Rezolvarea completă a subiectului de Bacalaureat 2019 la Matematică M_șt-nat , sesiunea august–septembrie, varianta 7 . 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. Suma primilor trei termeni ai progresiei aritmetice cu a 1 = 2 a_1 = 2 a 1 = 2 r = 2 r = 2 r = 2
a 2 = 4 a_2 = 4 a 2 = 4 a 3 = 6 a_3 = 6 a 3 = 6 = 12 = 12 = 12 Barem: 2p, 3p.
2. Intersecția graficului lui f ( x ) = x 2 − 10 x + 9 f(x) = x^2 - 10x + 9 f ( x ) = x 2 − 10 x + 9 O x Ox O x
x 2 − 10 x + 9 = 0 ⟺ ( x − 1 ) ( x − 9 ) = 0 x^2 - 10x + 9 = 0 \iff (x - 1)(x - 9) = 0 x 2 − 10 x + 9 = 0 ⟺ ( x − 1 ) ( x − 9 ) = 0
Răspuns: x ∈ { 1 , 9 } x \in \{1, 9\} x ∈ { 1 , 9 } Barem: 3p, 2p.
3. Ecuația 5 x + 1 − 3 ⋅ 5 x = 2 5^{x+1} - 3\cdot 5^x = 2 5 x + 1 − 3 ⋅ 5 x = 2
5 x ( 5 − 3 ) = 2 ⟺ 5 x = 1 ⟺ x = 0 5^x(5 - 3) = 2 \iff 5^x = 1 \iff x = 0 5 x ( 5 − 3 ) = 2 ⟺ 5 x = 1 ⟺ x = 0 Barem: 3p, 2p.
4. Probabilitatea ca x ∈ { 0 , … , 9 } x \in \{0, \dots, 9\} x ∈ { 0 , … , 9 } x 2 − 4 x + 4 = 0 x^2 - 4x + 4 = 0 x 2 − 4 x + 4 = 0
( x − 2 ) 2 = 0 ⟺ x = 2 (x - 2)^2 = 0 \iff x = 2 ( x − 2 ) 2 = 0 ⟺ x = 2
p = 1 10 . p = \frac1{10}. p = 10 1 .
Barem: 1p, 2p, 2p.
5. Lungimea vectorului A B → + A C → \overrightarrow{AB} + \overrightarrow{AC} A B + A C △ A B C \triangle ABC △ A B C A B = 2 AB = 2 A B = 2
Cu M M M B C BC B C A B → + A C → = 2 A M → \overrightarrow{AB} + \overrightarrow{AC} = 2\overrightarrow{AM} A B + A C = 2 A M A M = 2 3 2 = 3 AM = \frac{2\sqrt3}{2} = \sqrt3 A M = 2 2 3 = 3 2 3 2\sqrt3 2 3
Răspuns: 2 3 2\sqrt3 2 3 Barem: 3p, 2p.
6. sin 2 ( π 2 − x ) + sin 2 ( x + π ) = 1 \sin^2\left(\frac\pi2 - x\right) + \sin^2(x + \pi) = 1 sin 2 ( 2 π − x ) + sin 2 ( x + π ) = 1
sin ( π 2 − x ) = cos x \sin\left(\frac\pi2 - x\right) = \cos x sin ( 2 π − x ) = cos x sin ( x + π ) = − sin x \sin(x + \pi) = -\sin x sin ( x + π ) = − sin x cos 2 x + sin 2 x = 1 \cos^2x + \sin^2x = 1 cos 2 x + sin 2 x = 1 ■ \blacksquare ■ Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. A ( a ) = ( 1 + 4 a − 6 a 2 a 1 − 3 a ) A(a) = \begin{pmatrix}1 + 4a & -6a\\ 2a & 1 - 3a\end{pmatrix} A ( a ) = ( 1 + 4 a 2 a − 6 a 1 − 3 a )
a) A ( 1 ) = ( 5 − 6 2 − 2 ) A(1) = \begin{pmatrix}5 & -6\\ 2 & -2\end{pmatrix} A ( 1 ) = ( 5 2 − 6 − 2 ) det = − 10 + 12 = 2 \det = -10 + 12 = 2 det = − 10 + 12 = 2 ■ \blacksquare ■ Barem: 3p, 2p.
