Question

Let $S(x)=1+x-x^2-x^3+x^4+x^5-x^6-x^7+\ldots \ldots$; 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 \ldots \ldots\right)+\left(x-x^3+x^5-x^7 \ldots \ldots . .\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+2 x $
$\left.(\sqrt{2}+1) x^2-2 x+(\sqrt{2}-1)=0 \text { (divide by } \sqrt{2}+1\right) $
$x^2-2(\sqrt{2}-1) x+(\sqrt{2}-1)^2=0 $
$[x-(\sqrt{2}-1)]^2=0 \Rightarrow \quad x=\sqrt{2}-1$