EXERCICE 1 part a Soit la fonction définie sur l intervalle...

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--- ## Question 1 (EXERCICE 1, part a) **Question content**: Soit \( f \) la fonction définie sur l'intervalle \( I = [-1; 1] \) par \( f(x) = \frac{x^2 - 1}{x + 3} \). Pour tout réel \( x \) de \( I \), calculer \( f'(x) \). **Discipline**: Mathematics **Grade**: High School (11th-12th grade, or equivalent) ### Correct answer \( f'(x) = \frac{x^2 + 6x + 1}{(x + 3)^2} \) (pour \( x \in [-1, 1] \)) ### Detailed problem-solving steps 1. **Identify the quotient rule**: For a function \( f(x) = \frac{u(x)}{v(x)} \), the derivative is given by the quotient rule: \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \), where \( u(x) = x^2 - 1 \) and \( v(x) = x + 3 \). 2. **Compute \( u'(x) \) and \( v'(x) \)**: - \( u(x) = x^2 - 1 \implies u'(x) = 2x \) (using the power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \)). - \( v(x) = x + 3 \implies v'(x) = 1 \) (derivative of a linear function). 3. **Substitute into the quotient rule**: \[ f'(x) = \frac{(2x)(x + 3) - (x^2 - 1)(1)}{(x + 3)^2} \] 4. **Simplify the numerator**: Expand \( 2x(x + 3) \): \( 2x^2 + 6x \). Subtract \( (x^2 - 1) \): \( 2x^2 + 6x - (x^2 - 1) = 2x^2 + 6x - x^2 + 1 = x^2 + 6x + 1 \). 5. **Final form of \( f'(x) \)**: The denominator \( (x + 3)^2 \) is always positive on \( I = [-1, 1] \) (since \( x \geq -1 \implies x + 3 \geq 2 > 0 \)), so: \[ f'(x) = \frac{x^2 + 6x + 1}{(x + 3)^2} \] ### Knowledge points involved 1. **Quotient Rule for Differentiation** - **Definition**: For \( f(x) = \frac{u(x)}{v(x)} \) (with \( v(x) \neq 0 \)), \( f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \). - **Application**: Used to differentiate functions that are quotients of two differentiable functions (here, \( u(x) = x^2 - 1 \) and \( v(x) = x + 3 \)). 2. **Power Rule for Differentiation** - **Definition**: For \( f(x) = x^n \) (where \( n \) is a real number), \( f'(x) = nx^{n-1} \). - **Application**: Used to differentiate \( u(x) = x^2 - 1 \) (giving \( u'(x) = 2x \)) and \( v(x) = x + 3 \) (giving \( v'(x) = 1 \)). 3. **Simplification of Algebraic Expressions** - **Skill**: Expanding and combining like terms (e.g., \( 2x(x + 3) - (x^2 - 1) = x^2 + 6x + 1 \)) to simplify the numerator of the derivative. --- ## Question 2 (EXERCICE 1, part b) **Question content**: Soit \( g \) la fonction définie sur \( \mathbb{R} \) par \( g(x) = 2x^3 + 3x^2 + 1 \). Étudier les variations de la fonction \( g \) et en déduire son signe. **Discipline**: Mathematics **Grade**: High School (11th-12th grade, or equivalent) ### Correct answer - **Variations**: \( g \) est **décroissante** sur \( [-1, 0] \) et **croissante** sur \( [0, 1] \) (et globalement, \( g \) est décroissante sur \( (-\infty, -1) \), croissante sur \( (-1, +\infty) \), mais ici restreint à \( I = [-1, 1] \)). - **Signe**: \( g(x) > 0 \) pour \( x > -1 \), \( g(x) < 0 \) pour \( x < -1 \), et \( g(-1) = 0 \). ### Detailed problem-solving steps 1. **Compute the derivative \( g'(x) \)**: \( g(x) = 2x^3 + 3x^2 + 1 \implies g'(x) = 6x^2 + 6x \) (using the power rule). 2. **Factor \( g'(x) \) to find critical points**: \( g'(x) = 6x(x + 1) \). The critical points are \( x = 0 \) and \( x = -1 \) (where \( g'(x) = 0 \)). 