Rezolvarea completă a subiectului de Bacalaureat 2018 la Matematică M_pedagogic , sesiunea iunie–iulie, varianta 2 . 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 2 ( 0,1 ( 6 ) + 1 3 ) = 1 2\left(0{,}1(6) + \dfrac13\right) = 1 2 ( 0 , 1 ( 6 ) + 3 1 ) = 1
Transformăm fracția periodică mixtă: 0,1 ( 6 ) = 16 − 1 90 = 15 90 = 1 6 0{,}1(6) = \dfrac{16 - 1}{90} = \dfrac{15}{90} = \dfrac16 0 , 1 ( 6 ) = 90 16 − 1 = 90 15 = 6 1
2 ( 1 6 + 1 3 ) = 2 ⋅ 1 + 2 6 = 2 ⋅ 1 2 = 1. 2\left(\frac16 + \frac13\right) = 2\cdot\frac{1 + 2}{6} = 2\cdot\frac12 = 1. 2 ( 6 1 + 3 1 ) = 2 ⋅ 6 1 + 2 = 2 ⋅ 2 1 = 1.
Barem: 3p 0,1 ( 6 ) = 1 6 0{,}1(6) = \frac16 0 , 1 ( 6 ) = 6 1
2. f ( x ) = 2 x − 2 f(x) = 2x - 2 f ( x ) = 2 x − 2 a a a f ( a ) = a f(a) = a f ( a ) = a
f ( a ) = a ⟺ 2 a − 2 = a ⟺ a = 2. f(a) = a \iff 2a - 2 = a \iff a = 2. f ( a ) = a ⟺ 2 a − 2 = a ⟺ a = 2.
Răspuns: a = 2 a = 2 a = 2 Barem: 2p f ( a ) = 2 a − 2 f(a) = 2a-2 f ( a ) = 2 a − 2 a a a
3. Ecuația 5 x 2 + 6 = 5 5 x 5^{x^2+6} = 5^{5x} 5 x 2 + 6 = 5 5 x
Funcția exponențială este injectivă, deci egalăm exponenții:
x 2 + 6 = 5 x ⟺ x 2 − 5 x + 6 = 0 ⟺ ( x − 2 ) ( x − 3 ) = 0. x^2 + 6 = 5x \iff x^2 - 5x + 6 = 0 \iff (x-2)(x-3) = 0. x 2 + 6 = 5 x ⟺ x 2 − 5 x + 6 = 0 ⟺ ( x − 2 ) ( x − 3 ) = 0.
Răspuns: x ∈ { 2 , 3 } x \in \{2, 3\} x ∈ { 2 , 3 } Barem: 3p ecuația de gradul 2, 2p soluțiile.
4. Prețul de 900 900 900 10 % 10\% 10%
După prima ieftinire: 900 − 10 100 ⋅ 900 = 900 − 90 = 810 900 - \dfrac{10}{100}\cdot 900 = 900 - 90 = 810 900 − 100 10 ⋅ 900 = 900 − 90 = 810
După a doua: 810 − 10 100 ⋅ 810 = 810 − 81 = 729 810 - \dfrac{10}{100}\cdot 810 = 810 - 81 = 729 810 − 100 10 ⋅ 810 = 810 − 81 = 729
(Altfel: 900 ⋅ 0,9 2 = 900 ⋅ 0,81 = 729 900\cdot 0{,}9^2 = 900 \cdot 0{,}81 = 729 900 ⋅ 0 , 9 2 = 900 ⋅ 0 , 81 = 729
Răspuns: 729 729 729 Barem: 3p prima ieftinire, 2p a doua.
5. A ( 2 , − 1 ) A(2,-1) A ( 2 , − 1 ) B ( 1 , 2 ) B(1,2) B ( 1 , 2 ) C ( − 1 , − 2 ) C(-1,-2) C ( − 1 , − 2 ) A B C ABC A B C
A B = ( 1 − 2 ) 2 + ( 2 + 1 ) 2 = 10 , A C = ( − 1 − 2 ) 2 + ( − 2 + 1 ) 2 = 10 , AB = \sqrt{(1-2)^2 + (2+1)^2} = \sqrt{10}, \quad AC = \sqrt{(-1-2)^2 + (-2+1)^2} = \sqrt{10}, A B = ( 1 − 2 ) 2 + ( 2 + 1 ) 2 = 10 , A C = ( − 1 − 2 ) 2 + ( − 2 + 1 ) 2 = 10 ,
B C = ( − 1 − 1 ) 2 + ( − 2 − 2 ) 2 = 20 . BC = \sqrt{(-1-1)^2 + (-2-2)^2} = \sqrt{20}. B C = ( − 1 − 1 ) 2 + ( − 2 − 2 ) 2 = 20 .
