JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer Let\[f(x)=1/(x-1)\]and\[g(x)=1/({{x}^{2}}+x-2)\]. Then the set of points where \[(gof)(x)\] is discontinuous, is

    A) \[\{1\}\]                                   

    B) \[\{-2,\,\,1\}\]

    C) \[\{1/2,\,\,1,\,\,2\}\]                  

    D) \[\{1/2,\,\,1\}\]

    Correct Answer: C

    Solution :

     \[f(x)=\frac{1}{x-1}\]is discontinuous at\[x=1\]. \[(gof)(x)=g(f(x))=\frac{{{(x-1)}^{2}}}{(2x-1),\,\,(x-2)},\]which is not defined at\[x=1/2,\,\,2\].. Hence the set of points where \[(gof)(x)\] is discontinuous is\[\{1/2,\,\,1,\,\,2\}\]

adversite


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