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KVPY Sample Paper KVPY Stream-SX Model Paper-31

  • question_answer
    If \[\alpha \] and \[\beta \] are the roots of\[{{x}^{2}}-2x+2=0,\] then find minimum value of \[n\] such that \[{{\left( \frac{\alpha }{\beta } \right)}^{n}}=1:\]

    A) 4

    B) 3

    C) 2

    D) 5

    Correct Answer: A

    Solution :

    \[{{x}^{2}}-2x+2=0\]
    \[{{(x-1)}^{2}}=-1={{i}^{2}}\]
                \[x=1+i,1-i\]
    Let, \[\alpha =1+i,\beta =1-i\]
    \[{{\left( \frac{\alpha }{\beta } \right)}^{n}}=1\] \[\Rightarrow \]\[{{\left( \frac{1+i}{1-i} \right)}^{n}}=1\]\[\Rightarrow \]\[{{\left( \frac{\sqrt{2}{{e}^{i\pi /4}}}{\sqrt{2}{{e}^{-i\pi /4}}} \right)}^{n}}=1\]\[\Rightarrow \]\[{{e}^{1n\pi /2}}=1\]\[\Rightarrow \]\[n=4.\]


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