Rezolvarea completă a subiectului de Bacalaureat 2019 la Matematică M_mate-info , sesiunea iunie–iulie, varianta 6 . 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. n = ( 3 − i 2 ) ( 3 + i 2 ) n = (3 - i\sqrt2)(3 + i\sqrt2) n = ( 3 − i 2 ) ( 3 + i 2 )
n = 3 2 − ( i 2 ) 2 = 9 − 2 i 2 = 9 + 2 = 11 ∈ Z . ■ n = 3^2 - (i\sqrt2)^2 = 9 - 2i^2 = 9 + 2 = 11 \in \mathbb Z.\ \blacksquare n = 3 2 − ( i 2 ) 2 = 9 − 2 i 2 = 9 + 2 = 11 ∈ Z . ■
Barem: 2p, 3p.
2. A ( a , 3 ) A(a, 3) A ( a , 3 ) f ( x ) = 2 x + a f(x) = 2x + a f ( x ) = 2 x + a
f ( a ) = 3 ⟺ 2 a + a = 3 ⟺ a = 1 f(a) = 3 \iff 2a + a = 3 \iff a = 1 f ( a ) = 3 ⟺ 2 a + a = 3 ⟺ a = 1 Barem: 3p, 2p.
3. Ecuația 2019 x + 2019 − x = 2 2019^x + 2019^{-x} = 2 201 9 x + 201 9 − x = 2
Înmulțim cu 2019 x > 0 2019^x > 0 201 9 x > 0 ( 2019 x ) 2 − 2 ⋅ 2019 x + 1 = 0 ⟺ ( 2019 x − 1 ) 2 = 0 ⟺ 2019 x = 1 ⟺ x = 0 (2019^x)^2 - 2\cdot 2019^x + 1 = 0 \iff (2019^x - 1)^2 = 0 \iff 2019^x = 1 \iff x = 0 ( 201 9 x ) 2 − 2 ⋅ 201 9 x + 1 = 0 ⟺ ( 201 9 x − 1 ) 2 = 0 ⟺ 201 9 x = 1 ⟺ x = 0
Răspuns: x = 0 x = 0 x = 0 Barem: 3p, 2p.
4. Probabilitatea ca un număr de două cifre să aibă cifra unităților impară
90 de cazuri posibile. Prima cifră: 9 variante, ultima cifră din { 1 , 3 , 5 , 7 , 9 } \{1,3,5,7,9\} { 1 , 3 , 5 , 7 , 9 } 45 45 45
p = 45 90 = 1 2 . p = \frac{45}{90} = \frac12. p = 90 45 = 2 1 .
Barem: 2p, 2p, 1p.
5. Dreapta d d d A ( 3 , − 3 ) A(3,-3) A ( 3 , − 3 ) A B AB A B B ( 2 , − 2 ) B(2,-2) B ( 2 , − 2 )
m A B = − 2 + 3 2 − 3 = − 1 m_{AB} = \dfrac{-2 + 3}{2 - 3} = -1 m A B = 2 − 3 − 2 + 3 = − 1 m d = 1 m_d = 1 m d = 1 = − 1 = -1 = − 1
d : y + 3 = 1 ⋅ ( x − 3 ) ⟺ y = x − 6 d: y + 3 = 1\cdot(x - 3) \iff y = x - 6 d : y + 3 = 1 ⋅ ( x − 3 ) ⟺ y = x − 6
Răspuns: y = x − 6 y = x - 6 y = x − 6 Barem: 2p, 3p.
