Rezolvarea completă a subiectului de Bacalaureat 2019 la Matematică M_tehnologic , sesiunea august–septembrie, varianta 7 . 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 ( 3 2 − 2 3 ) : ( 3 2 + 2 3 ) ⋅ 13 5 = 1 \left(\frac32 - \frac23\right) : \left(\frac32 + \frac23\right)\cdot\frac{13}{5} = 1 ( 2 3 − 3 2 ) : ( 2 3 + 3 2 ) ⋅ 5 13 = 1
3 2 − 2 3 = 5 6 \frac32 - \frac23 = \frac56 2 3 − 3 2 = 6 5 3 2 + 2 3 = 13 6 \frac32 + \frac23 = \frac{13}{6} 2 3 + 3 2 = 6 13
5 6 ⋅ 6 13 ⋅ 13 5 = 1. ■ \frac56\cdot\frac{6}{13}\cdot\frac{13}{5} = 1.\ \blacksquare 6 5 ⋅ 13 6 ⋅ 5 13 = 1. ■
Barem: 3p, 2p.
2. f ( x ) = 2 x − 4 f(x) = 2x - 4 f ( x ) = 2 x − 4 f ( m + 1 ) = m f(m + 1) = m f ( m + 1 ) = m
2 ( m + 1 ) − 4 = 2 m − 2 = m ⟺ m = 2 2(m + 1) - 4 = 2m - 2 = m \iff m = 2 2 ( m + 1 ) − 4 = 2 m − 2 = m ⟺ m = 2 Barem: 3p, 2p.
3. Ecuația log 7 ( 2 x + 3 ) = log 7 9 \log_7(2x + 3) = \log_7 9 log 7 ( 2 x + 3 ) = log 7 9
2 x + 3 = 9 ⟺ x = 3 2x + 3 = 9 \iff x = 3 2 x + 3 = 9 ⟺ x = 3 Barem: 3p, 2p.
4. Probabilitatea ca un număr din { 10 , 20 , … , 90 } \{10, 20, \dots, 90\} { 10 , 20 , … , 90 }
9 cazuri posibile; multipli de 3: 30 , 60 , 90 30, 60, 90 30 , 60 , 90 p = 3 9 = 1 3 p = \frac39 = \frac13 p = 9 3 = 3 1 Barem: 2p, 2p, 1p.
5. Aria triunghiului M N P MNP M N P M ( 4 , 1 ) M(4,1) M ( 4 , 1 ) N ( 1 , 5 ) N(1,5) N ( 1 , 5 ) P ( 4 , 5 ) P(4,5) P ( 4 , 5 )
M P MP M P x = 4 x = 4 x = 4 N P NP N P y = 5 y = 5 y = 5 P P P M P = 4 MP = 4 M P = 4 N P = 3 NP = 3 N P = 3
A = 4 ⋅ 3 2 = 6. \mathcal A = \frac{4\cdot 3}{2} = 6. A = 2 4 ⋅ 3 = 6.
Barem: 2p, 3p.
6. Arătați că 1 3 sin 60 ∘ + sin 2 45 ∘ = 1 \frac{1}{\sqrt3}\sin60^\circ + \sin^245^\circ = 1 3 1 sin 6 0 ∘ + sin 2 4 5 ∘ = 1
1 3 ⋅ 3 2 + ( 2 2 ) 2 = 1 2 + 1 2 = 1 \frac1{\sqrt3}\cdot\frac{\sqrt3}2 + \left(\frac{\sqrt2}2\right)^2 = \frac12 + \frac12 = 1 3 1 ⋅ 2 3 + ( 2 2 ) 2 = 2 1 + 2 1 = 1 ■ \blacksquare ■ Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. A = ( 1 2 3 4 ) A = \begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix} A = ( 1 3 2 4 ) M ( a ) = ( 1 + a − a a 1 − a ) M(a) = \begin{pmatrix}1 + a & -a\\ a & 1 - a\end{pmatrix} M ( a ) = ( 1 + a a − a 1 − a )
a) det A = 4 − 6 = − 2 \det A = 4 - 6 = -2 det A = 4 − 6 = − 2 ■ \blacksquare ■ Barem: 3p, 2p.
b) M ( a ) M ( b ) = ( ( 1 + a ) ( 1 + b ) − a b − b ( 1 + a ) − a ( 1 − b ) a ( 1 + b ) + b ( 1 − a ) − a b + ( 1 − a ) ( 1 − b ) ) = ( 1 + a + b − ( a + b ) a + b 1 − ( a + b ) ) = M ( a + b ) . ■ M(a)M(b) = \begin{pmatrix}(1+a)(1+b) - ab & -b(1+a) - a(1-b)\\ a(1+b) + b(1 - a) & -ab + (1-a)(1-b)\end{pmatrix} = \begin{pmatrix}1 + a + b & -(a + b)\\ a + b & 1 - (a+b)\end{pmatrix} = M(a + b).\ \blacksquare M ( a ) M ( b ) = ( ( 1 + a ) ( 1 + b ) − ab a ( 1 + b ) + b ( 1 − a ) − b ( 1 + a ) − a ( 1 − b ) − ab + ( 1 − a ) ( 1 − b ) ) = ( 1 + a + b a + b − ( a + b ) 1 − ( a + b ) ) = M ( a + b ) . ■
Barem: 3p, 2p.
