Rezolvarea completă a subiectului de Bacalaureat 2020 la Matematică M_pedagogic , sesiunea august–septembrie, 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 96 ( 6 4 − 1 6 ) = 2 \sqrt{96}\left(\frac{\sqrt6}4 - \frac1{\sqrt6}\right) = 2 96 ( 4 6 − 6 1 ) = 2
96 = 4 6 \sqrt{96} = 4\sqrt6 96 = 4 6
4 6 ⋅ 6 4 − 4 6 ⋅ 1 6 = 6 − 4 = 2. 4\sqrt6\cdot\frac{\sqrt6}4 - 4\sqrt6\cdot\frac1{\sqrt6} = 6 - 4 = 2. 4 6 ⋅ 4 6 − 4 6 ⋅ 6 1 = 6 − 4 = 2.
Barem: 3p + 2p.
2. f ( x ) = 3 x + 1 f(x) = 3x + 1 f ( x ) = 3 x + 1 g ( x ) = x − 1 g(x) = x - 1 g ( x ) = x − 1 n n n f ( n ) ⋅ g ( n ) = 0 f(n)\cdot g(n) = 0 f ( n ) ⋅ g ( n ) = 0
( 3 n + 1 ) ( n − 1 ) = 0 ⟺ n ∈ { − 1 3 , 1 } (3n + 1)(n - 1) = 0 \iff n \in \left\{-\frac13, 1\right\} ( 3 n + 1 ) ( n − 1 ) = 0 ⟺ n ∈ { − 3 1 , 1 } n ∈ N n \in \mathbb N n ∈ N n = 1 n = 1 n = 1
Barem: 2p + 3p.
3. Ecuația 3 2 x + 5 = 3 6 x − 3 3^{2x+5} = 3^{6x-3} 3 2 x + 5 = 3 6 x − 3
2 x + 5 = 6 x − 3 ⟺ 4 x = 8 ⟺ x = 2 2x + 5 = 6x - 3 \iff 4x = 8 \iff x = 2 2 x + 5 = 6 x − 3 ⟺ 4 x = 8 ⟺ x = 2
Barem: 3p + 2p.
4. Probabilitatea ca un număr de două cifre să aibă cifra zecilor egală cu dublul cifrei unităților
90 cazuri posibile. Favorabile: 21 , 42 , 63 , 84 21, 42, 63, 84 21 , 42 , 63 , 84 1..4 1..4 1..4 00 ‾ \overline{00} 00
p = 4 90 = 2 45 . p = \frac4{90} = \frac2{45}. p = 90 4 = 45 2 .
Barem: 2p + 2p + 1p.
5. A ( 1 , 3 ) A(1,3) A ( 1 , 3 ) B ( 3 , 5 ) B(3,5) B ( 3 , 5 ) C ( 0 , a ) C(0,a) C ( 0 , a ) △ A B C \triangle ABC △ A B C A A A
m A B = 5 − 3 3 − 1 = 1 m_{AB} = \frac{5-3}{3-1} = 1 m A B = 3 − 1 5 − 3 = 1 m A C = a − 3 0 − 1 = 3 − a m_{AC} = \frac{a - 3}{0 - 1} = 3 - a m A C = 0 − 1 a − 3 = 3 − a A B ⊥ A C ⟺ 1 ⋅ ( 3 − a ) = − 1 ⟺ a = 4 AB \perp AC \iff 1\cdot(3 - a) = -1 \iff a = 4 A B ⊥ A C ⟺ 1 ⋅ ( 3 − a ) = − 1 ⟺ a = 4
Barem: 2p + 3p.
6. Arătați că 3 sin 60 ∘ + sin 30 ∘ − 4 sin 2 30 ∘ = 1 \sqrt3\sin60^\circ + \sin30^\circ - 4\sin^230^\circ = 1 3 sin 6 0 ∘ + sin 3 0 ∘ − 4 sin 2 3 0 ∘ = 1
3 ⋅ 3 2 + 1 2 − 4 ⋅ 1 4 = 3 2 + 1 2 − 1 = 1. \sqrt3\cdot\tfrac{\sqrt3}2 + \tfrac12 - 4\cdot\tfrac14 = \tfrac32 + \tfrac12 - 1 = 1. 3 ⋅ 2 3 + 2 1 − 4 ⋅ 4 1 = 2 3 + 2 1 − 1 = 1.
Barem: 2p + 3p.
Subiectul al II-lea (30 de puncte)
Legea asociativă pe R \mathbb R R x ∘ y = − x y + 4 x + 4 y − 12 x\circ y = -xy + 4x + 4y - 12 x ∘ y = − x y + 4 x + 4 y − 12
1. 2 ∘ 3 = 2 2\circ3 = 2 2 ∘ 3 = 2
2 ∘ 3 = − 6 + 8 + 12 − 12 = 2. 2\circ3 = -6 + 8 + 12 - 12 = 2. 2 ∘ 3 = − 6 + 8 + 12 − 12 = 2.
