Rezolvarea completă a modelului de subiect pentru Bacalaureat 2021 la Matematică M_mate-info . 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 = log 2 6 − 2 log 2 3 + log 2 24 N = \log_26 - 2\log_23 + \log_224 N = log 2 6 − 2 log 2 3 + log 2 24
N = log 2 6 − log 2 9 + log 2 24 = log 2 6 ⋅ 24 9 = log 2 16 = 4 ∈ N . N = \log_26 - \log_29 + \log_224 = \log_2\frac{6\cdot24}{9} = \log_216 = 4 \in \mathbb N. N = log 2 6 − log 2 9 + log 2 24 = log 2 9 6 ⋅ 24 = log 2 16 = 4 ∈ N .
Barem: 3p + 2p.
2. Dreapta y = 2 y = 2 y = 2 f ( x ) = x 2 − x + 2 f(x) = x^2 - x + 2 f ( x ) = x 2 − x + 2
f ( x ) = 2 ⟺ x 2 − x = 0 ⟺ x ( x − 1 ) = 0 ⟺ x ∈ { 0 , 1 } . f(x) = 2 \iff x^2 - x = 0 \iff x(x - 1) = 0 \iff x \in \{0, 1\}. f ( x ) = 2 ⟺ x 2 − x = 0 ⟺ x ( x − 1 ) = 0 ⟺ x ∈ { 0 , 1 } .
Ecuația are două soluții reale distincte, deci intersecția are două puncte: ( 0 , 2 ) (0,2) ( 0 , 2 ) ( 1 , 2 ) (1,2) ( 1 , 2 ) ■ \blacksquare ■ Barem: 2p + 3p.
3. Ecuația x 2 − 5 = 3 x − 1 \sqrt{x^2 - 5} = \sqrt{3x - 1} x 2 − 5 = 3 x − 1
Ridicăm la pătrat: x 2 − 5 = 3 x − 1 ⟺ x 2 − 3 x − 4 = 0 ⟺ x ∈ { − 1 , 4 } x^2 - 5 = 3x - 1 \iff x^2 - 3x - 4 = 0 \iff x \in \{-1, 4\} x 2 − 5 = 3 x − 1 ⟺ x 2 − 3 x − 4 = 0 ⟺ x ∈ { − 1 , 4 }
Verificare: x = − 1 x = -1 x = − 1 − 4 \sqrt{-4} − 4 x = 4 x = 4 x = 4 11 = 11 \sqrt{11} = \sqrt{11} 11 = 11
Răspuns: x = 4 x = 4 x = 4 Barem: 2p + 3p.
4. Câte elemente are A A A
C n 2 = n ( n − 1 ) 2 = 15 ⟺ n 2 − n − 30 = 0 ⟺ n = 6 ( n ≥ 2 ) . C_n^2 = \frac{n(n-1)}{2} = 15 \iff n^2 - n - 30 = 0 \iff n = 6\ (n \ge 2). C n 2 = 2 n ( n − 1 ) = 15 ⟺ n 2 − n − 30 = 0 ⟺ n = 6 ( n ≥ 2 ) .
Răspuns: 6 elemente. Barem: 2p + 3p.
5. M , N , P M, N, P M , N , P B C BC B C B M BM B M C M CM C M A M → + A N → + A P → = 3 2 ( A B → + A C → ) \overrightarrow{AM} + \overrightarrow{AN} + \overrightarrow{AP} = \frac32\left(\overrightarrow{AB} + \overrightarrow{AC}\right) A M + A N + A P = 2 3 ( A B + A C )
M M M B C BC B C A M → = 1 2 ( A B → + A C → ) \overrightarrow{AM} = \frac12\left(\overrightarrow{AB} + \overrightarrow{AC}\right) A M = 2 1 ( A B + A C ) B N = N M = M P = P C BN = NM = MP = PC B N = N M = M P = P C M M M N P NP N P A N → + A P → = 2 A M → \overrightarrow{AN} + \overrightarrow{AP} = 2\overrightarrow{AM} A N + A P = 2 A M
A M → + A N → + A P → = 3 A M → = 3 2 ( A B → + A C → ) . ■ \overrightarrow{AM} + \overrightarrow{AN} + \overrightarrow{AP} = 3\overrightarrow{AM} = \frac32\left(\overrightarrow{AB} + \overrightarrow{AC}\right).\ \blacksquare A M + A N + A P = 3 A M = 2 3 ( A B + A C ) . ■
Barem: 2p + 3p.
