Rezolvarea completă a subiectului de Bacalaureat 2022 la Matematică M_șt-nat , sesiunea specială, varianta 3 . 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 2 ( 2 − 1 ) ( 2 + 2 ) = 2 \sqrt2(\sqrt2 - 1)(2 + \sqrt2) = 2 2 ( 2 − 1 ) ( 2 + 2 ) = 2
2 ( 2 − 1 ) = 2 − 2 \sqrt2(\sqrt2 - 1) = 2 - \sqrt2 2 ( 2 − 1 ) = 2 − 2
( 2 − 2 ) ( 2 + 2 ) = 4 − 2 = 2. (2 - \sqrt2)(2 + \sqrt2) = 4 - 2 = 2. ( 2 − 2 ) ( 2 + 2 ) = 4 − 2 = 2.
Barem: 3p + 2p.
2. Abscisele punctelor de intersecție ale graficului lui f ( x ) = 2 x 2 − 4 x f(x) = 2x^2 - 4x f ( x ) = 2 x 2 − 4 x O x Ox O x
f ( x ) = 0 ⟺ 2 x ( x − 2 ) = 0 ⟺ x ∈ { 0 , 2 } . f(x) = 0 \iff 2x(x - 2) = 0 \iff x \in \{0, 2\}. f ( x ) = 0 ⟺ 2 x ( x − 2 ) = 0 ⟺ x ∈ { 0 , 2 } .
Barem: 3p + 2p.
3. Ecuația 2 x − 3 = 1 2 2 x 2^{x-3} = \dfrac{1}{2^{2x}} 2 x − 3 = 2 2 x 1
2 x − 3 = 2 − 2 x ⟺ x − 3 = − 2 x ⟺ x = 1 2^{x-3} = 2^{-2x} \iff x - 3 = -2x \iff x = 1 2 x − 3 = 2 − 2 x ⟺ x − 3 = − 2 x ⟺ x = 1
Răspuns: x = 1 x = 1 x = 1 Barem: 3p + 2p.
4. Probabilitatea ca un număr de două cifre să fie multiplu de 11
90 cazuri posibile; multipli de 11: 11 , 22 , … , 99 11, 22, \dots, 99 11 , 22 , … , 99
p = 9 90 = 1 10 . p = \frac{9}{90} = \frac1{10}. p = 90 9 = 10 1 .
Barem: 2p + 3p.
5. A ( − 1 , 0 ) A(-1,0) A ( − 1 , 0 ) B ( 0 , 3 ) B(0,3) B ( 0 , 3 ) C ( 4 , 0 ) C(4,0) C ( 4 , 0 ) A B C ABC A B C
A C = 4 − ( − 1 ) = 5 AC = 4 - (-1) = 5 A C = 4 − ( − 1 ) = 5 O x Ox O x B C = 4 2 + 3 2 = 5 BC = \sqrt{4^2 + 3^2} = 5 B C = 4 2 + 3 2 = 5 A C = B C AC = BC A C = B C ■ \blacksquare ■
Barem: 2p + 3p.
6. E ( x ) = tg x + sin 3 x 2 − 2 cos x 2 E(x) = \operatorname{tg}x + \sin\frac{3x}{2} - 2\cos\frac x2 E ( x ) = tg x + sin 2 3 x − 2 cos 2 x E ( π 3 ) = 1 E\left(\frac\pi3\right) = 1 E ( 3 π ) = 1
E ( π 3 ) = tg π 3 + sin π 2 − 2 cos π 6 = 3 + 1 − 2 ⋅ 3 2 = 1. E\left(\tfrac\pi3\right) = \operatorname{tg}\tfrac\pi3 + \sin\tfrac\pi2 - 2\cos\tfrac\pi6 = \sqrt3 + 1 - 2\cdot\tfrac{\sqrt3}{2} = 1. E ( 3 π ) = tg 3 π + sin 2 π − 2 cos 6 π = 3 + 1 − 2 ⋅ 2 3 = 1.
Barem: 3p + 2p.
Subiectul al II-lea (30 de puncte)
1. M ( x ) = ( x + 1 − x − 2 x 2 x + 1 ) M(x) = \begin{pmatrix}x + 1 & -x\\ -2x & 2x + 1\end{pmatrix} M ( x ) = ( x + 1 − 2 x − x 2 x + 1 )
a) det M ( 1 ) = 4 \det M(1) = 4 det M ( 1 ) = 4
M ( 1 ) = ( 2 − 1 − 2 3 ) , det M ( 1 ) = 6 − 2 = 4. M(1) = \begin{pmatrix}2&-1\\-2&3\end{pmatrix},\quad \det M(1) = 6 - 2 = 4. M ( 1 ) = ( 2 − 2 − 1 3 ) , det M ( 1 ) = 6 − 2 = 4.
