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

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

    A) \[\frac{{{x}^{2}}}{2}+\frac{{{x}^{4}}}{4}+\frac{{{x}^{6}}}{6}+....\infty \]

    B) \[\frac{{{x}^{2}}}{1.2}+\frac{{{x}^{4}}}{3.4}+\frac{{{x}^{6}}}{5.6}+....\infty \]

    C) \[2\,\,\left[ \frac{{{x}^{2}}}{1.2}+\frac{{{x}^{4}}}{3.4}+\frac{{{x}^{6}}}{5.6}+..\infty  \right]\]

    D) None of these

    Correct Answer: C

    Solution :

    \[{{\log }_{e}}\{{{(1+x)}^{1+x}}{{(1-x)}^{1-x}}\}\] \[=(1+x){{\log }_{e}}(1+x)+(1-x){{\log }_{e}}(1-x)\] \[=(1+x)\left\{ x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+...... \right\}\]\[+(1-x)\left\{ -x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}-....... \right\}\]       \[=2\left\{ -\frac{{{x}^{2}}}{2}-\frac{{{x}^{4}}}{4}-\frac{{{x}^{6}}}{6}-..... \right\}+2\left\{ {{x}^{2}}+\frac{{{x}^{4}}}{3}+\frac{{{x}^{6}}}{5}+...... \right\}\] \[=2\left[ {{x}^{2}}\left( 1-\frac{1}{2} \right)+{{x}^{4}}\left( \frac{1}{3}-\frac{1}{4} \right)+{{x}^{6}}\left( \frac{1}{5}-\frac{1}{6} \right)+...... \right]\] \[=2\left[ \frac{{{x}^{2}}}{1.2}+\frac{{{x}^{4}}}{3.4}+\frac{{{x}^{6}}}{5.6}+....... \right]\].


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