Rezolvarea completă a subiectului de Bacalaureat 2019 la Matematică M_pedagogic , 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. Arătați că 4 5 − 75 + 108 − 3 + 25 − 80 = 5 4\sqrt5 - \sqrt{75} + \sqrt{108} - \sqrt3 + \sqrt{25} - \sqrt{80} = 5 4 5 − 75 + 108 − 3 + 25 − 80 = 5
Scoatem factorii de sub radical: 75 = 5 3 \sqrt{75} = 5\sqrt3 75 = 5 3 108 = 6 3 \sqrt{108} = 6\sqrt3 108 = 6 3 80 = 4 5 \sqrt{80} = 4\sqrt5 80 = 4 5 25 = 5 \sqrt{25} = 5 25 = 5
( 4 5 − 4 5 ) + ( − 5 3 + 6 3 − 3 ) + 5 = 0 + 0 + 5 = 5. ■ (4\sqrt5 - 4\sqrt5) + (-5\sqrt3 + 6\sqrt3 - \sqrt3) + 5 = 0 + 0 + 5 = 5.\ \blacksquare ( 4 5 − 4 5 ) + ( − 5 3 + 6 3 − 3 ) + 5 = 0 + 0 + 5 = 5. ■
Barem: 3p, 2p.
2. f ( x ) = x + a f(x) = x + a f ( x ) = x + a f ( 1 ) = 8 f(1) = 8 f ( 1 ) = 8
1 + a = 8 ⟺ a = 7 1 + a = 8 \iff a = 7 1 + a = 8 ⟺ a = 7 Barem: 2p, 3p.
3. Ecuația 2 x − 1 = x \sqrt{2x - 1} = x 2 x − 1 = x
2 x − 1 = x 2 ⟺ ( x − 1 ) 2 = 0 ⟺ x = 1 2x - 1 = x^2 \iff (x - 1)^2 = 0 \iff x = 1 2 x − 1 = x 2 ⟺ ( x − 1 ) 2 = 0 ⟺ x = 1 1 = 1 \sqrt1 = 1 1 = 1
Răspuns: x = 1 x = 1 x = 1 Barem: 2p, 3p.
4. Câte numere de două cifre distincte se pot forma cu cifrele 1 , … , 9 1, \dots, 9 1 , … , 9
Cifra unităților: 9 moduri; cifra zecilor, diferită de ea: 8 moduri. Total: 9 ⋅ 8 = 72 9\cdot 8 = 72 9 ⋅ 8 = 72 A 9 2 A_9^2 A 9 2
Răspuns: 72 72 72 Barem: 2p, 3p.
5. Intersecția dreptei y = x − 2 y = x - 2 y = x − 2 O x Ox O x
Pe O x Ox O x y = 0 y = 0 y = 0 x − 2 = 0 ⟺ x = 2 x - 2 = 0 \iff x = 2 x − 2 = 0 ⟺ x = 2
Răspuns: ( 2 , 0 ) (2, 0) ( 2 , 0 ) Barem: 2p, 3p.
6. Aria triunghiului cu A B = 6 AB = 6 A B = 6 A C = 8 AC = 8 A C = 8 B C = 10 BC = 10 B C = 10
A B 2 + A C 2 = 36 + 64 = 100 = B C 2 AB^2 + AC^2 = 36 + 64 = 100 = BC^2 A B 2 + A C 2 = 36 + 64 = 100 = B C 2 A A A
A = 6 ⋅ 8 2 = 24. \mathcal A = \frac{6\cdot 8}{2} = 24. A = 2 6 ⋅ 8 = 24.
Barem: 3p, 2p.
