Rezolvarea completă a subiectului de la simularea Bacalaureatului 2019 (clasa a XII-a) la Matematică M_șt-nat . Pentru fiecare cerință găsești enunțul pe scurt, rezolvarea pas cu pas, răspunsul final și cum se punctează conform baremului oficial. Timp de lucru: 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. z z z 3 z + 2 z ˉ = 5 + 2 i 3z + 2\bar z = 5 + 2i 3 z + 2 z ˉ = 5 + 2 i
Fie z = a + b i z = a + bi z = a + bi 3 a + 3 b i + 2 a − 2 b i = 5 a + b i = 5 + 2 i ⇒ a = 1 3a + 3bi + 2a - 2bi = 5a + bi = 5 + 2i \Rightarrow a = 1 3 a + 3 bi + 2 a − 2 bi = 5 a + bi = 5 + 2 i ⇒ a = 1 b = 2 b = 2 b = 2
Răspuns: z = 1 + 2 i z = 1 + 2i z = 1 + 2 i Barem: 3p, 2p.
2. f ( x ) = x + a f(x) = x + a f ( x ) = x + a ( f ∘ f ∘ f ) ( x ) = x + 3 (f\circ f\circ f)(x) = x + 3 ( f ∘ f ∘ f ) ( x ) = x + 3
( f ∘ f ) ( x ) = x + 2 a (f\circ f)(x) = x + 2a ( f ∘ f ) ( x ) = x + 2 a ( f ∘ f ∘ f ) ( x ) = x + 3 a = x + 3 ⇒ a = 1 (f\circ f\circ f)(x) = x + 3a = x + 3 \Rightarrow a = 1 ( f ∘ f ∘ f ) ( x ) = x + 3 a = x + 3 ⇒ a = 1 Barem: 3p, 2p.
3. Ecuația log 3 ( 2 x + 3 ) − log 3 x = 1 \log_3(2x + 3) - \log_3 x = 1 log 3 ( 2 x + 3 ) − log 3 x = 1
Condiție x > 0 x > 0 x > 0 log 3 2 x + 3 x = 1 ⟺ 2 x + 3 = 3 x ⟺ x = 3 \log_3\frac{2x + 3}{x} = 1 \iff 2x + 3 = 3x \iff x = 3 log 3 x 2 x + 3 = 1 ⟺ 2 x + 3 = 3 x ⟺ x = 3 Barem: 3p, 2p.
4. Probabilitatea ca un număr de trei cifre să fie divizibil cu 10
900 de cazuri posibile; multiplii lui 10: 100 , 110 , … , 990 100, 110, \dots, 990 100 , 110 , … , 990 p = 90 900 = 1 10 p = \frac{90}{900} = \frac1{10} p = 900 90 = 10 1 Barem: 1p, 2p, 2p.
5. u ⃗ = ( a + 1 ) i ⃗ + ( 5 a − 1 ) j ⃗ \vec u = (a+1)\vec i + (5a - 1)\vec j u = ( a + 1 ) i + ( 5 a − 1 ) j v ⃗ = i ⃗ + 3 j ⃗ \vec v = \vec i + 3\vec j v = i + 3 j
a + 1 1 = 5 a − 1 3 ⟺ 3 a + 3 = 5 a − 1 ⟺ a = 2 \frac{a + 1}{1} = \frac{5a - 1}{3} \iff 3a + 3 = 5a - 1 \iff a = 2 1 a + 1 = 3 5 a − 1 ⟺ 3 a + 3 = 5 a − 1 ⟺ a = 2 Barem: 3p, 2p.
6. Aria lui △ A B C \triangle ABC △ A B C A B = 6 AB = 6 A B = 6 A C = 10 AC = 10 A C = 10 cos A = 3 5 \cos A = \frac35 cos A = 5 3
sin A = 1 − 9 25 = 4 5 \sin A = \sqrt{1 - \frac9{25}} = \frac45 sin A = 1 − 25 9 = 5 4
A = A B ⋅ A C ⋅ sin A 2 = 6 ⋅ 10 ⋅ 4 5 2 = 24. \mathcal A = \frac{AB\cdot AC\cdot\sin A}{2} = \frac{6\cdot10\cdot\frac45}{2} = 24. A = 2 A B ⋅ A C ⋅ s i n A = 2 6 ⋅ 10 ⋅ 5 4 = 24.
Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. M ( x ) = ( x + 1 0 x + 2 0 x 0 3 − x 0 4 − x ) M(x) = \begin{pmatrix}x + 1 & 0 & x + 2\\ 0 & x & 0\\ 3 - x & 0 & 4 - x\end{pmatrix} M ( x ) = x + 1 0 3 − x 0 x 0 x + 2 0 4 − x
a) M ( − 1 ) = ( 0 0 1 0 − 1 0 4 0 5 ) M(-1) = \begin{pmatrix}0 & 0 & 1\\ 0 & -1 & 0\\ 4 & 0 & 5\end{pmatrix} M ( − 1 ) = 0 0 4 0 − 1 0 1 0 5 det = − 1 ⋅ ( 0 ⋅ 5 − 1 ⋅ 4 ) = 4 \det = -1\cdot(0\cdot5 - 1\cdot 4) = 4 det = − 1 ⋅ ( 0 ⋅ 5 − 1 ⋅ 4 ) = 4
Răspuns: 4 4 4 Barem: 2p, 3p.
b) M ( x ) + M ( y ) = ( x + y + 2 0 x + y + 4 0 x + y 0 6 − x − y 0 8 − x − y ) = ( 1 0 2 0 0 0 3 0 4 ) + M ( x + y ) = M ( 0 ) + M ( x + y ) . ■ M(x) + M(y) = \begin{pmatrix}x + y + 2 & 0 & x + y + 4\\ 0 & x + y & 0\\ 6 - x - y & 0 & 8 - x - y\end{pmatrix} = \begin{pmatrix}1&0&2\\0&0&0\\3&0&4\end{pmatrix} + M(x + y) = M(0) + M(x+y).\ \blacksquare M ( x ) + M ( y ) = x + y + 2 0 6 − x − y 0 x + y 0 x + y + 4 0 8 − x − y = 1 0 3 0 0 0 2 0 4 + M ( x + y ) = M ( 0 ) + M ( x + y ) . ■
Barem: 3p, 2p.
c) Cu M ( 1 ) = ( 2 0 3 0 1 0 2 0 3 ) M(1) = \begin{pmatrix}2 & 0 & 3\\ 0 & 1 & 0\\ 2 & 0 & 3\end{pmatrix} M ( 1 ) = 2 0 2 0 1 0 3 0 3
M ( m ) M ( 1 ) = ( 4 m + 6 0 6 m + 9 0 m 0 14 − 4 m 0 21 − 6 m ) , M ( 1 ) M ( n ) = ( 11 − n 0 16 − n 0 n 0 11 − n 0 16 − n ) . M(m)M(1) = \begin{pmatrix}4m + 6 & 0 & 6m + 9\\ 0 & m & 0\\ 14 - 4m & 0 & 21 - 6m\end{pmatrix},\quad M(1)M(n) = \begin{pmatrix}11 - n & 0 & 16 - n\\ 0 & n & 0\\ 11 - n & 0 & 16 - n\end{pmatrix}. M ( m ) M ( 1 ) = 4 m + 6 0 14 − 4 m 0 m 0 6 m + 9 0 21 − 6 m , M ( 1 ) M ( n ) = 11 − n 0 11 − n 0 n 0 16 − n 0 16 − n .
Sumele elementelor: 50 + m 50 + m 50 + m 54 − 3 n 54 - 3n 54 − 3 n m + 3 n = 4 m + 3n = 4 m + 3 n = 4 m , n ∈ N m, n \in \mathbb N m , n ∈ N ( m , n ) ∈ { ( 4 , 0 ) , ( 1 , 1 ) } (m, n) \in \{(4, 0), (1, 1)\} ( m , n ) ∈ {( 4 , 0 ) , ( 1 , 1 )}
Barem: 2p, 3p.
