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JEE Main & Advanced Mathematics Functions Question Bank Self Evaluation Test - Relation and Functions-II

  • question_answer
    Let \[f(x)=\frac{x}{1+{{x}^{2}}}\] and \[g(x)=\frac{{{e}^{-x}}}{1+[x],}\], where \[[x]\]is the greatest integer less than or equal to x, then

    A) \[D(f+g)=R-[-2,0)\]

    B) \[D(f+g)=R-[-1,0)\]

    C) \[R(f)\cap R(g)=\left[ -2,\frac{1}{2} \right]\]

    D) None of these

    Correct Answer: D

    Solution :

    [d] \[D(f)=R;D(g)=R-[-1,0)\] \[\therefore D(f+g)=D(f)\cap D(g)=R\cap (R-[-1,0)=R\cap [-1,0)\]\[R(f)=\left[ -\frac{1}{2},\frac{1}{2} \right];R(g)=R-\{0\}\] \[\therefore \,\,\,\,R(f)\cap R(g)=\left[ -\frac{1}{2},\frac{1}{2} \right]\cap (R-\{0\})\] \[=\left[ -\frac{1}{2},\frac{1}{2} \right]-\{0\}\]


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