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JEE Mains · Maths · STD 11 - 4.1 complex nubers

ધારોકે સમીકરણ \(1+x^{2}+x^{4}=0\) નું એક બીજ \(\alpha\) છે. તો \(\alpha^{1011}+\alpha^{2022}-\alpha^{3033}\) નું મૂલ્ય \(\dots\dots\dots\) છે.

  1. A \(1\)
  2. B \(\alpha\)
  3. C \(1+\alpha\)
  4. D \(1+2 \alpha\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1\)

Step-by-step Solution

Detailed explanation

\(x^{4}+x^{2}+1=0\) \(\Rightarrow\left(x^{2}+x+1\right)\left(x^{2}-x+1\right)=0\) \(\Rightarrow x=\pm \;\omega, \pm \;\omega^{2}\) where \(\omega=1^{1 / 3}\) and imaginary. So \(\alpha^{1011}+\alpha^{2002}-\alpha^{3033}=1+1-1=1\)
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