c) Din b), M ( a ) M ( − a ) = M ( 0 ) = I 2 M(a)M(-a) = M(0) = I_2 M ( a ) M ( − a ) = M ( 0 ) = I 2 M ( a ) − 1 = M ( − a ) M(a)^{-1} = M(-a) M ( a ) − 1 = M ( − a )
X = M ( − 1 ) A M ( − 2 ) = ( 0 1 − 1 2 ) ( 1 2 3 4 ) ( − 1 2 − 2 3 ) = ( 3 4 5 6 ) ( − 1 2 − 2 3 ) = ( − 11 18 − 17 28 ) . X = M(-1)\,A\,M(-2) = \begin{pmatrix}0 & 1\\ -1 & 2\end{pmatrix}\begin{pmatrix}1 & 2\\ 3 & 4\end{pmatrix}\begin{pmatrix}-1 & 2\\ -2 & 3\end{pmatrix} = \begin{pmatrix}3 & 4\\ 5 & 6\end{pmatrix}\begin{pmatrix}-1 & 2\\ -2 & 3\end{pmatrix} = \begin{pmatrix}-11 & 18\\ -17 & 28\end{pmatrix}. X = M ( − 1 ) A M ( − 2 ) = ( 0 − 1 1 2 ) ( 1 3 2 4 ) ( − 1 − 2 2 3 ) = ( 3 5 4 6 ) ( − 1 − 2 2 3 ) = ( − 11 − 17 18 28 ) .
Barem: 2p, 3p.
2. f = 2 X 3 − 4 X 2 + 4 X − 3 f = 2X^3 - 4X^2 + 4X - 3 f = 2 X 3 − 4 X 2 + 4 X − 3
a) f ( 0 ) = − 3 f(0) = -3 f ( 0 ) = − 3 ■ \blacksquare ■ Barem: 3p, 2p.
b) Viète: x 1 + x 2 + x 3 = 2 x_1 + x_2 + x_3 = 2 x 1 + x 2 + x 3 = 2 x 1 x 2 + x 2 x 3 + x 3 x 1 = 2 x_1x_2 + x_2x_3 + x_3x_1 = 2 x 1 x 2 + x 2 x 3 + x 3 x 1 = 2 x 1 x 2 x 3 = 3 2 x_1x_2x_3 = \frac32 x 1 x 2 x 3 = 2 3
a = 3 ⋅ x 1 x 2 + x 2 x 3 + x 3 x 1 x 1 x 2 x 3 = 3 ⋅ 2 3 / 2 = 4 ∈ N . ■ a = 3\cdot\frac{x_1x_2 + x_2x_3 + x_3x_1}{x_1x_2x_3} = 3\cdot\frac{2}{3/2} = 4 \in \mathbb N.\ \blacksquare a = 3 ⋅ x 1 x 2 x 3 x 1 x 2 + x 2 x 3 + x 3 x 1 = 3 ⋅ 3/2 2 = 4 ∈ N . ■
Barem: 2p, 3p.
c) x 1 2 + x 2 2 + x 3 2 = 2 2 − 2 ⋅ 2 = 0 x_1^2 + x_2^2 + x_3^2 = 2^2 - 2\cdot 2 = 0 x 1 2 + x 2 2 + x 3 2 = 2 2 − 2 ⋅ 2 = 0 x 1 = x 2 = x 3 = 0 x_1 = x_2 = x_3 = 0 x 1 = x 2 = x 3 = 0 f ( 0 ) = − 3 ≠ 0 f(0) = -3 \ne 0 f ( 0 ) = − 3 = 0 ■ \blacksquare ■
Barem: 2p, 3p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x x 6 + 5 f(x) = \dfrac{x}{x^6 + 5} f ( x ) = x 6 + 5 x
a) f ′ ( x ) = x 6 + 5 − 6 x 6 ( x 6 + 5 ) 2 = 5 ( 1 − x 6 ) ( x 6 + 5 ) 2 = 5 ( 1 − x 3 ) ( 1 + x 3 ) ( x 6 + 5 ) 2 . ■ f'(x) = \frac{x^6 + 5 - 6x^6}{(x^6+5)^2} = \frac{5(1 - x^6)}{(x^6+5)^2} = \frac{5(1 - x^3)(1 + x^3)}{(x^6+5)^2}.\ \blacksquare f ′ ( x ) = ( x 6 + 5 ) 2 x 6 + 5 − 6 x 6 = ( x 6 + 5 ) 2 5 ( 1 − x 6 ) = ( x 6 + 5 ) 2 5 ( 1 − x 3 ) ( 1 + x 3 ) . ■
Barem: 3p, 2p.
