Rezolvarea completă a simulării pentru clasa a XII-a, Bacalaureat 2017, la Matematică M_pedagogic (filiera vocațională, profilul pedagogic). 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. Rația progresiei aritmetice cu a 1 + a 2 + a 3 + a 4 = 14 a_1 + a_2 + a_3 + a_4 = 14 a 1 + a 2 + a 3 + a 4 = 14 a 1 = 2 a_1 = 2 a 1 = 2
a 1 + ( a 1 + r ) + ( a 1 + 2 r ) + ( a 1 + 3 r ) = 4 a 1 + 6 r = 14 ⟺ 8 + 6 r = 14 ⟺ r = 1. a_1 + (a_1 + r) + (a_1 + 2r) + (a_1 + 3r) = 4a_1 + 6r = 14 \iff 8 + 6r = 14 \iff r = 1. a 1 + ( a 1 + r ) + ( a 1 + 2 r ) + ( a 1 + 3 r ) = 4 a 1 + 6 r = 14 ⟺ 8 + 6 r = 14 ⟺ r = 1.
Răspuns: r = 1 r = 1 r = 1 Barem: 3p, 2p.
2. f ( x ) = x 2 − 5 x + 4 f(x) = x^2 - 5x + 4 f ( x ) = x 2 − 5 x + 4 O x Ox O x
f ( x ) = 0 ⟺ x 2 − 5 x + 4 = 0 ⟺ ( x − 1 ) ( x − 4 ) = 0 f(x) = 0 \iff x^2 - 5x + 4 = 0 \iff (x - 1)(x - 4) = 0 f ( x ) = 0 ⟺ x 2 − 5 x + 4 = 0 ⟺ ( x − 1 ) ( x − 4 ) = 0 ( 1 , 0 ) (1, 0) ( 1 , 0 ) ( 4 , 0 ) (4, 0) ( 4 , 0 ) 4 − 1 = 3 4 - 1 = 3 4 − 1 = 3
Răspuns: 3 3 3 Barem: 2p, 3p.
3. Ecuația 2 x + 2 + 2 x + 1 + 2 x = 7 2^{x+2} + 2^{x+1} + 2^x = 7 2 x + 2 + 2 x + 1 + 2 x = 7
Dăm factor comun 2 x 2^x 2 x
2 x ( 2 2 + 2 1 + 1 ) = 7 ⟺ 7 ⋅ 2 x = 7 ⟺ 2 x = 1 ⟺ x = 0. 2^x(2^2 + 2^1 + 1) = 7 \iff 7\cdot 2^x = 7 \iff 2^x = 1 \iff x = 0. 2 x ( 2 2 + 2 1 + 1 ) = 7 ⟺ 7 ⋅ 2 x = 7 ⟺ 2 x = 1 ⟺ x = 0.
Răspuns: x = 0 x = 0 x = 0 Barem: 3p, 2p.
4. După două creșteri succesive cu câte 10 % 10\% 10%
După prima scumpire prețul devine p + 10 100 p = 1,1 p p + \frac{10}{100}p = 1{,}1p p + 100 10 p = 1 , 1 p 1,1 ⋅ 1,1 p = 1,21 p 1{,}1\cdot 1{,}1p = 1{,}21p 1 , 1 ⋅ 1 , 1 p = 1 , 21 p
1,21 p = 242 ⟺ p = 200. 1{,}21p = 242 \iff p = 200. 1 , 21 p = 242 ⟺ p = 200.
Răspuns: 200 200 200 Barem: 3p, 2p.
5. v ⃗ 1 = m i ⃗ + 6 j ⃗ \vec v_1 = m\vec i + 6\vec j v 1 = m i + 6 j v ⃗ 2 = 2 i ⃗ + 3 j ⃗ \vec v_2 = 2\vec i + 3\vec j v 2 = 2 i + 3 j
m 2 = 6 3 ⟺ m = 4. \frac m2 = \frac63 \iff m = 4. 2 m = 3 6 ⟺ m = 4.
