Rezolvarea completă a subiectului de Bacalaureat 2016 la Matematică M_tehnologic , sesiunea iunie–iulie, varianta 8 . 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 − 1 5 ) ⋅ 10 3 = 1 \left(\frac12 - \frac15\right)\cdot\frac{10}{3} = 1 ( 2 1 − 5 1 ) ⋅ 3 10 = 1
1 2 − 1 5 = 3 10 , 3 10 ⋅ 10 3 = 1. \frac12 - \frac15 = \frac{3}{10}, \qquad \frac{3}{10}\cdot\frac{10}{3} = 1. 2 1 − 5 1 = 10 3 , 10 3 ⋅ 3 10 = 1.
Barem: 3p + 2p.
2. A ( 1 , 0 ) A(1, 0) A ( 1 , 0 ) f ( x ) = x − a f(x) = x - a f ( x ) = x − a
f ( 1 ) = 0 ⟺ 1 − a = 0 ⟺ a = 1 f(1) = 0 \iff 1 - a = 0 \iff a = 1 f ( 1 ) = 0 ⟺ 1 − a = 0 ⟺ a = 1 Barem: 3p + 2p.
3. Ecuația x + 1 = 5 \sqrt{x + 1} = 5 x + 1 = 5
x + 1 = 25 ⟺ x = 24 x + 1 = 25 \iff x = 24 x + 1 = 25 ⟺ x = 24 Barem: 3p + 2p.
4. Probabilitatea ca un număr din M = { 10 , 20 , … , 90 } M = \{10, 20, \dots, 90\} M = { 10 , 20 , … , 90 }
Favorabile: 30 , 60 , 90 30, 60, 90 30 , 60 , 90 p = 1 3 p = \frac13 p = 3 1 Barem: 1p + 2p + 2p.
5. Mijlocul segmentului A B AB A B A ( 3 , 5 ) A(3,5) A ( 3 , 5 ) B ( 7 , 5 ) B(7,5) B ( 7 , 5 )
x M = 5 x_M = 5 x M = 5 y M = 5 y_M = 5 y M = 5 Răspuns: M ( 5 , 5 ) M(5, 5) M ( 5 , 5 ) Barem: 3p + 2p.
6. Dacă x ∈ ( 0 , π 2 ) x \in \left(0, \frac\pi2\right) x ∈ ( 0 , 2 π ) cos x = 5 13 \cos x = \frac{5}{13} cos x = 13 5 tg x = 12 5 \operatorname{tg} x = \frac{12}{5} tg x = 5 12
sin 2 x = 1 − 25 169 = 144 169 \sin^2 x = 1 - \frac{25}{169} = \frac{144}{169} sin 2 x = 1 − 169 25 = 169 144 sin x = 12 13 \sin x = \frac{12}{13} sin x = 13 12
tg x = sin x cos x = 12 13 ⋅ 13 5 = 12 5 . \operatorname{tg} x = \frac{\sin x}{\cos x} = \frac{12}{13}\cdot\frac{13}{5} = \frac{12}{5}. tg x = c o s x s i n x = 13 12 ⋅ 5 13 = 5 12 .
Barem: 3p + 2p.
Subiectul al II-lea (30 de puncte)
1. A = ( 1 1 − 1 0 ) A = \begin{pmatrix} 1 & 1\\ -1 & 0\end{pmatrix} A = ( 1 − 1 1 0 ) B = ( 0 − 1 1 1 ) B = \begin{pmatrix} 0 & -1\\ 1 & 1\end{pmatrix} B = ( 0 1 − 1 1 )
a) det A = 1 \det A = 1 det A = 1
det A = 1 ⋅ 0 − 1 ⋅ ( − 1 ) = 1. \det A = 1\cdot0 - 1\cdot(-1) = 1. det A = 1 ⋅ 0 − 1 ⋅ ( − 1 ) = 1.
Barem: 3p + 2p.
b) B ⋅ B + A = O 2 B\cdot B + A = O_2 B ⋅ B + A = O 2
B ⋅ B = ( 0 − 1 0 − 1 0 + 1 − 1 + 1 ) = ( − 1 − 1 1 0 ) = − A , B\cdot B = \begin{pmatrix} 0 - 1 & 0 - 1\\ 0 + 1 & -1 + 1\end{pmatrix} = \begin{pmatrix} -1 & -1\\ 1 & 0\end{pmatrix} = -A, B ⋅ B = ( 0 − 1 0 + 1 0 − 1 − 1 + 1 ) = ( − 1 1 − 1 0 ) = − A ,
deci B ⋅ B + A = O 2 B\cdot B + A = O_2 B ⋅ B + A = O 2 Barem: 3p + 2p.
c) x , y x, y x , y A + B = ( 2 x 0 0 4 y ) A + B = \begin{pmatrix} 2^x & 0\\ 0 & 4^y\end{pmatrix} A + B = ( 2 x 0 0 4 y )
A + B = ( 1 0 0 1 ) A + B = \begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix} A + B = ( 1 0 0 1 ) 2 x = 1 2^x = 1 2 x = 1 4 y = 1 4^y = 1 4 y = 1
Răspuns: x = 0 x = 0 x = 0 y = 0 y = 0 y = 0 Barem: 2p + 3p.