6. sin ( a − b ) sin ( a + b ) = ( sin a − sin b ) ( sin a + sin b ) \sin(a - b)\sin(a + b) = (\sin a - \sin b)(\sin a + \sin b) sin ( a − b ) sin ( a + b ) = ( sin a − sin b ) ( sin a + sin b )
sin ( a − b ) sin ( a + b ) = ( sin a cos b ) 2 − ( sin b cos a ) 2 = sin 2 a ( 1 − sin 2 b ) − sin 2 b ( 1 − sin 2 a ) = sin 2 a − sin 2 b , \sin(a-b)\sin(a+b) = (\sin a\cos b)^2 - (\sin b\cos a)^2 = \sin^2a(1 - \sin^2b) - \sin^2b(1 - \sin^2a) = \sin^2a - \sin^2b, sin ( a − b ) sin ( a + b ) = ( sin a cos b ) 2 − ( sin b cos a ) 2 = sin 2 a ( 1 − sin 2 b ) − sin 2 b ( 1 − sin 2 a ) = sin 2 a − sin 2 b ,
care este ( sin a − sin b ) ( sin a + sin b ) (\sin a - \sin b)(\sin a + \sin b) ( sin a − sin b ) ( sin a + sin b ) ■ \blacksquare ■ Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. A ( a ) = ( a 0 − a 0 2 0 − a 0 a ) A(a) = \begin{pmatrix}a & 0 & -a\\ 0 & 2 & 0\\ -a & 0 & a\end{pmatrix} A ( a ) = a 0 − a 0 2 0 − a 0 a
a) Regula lui Sarrus: det A ( a ) = 2 a 2 + 0 + 0 − 2 a 2 − 0 − 0 = 0 \det A(a) = 2a^2 + 0 + 0 - 2a^2 - 0 - 0 = 0 det A ( a ) = 2 a 2 + 0 + 0 − 2 a 2 − 0 − 0 = 0 ■ \blacksquare ■ Barem: 2p, 3p.
b) A ( a ) A ( b ) = ( a b + a b 0 − a b − a b 0 4 0 − a b − a b 0 a b + a b ) = 2 ( a b 0 − a b 0 2 0 − a b 0 a b ) = 2 A ( a b ) . ■ A(a)A(b) = \begin{pmatrix}ab + ab & 0 & -ab - ab\\ 0 & 4 & 0\\ -ab - ab & 0 & ab + ab\end{pmatrix} = 2\begin{pmatrix}ab & 0 & -ab\\ 0 & 2 & 0\\ -ab & 0 & ab\end{pmatrix} = 2A(ab).\ \blacksquare A ( a ) A ( b ) = ab + ab 0 − ab − ab 0 4 0 − ab − ab 0 ab + ab = 2 ab 0 − ab 0 2 0 − ab 0 ab = 2 A ( ab ) . ■
Barem: 3p, 2p.
c) B B B 14 14 14 2 , 3 , … , 15 2, 3, \dots, 15 2 , 3 , … , 15 13 13 13
B = 2 13 A ( log 2 3 ⋅ log 3 4 ⋯ log 15 16 ) = 2 13 A ( log 2 16 ) = 2 13 A ( 4 ) , B = 2^{13}A\left(\log_2 3\cdot\log_3 4\cdots\log_{15}16\right) = 2^{13}A(\log_2 16) = 2^{13}A(4), B = 2 13 A ( log 2 3 ⋅ log 3 4 ⋯ log 15 16 ) = 2 13 A ( log 2 16 ) = 2 13 A ( 4 ) ,
(produsul logaritmilor „în lanț” se simplifică: log a b ⋅ log b c = log a c \log_a b\cdot\log_b c = \log_a c log a b ⋅ log b c = log a c A ( 4 ) A(4) A ( 4 ) B B B ■ \blacksquare ■
Barem: 3p, 2p.
2. f = X 3 + X 2 + m X + n f = X^3 + X^2 + mX + n f = X 3 + X 2 + m X + n
a) f ( − 1 ) = − m + n f(-1) = -m + n f ( − 1 ) = − m + n f ( 0 ) = n f(0) = n f ( 0 ) = n f ( 1 ) = 2 + m + n f(1) = 2 + m + n f ( 1 ) = 2 + m + n
f ( − 1 ) − 2 f ( 0 ) + f ( 1 ) = − m + n − 2 n + 2 + m + n = 2. ■ f(-1) - 2f(0) + f(1) = -m + n - 2n + 2 + m + n = 2.\ \blacksquare f ( − 1 ) − 2 f ( 0 ) + f ( 1 ) = − m + n − 2 n + 2 + m + n = 2. ■
Barem: 2p, 3p.
