Rezolvarea completă a simulării pentru Bacalaureat 2022 la Matematică M_tehnologic (clasa a XII-a). Pentru fiecare cerință găsești enunțul pe scurt, rezolvarea pas cu pas, răspunsul final și cum se punctează conform baremului oficial. Timp de lucru: 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. Progresia aritmetică cu a 3 = 6 a_3 = 6 a 3 = 6 a 4 = 9 a_4 = 9 a 4 = 9 a 1 a_1 a 1
r = a 4 − a 3 = 3 r = a_4 - a_3 = 3 r = a 4 − a 3 = 3 a 3 = a 1 + 2 r ⇒ a 1 = 6 − 6 = 0 a_3 = a_1 + 2r \Rightarrow a_1 = 6 - 6 = 0 a 3 = a 1 + 2 r ⇒ a 1 = 6 − 6 = 0
Răspuns: a 1 = 0 a_1 = 0 a 1 = 0 Barem: 2p + 3p.
2. f ( x ) = x 2 + 2 x − 3 f(x) = x^2 + 2x - 3 f ( x ) = x 2 + 2 x − 3 g ( x ) = x − 3 g(x) = x - 3 g ( x ) = x − 3 a a a f ( a ) = g ( a ) f(a) = g(a) f ( a ) = g ( a )
a 2 + 2 a − 3 = a − 3 ⟺ a 2 + a = 0 ⟺ a ( a + 1 ) = 0. a^2 + 2a - 3 = a - 3 \iff a^2 + a = 0 \iff a(a + 1) = 0. a 2 + 2 a − 3 = a − 3 ⟺ a 2 + a = 0 ⟺ a ( a + 1 ) = 0.
Răspuns: a ∈ { − 1 , 0 } a \in \{-1, 0\} a ∈ { − 1 , 0 } Barem: 3p + 2p.
3. Ecuația log 3 ( x + 3 ) = 2 \log_3(x + 3) = 2 log 3 ( x + 3 ) = 2
x + 3 = 3 2 = 9 ⟺ x = 6 x + 3 = 3^2 = 9 \iff x = 6 x + 3 = 3 2 = 9 ⟺ x = 6 x + 3 > 0 x + 3 > 0 x + 3 > 0
Răspuns: x = 6 x = 6 x = 6 Barem: 3p + 2p.
4. După o scumpire cu 30%, prețul crește cu 60 de lei. Prețul după scumpire?
Fie x x x 30 100 x = 60 ⇒ x = 200 \frac{30}{100}x = 60 \Rightarrow x = 200 100 30 x = 60 ⇒ x = 200 200 + 60 = 260 200 + 60 = 260 200 + 60 = 260
Răspuns: 260 260 260 Barem: 3p + 2p.
5. A ( − 4 , 1 ) A(-4,1) A ( − 4 , 1 ) B ( 2 , 3 ) B(2,3) B ( 2 , 3 ) d : y = 2 x + a d: y = 2x + a d : y = 2 x + a A B AB A B d d d
Mijlocul: M ( − 4 + 2 2 , 1 + 3 2 ) = M ( − 1 , 2 ) M\left(\frac{-4+2}{2}, \frac{1+3}{2}\right) = M(-1, 2) M ( 2 − 4 + 2 , 2 1 + 3 ) = M ( − 1 , 2 ) M ∈ d ⟺ 2 = − 2 + a ⟺ a = 4 M \in d \iff 2 = -2 + a \iff a = 4 M ∈ d ⟺ 2 = − 2 + a ⟺ a = 4
Răspuns: a = 4 a = 4 a = 4 Barem: 2p + 3p.