2. f = X 3 − 2 X 2 − 2 X + 1 f = X^3 - 2X^2 - 2X + 1 f = X 3 − 2 X 2 − 2 X + 1
a) f ( 1 ) = − 2 f(1) = -2 f ( 1 ) = − 2
1 − 2 − 2 + 1 = − 2. 1 - 2 - 2 + 1 = -2. 1 − 2 − 2 + 1 = − 2.
Barem: 3p + 2p.
b) Câtul și restul împărțirii la X + 1 X + 1 X + 1
Horner cu − 1 -1 − 1
1 1 1 − 2 -2 − 2 − 2 -2 − 2 1 1 1 − 1 -1 − 1 1 1 1 − 3 -3 − 3 1 1 1 0 0 0
Răspuns: câtul X 2 − 3 X + 1 X^2 - 3X + 1 X 2 − 3 X + 1 0 0 0 Barem: 3p + 2p.
c) ( x 2 + x 3 ) ( x 3 + x 1 ) ( x 1 + x 2 ) = − 3 (x_2 + x_3)(x_3 + x_1)(x_1 + x_2) = -3 ( x 2 + x 3 ) ( x 3 + x 1 ) ( x 1 + x 2 ) = − 3
x 1 + x 2 + x 3 = 2 x_1 + x_2 + x_3 = 2 x 1 + x 2 + x 3 = 2 x 2 + x 3 = 2 − x 1 x_2 + x_3 = 2 - x_1 x 2 + x 3 = 2 − x 1
( 2 − x 1 ) ( 2 − x 2 ) ( 2 − x 3 ) = f ( 2 ) = 8 − 8 − 4 + 1 = − 3. ■ (2 - x_1)(2 - x_2)(2 - x_3) = f(2) = 8 - 8 - 4 + 1 = -3. \ \blacksquare ( 2 − x 1 ) ( 2 − x 2 ) ( 2 − x 3 ) = f ( 2 ) = 8 − 8 − 4 + 1 = − 3. ■
Barem: 2p + 3p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = − x 3 + 3 x + 2 f(x) = -x^3 + 3x + 2 f ( x ) = − x 3 + 3 x + 2
a) f ′ ( x ) = 3 ( 1 − x ) ( 1 + x ) f'(x) = 3(1 - x)(1 + x) f ′ ( x ) = 3 ( 1 − x ) ( 1 + x )
f ′ ( x ) = − 3 x 2 + 3 = 3 ( 1 − x 2 ) = 3 ( 1 − x ) ( 1 + x ) . f'(x) = -3x^2 + 3 = 3(1 - x^2) = 3(1 - x)(1 + x). f ′ ( x ) = − 3 x 2 + 3 = 3 ( 1 − x 2 ) = 3 ( 1 − x ) ( 1 + x ) .
Barem: 3p + 2p.
b) lim x → 2 f ( x ) x − 2 = − 9 \lim\limits_{x\to2}\dfrac{f(x)}{x - 2} = -9 x → 2 lim x − 2 f ( x ) = − 9
f ( 2 ) = − 8 + 6 + 2 = 0 f(2) = -8 + 6 + 2 = 0 f ( 2 ) = − 8 + 6 + 2 = 0 f ′ ( 2 ) = 3 ⋅ ( − 1 ) ⋅ 3 = − 9 f'(2) = 3\cdot(-1)\cdot3 = -9 f ′ ( 2 ) = 3 ⋅ ( − 1 ) ⋅ 3 = − 9 Barem: 3p + 2p.
c) f ( x ) ≤ 4 f(x) \le 4 f ( x ) ≤ 4 x ∈ [ − 1 , + ∞ ) x \in [-1, +\infty) x ∈ [ − 1 , + ∞ )
f ′ ( x ) = 0 ⟺ x = ± 1 f'(x) = 0 \iff x = \pm1 f ′ ( x ) = 0 ⟺ x = ± 1 [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] f ′ ≥ 0 f' \ge 0 f ′ ≥ 0 [ 1 , + ∞ ) [1, +\infty) [ 1 , + ∞ ) f ′ ≤ 0 f' \le 0 f ′ ≤ 0 [ − 1 , + ∞ ) [-1, +\infty) [ − 1 , + ∞ ) f ( 1 ) = − 1 + 3 + 2 = 4 f(1) = -1 + 3 + 2 = 4 f ( 1 ) = − 1 + 3 + 2 = 4 ■ \blacksquare ■ Barem: 2p + 1p + 1p + 1p.
