Rezolvarea completă a subiectului de Bacalaureat 2022 la Matematică M_șt-nat , sesiunea iunie–iulie, varianta 1 . 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. Progresia aritmetică cu a 2 = 6 a_2 = 6 a 2 = 6 a 3 = 12 a_3 = 12 a 3 = 12 a 1 a_1 a 1
Rația: r = a 3 − a 2 = 6 r = a_3 - a_2 = 6 r = a 3 − a 2 = 6 a 1 = a 2 − r = 6 − 6 = 0 a_1 = a_2 - r = 6 - 6 = 0 a 1 = a 2 − r = 6 − 6 = 0
Răspuns: a 1 = 0 a_1 = 0 a 1 = 0 Barem: 2p rația, 3p a 1 a_1 a 1
2. f ( x ) = x − 5 f(x) = x - 5 f ( x ) = x − 5 a a a f ( a ) + f ( 2 a ) = 2 f(a) + f(2a) = 2 f ( a ) + f ( 2 a ) = 2
( a − 5 ) + ( 2 a − 5 ) = 2 ⟺ 3 a = 12 ⟺ a = 4. (a - 5) + (2a - 5) = 2 \iff 3a = 12 \iff a = 4. ( a − 5 ) + ( 2 a − 5 ) = 2 ⟺ 3 a = 12 ⟺ a = 4.
Răspuns: a = 4 a = 4 a = 4 Barem: 3p + 2p.
3. Ecuația 5 x ⋅ 1 5 = 25 5^x\cdot\frac15 = 25 5 x ⋅ 5 1 = 25
5 x − 1 = 5 2 ⟺ x − 1 = 2 ⟺ x = 3 5^{x-1} = 5^2 \iff x - 1 = 2 \iff x = 3 5 x − 1 = 5 2 ⟺ x − 1 = 2 ⟺ x = 3
Răspuns: x = 3 x = 3 x = 3 Barem: 3p + 2p.
4. Probabilitatea ca un număr de două cifre să fie multiplu de 16
Cazuri posibile: 10 , … , 99 10, \dots, 99 10 , … , 99 90 . Multipli de 16 de două cifre: 16 , 32 , 48 , 64 , 80 , 96 16, 32, 48, 64, 80, 96 16 , 32 , 48 , 64 , 80 , 96 6 .
p = 6 90 = 1 15 . p = \frac{6}{90} = \frac{1}{15}. p = 90 6 = 15 1 .
Barem: 2p + 3p.
5. A ( 3 , 2 ) A(3,2) A ( 3 , 2 ) B ( 1 , 4 ) B(1,4) B ( 1 , 4 ) C C C A A A B C BC B C
1 + x C 2 = 3 ⇒ x C = 5 , 4 + y C 2 = 2 ⇒ y C = 0. \frac{1 + x_C}{2} = 3 \Rightarrow x_C = 5,\qquad \frac{4 + y_C}{2} = 2 \Rightarrow y_C = 0. 2 1 + x C = 3 ⇒ x C = 5 , 2 4 + y C = 2 ⇒ y C = 0.
Răspuns: C ( 5 , 0 ) C(5, 0) C ( 5 , 0 ) Barem: 3p + 2p.
6. E ( x ) = sin x + sin 3 x 2 − cos x 2 E(x) = \sin x + \sin\frac{3x}{2} - \cos\frac x2 E ( x ) = sin x + sin 2 3 x − cos 2 x E ( π 3 ) = 1 E\left(\frac\pi3\right) = 1 E ( 3 π ) = 1
E ( π 3 ) = sin π 3 + sin π 2 − cos π 6 = 3 2 + 1 − 3 2 = 1. E\left(\tfrac\pi3\right) = \sin\tfrac\pi3 + \sin\tfrac\pi2 - \cos\tfrac\pi6 = \tfrac{\sqrt3}{2} + 1 - \tfrac{\sqrt3}{2} = 1. E ( 3 π ) = sin 3 π + sin 2 π − cos 6 π = 2 3 + 1 − 2 3 = 1.
Barem: 3p valorile, 2p finalizarea.
