Rezolvarea completă a subiectului de Bacalaureat 2022 la Matematică M_pedagogic , 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 ( 3 + 1 ) 2 − ( 3 − 1 ) 2 = 48 (\sqrt3 + 1)^2 - (\sqrt3 - 1)^2 = \sqrt{48} ( 3 + 1 ) 2 − ( 3 − 1 ) 2 = 48
( 3 + 1 ) 2 = 4 + 2 3 (\sqrt3 + 1)^2 = 4 + 2\sqrt3 ( 3 + 1 ) 2 = 4 + 2 3 ( 3 − 1 ) 2 = 4 − 2 3 (\sqrt3 - 1)^2 = 4 - 2\sqrt3 ( 3 − 1 ) 2 = 4 − 2 3 4 3 = 16 ⋅ 3 = 48 4\sqrt3 = \sqrt{16\cdot 3} = \sqrt{48} 4 3 = 16 ⋅ 3 = 48 ■ \blacksquare ■
Barem: 3p pătratele, 2p finalizarea.
2. Intersecția graficelor lui f ( x ) = 2 x + 1 f(x) = 2x + 1 f ( x ) = 2 x + 1 g ( x ) = − 2 x + 5 g(x) = -2x + 5 g ( x ) = − 2 x + 5
f ( x ) = g ( x ) ⟺ 2 x + 1 = − 2 x + 5 ⟺ x = 1 , f ( 1 ) = 3. f(x) = g(x) \iff 2x + 1 = -2x + 5 \iff x = 1,\qquad f(1) = 3. f ( x ) = g ( x ) ⟺ 2 x + 1 = − 2 x + 5 ⟺ x = 1 , f ( 1 ) = 3.
Răspuns: punctul ( 1 , 3 ) (1, 3) ( 1 , 3 ) Barem: 2p abscisa, 3p punctul.
3. Ecuația 3 x − 2 = ( 1 3 ) − 2 x 3^{x-2} = \left(\frac13\right)^{-2x} 3 x − 2 = ( 3 1 ) − 2 x
( 1 3 ) − 2 x = 3 2 x \left(\frac13\right)^{-2x} = 3^{2x} ( 3 1 ) − 2 x = 3 2 x x − 2 = 2 x ⟺ x = − 2 x - 2 = 2x \iff x = -2 x − 2 = 2 x ⟺ x = − 2
Răspuns: x = − 2 x = -2 x = − 2 Barem: 3p + 2p.
4. Câte numere impare de două cifre se pot forma cu cifrele 1 , 2 , … , 8 1, 2, \dots, 8 1 , 2 , … , 8
Cifra unităților trebuie să fie impară: 1 , 3 , 5 , 7 1, 3, 5, 7 1 , 3 , 5 , 7 4 moduri . Cifra zecilor poate fi oricare dintre cele 8 cifre — 8 moduri . Regula produsului: 4 ⋅ 8 = 32 4\cdot 8 = 32 4 ⋅ 8 = 32
Răspuns: 32 32 32 Barem: 2p + 3p.
5. Distanța de la B ( 5 , 2 ) B(5,2) B ( 5 , 2 ) A C AC A C A ( 0 , 2 ) A(0,2) A ( 0 , 2 ) C ( 6 , 6 ) C(6,6) C ( 6 , 6 )
M ( 0 + 6 2 , 2 + 6 2 ) = M ( 3 , 4 ) M\left(\frac{0+6}{2}, \frac{2+6}{2}\right) = M(3, 4) M ( 2 0 + 6 , 2 2 + 6 ) = M ( 3 , 4 )
B M = ( 3 − 5 ) 2 + ( 4 − 2 ) 2 = 8 = 2 2 . BM = \sqrt{(3-5)^2 + (4-2)^2} = \sqrt{8} = 2\sqrt2. B M = ( 3 − 5 ) 2 + ( 4 − 2 ) 2 = 8 = 2 2 .
Răspuns: 2 2 2\sqrt2 2 2 Barem: 3p mijlocul, 2p distanța.
