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JEE Mains · Maths · STD 11 - 4.1 complex nubers

ધારો કે \(\alpha\) અને \(\beta\) એ સમીકરણ \(x^{2}+(2 i-1)=0\) નાં બીજ હોય,તો \(\left|\alpha^{8}+\beta^{8}\right|\) નું મૂલ્ય \(\dots\dots\dots\) છે..

  1. A \(50\)
  2. B \(250\)
  3. C \(1250\)
  4. D \(1500\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(50\)

Step-by-step Solution

Detailed explanation

\(X^{2}=1-2 i \quad \Rightarrow \alpha^{2}=1-2 i , \quad \beta^{2}=1-2 i\) Hence \(\alpha^{8}=\beta^{8}\) \(\left|\alpha^{8}+\beta^{8}\right|=\left|2 \alpha^{8}\right|=2\left|\alpha^{2}\right|^{4}\) \(=2 \sqrt{5}^{4}=50\)
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