Subiectul al II-lea (30 de puncte)
1. A = ( 1 − 1 − 1 1 ) A = \begin{pmatrix}1&-1\\-1&1\end{pmatrix} A = ( 1 − 1 − 1 1 ) B ( x ) = ( x 3 − x 2 − x x ) B(x) = \begin{pmatrix}x & 3 - x\\ 2 - x & x\end{pmatrix} B ( x ) = ( x 2 − x 3 − x x )
a) det A = 0 \det A = 0 det A = 0
det A = 1 ⋅ 1 − ( − 1 ) ( − 1 ) = 0. \det A = 1\cdot 1 - (-1)(-1) = 0. det A = 1 ⋅ 1 − ( − 1 ) ( − 1 ) = 0.
Barem: 3p + 2p.
b) B ( x ) − B ( 0 ) = x A B(x) - B(0) = xA B ( x ) − B ( 0 ) = x A
B ( 0 ) = ( 0 3 2 0 ) B(0) = \begin{pmatrix}0&3\\2&0\end{pmatrix} B ( 0 ) = ( 0 2 3 0 )
B ( x ) − B ( 0 ) = ( x − x − x x ) = x ( 1 − 1 − 1 1 ) = x A . B(x) - B(0) = \begin{pmatrix}x & -x\\ -x & x\end{pmatrix} = x\begin{pmatrix}1&-1\\-1&1\end{pmatrix} = xA. B ( x ) − B ( 0 ) = ( x − x − x x ) = x ( 1 − 1 − 1 1 ) = x A .
Barem: 3p + 2p.
c) C ( a ) = B ( a ) B ( 1 ) − B ( a + 1 ) C(a) = B(a)B(1) - B(a+1) C ( a ) = B ( a ) B ( 1 ) − B ( a + 1 ) a ∈ Z a \in \mathbb Z a ∈ Z
B ( 1 ) = ( 1 2 1 1 ) B(1) = \begin{pmatrix}1&2\\1&1\end{pmatrix} B ( 1 ) = ( 1 1 2 1 )
B ( a ) B ( 1 ) = ( 3 a + 3 2 4 − a ) , B ( a + 1 ) = ( a + 1 2 − a 1 − a a + 1 ) , B(a)B(1) = \begin{pmatrix}3 & a + 3\\ 2 & 4 - a\end{pmatrix},\quad B(a+1) = \begin{pmatrix}a+1 & 2 - a\\ 1 - a & a + 1\end{pmatrix}, B ( a ) B ( 1 ) = ( 3 2 a + 3 4 − a ) , B ( a + 1 ) = ( a + 1 1 − a 2 − a a + 1 ) ,
C ( a ) = ( 2 − a 2 a + 1 a + 1 3 − 2 a ) , det C ( a ) = ( 2 − a ) ( 3 − 2 a ) − ( 2 a + 1 ) ( a + 1 ) = 5 − 10 a . C(a) = \begin{pmatrix}2 - a & 2a + 1\\ a + 1 & 3 - 2a\end{pmatrix},\quad \det C(a) = (2-a)(3-2a) - (2a+1)(a+1) = 5 - 10a. C ( a ) = ( 2 − a a + 1 2 a + 1 3 − 2 a ) , det C ( a ) = ( 2 − a ) ( 3 − 2 a ) − ( 2 a + 1 ) ( a + 1 ) = 5 − 10 a .
det C ( a ) = 0 ⟺ a = 1 2 ∉ Z \det C(a) = 0 \iff a = \frac12 \notin \mathbb Z det C ( a ) = 0 ⟺ a = 2 1 ∈ / Z a a a det C ( a ) ≠ 0 \det C(a) \ne 0 det C ( a ) = 0 C ( a ) C(a) C ( a ) ■ \blacksquare ■ Barem: 3p determinantul, 2p concluzia.
2. Legea x ∗ y = ( 2 x − 1 ) ( 2 y − 1 ) + 1 x * y = (2x - 1)(2y - 1) + 1 x ∗ y = ( 2 x − 1 ) ( 2 y − 1 ) + 1
a) 1 ∗ 2 = 4 1 * 2 = 4 1 ∗ 2 = 4
1 ∗ 2 = ( 2 − 1 ) ( 4 − 1 ) + 1 = 3 + 1 = 4. 1 * 2 = (2 - 1)(4 - 1) + 1 = 3 + 1 = 4. 1 ∗ 2 = ( 2 − 1 ) ( 4 − 1 ) + 1 = 3 + 1 = 4.
