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<feed xmlns="http://www.w3.org/2005/Atom"><title>blog.bernatchez.net - calcul</title><link href="https://blog.bernatchez.net/lang-version.fr/" rel="alternate"/><link href="https://blog.bernatchez.net/lang-version.fr/feeds/calcul.atom.xml" rel="self"/><id>https://blog.bernatchez.net/lang-version.fr/</id><updated>2026-05-19T02:21:52+00:00</updated><entry><title>IB Calcul Problème 1</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq1-fr.html" rel="alternate"/><published>2026-05-19T02:21:52+00:00</published><updated>2026-05-19T02:21:52+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq1-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 1&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Soit la fonction &lt;span class="math"&gt;\(f(x) = 6x^2-3x\)&lt;/span&gt; representé ci-dessous.&lt;/p&gt;
&lt;p&gt;La figure n'est pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/6xsq-3x.png" /&gt;
&lt;/div&gt;
&lt;ol class="upperalpha"&gt;
&lt;li&gt;&lt;p class="first"&gt;Trouvez &lt;span class="math"&gt;\(\int \! (6x^2-3x) \, \mathrm{d}x\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p class="first"&gt;Trouvez l’aire de la région délimitée par la représentation graphique de &lt;span class="math"&gt;\(f(x)\)&lt;/span&gt;,&lt;/p&gt;
&lt;p&gt;l’axe des abscisses et les droites &lt;span class="math"&gt;\(x = 1\)&lt;/span&gt; et &lt;span class="math"&gt;\(x = 2\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;C'est à dire &lt;span class="math"&gt;\(\int_1^2 \! (6x^2-3x) \, \mathrm{d}x\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 2</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq2-fr.html" rel="alternate"/><published>2026-05-19T02:21:50+00:00</published><updated>2026-05-19T02:21:50+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq2-fr.html</id><summary type="html">&lt;p class="first last"&gt;IB Calcul Problème 2&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Une boîte en métal, cylindrique et fermée, de rayon égal à &lt;span class="math"&gt;\(r\)&lt;/span&gt; centimètres et de hauteur égale à &lt;span class="math"&gt;\(h\)&lt;/span&gt; centimètres possède un volume de &lt;span class="math"&gt;\(20\pi\, cm^3\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;La figure nest pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/cylindre.png" /&gt;
&lt;/div&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Exprimez &lt;span class="math"&gt;\(h\)&lt;/span&gt; en fonction de &lt;span class="math"&gt;\(r\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
&lt;div class="line-block"&gt;
&lt;div class="line"&gt;&lt;br /&gt;&lt;/div&gt;
&lt;/div&gt;
&lt;p&gt;Le métal pour la base et le couvercle de la boîte coûte 10 cents le &lt;span class="math"&gt;\(cm^2\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;le métal pour le côté incurvé coûte &lt;span class="math"&gt;\(8\)&lt;/span&gt; cents le &lt;span class="math"&gt;\(cm^2\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Le coût total du métal, en cents, est de &lt;span class="math"&gt;\(C\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple" start="2"&gt;
&lt;li&gt;Montrez que &lt;span class="math"&gt;\(C\,=\,20\pi{}r^2 + \frac{320\pi}{r}\)&lt;/span&gt;&lt;/li&gt;
&lt;li&gt;Sachant qu’il existe une valeur minimale pour &lt;span class="math"&gt;\(C\)&lt;/span&gt;, trouvez cette valeur minimale en fonction de &lt;span class="math"&gt;\(\pi\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 3</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq3-fr.html" rel="alternate"/><published>2026-05-19T02:21:48+00:00</published><updated>2026-05-19T02:21:48+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq3-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 3&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Considérez une fonction &lt;span class="math"&gt;\(f(x)\)&lt;/span&gt;. la droite &lt;span class="math"&gt;\(L_1\)&lt;/span&gt; d’équation &lt;span class="math"&gt;\(y = 3x + 1\)&lt;/span&gt; est une tangente à la représentation graphique de &lt;span class="math"&gt;\(f(x)\)&lt;/span&gt; lorsque &lt;span class="math"&gt;\(x = 2\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;&lt;ol class="first lowerroman"&gt;
