dfrac{\pi}{2} then xx is

(a) −12-\dfrac{1}{2}
(b) 11
(c) 00
(d) 12\dfrac{1}{2}
Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2016MCQ· 2mImportance★★★★★
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Test x=0x=0 directly; it satisfies the domain and the equation exactly.

Given sin⁡−1(1−x)−2sin⁡−1x=π2\sin^{-1}(1-x) - 2\sin^{-1}x = \dfrac{\pi}{2}.

Domain check: we need −1≤1−x≤1⇒0≤x≤2-1\le 1-x\le 1 \Rightarrow 0\le x\le 2, and −1≤x≤1-1\le x\le 1. Combined domain: 0≤x≤10\le x\le 1.

Try x=0x=0:

sin⁡−1(1−0)−2sin⁡−1(0)=sin⁡−1(1)−0=π2−0=π2\sin^{-1}(1-0) - 2\sin^{-1}(0) = \sin^{-1}(1) - 0 = \frac{\pi}{2} - 0 = \frac{\pi}{2}

This satisfies the equation exactly.

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