b) A ( a ) A ( b ) = ( ( 1 + 4 a ) ( 1 + 4 b ) − 12 a b − 6 b ( 1 + 4 a ) − 6 a ( 1 − 3 b ) 2 a ( 1 + 4 b ) + 2 b ( 1 − 3 a ) − 12 a b + ( 1 − 3 a ) ( 1 − 3 b ) ) = ( 1 + 4 a + 4 b + 4 a b − 6 a − 6 b − 6 a b 2 a + 2 b + 2 a b 1 − 3 a − 3 b − 3 a b ) , A(a)A(b) = \begin{pmatrix}(1+4a)(1+4b) - 12ab & -6b(1+4a) - 6a(1 - 3b)\\ 2a(1 + 4b) + 2b(1 - 3a) & -12ab + (1-3a)(1-3b)\end{pmatrix} = \begin{pmatrix}1 + 4a + 4b + 4ab & -6a - 6b - 6ab\\ 2a + 2b + 2ab & 1 - 3a - 3b - 3ab\end{pmatrix}, A ( a ) A ( b ) = ( ( 1 + 4 a ) ( 1 + 4 b ) − 12 ab 2 a ( 1 + 4 b ) + 2 b ( 1 − 3 a ) − 6 b ( 1 + 4 a ) − 6 a ( 1 − 3 b ) − 12 ab + ( 1 − 3 a ) ( 1 − 3 b ) ) = ( 1 + 4 a + 4 b + 4 ab 2 a + 2 b + 2 ab − 6 a − 6 b − 6 ab 1 − 3 a − 3 b − 3 ab ) ,
adică A ( a + b + a b ) A(a + b + ab) A ( a + b + ab ) ■ \blacksquare ■ Barem: 3p, 2p.
c) A ( m + n + m n ) = A ( 2 ) ⟺ m + n + m n = 2 ⟺ ( m + 1 ) ( n + 1 ) = 3 A(m + n + mn) = A(2) \iff m + n + mn = 2 \iff (m + 1)(n + 1) = 3 A ( m + n + mn ) = A ( 2 ) ⟺ m + n + mn = 2 ⟺ ( m + 1 ) ( n + 1 ) = 3 m , n m, n m , n { m + 1 , n + 1 } = { 1 , 3 } \{m + 1, n + 1\} = \{1, 3\} { m + 1 , n + 1 } = { 1 , 3 }
Răspuns: ( m , n ) ∈ { ( 0 , 2 ) , ( 2 , 0 ) } (m, n) \in \{(0, 2), (2, 0)\} ( m , n ) ∈ {( 0 , 2 ) , ( 2 , 0 )} Barem: 2p, 3p.
2. x ∘ y = 2 x y − 2 x − 2 y + 3 x \circ y = 2xy - 2x - 2y + 3 x ∘ y = 2 x y − 2 x − 2 y + 3
a) 2 x y − 2 x − 2 y + 2 + 1 = 2 ( x − 1 ) ( y − 1 ) + 1 2xy - 2x - 2y + 2 + 1 = 2(x - 1)(y - 1) + 1 2 x y − 2 x − 2 y + 2 + 1 = 2 ( x − 1 ) ( y − 1 ) + 1 ■ \blacksquare ■ Barem: 2p, 3p.
b) 2 ( x − 1 ) 2 + 1 ≤ 9 ⟺ ( x − 1 ) 2 ≤ 4 ⟺ − 2 ≤ x − 1 ≤ 2 ⟺ x ∈ [ − 1 , 3 ] 2(x - 1)^2 + 1 \le 9 \iff (x - 1)^2 \le 4 \iff -2 \le x - 1 \le 2 \iff x \in [-1, 3] 2 ( x − 1 ) 2 + 1 ≤ 9 ⟺ ( x − 1 ) 2 ≤ 4 ⟺ − 2 ≤ x − 1 ≤ 2 ⟺ x ∈ [ − 1 , 3 ] Barem: 2p, 3p.
c) 1 ∘ x = 2 ⋅ 0 ⋅ ( x − 1 ) + 1 = 1 1 \circ x = 2\cdot 0\cdot(x - 1) + 1 = 1 1 ∘ x = 2 ⋅ 0 ⋅ ( x − 1 ) + 1 = 1 x x x 1 n = 1 1^n = 1 1 n = 1
1 n ∘ ( 2 n ∘ ⋯ ∘ 2019 n ) = 1. 1^n \circ (2^n \circ \dots \circ 2019^n) = 1. 1 n ∘ ( 2 n ∘ ⋯ ∘ 201 9 n ) = 1.
Răspuns: 1 1 1 Barem: 2p, 3p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x − ln x e = x − e ln x f(x) = x - \ln x^e = x - e\ln x f ( x ) = x − ln x e = x − e ln x x > 0 x > 0 x > 0
a) f ′ ( x ) = 1 − e x = x − e x f'(x) = 1 - \dfrac ex = \dfrac{x - e}{x} f ′ ( x ) = 1 − x e = x x − e ■ \blacksquare ■ Barem: 2p, 3p.
b) Tangenta paralelă cu O x ⟺ f ′ ( a ) = 0 ⟺ a = e Ox \iff f'(a) = 0 \iff a = e O x ⟺ f ′ ( a ) = 0 ⟺ a = e Barem: 3p, 2p.
c) f ′ < 0 f' < 0 f ′ < 0 ( 0 , e ) (0, e) ( 0 , e ) f ′ > 0 f' > 0 f ′ > 0 ( e , + ∞ ) (e, +\infty) ( e , + ∞ ) e e e f ( e ) = e − e = 0 f(e) = e - e = 0 f ( e ) = e − e = 0 f ( x ) > 0 f(x) > 0 f ( x ) > 0 x ≠ e x \ne e x = e
e x = x e ⟺ x = e ln x ⟺ f ( x ) = 0 ⟺ x = e e^x = x^e \iff x = e\ln x \iff f(x) = 0 \iff x = e e x = x e ⟺ x = e ln x ⟺ f ( x ) = 0 ⟺ x = e x = e x = e x = e ■ \blacksquare ■
Barem: 2p, 3p.