&lt;li&gt;Écrivez &lt;span class="math"&gt;\(f^\prime(2)\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Trouvez &lt;span class="math"&gt;\(f(2)\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
&lt;/li&gt;
&lt;/ol&gt;
&lt;p&gt;Soit &lt;span class="math"&gt;\(g(x) = f(x^2 + 1)\)&lt;/span&gt; et &lt;span class="math"&gt;\(P\)&lt;/span&gt; le point sur la représentation graphique de &lt;span class="math"&gt;\(g\)&lt;/span&gt; où &lt;span class="math"&gt;\(x = 1\)&lt;/span&gt;&lt;/p&gt;
&lt;ol class="upperalpha simple" start="2"&gt;
&lt;li&gt;Montrez que la représentation graphique de &lt;span class="math"&gt;\(g\)&lt;/span&gt; a une pente de 6 au point &lt;span class="math"&gt;\(P\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
&lt;p&gt;Soit &lt;span class="math"&gt;\(L_2\)&lt;/span&gt; la tangente à la représentation graphique de &lt;span class="math"&gt;\(g\)&lt;/span&gt; au point &lt;span class="math"&gt;\(P\)&lt;/span&gt;. &lt;span class="math"&gt;\(L_1\)&lt;/span&gt; coupe &lt;span class="math"&gt;\(L_2\)&lt;/span&gt; au point &lt;span class="math"&gt;\(Q\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple" start="3"&gt;
&lt;li&gt;Trouvez l’ordonnée de &lt;span class="math"&gt;\(Q\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 4</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq4-fr.html" rel="alternate"/><published>2026-05-19T02:21:46+00:00</published><updated>2026-05-19T02:21:46+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq4-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 4&lt;/p&gt;
</summary><content type="html">&lt;p&gt;La figure suivante donne la représentation graphique de &lt;span class="math"&gt;\(f(x) = a\,Cos\,bx\)&lt;/span&gt;,
pour &lt;span class="math"&gt;\(0 \leq x \leq 4\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;La figure n'est pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/temp_cos_wave.png" /&gt;
&lt;/div&gt;
&lt;p&gt;Il y a un point minimum en &lt;span class="math"&gt;\(P( 2, -3 )\)&lt;/span&gt; et un point maximum en &lt;span class="math"&gt;\(Q( 4, 3 )\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;&lt;ol class="first lowerroman"&gt;
&lt;li&gt;Donnez la valeur de &lt;span class="math"&gt;\(a\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Trouvez la valeur de &lt;span class="math"&gt;\(b\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
&lt;/li&gt;
&lt;li&gt;Donnez la pente de la courbe en &lt;span class="math"&gt;\(P\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Donnez l'équation de la normale à la courbe en &lt;span class="math"&gt;\(P\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 5</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq5-fr.html" rel="alternate"/><published>2026-05-19T02:21:45+00:00</published><updated>2026-05-19T02:21:45+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq5-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 5&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Soit &lt;span class="math"&gt;\(h(x) = \frac{6x}{cos x}\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Trouvez &lt;span class="math"&gt;\(h^\prime(0)\)&lt;/span&gt;.&lt;/p&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 6</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq6-fr.html" rel="alternate"/><published>2026-05-19T02:21:44+00:00</published><updated>2026-05-19T02:21:44+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq6-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 6&lt;/p&gt;
</summary><content type="html">&lt;p&gt;La figure suivante représente une partie de la représentation graphique de &lt;span class="math"&gt;\(f(x) = 2x\sqrt[2]{a^2 - x^2}\)&lt;/span&gt;, pour &lt;span class="math"&gt;\(-1 \leq x \leq a\)&lt;/span&gt;, où &lt;span class="math"&gt;\(a &amp;gt; 1\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;La figure n'est pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/temp_2xsqrtasq-xsq.png" /&gt;
&lt;/div&gt;
&lt;p&gt;La droite &lt;span class="math"&gt;\(L\)&lt;/span&gt; est la tangente à la représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; à l'origine, &lt;span class="math"&gt;\(O\)&lt;/span&gt;.
