CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    If\[\omega \ne 1\]and\[{{\omega }^{3}}=1,\]then\[\frac{a\omega +b+c{{\omega }^{2}}}{a{{\omega }^{2}}+b\omega +c}+\frac{a{{\omega }^{2}}+b+c\omega }{a+b\omega +c{{\omega }^{2}}}\]is equal to

    A)  2  

    B)                                         \[\omega \]

    C)  \[2\omega \]                   

    D)         \[2{{\omega }^{2}}\]

    E)  \[a+b+c\]

    Correct Answer: C

    Solution :

    \[\frac{a\omega +b+c{{\omega }^{2}}}{a{{\omega }^{2}}+b\omega +c}+\frac{a{{\omega }^{2}}+b+c\omega }{a+b\omega +c{{\omega }^{2}}}\] \[=\frac{1}{\omega }\left[ \frac{a{{\omega }^{2}}+b\omega +c{{\omega }^{2}}}{a{{\omega }^{2}}+b\omega +c} \right]+{{\omega }^{2}}\left[ \frac{a{{\omega }^{2}}+b+c\omega }{a{{\omega }^{2}}+b{{\omega }^{3}}+c{{\omega }^{4}}} \right]\] \[=\frac{1}{\omega }+{{\omega }^{2}}=2{{\omega }^{2}}\]


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