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JEE Main & Advanced Sample Paper JEE Main Sample Paper-42

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

    A)  x = 0                         

    B)  x = 1

    C)  x = 4                         

    D)  x = 3

    Correct Answer: A

    Solution :

    Let\[f(x)={{x}^{2}}+\frac{1}{1+{{x}^{2}}}\] On differentiating w.r.t. x, we get \[f'(x)=2x-\frac{1}{{{(1+{{x}^{2}})}^{2}}}.2x\] For a minimum put \[f'(x)=0\Rightarrow x=0\] So, the function has minimum value at x = 0.


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