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\begin{document}



\title{Métropole - 19 juin 2024}
\author{Spécialité mathématiques - Baccalauréat}
\date{}



\begin{frame}
\titlepage
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}{Exercice 1}
\pause

\begin{enumerate}
 \item \textbf{Affirmation 1~: Vrai}\\ \pause
 Pour tout $x\in\mathbb{R}$, on a~:
 $$f(x)=5\times \dfrac{x}{\mathrm{e}^x}$$ \pause
 Et, par croissances comparées, \pause $\lim\limits_{x\to+\infty}\dfrac{x}{\mathrm{e}^x}=0$ \pause donc~:
 $$\lim\limits_{x\to+\infty}f(x)=0$$ \pause
 La courbe $\mathscr{C}_f$ admet donc la droite d'équation $y=0$ (soit l'axe des abscisses) \pause comme asymptote horizontale en $+\infty$.

\end{enumerate}


\end{frame}



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\begin{frame}




 \textbf{Affirmation 2~: Vrai}\\ \pause
 La fonction $f$ est dérivable sur $\mathbb{R}$ et, pour tout $\mathbb{R}$~: \pause
 \begin{eqnarray*}
  f'(x) & = & \pause 5\mathrm{e}^{-x}+5x\times\left(-\mathrm{e}^{-x}\right)\\ \pause
        & = & (5-5x)\mathrm{e}^{-x} \pause
 \end{eqnarray*}
Et donc, pour tout $\mathbb{R}$~: \pause
\begin{eqnarray*}
 f'(x)+f(x) & = & \pause (5-5x)\mathrm{e}^{-x}+5x\mathrm{e}^{-x}\\ \pause
            & = & 5\mathrm{e}^{-x}
\end{eqnarray*} \pause
La fonction $f$ est donc solution sur $\mathbb{R}$ de l'équation différentielle $y'+y=5\mathrm{e}^{-x}$.


\end{frame}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item \textbf{Affirmation 3~: Faux}\\ \pause
 Soit $(u_n)$ la suite constante égale à $1$, \pause $(w_n)$ la suite constante égale à $-1$ \pause et $(v_n)$ la suite définie, pour tout entier naturel $n$, par $v_n=(-1)^n$. \pause Toutes les hypothèses sont vérifiées mais la suite $(v_n)$ ne converge pas. \pause On a donc trouvé un contre-exemple, ce qui prouve que l'affirmation est fausse.
\end{enumerate}


\end{frame}



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\begin{frame}


 \textbf{Affirmation 4~: Vrai}\\ \pause
 La suite $(u_n)$ étant croissante, pour tout $n\in\mathbb{N}$, on a $u_0\leqslant u_n$. \pause De même, la suite $(w_n)$ étant décroissante, pour tout $n\in\mathbb{N}$, on a $w_0\geqslant w_n$. \pause On a alors, pour tout $n\in\mathbb{N}$~: \pause
 $$u_0\leqslant u_n\leqslant v_n\leqslant w_n\leqslant w_0$$


\end{frame}


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\begin{frame}{Exercice 2}
\pause

