Rezolvarea completă a modelului de subiect pentru Bacalaureat 2021 la Matematică M_tehnologic . 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 ( 0,25 ⋅ 10 − 1 2 ) ( 0,25 ⋅ 10 + 1 2 ) = 6 \left(0{,}25\cdot10 - \frac12\right)\left(0{,}25\cdot10 + \frac12\right) = 6 ( 0 , 25 ⋅ 10 − 2 1 ) ( 0 , 25 ⋅ 10 + 2 1 ) = 6
0,25 ⋅ 10 = 2,5 0{,}25\cdot10 = 2{,}5 0 , 25 ⋅ 10 = 2 , 5 ( 2,5 − 0,5 ) ( 2,5 + 0,5 ) = 2 ⋅ 3 = 6 (2{,}5 - 0{,}5)(2{,}5 + 0{,}5) = 2\cdot3 = 6 ( 2 , 5 − 0 , 5 ) ( 2 , 5 + 0 , 5 ) = 2 ⋅ 3 = 6
Barem: 2p + 3p.
2. f ( x ) = x 2 − a x + 1 f(x) = x^2 - ax + 1 f ( x ) = x 2 − a x + 1 a a a A ( 2 , 1 ) ∈ G f A(2,1) \in G_f A ( 2 , 1 ) ∈ G f
f ( 2 ) = 1 ⟺ 4 − 2 a + 1 = 1 ⟺ a = 2 f(2) = 1 \iff 4 - 2a + 1 = 1 \iff a = 2 f ( 2 ) = 1 ⟺ 4 − 2 a + 1 = 1 ⟺ a = 2
Barem: 3p + 2p.
3. Ecuația 3 x + 2 + 3 x = 30 3^{x+2} + 3^x = 30 3 x + 2 + 3 x = 30
3 x ( 9 + 1 ) = 30 ⟺ 3 x = 3 ⟺ x = 1 3^x(9 + 1) = 30 \iff 3^x = 3 \iff x = 1 3 x ( 9 + 1 ) = 30 ⟺ 3 x = 3 ⟺ x = 1
Barem: 3p + 2p.
4. Un obiect costă 500 de lei; prețul după o scumpire cu 20%
Scumpirea: 20 100 ⋅ 500 = 100 \frac{20}{100}\cdot500 = 100 100 20 ⋅ 500 = 100 600 600 600
Barem: 3p + 2p.
5. A ( 0 , 3 ) A(0,3) A ( 0 , 3 ) B ( 8 , 3 ) B(8,3) B ( 8 , 3 ) M M M A B AB A B O M OM O M
M ( 4 , 3 ) M(4, 3) M ( 4 , 3 ) O M = 4 2 + 3 2 = 5 OM = \sqrt{4^2 + 3^2} = 5 O M = 4 2 + 3 2 = 5
Barem: 2p + 3p.
6. △ A B C \triangle ABC △ A B C A A A A B = 5 AB = 5 A B = 5 A C = 10 AC = 10 A C = 10
A = A B ⋅ A C 2 = 5 ⋅ 10 2 = 25. \mathcal A = \frac{AB\cdot AC}2 = \frac{5\cdot10}2 = 25. A = 2 A B ⋅ A C = 2 5 ⋅ 10 = 25.
Barem: 2p + 3p.
Subiectul al II-lea (30 de puncte)
1. A = ( 1 3 1 2 ) A = \begin{pmatrix}1&3\\1&2\end{pmatrix} A = ( 1 1 3 2 )
a) det A = − 1 \det A = -1 det A = − 1
det A = 1 ⋅ 2 − 3 ⋅ 1 = − 1. \det A = 1\cdot2 - 3\cdot1 = -1. det A = 1 ⋅ 2 − 3 ⋅ 1 = − 1.
