JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer In  the expansion of \[{{\log }_{e}}\frac{1}{1-x-{{x}^{2}}+{{x}^{3}}}\],  the coefficient of \[x\] is

    A) 0

    B) 1

    C) ? 1

    D) 1/2

    Correct Answer: B

    Solution :

    \[{{\log }_{e}}\left[ \frac{1}{1-x-{{x}^{2}}+{{x}^{3}}} \right]={{\log }_{e}}\left[ \frac{1}{(1-x)-{{x}^{2}}(1-x)} \right]\] \[={{\log }_{e}}\left[ \frac{1}{(1-x)(1-{{x}^{2}})} \right]={{\log }_{e}}\left[ \frac{1}{{{(1-x)}^{2}}(1+x)} \right]\] \[={{\log }_{e}}\{{{(1-x)}^{-2}}{{(1+x)}^{-1}}\}\] ??(i) \[={{\log }_{e}}{{(1-x)}^{-2}}+{{\log }_{e}}{{(1+x)}^{-1}}\] \[=-2{{\log }_{e}}(1-x)-{{\log }_{e}}(1+x)\] \[=-2\,\left[ -x-\frac{{{x}^{2}}}{2}-\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}-.......\infty  \right]\]                 \[-\left[ x-\frac{{{x}^{2}}}{2}+\frac{{{x}^{3}}}{3}-\frac{{{x}^{4}}}{4}+.....\infty  \right]\], \[(\because \ -1<x\le 1)\] Hence coefficient of \[x=2-1=1\].

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