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JEE Main & Advanced Mathematics Complex Numbers and Quadratic Equations Question Bank Self Evaluation Test - Complex Numbers and Quadratic Equations

  • question_answer
    Let \[x+\frac{1}{x}=1\] and a, b and c are distinct positive integers such that \[\left( {{x}^{a}}+\frac{1}{{{x}^{a}}} \right)+\left( {{x}^{b}}+\frac{1}{{{x}^{b}}} \right)+\left( {{x}^{c}}+\frac{1}{{{x}^{c}}} \right)=0.\] Then the minimum value of \[(a+b+c)\] is

    A) 7                     

    B) 8

    C) 9                     

    D) 10

    Correct Answer: C

    Solution :

    \[x+\frac{1}{x}=1\]        or    \[{{x}^{2}}-x+1=0\] \[\therefore \,\,x=\frac{1}{2}\pm i\frac{\sqrt{3}}{2}\]    or  \[x={{e}^{\frac{i\pi }{3}}}\] \[\therefore \,\,\,{{x}^{a}}+{{x}^{-a}}={{e}^{\frac{ia\pi }{3}}}+{{e}^{\frac{-ia}{3}}}=2\cos \frac{a\pi }{3}\] Hence, \[\cos \frac{a\pi }{3}+\cos \frac{b\pi }{3}+\cos \frac{c\pi }{3}=0\] \[a,b,c\in I\,\,\,\,\therefore \,\,a+b+c{{|}_{\min }}=(1+3+5)=9\]


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