Le point &lt;span class="math"&gt;\(P(a; b)\)&lt;/span&gt; est sur &lt;span class="math"&gt;\(L\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Étant donné que &lt;span class="math"&gt;\(f^\prime(x) =\frac{2a^2 - 4x^2}{\sqrt{a^2-x^2}}\)&lt;/span&gt;, pour &lt;span class="math"&gt;\(-1 \leq x \leq a\)&lt;/span&gt;, trouvez l'équation de &lt;span class="math"&gt;\(L\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;À partir de là ou par toute autre méthode, trouvez l'expression pour &lt;span class="math"&gt;\(b\)&lt;/span&gt; en fonction de &lt;span class="math"&gt;\(a\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 7</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq7-fr.html" rel="alternate"/><published>2026-05-19T02:20:02+00:00</published><updated>2026-05-19T02:20:02+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq7-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 7&lt;/p&gt;
</summary><content type="html">&lt;p&gt;La figure suivante représente une partie de la représentation graphique de la fonction &lt;span class="math"&gt;\(f(x) = 2x^2\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;La figure n'est pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/temp_f_2xsquared.png" /&gt;
&lt;/div&gt;
&lt;p&gt;La droite &lt;span class="math"&gt;\(T\)&lt;/span&gt; est la tangente à la représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; en &lt;span class="math"&gt;\(x = 1\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Montrez que l'équation de &lt;span class="math"&gt;\(T\)&lt;/span&gt; est &lt;span class="math"&gt;\(y = 4x - 2\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Trouvez l'abscisse à l'origine de &lt;span class="math"&gt;\(T\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;La région grisée &lt;span class="math"&gt;\(R\)&lt;/span&gt; est limitée par la représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt;, la droite &lt;span class="math"&gt;\(T\)&lt;/span&gt; et l'axe des abscisses.&lt;ol class="lowerroman"&gt;
&lt;li&gt;Donnez une expression de l'aire de &lt;span class="math"&gt;\(R\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Trouvez l'aire de &lt;span class="math"&gt;\(R\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 8</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq8-fr.html" rel="alternate"/><published>2026-05-19T02:18:27+00:00</published><updated>2026-05-19T02:18:27+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq8-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 8&lt;/p&gt;
</summary><content type="html">&lt;p&gt;La figure suivante représente une partie de la représentation graphique de la fonction quadratique &lt;span class="math"&gt;\(f\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;La figure n'est pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/temp_down_parabola.png" /&gt;
&lt;/div&gt;
&lt;p&gt;Les abscisses à l'origine sont en &lt;span class="math"&gt;\(( -4; 0 )\)&lt;/span&gt; et &lt;span class="math"&gt;\(( 6; 0 )\)&lt;/span&gt; et l'ordonnée à l'origine est en &lt;span class="math"&gt;\(( 0; 240 )\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Donnez &lt;span class="math"&gt;\(f(x)\)&lt;/span&gt; sous la forme &lt;span class="math"&gt;\(f(x) = -10(x - p) (x - q)\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Trouvez une autre expression de &lt;span class="math"&gt;\(f(x)\)&lt;/span&gt; sous la forme &lt;span class="math"&gt;\(f(x) = -10(x - h)^2 + k\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Montrez que &lt;span class="math"&gt;\(f(x)\)&lt;/span&gt; peut aussi s'écrire sous la forme &lt;span class="math"&gt;\(f(x) = 240 + 20x -10x^2\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Une particule se déplace en ligne droite de telle sorte que sa vitesse &lt;span class="math"&gt;\(v\)&lt;/span&gt; ( en &lt;span class="math"&gt;\(ms^{-1}\)&lt;/span&gt; ),
au temps &lt;span class="math"&gt;\(t\)&lt;/span&gt; (en secondes), est donnée par &lt;span class="math"&gt;\(v = 240 + 20t -10t^2\)&lt;/span&gt; , avec &lt;span class="math"&gt;\(0 \leq t \leq 6\)&lt;/span&gt;.&lt;ol class="lowerroman"&gt;
&lt;li&gt;Trouvez la valeur de &lt;span class="math"&gt;\(t\)&lt;/span&gt; quand la vitesse de la particule est la plus grande.&lt;/li&gt;
&lt;li&gt;Trouvez l'accélération de la particule quand sa vitesse est nulle.&lt;/li&gt;
&lt;/ol&gt;
&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 9</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq9-fr.html" rel="alternate"/><published>2026-05-19T02:17:12+00:00</published><updated>2026-05-19T02:17:12+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq9-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 9&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Soit &lt;span class="math"&gt;\(f(x) = kx^4\)&lt;/span&gt;. Le point &lt;span class="math"&gt;\(P(1 ; k)\)&lt;/span&gt; est sur la courbe représentant &lt;span class="math"&gt;\(f\)&lt;/span&gt;.
