If alpha satisfies the equation x2+x+1=0 and (1+alpha)7=A+Balpha+Calpha2, A, B, C 0, then 5(3A-2B-C) is equal to:

Year	2026	2025	2024	Total
Questions	11	16	14	41

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

1 + ω + ω² = 0

$ω³ = 1$ $\omega^3 = 1$

Core Logic

The equation $x² + x + 1 = 0$ $x^2 + x + 1 = 0$ is the standard identity whose roots are the non-real cube roots of unity, $ω$ $\omega$ and $ω²$ $\omega^2$. Let us assign $α = ω$ $\alpha = \omega$. We are given the expression $(1+α)⁷$ $(1+\alpha)^7$. Substituting the root: $(1+ω)⁷$ $(1+\omega)^7$.

Step 1: Simplify using Unity Properties

From the identity $1 + ω + ω² = 0$ $1 + \omega + \omega^2 = 0$, we extract:

1 + ω = -ω²

Substitute this into the expression:

(1+ω)⁷ = (-ω²)⁷ = -ω¹⁴

Step 2: Cyclical Reduction

Using $ω³ = 1$ $\omega^3 = 1$, reduce the exponent 14 modulo 3:

14 = 3(4) + 2 \Rightarrow ω¹⁴ = (ω³)⁴ \cdot ω² = 1 \cdot ω² = ω²

Thus, the expression reduces to $-ω²$ $-\omega^2$. Rewrite this back to its linear form using $1 + ω + ω² = 0$ $1 + \omega + \omega^2 = 0$:

-ω² = 1 + ω = 1 + α

Step 3: Finding Co-efficients

We compare $1 + α$ $1 + \alpha$ with $A + Bα + Cα²$ $A + B\alpha + C\alpha^2$. Notice that $1 + α$ $1 + \alpha$ can be directly represented without any $α²$ $\alpha^2$ term (and we must keep A, B, C $\geq 0$ $\ge 0$). So, $A = 1$ $A = 1$, $B = 1$ $B = 1$, $C = 0$ $C = 0$.

Step 4: Final Output Evaluation

Substitute these constants into the required equation $5(3A - 2B - C)$ $5(3A - 2B - C)$:

5(3(1) - 2(1) - 0)

5(3 - 2) = 5(1) = 5

Pattern Recognition

The roots of $x²+x+1=0$ $x^2+x+1=0$ are always $ω, ω²$ $\omega, \omega^2$. Expressions of the form $(1+ω)^k$ $(1+\omega)^k$ rapidly collapse down to single variables via the $1+ω+ω²=0$ $1+\omega+\omega^2=0$ rule, making multi-variable polynomial equations instantly trivial.

Chapter Mix

Class 11 Maths: Complex Numbers

More Complex Numbers Previous-Year Questions — Page 9

Q26 jee_main_2024_31_jan_morning Properties of Modulus and Argument

α

denotes the number of solutions of

|1 - i|^x = 2^x

and

β = ((|z|)/( (z)))

, where

z = (π)/(4) (1 + i)⁴ ( 1 - \sqrt(π) i\sqrt(π) + i + \sqrt(π) - i1 + \sqrt(π) i), i = \sqrt(-1)

, then the distance of the point

(α, β)

from the line

4x - 3y = 7

Numerical Answer. Answer: 3 to 3

Solution

Core Logic

|1 - i|^x = 2^x (\sqrt(2))^x = 2^x 2x/2 = 2^x

This implies $(x)/(2) = x x = 0$ $\frac{x}{2} = x \implies x = 0$. There is exactly 1 solution, so $α = 1$ $\alpha = 1$.

Step 1: Simplify complex number z

(1+i)⁴ = ((1+i)²)² = (1 + i² + 2i)² = (2i)² = -4

Thus, $z = -π ( (1-\sqrt(π)i)(\sqrt(π)-i)π + 1 + (\sqrt(π)-i)(1-\sqrt(π)i)1 + π )$

Step 2: Simplify Bracket

Let's expand the terms directly: $z = (π)/(4)(-4) [ \sqrt(π) - π i - i - \sqrt(π)π + 1 + \sqrt(π) - i - π i - \sqrt(π)1 + π ]$

= -π [ (-i(π+1))/(π+1) + (-i(π+1))/(π+1) ]

= -π [ -i - i ] = 2π i

Step 3: Find beta

For $z = 2π i$ $z = 2\pi i$: $|z| = 2π$ $|z| = 2\pi$ and $(z) = (π)/(2)$ $\arg(z) = \frac{\pi}{2}$.

β = (|z|)/( (z)) = (2π)/(π/2) = 4

Step 4: Distance from Line

Distance of point $(α, β) = (1, 4)$ $(\alpha, \beta) = (1, 4)$ from the line $4x - 3y - 7 = 0$ $4x - 3y - 7 = 0$:

D = |4(1) - 3(4) - 7|\sqrt(4² + (-3)²) = (|4 - 12 - 7|)/(5) = (|-15|)/(5) = 3

Chapter Mix

Class 11 Maths: Complex Numbers and Quadratic Equations Class 11 Maths: Straight Lines

If $α$ $\alpha$ satisfies the equation $x²+x+1=0$ $x^{2}+x+1=0$ and $(1+α)⁷=A+Bα+Cα²$ $(1+\alpha)^{7}=A+B\alpha+C\alpha^{2}$, $A, B, C\geq 0$ $A, B, C\ge 0$, then $5(3A-2B-C)$ $5(3A-2B-C)$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify using Unity Properties

Step 2: Cyclical Reduction

Step 3: Finding Co-efficients

Step 4: Final Output Evaluation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Complex Numbers Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Simplify complex number z

Step 2: Simplify Bracket

Step 3: Find beta

Step 4: Distance from Line

Chapter Mix

More Complex Numbers Questions — jee_main_2024_27_jan_morning

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If α$\alpha$ satisfies the equation x²+x+1=0$x^{2}+x+1=0$ and (1+α)⁷=A+Bα+Cα²$(1+\alpha)^{7}=A+B\alpha+C\alpha^{2}$, A, B, C≥ 0$A, B, C\ge 0$, then 5(3A-2B-C)$5(3A-2B-C)$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify using Unity Properties

Step 2: Cyclical Reduction

Step 3: Finding Co-efficients

Step 4: Final Output Evaluation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Complex Numbers Previous-Year Questions — Page 9

Solution

Core Logic

Step 1: Simplify complex number z

Step 2: Simplify Bracket

Step 3: Find beta

Step 4: Distance from Line

Chapter Mix

More Complex Numbers Questions — jee_main_2024_27_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

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If $α$ $\alpha$ satisfies the equation $x²+x+1=0$ $x^{2}+x+1=0$ and $(1+α)⁷=A+Bα+Cα²$ $(1+\alpha)^{7}=A+B\alpha+C\alpha^{2}$, $A, B, C\geq 0$ $A, B, C\ge 0$, then $5(3A-2B-C)$ $5(3A-2B-C)$ is equal to: