Devoir 2 S02 en limites, dérivation et vecteurs l'espace v1
📅 April 15, 2023 | 👁️ Views: 359
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Prof MOSAID & Control 2 -- 1BACSF-1 & 2h \\
\bottomrule
\end{tabularx}\\
\textbf{\underline{Exercice 1:(7.5pts)}}\\
\begin{tabular}{@{}>{\centering\arraybackslash}m{0.05\textwidth}|p{0.92\textwidth}}
3 & 1)- Calculer les limites suivantes: (utiliser les nombres dérivés)\\
& \hspace*{0.5cm} $\displaystyle{\lim_{x \to \pi } \frac{cosx+1}{x-\pi} }$ \hspace*{0.2cm};\hspace*{0.2cm}
$\displaystyle{\lim_{x \to 1} \frac{x^3-\sqrt{x}+2x-2}{x-1} }$ \\
& \\
1.5 & 2)- Calculer le nombre dérivé: $f(x)=x^2+3x-2$ et $x_0=3$ \\
& \\
3 & 3)- Etudier la dérivablité au point $x_0=2$
$\begin{cases}
f(x) = - \frac{1}{2}x^2+5 \hspace*{0.2cm}; \hspace*{0.2cm} x \le 2 \\
f(x) = - \frac{x+1}{x-1} \hspace*{0.9cm};\hspace*{0.2cm} x > 2 \\
\end{cases}$\\
\bottomrule
\end{tabular}\\
\textbf{\underline{Exercice 2:(8pts)}}\\
\begin{tabular}{@{}>{\centering\arraybackslash}m{0.05\textwidth}|p{0.92\textwidth}}
2$\times$4 & Calculer les fonctions dérivées des fonctions:\\
&\\
& \hspace*{0.5cm} $f(x)=3x^4-7x^2+x-7$ \hspace*{0.2cm};\hspace*{0.2cm} $f(x) = \frac{2x-3}{x+1}$ \hspace*{0.2cm};
\hspace*{0.2cm} $f(x)=\sqrt{x}cosx$ \hspace*{0.2cm};\hspace*{0.2cm} $f(x) = \frac{x^2-3x-1}{\sqrt{x+1}}$ \\
&\\
\bottomrule
\end{tabular}\\
\textbf{\underline{Exercice 3:(4.5pts)}}\\
\begin{minipage}{0.60\textwidth}
\begin{tabular}{@{}>{\centering\arraybackslash}m{0.08\textwidth}|p{0.92\textwidth}}
& Soit le parallèlogramme $ABCDEFGH$ \\
3 & 1)- Simplifier les sommes :\\
& \\
& \hspace*{0.5cm}$\overrightarrow{HG}+\overrightarrow{CB}$ \hspace*{0.2cm};\hspace*{0.2cm}
$\overrightarrow{HE}+\overrightarrow{HG}+\overrightarrow{FH}$
\hspace*{0.2cm};\hspace*{0.2cm} $\overrightarrow{GH}-\overrightarrow{FG}+\overrightarrow{GC}$\\
&\\
1.5 & 2)- Montrer que les vecteurs $\overrightarrow{BE}$, $\overrightarrow{BC}$ et $\overrightarrow{BH}$ \\
& \hspace*{0.5cm} sont coplanaires. \\
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