3. **Study sign of \( g'(x) \) on intervals**: - **Interval \( [-1, 0] \)**: Test \( x = -0.5 \): \( g'(-0.5) = 6(-0.5)(-0.5 + 1) = 6(-0.5)(0.5) = -1.5 < 0 \). Thus, \( g'(x) \leq 0 \) on \( [-1, 0] \), so \( g \) is **décroissante** (decreasing) here. - **Interval \( [0, 1] \)**: Test \( x = 0.5 \): \( g'(0.5) = 6(0.5)(0.5 + 1) = 6(0.5)(1.5) = 4.5 > 0 \). Thus, \( g'(x) \geq 0 \) on \( [0, 1] \), so \( g \) is **croissante** (increasing) here. 4. **Deduce the sign of \( g(x) \)**: - For \( x < -1 \): \( x < -1 \implies x < 0 \) and \( x + 1 < 0 \implies x(x + 1) > 0 \implies g'(x) > 0 \), so \( g \) is increasing. But \( g(-1) = 2(-1)^3 + 3(-1)^2 + 1 = -2 + 3 + 1 = 0 \), so \( g(x) < 0 \) for \( x < -1 \) (since \( g \) increases from \( -\infty \) to \( g(-1) = 0 \)). - For \( x > -1 \): \( g \) is decreasing on \( [-1, 0] \) then increasing on \( [0, +\infty) \), and \( g(-1) = 0 \), so \( g(x) > 0 \) for \( x > -1 \) (since \( g \) decreases to \( g(0) = 1 > 0 \) then increases). ### Knowledge points involved 1. **Derivative of Polynomial Functions** - **Definition**: For \( f(x) = a_nx^n + \dots + a_1x + a_0 \), \( f'(x) = na_nx^{n-1} + \dots + a_1 \). - **Application**: Used to find \( g'(x) = 6x^2 + 6x \) from \( g(x) = 2x^3 + 3x^2 + 1 \). 2. **Monotonicity of Functions (Variations)** - **Definition**: A function \( f \) is decreasing on an interval if \( f'(x) \leq 0 \), and increasing if \( f'(x) \geq 0 \). - **Application**: Analyzed \( g \)’s monotonicity using the sign of \( g'(x) \) on subintervals of \( \mathbb{R} \). 3. **Sign Analysis of Functions** - **Skill**: Using critical points and monotonicity to determine when \( g(x) > 0 \), \( g(x) < 0 \), or \( g(x) = 0 \). --- ## Question 3 (EXERCICE 2, part 1) **Question content**: Dans un repère orthonormé du plan, on considère la parabole \( P \) d’équation \( y = x^2 \) et la droite \( d \) d’équation \( y = -2x + 3 \). On note leurs points d’intersection \( A \) et \( B \) (avec \( x_A < x_B \)). Calculer les coordonnées de \( A \) et \( B \). **Discipline**: Mathematics **Grade**: High School (10th-11th grade, or equivalent) ### Correct answer - \( A(-3, 9) \) - \( B(1, 1) \) ### Detailed problem-solving steps 1. **Find intersection points**: Solve the system \( \begin{cases} y = x^2 \\ y = -2x + 3 \end{cases} \). 2. **Set \( x^2 = -2x + 3 \)**: Rearrange: \( x^2 + 2x - 3 = 0 \). 3. **Solve the quadratic equation**: Factor: \( (x + 3)(x - 1) = 0 \implies x = -3 \) or \( x = 1 \). 4. **Find \( y \)-coordinates**: - For \( x = -3 \): \( y = (-3)^2 = 9 \implies A(-3, 9) \). - For \( x = 1 \): \( y = (1)^2 = 1 \implies B(1, 1) \). ### Knowledge points involved 1. **Intersection of a Parabola and a Line** - **Definition**: Intersection points satisfy both the parabola’s equation (\( y = x^2 \)) and the line’s equation (\( y = -2x + 3 \)). - **Application**: Solved the system by equating the two expressions for \( y \) and solving the resulting quadratic equation. 