CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    The domain of\[{{\sin }^{-1}}\left[ {{\log }_{2}}\left( \frac{x}{12} \right) \right]\]is

    A)  \[[2,12]\]           

    B)         \[[-1,1]\]

    C)  \[\left[ \frac{1}{3},24 \right]\]  

    D)         \[\left[ \frac{2}{3},24 \right]\]

    E)  \[[6,24]\]

    Correct Answer: E

    Solution :

    \[\because \] \[-1\le {{\sin }^{-1}}x\le 1\] \[\Rightarrow \]               \[-1\le {{\log }_{2}}\frac{x}{12}\le 1\] \[\Rightarrow \]               \[{{2}^{-1}}\le \frac{x}{12}\le {{2}^{1}}\] \[\Rightarrow \]               \[6\le x\le 24\] \[\Rightarrow \] Required domain = [6, 24]


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