Question

If $S$ denotes the set of all real values of $x$ such that $(x^{2010}+1)(1+x^2+x^4+\cdots +x^{2008})=2010x^{2009}$, then the number of elements in set $S$ is

Answer 1

Given ; $(x^{2010}+1)(1+x^2+x^4+\cdots +x^{2008})=2010x^{2009}$
As $x\ne 0$, so dividing by $x^{2009}$, we get
$\Rightarrow (x+\frac{1}{x^{2009}})(1+x^2+x^4+\cdots +x^{2008})=2010\Rightarrow x+x^3+x^5+\cdots +x^{2009}+\frac{1}{x^{2009}}+\frac{1}{x^{2007}}+\cdots +\frac{1}{x}=2010\Rightarrow x+\frac{1}{x}+x^3+\frac{1}{x^3}+\dots +x^{2009}+\frac{1}{x^{2009}}=2010$

We know that,
$x+\frac{1}{x}\in (-\infty ,-2]\cup [2,\infty )x^3+\frac{1}{x^3}\in (-\infty ,-2]\cup [2,\infty )⋮x^{2009}+\frac{1}{x^{2009}}\in (-\infty ,-2]\cup [2,\infty )$
So,
$x+\frac{1}{x}+x^3+\frac{1}{x^3}+\dots +x^{2009}+\frac{1}{x^{2009}}\ge 2\times 1005\Rightarrow x+\frac{1}{x}+x^3+\frac{1}{x^3}+\dots +x^{2009}+\frac{1}{x^{2009}}\ge 2010$
Therefore,
$x+\frac{1}{x}+x^3+\frac{1}{x^3}+\dots +x^{2009}+\frac{1}{x^{2009}}=2010$
When
$x+\frac{1}{x}=2\therefore x=1$

Hence, the number of elements in set $S$ is $1$.