CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    If\[2\alpha =-1-i\sqrt{3}\]and\[2\beta =-1-i\sqrt{3},\]then\[5{{\alpha }^{4}}+5{{\beta }^{4}}+7{{\alpha }^{-1}}{{\beta }^{-1}}\]is equal to

    A)  \[-1\]                                   

    B)  \[-2\]

    C)  0                            

    D)         1

    E)  2

    Correct Answer: E

    Solution :

    Given,\[2\alpha =-1-i\sqrt{3}\]and\[2\beta =-1-i\sqrt{3}\] \[\therefore \] \[\alpha +\beta =-1\] and \[\alpha \beta =1\] Now, \[5{{\alpha }^{4}}+5{{\beta }^{4}}+\frac{7}{\alpha \beta }\] \[=5[{{\{{{(\alpha +\beta )}^{2}}-2\alpha \beta \}}^{2}}-2{{(\alpha \beta )}^{2}}]+\frac{7}{\alpha \beta }\] \[=5[{{\{{{(-1)}^{2}}-2\times 1\}}^{2}}-2{{(1)}^{2}}]+\frac{7}{1}\] \[=5(1-2)+7=2\]


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