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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Rolle's theorem Lagrange's mean value theorem

  • question_answer
    From mean value theorem \[f(b)-f(a)=\] \[(b-a)f'({{x}_{1}});\] \[a<{{x}_{1}}<b\]if \[f(x)=\frac{1}{x}\], then \[{{x}_{1}}=\] [MP PET 1996]

    A)            \[\sqrt{ab}\]

    B)            \[\frac{a+b}{2}\]

    C)            \[\frac{2ab}{a+b}\]

    D)            \[\frac{b-a}{b+a}\]

    Correct Answer: A

    Solution :

               \[f'({{x}_{1}})=\frac{-1}{x_{1}^{2}}\]                    \[\therefore \frac{-1}{x_{1}^{2}}=\frac{\frac{1}{b}-\frac{1}{a}}{b-a}=-\frac{1}{ab}\Rightarrow {{x}_{1}}=\sqrt{ab}\].


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