We consider the function $f$ defined on the interval $]-\infty; 1[$ by
$$f(x) = \frac{\mathrm{e}^x}{x-1}$$
We admit that the function $f$ is differentiable on the interval $]-\infty; 1[$. We call $\mathscr{C}$ its representative curve in a coordinate system.

\begin{enumerate}
  \item \begin{enumerate}
    \item[a.] Determine the limit of the function $f$ at 1.
    \item[b.] Deduce from this a graphical interpretation.
  \end{enumerate}
  \item Determine the limit of the function $f$ at $-\infty$.
  \item \begin{enumerate}
    \item[a.] Show that for every real number $x$ in the interval $]-\infty; 1[$, we have
    $$f'(x) = \frac{(x-2)\mathrm{e}^x}{(x-1)^2}$$
    \item[b.] Draw up, by justifying, the table of variations of the function $f$ on the interval $]-\infty; 1[$.
  \end{enumerate}
  \item We admit that for every real number $x$ in the interval $]-\infty; 1[$, we have
  $$f''(x) = \frac{\left(x^2 - 4x + 5\right)\mathrm{e}^x}{(x-1)^3}.$$
  \begin{enumerate}
    \item[a.] Study the convexity of the function $f$ on the interval $]-\infty; 1[$.
    \item[b.] Determine the reduced equation of the tangent line $T$ to the curve $\mathscr{C}$ at the point with abscissa 0.
    \item[c.] Deduce from this that, for every real number $x$ in the interval $]-\infty; 1[$, we have:
    $$\mathrm{e}^x \geqslant (-2x-1)(x-1).$$
  \end{enumerate}
  \item \begin{enumerate}
    \item[a.] Justify that the equation $f(x) = -2$ admits a unique solution $\alpha$ on the interval $]-\infty; 1[$.
    \item[b.] Using a calculator, determine an interval containing $\alpha$ with amplitude $10^{-2}$.
  \end{enumerate}
\end{enumerate}