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JEE Main & Advanced Mathematics Functions Question Bank Self Evaluation Test - Relations and Functions-I

  • question_answer
    The domain of the function \[f(x)=lo{{g}_{2}}\left( -{{\log }_{1/2}}\left( 1+\frac{1}{{{x}^{1/4}}} \right)-1 \right)\] is

    A) \[\left( 0,\text{ }1 \right)\]

    B)        \[\left( 0,\text{ }1 \right]\]

    C) \[[1,\infty )\]

    D)        \[(1,\infty )\]

    Correct Answer: A

    Solution :

    [a] \[f(x)\] is defined if \[-{{\log }_{1/2}}\left( 1+\frac{1}{{{x}^{1/4}}} \right)-1>0\] \[\Rightarrow {{\log }_{1/2}}\left( 1+\frac{1}{{{x}^{1/4}}} \right)<-1\] \[\Rightarrow 1+\frac{1}{{{x}^{1/4}}}>{{\left( \frac{1}{2} \right)}^{-1}}\] \[\Rightarrow \frac{1}{{{x}^{1/4}}}>1\Rightarrow 0<x<1\]


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