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

  • question_answer
    \[{{\log }_{e}}(x+1)-{{\log }_{e}}(x-1)=\]

    A) \[2\,\left[ x+\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}+......\infty  \right]\]

    B)  \[\,\left[ x+\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}+......\infty  \right]\]

    C) \[2\,\left[ \frac{1}{x}+\frac{1}{3{{x}^{3}}}+\frac{1}{5{{x}^{5}}}+...\infty  \right]\]

    D)  \[\,\left[ \frac{1}{x}+\frac{1}{3{{x}^{3}}}+\frac{1}{5{{x}^{5}}}+...\infty  \right]\]

    Correct Answer: C

    Solution :

    \[{{\log }_{e}}(x+1)-{{\log }_{e}}(x-1)={{\log }_{e}}\frac{x+1}{x-1}\] \[={{\log }_{e}}\left( \frac{1+\frac{1}{x}}{1-\frac{1}{x}} \right)=2\left\{ \frac{1}{x}+\frac{1}{3{{x}^{3}}}+\frac{1}{5{{x}^{5}}}+....... \right\}\].


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