Rezolvarea completă a subiectului de Bacalaureat 2014 la Matematică M_mate-info , sesiunea de rezervă, varianta 3 . 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. z = 1 + i z = 1 + i z = 1 + i z 2 z^2 z 2
z 2 = 1 + 2 i + i 2 = 2 i z^2 = 1 + 2i + i^2 = 2i z 2 = 1 + 2 i + i 2 = 2 i Răspuns: 2 i 2i 2 i Barem: 2p, 3p.
2. Parabola lui f ( x ) = x 2 − 4 x + 6 f(x) = x^2 - 4x + 6 f ( x ) = x 2 − 4 x + 6 O x Ox O x
Δ = 16 − 24 = − 8 < 0 \Delta = 16 - 24 = -8 < 0 Δ = 16 − 24 = − 8 < 0 f ( x ) = 0 f(x) = 0 f ( x ) = 0 Barem: 3p, 2p.
3. log 2 ( 2 x − 3 ) = log 2 ( x + 1 ) \log_2(2x - 3) = \log_2(x + 1) log 2 ( 2 x − 3 ) = log 2 ( x + 1 )
2 x − 3 = x + 1 ⟺ x = 4 2x - 3 = x + 1 \iff x = 4 2 x − 3 = x + 1 ⟺ x = 4 5 = 5 > 0 5 = 5 > 0 5 = 5 > 0
Răspuns: x = 4 x = 4 x = 4 Barem: 2p, 3p.
4. Probabilitatea ca un număr de două cifre să fie impar
45 de numere impare (11 , 13 , … , 99 11, 13, \dots, 99 11 , 13 , … , 99
Răspuns: p = 1 2 p = \dfrac12 p = 2 1 Barem: 2p, 1p, 2p.
5. M , N , P M, N, P M , N , P A M ⃗ + B N ⃗ + C P ⃗ = 0 ⃗ \vec{AM} + \vec{BN} + \vec{CP} = \vec0 A M + B N + C P = 0
A M ⃗ = 1 2 A B ⃗ \vec{AM} = \frac12\vec{AB} A M = 2 1 A B B N ⃗ = 1 2 B C ⃗ \vec{BN} = \frac12\vec{BC} B N = 2 1 B C C P ⃗ = 1 2 C A ⃗ \vec{CP} = \frac12\vec{CA} C P = 2 1 C A
1 2 ( A B ⃗ + B C ⃗ + C A ⃗ ) = 1 2 A A ⃗ = 0 ⃗ . \frac12\left(\vec{AB} + \vec{BC} + \vec{CA}\right) = \frac12\vec{AA} = \vec0. 2 1 ( A B + B C + C A ) = 2 1 AA = 0 .
Barem: 2p, 3p.
6. tg a = 3 \operatorname{tg} a = \sqrt3 tg a = 3 sin a − cos a cos a + sin a = 2 − 3 \dfrac{\sin a - \cos a}{\cos a + \sin a} = 2 - \sqrt3 cos a + sin a sin a − cos a = 2 − 3
Împărțim numărătorul și numitorul la cos a ≠ 0 \cos a \ne 0 cos a = 0
tg a − 1 1 + tg a = 3 − 1 3 + 1 = ( 3 − 1 ) 2 2 = 4 − 2 3 2 = 2 − 3 . \frac{\operatorname{tg} a - 1}{1 + \operatorname{tg} a} = \frac{\sqrt3 - 1}{\sqrt3 + 1} = \frac{(\sqrt3 - 1)^2}{2} = \frac{4 - 2\sqrt3}{2} = 2 - \sqrt3. 1 + tg a tg a − 1 = 3 + 1 3 − 1 = 2 ( 3 − 1 ) 2 = 2 4 − 2 3 = 2 − 3 .
Barem: 3p, 2p.
Subiectul al II-lea (30 de puncte)
1. A ( a ) = ( 1 1 1 1 a 2 1 2 a ) A(a) = \begin{pmatrix} 1 & 1 & 1\\ 1 & a & 2\\ 1 & 2 & a\end{pmatrix} A ( a ) = 1 1 1 1 a 2 1 2 a
a) det ( A ( 1 ) ) = − 1 \det(A(1)) = -1 det ( A ( 1 )) = − 1
∣ 1 1 1 1 1 2 1 2 1 ∣ = 1 + 2 + 2 − 1 − 4 − 1 = − 1. \begin{vmatrix} 1 & 1 & 1\\ 1 & 1 & 2\\ 1 & 2 & 1\end{vmatrix} = 1 + 2 + 2 - 1 - 4 - 1 = -1. 1 1 1 1 1 2 1 2 1 = 1 + 2 + 2 − 1 − 4 − 1 = − 1.
