Rezolvarea completă a subiectului de Bacalaureat 2021 la Matematică M_tehnologic , sesiunea august–septembrie, varianta 4 . 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. Progresia aritmetică cu a 1 = 4 a_1 = 4 a 1 = 4 r = 5 r = 5 r = 5 a 3 a_3 a 3
a 3 = a 1 + 2 r = 4 + 10 = 14. a_3 = a_1 + 2r = 4 + 10 = 14. a 3 = a 1 + 2 r = 4 + 10 = 14.
Barem: 3p + 2p.
2. f ( x ) = x 2 − x − 1 f(x) = x^2 - x - 1 f ( x ) = x 2 − x − 1 f ( 0 ) = f ( 1 ) f(0) = f(1) f ( 0 ) = f ( 1 )
f ( 0 ) = − 1 f(0) = -1 f ( 0 ) = − 1 f ( 1 ) = 1 − 1 − 1 = − 1 f(1) = 1 - 1 - 1 = -1 f ( 1 ) = 1 − 1 − 1 = − 1 f ( 0 ) = f ( 1 ) f(0) = f(1) f ( 0 ) = f ( 1 ) ■ \blacksquare ■
Barem: 2p + 3p.
3. Ecuația log 4 ( 3 x + 4 ) = log 4 16 \log_4(3x + 4) = \log_416 log 4 ( 3 x + 4 ) = log 4 16
3 x + 4 = 16 ⟺ x = 4 3x + 4 = 16 \iff x = 4 3 x + 4 = 16 ⟺ x = 4
Barem: 3p + 2p.
4. După o scumpire cu 25%, produsul costă 350 de lei. Prețul inițial?
x + 25 100 x = 350 ⟺ 5 4 x = 350 ⟺ x = 280. x + \frac{25}{100}x = 350 \iff \frac54x = 350 \iff x = 280. x + 100 25 x = 350 ⟺ 4 5 x = 350 ⟺ x = 280.
Răspuns: 280 280 280 Barem: 3p + 2p.
5. A ( − 4 , 1 ) A(-4,1) A ( − 4 , 1 ) B ( a , b ) B(a,b) B ( a , b ) O O O A B AB A B
− 4 + a 2 = 0 ⇒ a = 4 , 1 + b 2 = 0 ⇒ b = − 1. \frac{-4 + a}{2} = 0 \Rightarrow a = 4,\qquad \frac{1 + b}{2} = 0 \Rightarrow b = -1. 2 − 4 + a = 0 ⇒ a = 4 , 2 1 + b = 0 ⇒ b = − 1.
Barem: 3p + 2p.
6. △ A B C \triangle ABC △ A B C A A A A B AB A B
A B = A C AB = AC A B = A C A = A B 2 2 = 8 ⟺ A B 2 = 16 ⟺ A B = 4 \mathcal A = \frac{AB^2}{2} = 8 \iff AB^2 = 16 \iff AB = 4 A = 2 A B 2 = 8 ⟺ A B 2 = 16 ⟺ A B = 4
Barem: 2p + 3p.
Subiectul al II-lea (30 de puncte)
1. A = ( 5 − 2 − 1 1 ) A = \begin{pmatrix}5&-2\\-1&1\end{pmatrix} A = ( 5 − 1 − 2 1 ) B ( x ) = ( x − 2 x 1 − 1 ) B(x) = \begin{pmatrix}x & -2x\\ 1 & -1\end{pmatrix} B ( x ) = ( x 1 − 2 x − 1 )
a) det A = 3 \det A = 3 det A = 3
det A = 5 ⋅ 1 − ( − 2 ) ( − 1 ) = 3. \det A = 5\cdot1 - (-2)(-1) = 3. det A = 5 ⋅ 1 − ( − 2 ) ( − 1 ) = 3.
