In the equation x^2021 x^2020 . x=1 then no of roots 1) all roots are ve 2) exactly one real root is positive 3) exactly one real root is negative 4) no real root is ve?

Question

In the equation x^2021 x^2020 .  x=1 then no of roots 1) all roots are  ve 2) exactly one real root is positive 3) exactly one real root is negative 4) no real root is  ve?

Answer

Question Analysis:
We are given the equation x^2021 x^2020 . x = 1 and we need to determine the number of roots based on the given conditions.

Solution:
Given equation: x^2021 x^2020 . x = 1

To solve this equation, we can simplify it by combining the exponents:
x^(2021+2020+1) = 1
x^4042 = 1

Roots of Unity:
In complex analysis, the roots of unity are the complex numbers that satisfy the equation x^n = 1, where n is a positive integer. The roots of unity form a regular polygon in the complex plane.

The equation x^4042 = 1 can be rewritten as (x^2021)^2 = 1. This means that x^2021 can be either 1 or -1.

Case 1: x^2021 = 1
If x^2021 = 1, then x can be any complex number that satisfies this equation. Since we are interested in real roots, we can conclude that all roots in this case are positive.

Case 2: x^2021 = -1
If x^2021 = -1, then x can also be any complex number that satisfies this equation. Again, since we are interested in real roots, we can conclude that all roots in this case are negative.

Therefore, the answer to the given question is:

Option 1: All roots are positive (x^2021 = 1)
Option 2: All roots are negative (x^2021 = -1)

Conclusion:
Based on the given equation x^2021 x^2020 . x = 1, we have determined that all roots are either positive or negative. Therefore, the correct answer is option 1: All roots are positive.

Answer

Problem: In the equation $x^{2021} \cdot x^{2020} \cdot x=1$, find the number of roots satisfying the given equation.

Solution:

To find the number of roots satisfying the equation $x^{2021} \cdot x^{2020} \cdot x=1$, we can simplify the equation first.

Simplifying the Equation:
We can simplify the equation by combining the exponents of x:
$x^{2021} \cdot x^{2020} \cdot x = x^{2021+2020+1} = x^{4041} = 1$

Understanding the Equation:
The equation $x^{4041} = 1$ is equivalent to finding the solutions of $x$ that, when raised to the power of 4041, equals 1.

Finding the Roots:
In general, for any real number $a$, we have $a^n = 1$ if and only if $a = 1$ or $n$ is an even integer. Therefore, the solutions to $x^{4041} = 1$ are:

1. $x = 1$: This is a valid solution since $1^{4041} = 1$.
2. $x = -1$: This is also a valid solution since $(-1)^{4041} = 1$.

Conclusion:
From our analysis, we can conclude that there are two roots satisfying the equation $x^{2021} \cdot x^{2020} \cdot x=1$. These roots are $x=1$ and $x=-1$. Therefore, none of the options provided (all roots are negative, exactly one real root is positive, exactly one real root is negative, no real root is negative) are correct.

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