6. A B = A C AB = AC A B = A C B C = 12 BC = 12 B C = 12 ∠ B = 45 ∘ \angle B = 45^\circ ∠ B = 4 5 ∘ = 36 = 36 = 36
Triunghi isoscel cu ∠ B = ∠ C = 45 ∘ \angle B = \angle C = 45^\circ ∠ B = ∠ C = 4 5 ∘ ∠ A = 90 ∘ \angle A = 90^\circ ∠ A = 9 0 ∘ A B = A C = B C 2 = 6 2 AB = AC = \frac{BC}{\sqrt2} = 6\sqrt2 A B = A C = 2 B C = 6 2
A = 6 2 ⋅ 6 2 2 = 72 2 = 36. \mathcal A = \frac{6\sqrt2\cdot6\sqrt2}{2} = \frac{72}{2} = 36. A = 2 6 2 ⋅ 6 2 = 2 72 = 36.
Barem: 3p + 2p.
Subiectul al II-lea (30 de puncte)
1. A ( x ) = ( x x 1 2 x + 1 ) A(x) = \begin{pmatrix}x & x\\ 1 & 2x + 1\end{pmatrix} A ( x ) = ( x 1 x 2 x + 1 )
a) det A ( 0 ) = 0 \det A(0) = 0 det A ( 0 ) = 0
A ( 0 ) = ( 0 0 1 1 ) , det A ( 0 ) = 0 ⋅ 1 − 0 ⋅ 1 = 0. A(0) = \begin{pmatrix}0&0\\1&1\end{pmatrix},\quad \det A(0) = 0\cdot1 - 0\cdot1 = 0. A ( 0 ) = ( 0 1 0 1 ) , det A ( 0 ) = 0 ⋅ 1 − 0 ⋅ 1 = 0.
Barem: 3p + 2p.
b) a a a 2 A ( 4 ) + A ( − 2 ) = a A ( 2 ) 2A(4) + A(-2) = aA(2) 2 A ( 4 ) + A ( − 2 ) = a A ( 2 )
2 ( 4 4 1 9 ) + ( − 2 − 2 1 − 3 ) = ( 6 6 3 15 ) = 3 ( 2 2 1 5 ) = 3 A ( 2 ) . 2\begin{pmatrix}4&4\\1&9\end{pmatrix} + \begin{pmatrix}-2&-2\\1&-3\end{pmatrix} = \begin{pmatrix}6&6\\3&15\end{pmatrix} = 3\begin{pmatrix}2&2\\1&5\end{pmatrix} = 3A(2). 2 ( 4 1 4 9 ) + ( − 2 1 − 2 − 3 ) = ( 6 3 6 15 ) = 3 ( 2 1 2 5 ) = 3 A ( 2 ) .
Răspuns: a = 3 a = 3 a = 3 Barem: 3p + 2p.
c) Dacă X ⋅ A ( 1 ) = A ( m ) X\cdot A(1) = A(m) X ⋅ A ( 1 ) = A ( m ) m ∈ Z m \in \mathbb Z m ∈ Z X X X
A ( 1 ) = ( 1 1 1 3 ) A(1) = \begin{pmatrix}1&1\\1&3\end{pmatrix} A ( 1 ) = ( 1 1 1 3 ) det A ( 1 ) = 2 \det A(1) = 2 det A ( 1 ) = 2 A ( 1 ) − 1 = 1 2 ( 3 − 1 − 1 1 ) A(1)^{-1} = \dfrac12\begin{pmatrix}3&-1\\-1&1\end{pmatrix} A ( 1 ) − 1 = 2 1 ( 3 − 1 − 1 1 )
X = A ( m ) A ( 1 ) − 1 = 1 2 ( m m 1 2 m + 1 ) ( 3 − 1 − 1 1 ) = 1 2 ( 2 m 0 2 − 2 m 2 m ) = ( m 0 1 − m m ) . X = A(m)A(1)^{-1} = \frac12\begin{pmatrix}m&m\\1&2m+1\end{pmatrix}\begin{pmatrix}3&-1\\-1&1\end{pmatrix} = \frac12\begin{pmatrix}2m & 0\\ 2 - 2m & 2m\end{pmatrix} = \begin{pmatrix}m & 0\\ 1 - m & m\end{pmatrix}. X = A ( m ) A ( 1 ) − 1 = 2 1 ( m 1 m 2 m + 1 ) ( 3 − 1 − 1 1 ) = 2 1 ( 2 m 2 − 2 m 0 2 m ) = ( m 1 − m 0 m ) .
