JEE Main & Advanced Sample Paper JEE Main - Mock Test - 23

  • question_answer
    Let \[\alpha ,{{\alpha }^{2}}\]be the roots of\[{{x}^{2}}+x+1=0\], then the equation whose roots are \[{{\alpha }^{31}},{{\alpha }^{62}}\] is

    A) \[{{x}^{2}}-x+1=0\]  

    B) \[{{x}^{2}}+x-1=0\]

    C) \[{{x}^{2}}+x+1=0\]

    D) none of these

    Correct Answer: C

    Solution :

    [c] : Given equation is \[{{x}^{2}}+x+1=0\] \[\therefore \] \[\alpha +{{\alpha }^{2}}=-1\] ...(i) and a3 = 1           ...(ii) We have to find the equation whose roots are \[{{\alpha }^{13}}\] and \[{{\alpha }^{62}}\]. Now,\[{{\alpha }^{31}}+{{\alpha }^{62}}={{\alpha }^{31}}(1+{{\alpha }^{31}})\] \[\Rightarrow \]\[{{\alpha }^{31}}+{{\alpha }^{62}}={{\alpha }^{31}}.\alpha (1+{{\alpha }^{30}}.\alpha )\] \[\Rightarrow \]\[{{\alpha }^{31}}+{{\alpha }^{62}}={{({{\alpha }^{3}})}^{10}}.\alpha \{1+{{({{\alpha }^{3}})}^{10}}.\alpha \}\] \[\Rightarrow \]\[{{\alpha }^{31}}+{{\alpha }^{62}}=\alpha (1+\alpha )\]                             [From(ii)] \[\Rightarrow \]\[{{\alpha }^{31}}+{{\alpha }^{62}}=-1\]                             [From(i)] Again, \[{{\alpha }^{31}}.{{\alpha }^{62}}={{\alpha }^{93}}\Rightarrow {{\alpha }^{31}}.{{\alpha }^{62}}={{[{{\alpha }^{3}}]}^{31}}=1\] Required equation is\[{{x}^{2}}-({{\alpha }^{31}}+{{\alpha }^{62}})x+{{\alpha }^{31}}.{{\alpha }^{62}}=0\] \[\Rightarrow \]\[{{x}^{2}}+x+1=0\]

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