Barem: 3p + 2p.
b) M ( x ) ⋅ M ( 1 ) = M ( 4 x + 1 ) M(x)\cdot M(1) = M(4x + 1) M ( x ) ⋅ M ( 1 ) = M ( 4 x + 1 )
M ( x ) M ( 1 ) = ( 2 ( x + 1 ) + 2 x − ( x + 1 ) − 3 x − 4 x − 2 ( 2 x + 1 ) 2 x + 3 ( 2 x + 1 ) ) = ( 4 x + 2 − 4 x − 1 − 8 x − 2 8 x + 3 ) . M(x)M(1) = \begin{pmatrix}2(x+1) + 2x & -(x+1) - 3x\\ -4x - 2(2x+1) & 2x + 3(2x+1)\end{pmatrix} = \begin{pmatrix}4x + 2 & -4x - 1\\ -8x - 2 & 8x + 3\end{pmatrix}. M ( x ) M ( 1 ) = ( 2 ( x + 1 ) + 2 x − 4 x − 2 ( 2 x + 1 ) − ( x + 1 ) − 3 x 2 x + 3 ( 2 x + 1 ) ) = ( 4 x + 2 − 8 x − 2 − 4 x − 1 8 x + 3 ) .
Cu t = 4 x + 1 t = 4x + 1 t = 4 x + 1 ( t + 1 − t − 2 t 2 t + 1 ) = M ( 4 x + 1 ) \begin{pmatrix}t + 1 & -t\\ -2t & 2t + 1\end{pmatrix} = M(4x+1) ( t + 1 − 2 t − t 2 t + 1 ) = M ( 4 x + 1 ) ■ \blacksquare ■ Barem: 3p + 2p.
c) x x x M ( x ) ⋅ M ( 1 ) ⋅ M ( 1 ) = M ( x + 2 ) M(x)\cdot M(1)\cdot M(1) = M(x + 2) M ( x ) ⋅ M ( 1 ) ⋅ M ( 1 ) = M ( x + 2 )
Aplicând de două ori b): M ( x ) M ( 1 ) M ( 1 ) = M ( 4 x + 1 ) M ( 1 ) = M ( 4 ( 4 x + 1 ) + 1 ) = M ( 16 x + 5 ) M(x)M(1)M(1) = M(4x + 1)M(1) = M\big(4(4x+1) + 1\big) = M(16x + 5) M ( x ) M ( 1 ) M ( 1 ) = M ( 4 x + 1 ) M ( 1 ) = M ( 4 ( 4 x + 1 ) + 1 ) = M ( 16 x + 5 )
M ( 16 x + 5 ) = M ( x + 2 ) ⟺ 16 x + 5 = x + 2 ⟺ x = − 1 5 M(16x + 5) = M(x + 2) \iff 16x + 5 = x + 2 \iff x = -\dfrac15 M ( 16 x + 5 ) = M ( x + 2 ) ⟺ 16 x + 5 = x + 2 ⟺ x = − 5 1
Răspuns: x = − 1 5 x = -\dfrac15 x = − 5 1 Barem: 3p + 2p.
2. Legea x ∘ y = 5 x y + 10 x + 10 y + 18 x\circ y = 5xy + 10x + 10y + 18 x ∘ y = 5 x y + 10 x + 10 y + 18
a) ( − 1 ) ∘ 0 = 8 (-1)\circ0 = 8 ( − 1 ) ∘ 0 = 8
( − 1 ) ∘ 0 = 0 − 10 + 0 + 18 = 8. (-1)\circ 0 = 0 - 10 + 0 + 18 = 8. ( − 1 ) ∘ 0 = 0 − 10 + 0 + 18 = 8.
Barem: 3p + 2p.
b) x ∘ y = 5 ( x + 2 ) ( y + 2 ) − 2 x\circ y = 5(x+2)(y+2) - 2 x ∘ y = 5 ( x + 2 ) ( y + 2 ) − 2
5 ( x + 2 ) ( y + 2 ) − 2 = 5 x y + 10 x + 10 y + 20 − 2 = x ∘ y . 5(x+2)(y+2) - 2 = 5xy + 10x + 10y + 20 - 2 = x\circ y. 5 ( x + 2 ) ( y + 2 ) − 2 = 5 x y + 10 x + 10 y + 20 − 2 = x ∘ y .
Barem: 3p + 2p.
c) m m m m ∘ m = m m\circ m = m m ∘ m = m
5 ( m + 2 ) 2 − 2 = m ⟺ 5 m 2 + 19 m + 18 = 0 ⟺ ( m + 2 ) ( 5 m + 9 ) = 0. 5(m + 2)^2 - 2 = m \iff 5m^2 + 19m + 18 = 0 \iff (m + 2)(5m + 9) = 0. 5 ( m + 2 ) 2 − 2 = m ⟺ 5 m 2 + 19 m + 18 = 0 ⟺ ( m + 2 ) ( 5 m + 9 ) = 0.