6. x ∈ ( 0 , π ) x \in (0,\pi) x ∈ ( 0 , π ) sin 2 x + 2 sin 2 x = 0 \sin2x + 2\sin^2x = 0 sin 2 x + 2 sin 2 x = 0
2 sin x cos x + 2 sin 2 x = 0 ⟺ 2 sin x ( cos x + sin x ) = 0. 2\sin x\cos x + 2\sin^2x = 0 \iff 2\sin x(\cos x + \sin x) = 0. 2 sin x cos x + 2 sin 2 x = 0 ⟺ 2 sin x ( cos x + sin x ) = 0.
Pe ( 0 , π ) (0,\pi) ( 0 , π ) sin x ≠ 0 \sin x \ne 0 sin x = 0 sin x = − cos x ⟺ tg x = − 1 ⟺ x = 3 π 4 \sin x = -\cos x \iff \operatorname{tg}x = -1 \iff x = \frac{3\pi}{4} sin x = − cos x ⟺ tg x = − 1 ⟺ x = 4 3 π
Răspuns: x = 3 π 4 x = \dfrac{3\pi}{4} x = 4 3 π Barem: 2p + 3p.
Subiectul al II-lea (30 de puncte)
1. A ( a , b , c ) = ( 1 1 1 a b c b c a c a b ) A(a,b,c) = \begin{pmatrix}1&1&1\\a&b&c\\bc&ac&ab\end{pmatrix} A ( a , b , c ) = 1 a b c 1 b a c 1 c ab
a) det A ( 0 , 1 , 2 ) = 2 \det A(0,1,2) = 2 det A ( 0 , 1 , 2 ) = 2
det ( 1 1 1 0 1 2 2 0 0 ) = 0 + 4 + 0 − 2 − 0 − 0 = 2. \det\begin{pmatrix}1&1&1\\0&1&2\\2&0&0\end{pmatrix} = 0 + 4 + 0 - 2 - 0 - 0 = 2. det 1 0 2 1 1 0 1 2 0 = 0 + 4 + 0 − 2 − 0 − 0 = 2.
Barem: 2p + 3p.
b) det A ( a , b , c ) = ( b − a ) ( c − a ) ( c − b ) \det A(a,b,c) = (b - a)(c - a)(c - b) det A ( a , b , c ) = ( b − a ) ( c − a ) ( c − b )
Scădem coloana 1 din coloanele 2 și 3:
det A = ∣ 1 0 0 a b − a c − a b c c ( a − b ) b ( a − c ) ∣ = ( b − a ) ( c − a ) ∣ 1 1 − c − b ∣ = ( b − a ) ( c − a ) ( c − b ) . \det A = \begin{vmatrix}1&0&0\\a&b - a&c - a\\bc&c(a - b)&b(a - c)\end{vmatrix} = (b - a)(c - a)\begin{vmatrix}1 & 1\\ -c & -b\end{vmatrix} = (b - a)(c - a)(c - b). det A = 1 a b c 0 b − a c ( a − b ) 0 c − a b ( a − c ) = ( b − a ) ( c − a ) 1 − c 1 − b = ( b − a ) ( c − a ) ( c − b ) .
Barem: 2p + 3p.
c) Dacă m < n < p m < n < p m < n < p det A ( m , n , p ) \det A(m,n,p) det A ( m , n , p ) m , n , p m, n, p m , n , p
det A ( m , n , p ) = ( n − m ) ( p − m ) ( p − n ) \det A(m,n,p) = (n - m)(p - m)(p - n) det A ( m , n , p ) = ( n − m ) ( p − m ) ( p − n ) p − m = ( p − n ) + ( n − m ) ≥ 2 p - m = (p - n) + (n - m) \ge 2 p − m = ( p − n ) + ( n − m ) ≥ 2
Ca produsul să fie prim, doi factori trebuie să fie 1 1 1 p − m ≥ 2 p - m \ge 2 p − m ≥ 2 n − m = 1 n - m = 1 n − m = 1 p − n = 1 p - n = 1 p − n = 1 det = 2 \det = 2 det = 2 n − m = p − n n - m = p - n n − m = p − n m , n , p m, n, p m , n , p ■ \blacksquare ■ Barem: 3p + 2p.