Subiectul al II-lea (30 de puncte)
Legea asociativă: x ∗ y = x y + x + y x * y = xy + x + y x ∗ y = x y + x + y
1. ( − 1 ) ∗ 3 = − 1 (-1) * 3 = -1 ( − 1 ) ∗ 3 = − 1
( − 1 ) ⋅ 3 + ( − 1 ) + 3 = − 3 − 1 + 3 = − 1 (-1)\cdot 3 + (-1) + 3 = -3 - 1 + 3 = -1 ( − 1 ) ⋅ 3 + ( − 1 ) + 3 = − 3 − 1 + 3 = − 1 ■ \blacksquare ■ Barem: 3p, 2p.
2. x ∗ y = ( x + 1 ) ( y + 1 ) − 1 x * y = (x+1)(y+1) - 1 x ∗ y = ( x + 1 ) ( y + 1 ) − 1
x y + x + y + 1 − 1 = x ( y + 1 ) + ( y + 1 ) − 1 = ( x + 1 ) ( y + 1 ) − 1 xy + x + y + 1 - 1 = x(y + 1) + (y + 1) - 1 = (x + 1)(y + 1) - 1 x y + x + y + 1 − 1 = x ( y + 1 ) + ( y + 1 ) − 1 = ( x + 1 ) ( y + 1 ) − 1 ■ \blacksquare ■ Barem: 2p, 3p.
3. e = 0 e = 0 e = 0
x ∗ 0 = ( x + 1 ) ⋅ 1 − 1 = x x * 0 = (x + 1)\cdot 1 - 1 = x x ∗ 0 = ( x + 1 ) ⋅ 1 − 1 = x 0 ∗ x = x 0 * x = x 0 ∗ x = x x x x Da , e = 0 e = 0 e = 0 Barem: 3p, 2p.
4. − 1 2 -\frac12 − 2 1 1 1 1
1 ∗ ( − 1 2 ) = 2 ⋅ 1 2 − 1 = 0 = e 1 * \left(-\frac12\right) = 2\cdot\frac12 - 1 = 0 = e 1 ∗ ( − 2 1 ) = 2 ⋅ 2 1 − 1 = 0 = e ( − 1 2 ) ∗ 1 = 0 \left(-\frac12\right) * 1 = 0 ( − 2 1 ) ∗ 1 = 0 Da. Barem: 2p, 3p.
5. x ∗ x ∗ x = x x * x * x = x x ∗ x ∗ x = x
x ∗ x ∗ x = ( x + 1 ) 3 − 1 x * x * x = (x + 1)^3 - 1 x ∗ x ∗ x = ( x + 1 ) 3 − 1 ( x + 1 ) 3 − ( x + 1 ) = 0 ⟺ ( x + 1 ) ( ( x + 1 ) 2 − 1 ) = 0 ⟺ x ( x + 1 ) ( x + 2 ) = 0 (x+1)^3 - (x + 1) = 0 \iff (x+1)\big((x+1)^2 - 1\big) = 0 \iff x(x+1)(x+2) = 0 ( x + 1 ) 3 − ( x + 1 ) = 0 ⟺ ( x + 1 ) ( ( x + 1 ) 2 − 1 ) = 0 ⟺ x ( x + 1 ) ( x + 2 ) = 0
Răspuns: x ∈ { − 2 , − 1 , 0 } x \in \{-2, -1, 0\} x ∈ { − 2 , − 1 , 0 } Barem: 2p, 3p.
6. Probabilitatea ca n n n n ∗ n = 3 n * n = 3 n ∗ n = 3
10 cazuri posibile. ( n + 1 ) 2 − 1 = 3 ⟺ ( n + 1 ) 2 = 4 ⟺ n = 1 (n + 1)^2 - 1 = 3 \iff (n+1)^2 = 4 \iff n = 1 ( n + 1 ) 2 − 1 = 3 ⟺ ( n + 1 ) 2 = 4 ⟺ n = 1
p = 1 10 . p = \frac1{10}. p = 10 1 .
Barem: 2p, 2p, 1p.
