JEE Main & Advanced Mathematics Other Series Question Bank Critical Thinking

  • question_answer ) If  \[y=2{{x}^{2}}-1\], then  \[\left[ \frac{1}{y}+\frac{1}{3{{y}^{3}}}+\frac{1}{5{{y}^{5}}}+.... \right]\] is equal to

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

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

    C) \[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}+..... \right]\]

    D) \[\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}-\frac{1}{3{{x}^{6}}}+\frac{1}{5{{x}^{10}}}-..... \right]\]

    Correct Answer: B

    Solution :

    Given that y = 2x2?1, then \[\left[ \frac{1}{y}+\frac{1}{3{{y}^{3}}}+\frac{1}{5{{y}^{5}}}+.... \right]\,=\frac{1}{2}\left[ {{\log }_{e}}\left( \frac{1+\frac{1}{y}}{1-\frac{1}{y}} \right) \right]\] \[=\frac{1}{2}{{\log }_{e}}\left( \frac{y+1}{y-1} \right)=-\frac{1}{2}{{\log }_{e}}\left( \frac{y-1}{y+1} \right)\] \[=-\frac{1}{2}{{\log }_{e}}\left( 1-\frac{2}{1+y} \right)=-\frac{1}{2}{{\log }_{e}}\left( 1-\frac{1}{{{x}^{2}}} \right)\] \[=\frac{1}{2}\left[ \frac{1}{{{x}^{2}}}+\frac{1}{2{{x}^{4}}}+\frac{1}{3{{x}^{6}}}+.... \right]\].

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