Soluții: m = − 2 m = -2 m = − 2 m = − 9 5 m = -\frac95 m = − 5 9
Răspuns: m = − 2 m = -2 m = − 2 Barem: 2p + 3p.
Subiectul al III-lea (30 de puncte)
1. f : ( 1 , ∞ ) → R f:(1,\infty)\to\mathbb R f : ( 1 , ∞ ) → R f ( x ) = x 2 + 1 x − 1 + ln ( x − 1 ) f(x) = \dfrac{x^2 + 1}{x - 1} + \ln(x - 1) f ( x ) = x − 1 x 2 + 1 + ln ( x − 1 )
a) f ′ ( x ) = x 2 − x − 2 ( x − 1 ) 2 f'(x) = \dfrac{x^2 - x - 2}{(x-1)^2} f ′ ( x ) = ( x − 1 ) 2 x 2 − x − 2
f ′ ( x ) = 2 x ( x − 1 ) − ( x 2 + 1 ) ( x − 1 ) 2 + 1 x − 1 = x 2 − 2 x − 1 + ( x − 1 ) ( x − 1 ) 2 = x 2 − x − 2 ( x − 1 ) 2 . f'(x) = \frac{2x(x-1) - (x^2 + 1)}{(x-1)^2} + \frac{1}{x-1} = \frac{x^2 - 2x - 1 + (x - 1)}{(x-1)^2} = \frac{x^2 - x - 2}{(x-1)^2}. f ′ ( x ) = ( x − 1 ) 2 2 x ( x − 1 ) − ( x 2 + 1 ) + x − 1 1 = ( x − 1 ) 2 x 2 − 2 x − 1 + ( x − 1 ) = ( x − 1 ) 2 x 2 − x − 2 .
Barem: 3p + 2p.
b) Tangenta în punctul de abscisă x = 2 x = 2 x = 2
f ( 2 ) = 5 1 + ln 1 = 5 f(2) = \frac{5}{1} + \ln 1 = 5 f ( 2 ) = 1 5 + ln 1 = 5 f ′ ( 2 ) = 4 − 2 − 2 1 = 0 f'(2) = \frac{4 - 2 - 2}{1} = 0 f ′ ( 2 ) = 1 4 − 2 − 2 = 0
y − 5 = 0 ⋅ ( x − 2 ) ⟺ y = 5. y - 5 = 0\cdot(x - 2) \iff y = 5. y − 5 = 0 ⋅ ( x − 2 ) ⟺ y = 5.
Răspuns: y = 5 y = 5 y = 5 Barem: 2p + 3p.
c) x 2 + 1 x − 1 + ln ( x − 1 ) ≥ 5 \dfrac{x^2+1}{x-1} + \ln(x - 1) \ge 5 x − 1 x 2 + 1 + ln ( x − 1 ) ≥ 5 ( 1 , ∞ ) (1,\infty) ( 1 , ∞ )
x 2 − x − 2 = ( x − 2 ) ( x + 1 ) x^2 - x - 2 = (x - 2)(x + 1) x 2 − x − 2 = ( x − 2 ) ( x + 1 ) f ′ ( x ) ≤ 0 f'(x) \le 0 f ′ ( x ) ≤ 0 ( 1 , 2 ] (1, 2] ( 1 , 2 ] f ′ ( x ) ≥ 0 f'(x) \ge 0 f ′ ( x ) ≥ 0 [ 2 , ∞ ) [2, \infty) [ 2 , ∞ ) f f f x = 2 x = 2 x = 2
Prin urmare f ( x ) ≥ f ( 2 ) = 5 f(x) \ge f(2) = 5 f ( x ) ≥ f ( 2 ) = 5 x > 1 x > 1 x > 1 ■ \blacksquare ■ Barem: 3p monotonia, 2p concluzia.
2. f ( x ) = x + 4 6 x 2 + 1 f(x) = \dfrac{x + 4}{6x^2 + 1} f ( x ) = 6 x 2 + 1 x + 4
a) ∫ 0 2 f ( x ) ( 6 x 2 + 1 ) d x = 10 \displaystyle\int_0^2 f(x)(6x^2+1)dx = 10 ∫ 0 2 f ( x ) ( 6 x 2 + 1 ) d x = 10
∫ 0 2 ( x + 4 ) d x = ( x 2 2 + 4 x ) ∣ 0 2 = 2 + 8 = 10. \int_0^2(x + 4)dx = \left(\frac{x^2}{2} + 4x\right)\Big|_0^2 = 2 + 8 = 10. ∫ 0 2 ( x + 4 ) d x = ( 2 x 2 + 4 x ) 0 2 = 2 + 8 = 10.