Barem: 3p + 2p.
b) A ⋅ A − 3 A = I 2 A\cdot A - 3A = I_2 A ⋅ A − 3 A = I 2
A ⋅ A = ( 4 9 3 7 ) , A ⋅ A − 3 A = ( 4 − 3 9 − 9 3 − 3 7 − 6 ) = I 2 . A\cdot A = \begin{pmatrix}4&9\\3&7\end{pmatrix},\quad A\cdot A - 3A = \begin{pmatrix}4 - 3 & 9 - 9\\ 3 - 3 & 7 - 6\end{pmatrix} = I_2. A ⋅ A = ( 4 3 9 7 ) , A ⋅ A − 3 A = ( 4 − 3 3 − 3 9 − 9 7 − 6 ) = I 2 .
Barem: 3p + 2p.
c) X = ( 1 x y 1 ) X = \begin{pmatrix}1&x\\y&1\end{pmatrix} X = ( 1 y x 1 ) A X − X A = ( − 2 1 − 1 2 ) AX - XA = \begin{pmatrix}-2&1\\-1&2\end{pmatrix} A X − X A = ( − 2 − 1 1 2 )
A X = ( 1 + 3 y x + 3 1 + 2 y x + 2 ) , X A = ( 1 + x 3 + 2 x 1 + y 3 y + 2 ) , A X − X A = ( 3 y − x − x y x − 3 y ) . AX = \begin{pmatrix}1 + 3y & x + 3\\ 1 + 2y & x + 2\end{pmatrix},\quad XA = \begin{pmatrix}1 + x & 3 + 2x\\ 1 + y & 3y + 2\end{pmatrix},\quad AX - XA = \begin{pmatrix}3y - x & -x\\ y & x - 3y\end{pmatrix}. A X = ( 1 + 3 y 1 + 2 y x + 3 x + 2 ) , X A = ( 1 + x 1 + y 3 + 2 x 3 y + 2 ) , A X − X A = ( 3 y − x y − x x − 3 y ) .
Egalând: − x = 1 ⇒ x = − 1 -x = 1 \Rightarrow x = -1 − x = 1 ⇒ x = − 1 y = − 1 y = -1 y = − 1 3 y − x = − 2 3y - x = -2 3 y − x = − 2 x − 3 y = 2 x - 3y = 2 x − 3 y = 2
Răspuns: x = y = − 1 x = y = -1 x = y = − 1 Barem: 2p + 3p.
2. Legea x ∘ y = 4 x y + x + y x\circ y = 4xy + x + y x ∘ y = 4 x y + x + y
a) 3 ∘ 2 = 29 3\circ2 = 29 3 ∘ 2 = 29
3 ∘ 2 = 24 + 3 + 2 = 29. 3\circ2 = 24 + 3 + 2 = 29. 3 ∘ 2 = 24 + 3 + 2 = 29.
Barem: 3p + 2p.
b) x ∘ y = ( 4 x + 1 ) ( 4 y + 1 ) − 1 4 x\circ y = \dfrac{(4x+1)(4y+1) - 1}{4} x ∘ y = 4 ( 4 x + 1 ) ( 4 y + 1 ) − 1
( 4 x + 1 ) ( 4 y + 1 ) − 1 4 = 16 x y + 4 x + 4 y + 1 − 1 4 = 4 x y + x + y = x ∘ y . \frac{(4x+1)(4y+1) - 1}4 = \frac{16xy + 4x + 4y + 1 - 1}4 = 4xy + x + y = x\circ y. 4 ( 4 x + 1 ) ( 4 y + 1 ) − 1 = 4 16 x y + 4 x + 4 y + 1 − 1 = 4 x y + x + y = x ∘ y .
