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JEE Main & Advanced Mathematics Applications of Derivatives Question Bank Increasing and Decreasing Function

  • question_answer
    Given function \[f(x)=\left( \frac{{{e}^{2x}}-1}{{{e}^{2x}}+1} \right)\] is [Orissa JEE 2005]

    A)            Increasing

    B)            Decreasing

    C)            Even

    D)            None of these

    Correct Answer: A

    Solution :

               \[f(x)=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}\]                    Þ(i)\[f(-x)=\frac{{{e}^{-2x}}-1}{{{e}^{-2x}}+1}=\frac{1-{{e}^{2x}}}{1+{{e}^{2x}}}\]Þ\[f(x)=-\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}=-f(x)\]                    \[f(x)\] is an odd function.                    Again \[f(x)=\frac{{{e}^{2x}}-1}{{{e}^{2x}}+1}\Rightarrow {f}'(x)=\frac{4{{e}^{2x}}}{{{(1+{{e}^{2x}})}^{2}}}>0\,\forall \,n\in R\]                    Þ \[f(x)\] is an increasing function.


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