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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2002

  • question_answer
    For all positive value of \[x\] and \[y,\]the value of \[\frac{(1+x+{{x}^{2}})(1+y+{{y}^{2}})}{xy}\]is:

    A) \[\le 9\]                                               

    B) \[<9\]

    C)  \[\ge 9\]                                            

    D) \[>9\]

    Correct Answer: C

    Solution :

    Taking \[x=1,y=2\]we get \[\frac{(1+x+{{x}^{2}})(1+y+{{y}^{2}})}{xy}=\frac{3\times 7}{2}>9\] for \[x=1,y=1,\]equality holds \[\therefore \,\,\,\,\ge 9\]


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