Rezolvarea completă a subiectului de Bacalaureat 2019 la Matematică M_tehnologic , sesiunea specială, 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 − 1 3 + 1 4 ) : ( 1 − 1 12 ) = 1 \left(1 - \frac13 + \frac14\right) : \left(1 - \frac1{12}\right) = 1 ( 1 − 3 1 + 4 1 ) : ( 1 − 12 1 ) = 1
1 − 1 3 + 1 4 = 12 − 4 + 3 12 = 11 12 1 - \frac13 + \frac14 = \frac{12 - 4 + 3}{12} = \frac{11}{12} 1 − 3 1 + 4 1 = 12 12 − 4 + 3 = 12 11 1 − 1 12 = 11 12 1 - \frac1{12} = \frac{11}{12} 1 − 12 1 = 12 11 1 1 1 ■ \blacksquare ■ Barem: 3p, 2p.
2. f ( x ) = x 2 + 4 f(x) = x^2 + 4 f ( x ) = x 2 + 4 f ( − 2 ) + f ( 2 ) = 4 f ( 0 ) f(-2) + f(2) = 4f(0) f ( − 2 ) + f ( 2 ) = 4 f ( 0 )
8 + 8 = 16 = 4 ⋅ 4 8 + 8 = 16 = 4\cdot 4 8 + 8 = 16 = 4 ⋅ 4 ■ \blacksquare ■ Barem: 2p, 3p.
3. Ecuația log 8 ( x 2 − 27 ) = log 8 ( x − 3 ) 2 \log_8(x^2 - 27) = \log_8(x - 3)^2 log 8 ( x 2 − 27 ) = log 8 ( x − 3 ) 2
x 2 − 27 = x 2 − 6 x + 9 ⟺ 6 x = 36 ⟺ x = 6 x^2 - 27 = x^2 - 6x + 9 \iff 6x = 36 \iff x = 6 x 2 − 27 = x 2 − 6 x + 9 ⟺ 6 x = 36 ⟺ x = 6 36 − 27 = 9 = 3 2 > 0 36 - 27 = 9 = 3^2 > 0 36 − 27 = 9 = 3 2 > 0
Răspuns: x = 6 x = 6 x = 6 Barem: 3p, 2p.
4. Probabilitatea ca un număr din { 10 , … , 19 } \{10, \dots, 19\} { 10 , … , 19 }
p = 5 10 = 1 2 p = \frac{5}{10} = \frac12 p = 10 5 = 2 1 Barem: 2p, 2p, 1p.
5. A ( 4 , 3 ) A(4,3) A ( 4 , 3 ) B ( 8 , 3 ) B(8,3) B ( 8 , 3 ) B B B A C AC A C
8 = 4 + x C 2 ⇒ x C = 12 8 = \frac{4 + x_C}{2} \Rightarrow x_C = 12 8 = 2 4 + x C ⇒ x C = 12 3 = 3 + y C 2 ⇒ y C = 3 3 = \frac{3 + y_C}{2} \Rightarrow y_C = 3 3 = 2 3 + y C ⇒ y C = 3
Răspuns: C ( 12 , 3 ) C(12, 3) C ( 12 , 3 ) Barem: 3p, 2p.
6. cos 2 30 ∘ + sin 2 60 ∘ − 2 cos 30 ∘ sin 60 ∘ = 0 \cos^230^\circ + \sin^260^\circ - 2\cos30^\circ\sin60^\circ = 0 cos 2 3 0 ∘ + sin 2 6 0 ∘ − 2 cos 3 0 ∘ sin 6 0 ∘ = 0
Expresia este ( cos 30 ∘ − sin 60 ∘ ) 2 = ( 3 2 − 3 2 ) 2 = 0 (\cos30^\circ - \sin60^\circ)^2 = \left(\frac{\sqrt3}2 - \frac{\sqrt3}2\right)^2 = 0 ( cos 3 0 ∘ − sin 6 0 ∘ ) 2 = ( 2 3 − 2 3 ) 2 = 0 ■ \blacksquare ■ Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. M = ( 2 1 1 2 ) M = \begin{pmatrix}2 & 1\\ 1 & 2\end{pmatrix} M = ( 2 1 1 2 ) A ( a ) = ( a 1 3 2 ) A(a) = \begin{pmatrix}a & 1\\ 3 & 2\end{pmatrix} A ( a ) = ( a 3 1 2 )
a) det M = 4 − 1 = 3 \det M = 4 - 1 = 3 det M = 4 − 1 = 3 ■ \blacksquare ■ Barem: 3p, 2p.
b) A ( a ) 2 = ( a 2 + 3 a + 2 3 a + 6 7 ) , 4 A ( a ) − I 2 = ( 4 a − 1 4 12 7 ) . A(a)^2 = \begin{pmatrix}a^2 + 3 & a + 2\\ 3a + 6 & 7\end{pmatrix},\qquad 4A(a) - I_2 = \begin{pmatrix}4a - 1 & 4\\ 12 & 7\end{pmatrix}. A ( a ) 2 = ( a 2 + 3 3 a + 6 a + 2 7 ) , 4 A ( a ) − I 2 = ( 4 a − 1 12 4 7 ) .
