Rezolvarea completă a modelului de subiect pentru Bacalaureat 2021 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 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. log 2 3 \log_23 log 2 3 log 2 6 \log_26 log 2 6 log 2 12 \log_212 log 2 12
Trebuie ca termenul din mijloc să fie media celorlalți:
log 2 3 + log 2 12 = log 2 36 = log 2 6 2 = 2 log 2 6. ■ \log_23 + \log_212 = \log_236 = \log_26^2 = 2\log_26.\ \blacksquare log 2 3 + log 2 12 = log 2 36 = log 2 6 2 = 2 log 2 6. ■
Barem: 3p + 2p.
2. f ( x ) = x 2 − x + 1 f(x) = x^2 - x + 1 f ( x ) = x 2 − x + 1 x x x f ( x ) = x f(x) = x f ( x ) = x
x 2 − x + 1 = x ⟺ x 2 − 2 x + 1 = 0 ⟺ x = 1. x^2 - x + 1 = x \iff x^2 - 2x + 1 = 0 \iff x = 1. x 2 − x + 1 = x ⟺ x 2 − 2 x + 1 = 0 ⟺ x = 1.
Barem: 3p + 2p.
3. Ecuația ∣ 2 x − 1 ∣ = 2 x + 1 |2x - 1| = 2x + 1 ∣2 x − 1∣ = 2 x + 1
Necesar 2 x + 1 ≥ 0 2x + 1 \ge 0 2 x + 1 ≥ 0 2 x − 1 = 2 x + 1 2x - 1 = 2x + 1 2 x − 1 = 2 x + 1 2 x − 1 = − ( 2 x + 1 ) ⟺ 4 x = 0 ⟺ x = 0 2x - 1 = -(2x + 1) \iff 4x = 0 \iff x = 0 2 x − 1 = − ( 2 x + 1 ) ⟺ 4 x = 0 ⟺ x = 0 ∣ − 1 ∣ = 1 |-1| = 1 ∣ − 1∣ = 1
Răspuns: x = 0 x = 0 x = 0 Barem: 3p + 2p.
4. Probabilitatea ca un număr de două cifre să aibă cifra zecilor pară și a unităților impară
90 cazuri posibile. Zecile: 2 , 4 , 6 , 8 2, 4, 6, 8 2 , 4 , 6 , 8 1 , 3 , 5 , 7 , 9 1, 3, 5, 7, 9 1 , 3 , 5 , 7 , 9 20 20 20
p = 20 90 = 2 9 . p = \frac{20}{90} = \frac29. p = 90 20 = 9 2 .
Barem: 2p + 2p + 1p.
5. a a a u ⃗ = a i ⃗ + j ⃗ \vec u = a\vec i + \vec j u = a i + j v ⃗ = 8 i ⃗ + 2 j ⃗ \vec v = 8\vec i + 2\vec j v = 8 i + 2 j
Coliniari ⟺ a 8 = 1 2 ⟺ a = 4 \iff \frac a8 = \frac12 \iff a = 4 ⟺ 8 a = 2 1 ⟺ a = 4
Barem: 3p + 2p.
6. △ A B C \triangle ABC △ A B C A A A B C = 12 BC = 12 B C = 12 B = π 6 B = \frac\pi6 B = 6 π = 18 3 = 18\sqrt3 = 18 3
cos B = A B B C ⇒ A B = 12 ⋅ 3 2 = 6 3 \cos B = \frac{AB}{BC} \Rightarrow AB = 12\cdot\frac{\sqrt3}2 = 6\sqrt3 cos B = B C A B ⇒ A B = 12 ⋅ 2 3 = 6 3
A = 1 2 A B ⋅ B C ⋅ sin B = 1 2 ⋅ 6 3 ⋅ 12 ⋅ 1 2 = 18 3 . \mathcal A = \frac12AB\cdot BC\cdot\sin B = \frac12\cdot6\sqrt3\cdot12\cdot\frac12 = 18\sqrt3. A = 2 1 A B ⋅ B C ⋅ sin B = 2 1 ⋅ 6 3 ⋅ 12 ⋅ 2 1 = 18 3 .
Barem: 3p + 2p.