Răspuns: m = 4 m = 4 m = 4 Barem: 3p, 2p.
6. Aria dreptunghiului A B C D ABCD A B C D A B = 3 AB = 3 A B = 3 A C = 5 AC = 5 A C = 5
În triunghiul dreptunghic A B C ABC A B C B B B B C = A C 2 − A B 2 = 25 − 9 = 4 BC = \sqrt{AC^2 - AB^2} = \sqrt{25 - 9} = 4 B C = A C 2 − A B 2 = 25 − 9 = 4
A A B C D = A B ⋅ B C = 3 ⋅ 4 = 12. \mathcal A_{ABCD} = AB\cdot BC = 3\cdot 4 = 12. A A B C D = A B ⋅ B C = 3 ⋅ 4 = 12.
Răspuns: 12 12 12 Barem: 3p, 2p.
Subiectul al II-lea (30 de puncte)
Legea asociativă: x ∗ y = x + y − x y x * y = x + y - xy x ∗ y = x + y − x y
1. ( − 1 ) ∗ 1 (-1) * 1 ( − 1 ) ∗ 1
( − 1 ) ∗ 1 = − 1 + 1 − ( − 1 ) ⋅ 1 = 0 + 1 = 1. (-1) * 1 = -1 + 1 - (-1)\cdot 1 = 0 + 1 = 1. ( − 1 ) ∗ 1 = − 1 + 1 − ( − 1 ) ⋅ 1 = 0 + 1 = 1.
Barem: 3p, 2p.
2. Legea este comutativă?
x ∗ y = x + y − x y = y + x − y x = y ∗ x x * y = x + y - xy = y + x - yx = y * x x ∗ y = x + y − x y = y + x − y x = y ∗ x x , y x, y x , y legea este comutativă . Barem: 3p, 2p.
3. x ∗ y = − ( x − 1 ) ( y − 1 ) + 1 x * y = -(x-1)(y-1) + 1 x ∗ y = − ( x − 1 ) ( y − 1 ) + 1
− ( x − 1 ) ( y − 1 ) + 1 = − ( x y − x − y + 1 ) + 1 = − x y + x + y = x ∗ y . -(x-1)(y-1) + 1 = -(xy - x - y + 1) + 1 = -xy + x + y = x * y. − ( x − 1 ) ( y − 1 ) + 1 = − ( x y − x − y + 1 ) + 1 = − x y + x + y = x ∗ y .
Barem: 2p, 3p.
4. x ∗ x = 0 x * x = 0 x ∗ x = 0
− ( x − 1 ) 2 + 1 = 0 ⟺ ( x − 1 ) 2 = 1 ⟺ x − 1 = ± 1 ⟺ x ∈ { 0 , 2 } . -(x-1)^2 + 1 = 0 \iff (x-1)^2 = 1 \iff x - 1 = \pm 1 \iff x \in \{0, 2\}. − ( x − 1 ) 2 + 1 = 0 ⟺ ( x − 1 ) 2 = 1 ⟺ x − 1 = ± 1 ⟺ x ∈ { 0 , 2 } .
Răspuns: x ∈ { 0 , 2 } x \in \{0, 2\} x ∈ { 0 , 2 } Barem: 3p, 2p.
5. a ∗ a ≥ 1 a * a \ge 1 a ∗ a ≥ 1
− ( a − 1 ) 2 + 1 ≥ 1 ⟺ ( a − 1 ) 2 ≤ 0 ⟺ a = 1. -(a-1)^2 + 1 \ge 1 \iff (a-1)^2 \le 0 \iff a = 1. − ( a − 1 ) 2 + 1 ≥ 1 ⟺ ( a − 1 ) 2 ≤ 0 ⟺ a = 1.
Răspuns: a = 1 a = 1 a = 1 Barem: 3p, 2p.
