همانی $f \to f(x) = x$

$g\left( x \right) = \frac{1}{{x - a}} \to \left| {g\left( x \right)} \right| - 2 = \left| {\frac{1}{{x - a}}} \right| - 2 = \frac{1}{{\left| x \right|}}$

$\xrightarrow{{x = \frac{{\sqrt 2 }}{2}}}\frac{1}{{\left| {\frac{{\sqrt 2 }}{2} - a} \right|}} = \frac{1}{{\frac{{\sqrt 2 }}{2}}} + 2 = 2 + \sqrt 2  \to \left| {\frac{{\sqrt 2 }}{2} - a} \right| = \frac{1}{{2 + \sqrt 2 }} = \frac{{2 - \sqrt 2 }}{2}$

$\left\{ \begin{gathered}
  \frac{{\sqrt 2 }}{2} - a = 1 - \frac{{\sqrt 2 }}{2} \to a = \sqrt 2  - 1 \hfill \\
  \frac{{\sqrt 2 }}{2} - a = \frac{{\sqrt 2 }}{2} - 1 \to a = 1 \hfill \\ 
\end{gathered}  \right.\xrightarrow{ - }2 - \sqrt 2 $