Pentru m m m X X X ■ \blacksquare ■ Barem: 2p + 3p.
2. Legea x ∗ y = ( x + y ) ( x − 1 ) ( y − 1 ) + 1 x * y = (x + y)(x - 1)(y - 1) + 1 x ∗ y = ( x + y ) ( x − 1 ) ( y − 1 ) + 1
a) 2 ∗ 1 = 1 2 * 1 = 1 2 ∗ 1 = 1
2 ∗ 1 = 3 ⋅ 1 ⋅ 0 + 1 = 1. 2 * 1 = 3\cdot1\cdot0 + 1 = 1. 2 ∗ 1 = 3 ⋅ 1 ⋅ 0 + 1 = 1.
Barem: 3p + 2p.
b) Legea este comutativă
x ∗ y = ( x + y ) ( x − 1 ) ( y − 1 ) + 1 = ( y + x ) ( y − 1 ) ( x − 1 ) + 1 = y ∗ x , x * y = (x + y)(x - 1)(y - 1) + 1 = (y + x)(y - 1)(x - 1) + 1 = y * x, x ∗ y = ( x + y ) ( x − 1 ) ( y − 1 ) + 1 = ( y + x ) ( y − 1 ) ( x − 1 ) + 1 = y ∗ x ,
din comutativitatea adunării și înmulțirii numerelor reale. ■ \blacksquare ■ Barem: 2p + 3p.
c) n n n n ∗ ( 1 − n ) ≥ n 2 n * (1 - n) \ge n^2 n ∗ ( 1 − n ) ≥ n 2
n ∗ ( 1 − n ) = 1 ⋅ ( n − 1 ) ( − n ) + 1 = − n 2 + n + 1. n * (1 - n) = 1\cdot(n - 1)(-n) + 1 = -n^2 + n + 1. n ∗ ( 1 − n ) = 1 ⋅ ( n − 1 ) ( − n ) + 1 = − n 2 + n + 1.
− n 2 + n + 1 ≥ n 2 ⟺ 2 n 2 − n − 1 ≤ 0 ⟺ ( 2 n + 1 ) ( n − 1 ) ≤ 0 ⟺ n ∈ [ − 1 2 , 1 ] -n^2 + n + 1 \ge n^2 \iff 2n^2 - n - 1 \le 0 \iff (2n + 1)(n - 1) \le 0 \iff n \in \left[-\frac12, 1\right] − n 2 + n + 1 ≥ n 2 ⟺ 2 n 2 − n − 1 ≤ 0 ⟺ ( 2 n + 1 ) ( n − 1 ) ≤ 0 ⟺ n ∈ [ − 2 1 , 1 ]
Răspuns: n ∈ { 0 , 1 } n \in \{0, 1\} n ∈ { 0 , 1 } Barem: 2p + 3p.
Subiectul al III-lea (30 de puncte)
1. f : ( 0 , ∞ ) → R f:(0,\infty)\to\mathbb R f : ( 0 , ∞ ) → R f ( x ) = x + 3 x 2 + ln x f(x) = \dfrac{x + 3}{x^2} + \ln x f ( x ) = x 2 x + 3 + ln x
a) f ′ ( x ) = x 2 − x − 6 x 3 f'(x) = \dfrac{x^2 - x - 6}{x^3} f ′ ( x ) = x 3 x 2 − x − 6
f ′ ( x ) = x 2 − 2 x ( x + 3 ) x 4 + 1 x = − x − 6 x 3 + x 2 x 3 = x 2 − x − 6 x 3 . f'(x) = \frac{x^2 - 2x(x + 3)}{x^4} + \frac1x = \frac{-x - 6}{x^3} + \frac{x^2}{x^3} = \frac{x^2 - x - 6}{x^3}. f ′ ( x ) = x 4 x 2 − 2 x ( x + 3 ) + x 1 = x 3 − x − 6 + x 3 x 2 = x 3 x 2 − x − 6 .