b) X 2 − 1 = ( X − 1 ) ( X + 1 ) ∣ f ⟺ f ( 1 ) = 0 X^2 - 1 = (X - 1)(X + 1) \mid f \iff f(1) = 0 X 2 − 1 = ( X − 1 ) ( X + 1 ) ∣ f ⟺ f ( 1 ) = 0 f ( − 1 ) = 0 f(-1) = 0 f ( − 1 ) = 0 2 + m + n = 0 2 + m + n = 0 2 + m + n = 0 − m + n = 0 ⇒ m = n = − 1 -m + n = 0 \Rightarrow m = n = -1 − m + n = 0 ⇒ m = n = − 1
Răspuns: m = − 1 m = -1 m = − 1 n = − 1 n = -1 n = − 1 Barem: 3p, 2p.
c) Viète: s 1 = − 1 s_1 = -1 s 1 = − 1 s 2 = m s_2 = m s 2 = m s 3 = − n s_3 = -n s 3 = − n
x 1 3 + x 2 3 + x 3 3 = s 1 3 − 3 s 1 s 2 + 3 s 3 = − 1 + 3 m − 3 n , x_1^3 + x_2^3 + x_3^3 = s_1^3 - 3s_1s_2 + 3s_3 = -1 + 3m - 3n, x 1 3 + x 2 3 + x 3 3 = s 1 3 − 3 s 1 s 2 + 3 s 3 = − 1 + 3 m − 3 n , 3 ( s 2 + s 3 ) − ( x 1 3 + x 2 3 + x 3 3 ) = 3 ( m − n ) − ( − 1 + 3 m − 3 n ) = 1. ■ 3(s_2 + s_3) - (x_1^3 + x_2^3 + x_3^3) = 3(m - n) - (-1 + 3m - 3n) = 1.\ \blacksquare 3 ( s 2 + s 3 ) − ( x 1 3 + x 2 3 + x 3 3 ) = 3 ( m − n ) − ( − 1 + 3 m − 3 n ) = 1. ■
Barem: 3p, 2p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x 2 e − x f(x) = x^2e^{-x} f ( x ) = x 2 e − x
a) f ′ ( x ) = 2 x e − x − x 2 e − x = x ( 2 − x ) e − x f'(x) = 2xe^{-x} - x^2e^{-x} = x(2 - x)e^{-x} f ′ ( x ) = 2 x e − x − x 2 e − x = x ( 2 − x ) e − x ■ \blacksquare ■ Barem: 3p, 2p.
b) f ′ ( x ) = 0 ⟺ x ∈ { 0 , 2 } f'(x) = 0 \iff x \in \{0, 2\} f ′ ( x ) = 0 ⟺ x ∈ { 0 , 2 } x ( 2 − x ) x(2 - x) x ( 2 − x )
f ′ ≤ 0 f' \le 0 f ′ ≤ 0 ( − ∞ , 0 ] (-\infty, 0] ( − ∞ , 0 ] f f f f ′ ≥ 0 f' \ge 0 f ′ ≥ 0 [ 0 , 2 ] [0, 2] [ 0 , 2 ] f f f f ′ ≤ 0 f' \le 0 f ′ ≤ 0 [ 2 , + ∞ ) [2, +\infty) [ 2 , + ∞ ) f f f
Barem: 2p, 3p.
c) Fie a ∈ ( 0 , 4 e − 2 ) a \in (0, 4e^{-2}) a ∈ ( 0 , 4 e − 2 ) f ( 0 ) = 0 < a f(0) = 0 < a f ( 0 ) = 0 < a f ( 2 ) = 4 e − 2 > a f(2) = 4e^{-2} > a f ( 2 ) = 4 e − 2 > a lim x → + ∞ f ( x ) = 0 < a \lim\limits_{x\to+\infty}f(x) = 0 < a x → + ∞ lim f ( x ) = 0 < a lim x → − ∞ f ( x ) = + ∞ \lim\limits_{x\to-\infty}f(x) = +\infty x → − ∞ lim f ( x ) = + ∞
f f f ( − ∞ , 0 ) (-\infty, 0) ( − ∞ , 0 ) ( 0 , 2 ) (0, 2) ( 0 , 2 ) ( 2 , + ∞ ) (2, +\infty) ( 2 , + ∞ ) a a a ( 0 , + ∞ ) (0, +\infty) ( 0 , + ∞ ) ( 0 , 4 e − 2 ) (0, 4e^{-2}) ( 0 , 4 e − 2 ) ( 0 , 4 e − 2 ) (0, 4e^{-2}) ( 0 , 4 e − 2 ) ■ \blacksquare ■
Barem: 3p, 2p.