2. x ∗ y = 4 x + 4 y − 4 x y − 3 x * y = 4x + 4y - 4xy - 3 x ∗ y = 4 x + 4 y − 4 x y − 3
a) 1 − 4 x y + 4 x + 4 y − 4 = 1 − 4 x ( y − 1 ) + 4 ( y − 1 ) = 1 − 4 ( x − 1 ) ( y − 1 ) 1 - 4xy + 4x + 4y - 4 = 1 - 4x(y - 1) + 4(y - 1) = 1 - 4(x - 1)(y - 1) 1 − 4 x y + 4 x + 4 y − 4 = 1 − 4 x ( y − 1 ) + 4 ( y − 1 ) = 1 − 4 ( x − 1 ) ( y − 1 ) ■ \blacksquare ■ Barem: 3p, 2p.
b) x ∗ 1 x = 1 − 4 ( x − 1 ) ⋅ 1 − x x = 1 + 4 ( x − 1 ) 2 x ≥ 1 x * \frac1x = 1 - 4(x - 1)\cdot\frac{1 - x}{x} = 1 + \frac{4(x-1)^2}{x} \ge 1 x ∗ x 1 = 1 − 4 ( x − 1 ) ⋅ x 1 − x = 1 + x 4 ( x − 1 ) 2 ≥ 1 x > 0 x > 0 x > 0 ■ \blacksquare ■ Barem: 3p, 2p.
c) Cu t = x − 1 t = x - 1 t = x − 1 x ∗ x = 1 − 4 t 2 x * x = 1 - 4t^2 x ∗ x = 1 − 4 t 2 x ∗ x ∗ x = 1 + 4 2 t 3 x * x * x = 1 + 4^2t^3 x ∗ x ∗ x = 1 + 4 2 t 3 x ∗ x ∗ x ∗ x = 1 − 4 3 t 4 x * x * x * x = 1 - 4^3t^4 x ∗ x ∗ x ∗ x = 1 − 4 3 t 4 1 − 64 t 4 = 1 + t ⟺ t ( 1 + 64 t 3 ) = 0 ⟺ t = 0 1 - 64t^4 = 1 + t \iff t(1 + 64t^3) = 0 \iff t = 0 1 − 64 t 4 = 1 + t ⟺ t ( 1 + 64 t 3 ) = 0 ⟺ t = 0 t = − 1 4 t = -\frac14 t = − 4 1
Răspuns: x ∈ { 3 4 , 1 } x \in \left\{\frac34, 1\right\} x ∈ { 4 3 , 1 } Barem: 3p, 2p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = 5 ln x − x 2 − 3 x f(x) = 5\ln x - x^2 - 3x f ( x ) = 5 ln x − x 2 − 3 x x > 0 x > 0 x > 0
a) f ′ ( x ) = 5 x − 2 x − 3 = − 2 x 2 − 3 x + 5 x = ( 1 − x ) ( 2 x + 5 ) x f'(x) = \frac5x - 2x - 3 = \frac{-2x^2 - 3x + 5}{x} = \frac{(1 - x)(2x + 5)}{x} f ′ ( x ) = x 5 − 2 x − 3 = x − 2 x 2 − 3 x + 5 = x ( 1 − x ) ( 2 x + 5 ) ■ \blacksquare ■ Barem: 3p, 2p.
b) f ′ ′ ( x ) = − 5 x 2 − 2 = − 2 x 2 − 5 x 2 < 0 f''(x) = -\frac{5}{x^2} - 2 = \frac{-2x^2 - 5}{x^2} < 0 f ′′ ( x ) = − x 2 5 − 2 = x 2 − 2 x 2 − 5 < 0 f f f ■ \blacksquare ■ Barem: 2p, 3p.
c) f ′ ≥ 0 f' \ge 0 f ′ ≥ 0 ( 0 , 1 ] (0, 1] ( 0 , 1 ] f ′ ≤ 0 f' \le 0 f ′ ≤ 0 [ 1 , + ∞ ) [1, +\infty) [ 1 , + ∞ ) f ( x ) ≤ f ( 1 ) = − 4 f(x) \le f(1) = -4 f ( x ) ≤ f ( 1 ) = − 4 5 ln x − x 2 − 3 x ≤ − 4 ⟺ 5 ln x ≤ x 2 + 3 x − 4 5\ln x - x^2 - 3x \le -4 \iff 5\ln x \le x^2 + 3x - 4 5 ln x − x 2 − 3 x ≤ − 4 ⟺ 5 ln x ≤ x 2 + 3 x − 4 ■ \blacksquare ■ Barem: 3p, 2p.
