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Q. The domain of $ {{\sin }^{-1}}\left( {{\log }_{3}}x \right) $ is:

KEAMKEAM 2005

Solution:

For domain of $ {{\sin }^{-1}}\left( {{\log }_{3}}x \right) $
$ \therefore $ $ -1\le {{\log }_{3}}x\le 1 $
$ \Rightarrow $ $ {{3}^{-1}}\le x\le 3 $
$ \therefore $ Domain of $ {{\sin }^{-1}}({{\log }_{3}}x)=\left[ \frac{1}{3},3 \right] $