\documentclass[12pt,a4paper]{article}
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\begin{document}\everymath{\displaystyle}
	
\thispagestyle{empty}
\boxit{4pt}{Minitest 13}\rightarrowfill\quad\boxit{3pt}{Corrigé}\bigskip

1) Les composantes du vecteur $\vect{AB}$ lorsque  $A(1,3)$ et $B(5,1)$\vskip 0.5cm

{\red On calcule $\vect{AB}\begin{bmatrix}5-1\\1-3\end{bmatrix} = \begin{bmatrix}4\\-2\end{bmatrix}$}

\vskip 0.5cm

2) Calculer le produit scalaire $\vect{u}\cdot\vect{v}$ \vskip 0.5cm

\qquad a) lorsque $\|\vect{u}|| = 3$ et $||\vect{v}|| = 5$ et $(\vect{u},\vect{v}) = \dfrac{\pi}{3}$\vskip 0.5cm

\qquad{\red $\vect{u}\cdot\vect{v} = \|\vect{u}\|\times\|\vect{v}\|\times cos(\widehat{\vect{u},\vect{v}}) = 3\times 5\times \dfrac12 = 7,5$}\vskip 0.5cm

\qquad b) $\vect{u}=\begin{pmatrix}-3\cr2\end{pmatrix}$ et $\vect{v}=\begin{pmatrix}4\cr1\end{pmatrix}$\vskip 0.5cm

\qquad {\red $\vect{u}\cdot\vect{v}=\begin{pmatrix}-3\cr2\end{pmatrix}\cdot\begin{pmatrix}4\cr1\end{pmatrix} = -3\times 4+2\times 1 = -12+2 = -10$}\vskip 0.5cm

3) Soit $f(x) = x^2-6x+1$

\qquad a) Calculer $f'(x)$\vskip 0.5cm

\qquad{\red $f'(x) = 2x-6$}\vskip 0.5cm

\qquad b) Résoudre $f'(x) = 0$\vskip 0.5cm

\qquad{\red $f'(x) = 0$ ssi $2x-6 = 0$ ssi $x=3$}\vskip 0.5cm

\qquad c) En déduire les variations de $f$

\qquad{\red $f'$ s'annule en 3 est est une fonction affine croissante, on a donc le tableau suivant : \medskip

\hfil\begin{tikzpicture}[scale=0.75]
	\tkzTabInit{$x$/1,$2x-6$/1,$f'(x)$/1,$f(x)$/2}{$-\infty$,$3$,$+\infty$}
	\tkzTabLine{ ,-,z,+}
	\tkzTabLine{ ,-,z,+}
	\tkzTabVar{+/,-/$f(3)$,+/}
\end{tikzpicture}


}\vskip 0.5cm

Simplifier les expressions : \bigskip

\(\overrightarrow{AC} + \overrightarrow{CB} + \overrightarrow{BA} = {\red \vect{AB} + \vect{BA} = \vect{AA} = \vect{0}}\).\vskip 0.5cm

 \(\overrightarrow{AB} - \overrightarrow{CB} = {\red \vect{AB} + \vect{BC}= \vect{AC}}\).\vskip 0.5cm

 \(\overrightarrow{AC} + \overrightarrow{CD} - \overrightarrow{AD} = {\red \vect{AD} + \vect{DA} = \vect{AA} = \vect{0}}\).

\newpage

\thispagestyle{empty}
\boxit{4pt}{Minitest 13}\rightarrowfill\quad\boxit{3pt}{Corrigé}\bigskip

1) Les composantes du vecteur $\vect{AB}$ lorsque  $A(7,3)$ et $B(1,8)$\vskip 0.5cm

{\red On calcule $\vect{AB}\begin{bmatrix}1-7\\8-3\end{bmatrix} = \begin{bmatrix}-6\\5\end{bmatrix}$}

\vskip 0.5cm

2) Calculer le produit scalaire $\vect{u}\cdot\vect{v}$ \vskip 0.5cm

\qquad a) lorsque $\|\vect{u}|| = 2$ et $||\vect{v}|| = 8$ et $(\vect{u},\vect{v}) = 120°$\vskip 0.5cm

\qquad{\red $\vect{u}\cdot\vect{v} = \|\vect{u}\|\times\|\vect{v}\|\times cos(\widehat{\vect{u},\vect{v}}) = 2\times 8\times \dfrac{-1}2 = -8$}\vskip 0.5cm

\qquad b) $\vect{u}=\begin{pmatrix}5\cr2\end{pmatrix}$ et $\vect{v}=\begin{pmatrix}-3\cr2\end{pmatrix}$\vskip 0.5cm

\qquad {\red $\vect{u}\cdot\vect{v}=\begin{pmatrix}5\cr2\end{pmatrix}\cdot\vect{v}=\begin{pmatrix}-3\cr2\end{pmatrix} = 5\times (-3)+2\times 2 = -15+4 = -11$}\vskip 0.5cm

3) Soit $f(x) = x^2+4x+2$

\qquad a) Calculer $f'(x)$\vskip 0.5cm

\qquad{\red $f'(x) = 2x+4$}\vskip 0.5cm

\qquad b) Résoudre $f'(x) = 0$\vskip 0.5cm

\qquad{\red $f'(x) = 0$ ssi $2x+4 = 0$ ssi $x=-2$}\vskip 0.5cm

\qquad c) En déduire les variations de $f$

\qquad{\red $f'$ s'annule en -2 est est une fonction affine croissante, on a donc le tableau suivant : \medskip
	
	\hfil\begin{tikzpicture}[scale=0.75]
		\tkzTabInit{$x$/1,$2x+4$/1,$f'(x)$/1,$f(x)$/2}{$-\infty$,$-2$,$+\infty$}
		\tkzTabLine{ ,-,z,+}
		\tkzTabLine{ ,-,z,+}
		\tkzTabVar{+/,-/$f(-2)$,+/}
	\end{tikzpicture}
	
	
}\vskip 0.5cm

Simplifier les expressions : \bigskip

\(\overrightarrow{AC} + \overrightarrow{CB} + \overrightarrow{BA} = {\red \vect{AB} + \vect{BA} = \vect{AA} = \vect{0}}\).\vskip 0.5cm

\(\overrightarrow{AB} - \overrightarrow{CB} = {\red \vect{AB} + \vect{BC}= \vect{AC}}\).\vskip 0.5cm

\(\overrightarrow{AC} + \overrightarrow{CD} - \overrightarrow{AD} = {\red \vect{AD} + \vect{DA} = \vect{AA} = \vect{0}}\).

\end{document} 