6. A B = 9 AB = 9 A B = 9 A C = 12 AC = 12 A C = 12 B C = 15 BC = 15 B C = 15 sin B + sin C = 7 5 \sin B + \sin C = \frac75 sin B + sin C = 5 7
A B 2 + A C 2 = 81 + 144 = 225 = B C 2 AB^2 + AC^2 = 81 + 144 = 225 = BC^2 A B 2 + A C 2 = 81 + 144 = 225 = B C 2 A A A
sin B = A C B C = 12 15 = 4 5 , sin C = A B B C = 9 15 = 3 5 , sin B + sin C = 7 5 . \sin B = \frac{AC}{BC} = \frac{12}{15} = \frac45,\quad \sin C = \frac{AB}{BC} = \frac{9}{15} = \frac35,\quad \sin B + \sin C = \frac75. sin B = B C A C = 15 12 = 5 4 , sin C = B C A B = 15 9 = 5 3 , sin B + sin C = 5 7 .
Barem: 3p unghiul drept, 2p finalizarea.
Subiectul al II-lea (30 de puncte)
Legea pe R \mathbb R R x ∗ y = − 1 3 x y + 1 3 ( x + y ) + 2 3 x * y = -\dfrac13xy + \dfrac13(x + y) + \dfrac23 x ∗ y = − 3 1 x y + 3 1 ( x + y ) + 3 2
1. 4 ∗ 2 = 0 4 * 2 = 0 4 ∗ 2 = 0
4 ∗ 2 = − 8 3 + 6 3 + 2 3 = 0. 4 * 2 = -\frac83 + \frac63 + \frac23 = 0. 4 ∗ 2 = − 3 8 + 3 6 + 3 2 = 0.
Barem: 3p + 2p.
2. x ∗ y = − 1 3 ( x − 1 ) ( y − 1 ) + 1 x * y = -\frac13(x - 1)(y - 1) + 1 x ∗ y = − 3 1 ( x − 1 ) ( y − 1 ) + 1
− 1 3 ( x − 1 ) ( y − 1 ) + 1 = − 1 3 ( x y − x − y + 1 ) + 1 = − 1 3 x y + 1 3 ( x + y ) − 1 3 + 1 = x ∗ y . -\tfrac13(x-1)(y-1) + 1 = -\tfrac13(xy - x - y + 1) + 1 = -\tfrac13xy + \tfrac13(x+y) - \tfrac13 + 1 = x * y. − 3 1 ( x − 1 ) ( y − 1 ) + 1 = − 3 1 ( x y − x − y + 1 ) + 1 = − 3 1 x y + 3 1 ( x + y ) − 3 1 + 1 = x ∗ y .
Barem: 3p + 2p.
3. x x x 4 ∗ x = x 4 * x = x 4 ∗ x = x
4 ∗ x = − 1 3 ⋅ 3 ( x − 1 ) + 1 = − x + 2 4 * x = -\tfrac13\cdot 3(x - 1) + 1 = -x + 2 4 ∗ x = − 3 1 ⋅ 3 ( x − 1 ) + 1 = − x + 2 − x + 2 = x ⟺ x = 1 -x + 2 = x \iff x = 1 − x + 2 = x ⟺ x = 1
Răspuns: x = 1 x = 1 x = 1 Barem: 2p + 3p.
4. e = − 2 e = -2 e = − 2
( − 2 ) ∗ x = − 1 3 ( − 3 ) ( x − 1 ) + 1 = x − 1 + 1 = x , (-2) * x = -\tfrac13(-3)(x - 1) + 1 = x - 1 + 1 = x, ( − 2 ) ∗ x = − 3 1 ( − 3 ) ( x − 1 ) + 1 = x − 1 + 1 = x ,
iar x ∗ ( − 2 ) = x x * (-2) = x x ∗ ( − 2 ) = x ■ \blacksquare ■ Barem: 2p + 3p.
5. x x x x ∗ x = − 2 x * x = -2 x ∗ x = − 2
− 1 3 ( x − 1 ) 2 + 1 = − 2 ⟺ ( x − 1 ) 2 = 9 ⟺ x − 1 = ± 3. -\tfrac13(x-1)^2 + 1 = -2 \iff (x - 1)^2 = 9 \iff x - 1 = \pm 3. − 3 1 ( x − 1 ) 2 + 1 = − 2 ⟺ ( x − 1 ) 2 = 9 ⟺ x − 1 = ± 3.