Barem: 2p, 3p.
b) m m m det ( A ( m ) ) = 0 \det(A(m)) = 0 det ( A ( m )) = 0
det ( A ( m ) ) = m 2 + 2 + 2 − m − 4 − m = m 2 − 2 m . \det(A(m)) = m^2 + 2 + 2 - m - 4 - m = m^2 - 2m. det ( A ( m )) = m 2 + 2 + 2 − m − 4 − m = m 2 − 2 m .
Răspuns: m ∈ { 0 , 2 } m \in \{0, 2\} m ∈ { 0 , 2 } Barem: 3p, 2p.
c) a a a A ( a ) A ( a ) − A ( a 2 ) = ( 2 1 1 1 5 − 5 1 − 5 5 ) A(a)A(a) - A(a^2) = \begin{pmatrix} 2 & 1 & 1\\ 1 & 5 & -5\\ 1 & -5 & 5\end{pmatrix} A ( a ) A ( a ) − A ( a 2 ) = 2 1 1 1 5 − 5 1 − 5 5
A ( a ) 2 = ( 3 a + 3 a + 3 a + 3 a 2 + 5 4 a + 1 a + 3 4 a + 1 a 2 + 5 ) , A ( a ) 2 − A ( a 2 ) = ( 2 a + 2 a + 2 a + 2 5 4 a − 1 a + 2 4 a − 1 5 ) . A(a)^2 = \begin{pmatrix} 3 & a + 3 & a + 3\\ a + 3 & a^2 + 5 & 4a + 1\\ a + 3 & 4a + 1 & a^2 + 5\end{pmatrix}, \quad A(a)^2 - A(a^2) = \begin{pmatrix} 2 & a + 2 & a + 2\\ a + 2 & 5 & 4a - 1\\ a + 2 & 4a - 1 & 5\end{pmatrix}. A ( a ) 2 = 3 a + 3 a + 3 a + 3 a 2 + 5 4 a + 1 a + 3 4 a + 1 a 2 + 5 , A ( a ) 2 − A ( a 2 ) = 2 a + 2 a + 2 a + 2 5 4 a − 1 a + 2 4 a − 1 5 .
Egalând: a + 2 = 1 a + 2 = 1 a + 2 = 1 4 a − 1 = − 5 4a - 1 = -5 4 a − 1 = − 5 a = − 1 a = -1 a = − 1
Răspuns: a = − 1 a = -1 a = − 1 Barem: 3p, 2p.
2. Legea x ∗ y = 3 x + 3 y − x y − 6 x * y = 3x + 3y - xy - 6 x ∗ y = 3 x + 3 y − x y − 6
a) 1 ∗ 3 1 * 3 1 ∗ 3
3 + 9 − 3 − 6 = 3 3 + 9 - 3 - 6 = 3 3 + 9 − 3 − 6 = 3 Răspuns: 3 3 3 Barem: 3p, 2p.
b) x ∗ y = 3 − ( x − 3 ) ( y − 3 ) x * y = 3 - (x - 3)(y - 3) x ∗ y = 3 − ( x − 3 ) ( y − 3 )
3 − ( x y − 3 x − 3 y + 9 ) = − x y + 3 x + 3 y − 6 = x ∗ y 3 - (xy - 3x - 3y + 9) = -xy + 3x + 3y - 6 = x * y 3 − ( x y − 3 x − 3 y + 9 ) = − x y + 3 x + 3 y − 6 = x ∗ y Barem: 2p, 3p.
c) x ∗ x ∗ … ∗ x ⏟ 2014 = x \underbrace{x * x * \ldots * x}_{2014} = x 2014 x ∗ x ∗ … ∗ x = x
Din b), ( x ∗ y ) − 3 = − ( x − 3 ) ( y − 3 ) (x * y) - 3 = -(x - 3)(y - 3) ( x ∗ y ) − 3 = − ( x − 3 ) ( y − 3 ) n n n x x x ( − 1 ) n − 1 ( x − 3 ) n + 3 (-1)^{n-1}(x - 3)^n + 3 ( − 1 ) n − 1 ( x − 3 ) n + 3 n = 2014 n = 2014 n = 2014 3 − ( x − 3 ) 2014 3 - (x - 3)^{2014} 3 − ( x − 3 ) 2014
3 − ( x − 3 ) 2014 = x ⟺ ( x − 3 ) 2014 = − ( x − 3 ) . 3 - (x - 3)^{2014} = x \iff (x - 3)^{2014} = -(x - 3). 3 − ( x − 3 ) 2014 = x ⟺ ( x − 3 ) 2014 = − ( x − 3 ) .