A B = A C AB = AC A B = A C A B 2 + A C 2 = 10 + 10 = 20 = B C 2 AB^2 + AC^2 = 10 + 10 = 20 = BC^2 A B 2 + A C 2 = 10 + 10 = 20 = B C 2 A A A Barem: 3p laturile și isoscel, 2p dreptunghic.
6. Arătați că sin 2 30 ∘ + sin 2 45 ∘ + sin 2 60 ∘ = 3 2 \sin^2 30^\circ + \sin^2 45^\circ + \sin^2 60^\circ = \dfrac32 sin 2 3 0 ∘ + sin 2 4 5 ∘ + sin 2 6 0 ∘ = 2 3
( 1 2 ) 2 + ( 2 2 ) 2 + ( 3 2 ) 2 = 1 4 + 2 4 + 3 4 = 6 4 = 3 2 . \left(\frac12\right)^2 + \left(\frac{\sqrt2}{2}\right)^2 + \left(\frac{\sqrt3}{2}\right)^2 = \frac14 + \frac24 + \frac34 = \frac64 = \frac32. ( 2 1 ) 2 + ( 2 2 ) 2 + ( 2 3 ) 2 = 4 1 + 4 2 + 4 3 = 4 6 = 2 3 .
Barem: 3p valorile, 2p calculul.
Subiectul al II-lea (30 de puncte)
Legea pe R \mathbb R R x ∗ y = 2 ( x + y ) + x y + 2 x * y = 2(x + y) + xy + 2 x ∗ y = 2 ( x + y ) + x y + 2
1. Arătați că 0 ∗ ( − 2 ) = − 2 0 * (-2) = -2 0 ∗ ( − 2 ) = − 2
0 ∗ ( − 2 ) = 2 ( 0 − 2 ) + 0 + 2 = − 4 + 2 = − 2. 0 * (-2) = 2(0 - 2) + 0 + 2 = -4 + 2 = -2. 0 ∗ ( − 2 ) = 2 ( 0 − 2 ) + 0 + 2 = − 4 + 2 = − 2.
Barem: 3p înlocuirea, 2p finalizarea.
2. x ∗ y = ( x + 2 ) ( y + 2 ) − 2 x * y = (x+2)(y+2) - 2 x ∗ y = ( x + 2 ) ( y + 2 ) − 2
x ∗ y = x y + 2 x + 2 y + 4 − 2 = x ( y + 2 ) + 2 ( y + 2 ) − 2 = ( x + 2 ) ( y + 2 ) − 2. x * y = xy + 2x + 2y + 4 - 2 = x(y+2) + 2(y+2) - 2 = (x+2)(y+2) - 2. x ∗ y = x y + 2 x + 2 y + 4 − 2 = x ( y + 2 ) + 2 ( y + 2 ) − 2 = ( x + 2 ) ( y + 2 ) − 2.
Barem: 2p adunarea și scăderea lui 2, 3p gruparea.
3. e = − 1 e = -1 e = − 1
x ∗ ( − 1 ) = ( x + 2 ) ( 1 ) − 2 = x , ( − 1 ) ∗ x = 1 ⋅ ( x + 2 ) − 2 = x , x * (-1) = (x+2)(1) - 2 = x, \qquad (-1) * x = 1\cdot(x+2) - 2 = x, x ∗ ( − 1 ) = ( x + 2 ) ( 1 ) − 2 = x , ( − 1 ) ∗ x = 1 ⋅ ( x + 2 ) − 2 = x ,
pentru orice x x x e = − 1 e = -1 e = − 1 Barem: 2p x ∗ ( − 1 ) x*(-1) x ∗ ( − 1 )
4. ( x + 1 ) ∗ ( x + 1 ) = 2 (x+1) * (x+1) = 2 ( x + 1 ) ∗ ( x + 1 ) = 2
( x + 3 ) 2 − 2 = 2 ⟺ ( x + 3 ) 2 = 4 ⟺ x 2 + 6 x + 5 = 0 ⟺ x ∈ { − 5 , − 1 } . (x+3)^2 - 2 = 2 \iff (x+3)^2 = 4 \iff x^2 + 6x + 5 = 0 \iff x \in \{-5, -1\}. ( x + 3 ) 2 − 2 = 2 ⟺ ( x + 3 ) 2 = 4 ⟺ x 2 + 6 x + 5 = 0 ⟺ x ∈ { − 5 , − 1 } .