Subiectul al III-lea (30 de puncte)
A ( a ) = ( 1 − 2 a 2 a − 3 a 1 + 3 a ) A(a) = \begin{pmatrix}1 - 2a & 2a\\ -3a & 1 + 3a\end{pmatrix} A ( a ) = ( 1 − 2 a − 3 a 2 a 1 + 3 a )
1. det A ( 0 ) = 1 \det A(0) = 1 det A ( 0 ) = 1
A ( 0 ) = I 2 A(0) = I_2 A ( 0 ) = I 2 det A ( 0 ) = 1 \det A(0) = 1 det A ( 0 ) = 1 ■ \blacksquare ■ Barem: 3p, 2p.
2. A ( 1 ) + A ( 5 ) = 2 A ( 3 ) A(1) + A(5) = 2A(3) A ( 1 ) + A ( 5 ) = 2 A ( 3 )
( − 1 2 − 3 4 ) + ( − 9 10 − 15 16 ) = ( − 10 12 − 18 20 ) = 2 ( − 5 6 − 9 10 ) = 2 A ( 3 ) . ■ \begin{pmatrix}-1 & 2\\ -3 & 4\end{pmatrix} + \begin{pmatrix}-9 & 10\\ -15 & 16\end{pmatrix} = \begin{pmatrix}-10 & 12\\ -18 & 20\end{pmatrix} = 2\begin{pmatrix}-5 & 6\\ -9 & 10\end{pmatrix} = 2A(3).\ \blacksquare ( − 1 − 3 2 4 ) + ( − 9 − 15 10 16 ) = ( − 10 − 18 12 20 ) = 2 ( − 5 − 9 6 10 ) = 2 A ( 3 ) . ■
Barem: 2p, 3p.
3. A ( 1 ) ⋅ A ( 2 ) = A ( 5 ) A(1)\cdot A(2) = A(5) A ( 1 ) ⋅ A ( 2 ) = A ( 5 )
( − 1 2 − 3 4 ) ( − 3 4 − 6 7 ) = ( 3 − 12 − 4 + 14 9 − 24 − 12 + 28 ) = ( − 9 10 − 15 16 ) = A ( 5 ) . ■ \begin{pmatrix}-1 & 2\\ -3 & 4\end{pmatrix}\begin{pmatrix}-3 & 4\\ -6 & 7\end{pmatrix} = \begin{pmatrix}3 - 12 & -4 + 14\\ 9 - 24 & -12 + 28\end{pmatrix} = \begin{pmatrix}-9 & 10\\ -15 & 16\end{pmatrix} = A(5).\ \blacksquare ( − 1 − 3 2 4 ) ( − 3 − 6 4 7 ) = ( 3 − 12 9 − 24 − 4 + 14 − 12 + 28 ) = ( − 9 − 15 10 16 ) = A ( 5 ) . ■
Barem: 2p, 3p.
4. Valorile lui a a a A ( a ) A(a) A ( a )
det A ( a ) = ( 1 − 2 a ) ( 1 + 3 a ) + 6 a 2 = 1 + a − 6 a 2 + 6 a 2 = 1 + a . \det A(a) = (1 - 2a)(1 + 3a) + 6a^2 = 1 + a - 6a^2 + 6a^2 = 1 + a. det A ( a ) = ( 1 − 2 a ) ( 1 + 3 a ) + 6 a 2 = 1 + a − 6 a 2 + 6 a 2 = 1 + a .
Inversabilă ⟺ 1 + a ≠ 0 \iff 1 + a \ne 0 ⟺ 1 + a = 0
Răspuns: a ∈ R ∖ { − 1 } a \in \mathbb R\setminus\{-1\} a ∈ R ∖ { − 1 } Barem: 3p, 2p.
