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

  • question_answer
    The maximum value of \[f(x)=\frac{x}{4+x+{{x}^{2}}}\] on \[[-1,\,1]\] is                                                                       [MP PET 2000]

    A)            \[-1/4\]

    B)            \[-1/3\]

    C)            \[1/6\]

    D)            \[1/5\]

    Correct Answer: C

    Solution :

               \[f(x)=\frac{x}{4+x+{{x}^{2}}}\]            Differentiate, \[{f}'(x)=\frac{4+x+{{x}^{2}}-x(1+2x)}{{{(4+x+{{x}^{2}})}^{2}}}\]            For maximum \[f'(x)=0\] Þ \[\frac{4-{{x}^{2}}}{{{(4+x+{{x}^{2}})}^{2}}}=0\]            Þ  \[x=2,\,-2\]            Both values of x are out of interval            \  \[f(-1)=\frac{-1}{4-1+1}=\frac{-1}{4}\],                \[f(1)=\frac{1}{4+1+1}=\frac{1}{6}\] (maximum).


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