Barem: 3p + 2p.
b) 3 B ( 2 ) + B ( 6 ) = 4 B ( 3 ) 3B(2) + B(6) = 4B(3) 3 B ( 2 ) + B ( 6 ) = 4 B ( 3 )
3 ( 2 − 4 1 − 1 ) + ( 6 − 12 1 − 1 ) = ( 12 − 24 4 − 4 ) = 4 ( 3 − 6 1 − 1 ) = 4 B ( 3 ) . 3\begin{pmatrix}2&-4\\1&-1\end{pmatrix} + \begin{pmatrix}6&-12\\1&-1\end{pmatrix} = \begin{pmatrix}12&-24\\4&-4\end{pmatrix} = 4\begin{pmatrix}3&-6\\1&-1\end{pmatrix} = 4B(3). 3 ( 2 1 − 4 − 1 ) + ( 6 1 − 12 − 1 ) = ( 12 4 − 24 − 4 ) = 4 ( 3 1 − 6 − 1 ) = 4 B ( 3 ) .
Barem: 3p + 2p.
c) x x x ( B ( − x ) − B ( x ) ) ( B ( − x ) + B ( x ) ) = A + B ( 3 ) \big(B(-x) - B(x)\big)\big(B(-x) + B(x)\big) = A + B(3) ( B ( − x ) − B ( x ) ) ( B ( − x ) + B ( x ) ) = A + B ( 3 )
B ( − x ) − B ( x ) = ( − 2 x 4 x 0 0 ) , B ( − x ) + B ( x ) = ( 0 0 2 − 2 ) , B(-x) - B(x) = \begin{pmatrix}-2x & 4x\\0&0\end{pmatrix},\quad B(-x) + B(x) = \begin{pmatrix}0&0\\2&-2\end{pmatrix}, B ( − x ) − B ( x ) = ( − 2 x 0 4 x 0 ) , B ( − x ) + B ( x ) = ( 0 2 0 − 2 ) ,
produsul = ( 8 x − 8 x 0 0 ) , A + B ( 3 ) = ( 8 − 8 0 0 ) . \text{produsul} = \begin{pmatrix}8x & -8x\\0&0\end{pmatrix},\qquad A + B(3) = \begin{pmatrix}8&-8\\0&0\end{pmatrix}. produsul = ( 8 x 0 − 8 x 0 ) , A + B ( 3 ) = ( 8 0 − 8 0 ) .
Răspuns: x = 1 x = 1 x = 1 Barem: 3p + 2p.
2. Legea x ∘ y = 3 x + 4 y − 25 x\circ y = 3x + 4y - 25 x ∘ y = 3 x + 4 y − 25
a) 3 ∘ 4 = 0 3\circ4 = 0 3 ∘ 4 = 0
3 ∘ 4 = 9 + 16 − 25 = 0. 3\circ4 = 9 + 16 - 25 = 0. 3 ∘ 4 = 9 + 16 − 25 = 0.
Barem: 3p + 2p.
b) x x x ( 2 x ) ∘ x = 5 (2x)\circ x = 5 ( 2 x ) ∘ x = 5
( 2 x ) ∘ x = 6 x + 4 x − 25 = 10 x − 25 = 5 ⟺ x = 3 (2x)\circ x = 6x + 4x - 25 = 10x - 25 = 5 \iff x = 3 ( 2 x ) ∘ x = 6 x + 4 x − 25 = 10 x − 25 = 5 ⟺ x = 3
Barem: 3p + 2p.
c) m m m m 2 ∘ 1 ≥ 1 ∘ m 2 m^2\circ1 \ge 1\circ m^2 m 2 ∘ 1 ≥ 1 ∘ m 2
m 2 ∘ 1 = 3 m 2 − 21 m^2\circ1 = 3m^2 - 21 m 2 ∘ 1 = 3 m 2 − 21 1 ∘ m 2 = 4 m 2 − 22 1\circ m^2 = 4m^2 - 22 1 ∘ m 2 = 4 m 2 − 22
3 m 2 − 21 ≥ 4 m 2 − 22 ⟺ m 2 ≤ 1 ⟺ m ∈ [ − 1 , 1 ] . 3m^2 - 21 \ge 4m^2 - 22 \iff m^2 \le 1 \iff m \in [-1, 1]. 3 m 2 − 21 ≥ 4 m 2 − 22 ⟺ m 2 ≤ 1 ⟺ m ∈ [ − 1 , 1 ] .
