JEE Mains · Maths · STD 11 - 4.1 complex nubers
If \(\alpha\) satisfies the equation \(x^2+x+1=0\) and \((1+\alpha)^7=\mathrm{A}+\mathrm{B} \alpha+\mathrm{C}^2, \mathrm{~A}, \mathrm{~B}, \mathrm{C} \geq 0\), then \(5(3 A-2 B-C)\) is equal to ...........
- A \(6\)
- B \(5\)
- C \(7\)
- D \(3\)
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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