Barem: 3p + 2p.
2. Legea este comutativă
x ∘ y = − x y + 4 x + 4 y − 12 = − y x + 4 y + 4 x − 12 = y ∘ x x\circ y = -xy + 4x + 4y - 12 = -yx + 4y + 4x - 12 = y\circ x x ∘ y = − x y + 4 x + 4 y − 12 = − y x + 4 y + 4 x − 12 = y ∘ x ■ \blacksquare ■ Barem: 2p + 3p.
3. x ∘ y = − ( x − 4 ) ( y − 4 ) + 4 x\circ y = -(x - 4)(y - 4) + 4 x ∘ y = − ( x − 4 ) ( y − 4 ) + 4
− ( x − 4 ) ( y − 4 ) + 4 = − x y + 4 x + 4 y − 16 + 4 = x ∘ y . -(x - 4)(y - 4) + 4 = -xy + 4x + 4y - 16 + 4 = x\circ y. − ( x − 4 ) ( y − 4 ) + 4 = − x y + 4 x + 4 y − 16 + 4 = x ∘ y .
Barem: 2p + 3p.
4. x ∘ 4 = 4 x\circ4 = 4 x ∘ 4 = 4
x ∘ 4 = − ( x − 4 ) ⋅ 0 + 4 = 4. x\circ4 = -(x - 4)\cdot0 + 4 = 4. x ∘ 4 = − ( x − 4 ) ⋅ 0 + 4 = 4.
Barem: 3p + 2p.
5. x x x x ∘ x = x x\circ x = x x ∘ x = x
− ( x − 4 ) 2 + 4 = x ⟺ − ( x − 4 ) 2 − ( x − 4 ) = 0 ⟺ − ( x − 4 ) ( x − 3 ) = 0. -(x - 4)^2 + 4 = x \iff -(x-4)^2 - (x - 4) = 0 \iff -(x - 4)(x - 3) = 0. − ( x − 4 ) 2 + 4 = x ⟺ − ( x − 4 ) 2 − ( x − 4 ) = 0 ⟺ − ( x − 4 ) ( x − 3 ) = 0.
Răspuns: x ∈ { 3 , 4 } x \in \{3, 4\} x ∈ { 3 , 4 } Barem: 2p + 3p.
6. Progresia aritmetică cu a 1 = − 5 a_1 = -5 a 1 = − 5 r = 3 r = 3 r = 3 a 1 ∘ a 2 ∘ a 3 ∘ a 4 = 4 a_1\circ a_2\circ a_3\circ a_4 = 4 a 1 ∘ a 2 ∘ a 3 ∘ a 4 = 4
a 4 = a 1 + 3 r = − 5 + 9 = 4 a_4 = a_1 + 3r = -5 + 9 = 4 a 4 = a 1 + 3 r = − 5 + 9 = 4
( a 1 ∘ a 2 ∘ a 3 ) ∘ 4 = 4. ■ (a_1\circ a_2\circ a_3)\circ4 = 4.\ \blacksquare ( a 1 ∘ a 2 ∘ a 3 ) ∘ 4 = 4. ■
Barem: 2p + 3p.
Subiectul al III-lea (30 de puncte)
A = ( 1 2 − 1 1 ) A = \begin{pmatrix}1&2\\-1&1\end{pmatrix} A = ( 1 − 1 2 1 ) B ( x ) = ( 1 x 0 1 ) B(x) = \begin{pmatrix}1&x\\0&1\end{pmatrix} B ( x ) = ( 1 0 x 1 )
1. det A = 3 \det A = 3 det A = 3
det A = 1 ⋅ 1 − 2 ⋅ ( − 1 ) = 3. \det A = 1\cdot1 - 2\cdot(-1) = 3. det A = 1 ⋅ 1 − 2 ⋅ ( − 1 ) = 3.
Barem: 3p + 2p.
2. B ( 1 ) + B ( 3 ) = 2 B ( 2 ) B(1) + B(3) = 2B(2) B ( 1 ) + B ( 3 ) = 2 B ( 2 )
( 1 1 0 1 ) + ( 1 3 0 1 ) = ( 2 4 0 2 ) = 2 ( 1 2 0 1 ) = 2 B ( 2 ) . \begin{pmatrix}1&1\\0&1\end{pmatrix} + \begin{pmatrix}1&3\\0&1\end{pmatrix} = \begin{pmatrix}2&4\\0&2\end{pmatrix} = 2\begin{pmatrix}1&2\\0&1\end{pmatrix} = 2B(2). ( 1 0 1 1 ) + ( 1 0 3 1 ) = ( 2 0 4 2 ) = 2 ( 1 0 2 1 ) = 2 B ( 2 ) .
