Source Latex
de la correction du devoir
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pdfauthor={Yoann Morel},
pdfsubject={Devoir de mathématiques: probabilités, fonction, fonction affine, système d'équation, courbe représentative d'une fonction},
pdftitle={Devoir de mathématiques},
pdfkeywords={probabilités, fonction, fonction affine, système d'équation, courbe représentative d'une fonction, mathématiques, seconde, 2nde}
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\begin{document}
\ct{\bf\LARGE{Correction du devoir de math\'ematiques}}
\bgex
\bgen
\item cf. cours.
\bgen[a)]
\item $f(-1)=-a(-1)^2+b(-1)+a+b=-a-b+a+b=0$, et donc
on a $A\in\mathcal{C}_f$.
\item $B(1;2)\in\mathcal{C}_f\iff f(1)=-a+b+a+b=2\iff 2b=2\iff b=1$ \\
et
$C(0;3)\in\mathcal{C}_f\iff f(0)=a+b=3$,
soit, avec $b=2$, on trouve $a=1$.
\enen
\enen
\enex
\bgex
\bgen
\item Soit $A(x;y)$ alors comme $A$ appartient \`a l'axe des ordonn\'ees,
on a $x=0$, et, comme $A\in\mathcal{C}_f$, $y=-2x+5$ soit, avec $x=0$,
$y=5$.
Ainsi on trouve $A(0;5)$.
Soit $B(x;y)$ alors comme $B$ appartient \`a l'axe des abscisses,
on a $y=0$, puis comme $B\in\mathcal{C}_f$, $y=-2x+5$ soit, avec $y=0$,
$2x=5\iff x=\dfrac52$.
Ainsi, on trouve $B\lp\dfrac52;0\rp$.
\item
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\item Soit $M(x;y)$ le point d'intersection de $\mathcal{C}_f$
et $\mathcal{C}_g$, alors on a
$\la\bgar{ccr}
M(x;y)\in\mathcal{C}_f\iff y&=&-2x+5\\
M(x;y)\in\mathcal{C}_g\iff y&=&3x-1
\enar\right.$\\
d'o\`u $y=-2x+5=3x-1$ donc
$5x=6\iff x=\dfrac65$,
et alors $y=-2x+5=-2\tm\dfrac65+5=\dfrac{13}{5}$.
Ainsi, le point d'intersection est
$M\lp\dfrac65;\dfrac{13}{5}\rp$.
\enen
\enex
\bgex
Il y 6 issues possibles: 1, 2, 3 ou 4.
Avec un arbre, en lan\c cant deux dés, on dénombre
$4\tm4=16$ tirages possibles.
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La loi de probabilité de ce jeux est:
\[\begin{tabular}{|c|c|c|c|c|}\hline
Issues & 1 & 2 & 3 & 4 \\\hline
\rule[-1em]{0cm}{3em}
Probabilités & $\dfrac{7}{16}$ & $\dfrac{5}{16}$
& $\dfrac{3}{16}$ & $\dfrac{1}{16}$\\
\hline
\end{tabular}\]
\enex
\label{LastPage}
\end{document}
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