Răspuns: x ∈ { − 5 , − 1 } x \in \{-5, -1\} x ∈ { − 5 , − 1 } Barem: 3p ecuația, 2p soluțiile.
5. x > 0 x > 0 x > 0 lg x ∗ lg ( 2 x ) = − 2 \lg x * \lg(2x) = -2 lg x ∗ lg ( 2 x ) = − 2
( lg x + 2 ) ( lg ( 2 x ) + 2 ) − 2 = − 2 ⟺ ( lg x + 2 ) ( lg ( 2 x ) + 2 ) = 0. (\lg x + 2)(\lg(2x) + 2) - 2 = -2 \iff (\lg x + 2)(\lg(2x) + 2) = 0. ( lg x + 2 ) ( lg ( 2 x ) + 2 ) − 2 = − 2 ⟺ ( lg x + 2 ) ( lg ( 2 x ) + 2 ) = 0.
lg x = − 2 ⇒ x = 10 − 2 = 1 100 \lg x = -2 \Rightarrow x = 10^{-2} = \dfrac{1}{100} lg x = − 2 ⇒ x = 1 0 − 2 = 100 1 lg ( 2 x ) = − 2 ⇒ 2 x = 1 100 ⇒ x = 1 200 \lg(2x) = -2 \Rightarrow 2x = \dfrac1{100} \Rightarrow x = \dfrac{1}{200} lg ( 2 x ) = − 2 ⇒ 2 x = 100 1 ⇒ x = 200 1
Ambele sunt pozitive, deci convin. Răspuns: x ∈ { 1 100 , 1 200 } x \in \left\{\dfrac1{100}, \dfrac1{200}\right\} x ∈ { 100 1 , 200 1 } Barem: 3p descompunerea, 2p soluțiile.
6. a , b ∈ Q ∖ Z a, b \in \mathbb Q \setminus \mathbb Z a , b ∈ Q ∖ Z a ∗ b ∈ Z a * b \in \mathbb Z a ∗ b ∈ Z
a ∗ b ∈ Z ⟺ ( a + 2 ) ( b + 2 ) ∈ Z a * b \in \mathbb Z \iff (a+2)(b+2) \in \mathbb Z a ∗ b ∈ Z ⟺ ( a + 2 ) ( b + 2 ) ∈ Z a + 2 = 2 3 a + 2 = \dfrac23 a + 2 = 3 2 b + 2 = 3 2 b + 2 = \dfrac32 b + 2 = 2 3
a = − 4 3 , b = − 1 2 , a ∗ b = 2 3 ⋅ 3 2 − 2 = 1 − 2 = − 1 ∈ Z . a = -\frac43, \quad b = -\frac12, \qquad a * b = \frac23\cdot\frac32 - 2 = 1 - 2 = -1 \in \mathbb Z. a = − 3 4 , b = − 2 1 , a ∗ b = 3 2 ⋅ 2 3 − 2 = 1 − 2 = − 1 ∈ Z .
Răspuns (un exemplu): a = − 4 3 a = -\dfrac43 a = − 3 4 b = − 1 2 b = -\dfrac12 b = − 2 1 Barem: 2p condiția, 3p exemplul.
Subiectul al III-lea (30 de puncte)
A = ( 1 1 0 2 ) A = \begin{pmatrix}1&1\\0&2\end{pmatrix} A = ( 1 0 1 2 ) I 2 = ( 1 0 0 1 ) I_2 = \begin{pmatrix}1&0\\0&1\end{pmatrix} I 2 = ( 1 0 0 1 ) M ( a ) = a A + I 2 M(a) = aA + I_2 M ( a ) = a A + I 2
1. det A = 2 \det A = 2 det A = 2
det A = 1 ⋅ 2 − 0 ⋅ 1 = 2. \det A = 1\cdot 2 - 0\cdot 1 = 2. det A = 1 ⋅ 2 − 0 ⋅ 1 = 2.