2. f ( x ) = x 2 + ln x f(x) = x^2 + \ln x f ( x ) = x 2 + ln x
a) ∫ 1 2 x 2 d x = x 3 3 ∣ 1 2 = 8 3 − 1 3 = 7 3 . ■ \int_1^2 x^2dx = \frac{x^3}{3}\Big|_1^2 = \frac83 - \frac13 = \frac73.\ \blacksquare ∫ 1 2 x 2 d x = 3 x 3 1 2 = 3 8 − 3 1 = 3 7 . ■
Barem: 3p, 2p.
b) g ( x ) = 2 x + ln x > 0 g(x) = 2x + \ln x > 0 g ( x ) = 2 x + ln x > 0 [ 1 , e ] [1, e] [ 1 , e ] ∫ ln x d x = x ln x − x \int\ln x\,dx = x\ln x - x ∫ ln x d x = x ln x − x
A = ∫ 1 e ( 2 x + ln x ) d x = ( x 2 + x ln x − x ) ∣ 1 e = ( e 2 + e − e ) − ( 1 + 0 − 1 ) = e 2 . ■ \mathcal A = \int_1^e(2x + \ln x)\,dx = \left(x^2 + x\ln x - x\right)\Big|_1^e = (e^2 + e - e) - (1 + 0 - 1) = e^2.\ \blacksquare A = ∫ 1 e ( 2 x + ln x ) d x = ( x 2 + x ln x − x ) 1 e = ( e 2 + e − e ) − ( 1 + 0 − 1 ) = e 2 . ■
Barem: 3p, 2p.
c) f ( x ) − x 2 = ln x f(x) - x^2 = \ln x f ( x ) − x 2 = ln x
∫ 1 / e 1 x n ln x d x = ( x n + 1 n + 1 ln x − x n + 1 ( n + 1 ) 2 ) ∣ 1 / e 1 = − 1 ( n + 1 ) 2 + 1 ( n + 1 ) e n + 1 + 1 ( n + 1 ) 2 e n + 1 . \int_{1/e}^1 x^n\ln x\,dx = \left(\frac{x^{n+1}}{n+1}\ln x - \frac{x^{n+1}}{(n+1)^2}\right)\Big|_{1/e}^1 = -\frac{1}{(n+1)^2} + \frac{1}{(n+1)e^{n+1}} + \frac{1}{(n+1)^2e^{n+1}}. ∫ 1/ e 1 x n ln x d x = ( n + 1 x n + 1 ln x − ( n + 1 ) 2 x n + 1 ) 1/ e 1 = − ( n + 1 ) 2 1 + ( n + 1 ) e n + 1 1 + ( n + 1 ) 2 e n + 1 1 .
Fiecare termen tinde la 0 0 0 n → + ∞ n\to+\infty n → + ∞ 0 0 0 ■ \blacksquare ■
Barem: 3p, 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 n = 11 n = 11 n = 11 I.2 a = 1 a = 1 a = 1 I.3 x = 0 x = 0 x = 0 I.4 p = 1 2 p = \tfrac12 p = 2 1 I.5 y = x − 6 y = x - 6 y = x − 6 II.1.c B = 2 13 A ( 4 ) B = 2^{13}A(4) B = 2 13 A ( 4 ) II.2.b m = n = − 1 m = n = -1 m = n = − 1 III.1.b ↘ \searrow ↘ ( − ∞ , 0 ] (-\infty,0] ( − ∞ , 0 ] ↗ \nearrow ↗ [ 0 , 2 ] [0,2] [ 0 , 2 ] ↘ \searrow ↘ [ 2 , + ∞ ) [2,+\infty) [ 2 , + ∞ )
Ce merită reținut din varianta 6
a x + a − x ≥ 2 a^x + a^{-x} \ge 2 a x + a − x ≥ 2 x = 0 x = 0 x = 0 Produs de logaritmi în lanț: log 2 3 ⋅ log 3 4 ⋯ log 15 16 = log 2 16 = 4 \log_2 3\cdot\log_3 4\cdots\log_{15}16 = \log_2 16 = 4 log 2 3 ⋅ log 3 4 ⋯ log 15 16 = log 2 16 = 4 Numărul de soluții: monotonie pe intervale + continuitate + valorile la capete.