2. f = X 4 + a X 3 + 2 ^ X + b ∈ Z 3 [ X ] f = X^4 + aX^3 + \hat2X + b \in \mathbb Z_3[X] f = X 4 + a X 3 + 2 ^ X + b ∈ Z 3 [ X ]
a) a = 1 ^ a = \hat1 a = 1 ^ b = 2 ^ b = \hat2 b = 2 ^ f ( 0 ^ ) + f ( 2 ^ ) = 2 ^ f(\hat0) + f(\hat2) = \hat2 f ( 0 ^ ) + f ( 2 ^ ) = 2 ^
f = X 4 + X 3 + 2 ^ X + 2 ^ f = X^4 + X^3 + \hat2X + \hat2 f = X 4 + X 3 + 2 ^ X + 2 ^ f ( 0 ^ ) = 2 ^ f(\hat0) = \hat2 f ( 0 ^ ) = 2 ^ f ( 2 ^ ) = 16 + 8 + 4 + 2 = 30 = 0 ^ f(\hat2) = 16 + 8 + 4 + 2 = 30 = \hat0 f ( 2 ^ ) = 16 + 8 + 4 + 2 = 30 = 0 ^ Z 3 \mathbb Z_3 Z 3
f ( 0 ^ ) + f ( 2 ^ ) = 2 ^ + 0 ^ = 2 ^ . f(\hat0) + f(\hat2) = \hat2 + \hat0 = \hat2. f ( 0 ^ ) + f ( 2 ^ ) = 2 ^ + 0 ^ = 2 ^ .
Barem: 3p + 2p.
b) Perechile ( a , b ) (a,b) ( a , b ) f f f X + 2 ^ X + \hat2 X + 2 ^
X + 2 ^ = X − 1 ^ X + \hat2 = X - \hat1 X + 2 ^ = X − 1 ^ f f f ⟺ f ( 1 ^ ) = 0 ^ \iff f(\hat1) = \hat0 ⟺ f ( 1 ^ ) = 0 ^
1 ^ + a + 2 ^ + b = 0 ^ ⟺ a + b = 0 ^ . \hat1 + a + \hat2 + b = \hat0 \iff a + b = \hat0. 1 ^ + a + 2 ^ + b = 0 ^ ⟺ a + b = 0 ^ .
Răspuns: ( 0 ^ , 0 ^ ) (\hat0,\hat0) ( 0 ^ , 0 ^ ) ( 1 ^ , 2 ^ ) (\hat1,\hat2) ( 1 ^ , 2 ^ ) ( 2 ^ , 1 ^ ) (\hat2,\hat1) ( 2 ^ , 1 ^ ) Barem: 3p + 2p.
c) Pentru orice a , b a, b a , b x ≠ y x \ne y x = y f ( x ) = f ( y ) f(x) = f(y) f ( x ) = f ( y )
f ( 0 ^ ) = b f(\hat0) = b f ( 0 ^ ) = b f ( 1 ^ ) = 1 ^ + a + 2 ^ + b = a + b f(\hat1) = \hat1 + a + \hat2 + b = a + b f ( 1 ^ ) = 1 ^ + a + 2 ^ + b = a + b f ( 2 ^ ) = 1 ^ + 2 ^ a + 1 ^ + b = 2 ^ + 2 ^ a + b f(\hat2) = \hat1 + \hat2a + \hat1 + b = \hat2 + \hat2a + b f ( 2 ^ ) = 1 ^ + 2 ^ a + 1 ^ + b = 2 ^ + 2 ^ a + b
f ( 0 ^ ) + f ( 1 ^ ) + f ( 2 ^ ) = 3 ^ a + 3 ^ b + 2 ^ = 2 ^ . f(\hat0) + f(\hat1) + f(\hat2) = \hat3a + \hat3b + \hat2 = \hat2. f ( 0 ^ ) + f ( 1 ^ ) + f ( 2 ^ ) = 3 ^ a + 3 ^ b + 2 ^ = 2 ^ .