2. f ( x ) = ( x − 1 ) ( x + 1 ) e x = ( x 2 − 1 ) e x f(x) = (x - 1)(x + 1)e^x = (x^2 - 1)e^x f ( x ) = ( x − 1 ) ( x + 1 ) e x = ( x 2 − 1 ) e x
a) ∫ 0 3 f ( x ) e x d x = ∫ 0 3 ( x 2 − 1 ) d x = ( x 3 3 − x ) ∣ 0 3 = 9 − 3 = 6. ■ \int_0^3\frac{f(x)}{e^x}dx = \int_0^3(x^2 - 1)\,dx = \left(\frac{x^3}{3} - x\right)\Big|_0^3 = 9 - 3 = 6.\ \blacksquare ∫ 0 3 e x f ( x ) d x = ∫ 0 3 ( x 2 − 1 ) d x = ( 3 x 3 − x ) 0 3 = 9 − 3 = 6. ■
Barem: 3p, 2p.
b) Pe [ 1 , 2 ] [1, 2] [ 1 , 2 ] f ≥ 0 f \ge 0 f ≥ 0
A = ∫ 1 2 ( x 2 − 1 ) e x d x = ( x 2 − 1 ) e x ∣ 1 2 − ∫ 1 2 2 x e x d x = 3 e 2 − 2 ( x − 1 ) e x ∣ 1 2 = 3 e 2 − 2 e 2 = e 2 . \mathcal A = \int_1^2(x^2 - 1)e^xdx = (x^2 - 1)e^x\Big|_1^2 - \int_1^2 2xe^xdx = 3e^2 - 2(x - 1)e^x\Big|_1^2 = 3e^2 - 2e^2 = e^2. A = ∫ 1 2 ( x 2 − 1 ) e x d x = ( x 2 − 1 ) e x 1 2 − ∫ 1 2 2 x e x d x = 3 e 2 − 2 ( x − 1 ) e x 1 2 = 3 e 2 − 2 e 2 = e 2 .
Răspuns: e 2 e^2 e 2 Barem: 2p, 3p.
c) 2 x e x f ( x ) = 2 x x 2 − 1 \dfrac{2xe^x}{f(x)} = \dfrac{2x}{x^2 - 1} f ( x ) 2 x e x = x 2 − 1 2 x
∫ 2 a 2 x x 2 − 1 d x = ln ( x 2 − 1 ) ∣ 2 a = ln a 2 − 1 3 . \int_2^a\frac{2x}{x^2-1}dx = \ln(x^2 - 1)\Big|_2^a = \ln\frac{a^2 - 1}{3}. ∫ 2 a x 2 − 1 2 x d x = ln ( x 2 − 1 ) 2 a = ln 3 a 2 − 1 .
ln a 2 − 1 3 = 3 ln 2 = ln 8 ⟺ a 2 − 1 = 24 ⟺ a 2 = 25 \ln\dfrac{a^2-1}{3} = 3\ln2 = \ln 8 \iff a^2 - 1 = 24 \iff a^2 = 25 ln 3 a 2 − 1 = 3 ln 2 = ln 8 ⟺ a 2 − 1 = 24 ⟺ a 2 = 25 a > 2 a > 2 a > 2 a = 5 a = 5 a = 5
Răspuns: a = 5 a = 5 a = 5 Barem: 3p, 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 12 12 12 I.2 x ∈ { 1 , 9 } x \in \{1, 9\} x ∈ { 1 , 9 } I.3 x = 0 x = 0 x = 0 I.4 p = 1 10 p = \tfrac1{10} p = 10 1 I.5 2 3 2\sqrt3 2 3 II.1.c ( 0 , 2 ) (0,2) ( 0 , 2 ) ( 2 , 0 ) (2,0) ( 2 , 0 ) II.2.b x ∈ [ − 1 , 3 ] x \in [-1, 3] x ∈ [ − 1 , 3 ] II.2.c 1 1 1 III.1.b x = e x = e x = e III.2.b e 2 e^2 e 2 III.2.c a = 5 a = 5 a = 5
Ce merită reținut din varianta 7
A ( a ) A ( b ) = A ( a + b + a b ) A(a)A(b) = A(a + b + ab) A ( a ) A ( b ) = A ( a + b + ab ) a + 1 a + 1 a + 1 ( m + 1 ) ( n + 1 ) = 3 (m+1)(n+1) = 3 ( m + 1 ) ( n + 1 ) = 3 e x = x e e^x = x^e e x = x e x − e ln x x - e\ln x x − e ln x Suma a doi vectori cu origine comună: de două ori vectorul median.