\begin{enumerate}
 \item On complète l'arbre de la façon suivante~: \pause

 \begin{center}
\psset{nodesep=0mm,levelsep=30mm,treesep=10mm}
\pstree[treemode=R]{\Tdot}
{
\pstree
{\Tdot~[tnpos=a]{$I$}\taput{\footnotesize $0,6$}}
{
\Tdot~[tnpos=r]{$S$}\taput{\footnotesize $0,75$}
\Tdot~[tnpos=r]{$\overline{S}$}\tbput{\footnotesize $0,25$}
}
\pstree
{\Tdot~[tnpos=a]{$M$}\tbput{\footnotesize $0,3$}}
{
\Tdot~[tnpos=r]{$S$}\taput{\footnotesize $0,9$}
\Tdot~[tnpos=r]{$\overline{S}$}\tbput{\footnotesize $0,1$}
}
\pstree
{\Tdot~[tnpos=a]{$G$}\tbput{\footnotesize $0,1$}}
{
\Tdot~[tnpos=r]{$S$}\taput{\footnotesize $0,8$}
\Tdot~[tnpos=r]{$\overline{S}$}\tbput{\footnotesize $0,2$}
}
}
\end{center}
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item Il s'agit de calculer $P(I\cap S)$~: \pause
 \begin{eqnarray*}
  P(I\cap S) & = & \pause P(I)\times P_I(S)\\ \pause
             & = & 0,6\times 0,75\\ \pause
             & = & 0,45
 \end{eqnarray*} \pause
La probabilité que le client ait réalisé son achat sur internet et soit satisfait du service clientèle est donc~: \pause
$$\boxed{P(I\cap S)=0,45}$$
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{2}
 \item Les événements $I$, $M$ et $G$ forment une partition de l'univers. \pause On a donc, d'après la formule des probabilités totales~: \pause
 \begin{eqnarray*}
  P(S) & = & \pause P(I)\times P_I(S)+P(M)\times P_M(S)+P(G)\times P_G(S)\\ \pause
       & = & 0,6\times 0,75+0,3\times 0,9+0,1\times 0,8\\ \pause
       & = & 0,45+0,27+0,08\\ \pause
       & = & 0,8 \pause
 \end{eqnarray*}
On a donc bien~:
$$\boxed{P(S)=0,8}$$
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{3}
 \item Il s'agit de calculer $P_S(I)$~: \pause
 \begin{eqnarray*}
  P_S(I) & = & \pause \dfrac{P(I\cap S)}{P(S)}\\ \pause
         & = & \dfrac{0,45}{0,8}\\ \pause
         & \approx & 0,563
 \end{eqnarray*} \pause
La probabilité que le client ait effectué son achat sur internet sachant qu'il est satisfait de son achat est donc~: \pause
$$\boxed{P_S(I)\approx 0,563}$$
\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(a)]
  \item On répète 30 fois, de façon identique et indépendante, \pause une épreuve de Bernoulli dont la probabilité de succès est égale à $0,8$. \pause La variable aléatoire $X$ compte le nombre de succès donc~: \pause
  $$\boxed{\text{$X$ suit une loi binomiale de paramètres $n=30$ et $p=0,8$}}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{4}
 \item
 \begin{enumerate}[(b)]
  \item On obtient, à l'aide de la calculatrice, \pause que la probabilité qu'au moins 25 clients soient satisfaits est~: \pause
  $$\boxed{P(X\geqslant 25)\approx 0,428}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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\begin{frame}

\begin{small}
\begin{enumerate}
 \setcounter{enumi}{5}
 \item Soit $n$ le nombre de clients contactés et $Y$ la variable aléatoire égale au nombre de clients satisfaits parmi eux. \pause La variable aléatoire $Y$ suit alors une loi binomiale de paramètres $n$ et $p=0,8$. \pause La probabilité qu'au moins un client ne soit pas satisfait est donc égale à $P(Y\leqslant n-1)$. \pause Or on a~:
 $$P(Y\leqslant n-1)= \pause 1-P(Y=n) \pause =1-0,8^n$$ \pause
 On a alors~:
 \begin{eqnarray*}
  P(Y\leqslant n-1)\geqslant 0,99 & \Longleftrightarrow & \pause 1-0,8^n\geqslant 0,99 \\ \pause
                                  & \Longleftrightarrow & 0,8^n\leqslant 0,01 \\ \pause
                                  & \Longleftrightarrow & \ln\left(0,8^n\right)\leqslant \ln(0,01) \quad \textsl{(croissance de $\ln$)}\\ \pause
                                  & \Longleftrightarrow & n\ln(0,8)\leqslant \ln(0,01)  \\ \pause
                                  & \Longleftrightarrow & n\geqslant \dfrac{\ln(0,01)}{\ln(0,8)} \quad \textsl{(car $\ln(0,8)<0$)} \pause
 \end{eqnarray*}