Dacă valorile ar fi distincte două câte două, ar fi exact 0 ^ , 1 ^ , 2 ^ \hat0, \hat1, \hat2 0 ^ , 1 ^ , 2 ^ 3 ^ = 0 ^ ≠ 2 ^ \hat3 = \hat0 \ne \hat2 3 ^ = 0 ^ = 2 ^ ■ \blacksquare ■ Barem: 3p + 2p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x e x e x + 2 f(x) = \dfrac{xe^x}{e^x + 2} f ( x ) = e x + 2 x e x
a) f ′ ( x ) = e x ( e x + 2 x + 2 ) ( e x + 2 ) 2 f'(x) = \dfrac{e^x(e^x + 2x + 2)}{(e^x + 2)^2} f ′ ( x ) = ( e x + 2 ) 2 e x ( e x + 2 x + 2 )
f ′ ( x ) = ( e x + x e x ) ( e x + 2 ) − x e x ⋅ e x ( e x + 2 ) 2 = e 2 x + 2 e x + 2 x e x ( e x + 2 ) 2 = e x ( e x + 2 x + 2 ) ( e x + 2 ) 2 . f'(x) = \frac{(e^x + xe^x)(e^x + 2) - xe^x\cdot e^x}{(e^x+2)^2} = \frac{e^{2x} + 2e^x + 2xe^x}{(e^x+2)^2} = \frac{e^x(e^x + 2x + 2)}{(e^x+2)^2}. f ′ ( x ) = ( e x + 2 ) 2 ( e x + x e x ) ( e x + 2 ) − x e x ⋅ e x = ( e x + 2 ) 2 e 2 x + 2 e x + 2 x e x = ( e x + 2 ) 2 e x ( e x + 2 x + 2 ) .
Barem: 3p + 2p.
b) y = x y = x y = x + ∞ +\infty + ∞
m = lim x → ∞ f ( x ) x = lim x → ∞ e x e x + 2 = 1 , n = lim x → ∞ ( f ( x ) − x ) = lim x → ∞ − 2 x e x + 2 = 0. m = \lim_{x\to\infty}\frac{f(x)}{x} = \lim_{x\to\infty}\frac{e^x}{e^x + 2} = 1,\qquad n = \lim_{x\to\infty}(f(x) - x) = \lim_{x\to\infty}\frac{-2x}{e^x + 2} = 0. m = lim x → ∞ x f ( x ) = lim x → ∞ e x + 2 e x = 1 , n = lim x → ∞ ( f ( x ) − x ) = lim x → ∞ e x + 2 − 2 x = 0.
Deci y = x y = x y = x + ∞ +\infty + ∞ ■ \blacksquare ■ Barem: 2p + 3p.
c) f f f
Semnul lui f ′ f' f ′ g ( x ) = e x + 2 x + 2 g(x) = e^x + 2x + 2 g ( x ) = e x + 2 x + 2 g g g lim x → − ∞ g = − ∞ \lim\limits_{x\to-\infty}g = -\infty x → − ∞ lim g = − ∞ lim x → ∞ g = + ∞ \lim\limits_{x\to\infty}g = +\infty x → ∞ lim g = + ∞ c c c g ( c ) = 0 g(c) = 0 g ( c ) = 0 g < 0 g < 0 g < 0 ( − ∞ , c ) (-\infty, c) ( − ∞ , c ) g > 0 g > 0 g > 0 ( c , ∞ ) (c, \infty) ( c , ∞ )
Așadar f f f ( − ∞ , c ] (-\infty, c] ( − ∞ , c ] [ c , ∞ ) [c,\infty) [ c , ∞ ) c c c ■ \blacksquare ■ Barem: 3p + 2p.
2. f ( x ) = 2 ( x + 3 ) x 2 + 4 x + 5 f(x) = \dfrac{2(x + 3)}{\sqrt{x^2 + 4x + 5}} f ( x ) = x 2 + 4 x + 5 2 ( x + 3 )
a) ∫ 0 1 f ( x ) x 2 + 4 x + 5 d x = 7 \displaystyle\int_0^1 f(x)\sqrt{x^2 + 4x + 5}\,dx = 7 ∫ 0 1 f ( x ) x 2 + 4 x + 5 d x = 7
∫ 0 1 2 ( x + 3 ) d x = ( x 2 + 6 x ) ∣ 0 1 = 7. \int_0^1 2(x + 3)dx = (x^2 + 6x)\Big|_0^1 = 7. ∫ 0 1 2 ( x + 3 ) d x = ( x 2 + 6 x ) 0 1 = 7.
