Let $f$ and $g$ be the functions defined on $] 0 ; + \infty [$ by

$$f ( x ) = \mathrm { e } ^ { - x } \quad \text { and } \quad g ( x ) = \frac { 1 } { x ^ { 2 } } \mathrm { e } ^ { - \frac { 1 } { x } } .$$

We admit that $f$ and $g$ are differentiable on $] 0 ; + \infty [$. We denote $f ^ { \prime }$ and $g ^ { \prime }$ their respective derivative functions.

\section*{Part A - Graphical conjectures}
In each of the questions in this part, no explanation is required.

\begin{enumerate}
  \item Conjecture graphically a solution to the equation $f ( x ) = g ( x )$ on $] 0 ; + \infty [$.
  \item Conjecture graphically a solution to the equation $g ^ { \prime } ( x ) = 0$ on $] 0 ; + \infty [$.
\end{enumerate}

\section*{Part B - Study of the function $g$}
\begin{enumerate}
  \item Calculate the limit of $g ( x )$ as $x$ tends to $+ \infty$.
  \item We admit that the function $g$ is strictly positive on $] 0 ; + \infty [$.
\end{enumerate}

Let $h$ be the function defined on $] 0 ; + \infty [$ by $h ( x ) = \ln ( g ( x ) )$.\\
a. Prove that, for every strictly positive real number $x$,

$$h ( x ) = \frac { - 1 - 2 x \ln x } { x } .$$

b. Calculate the limit of $h ( x )$ as $x$ tends to 0.\\
c. Deduce the limit of $g ( x )$ as $x$ tends to 0.\\
3. Prove that, for every strictly positive real number $x$,

$$g ^ { \prime } ( x ) = \frac { \mathrm { e } ^ { - \frac { 1 } { x } } ( 1 - 2 x ) } { x ^ { 4 } } .$$

\begin{enumerate}
  \setcounter{enumi}{3}
  \item Deduce the variations of the function $g$ on $] 0 ; + \infty [$.
\end{enumerate}

\section*{Part C - Area of the two regions between the curves $\mathscr { C } _ { f }$ and $\mathscr { C } _ { g }$}
\begin{enumerate}
  \item Prove that point A with coordinates $( 1 ; \mathrm { e } ^ { - 1 } )$ is an intersection point of $\mathscr { C } _ { f }$ and $\mathscr { C } _ { g }$.\\
We admit that this point is the unique intersection point of $\mathscr { C } _ { f }$ and $\mathscr { C } _ { g }$, and that $\mathscr { C } _ { f }$ is above $\mathscr { C } _ { g }$ on the interval $] 0 ; 1 [$ and below on the interval $] 1 ; + \infty [$.
  \item Let $a$ and $b$ be two strictly positive real numbers. Prove that
\end{enumerate}

$$\int _ { a } ^ { b } ( f ( x ) - g ( x ) ) \mathrm { d } x = \mathrm { e } ^ { - a } + \mathrm { e } ^ { - \frac { 1 } { a } } - \mathrm { e } ^ { - b } - \mathrm { e } ^ { - \frac { 1 } { b } } .$$

\begin{enumerate}
  \setcounter{enumi}{2}
  \item Prove that
\end{enumerate}

$$\lim _ { a \rightarrow 0 } \int _ { a } ^ { 1 } ( f ( x ) - g ( x ) ) \mathrm { d } x = 1 - 2 \mathrm { e } ^ { - 1 } .$$

\begin{enumerate}
  \setcounter{enumi}{3}
  \item We admit that
\end{enumerate}

$$\lim _ { a \rightarrow 0 } \int _ { a } ^ { 1 } ( f ( x ) - g ( x ) ) \mathrm { d } x = \lim _ { b \rightarrow + \infty } \int _ { 1 } ^ { b } ( g ( x ) - f ( x ) ) \mathrm { d } x .$$

Give a graphical interpretation of this equality.