JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2012

  • question_answer
        If\[y=x-{{x}^{2}}+{{x}^{3}}-{{x}^{4}}+.....\infty ,\]then the value of \[x\]will be\[(-1<x<1)\]

    A)  \[y+\frac{1}{y}\]                             

    B)  \[\frac{y}{1+y}\]

    C)  \[y-\frac{1}{y}\]                              

    D)  \[\frac{y}{1-y}\]

    Correct Answer: D

    Solution :

                    We have the equation \[2{{x}^{2}}+{{y}^{2}}-4x+4y=0\]                        ...(i) Since, axes are to be shifted to the point\[(1,-2)\] therefore substitute\[x=X+1\]and\[y=Y-2\]in Eq. (i), we get \[2{{(X+1\text{ })}^{2}}+{{(Y-2)}^{2}}-4(X+1)+4(Y-2)=0\] \[\Rightarrow \]\[2({{X}^{2}}+1+2X)+{{y}^{2}}+4-4Y-4X\]\[-4+4Y-8=0\] \[\Rightarrow \]               \[2{{X}^{2}}+{{Y}^{2}}-6=0\] \[\Rightarrow \]               \[2{{X}^{2}}+{{Y}^{2}}=6\] which is the required equation.


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