Subiectul al II-lea (30 de puncte)
1. A ( a ) = ( 2 a − 1 4 a − 4 1 − a 3 − 2 a ) A(a) = \begin{pmatrix}2a - 1 & 4a - 4\\ 1 - a & 3 - 2a\end{pmatrix} A ( a ) = ( 2 a − 1 1 − a 4 a − 4 3 − 2 a )
a) det A ( 2 ) = 1 \det A(2) = 1 det A ( 2 ) = 1
A ( 2 ) = ( 3 4 − 1 − 1 ) , det = − 3 + 4 = 1. A(2) = \begin{pmatrix}3&4\\-1&-1\end{pmatrix},\quad \det = -3 + 4 = 1. A ( 2 ) = ( 3 − 1 4 − 1 ) , det = − 3 + 4 = 1.
Barem: 3p + 2p.
b) A ( a ) A ( b ) = A ( a + b − 1 ) A(a)A(b) = A(a + b - 1) A ( a ) A ( b ) = A ( a + b − 1 )
A ( a ) A ( b ) = ( ( 2 a − 1 ) ( 2 b − 1 ) + ( 4 a − 4 ) ( 1 − b ) ( 2 a − 1 ) ( 4 b − 4 ) + ( 4 a − 4 ) ( 3 − 2 b ) ( 1 − a ) ( 2 b − 1 ) + ( 3 − 2 a ) ( 1 − b ) ( 1 − a ) ( 4 b − 4 ) + ( 3 − 2 a ) ( 3 − 2 b ) ) = ( 2 a + 2 b − 3 4 a + 4 b − 8 2 − a − b 5 − 2 a − 2 b ) . A(a)A(b) = \begin{pmatrix}(2a-1)(2b-1) + (4a-4)(1-b) & (2a-1)(4b-4) + (4a-4)(3-2b)\\ (1-a)(2b-1) + (3-2a)(1-b) & (1-a)(4b-4) + (3-2a)(3-2b)\end{pmatrix} = \begin{pmatrix}2a + 2b - 3 & 4a + 4b - 8\\ 2 - a - b & 5 - 2a - 2b\end{pmatrix}. A ( a ) A ( b ) = ( ( 2 a − 1 ) ( 2 b − 1 ) + ( 4 a − 4 ) ( 1 − b ) ( 1 − a ) ( 2 b − 1 ) + ( 3 − 2 a ) ( 1 − b ) ( 2 a − 1 ) ( 4 b − 4 ) + ( 4 a − 4 ) ( 3 − 2 b ) ( 1 − a ) ( 4 b − 4 ) + ( 3 − 2 a ) ( 3 − 2 b ) ) = ( 2 a + 2 b − 3 2 − a − b 4 a + 4 b − 8 5 − 2 a − 2 b ) .
Cu t = a + b − 1 t = a + b - 1 t = a + b − 1 2 t − 1 = 2 a + 2 b − 3 2t - 1 = 2a + 2b - 3 2 t − 1 = 2 a + 2 b − 3 4 t − 4 = 4 a + 4 b − 8 4t - 4 = 4a + 4b - 8 4 t − 4 = 4 a + 4 b − 8 1 − t = 2 − a − b 1 - t = 2 - a - b 1 − t = 2 − a − b 3 − 2 t = 5 − 2 a − 2 b 3 - 2t = 5 - 2a - 2b 3 − 2 t = 5 − 2 a − 2 b A ( a + b − 1 ) A(a + b - 1) A ( a + b − 1 ) ■ \blacksquare ■ Barem: 3p + 2p.
c) n n n A ( 1 ) A ( 2 ) A ( 2 2 ) A ( 2 3 ) A ( 2 4 ) = A ( 32 ) A ( − n ) A(1)A(2)A(2^2)A(2^3)A(2^4) = A(32)A(-n) A ( 1 ) A ( 2 ) A ( 2 2 ) A ( 2 3 ) A ( 2 4 ) = A ( 32 ) A ( − n )
Aplicând repetat b), parametrii se adună și se scade câte 1 la fiecare înmulțire (4 înmulțiri):
A ( 1 + 2 + 4 + 8 + 16 − 4 ) = A ( 27 ) , A ( 32 ) A ( − n ) = A ( 31 − n ) . A(1 + 2 + 4 + 8 + 16 - 4) = A(27),\qquad A(32)A(-n) = A(31 - n). A ( 1 + 2 + 4 + 8 + 16 − 4 ) = A ( 27 ) , A ( 32 ) A ( − n ) = A ( 31 − n ) .
