Rezolvarea completă a subiectului de Bacalaureat 2015 la Matematică M_tehnologic , sesiunea august–septembrie, varianta 1 . 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 1 2 : 0,5 − 1 = 0 \frac12 : 0{,}5 - 1 = 0 2 1 : 0 , 5 − 1 = 0
0,5 = 1 2 0{,}5 = \frac12 0 , 5 = 2 1 1 2 : 1 2 − 1 = 1 − 1 = 0 \frac12 : \frac12 - 1 = 1 - 1 = 0 2 1 : 2 1 − 1 = 1 − 1 = 0 Barem: 2p + 3p.
2. f ( x ) = x 2 + x f(x) = x^2 + x f ( x ) = x 2 + x f ( − 1 ) + f ( 0 ) + f ( 1 ) f(-1) + f(0) + f(1) f ( − 1 ) + f ( 0 ) + f ( 1 )
f ( − 1 ) = 0 f(-1) = 0 f ( − 1 ) = 0 f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 f ( 1 ) = 2 f(1) = 2 f ( 1 ) = 2 2 2 2 Barem: 3p + 2p.
3. Ecuația 3 x + 1 = 5 \sqrt{3x + 1} = 5 3 x + 1 = 5
3 x + 1 = 25 ⟺ x = 8 3x + 1 = 25 \iff x = 8 3 x + 1 = 25 ⟺ x = 8 25 = 5 \sqrt{25} = 5 25 = 5 Barem: 3p + 2p.
4. Prețul de 150 lei după o scumpire cu 30%
30 100 ⋅ 150 = 45 \frac{30}{100}\cdot150 = 45 100 30 ⋅ 150 = 45 195 195 195 Barem: 3p + 2p.
5. Distanța dintre A ( 1 , 5 ) A(1,5) A ( 1 , 5 ) B ( 3 , 5 ) B(3,5) B ( 3 , 5 )
A B = ( 3 − 1 ) 2 + 0 2 = 2. AB = \sqrt{(3-1)^2 + 0^2} = 2. A B = ( 3 − 1 ) 2 + 0 2 = 2.
Barem: 3p + 2p.
6. Triunghi dreptunghic în A A A A C = 5 AC = 5 A C = 5 B = 45 ∘ B = 45^\circ B = 4 5 ∘ A B AB A B
C = 90 ∘ − 45 ∘ = 45 ∘ = B C = 90^\circ - 45^\circ = 45^\circ = B C = 9 0 ∘ − 4 5 ∘ = 4 5 ∘ = B A B = A C = 5 AB = AC = 5 A B = A C = 5 Barem: 3p + 2p.
Subiectul al II-lea (30 de puncte)
1. M = ( − 2 2 − 1 − 1 ) M = \begin{pmatrix} -2 & 2\\ -1 & -1\end{pmatrix} M = ( − 2 − 1 2 − 1 )
a) det M = 4 \det M = 4 det M = 4
det M = ( − 2 ) ( − 1 ) − 2 ⋅ ( − 1 ) = 2 + 2 = 4. \det M = (-2)(-1) - 2\cdot(-1) = 2 + 2 = 4. det M = ( − 2 ) ( − 1 ) − 2 ⋅ ( − 1 ) = 2 + 2 = 4.
Barem: 3p + 2p.
b) M ⋅ M + 3 M + 4 I 2 = O 2 M\cdot M + 3M + 4I_2 = O_2 M ⋅ M + 3 M + 4 I 2 = O 2
M ⋅ M = ( 4 − 2 − 4 − 2 2 + 1 − 2 + 1 ) = ( 2 − 6 3 − 1 ) , 3 M = ( − 6 6 − 3 − 3 ) , 4 I 2 = ( 4 0 0 4 ) . M\cdot M = \begin{pmatrix} 4 - 2 & -4 - 2\\ 2 + 1 & -2 + 1\end{pmatrix} = \begin{pmatrix} 2 & -6\\ 3 & -1\end{pmatrix}, \quad 3M = \begin{pmatrix} -6 & 6\\ -3 & -3\end{pmatrix}, \quad 4I_2 = \begin{pmatrix} 4 & 0\\ 0 & 4\end{pmatrix}. M ⋅ M = ( 4 − 2 2 + 1 − 4 − 2 − 2 + 1 ) = ( 2 3 − 6 − 1 ) , 3 M = ( − 6 − 3 6 − 3 ) , 4 I 2 = ( 4 0 0 4 ) .