En &lt;span class="math"&gt;\(P\)&lt;/span&gt;, la normale à la courbe est parallèle à &lt;span class="math"&gt;\(y = -\frac{1}{8}x\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Trouvez la valeur de &lt;span class="math"&gt;\(k\)&lt;/span&gt;.&lt;/p&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 10</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq10-fr.html" rel="alternate"/><published>2026-05-19T02:14:08+00:00</published><updated>2026-05-19T02:14:08+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq10-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 10&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Une fonction &lt;span class="math"&gt;\(f\)&lt;/span&gt; est définie pour &lt;span class="math"&gt;\(-4 \leq x \leq 3\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;La représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; est donnée ci-dessous.&lt;/p&gt;
&lt;p&gt;La figure n'est pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/temp_polynom_minus4_to_3.png" /&gt;
&lt;/div&gt;
&lt;p&gt;La représentation graphique présente un maximum relatif en &lt;span class="math"&gt;\(x = 0\)&lt;/span&gt; et des minimums relatifs en &lt;span class="math"&gt;\(x = -3\)&lt;/span&gt; et &lt;span class="math"&gt;\(x = 2\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha"&gt;
&lt;li&gt;&lt;p class="first"&gt;Donnez les abscisses à l'origine de la représentation graphique de la fonction dérivée, &lt;span class="math"&gt;\(f^\prime\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p class="first"&gt;Donnez toutes les valeurs de &lt;span class="math"&gt;\(x\)&lt;/span&gt; pour lesquelles &lt;span class="math"&gt;\(f^\prime(x)\)&lt;/span&gt; est positive.&lt;/p&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p class="first"&gt;Au point &lt;span class="math"&gt;\(D\)&lt;/span&gt; sur la représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt;, l'abscisse est &lt;span class="math"&gt;\(-0,5\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Expliquez pourquoi &lt;span class="math"&gt;\(f^{\prime\prime}(x) &amp;lt; 0\)&lt;/span&gt; en &lt;span class="math"&gt;\(D\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 11</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq11-fr.html" rel="alternate"/><published>2026-05-19T02:12:44+00:00</published><updated>2026-05-19T02:12:44+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq11-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 11&lt;/p&gt;
</summary><content type="html">&lt;p&gt;On considère la fonction &lt;span class="math"&gt;\(f\)&lt;/span&gt; dont la dérivée seconde est &lt;span class="math"&gt;\(f^{\prime\prime}(x) = 3x -1\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;La représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; présente un point minimum en &lt;span class="math"&gt;\(A(2 ; 4)\)&lt;/span&gt; et un point maximum en &lt;span class="math"&gt;\(B(-\frac{4}{3}; \frac{358}{27})\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Utilisez la dérivée seconde pour justifier que &lt;span class="math"&gt;\(B\)&lt;/span&gt; est un maximum.&lt;/li&gt;
&lt;li&gt;Étant donné que &lt;span class="math"&gt;\(f^\prime(x) = \frac{3}{2}x^2 - x + p\)&lt;/span&gt;, montrez que &lt;span class="math"&gt;\(p = -4\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Trouvez &lt;span class="math"&gt;\(f(x)\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 12</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq12-fr.html" rel="alternate"/><published>2026-05-19T02:07:34+00:00</published><updated>2026-05-19T02:07:34+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq12-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 12&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Soit &lt;span class="math"&gt;\(f(x) = 6 + 6\,sin\,x\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Une partie de la représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; est donnée ci-dessous.&lt;/p&gt;
&lt;p&gt;La figure n'est pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/temp_6_plus_6sinx.png" /&gt;
&lt;/div&gt;
&lt;p&gt;La région grisée est limitée par la courbe représentant &lt;span class="math"&gt;\(f\)&lt;/span&gt;, l'axe des abscisses et l'axe des ordonnées.&lt;/p&gt;