Cum A ( a ) = A ( b ) ⇒ 2 a − 1 = 2 b − 1 ⇒ a = b A(a) = A(b) \Rightarrow 2a - 1 = 2b - 1 \Rightarrow a = b A ( a ) = A ( b ) ⇒ 2 a − 1 = 2 b − 1 ⇒ a = b 27 = 31 − n ⟺ n = 4 27 = 31 - n \iff n = 4 27 = 31 − n ⟺ n = 4
Barem: 3p + 2p.
2. Legea x ∘ y = 3 x 2 − 5 x y + 2 y 2 x\circ y = 3x^2 - 5xy + 2y^2 x ∘ y = 3 x 2 − 5 x y + 2 y 2
a) 1 ∘ 2 = 1 1\circ2 = 1 1 ∘ 2 = 1
1 ∘ 2 = 3 − 10 + 8 = 1. 1\circ2 = 3 - 10 + 8 = 1. 1 ∘ 2 = 3 − 10 + 8 = 1.
Barem: 3p + 2p.
b) x ∘ x = 0 x\circ x = 0 x ∘ x = 0
x ∘ x = 3 x 2 − 5 x 2 + 2 x 2 = 0. x\circ x = 3x^2 - 5x^2 + 2x^2 = 0. x ∘ x = 3 x 2 − 5 x 2 + 2 x 2 = 0.
Barem: 3p + 2p.
c) x x x 2 x ∘ 3 x = 0 2^x\circ3^x = 0 2 x ∘ 3 x = 0
Legea se factorizează: x ∘ y = ( x − y ) ( 3 x − 2 y ) x\circ y = (x - y)(3x - 2y) x ∘ y = ( x − y ) ( 3 x − 2 y )
2 x ∘ 3 x = ( 2 x − 3 x ) ( 3 ⋅ 2 x − 2 ⋅ 3 x ) = 0. 2^x\circ3^x = (2^x - 3^x)(3\cdot2^x - 2\cdot3^x) = 0. 2 x ∘ 3 x = ( 2 x − 3 x ) ( 3 ⋅ 2 x − 2 ⋅ 3 x ) = 0.
2 x = 3 x ⟺ ( 2 3 ) x = 1 ⟺ x = 0 2^x = 3^x \iff \left(\frac23\right)^x = 1 \iff x = 0 2 x = 3 x ⟺ ( 3 2 ) x = 1 ⟺ x = 0 3 ⋅ 2 x = 2 ⋅ 3 x ⟺ 2 x − 1 = 3 x − 1 ⟺ x = 1 3\cdot2^x = 2\cdot3^x \iff 2^{x-1} = 3^{x-1} \iff x = 1 3 ⋅ 2 x = 2 ⋅ 3 x ⟺ 2 x − 1 = 3 x − 1 ⟺ x = 1
Răspuns: x ∈ { 0 , 1 } x \in \{0, 1\} x ∈ { 0 , 1 } Barem: 3p + 2p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x 2 + 2 x + 2 − x f(x) = \sqrt{x^2 + 2x + 2} - x f ( x ) = x 2 + 2 x + 2 − x
a) f ′ ( x ) = x + 1 x 2 + 2 x + 2 − 1 f'(x) = \dfrac{x+1}{\sqrt{x^2+2x+2}} - 1 f ′ ( x ) = x 2 + 2 x + 2 x + 1 − 1
f ′ ( x ) = 2 x + 2 2 x 2 + 2 x + 2 − 1 = x + 1 x 2 + 2 x + 2 − 1. f'(x) = \frac{2x + 2}{2\sqrt{x^2+2x+2}} - 1 = \frac{x + 1}{\sqrt{x^2+2x+2}} - 1. f ′ ( x ) = 2 x 2 + 2 x + 2 2 x + 2 − 1 = x 2 + 2 x + 2 x + 1 − 1.
