We consider the function $g$ defined on the interval $] 0 ; + \infty [$, by

$$g ( x ) = \frac { \ln ( x ) } { 1 + x ^ { 2 } }$$

We admit that $g$ is differentiable on the interval $] 0 ; + \infty \left[ \right.$ and we denote $g ^ { \prime }$ its derivative function. We denote $\mathscr { C } _ { g }$ the representative curve of the function $g$ in the plane with respect to a coordinate system $( \mathrm { O } ; \vec { \imath } , \vec { \jmath } )$.

We also consider the function $f$ defined on $]0;+\infty[$ by $f(x) = 1 + x^2 - 2x^2\ln(x)$, and $\alpha$ denotes the unique solution of $f(x)=0$ in $[1;+\infty[$. We admit that $g(\alpha) = \frac{1}{2\alpha^2}$.

\begin{enumerate}
  \item Prove that for all real $x$ in the interval $] 0 ; + \infty \left[ , \quad g ^ { \prime } ( x ) = \frac { f ( x ) } { x \left( 1 + x ^ { 2 } \right) ^ { 2 } } \right.$.
  \item Prove that the function $g$ admits a maximum at $x = \alpha$.
  \item We denote $T _ { 1 }$ the tangent line to $\mathscr { C } _ { g }$ at the point with abscissa 1 and we denote $T _ { \alpha }$ the tangent line to $\mathscr { C } _ { g }$ at the point with abscissa $\alpha$.\\
Determine, as a function of $\alpha$, the coordinates of the intersection point of the lines $T _ { 1 }$ and $T _ { \alpha }$.
\end{enumerate}