Răspuns: x ∈ { − 2 , 4 } x \in \{-2, 4\} x ∈ { − 2 , 4 } Barem: 2p + 3p.
6. 1 x ∗ 1 x ≤ 1 \frac1x * \frac1x \le 1 x 1 ∗ x 1 ≤ 1 x ≠ 0 x \ne 0 x = 0
1 x ∗ 1 x = − 1 3 ( 1 x − 1 ) 2 + 1 ≤ 1 ⟺ − 1 3 ( 1 x − 1 ) 2 ≤ 0 , \tfrac1x * \tfrac1x = -\tfrac13\left(\tfrac1x - 1\right)^2 + 1 \le 1 \iff -\tfrac13\left(\tfrac1x - 1\right)^2 \le 0, x 1 ∗ x 1 = − 3 1 ( x 1 − 1 ) 2 + 1 ≤ 1 ⟺ − 3 1 ( x 1 − 1 ) 2 ≤ 0 ,
adevărat, deoarece un pătrat este nenegativ. ■ \blacksquare ■ Barem: 3p + 2p.
Subiectul al III-lea (30 de puncte)
A ( x , y ) = ( x y 1 2 ) A(x,y) = \begin{pmatrix}x&y\\1&2\end{pmatrix} A ( x , y ) = ( x 1 y 2 ) B = ( 1 2 − 1 0 ) B = \begin{pmatrix}1&2\\-1&0\end{pmatrix} B = ( 1 − 1 2 0 ) I 2 I_2 I 2
1. det B = 2 \det B = 2 det B = 2
det B = 1 ⋅ 0 − 2 ⋅ ( − 1 ) = 2. \det B = 1\cdot 0 - 2\cdot(-1) = 2. det B = 1 ⋅ 0 − 2 ⋅ ( − 1 ) = 2.
Barem: 3p + 2p.
2. det A ( 2 n , 2 n + 1 ) \det A(2n, 2n+1) det A ( 2 n , 2 n + 1 ) n ≥ 1 n \ge 1 n ≥ 1
det A ( 2 n , 2 n + 1 ) = ∣ 2 n 2 n + 1 1 2 ∣ = 4 n − 2 n − 1 = 2 n − 1 , \det A(2n, 2n+1) = \begin{vmatrix}2n & 2n+1\\1 & 2\end{vmatrix} = 4n - 2n - 1 = 2n - 1, det A ( 2 n , 2 n + 1 ) = 2 n 1 2 n + 1 2 = 4 n − 2 n − 1 = 2 n − 1 ,
care pentru n ≥ 1 n \ge 1 n ≥ 1 Barem: 3p + 2p.
3. A ( 2 x , 0 ) + A ( 0 , 2 x ) = 2 A ( x , x ) A(2x, 0) + A(0, 2x) = 2A(x,x) A ( 2 x , 0 ) + A ( 0 , 2 x ) = 2 A ( x , x )
( 2 x 0 1 2 ) + ( 0 2 x 1 2 ) = ( 2 x 2 x 2 4 ) = 2 ( x x 1 2 ) = 2 A ( x , x ) . \begin{pmatrix}2x&0\\1&2\end{pmatrix} + \begin{pmatrix}0&2x\\1&2\end{pmatrix} = \begin{pmatrix}2x&2x\\2&4\end{pmatrix} = 2\begin{pmatrix}x&x\\1&2\end{pmatrix} = 2A(x,x). ( 2 x 1 0 2 ) + ( 0 1 2 x 2 ) = ( 2 x 2 2 x 4 ) = 2 ( x 1 x 2 ) = 2 A ( x , x ) .
Barem: 3p + 2p.