Barem: 3p + 2p.
b) Tangenta în punctul de abscisă x = − 1 x = -1 x = − 1
f ( − 1 ) = 1 + 1 = 2 f(-1) = \sqrt1 + 1 = 2 f ( − 1 ) = 1 + 1 = 2 f ′ ( − 1 ) = 0 − 1 = − 1 f'(-1) = 0 - 1 = -1 f ′ ( − 1 ) = 0 − 1 = − 1
y − 2 = − ( x + 1 ) ⟺ y = − x + 1. y - 2 = -(x + 1) \iff y = -x + 1. y − 2 = − ( x + 1 ) ⟺ y = − x + 1.
Barem: 2p + 3p.
c) ( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) + 2 − 4 x 2 + 4 x + 2 ≤ ( x − 1 ) 2 \sqrt{(x^2+1)^2 + 2(x^2+1) + 2} - \sqrt{4x^2 + 4x + 2} \le (x - 1)^2 ( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) + 2 − 4 x 2 + 4 x + 2 ≤ ( x − 1 ) 2
Cum ( x + 1 ) 2 + 1 > ∣ x + 1 ∣ ≥ x + 1 \sqrt{(x+1)^2 + 1} > |x + 1| \ge x + 1 ( x + 1 ) 2 + 1 > ∣ x + 1∣ ≥ x + 1 f ′ ( x ) < 0 f'(x) < 0 f ′ ( x ) < 0 f f f R \mathbb R R x 2 + 1 ≥ 2 x x^2 + 1 \ge 2x x 2 + 1 ≥ 2 x f ( x 2 + 1 ) ≤ f ( 2 x ) f(x^2 + 1) \le f(2x) f ( x 2 + 1 ) ≤ f ( 2 x )
( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) + 2 − ( x 2 + 1 ) ≤ 4 x 2 + 4 x + 2 − 2 x , \sqrt{(x^2+1)^2 + 2(x^2+1) + 2} - (x^2 + 1) \le \sqrt{4x^2 + 4x + 2} - 2x, ( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) + 2 − ( x 2 + 1 ) ≤ 4 x 2 + 4 x + 2 − 2 x ,
adică ( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) + 2 − 4 x 2 + 4 x + 2 ≤ x 2 − 2 x + 1 = ( x − 1 ) 2 \sqrt{(x^2+1)^2 + 2(x^2+1) + 2} - \sqrt{4x^2+4x+2} \le x^2 - 2x + 1 = (x - 1)^2 ( x 2 + 1 ) 2 + 2 ( x 2 + 1 ) + 2 − 4 x 2 + 4 x + 2 ≤ x 2 − 2 x + 1 = ( x − 1 ) 2 ■ \blacksquare ■ Barem: 3p + 2p.
2. f ( x ) = 2 x x 2 + 1 f(x) = \dfrac{2x}{\sqrt{x^2+1}} f ( x ) = x 2 + 1 2 x
a) ∫ 0 1 ( 4 − f 2 ( x ) ) d x = π \displaystyle\int_0^1\left(4 - f^2(x)\right)dx = \pi ∫ 0 1 ( 4 − f 2 ( x ) ) d x = π
4 − f 2 ( x ) = 4 − 4 x 2 x 2 + 1 = 4 x 2 + 1 4 - f^2(x) = 4 - \dfrac{4x^2}{x^2+1} = \dfrac{4}{x^2+1} 4 − f 2 ( x ) = 4 − x 2 + 1 4 x 2 = x 2 + 1 4
∫ 0 1 4 x 2 + 1 d x = 4 arctg x ∣ 0 1 = 4 ⋅ π 4 = π . \int_0^1\frac4{x^2+1}dx = 4\operatorname{arctg}x\Big|_0^1 = 4\cdot\frac\pi4 = \pi. ∫ 0 1 x 2 + 1 4 d x = 4 arctg x 0 1 = 4 ⋅ 4 π = π .