Suma: ( 2 − 6 + 4 − 6 + 6 3 − 3 − 1 − 3 + 4 ) = O 2 \begin{pmatrix} 2 - 6 + 4 & -6 + 6\\ 3 - 3 & -1 - 3 + 4\end{pmatrix} = O_2 ( 2 − 6 + 4 3 − 3 − 6 + 6 − 1 − 3 + 4 ) = O 2 ■ \blacksquare ■ Barem: 3p + 2p.
c) a , b a, b a , b M ⋅ M ⋅ M = a M + b I 2 M\cdot M\cdot M = aM + bI_2 M ⋅ M ⋅ M = a M + b I 2
Din b), M 2 = − 3 M − 4 I 2 M^2 = -3M - 4I_2 M 2 = − 3 M − 4 I 2 M M M
M 3 = − 3 M 2 − 4 M = − 3 ( − 3 M − 4 I 2 ) − 4 M = 5 M + 12 I 2 . M^3 = -3M^2 - 4M = -3(-3M - 4I_2) - 4M = 5M + 12I_2. M 3 = − 3 M 2 − 4 M = − 3 ( − 3 M − 4 I 2 ) − 4 M = 5 M + 12 I 2 .
Verificare directă: M 3 = ( 2 10 − 5 7 ) M^3 = \begin{pmatrix} 2 & 10\\ -5 & 7\end{pmatrix} M 3 = ( 2 − 5 10 7 ) 5 M + 12 I 2 = ( − 10 + 12 10 − 5 − 5 + 12 ) 5M + 12I_2 = \begin{pmatrix} -10 + 12 & 10\\ -5 & -5 + 12\end{pmatrix} 5 M + 12 I 2 = ( − 10 + 12 − 5 10 − 5 + 12 )
Răspuns: a = 5 a = 5 a = 5 b = 12 b = 12 b = 12 Barem: 3p + 2p.
2. f = X 3 − 5 X 2 + 5 X − 1 f = X^3 - 5X^2 + 5X - 1 f = X 3 − 5 X 2 + 5 X − 1
a) f ( 1 ) = 0 f(1) = 0 f ( 1 ) = 0
f ( 1 ) = 1 − 5 + 5 − 1 = 0. f(1) = 1 - 5 + 5 - 1 = 0. f ( 1 ) = 1 − 5 + 5 − 1 = 0.
Barem: 3p + 2p.
b) f ( a ) + f ( − a ) + 2 ≤ 0 f(a) + f(-a) + 2 \le 0 f ( a ) + f ( − a ) + 2 ≤ 0
f ( a ) = a 3 − 5 a 2 + 5 a − 1 f(a) = a^3 - 5a^2 + 5a - 1 f ( a ) = a 3 − 5 a 2 + 5 a − 1 f ( − a ) = − a 3 − 5 a 2 − 5 a − 1 f(-a) = -a^3 - 5a^2 - 5a - 1 f ( − a ) = − a 3 − 5 a 2 − 5 a − 1
f ( a ) + f ( − a ) + 2 = − 10 a 2 − 2 + 2 = − 10 a 2 ≤ 0. ■ f(a) + f(-a) + 2 = -10a^2 - 2 + 2 = -10a^2 \le 0. \ \blacksquare f ( a ) + f ( − a ) + 2 = − 10 a 2 − 2 + 2 = − 10 a 2 ≤ 0. ■
Barem: 2p + 3p.
c) x 1 2 + x 2 2 + x 3 2 = 15 x 1 x 2 x 3 x_1^2 + x_2^2 + x_3^2 = 15x_1x_2x_3 x 1 2 + x 2 2 + x 3 2 = 15 x 1 x 2 x 3
Viète: ∑ x i = 5 \sum x_i = 5 ∑ x i = 5 ∑ x i x j = 5 \sum x_ix_j = 5 ∑ x i x j = 5 x 1 x 2 x 3 = 1 x_1x_2x_3 = 1 x 1 x 2 x 3 = 1
x 1 2 + x 2 2 + x 3 2 = 25 − 10 = 15 = 15 ⋅ 1 = 15 x 1 x 2 x 3 . ■ x_1^2 + x_2^2 + x_3^2 = 25 - 10 = 15 = 15\cdot1 = 15x_1x_2x_3. \ \blacksquare x 1 2 + x 2 2 + x 3 2 = 25 − 10 = 15 = 15 ⋅ 1 = 15 x 1 x 2 x 3 . ■
Barem: 3p + 2p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = 2 x 3 − 6 x + 1 f(x) = 2x^3 - 6x + 1 f ( x ) = 2 x 3 − 6 x + 1
a) f ′ ( x ) = 6 ( x − 1 ) ( x + 1 ) f'(x) = 6(x-1)(x+1) f ′ ( x ) = 6 ( x − 1 ) ( x + 1 )
f ′ ( x ) = 6 x 2 − 6 = 6 ( x 2 − 1 ) = 6 ( x − 1 ) ( x + 1 ) . f'(x) = 6x^2 - 6 = 6(x^2 - 1) = 6(x-1)(x+1). f ′ ( x ) = 6 x 2 − 6 = 6 ( x 2 − 1 ) = 6 ( x − 1 ) ( x + 1 ) .
