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JEE Main & Advanced Mathematics Functions Question Bank Critical Thinking

  • question_answer
    The function \[f(x)={{[x]}^{2}}-[{{x}^{2}}]\], (where [y] is the greatest integer less than or equal to y),is discontinuous at [IIT 1999]

    A) All integers                               

    B) All integers except 0 and 1

    C) All integers except 0

    D) All integers except 1

    Correct Answer: D

    Solution :

    • Given \[f(x)={{[x]}^{2}}-[{{x}^{2}}]\]                   
    • \[-1<x<0,\,\,f(x)={{(-1)}^{2}}-0=1\]           
    • \[x=0,\,\,f(x)={{0}^{2}}-0=0\]           
    • \[0<x<1,\,\,f(x)={{0}^{2}}-0=0\]           
    • \[x=1,\,\,f(x)={{1}^{2}}-1=0\]           
    • \[1<x<\sqrt{2},\,\,f(x)={{1}^{2}}-1=0\]           
    • \[x=\sqrt{2},\,\,f(x)={{1}^{2}}-2=-1\]           
    • \[\sqrt{2}<x<\sqrt{3},\,\,f(x)={{1}^{2}}-2=-1\]           
    • \[x=\sqrt{3},\,\,f(x)={{1}^{2}}-3=-2\]          
    • \[\sqrt{3}<x<2,\,\,f(x)={{1}^{2}}-3=-2\]           
    • \[x=2,\,\,f(x)=4-4=0\]; \[2<x<\sqrt{5},\,\,f(x)=4-4=0\]           
    • \[x=\sqrt{5},\,\,f(x)=4-5=-1\]                   
    • Hence function is discontinuous at all integers except 1.


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