A) 1, 0
B) 1,-1
C) \[0,\frac{1}{2}\]
D) None of these
Correct Answer: C
Solution :
\[{{\sin }^{-1}}(1-x)=\left( \frac{\pi }{2}-{{\sin }^{-1}}x \right)-{{\sin }^{-1}}x\] \[(\because {{\cos }^{-1}}x=\frac{\pi }{2}-{{\sin }^{-1}}x)\] \[{{\sin }^{-1}}(1-x)=\frac{\pi }{2}-2{{\sin }^{-1}}x\] Taking sum of both sides \[1-x=\sin \left( \frac{\pi }{2}-2{{\sin }^{-1}}x \right)=(2{{\sin }^{-1}}x)\] \[=\cos 2\theta ,\]where \[{{\sin }^{-1}}x=\theta \] \[1-x=1-2{{\sin }^{2}}\theta =1-2{{x}^{2}}\]or \[x(1-2x)=0\]or \[x=0,\frac{1}{2}\]
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