Barem: 3p + 2p.
3. det B ( x ) = 1 \det B(x) = 1 det B ( x ) = 1
det B ( x ) = 1 ⋅ 1 − x ⋅ 0 = 1. \det B(x) = 1\cdot1 - x\cdot0 = 1. det B ( x ) = 1 ⋅ 1 − x ⋅ 0 = 1.
Barem: 2p + 3p.
4. B ( x ) B ( y ) = B ( x + y ) B(x)B(y) = B(x + y) B ( x ) B ( y ) = B ( x + y )
( 1 x 0 1 ) ( 1 y 0 1 ) = ( 1 y + x 0 1 ) = B ( x + y ) . \begin{pmatrix}1&x\\0&1\end{pmatrix}\begin{pmatrix}1&y\\0&1\end{pmatrix} = \begin{pmatrix}1 & y + x\\0&1\end{pmatrix} = B(x + y). ( 1 0 x 1 ) ( 1 0 y 1 ) = ( 1 0 y + x 1 ) = B ( x + y ) .
Barem: 3p + 2p.
5. x x x A ⋅ B ( x ) = B ( x ) ⋅ A A\cdot B(x) = B(x)\cdot A A ⋅ B ( x ) = B ( x ) ⋅ A
A B ( x ) = ( 1 x + 2 − 1 − x + 1 ) , B ( x ) A = ( 1 − x 2 + x − 1 1 ) . AB(x) = \begin{pmatrix}1 & x + 2\\ -1 & -x + 1\end{pmatrix},\quad B(x)A = \begin{pmatrix}1 - x & 2 + x\\ -1 & 1\end{pmatrix}. A B ( x ) = ( 1 − 1 x + 2 − x + 1 ) , B ( x ) A = ( 1 − x − 1 2 + x 1 ) .
Egalând elementele ( 1 , 1 ) (1,1) ( 1 , 1 ) 1 = 1 − x ⟺ x = 0 1 = 1 - x \iff x = 0 1 = 1 − x ⟺ x = 0
Răspuns: x = 0 x = 0 x = 0 Barem: 3p + 2p.
6. Perechi ( m , n ) (m, n) ( m , n ) B ( 2 m ) B ( 2 n ) = B ( 2 m + n − 2 ) B(2^m)B(2^n) = B(2^{m+n} - 2) B ( 2 m ) B ( 2 n ) = B ( 2 m + n − 2 )
Din 4: B ( 2 m + 2 n ) = B ( 2 m + n − 2 ) B(2^m + 2^n) = B(2^{m+n} - 2) B ( 2 m + 2 n ) = B ( 2 m + n − 2 ) 2 m + 2 n = 2 m + n − 2 2^m + 2^n = 2^{m+n} - 2 2 m + 2 n = 2 m + n − 2
2 m + n − 2 m − 2 n + 1 = 3 ⟺ ( 2 m − 1 ) ( 2 n − 1 ) = 3. 2^{m+n} - 2^m - 2^n + 1 = 3 \iff (2^m - 1)(2^n - 1) = 3. 2 m + n − 2 m − 2 n + 1 = 3 ⟺ ( 2 m − 1 ) ( 2 n − 1 ) = 3.
Factorii sunt naturali: { 2 m − 1 , 2 n − 1 } = { 1 , 3 } \{2^m - 1, 2^n - 1\} = \{1, 3\} { 2 m − 1 , 2 n − 1 } = { 1 , 3 } { m , n } = { 1 , 2 } \{m, n\} = \{1, 2\} { m , n } = { 1 , 2 }
Răspuns: ( 1 , 2 ) (1, 2) ( 1 , 2 ) ( 2 , 1 ) (2, 1) ( 2 , 1 ) Barem: 3p + 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 n = 1 n = 1 n = 1 I.3 x = 2 x = 2 x = 2 I.4 p = 2 45 p = \frac2{45} p = 45 2 I.5 a = 4 a = 4 a = 4 II.5 x ∈ { 3 , 4 } x \in \{3,4\} x ∈ { 3 , 4 } III.5 x = 0 x = 0 x = 0 III.6 ( 1 , 2 ) , ( 2 , 1 ) (1,2), (2,1) ( 1 , 2 ) , ( 2 , 1 )
Ce merită reținut din varianta 3
Element absorbant (4 4 4 4 4 4 Perpendicularitate: produsul pantelor este − 1 -1 − 1 ( 2 m − 1 ) ( 2 n − 1 ) = k (2^m - 1)(2^n - 1) = k ( 2 m − 1 ) ( 2 n − 1 ) = k