\leftskip=1.5cm

Or $\dfrac{\ln(0,01)}{\ln(0,8)}\approx 20,6$ \pause donc la taille minimale de l'échantillon de clients à contacter pour que la probabilité qu'au moins l'un d'entre eux ne soit pas satisfait soit supérieure à $0,99$ est~: \pause
$$\boxed{n=21}$$
\end{enumerate}
\end{small}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{6}
 \item
 \begin{enumerate}[(a)]
  \item Par linéarité de l'espérance, on a~: \pause
 $$E(T)= \pause E(T_1)+E(T_2) \pause =4+3 \pause =7$$ \pause
 Et comme les variables aléatoires $T_1$ et $T_2$ sont indépendantes~: \pause
 $$V(T)=\pause V(T_1)+V(T_2) \pause =2+1 \pause =3$$ \pause
Soit~:
$$\boxed{E(T)=7} \quad \text{et} \quad \boxed{V(T)=3}$$
 \end{enumerate}
\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{6}
 \item
 \begin{enumerate}[(b)]
  \item On s'intéresse ici à la probabilité de l'événement $\left(5\leqslant T\leqslant 9\right)$. \pause Son événement contraire est $(T<5)\cup(T>9)$ \pause ou encore, comme $T$ ne prend que des valeurs entières, $(T\leqslant 4)\cup(T\geqslant 10)$. \pause Cet événement s'écrit également $(|T-7|\geqslant 3)$. \pause Or , d'après l'inégalité de Bienaymé-Tchebychev~: \pause
 $$P\left(|T-E(T)|\geqslant 3|\right)\leqslant \dfrac{V(T)}{3^2}$$ \pause
 Soit~:
 $$P\left(|T-7|\geqslant 3|\right)\leqslant \dfrac{3}{3^2}$$ \pause
 Et donc~:
 $$P\left(|T-7|\geqslant 3|\right)\leqslant \dfrac{1}{3}$$ \pause
 Et donc, en passant à l'événement contraire~:
 $$P\left(|T-7|< 3|\right)\geqslant 1-\dfrac{1}{3} \pause \quad \text{soit} \quad P\left(4<T<10\right)\geqslant \dfrac{2}{3}$$ \pause

\leftskip=1.5cm

 Et donc, la probabilité que le client reçoive son téléviseur entre $5$ et $9$ jours après sa commande est~: \pause
 $$\boxed{P\left(5\leqslant T\leqslant 9\right)\geqslant \dfrac{2}{3}}$$
 \end{enumerate}
\end{enumerate}



\end{frame}




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\begin{frame}{Exercice 3}

\pause
\begin{enumerate}
 \item
 \begin{enumerate}[(a)]
  \item On a $\overrightarrow{CA}\begin{pmatrix}5\\5\\-10\end{pmatrix}$ \pause et $\overrightarrow{CD}\begin{pmatrix}0\\0\\-\mfrac{25}{2} \end{pmatrix}$ \pause et donc~:
  \begin{itemize}
   \item $\overrightarrow{n_1}.\overrightarrow{CA}= \pause 1\times 5+(-1)\times 5+0\times(-10) \pause =5-5+0 \pause =0$. \pause
   \item $\overrightarrow{n_1}.\overrightarrow{CD}= \pause 1\times 0+(-1)\times 0+0\times\left(-\dfrac{25}{2}\right) \pause =0$. \pause
  \end{itemize}
  Le vecteur $\overrightarrow{n_1}$ est donc orthogonal à deux vecteurs non colinéaires du plan $(CAD)$. \pause On en déduit que~:
  $$\boxed{\text{Le vecteur $\overrightarrow{n_1}$ est normal au plan $(CAD)$}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \item
 \begin{enumerate}[(b)]
  \item D'après la question précédente, le plan $(CAD)$ admet une équation cartésienne de la forme~: \pause
  $$x-y+d=0 \quad \text{avec $d\in\mathbb{R}$}$$ \pause
  De plus, le point $C(0\,;\,0\,;\,10)$ appartient à ce plan donc ses coordonnées vérifient l'équation, \pause soit $0-0+d=0$ \pause et donc $d=0$. \pause Le plan $(CAD)$ a donc pour équation cartésienne~:
  $$\boxed{x-y=0}$$
 \end{enumerate}

