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CLAT CLAT Solved Paper-2012

  • question_answer
    The value of \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}\], where \[x=2+\sqrt{3}\] and \[y=2-\sqrt{3}\], is

    A)  12                              

    B)  16

    C)  14                              

    D)  10

    Correct Answer: C

    Solution :

    Ans.    \[x=2+\sqrt{3,}\,\,\,y=2-\sqrt{3}\] \[\therefore \]    \[x+y=2+\sqrt{3}+2-\sqrt{3}=4\] and      \[xy=(2+\sqrt{3})\,(2-\sqrt{3})\,=4-3=1\] now, \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=\frac{{{x}^{2}}+{{y}^{2}}}{{{(xy)}^{2}}}\] \[=\frac{{{(x+y)}^{2}}-2xy}{{{(xy)}^{2}}}\] \[=\frac{{{(4)}^{2}}-2(1)}{{{(1)}^{2}}}=\frac{16-2}{1}=14\]


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