JEE Main & Advanced Mathematics Sequence & Series Question Bank Exponential series

  • question_answer
    If   \[y=1+\frac{x}{1\,!}+\frac{{{x}^{2}}}{2\,!}+\frac{{{x}^{3}}}{3\,!}+......\infty \],  then \[x=\]

    A) \[{{\log }_{e}}y\]

    B) \[{{\log }_{e}}\frac{1}{y}\]

    C) \[{{e}^{y}}\]

    D) \[{{e}^{-y}}\]!

    Correct Answer: A

    Solution :

    \[y=1+\frac{x}{1\ !}+\frac{{{x}^{2}}}{2\ !}+\frac{{{x}^{3}}}{3\ !}+......={{e}^{x}}\Rightarrow x={{\log }_{e}}y\].


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