Barem: 3p + 2p.
c) x x x x ∘ x ≤ 2 x\circ x \le 2 x ∘ x ≤ 2
( 4 x + 1 ) 2 − 1 4 ≤ 2 ⟺ ( 4 x + 1 ) 2 ≤ 9 ⟺ − 3 ≤ 4 x + 1 ≤ 3 ⟺ − 1 ≤ x ≤ 1 2 . \frac{(4x+1)^2 - 1}4 \le 2 \iff (4x + 1)^2 \le 9 \iff -3 \le 4x + 1 \le 3 \iff -1 \le x \le \frac12. 4 ( 4 x + 1 ) 2 − 1 ≤ 2 ⟺ ( 4 x + 1 ) 2 ≤ 9 ⟺ − 3 ≤ 4 x + 1 ≤ 3 ⟺ − 1 ≤ x ≤ 2 1 .
Răspuns: x ∈ [ − 1 , 1 2 ] x \in \left[-1, \frac12\right] x ∈ [ − 1 , 2 1 ] Barem: 2p + 3p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = e x + x x 2 + 1 f(x) = e^x + \dfrac{x}{x^2 + 1} f ( x ) = e x + x 2 + 1 x
a) f ′ ( x ) = e x + 1 − x 2 ( x 2 + 1 ) 2 f'(x) = e^x + \dfrac{1 - x^2}{(x^2+1)^2} f ′ ( x ) = e x + ( x 2 + 1 ) 2 1 − x 2
f ′ ( x ) = e x + ( x 2 + 1 ) − x ⋅ 2 x ( x 2 + 1 ) 2 = e x + 1 − x 2 ( x 2 + 1 ) 2 . f'(x) = e^x + \frac{(x^2 + 1) - x\cdot2x}{(x^2+1)^2} = e^x + \frac{1 - x^2}{(x^2+1)^2}. f ′ ( x ) = e x + ( x 2 + 1 ) 2 ( x 2 + 1 ) − x ⋅ 2 x = e x + ( x 2 + 1 ) 2 1 − x 2 .
Barem: 3p + 2p.
b) Asimptota orizontală spre − ∞ -\infty − ∞
lim x → − ∞ f ( x ) = lim x → − ∞ e x + lim x → − ∞ x x 2 + 1 = 0 + 0 = 0. \lim_{x\to-\infty}f(x) = \lim_{x\to-\infty}e^x + \lim_{x\to-\infty}\frac{x}{x^2+1} = 0 + 0 = 0. lim x → − ∞ f ( x ) = lim x → − ∞ e x + lim x → − ∞ x 2 + 1 x = 0 + 0 = 0.
Răspuns: y = 0 y = 0 y = 0 Barem: 3p + 2p.
c) 2 − e 2 e ≤ f ( x ) ≤ 2 e + 1 2 \dfrac{2 - e}{2e} \le f(x) \le \dfrac{2e + 1}2 2 e 2 − e ≤ f ( x ) ≤ 2 2 e + 1 [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ]
Pe [ − 1 , 1 ] [-1,1] [ − 1 , 1 ] 1 − x 2 ≥ 0 1 - x^2 \ge 0 1 − x 2 ≥ 0 e x > 0 e^x > 0 e x > 0 f ′ ( x ) > 0 f'(x) > 0 f ′ ( x ) > 0 f f f f ( − 1 ) ≤ f ( x ) ≤ f ( 1 ) f(-1) \le f(x) \le f(1) f ( − 1 ) ≤ f ( x ) ≤ f ( 1 )
f ( − 1 ) = 1 e − 1 2 = 2 − e 2 e , f ( 1 ) = e + 1 2 = 2 e + 1 2 . ■ f(-1) = \frac1e - \frac12 = \frac{2 - e}{2e},\qquad f(1) = e + \frac12 = \frac{2e + 1}2.\ \blacksquare f ( − 1 ) = e 1 − 2 1 = 2 e 2 − e , f ( 1 ) = e + 2 1 = 2 2 e + 1 . ■
Barem: 3p + 2p.
