✅ Corrigé détaillé — Exercice 2
Partie A
| Équation | Résolution | Solution(s) |
|---|---|---|
| a) e^(2x+1) = e^(x+4) | 2x+1 = x+4 → x = 3 | S = {3} |
| b) e^(x²) = e^(3x−2) | x² = 3x−2 → x²−3x+2 = 0 → (x−1)(x−2) = 0 | S = {1, 2} |
| c) e^(2x) = 1 = e⁰ | 2x = 0 → x = 0 | S = {0} |
| d) eˣ = e^(−x+6) | x = −x+6 → 2x = 6 → x = 3 | S = {3} |
Partie B
| Inéquation | Résolution | Ensemble solution |
|---|---|---|
| a) e^(x+2) < e⁵ | x+2 < 5 → x < 3 | ]−∞ ; 3[ |
| b) e^(3x−1) ≥ e^(x+3) | 3x−1 ≥ x+3 → 2x ≥ 4 → x ≥ 2 | [2 ; +∞[ |
| c) eˣ > 1 = e⁰ | x > 0 | ]0 ; +∞[ |
| d) e^(2x+1) ≤ e^(4−x) | 2x+1 ≤ 4−x → 3x ≤ 3 → x ≤ 1 | ]−∞ ; 1] |
Partie C
3a) e^(3x+5) = e^(x²−x+3) ↔ 3x + 5 = x² − x + 3 ↔ x² − 4x − 2 = 0.
3b) Δ = b² − 4ac = (−4)² − 4(1)(−2) = 16 + 8 = 24 > 0.
x₁ = (4 − √24)/2 = (4 − 2√6)/2 = 2 − √6.
x₂ = (4 + 2√6)/2 = 2 + √6.