2. f ( x ) = ( x 2 + 4 x + 5 ) e x f(x) = (x^2 + 4x + 5)e^x f ( x ) = ( x 2 + 4 x + 5 ) e x
a) x 2 + 4 x + 5 = ( x + 2 ) 2 + 1 > 0 x^2 + 4x + 5 = (x + 2)^2 + 1 > 0 x 2 + 4 x + 5 = ( x + 2 ) 2 + 1 > 0 F ′ = f > 0 F' = f > 0 F ′ = f > 0 ■ \blacksquare ■ Barem: 2p, 3p.
b) f ( x ) − x 2 e x − 5 e x = 4 x e x f(x) - x^2e^x - 5e^x = 4xe^x f ( x ) − x 2 e x − 5 e x = 4 x e x ∫ x e x = ( x − 1 ) e x \int xe^x = (x - 1)e^x ∫ x e x = ( x − 1 ) e x
∫ 0 1 4 x e x d x = 4 ( x − 1 ) e x ∣ 0 1 = 0 − 4 ⋅ ( − 1 ) = 4. \int_0^1 4xe^xdx = 4(x - 1)e^x\Big|_0^1 = 0 - 4\cdot(-1) = 4. ∫ 0 1 4 x e x d x = 4 ( x − 1 ) e x 0 1 = 0 − 4 ⋅ ( − 1 ) = 4.
Răspuns: 4 4 4 Barem: 3p, 2p.
c) Pe [ − 3 , − 1 ] [-3, -1] [ − 3 , − 1 ] ( x + 2 ) 2 ∈ [ 0 , 1 ] (x + 2)^2 \in [0, 1] ( x + 2 ) 2 ∈ [ 0 , 1 ] 1 ≤ x 2 + 4 x + 5 ≤ 2 1 \le x^2 + 4x + 5 \le 2 1 ≤ x 2 + 4 x + 5 ≤ 2 e x ≤ f ( x ) ≤ 2 e x e^x \le f(x) \le 2e^x e x ≤ f ( x ) ≤ 2 e x
∫ − 3 − 1 e x d x = e − 1 − e − 3 = e 2 − 1 e 3 ⇒ e 2 − 1 e 3 ≤ ∫ − 3 − 1 f ( x ) d x ≤ 2 ( e 2 − 1 ) e 3 . ■ \int_{-3}^{-1}e^xdx = e^{-1} - e^{-3} = \frac{e^2 - 1}{e^3} \;\Rightarrow\; \frac{e^2 - 1}{e^3} \le \int_{-3}^{-1}f(x)\,dx \le \frac{2(e^2 - 1)}{e^3}.\ \blacksquare ∫ − 3 − 1 e x d x = e − 1 − e − 3 = e 3 e 2 − 1 ⇒ e 3 e 2 − 1 ≤ ∫ − 3 − 1 f ( x ) d x ≤ e 3 2 ( e 2 − 1 ) . ■
Barem: 2p, 3p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 z = 1 + 2 i z = 1 + 2i z = 1 + 2 i I.2 a = 1 a = 1 a = 1 I.3 x = 3 x = 3 x = 3 I.4 p = 1 10 p = \tfrac1{10} p = 10 1 I.5 a = 2 a = 2 a = 2 I.6 24 24 24 II.1.a 4 4 4 II.1.c ( 4 , 0 ) (4,0) ( 4 , 0 ) ( 1 , 1 ) (1,1) ( 1 , 1 ) II.2.c x ∈ { 3 4 , 1 } x \in \{\tfrac34, 1\} x ∈ { 4 3 , 1 } III.2.b 4 4 4
Ce merită reținut din simulare
Ecuații cu z ˉ \bar z z ˉ scrie z = a + b i z = a + bi z = a + bi Concavitate: f ′ ′ < 0 f'' < 0 f ′′ < 0 inegalitate din maximul global al funcției.Încadrarea integralei: încadrează integrandul și integrează inegalitățile.