&lt;p&gt;Le chemin fait un angle de &lt;span class="math"&gt;\(4^\circ\)&lt;/span&gt; avec l'horizontale.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Résolvez, avec &lt;span class="math"&gt;\(0 \leq x \leq 2\pi\)&lt;/span&gt;&lt;ol class="lowerroman"&gt;
&lt;li&gt;&lt;span class="math"&gt;\(6 + 6\,sin\,x = 6\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;&lt;span class="math"&gt;\(6 + 6\,sin\,x = 0\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
&lt;/li&gt;
&lt;li&gt;Donnez la valeur exacte de l'abscisse à l'origine de &lt;span class="math"&gt;\(f\)&lt;/span&gt;, avec &lt;span class="math"&gt;\(0 \leq x \leq 2\pi\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;L'aire de la région grisée est &lt;span class="math"&gt;\(k\)&lt;/span&gt;. Trouvez la valeur de &lt;span class="math"&gt;\(k\)&lt;/span&gt;, en donnant votre réponse en fonction de &lt;span class="math"&gt;\(\pi\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
&lt;p&gt;Soit &lt;span class="math"&gt;\(g(x) = 6 + 6\,sin\,(x - \frac{\pi}{2})\)&lt;/span&gt;. La représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; est transformée en celle de &lt;span class="math"&gt;\(g\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple" start="4"&gt;
&lt;li&gt;Donnez une description géométrique complète de cette transformation.&lt;/li&gt;
&lt;li&gt;Étant donné que &lt;span class="math"&gt;\(\int_p^{p+\frac{3\pi}{2}}g(x)\,dx = k\)&lt;/span&gt; et &lt;span class="math"&gt;\(0 \leq p &amp;lt; 2\pi\)&lt;/span&gt;, donnez les deux valeurs de &lt;span class="math"&gt;\(p\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 13</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq13-fr.html" rel="alternate"/><published>2026-05-19T02:04:07+00:00</published><updated>2026-05-19T02:04:07+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq13-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 13&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Soit &lt;span class="math"&gt;\(f^\prime(x) = 12x^2 - 2\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Sachant que &lt;span class="math"&gt;\(f(-1) = -1\)&lt;/span&gt;, trouvez &lt;span class="math"&gt;\(f(x)\)&lt;/span&gt;.&lt;/p&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 14</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq14-fr.html" rel="alternate"/><published>2026-05-19T02:00:55+00:00</published><updated>2026-05-19T02:00:55+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-19:/lang-version.fr/calculusq14-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 14&lt;/p&gt;
</summary><content type="html">&lt;p&gt;La vitesse &lt;span class="math"&gt;\(v\)&lt;/span&gt;, en &lt;span class="math"&gt;\(ms^{-1}\)&lt;/span&gt;, d'une particule se déplaçant en ligne droite est donnée par &lt;span class="math"&gt;\(v=e^{3t-2}\)&lt;/span&gt;, où &lt;span class="math"&gt;\(t\)&lt;/span&gt; est le temps en secondes.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Trouvez l'accélération de la particule à l'instant &lt;span class="math"&gt;\(t = 1\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Pour quelle valeur de &lt;span class="math"&gt;\(t\)&lt;/span&gt; la particule a-t-elle une vitesse de &lt;span class="math"&gt;\(22,3ms^{-1}\)&lt;/span&gt; ?&lt;/li&gt;
&lt;li&gt;Trouvez la distance parcourue pendant la première seconde.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 15</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq15-fr.html" rel="alternate"/><published>2026-05-18T20:11:31+00:00</published><updated>2026-05-18T20:11:31+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-18:/lang-version.fr/calculusq15-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 15&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Soit &lt;span class="math"&gt;\(f(x) = 3 Cos2x + Sin^2x\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha"&gt;
&lt;li&gt;&lt;p class="first"&gt;Montrez que &lt;span class="math"&gt;\(f^\prime(x) = -5Sin2x\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p class="first"&gt;Dans l'intervalle &lt;span class="math"&gt;\(f^\frac{\pi}{4} leq x leq \frac{3\pi}{4}\)&lt;/span&gt;,