Barem: 3p + 2p.
b) ∫ 0 2 ( f ( x ) − 4 6 x 2 + 1 ) d x = ln 5 6 \displaystyle\int_0^2\left(f(x) - \frac4{6x^2+1}\right)dx = \frac{\ln5}{6} ∫ 0 2 ( f ( x ) − 6 x 2 + 1 4 ) d x = 6 ln 5
∫ 0 2 x 6 x 2 + 1 d x = 1 12 ∫ 0 2 ( 6 x 2 + 1 ) ′ 6 x 2 + 1 d x = 1 12 ln ( 6 x 2 + 1 ) ∣ 0 2 = ln 25 12 = ln 5 6 . \int_0^2\frac{x}{6x^2+1}dx = \frac1{12}\int_0^2\frac{(6x^2+1)'}{6x^2+1}dx = \frac1{12}\ln(6x^2+1)\Big|_0^2 = \frac{\ln 25}{12} = \frac{\ln5}{6}. ∫ 0 2 6 x 2 + 1 x d x = 12 1 ∫ 0 2 6 x 2 + 1 ( 6 x 2 + 1 ) ′ d x = 12 1 ln ( 6 x 2 + 1 ) 0 2 = 12 l n 25 = 6 l n 5 .
Barem: 3p + 2p.
c) m m m ∫ 0 1 x + 4 f ( x ) e 2 x d x = m ( e 2 − 1 ) \displaystyle\int_0^1\frac{x+4}{f(x)}e^{2x}dx = m(e^2 - 1) ∫ 0 1 f ( x ) x + 4 e 2 x d x = m ( e 2 − 1 )
x + 4 f ( x ) = 6 x 2 + 1 \dfrac{x + 4}{f(x)} = 6x^2 + 1 f ( x ) x + 4 = 6 x 2 + 1 e 2 x ( 3 x 2 − 3 x + 2 ) e^{2x}(3x^2 - 3x + 2) e 2 x ( 3 x 2 − 3 x + 2 )
( e 2 x ( 3 x 2 − 3 x + 2 ) ) ′ = e 2 x ( 6 x 2 − 6 x + 4 + 6 x − 3 ) = e 2 x ( 6 x 2 + 1 ) . \left(e^{2x}(3x^2 - 3x + 2)\right)' = e^{2x}\left(6x^2 - 6x + 4 + 6x - 3\right) = e^{2x}(6x^2 + 1). ( e 2 x ( 3 x 2 − 3 x + 2 ) ) ′ = e 2 x ( 6 x 2 − 6 x + 4 + 6 x − 3 ) = e 2 x ( 6 x 2 + 1 ) .
∫ 0 1 ( 6 x 2 + 1 ) e 2 x d x = e 2 x ( 3 x 2 − 3 x + 2 ) ∣ 0 1 = 2 e 2 − 2 = 2 ( e 2 − 1 ) . \int_0^1(6x^2 + 1)e^{2x}dx = e^{2x}(3x^2 - 3x + 2)\Big|_0^1 = 2e^2 - 2 = 2(e^2 - 1). ∫ 0 1 ( 6 x 2 + 1 ) e 2 x d x = e 2 x ( 3 x 2 − 3 x + 2 ) 0 1 = 2 e 2 − 2 = 2 ( e 2 − 1 ) .
Răspuns: m = 2 m = 2 m = 2 Barem: 3p + 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 x ∈ { 0 , 2 } x \in \{0, 2\} x ∈ { 0 , 2 } I.3 x = 1 x = 1 x = 1 I.4 p = 1 10 p = \frac1{10} p = 10 1 II.1.c x = − 1 5 x = -\frac15 x = − 5 1 II.2.c m = − 2 m = -2 m = − 2 III.1.b y = 5 y = 5 y = 5 III.2.c m = 2 m = 2 m = 2
Ce merită reținut din varianta 3
Inegalitate cu funcții = minimul funcției: studiază monotonia și compară cu valoarea în punctul de minim.Relația M ( x ) M ( 1 ) = M ( 4 x + 1 ) M(x)M(1) = M(4x+1) M ( x ) M ( 1 ) = M ( 4 x + 1 ) se poate itera fără a mai înmulți matrice.∫ x a x 2 + b d x = 1 2 a ln ( a x 2 + b ) \int\frac{x}{ax^2+b}dx = \frac{1}{2a}\ln(ax^2+b) ∫ a x 2 + b x d x = 2 a 1 ln ( a x 2 + b )