\end{enumerate}


\end{frame}





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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item On injecte les coordonnées de la représentation paramétrique de $\mathscr{D}$ dans l'équation cartésienne de $(CAD)$~: \pause
 \begin{eqnarray*}
  \dfrac{5}{2}t-\left(5-\dfrac{5}{2}t\right)=0 & \Longleftrightarrow & \pause \dfrac{5}{2}t-5+\dfrac{5}{2}t=0\\ \pause
      & \Longleftrightarrow & 5t-5=0\\ \pause
       & \Longleftrightarrow & 5t=5\\ \pause
        & \Longleftrightarrow & t=1 \pause
 \end{eqnarray*}
Le point $H$ est donc le point de paramètre $t=1$ dans la représentation paramétrique de $\mathscr{D}$, soit~: \pause
$$\boxed{H\left(\dfrac{5}{2}\,;\,\dfrac{5}{2}\,;\,0\right)}$$
 \end{enumerate}

\end{enumerate}


\end{frame}




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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item Le point $B$ appartient à la droite $\mathscr{D}$ \pause car c'est le point de paramètre $t=0$ dans la représentation paramétrique précédente. \pause Et la droite $\mathscr{D}$ est orthogonale au plan $(CAD)$ \pause car elle est dirigée par le vecteur $\renewcommand{\arraystretch}{1.7} \begin{pmatrix}\mfrac{5}{2}\\-\mfrac{5}{2}\\0 \end{pmatrix}$ \pause qui est colinéaire au vecteur $\overrightarrow{n_1}$, normal au plan $(CAD)$. \pause On en déduit que~: \pause
$$\boxed{\text{$H$ est le projeté orthogonal de $B$ sur le plan $(CAD)$}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(a)]
  \item On a $\renewcommand{\arraystretch}{1.7}\overrightarrow{HB}\begin{pmatrix}-\mfrac{5}{2}\\\mfrac{5}{2}\\0 \end{pmatrix}$ \pause et $\renewcommand{\arraystretch}{1.7}\overrightarrow{HA}\begin{pmatrix}\mfrac{5}{2}\\\mfrac{5}{2}\\0 \end{pmatrix}$ \pause et donc~:
  $$\overrightarrow{HB}.\overrightarrow{HA}= \pause -\dfrac{5}{2}\times \dfrac{5}{2}+\dfrac{5}{2}\times \dfrac{5}{2}+0\times 0 \pause =-\dfrac{25}{4}+\dfrac{25}{4} \pause =0$$ \pause
  Les vecteurs $\overrightarrow{HB}$ et $\overrightarrow{HA}$ sont donc orthogonaux. \pause On en déduit que~:
  $$\boxed{\text{Le triangle $ABH$ est rectangle en $H$}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}





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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{2}
 \item
 \begin{enumerate}[(b)]
  \item On a~:
  $$HB= \pause \sqrt{\left(-\dfrac{5}{2}\right)^2+\left(\dfrac{5}{2}\right)^2+0^2} \pause =\sqrt{\dfrac{25}{4}+\dfrac{25}{4}} \pause =\sqrt{\dfrac{25}{2}}$$ \pause
  De même, $HA= \pause \sqrt{\dfrac{25}{2}}$ \pause et comme le triangle $ABH$ est rectangle en $H$, son aire est~: \pause
  $$\mathscr{A}_{ABH}= \pause \dfrac{HA\times HB}{2} \pause =\dfrac{\sqrt{\mfrac{25}{2}}\times \sqrt{\mfrac{25}{2}}}{2} \pause =\dfrac{\mfrac{25}{2}}{2} \pause =\dfrac{25}{4}$$ \pause
  L'aire du triangle $ABH$ est donc bien~: \pause
  $$\boxed{\mathscr{A}_{ABH}=\dfrac{25}{4}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(a)]
  \item On a $\overrightarrow{CO}\begin{pmatrix}0\\0\\-10\end{pmatrix}$ \pause et donc~:
  \begin{itemize}
   \item $\overrightarrow{CO}.\overrightarrow{HB}= \pause 0\times \left(-\dfrac{5}{2}\right)+0\times \left(\dfrac{5}{2}\right)+(-10)\times 0 \pause =0$. \pause
   \item $\overrightarrow{CO}.\overrightarrow{HA}= \pause 0\times \left(\dfrac{5}{2}\right)+0\times \left(\dfrac{5}{2}\right)+(-10)\times 0 \pause =0$. \pause
  \end{itemize}
La droite $(CO)$ est donc orthogonale au plan $(ABH)$, \pause on en déduit que~:
$$\boxed{\text{$(CO)$ est la hauteur issue de $C$ dans le tétraèdre $ABCH$}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}




