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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If one root of the equation\[l{{x}^{2}}+mx+n=0\]is \[\frac{9}{2}\] \[(l,m\]and n are positive integers) and \[\frac{m}{4n}=\frac{l}{m},\]then\[\frac{1}{x}+\frac{1}{y}\]is equal to

    A)  80                                         

    B)  85

    C)  90                         

    D)         95

    E)  100

    Correct Answer: B

    Solution :

    Given, \[l{{x}^{2}}+mx+n=0\]                                     ...(i) Now,         \[D={{m}^{2}}-4ln\] \[=0\]    (\[\because \]\[{{m}^{2}}=4ln\]given) It means roots of given equation are equal. \[\therefore \]  \[{{\left( x-\frac{9}{2} \right)}^{2}}=0\] \[\Rightarrow \]      \[4{{x}^{2}}+81-36x=0\]                 ...(ii) On comparing Eqs. (i) and (ii), we get \[l=4,\text{ }m=-36,n=81\] \[\therefore \]  \[l+n=4+81=85\]


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