Question

let s denote the set of all real values of x for which (x ^ 2022 + 1)× ( 1 + x^2 + x ^ 4 +......+ x^2020) is equal to 2022 . X ^2021 then find no of elements in s

Answer 1

let s denote the set of all real values of x for which (x ^ 2022 + 1)× ( 1 + x^2 + x ^ 4 +......+ x^2020) is equal to 2022 . X ^2021 then find no of elements in s

Step-by-step explanation:

The left side is positive, thus any real root will be positive. x=0 is not a solution.

If you multiply everything out you find that the polynomial is symmetric under degree reversion, so with some root x also x−1 is a root.

If you divide by x2021, the left side will turn into a convex function with a minimum at x=1, while the right side is constant.

By inspection, x=1 is a solution.

Being a minimizer makes the root at x=1 into a double root. There are no other roots.

The polynomial equation can also be written as

0=∑k=01010(x2k−2x2021+x4042−2k)=∑k=01010x2k(x2021−2k−1)2=(x−1)2∑k=01010x2k(x2020−2k+x2019−2k+...+x+1)2

which again is twice the linear factor for x=1 and then a positive factor with no additional root.