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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{3}
 \item
 \begin{enumerate}[(b)]
  \item Le volume du tétraèdre $ABCH$ est donc~: \pause
  \begin{eqnarray*}
   \mathscr{V}_{ABCH} & = & \pause \dfrac{1}{3}\times \mathscr{A}_{ABH}\times CO\\ \pause
                      & = & \dfrac{1}{3}\times \dfrac{25}{4}\times 10\\ \pause
                      & = & \dfrac{25\times 10}{3\times 4}\\ \pause
                      & = & \dfrac{125}{6} \pause
  \end{eqnarray*}
  Soit~:
  $$\boxed{\mathscr{V}_{ABCH}=\dfrac{125}{6}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{4}
 \item Soit $h$ la distance du point $H$ au plan $(ABC)$, \pause il s'agit de la distance entre le point $H$ et le projeté orthogonal de $H$ sur le plan $(ABC)$, \pause autrement dit de la hauteur issue de $H$ dans le tétraèdre $ABCH$. \pause En prenant le triangle $ABC$ comme base, le volume du tétraèdre $ABCH$ est égal à~: \pause
 $$\mathscr{V}_{ABCH}= \pause \dfrac{1}{3}\times \mathscr{A}_{ABC}\times h$$ \pause
 où $\mathscr{A}_{ABC}$ désigne l'aire du triangle $ABC$. \pause Or $BA= \pause \sqrt{5^2+0^2+0^2} \pause =5$ \pause et $BC= \pause \sqrt{0^2+5^2+10^2} \pause =\sqrt{125} \pause =5\sqrt{5}$, \pause et comme le triangle $ABC$ est rectangle en $B$, on a~: \pause
 $$\mathscr{A}_{ABC}= \pause \dfrac{BA\times BC}{2} \pause =\dfrac{5\times 5\sqrt{5}}{2} \pause =\dfrac{25\sqrt{5}}{2}$$ \pause
 Ainsi~:
 $$\mathscr{V}_{ABCH}= \pause \dfrac{1}{3}\times \dfrac{25\sqrt{5}}{2} \times h=\dfrac{25\sqrt{5}}{6}\times h$$
\end{enumerate}


\end{frame}



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\begin{frame}


 Et d'autre part, d'après la question précédente $\mathscr{V}_{ABCH}=\dfrac{125}{6}$ \pause d'où~:
 $$\dfrac{25\sqrt{5}}{6}\times h=\dfrac{125}{6}$$ \pause
 Et donc~:
 $$h= \pause \dfrac{125}{6}\times \dfrac{6}{25\sqrt{5}}$$ \pause
Finalement, la distance du point $H$ au plan $(ABC)$ est~: \pause
$$\boxed{h=\sqrt{5}}$$




\end{frame}





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\begin{frame}{Exercice 4 - Partie A}