Barem: 3p + 2p.
b) x x x x ∗ x = 2 x * x = 2 x ∗ x = 2
( 2 x − 1 ) 2 + 1 = 2 ⟺ ( 2 x − 1 ) 2 = 1 ⟺ 4 x 2 − 4 x = 0 ⟺ 4 x ( x − 1 ) = 0. (2x - 1)^2 + 1 = 2 \iff (2x - 1)^2 = 1 \iff 4x^2 - 4x = 0 \iff 4x(x - 1) = 0. ( 2 x − 1 ) 2 + 1 = 2 ⟺ ( 2 x − 1 ) 2 = 1 ⟺ 4 x 2 − 4 x = 0 ⟺ 4 x ( x − 1 ) = 0.
Răspuns: x ∈ { 0 , 1 } x \in \{0, 1\} x ∈ { 0 , 1 } Barem: 2p + 3p.
c) m ∈ Z ∗ m \in \mathbb Z^* m ∈ Z ∗ m ∗ ( 1 + 1 m ) = 1 m * \left(1 + \frac1m\right) = 1 m ∗ ( 1 + m 1 ) = 1
2 ( 1 + 1 m ) − 1 = 1 + 2 m 2\left(1 + \frac1m\right) - 1 = 1 + \frac2m 2 ( 1 + m 1 ) − 1 = 1 + m 2
m ∗ ( 1 + 1 m ) = ( 2 m − 1 ) ( 1 + 2 m ) + 1 = 1 ⟺ ( 2 m − 1 ) ⋅ m + 2 m = 0. m * \left(1 + \tfrac1m\right) = (2m - 1)\left(1 + \tfrac2m\right) + 1 = 1 \iff (2m - 1)\cdot\frac{m + 2}{m} = 0. m ∗ ( 1 + m 1 ) = ( 2 m − 1 ) ( 1 + m 2 ) + 1 = 1 ⟺ ( 2 m − 1 ) ⋅ m m + 2 = 0.
Soluțiile sunt m = 1 2 m = \frac12 m = 2 1 m = − 2 m = -2 m = − 2
Răspuns: m = − 2 m = -2 m = − 2 Barem: 2p + 3p.
Subiectul al III-lea (30 de puncte)
1. f : ( 0 , ∞ ) → R f:(0,\infty)\to\mathbb R f : ( 0 , ∞ ) → R f ( x ) = 2 x 2 + 1 + ln x f(x) = 2x^2 + 1 + \ln x f ( x ) = 2 x 2 + 1 + ln x
a) f ′ ( x ) = 4 x 2 + 1 x f'(x) = \dfrac{4x^2 + 1}{x} f ′ ( x ) = x 4 x 2 + 1
f ′ ( x ) = 4 x + 1 x = 4 x 2 + 1 x , x > 0. f'(x) = 4x + \frac1x = \frac{4x^2 + 1}{x},\quad x > 0. f ′ ( x ) = 4 x + x 1 = x 4 x 2 + 1 , x > 0.
Barem: 2p + 3p.
b) lim x → + ∞ f ( x ) − ln x x 2 + x + 4 = 2 \lim\limits_{x\to+\infty}\dfrac{f(x) - \ln x}{x^2 + x + 4} = 2 x → + ∞ lim x 2 + x + 4 f ( x ) − ln x = 2
lim x → ∞ 2 x 2 + 1 x 2 + x + 4 = lim x → ∞ x 2 ( 2 + 1 x 2 ) x 2 ( 1 + 1 x + 4 x 2 ) = 2. \lim_{x\to\infty}\frac{2x^2 + 1}{x^2 + x + 4} = \lim_{x\to\infty}\frac{x^2\left(2 + \frac1{x^2}\right)}{x^2\left(1 + \frac1x + \frac4{x^2}\right)} = 2. lim x → ∞ x 2 + x + 4 2 x 2 + 1 = lim x → ∞ x 2 ( 1 + x 1 + x 2 4 ) x 2 ( 2 + x 2 1 ) = 2.
Barem: 3p + 2p.
c) f f f
Injectivitate: f ′ ( x ) > 0 f'(x) > 0 f ′ ( x ) > 0 ( 0 , ∞ ) (0, \infty) ( 0 , ∞ ) f f f
Surjectivitate: f f f lim x ↘ 0 f ( x ) = 1 + ( − ∞ ) = − ∞ \lim\limits_{x\searrow0} f(x) = 1 + (-\infty) = -\infty x ↘ 0 lim f ( x ) = 1 + ( − ∞ ) = − ∞ lim x → ∞ f ( x ) = + ∞ \lim\limits_{x\to\infty} f(x) = +\infty x → ∞ lim f ( x ) = + ∞ f f f
Deci f f f ■ \blacksquare ■ Barem: 2p + 3p.