5. A ( a ) A ( − 1 ) = A ( − 1 ) A ( a ) = A ( − 1 ) A(a)A(-1) = A(-1)A(a) = A(-1) A ( a ) A ( − 1 ) = A ( − 1 ) A ( a ) = A ( − 1 )
A ( − 1 ) = ( 3 − 2 3 − 2 ) A(-1) = \begin{pmatrix}3 & -2\\ 3 & -2\end{pmatrix} A ( − 1 ) = ( 3 3 − 2 − 2 )
A ( a ) A ( − 1 ) = ( 3 ( 1 − 2 a ) + 6 a − 2 ( 1 − 2 a ) − 4 a − 9 a + 3 ( 1 + 3 a ) 6 a − 2 ( 1 + 3 a ) ) = ( 3 − 2 3 − 2 ) , A(a)A(-1) = \begin{pmatrix}3(1 - 2a) + 6a & -2(1-2a) - 4a\\ -9a + 3(1 + 3a) & 6a - 2(1 + 3a)\end{pmatrix} = \begin{pmatrix}3 & -2\\ 3 & -2\end{pmatrix}, A ( a ) A ( − 1 ) = ( 3 ( 1 − 2 a ) + 6 a − 9 a + 3 ( 1 + 3 a ) − 2 ( 1 − 2 a ) − 4 a 6 a − 2 ( 1 + 3 a ) ) = ( 3 3 − 2 − 2 ) ,
A ( − 1 ) A ( a ) = ( 3 ( 1 − 2 a ) + 6 a 6 a − 2 ( 1 + 3 a ) 3 ( 1 − 2 a ) + 6 a 6 a − 2 ( 1 + 3 a ) ) = ( 3 − 2 3 − 2 ) . ■ A(-1)A(a) = \begin{pmatrix}3(1-2a) + 6a & 6a - 2(1 + 3a)\\ 3(1-2a) + 6a & 6a - 2(1+3a)\end{pmatrix} = \begin{pmatrix}3 & -2\\ 3 & -2\end{pmatrix}.\ \blacksquare A ( − 1 ) A ( a ) = ( 3 ( 1 − 2 a ) + 6 a 3 ( 1 − 2 a ) + 6 a 6 a − 2 ( 1 + 3 a ) 6 a − 2 ( 1 + 3 a ) ) = ( 3 3 − 2 − 2 ) . ■
Barem: 3p, 2p.
6. Numerele naturale nenule n n n det A ( n 4 ) < 32 \det A(n^4) < 32 det A ( n 4 ) < 32
det A ( n 4 ) = 1 + n 4 < 32 ⟺ n 4 < 31 \det A(n^4) = 1 + n^4 < 32 \iff n^4 < 31 det A ( n 4 ) = 1 + n 4 < 32 ⟺ n 4 < 31 1 4 = 1 1^4 = 1 1 4 = 1 2 4 = 16 2^4 = 16 2 4 = 16 3 4 = 81 3^4 = 81 3 4 = 81
Răspuns: n ∈ { 1 , 2 } n \in \{1, 2\} n ∈ { 1 , 2 } Barem: 2p, 3p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 a = 7 a = 7 a = 7 I.3 x = 1 x = 1 x = 1 I.4 72 72 72 I.5 ( 2 , 0 ) (2, 0) ( 2 , 0 ) I.6 24 24 24 II.5 x ∈ { − 2 , − 1 , 0 } x \in \{-2, -1, 0\} x ∈ { − 2 , − 1 , 0 } II.6 p = 1 10 p = \tfrac1{10} p = 10 1 III.4 a ≠ − 1 a \ne -1 a = − 1 III.6 n ∈ { 1 , 2 } n \in \{1, 2\} n ∈ { 1 , 2 }
Ce merită reținut din varianta 7
Legea x y + x + y xy + x + y x y + x + y este ( x + 1 ) ( y + 1 ) − 1 (x+1)(y+1) - 1 ( x + 1 ) ( y + 1 ) − 1 0 0 0 n n n ( x + 1 ) n − 1 (x+1)^n - 1 ( x + 1 ) n − 1 Reciproca teoremei lui Pitagora recunoaște triunghiul dreptunghic din laturi.Numere cu cifre distincte: principiul produsului.