une normale à la représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; a pour équation &lt;span class="math"&gt;\(x = k\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Trouvez la valeur de &lt;span class="math"&gt;\(k\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 16</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq16-fr.html" rel="alternate"/><published>2026-05-18T20:09:52+00:00</published><updated>2026-05-18T20:09:52+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-18:/lang-version.fr/calculusq16-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 16&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Une particule &lt;span class="math"&gt;\(P\)&lt;/span&gt; se déplace le long d'une droite. Le vecteur vitesse &lt;span class="math"&gt;\(v\)&lt;/span&gt; &lt;span class="math"&gt;\(ms^{-1}\)&lt;/span&gt; de &lt;span class="math"&gt;\(P\)&lt;/span&gt; après &lt;span class="math"&gt;\(t\)&lt;/span&gt; secondes est donnée par &lt;span class="math"&gt;\(v(t) = 7 cos t - 5t^{cos t}\)&lt;/span&gt;, pour &lt;span class="math"&gt;\(0 \leq t \leq 7\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Le diagramme suivant montre la représentation graphique de &lt;span class="math"&gt;\(v\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;La figure n'est pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/temp_graph_v_vs_t.png" /&gt;
&lt;/div&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Trouvez le vecteur vitesse initiale de &lt;span class="math"&gt;\(P\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Trouvez la vitesse maximale de &lt;span class="math"&gt;\(P\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Écrivez le nombre de fois où l'accélération de &lt;span class="math"&gt;\(P\)&lt;/span&gt; est &lt;span class="math"&gt;\(0\)&lt;/span&gt; &lt;span class="math"&gt;\(ms^{-2}\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Trouvez l'accélération de &lt;span class="math"&gt;\(P\)&lt;/span&gt; lorsque la particule change de direction.&lt;/li&gt;
&lt;li&gt;Trouvez la distance totale parcourue par &lt;span class="math"&gt;\(P\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 17</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq17-fr.html" rel="alternate"/><published>2026-05-18T20:07:39+00:00</published><updated>2026-05-18T20:07:39+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-18:/lang-version.fr/calculusq17-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 17&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Soit &lt;span class="math"&gt;\(f(x) = Cos(e^x)\)&lt;/span&gt;, pour &lt;span class="math"&gt;\(-2 \leq x \leq 2\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Trouvez &lt;span class="math"&gt;\(f^\prime(x)\)&lt;/span&gt;.&lt;/li&gt;
&lt;li&gt;Sur le système d'axes ci-dessous, esquissez la représentation graphique de &lt;span class="math"&gt;\(f^\prime(x)\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/grid_5_12.png" /&gt;
&lt;/div&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 18</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq18-fr.html" rel="alternate"/><published>2026-05-18T20:04:46+00:00</published><updated>2026-05-18T20:04:46+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-18:/lang-version.fr/calculusq18-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 18&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Une particule se déplace sur une ligne droite avec une vitesse &lt;span class="math"&gt;\(v = 12t - 2t^3 - 1\)&lt;/span&gt;,
pour &lt;span class="math"&gt;\(t \geq 0\)&lt;/span&gt;, où &lt;span class="math"&gt;\(v\)&lt;/span&gt; est en centimètres par seconde et &lt;span class="math"&gt;\(t\)&lt;/span&gt; est en secondes.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Trouvez l'accélération de la particule à l'instant &lt;span class="math"&gt;\(t = 2,7\)&lt;/span&gt; secondes.&lt;/li&gt;
&lt;li&gt;Trouvez le déplacement de la particule après &lt;span class="math"&gt;\(1,3\)&lt;/span&gt; secondes.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 19</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq19-fr.html" rel="alternate"/><published>2026-05-18T20:02:49+00:00</published><updated>2026-05-18T20:02:49+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-18:/lang-version.fr/calculusq19-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 19&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Soit &lt;span class="math"&gt;\(f(x) = ax^3 + bx^2 + c\)&lt;/span&gt;, où &lt;span class="math"&gt;\(a\)&lt;/span&gt;, &lt;span class="math"&gt;\(b\)&lt;/span&gt; et &lt;span class="math"&gt;\(c\)&lt;/span&gt; sont des nombres réels.
