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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If\[x=8+3\sqrt{7}\]and\[xy=1,\]then the value of\[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}\]is:

    A)  254                       

    B)         192

    C)  292                       

    D)         66

    E)  62

    Correct Answer: A

    Solution :

    \[\because \]     \[x=8+3\sqrt{7}\] \[\therefore \]  \[y=\frac{1}{8+3\sqrt{7}}=8-3\sqrt{7}\] Now, \[\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=\frac{{{x}^{2}}+{{y}^{2}}}{{{(xy)}^{2}}}=\frac{{{(x+y)}^{2}}-2xy}{{{(xy)}^{2}}}\]                 \[={{(x+y)}^{2}}-2\]                 \[={{(8+3\sqrt{7}+8-3\sqrt{7})}^{2}}-2\]                 \[={{16}^{2}}-2=254\]


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