a + 2 = 4 ⇒ a = 2 a + 2 = 4 \Rightarrow a = 2 a + 2 = 4 ⇒ a = 2 a 2 + 3 = 7 = 4 a − 1 a^2 + 3 = 7 = 4a - 1 a 2 + 3 = 7 = 4 a − 1 3 a + 6 = 12 3a + 6 = 12 3 a + 6 = 12
Răspuns: a = 2 a = 2 a = 2 Barem: 2p, 3p.
c) a A ( a ) + M = ( a 2 + 2 a + 1 3 a + 1 2 a + 2 ) , aA(a) + M = \begin{pmatrix}a^2 + 2 & a + 1\\ 3a + 1 & 2a + 2\end{pmatrix}, a A ( a ) + M = ( a 2 + 2 3 a + 1 a + 1 2 a + 2 ) , det = ( a 2 + 2 ) ⋅ 2 ( a + 1 ) − ( a + 1 ) ( 3 a + 1 ) = ( a + 1 ) ( 2 a 2 − 3 a + 3 ) . \det = (a^2 + 2)\cdot 2(a + 1) - (a + 1)(3a + 1) = (a + 1)(2a^2 - 3a + 3). det = ( a 2 + 2 ) ⋅ 2 ( a + 1 ) − ( a + 1 ) ( 3 a + 1 ) = ( a + 1 ) ( 2 a 2 − 3 a + 3 ) .
2 a 2 − 3 a + 3 2a^2 - 3a + 3 2 a 2 − 3 a + 3 Δ = 9 − 24 < 0 \Delta = 9 - 24 < 0 Δ = 9 − 24 < 0 a = − 1 a = -1 a = − 1
Răspuns: a = − 1 a = -1 a = − 1 Barem: 3p, 2p.
2. f = X 3 − 4 X 2 + m X + 2 f = X^3 - 4X^2 + mX + 2 f = X 3 − 4 X 2 + m X + 2
a) f ( 2 ) = 8 − 16 + 2 m + 2 = 2 m − 6 f(2) = 8 - 16 + 2m + 2 = 2m - 6 f ( 2 ) = 8 − 16 + 2 m + 2 = 2 m − 6 ■ \blacksquare ■ Barem: 3p, 2p.
b) E = x 1 x 2 x 3 ( x 1 + x 2 + x 3 ) = ( − 2 ) ⋅ 4 = − 8 ∈ Z E = x_1x_2x_3(x_1 + x_2 + x_3) = (-2)\cdot 4 = -8 \in \mathbb Z E = x 1 x 2 x 3 ( x 1 + x 2 + x 3 ) = ( − 2 ) ⋅ 4 = − 8 ∈ Z s 1 = 4 s_1 = 4 s 1 = 4 s 3 = − 2 s_3 = -2 s 3 = − 2 m m m ■ \blacksquare ■ Barem: 2p, 3p.
c) Pentru m = 3 m = 3 m = 3 f ( 2 ) = 0 f(2) = 0 f ( 2 ) = 0 f = ( X − 2 ) ( X 2 − 2 X − 1 ) f = (X - 2)(X^2 - 2X - 1) f = ( X − 2 ) ( X 2 − 2 X − 1 ) x 2 − 2 x − 1 = 0 ⇒ x = 1 ± 2 x^2 - 2x - 1 = 0 \Rightarrow x = 1 \pm\sqrt2 x 2 − 2 x − 1 = 0 ⇒ x = 1 ± 2
Răspuns: 1 − 2 1 - \sqrt2 1 − 2 2 2 2 1 + 2 1 + \sqrt2 1 + 2 Barem: 2p, 3p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = 7 x 3 − 5 x 2 + x + 1 f(x) = 7x^3 - 5x^2 + x + 1 f ( x ) = 7 x 3 − 5 x 2 + x + 1
a) f ′ ( x ) = 21 x 2 − 10 x + 1 = ( 3 x − 1 ) ( 7 x − 1 ) f'(x) = 21x^2 - 10x + 1 = (3x - 1)(7x - 1) f ′ ( x ) = 21 x 2 − 10 x + 1 = ( 3 x − 1 ) ( 7 x − 1 ) ■ \blacksquare ■ Barem: 2p, 3p.
b) lim x → + ∞ x f ′ ( x ) f ( x ) = lim x → + ∞ 21 x 3 − 10 x 2 + x 7 x 3 − 5 x 2 + x + 1 = 21 7 = 3. \lim_{x\to+\infty}\frac{xf'(x)}{f(x)} = \lim_{x\to+\infty}\frac{21x^3 - 10x^2 + x}{7x^3 - 5x^2 + x + 1} = \frac{21}{7} = 3. lim x → + ∞ f ( x ) x f ′ ( x ) = lim x → + ∞ 7 x 3 − 5 x 2 + x + 1 21 x 3 − 10 x 2 + x = 7 21 = 3.
