Rezolvarea completă a subiectului de Bacalaureat 2019 la Matematică M_șt-nat , sesiunea specială, varianta 8 . Pentru fiecare cerință găsești enunțul pe scurt, rezolvarea pas cu pas, răspunsul final și cum se punctează conform baremului oficial. Timp de lucru la examen: 3 ore; se acordă 10 puncte din oficiu.
Cum să folosești rezolvarea: lucrează întâi singur subiectul (deschide PDF-ul de mai sus), apoi compară fiecare pas. Dacă ai ajuns la alt drum corect, primești tot punctajul — baremul acceptă orice soluție corectă.
Subiectul I (30 de puncte)
1. Arătați că ( 1 + i ) 2 − 2 i = 0 (1 + i)^2 - 2i = 0 ( 1 + i ) 2 − 2 i = 0
( 1 + i ) 2 = 1 + 2 i + i 2 = 2 i (1 + i)^2 = 1 + 2i + i^2 = 2i ( 1 + i ) 2 = 1 + 2 i + i 2 = 2 i 2 i − 2 i = 0 2i - 2i = 0 2 i − 2 i = 0 ■ \blacksquare ■ Barem: 3p, 2p.
2. f ( x ) = m x 2 + 8 x − 7 f(x) = mx^2 + 8x - 7 f ( x ) = m x 2 + 8 x − 7 12 12 12
x V = − b 2 a = − 8 2 m = 12 ⟺ m = − 1 3 x_V = -\frac{b}{2a} = -\frac{8}{2m} = 12 \iff m = -\frac13 x V = − 2 a b = − 2 m 8 = 12 ⟺ m = − 3 1
Răspuns: m = − 1 3 m = -\dfrac13 m = − 3 1 Barem: 3p, 2p.
3. Ecuația log 4 ( x 2 − 10 x + 40 ) = 2 \log_4(x^2 - 10x + 40) = 2 log 4 ( x 2 − 10 x + 40 ) = 2
x 2 − 10 x + 40 = 16 ⟺ x 2 − 10 x + 24 = 0 ⟺ x ∈ { 4 , 6 } x^2 - 10x + 40 = 16 \iff x^2 - 10x + 24 = 0 \iff x \in \{4, 6\} x 2 − 10 x + 40 = 16 ⟺ x 2 − 10 x + 24 = 0 ⟺ x ∈ { 4 , 6 } 16 > 0 16 > 0 16 > 0
Răspuns: x ∈ { 4 , 6 } x \in \{4, 6\} x ∈ { 4 , 6 } Barem: 3p, 2p.
4. Probabilitatea ca un număr din { 1 , 2 , … , 100 } \{1, 2, \dots, 100\} { 1 , 2 , … , 100 }
p = 50 100 = 1 2 p = \frac{50}{100} = \frac12 p = 100 50 = 2 1 Barem: 2p, 2p, 1p.
5. A B C D ABCD A B C D A ( 2 , − 1 ) A(2,-1) A ( 2 , − 1 ) B ( − 2 , 0 ) B(-2,0) B ( − 2 , 0 ) C ( 0 , 3 ) C(0,3) C ( 0 , 3 ) B D → \overrightarrow{BD} B D
Regula paralelogramului: B D → = B A → + B C → = ( 4 i ⃗ − j ⃗ ) + ( 2 i ⃗ + 3 j ⃗ ) = 6 i ⃗ + 2 j ⃗ \overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{BC} = (4\vec i - \vec j) + (2\vec i + 3\vec j) = 6\vec i + 2\vec j B D = B A + B C = ( 4 i − j ) + ( 2 i + 3 j ) = 6 i + 2 j
∣ B D → ∣ = 36 + 4 = 2 10 |\overrightarrow{BD}| = \sqrt{36 + 4} = 2\sqrt{10} ∣ B D ∣ = 36 + 4 = 2 10
Răspuns: 2 10 2\sqrt{10} 2 10 Barem: 2p, 3p.
