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JEE Main & Advanced Mathematics Functions Question Bank Continuity

  • question_answer
    The function \[f(x)=\frac{2{{x}^{2}}+7}{{{x}^{3}}+3{{x}^{2}}-x-3}\]is discontinuous for [J&K 2005]

    A)            \[x=1\] only                               

    B)            \[x=1\] and \[x=-1\] only

    C)            \[x=1,x=-1,x=-3\] only

    D)            \[x=1,x=-1,x=-3\] and other values of x

    Correct Answer: C

    Solution :

               \[f(x)=\frac{2{{x}^{2}}+7}{{{x}^{2}}(x+3)-1(x+3)}\frac{9{{x}^{2}}+7}{({{x}^{2}}-1)(x+3)}\]                           \[=\frac{2{{x}^{2}}+7}{(x-1)(x+1)(x+3)}\]                    Hence points of discontinuity are                    \[x=1\], \[x=-1\] and \[x=-3\] only.


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