6. 1 2016 ∗ 2 2016 ∗ ⋯ ∗ 2017 2016 \dfrac{1}{2016} * \dfrac{2}{2016} * \dots * \dfrac{2017}{2016} 2016 1 ∗ 2016 2 ∗ ⋯ ∗ 2016 2017
Din 3): x ∗ 1 = − ( x − 1 ) ⋅ 0 + 1 = 1 x * 1 = -(x-1)\cdot 0 + 1 = 1 x ∗ 1 = − ( x − 1 ) ⋅ 0 + 1 = 1 1 ∗ y = 1 1 * y = 1 1 ∗ y = 1 1 1 1
Printre factori se află 2016 2016 = 1 \frac{2016}{2016} = 1 2016 2016 = 1
( 1 2016 ∗ ⋯ ∗ 2015 2016 ) ∗ 1 ∗ ( 2017 2016 ) = 1 ∗ 2017 2016 = 1. \left(\frac{1}{2016} * \dots * \frac{2015}{2016}\right) * 1 * \left(\frac{2017}{2016}\right) = 1 * \frac{2017}{2016} = 1. ( 2016 1 ∗ ⋯ ∗ 2016 2015 ) ∗ 1 ∗ ( 2016 2017 ) = 1 ∗ 2016 2017 = 1.
Răspuns: 1 1 1 Barem: 2p, 3p.
Subiectul al III-lea (30 de puncte)
Matricea A ( n ) = ( 1 n 0 1 ) A(n) = \begin{pmatrix} 1 & n\\ 0 & 1\end{pmatrix} A ( n ) = ( 1 0 n 1 ) n ∈ Z n \in \mathbb Z n ∈ Z
1. det ( A ( 2017 ) ) \det(A(2017)) det ( A ( 2017 ))
det A ( 2017 ) = ∣ 1 2017 0 1 ∣ = 1 ⋅ 1 − 2017 ⋅ 0 = 1. \det A(2017) = \begin{vmatrix} 1 & 2017\\ 0 & 1\end{vmatrix} = 1\cdot 1 - 2017\cdot 0 = 1. det A ( 2017 ) = 1 0 2017 1 = 1 ⋅ 1 − 2017 ⋅ 0 = 1.
Barem: 2p, 3p.
2. A ( − 2017 ) + A ( 2017 ) = 2 I 2 A(-2017) + A(2017) = 2I_2 A ( − 2017 ) + A ( 2017 ) = 2 I 2
( 1 − 2017 0 1 ) + ( 1 2017 0 1 ) = ( 2 0 0 2 ) = 2 I 2 . \begin{pmatrix} 1 & -2017\\ 0 & 1\end{pmatrix} + \begin{pmatrix} 1 & 2017\\ 0 & 1\end{pmatrix} = \begin{pmatrix} 2 & 0\\ 0 & 2\end{pmatrix} = 2I_2. ( 1 0 − 2017 1 ) + ( 1 0 2017 1 ) = ( 2 0 0 2 ) = 2 I 2 .
Barem: 3p, 2p.
3. A ( m ) ⋅ A ( n ) = A ( m + n ) A(m)\cdot A(n) = A(m + n) A ( m ) ⋅ A ( n ) = A ( m + n )
( 1 m 0 1 ) ( 1 n 0 1 ) = ( 1 n + m 0 1 ) = A ( m + n ) . \begin{pmatrix} 1 & m\\ 0 & 1\end{pmatrix}\begin{pmatrix} 1 & n\\ 0 & 1\end{pmatrix} = \begin{pmatrix} 1 & n + m\\ 0 & 1\end{pmatrix} = A(m + n). ( 1 0 m 1 ) ( 1 0 n 1 ) = ( 1 0 n + m 1 ) = A ( m + n ) .
Barem: 3p, 2p.