2. **Solving Quadratic Equations** - **Method**: Factoring (or quadratic formula). Here, \( x^2 + 2x - 3 = (x + 3)(x - 1) \). - **Application**: Found \( x \)-coordinates of intersection, then substituted back to find \( y \)-coordinates. --- ## Question 4 (EXERCICE 2 Bis) **Question content**: Soit \( f \) la fonction définie sur \( \mathbb{R} \) par \( f(x) = -\frac{4}{9}x^2 + 4 \) et \( P \) sa représentation graphique (une parabole ouverte vers le bas, sommet en \( (0, 4) \), intersectant l’axe des abscisses en \( x = \pm 3 \)). Un rectangle hachuré, symétrique par rapport à l’axe des ordonnées, a pour sommet en haut le graphe de \( f \) et en bas l’axe des abscisses. Quelle est son aire maximale ? Quelles sont ses dimensions ? **Discipline**: Mathematics **Grade**: High School (11th-12th grade, or equivalent) ### Correct answer - **Aire maximale**: \( \frac{16\sqrt{3}}{3} \) (ou \( \frac{16}{3}\sqrt{3} \)) - **Dimensions**: Largeur \( 2\sqrt{3} \) (de \( x = -\sqrt{3} \) à \( x = \sqrt{3} \)) et hauteur \( \frac{8}{3} \). ### Detailed problem-solving steps 1. **Define the rectangle’s dimensions**: Let the right vertex of the rectangle be at \( (x, 0) \) (since it’s symmetric about the \( y \)-axis, the left vertex is at \( (-x, 0) \)). The top vertex is at \( (x, f(x)) \), so: - Width: \( 2x \) (distance from \( -x \) to \( x \)). - Height: \( f(x) = -\frac{4}{9}x^2 + 4 \) (vertical distance from \( x \)-axis to the parabola). 2. **Express the area \( A(x) \)**: \( A(x) = \text{width} \times \text{height} = 2x \cdot \left(-\frac{4}{9}x^2 + 4\right) = -\frac{8}{9}x^3 + 8x \). 3. **Find the maximum area (using calculus)**: - Compute the derivative: \( A'(x) = -\frac{24}{9}x^2 + 8 = -\frac{8}{3}x^2 + 8 \). - Set \( A'(x) = 0 \): \( -\frac{8}{3}x^2 + 8 = 0 \implies \frac{8}{3}x^2 = 8 \implies x^2 = 3 \implies x = \sqrt{3} \) (since \( x > 0 \)). 4. **Verify it’s a maximum**: The second derivative \( A''(x) = -\frac{16}{3}x \), which is negative at \( x = \sqrt{3} \) (since \( x > 0 \)), so \( x = \sqrt{3} \) is a maximum. 5. **Compute dimensions and area**: - Width: \( 2x = 2\sqrt{3} \). - Height: \( f(\sqrt{3}) = -\frac{4}{9}(\sqrt{3})^2 + 4 = -\frac{4}{9}(3) + 4 = -\frac{4}{3} + 4 = \frac{8}{3} \). - Aire maximale: \( A(\sqrt{3}) = 2\sqrt{3} \cdot \frac{8}{3} = \frac{16\sqrt{3}}{3} \). ### Knowledge points involved 1. **Optimization of Geometric Figures (Maximizing Area)** - **Method**: Express the area as a function of a single variable (here, \( x \)), then use calculus (derivatives) to find critical points and determine maxima/minima. - **Application**: Modeled the rectangle’s area using the parabola’s equation, then optimized using differentiation. 2. **Symmetry of Functions (Even Functions)** - **Definition**: A function \( f \) is even if \( f(-x) = f(x) \) (symmetric about the \( y \)-axis). Here, \( f(x) = -\frac{4}{9}x^2 + 4 \) is even, so the rectangle is symmetric about the \( y \)-axis. - **Application**: Simplified the problem by considering only \( x > 0 \) and doubling the width. 3. **Derivatives for Optimization** - **Critical Points**: Found by setting \( A'(x) = 0 \). - **Second Derivative Test**: Confirmed the critical point is a maximum (since \( A''(x) < 0 \)). (Note: Additional questions from the image can be structured similarly, but the above examples cover key problems in differentiation, optimization, and geometric applications.)

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