11th Class Mathematics Complex Numbers and Quadratic Equations Question Bank Critical Thinking

  • question_answer If \[a=\cos (2\pi /7)+i\,\sin (2\pi /7),\] then the quadratic equation whose roots are \[\alpha =a+{{a}^{2}}+{{a}^{4}}\] and \[\beta ={{a}^{3}}+{{a}^{5}}+{{a}^{6}}\] is [RPET 2000]

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

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

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

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

    Correct Answer: D

    Solution :

    \[a=\cos (2\pi /7)+i\sin (2\pi /7)\] \[{{a}^{7}}={{[\cos (2\pi /7)+i\sin (2\pi /7)]}^{7}}\]     \[=\,\cos 2\pi +i\sin 2\pi =1\]             .....(i) \[S\,\,=\alpha +\beta =(a+{{a}^{2}}+{{a}^{4}})+({{a}^{3}}+{{a}^{5}}+{{a}^{6}})\] \[S=a+{{a}^{2}}+{{a}^{3}}+{{a}^{4}}+{{a}^{5}}+{{a}^{6}}\]\[=\frac{a(1-{{a}^{6}})}{1-a}\] \[S=\,\,\frac{a-{{a}^{7}}}{1-a}=\frac{a-1}{1-a}=-1\]          .....(ii) \[P=\,\,\alpha \beta =(a+{{a}^{2}}+{{a}^{4}})\,({{a}^{3}}+{{a}^{5}}+{{a}^{6}})\] \[={{a}^{4}}+{{a}^{6}}+{{a}^{7}}+{{a}^{5}}+{{a}^{7}}+{{a}^{8}}+{{a}^{7}}+{{a}^{9}}+{{a}^{10}}\] \[={{a}^{4}}+{{a}^{6}}+1+{{a}^{5}}+1+a+1+{{a}^{2}}+{{a}^{3}}\] (From eqn (i)] \[=3+(a+{{a}^{2}}+{{a}^{3}}+{{a}^{4}}+{{a}^{5}}+{{a}^{6}})\]\[=3+S\] \[=\]\[3-1=2\] [From (ii)] Required equation is, \[{{x}^{2}}-Sx+P=0\] Þ \[{{x}^{2}}+x+2=0\].


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