4. Suma elementelor matricei B = A ( 0 ) + A ( 1 ) + ⋯ + A ( 6 ) B = A(0) + A(1) + \dots + A(6) B = A ( 0 ) + A ( 1 ) + ⋯ + A ( 6 )
B = ( 7 0 + 1 + ⋯ + 6 0 7 ) = ( 7 21 0 7 ) . B = \begin{pmatrix} 7 & 0 + 1 + \dots + 6\\ 0 & 7\end{pmatrix} = \begin{pmatrix} 7 & 21\\ 0 & 7\end{pmatrix}. B = ( 7 0 0 + 1 + ⋯ + 6 7 ) = ( 7 0 21 7 ) .
Suma elementelor: 7 + 21 + 0 + 7 = 35 = 7 ⋅ 5 7 + 21 + 0 + 7 = 35 = 7\cdot 5 7 + 21 + 0 + 7 = 35 = 7 ⋅ 5 Barem: 3p, 2p.
5. A ( n ) A(n) A ( n )
det A ( n ) = 1 ≠ 0 \det A(n) = 1 \ne 0 det A ( n ) = 1 = 0 n n n A ( n ) A(n) A ( n ) A ( n ) ⋅ A ( − n ) = A ( 0 ) = I 2 A(n)\cdot A(-n) = A(0) = I_2 A ( n ) ⋅ A ( − n ) = A ( 0 ) = I 2 ( A ( n ) ) − 1 = A ( − n ) (A(n))^{-1} = A(-n) ( A ( n ) ) − 1 = A ( − n ) Barem: 2p, 3p.
6. X ∈ M 2 ( Z ) X \in \mathcal M_2(\mathbb Z) X ∈ M 2 ( Z ) A ( 2017 ) ⋅ X = A ( 2018 ) A(2017)\cdot X = A(2018) A ( 2017 ) ⋅ X = A ( 2018 )
Înmulțim la stânga cu ( A ( 2017 ) ) − 1 = A ( − 2017 ) (A(2017))^{-1} = A(-2017) ( A ( 2017 ) ) − 1 = A ( − 2017 )
X = A ( − 2017 ) ⋅ A ( 2018 ) = A ( − 2017 + 2018 ) = A ( 1 ) = ( 1 1 0 1 ) . X = A(-2017)\cdot A(2018) = A(-2017 + 2018) = A(1) = \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}. X = A ( − 2017 ) ⋅ A ( 2018 ) = A ( − 2017 + 2018 ) = A ( 1 ) = ( 1 0 1 1 ) .
Răspuns: X = ( 1 1 0 1 ) X = \begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix} X = ( 1 0 1 1 ) Barem: 2p, 3p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 r = 1 r = 1 r = 1 I.2 3 3 3 I.3 x = 0 x = 0 x = 0 I.4 200 200 200 I.5 m = 4 m = 4 m = 4 I.6 12 12 12 II.1 1 1 1 II.4 x ∈ { 0 , 2 } x \in \{0, 2\} x ∈ { 0 , 2 } II.5 a = 1 a = 1 a = 1 II.6 1 1 1 III.1 1 1 1 III.6 X = ( 1 1 0 1 ) X = \begin{pmatrix}1&1\\0&1\end{pmatrix} X = ( 1 0 1 1 )
Ce merită reținut din simulare
Creșteri procentuale succesive se înmulțesc: + 10 % +10\% + 10% × 1,21 \times 1{,}21 × 1 , 21 + 20 % +20\% + 20% Factor comun la exponențiale : 2 x + 2 + 2 x + 1 + 2 x = 2 x ⋅ 7 2^{x+2} + 2^{x+1} + 2^x = 2^x\cdot 7 2 x + 2 + 2 x + 1 + 2 x = 2 x ⋅ 7 Element absorbant : dacă apare printre factori, întregul produs este egal cu el.Matrice de forma A ( m ) A ( n ) = A ( m + n ) A(m)A(n) = A(m+n) A ( m ) A ( n ) = A ( m + n ) : inversa este A ( − n ) A(-n) A ( − n )