6. x ∈ ( 0 , π 2 ) x \in \left(0, \frac\pi2\right) x ∈ ( 0 , 2 π ) sin x = 3 2 \sin x = \frac{\sqrt3}{2} sin x = 2 3 sin 3 x + sin 2 x + sin x = 3 \sin3x + \sin2x + \sin x = \sqrt3 sin 3 x + sin 2 x + sin x = 3
x = π 3 x = \frac\pi3 x = 3 π sin π + sin 2 π 3 + sin π 3 = 0 + 3 2 + 3 2 = 3 \sin\pi + \sin\frac{2\pi}{3} + \sin\frac\pi3 = 0 + \frac{\sqrt3}2 + \frac{\sqrt3}2 = \sqrt3 sin π + sin 3 2 π + sin 3 π = 0 + 2 3 + 2 3 = 3 ■ \blacksquare ■ Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. A = ( 1 1 3 3 ) A = \begin{pmatrix}1 & 1\\ 3 & 3\end{pmatrix} A = ( 1 3 1 3 ) M ( a ) = I 2 + a A M(a) = I_2 + aA M ( a ) = I 2 + a A
a) M ( 1 ) = ( 2 1 3 4 ) M(1) = \begin{pmatrix}2 & 1\\ 3 & 4\end{pmatrix} M ( 1 ) = ( 2 3 1 4 ) det = 8 − 3 = 5 \det = 8 - 3 = 5 det = 8 − 3 = 5 ■ \blacksquare ■ Barem: 3p, 2p.
b) A 2 = ( 4 4 12 12 ) = 4 A A^2 = \begin{pmatrix}4 & 4\\ 12 & 12\end{pmatrix} = 4A A 2 = ( 4 12 4 12 ) = 4 A
M ( a ) M ( b ) = I 2 + ( a + b ) A + a b A 2 = I 2 + ( a + b + 4 a b ) A = M ( a + b + 4 a b ) . ■ M(a)M(b) = I_2 + (a + b)A + abA^2 = I_2 + (a + b + 4ab)A = M(a + b + 4ab).\ \blacksquare M ( a ) M ( b ) = I 2 + ( a + b ) A + ab A 2 = I 2 + ( a + b + 4 ab ) A = M ( a + b + 4 ab ) . ■
Barem: 2p, 3p.
c) M ( a ) 2 = M ( 2 a + 4 a 2 ) = M ( 2 ) M(a)^2 = M(2a + 4a^2) = M(2) M ( a ) 2 = M ( 2 a + 4 a 2 ) = M ( 2 ) A ≠ O 2 A \ne O_2 A = O 2 M M M 4 a 2 + 2 a − 2 = 0 ⟺ 2 a 2 + a − 1 = 0 ⟺ a ∈ { − 1 , 1 2 } 4a^2 + 2a - 2 = 0 \iff 2a^2 + a - 1 = 0 \iff a \in \left\{-1, \frac12\right\} 4 a 2 + 2 a − 2 = 0 ⟺ 2 a 2 + a − 1 = 0 ⟺ a ∈ { − 1 , 2 1 }
Răspuns: a ∈ { − 1 , 1 2 } a \in \left\{-1, \frac12\right\} a ∈ { − 1 , 2 1 } Barem: 3p, 2p.
2. x ∗ y = 5 x + 5 y − x y − 20 x * y = 5x + 5y - xy - 20 x ∗ y = 5 x + 5 y − x y − 20
a) − x y + 5 x + 5 y − 25 + 5 = − x ( y − 5 ) + 5 ( y − 5 ) + 5 = − ( x − 5 ) ( y − 5 ) + 5 -xy + 5x + 5y - 25 + 5 = -x(y - 5) + 5(y - 5) + 5 = -(x - 5)(y - 5) + 5 − x y + 5 x + 5 y − 25 + 5 = − x ( y − 5 ) + 5 ( y − 5 ) + 5 = − ( x − 5 ) ( y − 5 ) + 5 ■ \blacksquare ■ Barem: 3p, 2p.