Răspuns: 3 3 3 Barem: 2p, 3p.
c) f ′ ≥ 0 f' \ge 0 f ′ ≥ 0 ( − ∞ , 1 7 ] \left(-\infty, \frac17\right] ( − ∞ , 7 1 ] f ′ ≤ 0 f' \le 0 f ′ ≤ 0 [ 1 7 , 1 3 ] \left[\frac17, \frac13\right] [ 7 1 , 3 1 ] ( − ∞ , 1 3 ] \left(-\infty, \frac13\right] ( − ∞ , 3 1 ]
f ( 1 7 ) = 7 343 − 5 49 + 1 7 + 1 = 1 − 5 + 7 + 49 49 = 52 49 . ■ f\left(\tfrac17\right) = \frac{7}{343} - \frac{5}{49} + \frac17 + 1 = \frac{1 - 5 + 7 + 49}{49} = \frac{52}{49}.\ \blacksquare f ( 7 1 ) = 343 7 − 49 5 + 7 1 + 1 = 49 1 − 5 + 7 + 49 = 49 52 . ■
Barem: 2p, 3p.
2. f ( x ) = { x 2 + 8 x − 2 , x ≤ 0 x − 2 , x > 0 f(x) = \begin{cases}x^2 + 8x - 2, & x \le 0\\ x - 2, & x > 0\end{cases} f ( x ) = { x 2 + 8 x − 2 , x − 2 , x ≤ 0 x > 0
a) Pe [ 1 , 2 ] [1, 2] [ 1 , 2 ] f ( x ) = x − 2 f(x) = x - 2 f ( x ) = x − 2 ∫ 1 2 ( x − 2 ) d x = ( x 2 2 − 2 x ) ∣ 1 2 = − 2 + 3 2 = − 1 2 \int_1^2(x - 2)\,dx = \left(\frac{x^2}{2} - 2x\right)\Big|_1^2 = -2 + \frac32 = -\frac12 ∫ 1 2 ( x − 2 ) d x = ( 2 x 2 − 2 x ) 1 2 = − 2 + 2 3 = − 2 1 ■ \blacksquare ■ Barem: 3p, 2p.
b) Limitele laterale în 0 0 0 − 2 -2 − 2 − 2 -2 − 2 f ( 0 ) = − 2 f(0) = -2 f ( 0 ) = − 2 f f f 0 0 0 ( − ∞ , 0 ) (-\infty, 0) ( − ∞ , 0 ) ( 0 , + ∞ ) (0, +\infty) ( 0 , + ∞ ) R \mathbb R R ■ \blacksquare ■ Barem: 3p, 2p.
c) Pe [ − 1 , 0 ] [-1, 0] [ − 1 , 0 ] x 2 + 8 x − 2 = ( x + 4 ) 2 − 18 ≤ 16 − 18 < 0 x^2 + 8x - 2 = (x + 4)^2 - 18 \le 16 - 18 < 0 x 2 + 8 x − 2 = ( x + 4 ) 2 − 18 ≤ 16 − 18 < 0
A = ∫ − 1 0 ( − x 2 − 8 x + 2 ) d x = ( − x 3 3 − 4 x 2 + 2 x ) ∣ − 1 0 = 0 − ( 1 3 − 4 − 2 ) = 17 3 . ■ \mathcal A = \int_{-1}^0(-x^2 - 8x + 2)\,dx = \left(-\frac{x^3}{3} - 4x^2 + 2x\right)\Big|_{-1}^0 = 0 - \left(\frac13 - 4 - 2\right) = \frac{17}{3}.\ \blacksquare A = ∫ − 1 0 ( − x 2 − 8 x + 2 ) d x = ( − 3 x 3 − 4 x 2 + 2 x ) − 1 0 = 0 − ( 3 1 − 4 − 2 ) = 3 17 . ■
Barem: 2p, 3p.
Rezumatul răspunsurilor
Cerință Răspuns I.3 x = 6 x = 6 x = 6 I.4 p = 1 2 p = \tfrac12 p = 2 1 I.5 C ( 12 , 3 ) C(12, 3) C ( 12 , 3 ) II.1.b a = 2 a = 2 a = 2 II.1.c a = − 1 a = -1 a = − 1 II.2.b E = − 8 E = -8 E = − 8 II.2.c 1 ± 2 1 \pm\sqrt2 1 ± 2 2 2 2 III.1.b 3 3 3
Ce merită reținut din varianta 8
Existența primitivelor: o funcție continuă pe un interval admite primitive — verifică continuitatea în punctul de racord.Aria pentru f < 0 f < 0 f < 0 A = − ∫ f \mathcal A = -\int f A = − ∫ f Expresii de tip x 1 x 2 x 3 ( x 1 + x 2 + x 3 ) x_1x_2x_3(x_1 + x_2 + x_3) x 1 x 2 x 3 ( x 1 + x 2 + x 3 ) se calculează direct cu Viète.