\section*{Exercise 2 - 081042}

Given two real numbers \(a\) and \(b\) with \(a \neq b\) and \(a b > 0\). Investigate whether for

\[
x=\frac{1}{2}\left(\frac{a}{b}+\frac{b}{a}\right) \quad \text { the expression } \quad s=\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}
\]

exists! If this is the case, express \(s\) as simplified as possible in terms of \(a\) and \(b\), in this case rationally!
}

Due to \(a b>0\), \(a\) and \(b\) are either both positive or both negative, so the fractions \(\frac{a}{b}\) and \(\frac{b}{a}\) are positive in either case. Then, because \(a \neq b\), their sum is strictly greater than 2 and \(x>1\), making \(\sqrt{x-1}\) well-defined. Due to the strict monotonicity of the square root function, \(\sqrt{x+1}>\sqrt{x-1}\), so the denominator of \(s\) never vanishes, and this term is thus well-defined for any such choice of \(a\) and \(b\).

To simplify \(s\), we multiply the fraction by \(\sqrt{x+1}+\sqrt{x-1}\) and obtain

\[
s=\frac{(\sqrt{x+1}+\sqrt{x-1})^{2}}{\sqrt{x+1}^{2}-\sqrt{x-1}^{2}}=\frac{x+1+2 \sqrt{x^{2}-1}+x-1}{x+1-x+1}=\frac{2 x+2 \sqrt{x^{2}-1}}{2}=x+\sqrt{x^{2}-1}
\]

Here,

\[
x^{2}-1=\frac{1}{4} \cdot\left(\frac{a}{b}+\frac{b}{a}\right)^{2}-1=\frac{1}{4} \cdot\left(\frac{a^{2}}{b^{2}}+2+\frac{b^{2}}{a^{2}}-4\right)=\frac{1}{4} \cdot\left(\frac{a}{b}-\frac{b}{a}\right)^{2}
\]

Substituting this in, we get

\[
s=x+\sqrt{x^{2}-1}=\frac{1}{2}\left(\frac{a}{b}+\frac{b}{a}\right)+\frac{1}{2}\left|\frac{a}{b}-\frac{b}{a}\right|
\]

If \(|a|>|b|\), the difference inside the absolute value is positive, and we obtain \(s=\frac{a}{b}\); otherwise, it is negative, and we obtain \(s=\frac{b}{a}\).