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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    If\[\alpha \]and\[\beta \]be the roots of the equation \[{{x}^{2}}-2x+2=0\], then the least value of n for which \[{{\left( \frac{\alpha }{\beta } \right)}^{n}}=1\]is : [JEE Main 8-4-2019 Morning]

    A) 2                                 

    B) 3

    C) 4         

    D) 5

    Correct Answer: C

    Solution :

    \[{{(x-1)}^{2}}+1=0\Rightarrow x=1+i,1-i\]           \[\therefore \]\[{{\left( \frac{\alpha }{\beta } \right)}^{n}}=1\Rightarrow {{(\pm i)}^{n}}=1\] \[\therefore \] n (least natural number) = 4                        


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