If α satisfies the equation x 2 +x+1=0 and (1+α) 7 = A + B α+ C α 2 , A , B , C geq 0 then 5(3 A -2 B - C ) is equal to

Mathematics · Complex Numbers

JEE Main 2024 — 27 January, Shift 1 — Question 28

If $\alpha$ satisfies the equation $x^{2}+x+1=0$ and $(1+\alpha)^{7}=\mathrm{A}+\mathrm{B} \alpha+\mathrm{C} \alpha^{2}, \mathrm{~A}, \mathrm{~B}, \mathrm{C} \geq 0$ then $5(3 \mathrm{~A}-2 \mathrm{~B}-\mathrm{C})$ is equal to

Answer: 5

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{x}^{2}+\mathrm{x}+1=0 \Rightarrow \mathrm{x}=\omega, \omega^{2}=\alpha$

Let $\alpha=\omega$

Now $(1+\alpha)^{7}=-\alpha^{14}=-\alpha^{2}=1+\alpha$

$\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=0$

$\therefore 5(3 \mathrm{~A}-2 \mathrm{~B}-\mathrm{C})=5(3-2-0)=5$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Mathematics
Chapter: Complex Numbers
Topic: Demoivre's Theorem and Roots of Unity

If S=\z in C: z-i = z+i = z-1 \ , then, n(S) is:

Answer: 5

Step-by-step solution

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