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

  • question_answer
    The minimum value of \[2{{x}^{2}}+x-1\]is    [EAMCET 2003]

    A)  \[-\frac{1}{4}\]

    B)            \[\frac{3}{2}\]

    C)   \[\frac{-9}{8}\]

    D)            \[\frac{9}{4}\]

    Correct Answer: C

    Solution :

               \[f(x)=2{{x}^{2}}+x-1\]                    Þ \[f'(x)=4x+1\Rightarrow f'(x)=0\Rightarrow x=-\frac{1}{4}\]                     \[{f}''\,(x)=4=+ve\]                    \[\therefore {{[f(-1/4)]}_{\min }}=\frac{2}{16}-\frac{1}{4}-1=\frac{-9}{8}\].


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