Question

Let $S(x)=1+x-x^2-x^3+x^4+x^5-x^6-x^7+\dots \dots$; where $0<x<1$ If $S(x)=\frac{\sqrt{2}+1}{2}$ then the value of $x$ equals

• A. $2-\sqrt{2}$
• B. $\sqrt{2}-1$
• C. $\frac{1}{\sqrt{2}}$
• D. $\left(1-\frac{1}{\sqrt{2}}\right)$

Answer 1

Answer: (B)

$\frac{\sqrt{2}+1}{2}=\left(1-x^2+x^4-x^6+x^8\dots \dots \right)+\left(x-x^3+x^5-x^7\dots \dots ..\right)$
$\frac{\sqrt{2}+1}{2}=\frac{1}{1+x^2}+\frac{x}{1+x^2}=\frac{1+x}{1+x^2}$
or $(\sqrt{2}+1)x^2+(\sqrt{2}+1)=2+2x$
$(\sqrt{2}+1)x^2-2x+(\sqrt{2}-1)=0\text{ (divide by }\sqrt{2}+1)$
$x^2-2(\sqrt{2}-1)x+(\sqrt{2}-1{)}^2=0$
$[x-(\sqrt{2}-1){]}^2=0\Rightarrow x=\sqrt{2}-1$