2. f ( x ) = x + 2 f(x) = x + 2 f ( x ) = x + 2
a) ∫ − 1 1 ( f ( x ) − 2 ) d x = 0 \displaystyle\int_{-1}^1(f(x) - 2)\,dx = 0 ∫ − 1 1 ( f ( x ) − 2 ) d x = 0
∫ − 1 1 x d x = x 2 2 ∣ − 1 1 = 1 2 − 1 2 = 0. \int_{-1}^1 x\,dx = \frac{x^2}{2}\Big|_{-1}^1 = \frac12 - \frac12 = 0. ∫ − 1 1 x d x = 2 x 2 − 1 1 = 2 1 − 2 1 = 0.
Barem: 2p + 3p.
b) ∫ 0 1 e x f ( x ) d x = 2 e − 1 \displaystyle\int_0^1 e^xf(x)\,dx = 2e - 1 ∫ 0 1 e x f ( x ) d x = 2 e − 1
Prin părți:
∫ 0 1 e x ( x + 2 ) d x = e x ( x + 2 ) ∣ 0 1 − ∫ 0 1 e x d x = ( 3 e − 2 ) − ( e − 1 ) = 2 e − 1. \int_0^1 e^x(x+2)\,dx = e^x(x+2)\Big|_0^1 - \int_0^1 e^x\,dx = (3e - 2) - (e - 1) = 2e - 1. ∫ 0 1 e x ( x + 2 ) d x = e x ( x + 2 ) 0 1 − ∫ 0 1 e x d x = ( 3 e − 2 ) − ( e − 1 ) = 2 e − 1.
Barem: 3p + 2p.
c) a a a ∫ 0 a f ( x ) d x = ∫ 0 6 − a ( f ( x ) − 4 ) d x \displaystyle\int_0^a f(x)\,dx = \int_0^{6-a}(f(x) - 4)\,dx ∫ 0 a f ( x ) d x = ∫ 0 6 − a ( f ( x ) − 4 ) d x
∫ 0 a ( x + 2 ) d x = a 2 2 + 2 a , ∫ 0 6 − a ( x − 2 ) d x = ( 6 − a ) 2 2 − 2 ( 6 − a ) . \int_0^a(x + 2)\,dx = \frac{a^2}{2} + 2a, \qquad \int_0^{6-a}(x - 2)\,dx = \frac{(6-a)^2}{2} - 2(6 - a). ∫ 0 a ( x + 2 ) d x = 2 a 2 + 2 a , ∫ 0 6 − a ( x − 2 ) d x = 2 ( 6 − a ) 2 − 2 ( 6 − a ) .
Înmulțind cu 2: a 2 + 4 a = 36 − 12 a + a 2 − 24 + 4 a ⟺ 4 a = 12 − 8 a ⟺ a = 1 a^2 + 4a = 36 - 12a + a^2 - 24 + 4a \iff 4a = 12 - 8a \iff a = 1 a 2 + 4 a = 36 − 12 a + a 2 − 24 + 4 a ⟺ 4 a = 12 − 8 a ⟺ a = 1
Răspuns: a = 1 a = 1 a = 1 Barem: 2p + 2p + 1p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 a = 1 a = 1 a = 1 I.3 x = 24 x = 24 x = 24 I.4 p = 1 3 p = \frac13 p = 3 1 I.5 M ( 5 , 5 ) M(5, 5) M ( 5 , 5 ) II.1.c x = y = 0 x = y = 0 x = y = 0 II.2.b cât X 2 − 3 X + 1 X^2 - 3X + 1 X 2 − 3 X + 1 0 0 0 III.1.c maxim f ( 1 ) = 4 f(1) = 4 f ( 1 ) = 4 III.2.b 2 e − 1 2e - 1 2 e − 1 III.2.c a = 1 a = 1 a = 1
Ce merită reținut din varianta 8
( x 2 + x 3 ) ( x 3 + x 1 ) ( x 1 + x 2 ) = f ( S ) (x_2 + x_3)(x_3 + x_1)(x_1 + x_2) = f(S) ( x 2 + x 3 ) ( x 3 + x 1 ) ( x 1 + x 2 ) = f ( S ) S = x 1 + x 2 + x 3 S = x_1 + x_2 + x_3 S = x 1 + x 2 + x 3 f f f Integrala unei funcții impare pe un interval simetric este 0 0 0 ∫ e x ( x + c ) d x = e x ( x + c − 1 ) + C \int e^x(x + c)\,dx = e^x(x + c - 1) + C ∫ e x ( x + c ) d x = e x ( x + c − 1 ) + C Maximul pe un interval se citește din tabelul de variație.