This question was uploaded on 01/02/22 on social media accounts.
Solution:
Each term in f(x) is increasing but it has a finite value given so the absolute value of x must be less than 1.
Given series:
\[\color{blue} {f(x)=\frac{\color{red}{(1-x)}(1+x)}{\color{red}{(1-x)}}(1+x^2)(1+x^4)(1+x^8)......}\]
\[\Rightarrow\color{blue} {f(x)=\frac{\color{red}{(1-x^2)}(1+x^2)}{\color{red}{(1-x)}}(1+x^4)(1+x^8)......}\]
Similarly:
\[\Rightarrow\color{blue} {f(x)=\frac{1-x^{\infty}}{1-x}=\frac1{1-x}}\]
Here, higher power of x will tends to zero.
Now:
\[\color{green} {2022=\frac1{1-x}}\]
\[\Rightarrow\color{green} {x=\frac{2021}{2022}}\]