2. f : R → R f:\mathbb R\to\mathbb R f : R → R f ( x ) = x ( e x + 2 x 2 ) f(x) = x\left(e^x + 2x^2\right) f ( x ) = x ( e x + 2 x 2 )
a) ∫ 0 4 f ( x ) e x + 2 x 2 d x = 8 \displaystyle\int_0^4 \frac{f(x)}{e^x + 2x^2}dx = 8 ∫ 0 4 e x + 2 x 2 f ( x ) d x = 8
∫ 0 4 x d x = x 2 2 ∣ 0 4 = 8. \int_0^4 x\,dx = \frac{x^2}{2}\Big|_0^4 = 8. ∫ 0 4 x d x = 2 x 2 0 4 = 8.
Barem: 3p + 2p.
b) ∫ 0 1 ( f ( x ) − 2 x 3 ) d x = 1 \displaystyle\int_0^1\left(f(x) - 2x^3\right)dx = 1 ∫ 0 1 ( f ( x ) − 2 x 3 ) d x = 1
f ( x ) − 2 x 3 = x e x f(x) - 2x^3 = xe^x f ( x ) − 2 x 3 = x e x u = x u = x u = x v ′ = e x v' = e^x v ′ = e x
∫ 0 1 x e x d x = x e x ∣ 0 1 − ∫ 0 1 e x d x = e − ( e − 1 ) = 1. \int_0^1 xe^x dx = xe^x\Big|_0^1 - \int_0^1 e^x dx = e - (e - 1) = 1. ∫ 0 1 x e x d x = x e x 0 1 − ∫ 0 1 e x d x = e − ( e − 1 ) = 1.
Barem: 3p + 2p.
c) a a a ∫ 1 2 1 x f ( x 2 ) d x = e 4 − e 2 + a \displaystyle\int_1^2\frac1x f(x^2)\,dx = \frac{e^4 - e}{2} + a ∫ 1 2 x 1 f ( x 2 ) d x = 2 e 4 − e + a
f ( x 2 ) = x 2 ( e x 2 + 2 x 4 ) f(x^2) = x^2\left(e^{x^2} + 2x^4\right) f ( x 2 ) = x 2 ( e x 2 + 2 x 4 ) 1 x f ( x 2 ) = x e x 2 + 2 x 5 \frac1x f(x^2) = xe^{x^2} + 2x^5 x 1 f ( x 2 ) = x e x 2 + 2 x 5
∫ 1 2 x e x 2 d x = 1 2 e x 2 ∣ 1 2 = e 4 − e 2 , ∫ 1 2 2 x 5 d x = x 6 3 ∣ 1 2 = 64 − 1 3 = 21. \int_1^2 xe^{x^2}dx = \frac12 e^{x^2}\Big|_1^2 = \frac{e^4 - e}{2},\qquad \int_1^2 2x^5dx = \frac{x^6}{3}\Big|_1^2 = \frac{64 - 1}{3} = 21. ∫ 1 2 x e x 2 d x = 2 1 e x 2 1 2 = 2 e 4 − e , ∫ 1 2 2 x 5 d x = 3 x 6 1 2 = 3 64 − 1 = 21.
Răspuns: a = 21 a = 21 a = 21 Barem: 3p calculul integralei, 2p valoarea lui a a a
Rezumatul răspunsurilor
Cerință Răspuns I.1 a 1 = 0 a_1 = 0 a 1 = 0 I.2 a = 4 a = 4 a = 4 I.3 x = 3 x = 3 x = 3 I.4 p = 1 15 p = \frac1{15} p = 15 1 I.5 C ( 5 , 0 ) C(5,0) C ( 5 , 0 ) II.2.b x ∈ { 0 , 1 } x \in \{0,1\} x ∈ { 0 , 1 } II.2.c m = − 2 m = -2 m = − 2 III.2.c a = 21 a = 21 a = 21
Ce merită reținut din varianta 1
Bijectivitate = strict monotonă (injectivă) + continuă cu limitele − ∞ -\infty − ∞ + ∞ +\infty + ∞ ∫ x e x 2 d x = 1 2 e x 2 \int xe^{x^2}dx = \frac12e^{x^2} ∫ x e x 2 d x = 2 1 e x 2 t = x 2 t = x^2 t = x 2 Inversabilitate pentru valori întregi: determinantul se anulează doar într-o valoare neîntreagă.