b) − ( x − 5 ) 2 + 5 ≥ x ⟺ ( x − 5 ) 2 + ( x − 5 ) ≤ 0 ⟺ ( x − 5 ) ( x − 4 ) ≤ 0 ⟺ x ∈ [ 4 , 5 ] -(x - 5)^2 + 5 \ge x \iff (x - 5)^2 + (x - 5) \le 0 \iff (x - 5)(x - 4) \le 0 \iff x \in [4, 5] − ( x − 5 ) 2 + 5 ≥ x ⟺ ( x − 5 ) 2 + ( x − 5 ) ≤ 0 ⟺ ( x − 5 ) ( x − 4 ) ≤ 0 ⟺ x ∈ [ 4 , 5 ] Barem: 3p, 2p.
c) x ∗ 5 = 5 ∗ x = 5 x * 5 = 5 * x = 5 x ∗ 5 = 5 ∗ x = 5 x x x − ( x − 5 ) ⋅ 0 + 5 = 5 -(x - 5)\cdot 0 + 5 = 5 − ( x − 5 ) ⋅ 0 + 5 = 5
( 1 ∗ ( − 2 ) ∗ 3 ∗ ( − 4 ) ) ∗ 5 ∗ ( ( − 6 ) ∗ ⋯ ∗ 2019 ) = 5. \big(1 * (-2) * 3 * (-4)\big) * 5 * \big((-6) * \dots * 2019\big) = 5. ( 1 ∗ ( − 2 ) ∗ 3 ∗ ( − 4 ) ) ∗ 5 ∗ ( ( − 6 ) ∗ ⋯ ∗ 2019 ) = 5.
Răspuns: 5 5 5 Barem: 2p, 3p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x 2 + 6 x + 9 e x = ( x + 3 ) 2 e − x f(x) = \dfrac{x^2 + 6x + 9}{e^x} = (x + 3)^2e^{-x} f ( x ) = e x x 2 + 6 x + 9 = ( x + 3 ) 2 e − x
a) f ′ ( x ) = ( 2 x + 6 ) e x − ( x 2 + 6 x + 9 ) e x e 2 x = − x 2 − 4 x − 3 e x = − ( x + 1 ) ( x + 3 ) e x . ■ f'(x) = \frac{(2x + 6)e^x - (x^2 + 6x + 9)e^x}{e^{2x}} = \frac{-x^2 - 4x - 3}{e^x} = \frac{-(x + 1)(x + 3)}{e^x}.\ \blacksquare f ′ ( x ) = e 2 x ( 2 x + 6 ) e x − ( x 2 + 6 x + 9 ) e x = e x − x 2 − 4 x − 3 = e x − ( x + 1 ) ( x + 3 ) . ■
Barem: 3p, 2p.
b) L'Hospital de două ori: lim x → + ∞ x 2 + 6 x + 9 e x = lim 2 x + 6 e x = lim 2 e x = 0 \lim\limits_{x\to+\infty}\frac{x^2 + 6x + 9}{e^x} = \lim\frac{2x + 6}{e^x} = \lim\frac{2}{e^x} = 0 x → + ∞ lim e x x 2 + 6 x + 9 = lim e x 2 x + 6 = lim e x 2 = 0 y = 0 y = 0 y = 0 Barem: 3p, 2p.
c) f ′ ≥ 0 f' \ge 0 f ′ ≥ 0 [ − 3 , − 1 ] [-3, -1] [ − 3 , − 1 ] f ′ ≤ 0 f' \le 0 f ′ ≤ 0 [ − 1 , + ∞ ) [-1, +\infty) [ − 1 , + ∞ ) [ − 3 , + ∞ ) [-3, +\infty) [ − 3 , + ∞ ) f ( x ) ≤ f ( − 1 ) = 4 e f(x) \le f(-1) = 4e f ( x ) ≤ f ( − 1 ) = 4 e ( x + 3 ) 2 ≤ 4 e x + 1 (x + 3)^2 \le 4e^{x+1} ( x + 3 ) 2 ≤ 4 e x + 1 x + 3 ≥ 0 x + 3 \ge 0 x + 3 ≥ 0
0 ≤ x + 3 ≤ 2 e x + 1 2 , 0 ≤ y + 3 ≤ 2 e y + 1 2 . 0 \le x + 3 \le 2e^{\frac{x+1}{2}},\qquad 0 \le y + 3 \le 2e^{\frac{y+1}{2}}. 0 ≤ x + 3 ≤ 2 e 2 x + 1 , 0 ≤ y + 3 ≤ 2 e 2 y + 1 .