Barem: 3p formula, 2p calculul.
2. det ( M ( a ) ) = ( a + 1 ) ( 2 a + 1 ) \det(M(a)) = (a+1)(2a+1) det ( M ( a )) = ( a + 1 ) ( 2 a + 1 )
M ( a ) = ( a a 0 2 a ) + ( 1 0 0 1 ) = ( a + 1 a 0 2 a + 1 ) , M(a) = \begin{pmatrix}a & a\\0 & 2a\end{pmatrix} + \begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}a+1 & a\\0 & 2a+1\end{pmatrix}, M ( a ) = ( a 0 a 2 a ) + ( 1 0 0 1 ) = ( a + 1 0 a 2 a + 1 ) ,
det ( M ( a ) ) = ( a + 1 ) ( 2 a + 1 ) − 0 = ( a + 1 ) ( 2 a + 1 ) . \det(M(a)) = (a+1)(2a+1) - 0 = (a+1)(2a+1). det ( M ( a )) = ( a + 1 ) ( 2 a + 1 ) − 0 = ( a + 1 ) ( 2 a + 1 ) .
Barem: 3p matricea M ( a ) M(a) M ( a )
3. Inversa matricei M ( − 2 ) M(-2) M ( − 2 )
M ( − 2 ) = ( − 1 − 2 0 − 3 ) M(-2) = \begin{pmatrix}-1 & -2\\0 & -3\end{pmatrix} M ( − 2 ) = ( − 1 0 − 2 − 3 ) det = ( − 1 ) ( − 3 ) = 3 ≠ 0 \det = (-1)(-3) = 3 \ne 0 det = ( − 1 ) ( − 3 ) = 3 = 0
M − 1 ( − 2 ) = 1 3 ( − 3 2 0 − 1 ) = ( − 1 2 3 0 − 1 3 ) . M^{-1}(-2) = \frac13\begin{pmatrix}-3 & 2\\0 & -1\end{pmatrix} = \begin{pmatrix}-1 & \frac23\\[2pt] 0 & -\frac13\end{pmatrix}. M − 1 ( − 2 ) = 3 1 ( − 3 0 2 − 1 ) = ( − 1 0 3 2 − 3 1 ) .
(Verificare: M ( − 2 ) ⋅ M − 1 ( − 2 ) = I 2 M(-2)\cdot M^{-1}(-2) = I_2 M ( − 2 ) ⋅ M − 1 ( − 2 ) = I 2 Barem: 2p determinantul, 3p inversa.
4. M ( 1 ) ⋅ M ( 2 ) = 3 ( A ⋅ A + I 2 ) M(1)\cdot M(2) = 3(A\cdot A + I_2) M ( 1 ) ⋅ M ( 2 ) = 3 ( A ⋅ A + I 2 )
M ( 1 ) M ( 2 ) = ( 2 1 0 3 ) ( 3 2 0 5 ) = ( 6 9 0 15 ) = 3 ( 2 3 0 5 ) , M(1)M(2) = \begin{pmatrix}2&1\\0&3\end{pmatrix}\begin{pmatrix}3&2\\0&5\end{pmatrix} = \begin{pmatrix}6&9\\0&15\end{pmatrix} = 3\begin{pmatrix}2&3\\0&5\end{pmatrix}, M ( 1 ) M ( 2 ) = ( 2 0 1 3 ) ( 3 0 2 5 ) = ( 6 0 9 15 ) = 3 ( 2 0 3 5 ) ,
A ⋅ A + I 2 = ( 1 3 0 4 ) + ( 1 0 0 1 ) = ( 2 3 0 5 ) . A\cdot A + I_2 = \begin{pmatrix}1&3\\0&4\end{pmatrix} + \begin{pmatrix}1&0\\0&1\end{pmatrix} = \begin{pmatrix}2&3\\0&5\end{pmatrix}. A ⋅ A + I 2 = ( 1 0 3 4 ) + ( 1 0 0 1 ) = ( 2 0 3 5 ) .
Deci egalitatea are loc. Barem: 3p produsul, 2p finalizarea.