\pause
\begin{enumerate}
 \item
 \begin{enumerate}[(a)]
  \item On a~:
 $$\boxed{\lim\limits_{x\to 0}f(x)=-\infty} \pause \quad \text{car} \quad \begin{dcases}\lim\limits_{x\to 0}(x-2)=-2\\\lim\limits_{x\to 0}\left(\dfrac{1}{2}\ln(x)\right)=-\infty \end{dcases}$$ \pause
 Et~:
 $$\boxed{\lim\limits_{x\to +\infty}f(x)=+\infty} \pause \quad \text{car} \quad \begin{dcases}\lim\limits_{x\to +\infty}(x-2)=+\infty\\\lim\limits_{x\to +\infty}\left(\dfrac{1}{2}\ln(x)\right)=+\infty \end{dcases}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \item
 \begin{enumerate}[(b)]
  \item La fonction $f$ est dérivable sur $\left]0\,;\,+\infty\right[$ et, pour tout $x\in\left]0\,;\,+\infty\right[$, on a~: \pause
 \begin{eqnarray*}
  f'(x) & = & \pause 1+\dfrac{1}{2}\times \dfrac{1}{x}\\ \pause
        & = & 1+\dfrac{1}{2x}\\ \pause
        & = & \dfrac{2x+1}{2x} \pause
 \end{eqnarray*}
Soit~:
$$\boxed{f'(x)=\dfrac{2x+1}{2x}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \item
 \begin{enumerate}[(c)]
  \item Sur l'intervalle $\left]0\,;\,+\infty\right[$, \pause on a $2x+1>0$ \pause et $2x>0$ \pause donc $f'(x)>0$. \pause On en déduit que~:
 $$\boxed{\text{La fonction $f$ est strictement croissante sur $\left]0\,;\,+\infty\right[$}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \item
 \begin{enumerate}[(d)]
  \item La fonction $f'$ est dérivable sur $\left]0\,;\,+\infty\right[$ et, pour tout $x\in\left]0\,;\,+\infty\right[$, on a~: \pause
 \begin{eqnarray*}
  f''(x) & = & \pause \dfrac{2\times 2x-(2x+1)\times 2}{\left(2x\right)^2}\\ \pause
         & = & \dfrac{4x-4x-2}{4x^2}\\ \pause
         & = & \dfrac{-2}{4x^2}\\ \pause
         & = & \dfrac{-1}{2x^2} \pause
 \end{eqnarray*}
Pour tout $x\in\left]0\,;\,+\infty\right[$, \pause $2x^2>0$ \pause donc $f''(x)<0$. \pause On en déduit que~: \pause
 $$\boxed{\text{La fonction $f$ est concave sur $\left]0\,;\,+\infty\right[$}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item Sur l'intervalle $\left]0\,;\,+\infty\right[$, \pause la fonction $f$ est continue et strictement croissante. \pause De plus $\lim\limits_{x\to 0}f(x)=-\infty$ \pause et $\lim\limits_{x\to +\infty}f(x)=+\infty$. \pause Or $0\in\left]-\infty\,;\,+\infty\right[$ \pause donc, d'après le théorème des valeurs intermédiaires, \pause l'équation $f(x)=0$ admet une unique solution $\alpha$ sur $\left]0\,;\,+\infty\right[$. \pause De plus, on a $f(1)=-1<0$ \pause et $f(2)=\dfrac{1}{2}\ln(2)>0$ \pause donc~:
  $$\boxed{1<\alpha<2}$$
 \end{enumerate}

\end{enumerate}


\end{frame}




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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item La fonction $f$ est strictement croissante sur $\left]0\,;\,+\infty\right[$ et elle s'annule en $\alpha$. \pause On en déduit le tableau de signes~: \pause

\begin{center}
\begin{tikzpicture}[scale=1]
\tkzTabInit[ lgt=2, deltacl=0.6, espcl=2.2]%
{$x$ /0.7,
$f(x)$ /0.7}
{ $0$ , $\alpha$ , $+\infty$}
\tkzTabLine { d , - , z , + , }
\end{tikzpicture}
\end{center}
 \end{enumerate}