La courbe de &lt;span class="math"&gt;\(f\)&lt;/span&gt; passe par le point &lt;span class="math"&gt;\(( 2; 9 )\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple"&gt;
&lt;li&gt;Montrez que &lt;span class="math"&gt;\(8a + 4b +c = 9\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
&lt;p&gt;La courbe de &lt;span class="math"&gt;\(f\)&lt;/span&gt; a un minimum relatif en &lt;span class="math"&gt;\((1; 4)\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha simple" start="2"&gt;
&lt;li&gt;Trouvez deux autres équations en termes de &lt;span class="math"&gt;\(a\)&lt;/span&gt; , &lt;span class="math"&gt;\(b\)&lt;/span&gt; et &lt;span class="math"&gt;\(c\)&lt;/span&gt;,
en donnant vos réponses sous une forme similaire à celle de la partie A.&lt;/li&gt;
&lt;li&gt;Trouvez la valeur de &lt;span class="math"&gt;\(a\)&lt;/span&gt;, celle de &lt;span class="math"&gt;\(b\)&lt;/span&gt; et celle de &lt;span class="math"&gt;\(c\)&lt;/span&gt;.&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 20</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq20-fr.html" rel="alternate"/><published>2026-05-18T20:00:06+00:00</published><updated>2026-05-18T20:00:06+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-18:/lang-version.fr/calculusq20-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 20&lt;/p&gt;
</summary><content type="html">&lt;p&gt;La pente d'une fonction est donnée par &lt;span class="math"&gt;\(\dfrac{dy}{dx} = 10e^{2x - 5}\)&lt;/span&gt;. Quand &lt;span class="math"&gt;\(x = 0\)&lt;/span&gt;, &lt;span class="math"&gt;\(y = 8\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Trouvez la valeur de &lt;span class="math"&gt;\(y\)&lt;/span&gt; quand &lt;span class="math"&gt;\(x = 1\)&lt;/span&gt;.&lt;/p&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 21</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq21-fr.html" rel="alternate"/><published>2026-05-18T19:58:04+00:00</published><updated>2026-05-18T19:58:04+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-18:/lang-version.fr/calculusq21-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 21&lt;/p&gt;
</summary><content type="html">&lt;p&gt;Soit &lt;span class="math"&gt;\(f(x) = e^x Sin 2x + 10\)&lt;/span&gt;, avec &lt;span class="math"&gt;\(0 \leq x \leq 4\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Une partie de la représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; est donnée ci-dessous.&lt;/p&gt;
&lt;p&gt;La figure n'est pas à l'échelle.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/etox_sin2x.png" /&gt;
&lt;/div&gt;
&lt;p&gt;Sont représentés une abscisse à l'origine au point &lt;span class="math"&gt;\(A\)&lt;/span&gt;, un maximum relatif au point &lt;span class="math"&gt;\(M\)&lt;/span&gt; avec &lt;span class="math"&gt;\(x = p\)&lt;/span&gt; et un minimum relatif au point &lt;span class="math"&gt;\(N\)&lt;/span&gt; avec &lt;span class="math"&gt;\(x = q\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha"&gt;
&lt;li&gt;&lt;p class="first"&gt;Donnez l'abscisse de &lt;span class="math"&gt;\(A\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p class="first"&gt;Trouvez la valeur de&lt;/p&gt;
&lt;ol class="lowerroman simple"&gt;
