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

  • question_answer
    The minimum value of \[4{{e}^{2x}}+9{{e}^{-2x}}\] is  [J & K 2005]

    A)            11

    B)            12

    C)            10

    D)            14

    Correct Answer: B

    Solution :

               Let \[f(x)=4{{e}^{2x}}+9{{e}^{-2x}}\]                    \[\therefore \] \[{f}'(x)=8{{e}^{2x}}-18{{e}^{-2x}}\]                    Put \[{f}'(x)=0\Rightarrow 8{{e}^{2x}}-18{{e}^{-2x}}=0\]                    \[{{e}^{2x}}=3/2\Rightarrow x=\log {{(3/2)}^{1/2}}\]                    Again \[{f}''(x)=16{{e}^{2x}}+36{{e}^{-2x}}>0\]                    Now \[f(\log {{(3/2)}^{1/2}})=4{{e}^{2.(\log {{(3/2)}^{1/2}})}}+9{{e}^{-2(\log {{(3/2)}^{1/2}})}}\]                    = \[4\times \frac{3}{2}+9\times \frac{2}{3}\] = \[6+6=12\]                    Hence minimum value = 12.


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