Barem: 3p + 2p.
b) Tangenta în punctul de abscisă x = 1 x = 1 x = 1
f ( 1 ) = 4 f(1) = 4 f ( 1 ) = 4 f ′ ( 1 ) = − 6 f'(1) = -6 f ′ ( 1 ) = − 6
y − 4 = − 6 ( x − 1 ) ⟺ y = − 6 x + 10. y - 4 = -6(x - 1) \iff y = -6x + 10. y − 4 = − 6 ( x − 1 ) ⟺ y = − 6 x + 10.
Barem: 2p + 3p.
c) ln x 3 ≥ 2 3 − 1 x − 3 x 2 \ln\dfrac x3 \ge \dfrac23 - \dfrac1x - \dfrac3{x^2} ln 3 x ≥ 3 2 − x 1 − x 2 3 ( 0 , ∞ ) (0,\infty) ( 0 , ∞ )
x 2 − x − 6 = ( x − 3 ) ( x + 2 ) x^2 - x - 6 = (x - 3)(x + 2) x 2 − x − 6 = ( x − 3 ) ( x + 2 ) f ′ ≤ 0 f' \le 0 f ′ ≤ 0 ( 0 , 3 ] (0, 3] ( 0 , 3 ] f ′ ≥ 0 f' \ge 0 f ′ ≥ 0 [ 3 , ∞ ) [3,\infty) [ 3 , ∞ ) f ( 3 ) = 6 9 + ln 3 = 2 3 + ln 3 f(3) = \frac69 + \ln3 = \frac23 + \ln3 f ( 3 ) = 9 6 + ln 3 = 3 2 + ln 3
x + 3 x 2 + ln x ≥ 2 3 + ln 3 ⟺ ln x − ln 3 ≥ 2 3 − 1 x − 3 x 2 ⟺ ln x 3 ≥ 2 3 − 1 x − 3 x 2 . ■ \frac{x+3}{x^2} + \ln x \ge \frac23 + \ln3 \iff \ln x - \ln 3 \ge \frac23 - \frac1x - \frac3{x^2} \iff \ln\frac x3 \ge \frac23 - \frac1x - \frac3{x^2}.\ \blacksquare x 2 x + 3 + ln x ≥ 3 2 + ln 3 ⟺ ln x − ln 3 ≥ 3 2 − x 1 − x 2 3 ⟺ ln 3 x ≥ 3 2 − x 1 − x 2 3 . ■
Barem: 3p + 2p.
2. f ( x ) = x + e x 2 + 1 f(x) = x + \dfrac{e^x}{2} + 1 f ( x ) = x + 2 e x + 1
a) ∫ 0 2 ( f ( x ) − e x 2 ) d x = 4 \displaystyle\int_0^2\left(f(x) - \frac{e^x}{2}\right)dx = 4 ∫ 0 2 ( f ( x ) − 2 e x ) d x = 4
∫ 0 2 ( x + 1 ) d x = ( x 2 2 + x ) ∣ 0 2 = 2 + 2 = 4. \int_0^2(x + 1)dx = \left(\frac{x^2}{2} + x\right)\Big|_0^2 = 2 + 2 = 4. ∫ 0 2 ( x + 1 ) d x = ( 2 x 2 + x ) 0 2 = 2 + 2 = 4.
