JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer The value of  \[{{\log }_{e}}\left( 1+a{{x}^{2}}+{{a}^{2}}+\frac{a}{{{x}^{2}}} \right)\] is

    A) \[a\,\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)-\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}-\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}-\frac{1}{{{x}^{6}}} \right)-.....\]

    B)  \[a\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)-\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}+\frac{1}{{{x}^{6}}} \right)-.....\]

    C) \[a\,\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)+\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}+\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}+\frac{1}{{{x}^{6}}} \right)+.....\]

    D)  \[a\,\left( {{x}^{2}}-\frac{1}{{{x}^{2}}} \right)+\frac{{{a}^{2}}}{2}\,\left( {{x}^{4}}-\frac{1}{{{x}^{4}}} \right)+\frac{{{a}^{3}}}{3}\,\left( {{x}^{6}}-\frac{1}{{{x}^{6}}} \right)+.....\]

    Correct Answer: B

    Solution :

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

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