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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2008

  • question_answer
        If\[y=\frac{1}{x}+\frac{1}{{{x}^{2}}}+\frac{1}{{{x}^{3}}}+....\]with\[|x|>1,\]then\[\frac{dy}{dx}\]is

    A)  \[\frac{{{x}^{2}}}{{{y}^{2}}}\]                                    

    B)  \[{{x}^{2}}{{y}^{2}}\]

    C)  \[\frac{{{y}^{2}}}{{{x}^{2}}}\]                                    

    D)  \[\frac{-{{y}^{2}}}{{{x}^{2}}}\]

    Correct Answer: D

    Solution :

                    Given that, \[y=1+\frac{1}{x}+\frac{1}{{{x}^{2}}}+\frac{1}{{{x}^{3}}}+....\] \[\therefore \]  \[y=\frac{1}{1-\frac{1}{x}}\]                        (GP series) \[=\frac{x}{x-1}\]                                  ...(i) On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=\frac{1.(x-1)-x.1}{{{(x-1)}^{2}}}=-\frac{1}{{{(x-1)}^{2}}}\] \[\Rightarrow \]               \[\frac{dy}{dx}=-\frac{{{y}^{2}}}{{{x}^{2}}}\]      [from(i)]


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