Barem: 3p + 2p.
b) Tangenta în x = 1 x = 1 x = 1
f ( 1 ) = 2 − 6 + 1 = − 3 f(1) = 2 - 6 + 1 = -3 f ( 1 ) = 2 − 6 + 1 = − 3 f ′ ( 1 ) = 0 f'(1) = 0 f ′ ( 1 ) = 0 y = − 3 y = -3 y = − 3 Barem: 2p + 3p.
c) f ( 2012 ) + f ( 2014 ) ≤ f ( 2013 ) + f ( 2015 ) f(2012) + f(2014) \le f(2013) + f(2015) f ( 2012 ) + f ( 2014 ) ≤ f ( 2013 ) + f ( 2015 )
Pe [ 1 , + ∞ ) [1, +\infty) [ 1 , + ∞ ) f ′ ( x ) ≥ 0 f'(x) \ge 0 f ′ ( x ) ≥ 0 f f f f ( 2012 ) ≤ f ( 2013 ) f(2012) \le f(2013) f ( 2012 ) ≤ f ( 2013 ) f ( 2014 ) ≤ f ( 2015 ) f(2014) \le f(2015) f ( 2014 ) ≤ f ( 2015 ) ■ \blacksquare ■ Barem: 3p + 2p.
2. f ( x ) = x 2 − 4 f(x) = x^2 - 4 f ( x ) = x 2 − 4
a) ∫ 0 1 ( f ( x ) + 4 ) d x = 1 3 \displaystyle\int_0^1(f(x) + 4)\,dx = \frac13 ∫ 0 1 ( f ( x ) + 4 ) d x = 3 1
∫ 0 1 x 2 d x = 1 3 . \int_0^1 x^2\,dx = \frac13. ∫ 0 1 x 2 d x = 3 1 .
Barem: 2p + 3p.
b) Aria sub g ( x ) = 1 f ( x ) + 5 = 1 x 2 + 1 g(x) = \dfrac{1}{f(x) + 5} = \dfrac{1}{x^2 + 1} g ( x ) = f ( x ) + 5 1 = x 2 + 1 1 [ 0 , 1 ] [0,1] [ 0 , 1 ]
A = ∫ 0 1 1 x 2 + 1 d x = arctg x ∣ 0 1 = π 4 . \mathcal A = \int_0^1\frac{1}{x^2 + 1}\,dx = \operatorname{arctg} x\Big|_0^1 = \frac\pi4. A = ∫ 0 1 x 2 + 1 1 d x = arctg x 0 1 = 4 π .
Barem: 3p + 2p.
c) a > 1 a > 1 a > 1 ∫ 1 a f ( x ) + 4 x d x = 12 \displaystyle\int_1^a\frac{f(x) + 4}{x}\,dx = 12 ∫ 1 a x f ( x ) + 4 d x = 12
x 2 x = x \dfrac{x^2}{x} = x x x 2 = x
∫ 1 a x d x = a 2 − 1 2 = 12 ⟺ a 2 = 25 , \int_1^a x\,dx = \frac{a^2 - 1}{2} = 12 \iff a^2 = 25, ∫ 1 a x d x = 2 a 2 − 1 = 12 ⟺ a 2 = 25 ,
iar a > 1 a > 1 a > 1 a = 5 a = 5 a = 5 Barem: 3p + 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 2 2 2 I.3 x = 8 x = 8 x = 8 I.4 195 lei I.5 2 2 2 I.6 A B = 5 AB = 5 A B = 5 II.1.c a = 5 a = 5 a = 5 b = 12 b = 12 b = 12 III.1.b y = − 3 y = -3 y = − 3 III.2.b π 4 \frac\pi4 4 π III.2.c a = 5 a = 5 a = 5
Ce merită reținut din varianta 1
O relație M 2 = p M + q I M^2 = pM + qI M 2 = pM + q I permite exprimarea oricărei puteri M n M^n M n M M M I I I f ( a ) + f ( − a ) f(a) + f(-a) f ( a ) + f ( − a ) Comparații de valori ale funcției se obțin din monotonie.∫ 1 x 2 + 1 d x = arctg x \int\frac{1}{x^2 + 1}dx = \operatorname{arctg} x ∫ x 2 + 1 1 d x = arctg x