Barem: 3p + 2p.
b) Aria suprafeței dintre G f G_f G f O x Ox O x x = 0 x = 0 x = 0 x = 1 x = 1 x = 1 2 ( 2 − 1 ) 2(\sqrt2 - 1) 2 ( 2 − 1 )
f ( x ) ≥ 0 f(x) \ge 0 f ( x ) ≥ 0 [ 0 , 1 ] [0,1] [ 0 , 1 ]
A = ∫ 0 1 2 x x 2 + 1 d x = 2 x 2 + 1 ∣ 0 1 = 2 2 − 2 = 2 ( 2 − 1 ) . \mathcal A = \int_0^1\frac{2x}{\sqrt{x^2+1}}dx = 2\sqrt{x^2+1}\Big|_0^1 = 2\sqrt2 - 2 = 2(\sqrt2 - 1). A = ∫ 0 1 x 2 + 1 2 x d x = 2 x 2 + 1 0 1 = 2 2 − 2 = 2 ( 2 − 1 ) .
Barem: 3p + 2p.
c) ∫ 1 2 f ( x 2 ) x d x = ln ( 34 − 17 + 4 2 − 4 ) \displaystyle\int_1^2\frac{f(x^2)}{x}dx = \ln\left(\sqrt{34} - \sqrt{17} + 4\sqrt2 - 4\right) ∫ 1 2 x f ( x 2 ) d x = ln ( 34 − 17 + 4 2 − 4 )
f ( x 2 ) x = 2 x x 4 + 1 \dfrac{f(x^2)}{x} = \dfrac{2x}{\sqrt{x^4+1}} x f ( x 2 ) = x 4 + 1 2 x t = x 2 t = x^2 t = x 2 d t = 2 x d x dt = 2x\,dx d t = 2 x d x
∫ 1 4 d t t 2 + 1 = ln ( t + t 2 + 1 ) ∣ 1 4 = ln 4 + 17 1 + 2 . \int_1^4\frac{dt}{\sqrt{t^2+1}} = \ln\left(t + \sqrt{t^2+1}\right)\Big|_1^4 = \ln\frac{4 + \sqrt{17}}{1 + \sqrt2}. ∫ 1 4 t 2 + 1 d t = ln ( t + t 2 + 1 ) 1 4 = ln 1 + 2 4 + 17 .
Amplificăm cu 2 − 1 \sqrt2 - 1 2 − 1 4 + 17 1 + 2 = ( 4 + 17 ) ( 2 − 1 ) = 4 2 − 4 + 34 − 17 \dfrac{4 + \sqrt{17}}{1 + \sqrt2} = (4 + \sqrt{17})(\sqrt2 - 1) = 4\sqrt2 - 4 + \sqrt{34} - \sqrt{17} 1 + 2 4 + 17 = ( 4 + 17 ) ( 2 − 1 ) = 4 2 − 4 + 34 − 17 ■ \blacksquare ■
Barem: 3p + 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 x = 1 x = 1 x = 1 I.3 x = 0 x = 0 x = 0 I.4 p = 2 9 p = \frac29 p = 9 2 I.5 a = 4 a = 4 a = 4 II.1.c n = 4 n = 4 n = 4 II.2.c x ∈ { 0 , 1 } x \in \{0,1\} x ∈ { 0 , 1 } III.1.b y = − x + 1 y = -x + 1 y = − x + 1
Ce merită reținut din model
Lege de forma A ( a ) A ( b ) = A ( a + b − 1 ) A(a)A(b) = A(a + b - 1) A ( a ) A ( b ) = A ( a + b − 1 ) la k k k ( k − 1 ) (k - 1) ( k − 1 ) Formă pătratică omogenă 3 x 2 − 5 x y + 2 y 2 = ( x − y ) ( 3 x − 2 y ) 3x^2 - 5xy + 2y^2 = (x - y)(3x - 2y) 3 x 2 − 5 x y + 2 y 2 = ( x − y ) ( 3 x − 2 y ) ∫ d t t 2 + 1 = ln ( t + t 2 + 1 ) \int\frac{dt}{\sqrt{t^2+1}} = \ln\left(t + \sqrt{t^2+1}\right) ∫ t 2 + 1 d t = ln ( t + t 2 + 1 )