Barem: 3p + 2p.
b) ∫ 0 1 ( f 2 ( x ) − 4 ) d x = 4 ln 2 \displaystyle\int_0^1\left(f^2(x) - 4\right)dx = 4\ln2 ∫ 0 1 ( f 2 ( x ) − 4 ) d x = 4 ln 2
f 2 ( x ) − 4 = 4 ( x + 3 ) 2 − 4 ( x 2 + 4 x + 5 ) x 2 + 4 x + 5 = 4 ( 2 x + 4 ) x 2 + 4 x + 5 = 4 ⋅ ( x 2 + 4 x + 5 ) ′ x 2 + 4 x + 5 . f^2(x) - 4 = \frac{4(x+3)^2 - 4(x^2 + 4x + 5)}{x^2 + 4x + 5} = \frac{4(2x + 4)}{x^2 + 4x + 5} = 4\cdot\frac{(x^2 + 4x + 5)'}{x^2 + 4x + 5}. f 2 ( x ) − 4 = x 2 + 4 x + 5 4 ( x + 3 ) 2 − 4 ( x 2 + 4 x + 5 ) = x 2 + 4 x + 5 4 ( 2 x + 4 ) = 4 ⋅ x 2 + 4 x + 5 ( x 2 + 4 x + 5 ) ′ .
∫ 0 1 ( f 2 − 4 ) d x = 4 ln ( x 2 + 4 x + 5 ) ∣ 0 1 = 4 ln 10 5 = 4 ln 2. \int_0^1\left(f^2 - 4\right)dx = 4\ln(x^2 + 4x + 5)\Big|_0^1 = 4\ln\frac{10}{5} = 4\ln2. ∫ 0 1 ( f 2 − 4 ) d x = 4 ln ( x 2 + 4 x + 5 ) 0 1 = 4 ln 5 10 = 4 ln 2.
Barem: 2p + 3p.
c) 0 ≤ a < b 0 \le a < b 0 ≤ a < b I n = ∫ a b f n ( x ) d x I_n = \displaystyle\int_a^bf^n(x)dx I n = ∫ a b f n ( x ) d x lim I n = + ∞ \lim I_n = +\infty lim I n = + ∞
Din b), f 2 ( x ) − 4 = 4 ( 2 x + 4 ) x 2 + 4 x + 5 > 0 f^2(x) - 4 = \frac{4(2x+4)}{x^2+4x+5} > 0 f 2 ( x ) − 4 = x 2 + 4 x + 5 4 ( 2 x + 4 ) > 0 x ≥ 0 x \ge 0 x ≥ 0 f ( x ) > 0 f(x) > 0 f ( x ) > 0 x ≥ 0 x \ge 0 x ≥ 0 f ( x ) > 2 f(x) > 2 f ( x ) > 2 f n ( x ) ≥ 2 n f^n(x) \ge 2^n f n ( x ) ≥ 2 n [ a , b ] [a, b] [ a , b ]
I n ≥ ∫ a b 2 n d x = 2 n ( b − a ) → n → ∞ + ∞ , I_n \ge \int_a^b2^ndx = 2^n(b - a) \xrightarrow[n\to\infty]{} +\infty, I n ≥ ∫ a b 2 n d x = 2 n ( b − a ) n → ∞ + ∞ ,
deci lim n → ∞ I n = + ∞ \lim\limits_{n\to\infty}I_n = +\infty n → ∞ lim I n = + ∞ ■ \blacksquare ■ Barem: 2p + 3p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 N = 4 N = 4 N = 4 I.3 x = 4 x = 4 x = 4 I.4 6 elemente I.6 x = 3 π 4 x = \frac{3\pi}{4} x = 4 3 π II.1.b ( b − a ) ( c − a ) ( c − b ) (b-a)(c-a)(c-b) ( b − a ) ( c − a ) ( c − b ) II.2.b ( 0 ^ , 0 ^ ) , ( 1 ^ , 2 ^ ) , ( 2 ^ , 1 ^ ) (\hat0,\hat0), (\hat1,\hat2), (\hat2,\hat1) ( 0 ^ , 0 ^ ) , ( 1 ^ , 2 ^ ) , ( 2 ^ , 1 ^ ) III.1.b y = x y = x y = x
Ce merită reținut din model
Determinant de tip Vandermonde: scade coloanele și scoate factori comuni.În Z 3 \mathbb Z_3 Z 3 , suma 0 ^ + 1 ^ + 2 ^ = 0 ^ \hat0 + \hat1 + \hat2 = \hat0 0 ^ + 1 ^ + 2 ^ = 0 ^ Limite de ∫ f n \int f^n ∫ f n dacă f ≥ q > 1 f \ge q > 1 f ≥ q > 1 + ∞ +\infty + ∞ 0 ≤ f ≤ q < 1 0 \le f \le q < 1 0 ≤ f ≤ q < 1 0 0 0