Barem: 3p + 2p.
b) ∫ 0 1 2 x ( f ( x ) − 1 ) d x = 5 3 \displaystyle\int_0^1 2x\left(f(x) - 1\right)dx = \frac53 ∫ 0 1 2 x ( f ( x ) − 1 ) d x = 3 5
2 x ( f ( x ) − 1 ) = 2 x 2 + x e x 2x(f(x) - 1) = 2x^2 + xe^x 2 x ( f ( x ) − 1 ) = 2 x 2 + x e x x e x xe^x x e x ( x − 1 ) e x (x - 1)e^x ( x − 1 ) e x
∫ 0 1 ( 2 x 2 + x e x ) d x = 2 x 3 3 ∣ 0 1 + ( x − 1 ) e x ∣ 0 1 = 2 3 + ( 0 − ( − 1 ) ) = 5 3 . \int_0^1(2x^2 + xe^x)dx = \frac{2x^3}{3}\Big|_0^1 + (x - 1)e^x\Big|_0^1 = \frac23 + (0 - (-1)) = \frac53. ∫ 0 1 ( 2 x 2 + x e x ) d x = 3 2 x 3 0 1 + ( x − 1 ) e x 0 1 = 3 2 + ( 0 − ( − 1 )) = 3 5 .
Barem: 3p + 2p.
c) a a a ∫ − 1 0 ( f ( x ) − x ) f ( x ) d x = ( 3 e + 1 ) ( 3 e + a ) 8 e 2 \displaystyle\int_{-1}^0\big(f(x) - x\big)f(x)dx = \frac{(3e+1)(3e+a)}{8e^2} ∫ − 1 0 ( f ( x ) − x ) f ( x ) d x = 8 e 2 ( 3 e + 1 ) ( 3 e + a )
f ′ ( x ) = 1 + e x 2 = f ( x ) − x f'(x) = 1 + \frac{e^x}{2} = f(x) - x f ′ ( x ) = 1 + 2 e x = f ( x ) − x
∫ − 1 0 f ′ ( x ) f ( x ) d x = f 2 ( x ) 2 ∣ − 1 0 = 1 2 ( 9 4 − 1 4 e 2 ) = 9 e 2 − 1 8 e 2 = ( 3 e + 1 ) ( 3 e − 1 ) 8 e 2 , \int_{-1}^0 f'(x)f(x)dx = \frac{f^2(x)}{2}\Big|_{-1}^0 = \frac12\left(\frac94 - \frac1{4e^2}\right) = \frac{9e^2 - 1}{8e^2} = \frac{(3e+1)(3e-1)}{8e^2}, ∫ − 1 0 f ′ ( x ) f ( x ) d x = 2 f 2 ( x ) − 1 0 = 2 1 ( 4 9 − 4 e 2 1 ) = 8 e 2 9 e 2 − 1 = 8 e 2 ( 3 e + 1 ) ( 3 e − 1 ) ,
unde f ( 0 ) = 3 2 f(0) = \frac32 f ( 0 ) = 2 3 f ( − 1 ) = 1 2 e f(-1) = \frac1{2e} f ( − 1 ) = 2 e 1
Răspuns: a = − 1 a = -1 a = − 1 Barem: 3p + 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 a 1 = 0 a_1 = 0 a 1 = 0 I.2 a ∈ { − 1 , 0 } a \in \{-1, 0\} a ∈ { − 1 , 0 } I.3 x = 6 x = 6 x = 6 I.4 260 260 260 I.5 a = 4 a = 4 a = 4 II.1.b a = 3 a = 3 a = 3 II.2.c n ∈ { 0 , 1 } n \in \{0,1\} n ∈ { 0 , 1 } III.1.b y = − 6 x + 10 y = -6x + 10 y = − 6 x + 10 III.2.c a = − 1 a = -1 a = − 1
Ce merită reținut din simulare
Ecuații matriceale X A = B XA = B X A = B : X = B A − 1 X = BA^{-1} X = B A − 1 Observă relația f ′ = f − x f' = f - x f ′ = f − x — transformă integrala în ∫ f ′ f = f 2 2 \int f'f = \frac{f^2}{2} ∫ f ′ f = 2 f 2 Procente: „crește cu 60 de lei” este suma scumpirii, nu prețul final.