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JEE Main & Advanced Mathematics Functions Question Bank Functions

  • question_answer
    \[f(x)=x+\sqrt{{{x}^{2}}}\] is a function from R\[\to \]R , then \[f(x)\] is [Orissa JEE 2004]           

    A)            Injective

    B)            Surjective

    C)            Bijective

    D)            None of these

    Correct Answer: D

    Solution :

               We have \[f(x)=x+\sqrt{{{x}^{2}}}=x+|x|\]            Clearly f is not one-one as \[f(-1)=f(-2)=0\] but \[-1\ne 2\].            Also f is not onto as \[f(x)\ge 0,\,\forall x\in R,\]            Also, range of \[f=(0,\infty )\subset R\].


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