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JEE Mains · Maths · STD 11 - 4.1 complex nubers

જે \(\alpha\) એ સમીકરણ \(x^2+x+1=0\) નું સમાધાન કરે અને \((1+\alpha)^7=\mathrm{A}+\mathrm{B} \alpha+\mathrm{C} \alpha^2, \mathrm{~A}, \mathrm{~B}, \mathrm{C} \geqslant 0\) હોય, તો \(5(3 A-2 B-C)=\) ...........

  1. A \(6\)
  2. B \(5\)
  3. C \(7\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(5\)

Step-by-step Solution

Detailed explanation

\(x^2+x+1=0 \Rightarrow x=\omega, \omega^2=\alpha\) Let \(\alpha=\omega\) Now \((1+\alpha)^7=-\omega^{14}=-\omega^2=1+\omega\) \( A=1, B=1, C=0 \) \( \therefore 5(3 A-2 B-C)=5(3-2-0)=5\)
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