Răspuns: m ∈ { − 1 , 0 , 1 } m \in \{-1, 0, 1\} m ∈ { − 1 , 0 , 1 } Barem: 2p + 3p.
Subiectul al III-lea (30 de puncte)
1. f : ( − 1 3 , ∞ ) → R f:\left(-\frac13,\infty\right)\to\mathbb R f : ( − 3 1 , ∞ ) → R f ( x ) = 2 x 3 x + 1 f(x) = \dfrac{2x}{3x + 1} f ( x ) = 3 x + 1 2 x
a) f ′ ( x ) = 2 ( 3 x + 1 ) 2 f'(x) = \dfrac{2}{(3x+1)^2} f ′ ( x ) = ( 3 x + 1 ) 2 2
f ′ ( x ) = 2 ( 3 x + 1 ) − 2 x ⋅ 3 ( 3 x + 1 ) 2 = 2 ( 3 x + 1 ) 2 . f'(x) = \frac{2(3x + 1) - 2x\cdot3}{(3x+1)^2} = \frac2{(3x+1)^2}. f ′ ( x ) = ( 3 x + 1 ) 2 2 ( 3 x + 1 ) − 2 x ⋅ 3 = ( 3 x + 1 ) 2 2 .
Barem: 3p + 2p.
b) Asimptota orizontală spre + ∞ +\infty + ∞
lim x → ∞ 2 x 3 x + 1 = lim x → ∞ 2 3 + 1 x = 2 3 . \lim_{x\to\infty}\frac{2x}{3x + 1} = \lim_{x\to\infty}\frac2{3 + \frac1x} = \frac23. lim x → ∞ 3 x + 1 2 x = lim x → ∞ 3 + x 1 2 = 3 2 .
Răspuns: y = 2 3 y = \frac23 y = 3 2 Barem: 3p + 2p.
c) f f f
f ′ ′ ( x ) = ( 2 ( 3 x + 1 ) − 2 ) ′ = − 4 ⋅ 3 ( 3 x + 1 ) − 3 = − 12 ( 3 x + 1 ) 3 . f''(x) = \left(2(3x+1)^{-2}\right)' = -4\cdot3(3x+1)^{-3} = -\frac{12}{(3x+1)^3}. f ′′ ( x ) = ( 2 ( 3 x + 1 ) − 2 ) ′ = − 4 ⋅ 3 ( 3 x + 1 ) − 3 = − ( 3 x + 1 ) 3 12 .
Pe ( − 1 3 , ∞ ) \left(-\frac13,\infty\right) ( − 3 1 , ∞ ) 3 x + 1 > 0 3x + 1 > 0 3 x + 1 > 0 f ′ ′ ( x ) < 0 f''(x) < 0 f ′′ ( x ) < 0 f f f ■ \blacksquare ■ Barem: 3p + 2p.
2. f : ( 0 , ∞ ) → R f:(0,\infty)\to\mathbb R f : ( 0 , ∞ ) → R f ( x ) = x 2 + ln x − 1 f(x) = x^2 + \ln x - 1 f ( x ) = x 2 + ln x − 1
a) ∫ 1 4 ( f ( x ) − ln x + 1 ) d x = 21 \displaystyle\int_1^4\left(f(x) - \ln x + 1\right)dx = 21 ∫ 1 4 ( f ( x ) − ln x + 1 ) d x = 21
∫ 1 4 x 2 d x = x 3 3 ∣ 1 4 = 64 − 1 3 = 21. \int_1^4x^2dx = \frac{x^3}3\Big|_1^4 = \frac{64 - 1}3 = 21. ∫ 1 4 x 2 d x = 3 x 3 1 4 = 3 64 − 1 = 21.