4. x , y x, y x , y A ( x , y ) ⋅ B = B ⋅ A ( x , y ) A(x,y)\cdot B = B\cdot A(x,y) A ( x , y ) ⋅ B = B ⋅ A ( x , y )
A ( x , y ) B = ( x − y 2 x − 1 2 ) , B A ( x , y ) = ( x + 2 y + 4 − x − y ) . A(x,y)B = \begin{pmatrix}x - y & 2x\\ -1 & 2\end{pmatrix},\qquad BA(x,y) = \begin{pmatrix}x + 2 & y + 4\\ -x & -y\end{pmatrix}. A ( x , y ) B = ( x − y − 1 2 x 2 ) , B A ( x , y ) = ( x + 2 − x y + 4 − y ) .
Egalând: x − y = x + 2 ⇒ y = − 2 x - y = x + 2 \Rightarrow y = -2 x − y = x + 2 ⇒ y = − 2 − 1 = − x ⇒ x = 1 -1 = -x \Rightarrow x = 1 − 1 = − x ⇒ x = 1 2 x = 2 = y + 4 2x = 2 = y + 4 2 x = 2 = y + 4 2 = − y 2 = -y 2 = − y
Răspuns: x = 1 x = 1 x = 1 y = − 2 y = -2 y = − 2 Barem: 3p + 2p.
5. x > 0 x > 0 x > 0 A ( log 3 x , 1 ) A(\log_3 x, 1) A ( log 3 x , 1 )
Suma elementelor: log 3 x + 1 + 1 + 2 = 4 + log 3 x = 5 ⟺ log 3 x = 1 ⟺ x = 3 \log_3 x + 1 + 1 + 2 = 4 + \log_3 x = 5 \iff \log_3 x = 1 \iff x = 3 log 3 x + 1 + 1 + 2 = 4 + log 3 x = 5 ⟺ log 3 x = 1 ⟺ x = 3
Răspuns: x = 3 x = 3 x = 3 Barem: 3p + 2p.
6. x , y x, y x , y A ( x , y ) ⋅ A ( x , y ) = 2 I 2 A(x,y)\cdot A(x,y) = 2I_2 A ( x , y ) ⋅ A ( x , y ) = 2 I 2
A ( x , y ) 2 = ( x 2 + y x y + 2 y x + 2 y + 4 ) = ( 2 0 0 2 ) . A(x,y)^2 = \begin{pmatrix}x^2 + y & xy + 2y\\ x + 2 & y + 4\end{pmatrix} = \begin{pmatrix}2&0\\0&2\end{pmatrix}. A ( x , y ) 2 = ( x 2 + y x + 2 x y + 2 y y + 4 ) = ( 2 0 0 2 ) .
Din x + 2 = 0 x + 2 = 0 x + 2 = 0 y + 4 = 2 y + 4 = 2 y + 4 = 2 x = − 2 x = -2 x = − 2 y = − 2 y = -2 y = − 2 x 2 + y = 2 x^2 + y = 2 x 2 + y = 2 x y + 2 y = 4 − 4 = 0 xy + 2y = 4 - 4 = 0 x y + 2 y = 4 − 4 = 0
Răspuns: x = y = − 2 x = y = -2 x = y = − 2 Barem: 2p + 3p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 ( 1 , 3 ) (1,3) ( 1 , 3 ) I.3 x = − 2 x = -2 x = − 2 I.4 32 32 32 I.5 2 2 2\sqrt2 2 2 II.3 x = 1 x = 1 x = 1 II.5 x ∈ { − 2 , 4 } x \in \{-2, 4\} x ∈ { − 2 , 4 } III.4 x = 1 , y = − 2 x = 1,\ y = -2 x = 1 , y = − 2 III.5 x = 3 x = 3 x = 3 III.6 x = y = − 2 x = y = -2 x = y = − 2
Ce merită reținut din varianta 3
Forma − 1 3 ( x − 1 ) ( y − 1 ) + 1 -\frac13(x-1)(y-1) + 1 − 3 1 ( x − 1 ) ( y − 1 ) + 1 face imediate calculele de tip x ∗ x x * x x ∗ x Reciproca teoremei lui Pitagora identifică triunghiul dreptunghic din lungimile laturilor.Egalitatea de matrice = egalitatea element cu element; alege întâi ecuațiile cele mai simple.