CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    If\[y={{2}^{x}}{{.3}^{2x-1}},\]then\[\frac{{{d}^{2}}y}{d{{x}^{2}}}\]is equal to:

    A)  \[(log\text{ }2)\text{ }(log\text{ }3)\]  

    B) \[(log\text{ 18})\]           

    C)         \[(log\text{ 1}{{\text{8}}^{2}}){{y}^{2}}\]   

    D)         \[(log\text{ 18})y\]

    E)  \[{{(log\text{ 18})}^{2}}y\]

    Correct Answer: D

    Solution :

    \[y={{2}^{x}}{{.3}^{2x-1}}\] On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}={{2}^{x}}{{.3}^{2x-1}}2\log 3+{{3}^{2x-1}}{{.2}^{x}}\log 2\] \[={{2}^{x}}{{3}^{2x-1}}\log 18=y\log 18\]


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