Barem: 3p + 2p.
b) ∫ 2 4 x f ( x ) − ln x d x = 1 2 ln 5 \displaystyle\int_2^4\frac{x}{f(x) - \ln x}dx = \frac12\ln5 ∫ 2 4 f ( x ) − ln x x d x = 2 1 ln 5
∫ 2 4 x x 2 − 1 d x = 1 2 ln ( x 2 − 1 ) ∣ 2 4 = 1 2 ( ln 15 − ln 3 ) = 1 2 ln 5. \int_2^4\frac{x}{x^2 - 1}dx = \frac12\ln(x^2 - 1)\Big|_2^4 = \frac12(\ln15 - \ln3) = \frac12\ln5. ∫ 2 4 x 2 − 1 x d x = 2 1 ln ( x 2 − 1 ) 2 4 = 2 1 ( ln 15 − ln 3 ) = 2 1 ln 5.
Barem: 3p + 2p.
c) a > 1 a > 1 a > 1 ∫ 1 a f ( x ) x 2 d x = a − ln a a \displaystyle\int_1^a\frac{f(x)}{x^2}dx = \frac{a - \ln a}a ∫ 1 a x 2 f ( x ) d x = a a − ln a
f ( x ) x 2 = 1 + ln x x 2 − 1 x 2 \dfrac{f(x)}{x^2} = 1 + \dfrac{\ln x}{x^2} - \dfrac1{x^2} x 2 f ( x ) = 1 + x 2 ln x − x 2 1 ∫ 1 a ln x x 2 d x = − ln x x ∣ 1 a + ∫ 1 a 1 x 2 d x \displaystyle\int_1^a\frac{\ln x}{x^2}dx = -\frac{\ln x}{x}\Big|_1^a + \int_1^a\frac1{x^2}dx ∫ 1 a x 2 ln x d x = − x ln x 1 a + ∫ 1 a x 2 1 d x
∫ 1 a f ( x ) x 2 d x = ( a − 1 ) − ln a a . \int_1^a\frac{f(x)}{x^2}dx = (a - 1) - \frac{\ln a}a. ∫ 1 a x 2 f ( x ) d x = ( a − 1 ) − a l n a .
a − 1 − ln a a = 1 − ln a a ⟺ a = 2 a - 1 - \dfrac{\ln a}a = 1 - \dfrac{\ln a}a \iff a = 2 a − 1 − a ln a = 1 − a ln a ⟺ a = 2 a > 1 a > 1 a > 1
Răspuns: a = 2 a = 2 a = 2 Barem: 3p + 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 a 3 = 14 a_3 = 14 a 3 = 14 I.3 x = 4 x = 4 x = 4 I.4 280 280 280 I.5 a = 4 a = 4 a = 4 b = − 1 b = -1 b = − 1 I.6 A B = 4 AB = 4 A B = 4 II.1.c x = 1 x = 1 x = 1 II.2.b x = 3 x = 3 x = 3 II.2.c m ∈ { − 1 , 0 , 1 } m \in \{-1,0,1\} m ∈ { − 1 , 0 , 1 } III.1.b y = 2 3 y = \frac23 y = 3 2 III.2.c a = 2 a = 2 a = 2
Ce merită reținut din varianta 4
Concavitate: semnul lui f ′ ′ f'' f ′′ „După o scumpire cu p % p\% p % S S S ⇒ \Rightarrow ⇒ = S 1 + p / 100 = \frac{S}{1 + p/100} = 1 + p /100 S ∫ ln x x 2 d x \int\frac{\ln x}{x^2}dx ∫ x 2 l n x d x ( − 1 x ) ′ = 1 x 2 \left(-\frac1x\right)' = \frac1{x^2} ( − x 1 ) ′ = x 2 1