2. f : [ 0 , ∞ ) → R f:[0,\infty)\to\mathbb R f : [ 0 , ∞ ) → R f ( x ) = ( x + 1 ) x f(x) = (x + 1)\sqrt x f ( x ) = ( x + 1 ) x
a) ∫ 1 2 f ( x ) x d x = 5 2 \displaystyle\int_1^2\frac{f(x)}{\sqrt x}dx = \frac52 ∫ 1 2 x f ( x ) d x = 2 5
∫ 1 2 ( x + 1 ) d x = ( x 2 2 + x ) ∣ 1 2 = 4 − 3 2 = 5 2 . \int_1^2(x + 1)dx = \left(\frac{x^2}2 + x\right)\Big|_1^2 = 4 - \frac32 = \frac52. ∫ 1 2 ( x + 1 ) d x = ( 2 x 2 + x ) 1 2 = 4 − 2 3 = 2 5 .
Barem: 3p + 2p.
b) Volumul corpului de rotație al lui g = f ∣ [ 0 , 1 ] g = f|_{[0,1]} g = f ∣ [ 0 , 1 ] 17 π 12 \frac{17\pi}{12} 12 17 π
V = π ∫ 0 1 g 2 ( x ) d x = π ∫ 0 1 ( x + 1 ) 2 x d x = π ∫ 0 1 ( x 3 + 2 x 2 + x ) d x = π ( 1 4 + 2 3 + 1 2 ) = 17 π 12 . V = \pi\int_0^1g^2(x)dx = \pi\int_0^1(x+1)^2x\,dx = \pi\int_0^1(x^3 + 2x^2 + x)dx = \pi\left(\frac14 + \frac23 + \frac12\right) = \frac{17\pi}{12}. V = π ∫ 0 1 g 2 ( x ) d x = π ∫ 0 1 ( x + 1 ) 2 x d x = π ∫ 0 1 ( x 3 + 2 x 2 + x ) d x = π ( 4 1 + 3 2 + 2 1 ) = 12 17 π .
Barem: 3p + 2p.
c) a a a ∫ 1 e f ( x ) x ln x x + 1 d x = e 2 + a 4 \displaystyle\int_1^e\frac{f(x)\sqrt x\ln x}{x + 1}dx = \frac{e^2 + a}4 ∫ 1 e x + 1 f ( x ) x ln x d x = 4 e 2 + a
Integrandul se simplifică la x ln x x\ln x x ln x
∫ 1 e x ln x d x = x 2 2 ln x ∣ 1 e − ∫ 1 e x 2 d x = e 2 2 − e 2 − 1 4 = e 2 + 1 4 . \int_1^ex\ln x\,dx = \frac{x^2}2\ln x\Big|_1^e - \int_1^e\frac x2dx = \frac{e^2}2 - \frac{e^2 - 1}4 = \frac{e^2 + 1}4. ∫ 1 e x ln x d x = 2 x 2 ln x 1 e − ∫ 1 e 2 x d x = 2 e 2 − 4 e 2 − 1 = 4 e 2 + 1 .
Răspuns: a = 1 a = 1 a = 1 Barem: 3p + 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 a = 2 a = 2 a = 2 I.3 x = 1 x = 1 x = 1 I.4 600 600 600 I.5 O M = 5 OM = 5 O M = 5 I.6 25 25 25 II.1.c x = y = − 1 x = y = -1 x = y = − 1 II.2.c x ∈ [ − 1 , 1 2 ] x \in [-1, \frac12] x ∈ [ − 1 , 2 1 ] III.1.b y = 0 y = 0 y = 0 III.2.c a = 1 a = 1 a = 1
Ce merită reținut din model
Volumul corpului de rotație: V = π ∫ a b f 2 ( x ) d x V = \pi\int_a^bf^2(x)dx V = π ∫ a b f 2 ( x ) d x Forma ( 4 x + 1 ) ( 4 y + 1 ) − 1 4 \frac{(4x+1)(4y+1) - 1}4 4 ( 4 x + 1 ) ( 4 y + 1 ) − 1 transformă inecuația x ∘ x ≤ 2 x\circ x \le 2 x ∘ x ≤ 2 Încadrare pe interval: monotonie + valorile din capete.