A) \[-\frac{1}{\sqrt{2}}<a<\frac{1}{\sqrt{2}}\]
B) all real values of a
C) \[\left| a \right|<\frac{1}{2}\]
D) \[\left| a \right|\ge \frac{1}{\sqrt{2}}\]
Correct Answer: A
Solution :
The range of \[{{\sin }^{-1}}x\] is \[\left[ -\frac{\pi }{2},\frac{\pi }{2} \right]\] Given, \[{{\sin }^{-1}}x=2{{\sin }^{-1}}a\] Since, \[-\frac{\pi }{2}2{{\sin }^{-1}}a\le \frac{\pi }{2}\] \[\Rightarrow \] \[-\frac{\pi }{2}\le 2{{\sin }^{-1}}a\le \frac{\pi }{2}\] \[\Rightarrow \] \[-\frac{\pi }{4}\le {{\sin }^{-1}}a\le \frac{\pi }{4}\] \[\Rightarrow \] \[\sin \left( -\frac{\pi }{4} \right)\le a\le \sin \frac{\pi }{4}\] \[\Rightarrow \] \[-\frac{1}{\sqrt{2}}\le a\le \frac{1}{\sqrt{2}}\Rightarrow \] \[\left| a \right|\le \frac{1}{\sqrt{2}}\]
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