We consider the function $f$ defined on the set $] 0 ; + \infty [$ by

$$f ( x ) = 1 + x ^ { 2 } - 2 x ^ { 2 } \ln ( x )$$

We admit that $f$ is differentiable on the interval and we denote $f ^ { \prime }$ its derivative function.

\begin{enumerate}
  \item Justify that $\lim _ { x \rightarrow 0 } f ( x ) = 1$ and, by noting that $f ( x ) = 1 + x ^ { 2 } [ 1 - 2 \ln ( x ) ]$, justify that $\lim _ { x \rightarrow + \infty } f ( x ) = - \infty$.
  \item Show that for all real $x$ in the interval $] 0 ; + \infty \left[ , f ^ { \prime } ( x ) = - 4 x \ln ( x ) \right.$.
  \item Study the sign of $f ^ { \prime } ( x )$ on the interval $] 0$; $+ \infty [$, then draw up the table of variations of the function on the interval $] 0 ; + \infty [$.
  \item Prove that the equation $f ( x ) = 0$ admits a unique solution $\alpha$ in the interval $[ 1 ; + \infty [$ and that $\alpha \in [ 1 ; \mathrm { e } ]$.
\end{enumerate}

We admit in the rest of the exercise that the equation $f ( x ) = 0$ has no solution on the interval $] 0 ; 1]$.

5. We are given the function below written in Python. The instruction from lycee import* allows access to the function $\ln$.

\begin{verbatim}
from lycee import *
def f(x) :
    return 1 + x**2 - 2*x**2*ln(x)
def dichotomie(p)
    a=1
    b=2.7
    while b - a > 10**(-p) :
        if f(a)*f((a+b)/2) < 0 :
        | b = (a+b)/2
        else :
        |a =(a+b)/2
    return (a,b)
\end{verbatim}

It writes in the execution console:

\begin{verbatim}
>>> dichotomie(1)
\end{verbatim}

Among the four propositions below, copy the one displayed by the previous instruction. Justify your answer (you may proceed by elimination).

Proposition A: $\quad ( 1.75,1.9031250000000002 )$\\
Proposition B : ( $1.85,1.9031250000000002 )$\\
Proposition C : $\quad ( 2.75,2.9031250000000002 )$\\
Proposition D : (2.85, 2.9031250000000002)