Înmulțind (numere nenegative): 0 ≤ ( x + 3 ) ( y + 3 ) ≤ 4 e x + y + 2 2 0 \le (x + 3)(y + 3) \le 4e^{\frac{x + y + 2}{2}} 0 ≤ ( x + 3 ) ( y + 3 ) ≤ 4 e 2 x + y + 2 ■ \blacksquare ■
Barem: 3p, 2p.
2. f ( x ) = x 3 x + 1 f(x) = \dfrac{x^3}{x + 1} f ( x ) = x + 1 x 3 x > − 1 x > -1 x > − 1
a) ∫ 0 2 ( x + 1 ) f ( x ) d x = ∫ 0 2 x 3 d x = 16 4 = 4 \int_0^2(x+1)f(x)\,dx = \int_0^2 x^3dx = \frac{16}{4} = 4 ∫ 0 2 ( x + 1 ) f ( x ) d x = ∫ 0 2 x 3 d x = 4 16 = 4 ■ \blacksquare ■ Barem: 3p, 2p.
b) F ′ ( x ) = x 2 − x + 1 − 1 x + 1 = ( x 2 − x + 1 ) ( x + 1 ) − 1 x + 1 = x 3 + 1 − 1 x + 1 = x 3 x + 1 = f ( x ) . ■ F'(x) = x^2 - x + 1 - \frac{1}{x + 1} = \frac{(x^2 - x + 1)(x + 1) - 1}{x + 1} = \frac{x^3 + 1 - 1}{x + 1} = \frac{x^3}{x+1} = f(x).\ \blacksquare F ′ ( x ) = x 2 − x + 1 − x + 1 1 = x + 1 ( x 2 − x + 1 ) ( x + 1 ) − 1 = x + 1 x 3 + 1 − 1 = x + 1 x 3 = f ( x ) . ■
Barem: 3p, 2p.
c) g ( x ) = 1 x 3 f ( x ) = 1 x + 1 > 0 g(x) = \frac{1}{x^3}f(x) = \frac{1}{x + 1} > 0 g ( x ) = x 3 1 f ( x ) = x + 1 1 > 0
A = ∫ 1 a 2 d x x + 1 = ln ( a 2 + 1 ) − ln 2 = ln 5 ⟺ a 2 + 1 = 10 ⟺ a = 3 ( a > 1 ) . \mathcal A = \int_1^{a^2}\frac{dx}{x + 1} = \ln(a^2 + 1) - \ln 2 = \ln 5 \iff a^2 + 1 = 10 \iff a = 3\ (a > 1). A = ∫ 1 a 2 x + 1 d x = ln ( a 2 + 1 ) − ln 2 = ln 5 ⟺ a 2 + 1 = 10 ⟺ a = 3 ( a > 1 ) .
Răspuns: a = 3 a = 3 a = 3 Barem: 3p, 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 m = − 1 3 m = -\tfrac13 m = − 3 1 I.3 x ∈ { 4 , 6 } x \in \{4, 6\} x ∈ { 4 , 6 } I.4 p = 1 2 p = \tfrac12 p = 2 1 I.5 2 10 2\sqrt{10} 2 10 II.1.c a ∈ { − 1 , 1 2 } a \in \{-1, \tfrac12\} a ∈ { − 1 , 2 1 } II.2.b x ∈ [ 4 , 5 ] x \in [4, 5] x ∈ [ 4 , 5 ] II.2.c 5 5 5 III.1.b y = 0 y = 0 y = 0 III.2.c a = 3 a = 3 a = 3
Ce merită reținut din varianta 8
A 2 = k A A^2 = kA A 2 = k A M ( a ) = I 2 + a A M(a) = I_2 + aA M ( a ) = I 2 + a A a + b + k a b a + b + kab a + b + k ab Element absorbant (x ∗ 5 = 5 x * 5 = 5 x ∗ 5 = 5 5 5 5 Inegalități cu două variabile: obține-o pentru o variabilă din maximul funcției, apoi înmulțește.