\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(c)]
  \item On sait que $f(\alpha)=0$. \pause Or~:
  \begin{eqnarray*}
   f(\alpha)=0 & \Longleftrightarrow & \pause \alpha-2+\dfrac{1}{2}\ln(\alpha)=0\\ \pause
               & \Longleftrightarrow & \dfrac{1}{2}\ln(\alpha)=2-\alpha\\ \pause
               & \Longleftrightarrow & \ln(\alpha)=2(2-\alpha) \pause
  \end{eqnarray*}
On a donc bien~:
$$\boxed{\ln(\alpha)=2(2-\alpha)}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{frame}{Partie B}

\pause
\begin{enumerate}
 \item La fonction $g$ est dérivable sur $\left]0\,;\,1\right]$ et, pour tout $x\in \left]0\,;\,1\right]$, on a~: \pause
 \begin{eqnarray*}
  g'(x) & = & \pause -\dfrac{7}{4}x+1-\dfrac{1}{4}\left(2x\ln(x)+x^2\times\dfrac{1}{x}\right)\\ \pause
        & = & -\dfrac{7}{4}x+1-\dfrac{1}{2}x\ln(x)-\dfrac{1}{4}x\\ \pause
        & = & -2x+1-\dfrac{1}{2}x\ln(x) \pause
 \end{eqnarray*}
Et d'autre part, pour tout $x\in \left]0\,;\,1\right]$~: \pause
\begin{eqnarray*}
 xf\left(\dfrac{1}{x}\right) & = & \pause x\left(\dfrac{1}{x}-2+\dfrac{1}{2}\ln\left(\dfrac{1}{x}\right)\right)\\ \pause
    & = & 1-2x+\dfrac{1}{2}x\ln\left(\dfrac{1}{x}\right)\\ \pause
    & = & -2x+1-\dfrac{1}{2}x\ln(x) \qquad \textsl{(car $\ln\left(\mfrac{1}{x}\right)=-\ln(x)$)} \pause
\end{eqnarray*}

\leftskip=1.5cm

On a donc bien~:
$$\boxed{g'(x)=xf\left(\dfrac{1}{x}\right)}$$
\end{enumerate}


\end{frame}



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\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(a)]
  \item Si $x\in\left]0\,;\,\dfrac{1}{\alpha}\right[$ \pause alors $\dfrac{1}{x}\in\left]\alpha\,;\,+\infty\right[$ \pause donc, d'après le tableau de signes de $f$~: \pause
  $$\boxed{f\left(\dfrac{1}{x}\right)>0}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item
 \begin{enumerate}[(b)]
  \item On a vu que pour tout $x\in \left]0\,;\,1\right]$, \pause $g'(x)=xf\left(\dfrac{1}{x}\right)$. \pause On en déduit le tableau~: \pause

  \begin{center}
\begin{tikzpicture}[scale=0.8]
\tkzTabInit[ lgt=3.5, deltacl=0.6, espcl=3]%
{$x$ /1,
$x$ /1,
$f\left(\mfrac{1}{x}\right)$ /1,
$g'(x)$ /1,
$g(x)$ /1.8}
{ $0$ , $\mfrac{1}{\alpha}$ , $1$}
\tkzTabLine { , + , t , + , }
\tkzTabLine { d , + , z , - , }
\tkzTabLine { d , + , z , - , }
\tkzTabVar {D-/ , +/ , -/}
\end{tikzpicture}
\end{center}
 \end{enumerate}

\end{enumerate}


\end{frame}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{frame}{Partie C}


\begin{enumerate}
 \item
 \begin{enumerate}[(a)]
  \item Pour tout $x\in \left]0\,;\,1\right]$, on a~: \pause
   $$g(x)-\left(-\dfrac{7}{8}x^2+x\right)= \pause -\dfrac{1}{4}x^2\ln(x)$$ \pause
   Or, pour tout $x\in \left]0\,;\,1\right]$, \pause $\ln(x)\leqslant 0$ \pause donc~:
   $$g(x)-\left(-\dfrac{7}{8}x^2+x\right)\geqslant 0$$ \pause
On en déduit que~: \pause
$$\boxed{\text{$\mathscr{C}_g$ est au-dessus de $\mathscr{P}$ sur $\left]0\,;\,1\right]$}}$$
 \end{enumerate}

