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Manipal Engineering Manipal Engineering Solved Paper-2011

  • question_answer
    If \[S=\sum\limits_{n=0}^{\infty }{\frac{{{(\log x)}^{2n}}}{(2n)!}}\], then\[S\]is equal to

    A) \[x+{{x}^{-1}}\]                

    B) \[x-{{x}^{-1}}\]

    C) \[\frac{1}{2}(x+{{x}^{-1}})\]       

    D)         None of these

    Correct Answer: C

    Solution :

    Given that,                 \[S=\sum\limits_{n=0}^{\infty }{\frac{{{(\log x)}^{2n}}}{(2n)!}}\]                 \[=\left( \frac{{{e}^{\log x}}+{{e}^{-\log x}}}{2} \right)=\frac{x+{{x}^{-1}}}{2}\]


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