5. det ( M ( a ) − 2 a A ) ≠ 1 \det(M(a) - 2aA) \ne 1 det ( M ( a ) − 2 a A ) = 1 a a a
M ( a ) − 2 a A = I 2 − a A = M ( − a ) M(a) - 2aA = I_2 - aA = M(-a) M ( a ) − 2 a A = I 2 − a A = M ( − a )
det ( M ( a ) − 2 a A ) = ( 1 − a ) ( 1 − 2 a ) . \det(M(a) - 2aA) = (1-a)(1-2a). det ( M ( a ) − 2 a A ) = ( 1 − a ) ( 1 − 2 a ) .
( 1 − a ) ( 1 − 2 a ) = 1 ⟺ 2 a 2 − 3 a = 0 ⟺ a ( 2 a − 3 ) = 0 ⟺ a ∈ { 0 , 3 2 } (1-a)(1-2a) = 1 \iff 2a^2 - 3a = 0 \iff a(2a - 3) = 0 \iff a \in \left\{0, \tfrac32\right\} ( 1 − a ) ( 1 − 2 a ) = 1 ⟺ 2 a 2 − 3 a = 0 ⟺ a ( 2 a − 3 ) = 0 ⟺ a ∈ { 0 , 2 3 } ■ \blacksquare ■ Barem: 2p determinantul, 3p concluzia.
6. X ∈ M 2 , 1 ( R ) X \in \mathcal M_{2,1}(\mathbb R) X ∈ M 2 , 1 ( R ) A ⋅ X = ( 0 − 4 ) A\cdot X = \begin{pmatrix}0\\-4\end{pmatrix} A ⋅ X = ( 0 − 4 )
Fie X = ( x y ) X = \begin{pmatrix}x\\y\end{pmatrix} X = ( x y )
( x + y 2 y ) = ( 0 − 4 ) ⟺ { x + y = 0 2 y = − 4 ⟺ y = − 2 , x = 2. \begin{pmatrix}x + y\\2y\end{pmatrix} = \begin{pmatrix}0\\-4\end{pmatrix} \iff \begin{cases}x + y = 0\\2y = -4\end{cases} \iff y = -2,\ x = 2. ( x + y 2 y ) = ( 0 − 4 ) ⟺ { x + y = 0 2 y = − 4 ⟺ y = − 2 , x = 2.
Răspuns: X = ( 2 − 2 ) X = \begin{pmatrix}2\\-2\end{pmatrix} X = ( 2 − 2 ) Barem: 2p sistemul, 3p soluția.
Rezumatul răspunsurilor
Cerință Răspuns I.2 a = 2 a = 2 a = 2 I.3 x ∈ { 2 , 3 } x \in \{2, 3\} x ∈ { 2 , 3 } I.4 729 729 729 II.4 x ∈ { − 5 , − 1 } x \in \{-5, -1\} x ∈ { − 5 , − 1 } II.5 x ∈ { 1 100 , 1 200 } x \in \{\frac1{100}, \frac1{200}\} x ∈ { 100 1 , 200 1 } II.6 ex. a = − 4 3 a = -\frac43 a = − 3 4 b = − 1 2 b = -\frac12 b = − 2 1 III.3 M − 1 ( − 2 ) = ( − 1 2 3 0 − 1 3 ) M^{-1}(-2) = \begin{pmatrix}-1&\frac23\\0&-\frac13\end{pmatrix} M − 1 ( − 2 ) = ( − 1 0 3 2 − 3 1 ) III.6 X = ( 2 − 2 ) X = \begin{pmatrix}2\\-2\end{pmatrix} X = ( 2 − 2 )
Ce merită reținut din varianta 2
Legea x y + 2 x + 2 y + 2 xy + 2x + 2y + 2 x y + 2 x + 2 y + 2 se scrie ( x + 2 ) ( y + 2 ) − 2 (x+2)(y+2) - 2 ( x + 2 ) ( y + 2 ) − 2 Ieftiniri succesive cu p % p\% p % : înmulțește cu ( 1 − p 100 ) \left(1 - \frac p{100}\right) ( 1 − 100 p ) 2 p % 2p\% 2 p % Triunghi dreptunghic din coordonate : compară pătratele lungimilor (reciproca lui Pitagora).Inversa unei matrice 2 × 2 2\times2 2 × 2 : 1 det ( d − b − c a ) \frac{1}{\det}\begin{pmatrix}d & -b\\-c & a\end{pmatrix} d e t 1 ( d − c − b a )