\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}


\begin{enumerate}
 \item
 \begin{enumerate}[(b)]
  \item On pose~:
  $$\begin{dcases}u(x)=\ln(x)\\v'(x)=x^2\end{dcases} \pause \quad \text{et} \quad \begin{dcases}u'(x)=\dfrac{1}{x}\\v(x)=\dfrac{1}{3}x^3\end{dcases}$$ \pause
  On a alors, par intégration par parties~:
 \end{enumerate}

\end{enumerate}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

\begin{eqnarray*}
   \int_{\text{$\frac{1}{\alpha}$}}^1x^2\ln x\,\mathrm{d}x & = & \pause \left[\dfrac{1}{3}x^3\ln(x)\right]_{\text{$\frac{1}{\alpha}$}}^1-\int_{\text{$\frac{1}{\alpha}$}}^1 \dfrac{1}{x}\times \dfrac{1}{3}x^3 \,\mathrm{d}x\\ \pause
   & = & \left[\dfrac{1}{3}x^3\ln(x)\right]_{\text{$\frac{1}{\alpha}$}}^1-\int_{\text{$\frac{1}{\alpha}$}}^1 \dfrac{1}{3}x^2 \,\mathrm{d}x\\ \pause
    & = & \dfrac{1}{3}\times 1^3\times\underbrace{\ln(1)}_{0}-\dfrac{1}{3}\times \left(\dfrac{1}{\alpha}\right)^3\underbrace{\ln\left(\dfrac{1}{\alpha}\right)}_{-\ln(\alpha)}-\left[\dfrac{1}{9}x^3\right]_{\text{$\frac{1}{\alpha}$}}^1\\ \pause
    & = & \dfrac{\ln(\alpha)}{3\alpha^3}-\left(\dfrac{1}{9}-\dfrac{1}{9}\times \dfrac{1}{\alpha^3}\right)\\ \pause
    & = & \dfrac{2(2-\alpha)}{3\alpha^3}-\dfrac{1}{9}+\dfrac{1}{9\alpha^3} \qquad \textsl{(car $\ln(\alpha)=2(2-\alpha)$)}\\ \pause
    & = & \dfrac{12-6\alpha-\alpha^3+1}{9\alpha^3}\\ \pause
    & = & \dfrac{-\alpha^3-6\alpha+13}{9\alpha^3}
  \end{eqnarray*}


\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\begin{frame}

On a donc bien~:
  $$\boxed{\int_{\text{$\frac{1}{\alpha}$}}^1x^2\ln x\,\mathrm{d}x=\dfrac{-\alpha^3-6\alpha+13}{9\alpha^3}}$$

\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}


\begin{enumerate}
 \setcounter{enumi}{1}
 \item Sur l'intervalle $\left]0\,;\,1\right]$, $\mathscr{C}_g$ est au-dessus de $\mathscr{P}$ donc~: \pause
  \begin{eqnarray*}
   \mathscr{A} & = & \pause \int_{\text{$\frac{1}{\alpha}$}}^1 g(x)-\left(-\dfrac{7}{8}x^2+x\right)\,\mathrm{d}x\\ \pause
    & = & \int_{\text{$\frac{1}{\alpha}$}}^1 -\dfrac{1}{4}x^2\ln(x)\,\mathrm{d}x\\ \pause
    & = & -\dfrac{1}{4}\int_{\text{$\frac{1}{\alpha}$}}^1 x^2\ln(x)\,\mathrm{d}x\\ \pause
    & = & -\dfrac{1}{4}\times \dfrac{-\alpha^3-6\alpha+13}{9\alpha^3} \pause
  \end{eqnarray*}
Soit~:
  $$\boxed{\mathscr{A}=\dfrac{\alpha^3+6\alpha-13}{36\alpha^3}}$$
\end{enumerate}


\end{frame}







\end{document}









