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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Maxima and Minima

  • question_answer
    The minimum value of \[{{x}^{2}}+\frac{1}{1+{{x}^{2}}}\] is at    [UPSEAT 2003]

    A)            \[x=0\]

    B)            \[x=1\]

    C)            \[x=4\]

    D)            \[x=3\]

    Correct Answer: A

    Solution :

               \[f(x)={{x}^{2}}+\frac{1}{1+{{x}^{2}}}\], \[{f}'(x)=2x-\frac{1}{{{(1+{{x}^{2}})}^{2}}}\,.\,2x\]            Now \[{f}'(x)=0\] Þ \[x=0\]                    So the function has minimum value at \[x=0\].


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