&lt;li&gt;&lt;span class="math"&gt;\(p\)&lt;/span&gt;&lt;/li&gt;
&lt;li&gt;&lt;span class="math"&gt;\(q\)&lt;/span&gt;&lt;/li&gt;
&lt;/ol&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p class="first"&gt;Trouvez &lt;span class="math"&gt;\(\int_p^qf(x)dx\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Expliquez pourquoi ceci n'est pas l'aire de la région grisée.&lt;/p&gt;
&lt;/li&gt;
&lt;/ol&gt;
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&lt;/script&gt;</content><category term="calcul"/><category term="IB"/><category term="question"/></entry><entry><title>IB Calcul Problème 22</title><link href="https://blog.bernatchez.net/lang-version.fr/calculusq22-fr.html" rel="alternate"/><published>2026-05-18T19:55:55+00:00</published><updated>2026-05-18T19:55:55+00:00</updated><author><name>Annie Bernatchez</name></author><id>tag:blog.bernatchez.net,2026-05-18:/lang-version.fr/calculusq22-fr.html</id><summary type="html">&lt;p class="first last"&gt;Problème de calcul 22&lt;/p&gt;
</summary><content type="html">&lt;p&gt;On considère &lt;span class="math"&gt;\(f(x) = x\ln(4 - x^2)\)&lt;/span&gt;, avec &lt;span class="math"&gt;\(-2 &amp;lt; x &amp;lt; 2\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Une partie de la représentation graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; est donnée ci-dessous.&lt;/p&gt;
&lt;div class="figure"&gt;
&lt;img alt="image indispensable à la compréhension de la question" src="../images/xln4minuxsqr.png" /&gt;
&lt;/div&gt;
&lt;p&gt;Soient &lt;span class="math"&gt;\(P\)&lt;/span&gt; et &lt;span class="math"&gt;\(Q\)&lt;/span&gt; les points de la courbe représentant &lt;span class="math"&gt;\(f\)&lt;/span&gt;
où la tangente à la représentante graphique de &lt;span class="math"&gt;\(f\)&lt;/span&gt; est
parallèle à l'axe des abscisses.&lt;/p&gt;
&lt;ol class="upperalpha"&gt;
&lt;li&gt;&lt;ol class="first lowerroman"&gt;
&lt;li&gt;&lt;p class="first"&gt;Trouvez l'abscisse de &lt;span class="math"&gt;\(p\)&lt;/span&gt; et &lt;span class="math"&gt;\(q\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p class="first"&gt;On considère &lt;span class="math"&gt;\(f(x) = k\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Donnez toutes les valeurs de &lt;span class="math"&gt;\(k\)&lt;/span&gt; pour les-quelles il y a exactement deux solutions.&lt;/p&gt;
&lt;/li&gt;
&lt;/ol&gt;
&lt;/li&gt;
&lt;/ol&gt;
&lt;p&gt;Soit &lt;span class="math"&gt;\(g(x) = x^3\ln(4 - x^2)\)&lt;/span&gt;, avec &lt;span class="math"&gt;\(-2 &amp;lt; x &amp;lt; 2\)&lt;/span&gt;.&lt;/p&gt;
&lt;ol class="upperalpha" start="2"&gt;
&lt;li&gt;&lt;p class="first"&gt;Montrez que &lt;span class="math"&gt;\(g^\prime(x)=\frac{-2x^4}{4 - x^2} + 3x^2\ln(4 - x^2)\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p class="first"&gt;Esquissez la représentation graphique de &lt;span class="math"&gt;\(g^\prime\)&lt;/span&gt;.&lt;/p&gt;
&lt;/li&gt;
&lt;li&gt;&lt;p class="first"&gt;On considère &lt;span class="math"&gt;\(g^\prime(x) = w\)&lt;/span&gt;.&lt;/p&gt;
&lt;p&gt;Donnez toutes les valeurs de &lt;span class="math"&gt;\(w\)&lt;/span&gt; pour les-quelles il y a exactement deux solutions.&lt;/p&gt;
&lt;/li&gt;
&lt;/ol&gt;
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