b) f ( 0 ) = 0 f(0) = 0 f ( 0 ) = 0 f ′ ( 0 ) = 5 25 = 1 5 f'(0) = \frac{5}{25} = \frac15 f ′ ( 0 ) = 25 5 = 5 1 y = 1 5 x y = \frac15x y = 5 1 x Barem: 2p, 3p.
c) f ′ ≤ 0 f' \le 0 f ′ ≤ 0 ( − ∞ , − 1 ] (-\infty, -1] ( − ∞ , − 1 ] f ′ ≥ 0 f' \ge 0 f ′ ≥ 0 [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] f ′ ≤ 0 f' \le 0 f ′ ≤ 0 [ 1 , + ∞ ) [1, +\infty) [ 1 , + ∞ ) lim x → ± ∞ f ( x ) = 0 \lim\limits_{x\to\pm\infty}f(x) = 0 x → ± ∞ lim f ( x ) = 0 f ( − 1 ) = − 1 6 f(-1) = -\frac16 f ( − 1 ) = − 6 1 f ( 1 ) = 1 6 f(1) = \frac16 f ( 1 ) = 6 1
Răspuns: I m f = [ − 1 6 , 1 6 ] \mathrm{Im}\,f = \left[-\frac16, \frac16\right] Im f = [ − 6 1 , 6 1 ] Barem: 2p, 3p.
2. f ( x ) = ( x − 1 ) e x f(x) = (x - 1)e^x f ( x ) = ( x − 1 ) e x
a) ∫ 0 1 f ( x ) e x d x = ∫ 0 1 ( x − 1 ) d x = 1 2 − 1 = − 1 2 \int_0^1\frac{f(x)}{e^x}dx = \int_0^1(x - 1)\,dx = \frac12 - 1 = -\frac12 ∫ 0 1 e x f ( x ) d x = ∫ 0 1 ( x − 1 ) d x = 2 1 − 1 = − 2 1 ■ \blacksquare ■ Barem: 3p, 2p.
b) F ′ ( x ) = e x + ( x − 2 ) e x = ( x − 1 ) e x = f ( x ) F'(x) = e^x + (x - 2)e^x = (x - 1)e^x = f(x) F ′ ( x ) = e x + ( x − 2 ) e x = ( x − 1 ) e x = f ( x ) ■ \blacksquare ■ Barem: 3p, 2p.
c) f 2 f ′ = ( f 3 3 ) ′ f^2f' = \left(\frac{f^3}{3}\right)' f 2 f ′ = ( 3 f 3 ) ′
∫ 0 1 f 2 ( x ) f ′ ( x ) d x = f 3 ( 1 ) − f 3 ( 0 ) 3 = 0 − ( − 1 ) 3 3 = 1 3 . \int_0^1 f^2(x)f'(x)\,dx = \frac{f^3(1) - f^3(0)}{3} = \frac{0 - (-1)^3}{3} = \frac13. ∫ 0 1 f 2 ( x ) f ′ ( x ) d x = 3 f 3 ( 1 ) − f 3 ( 0 ) = 3 0 − ( − 1 ) 3 = 3 1 .
Răspuns: 1 3 \dfrac13 3 1 Barem: 3p, 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 m = 2 m = 2 m = 2 I.3 x = 3 x = 3 x = 3 I.4 p = 1 3 p = \tfrac13 p = 3 1 I.5 6 6 6 II.1.c X = ( − 11 18 − 17 28 ) X = \begin{pmatrix}-11&18\\-17&28\end{pmatrix} X = ( − 11 − 17 18 28 ) II.2.b a = 4 a = 4 a = 4 III.1.b y = 1 5 x y = \tfrac15x y = 5 1 x III.1.c [ − 1 6 , 1 6 ] [-\tfrac16, \tfrac16] [ − 6 1 , 6 1 ] III.2.c 1 3 \tfrac13 3 1
Ce merită reținut din varianta 7
M ( a ) M ( b ) = M ( a + b ) M(a)M(b) = M(a+b) M ( a ) M ( b ) = M ( a + b ) M ( a ) − 1 = M ( − a ) M(a)^{-1} = M(-a) M ( a ) − 1 = M ( − a ) Suma pătratelor rădăcinilor = 0 = 0 = 0 cu rădăcini reale implică toate nule.∫ f 2 f ′ = f 3 3 \int f^2f' = \frac{f^3}{3} ∫ f 2 f ′ = 3 f 3