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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If\[y={{e}^{x}}.{{e}^{{{x}^{2}}}}.{{e}^{{{x}^{3}}}}.....{{e}^{{{x}^{n}}}}....,\]for\[0<x<1,\]then \[\frac{dy}{dx}\]at\[x=\frac{1}{2}\]is

    A)  e                                           

    B)  4e

    C)  2e                         

    D)         3e

    E)  5e

    Correct Answer: B

    Solution :

    Given, \[y={{e}^{x}}.{{e}^{{{x}^{2}}}}.{{e}^{{{x}^{3}}}}.....{{e}^{{{x}^{n}}}}.....\] \[\Rightarrow \]               \[y={{e}^{(x+{{x}^{2}}+....+\infty )}}\] \[\Rightarrow \]               \[y={{e}^{\frac{x}{1-x}}}\] \[\Rightarrow \]               \[\frac{dy}{dx}={{e}^{\frac{x}{1-x}}}\left[ \frac{(1-x)1-x(-1)}{{{(1-x)}^{2}}} \right]\]                 \[={{e}^{\frac{x}{1-x}}}.\frac{1}{{{(1-x)}^{2}}}\] At           \[x=\frac{1}{2},\]                 \[{{\left( \frac{dy}{dx} \right)}_{x=\frac{1}{2}}}={{e}^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}}.\frac{1}{{{\left( 1-\frac{1}{2} \right)}^{2}}}\]                 \[=4e\]


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