Cu t = x − 3 t = x - 3 t = x − 3 t ( t 2013 + 1 ) = 0 ⟺ t = 0 t(t^{2013} + 1) = 0 \iff t = 0 t ( t 2013 + 1 ) = 0 ⟺ t = 0 t = − 1 t = -1 t = − 1
Răspuns: x ∈ { 2 , 3 } x \in \{2, 3\} x ∈ { 2 , 3 } Barem: 3p, 2p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x − 2 x 2 − 4 x + 5 f(x) = \dfrac{x - 2}{x^2 - 4x + 5} f ( x ) = x 2 − 4 x + 5 x − 2
a) f ′ ( x ) = ( 1 − x ) ( x − 3 ) ( x 2 − 4 x + 5 ) 2 f'(x) = \dfrac{(1 - x)(x - 3)}{(x^2 - 4x + 5)^2} f ′ ( x ) = ( x 2 − 4 x + 5 ) 2 ( 1 − x ) ( x − 3 )
f ′ ( x ) = ( x 2 − 4 x + 5 ) − ( x − 2 ) ( 2 x − 4 ) ( x 2 − 4 x + 5 ) 2 = − x 2 + 4 x − 3 ( x 2 − 4 x + 5 ) 2 = ( 1 − x ) ( x − 3 ) ( x 2 − 4 x + 5 ) 2 . f'(x) = \frac{(x^2 - 4x + 5) - (x - 2)(2x - 4)}{(x^2 - 4x + 5)^2} = \frac{-x^2 + 4x - 3}{(x^2 - 4x + 5)^2} = \frac{(1 - x)(x - 3)}{(x^2 - 4x + 5)^2}. f ′ ( x ) = ( x 2 − 4 x + 5 ) 2 ( x 2 − 4 x + 5 ) − ( x − 2 ) ( 2 x − 4 ) = ( x 2 − 4 x + 5 ) 2 − x 2 + 4 x − 3 = ( x 2 − 4 x + 5 ) 2 ( 1 − x ) ( x − 3 ) .
Barem: 2p, 3p.
b) Asimptota spre + ∞ +\infty + ∞
lim x → + ∞ x − 2 x 2 − 4 x + 5 = 0 \lim\limits_{x\to+\infty}\dfrac{x - 2}{x^2 - 4x + 5} = 0 x → + ∞ lim x 2 − 4 x + 5 x − 2 = 0
Răspuns: y = 0 y = 0 y = 0 Barem: 3p, 2p.
c) Monotonia
f ′ ( x ) = 0 ⟺ x ∈ { 1 , 3 } f'(x) = 0 \iff x \in \{1, 3\} f ′ ( x ) = 0 ⟺ x ∈ { 1 , 3 } ( 1 − x ) ( x − 3 ) ≥ 0 (1 - x)(x - 3) \ge 0 ( 1 − x ) ( x − 3 ) ≥ 0
Răspuns: f f f ( − ∞ , 1 ] (-\infty, 1] ( − ∞ , 1 ] [ 1 , 3 ] [1, 3] [ 1 , 3 ] [ 3 , + ∞ ) [3, +\infty) [ 3 , + ∞ ) Barem: 2p, 1p, 1p, 1p.
2. I n = ∫ 1 e x ln n x d x I_n = \displaystyle\int_1^e x\ln^n x\,dx I n = ∫ 1 e x ln n x d x
a) I 1 = e 2 + 1 4 I_1 = \dfrac{e^2 + 1}{4} I 1 = 4 e 2 + 1
I 1 = x 2 2 ln x ∣ 1 e − ∫ 1 e x 2 d x = e 2 2 − e 2 − 1 4 = e 2 + 1 4 . I_1 = \frac{x^2}{2}\ln x\Big|_1^e - \int_1^e\frac{x}{2}dx = \frac{e^2}{2} - \frac{e^2 - 1}{4} = \frac{e^2 + 1}{4}. I 1 = 2 x 2 ln x 1 e − ∫ 1 e 2 x d x = 2 e 2 − 4 e 2 − 1 = 4 e 2 + 1 .
Barem: 3p, 2p.
b) I n + 1 ≤ I n I_{n+1} \le I_n I n + 1 ≤ I n
I n + 1 − I n = ∫ 1 e x ln n x ( ln x − 1 ) d x ≤ 0 , I_{n+1} - I_n = \int_1^e x\ln^n x(\ln x - 1)\,dx \le 0, I n + 1 − I n = ∫ 1 e x ln n x ( ln x − 1 ) d x ≤ 0 ,
căci pe [ 1 , e ] [1, e] [ 1 , e ] ln x ∈ [ 0 , 1 ] \ln x \in [0, 1] ln x ∈ [ 0 , 1 ] ln n x ≥ 0 \ln^n x \ge 0 ln n x ≥ 0 ln x − 1 ≤ 0 \ln x - 1 \le 0 ln x − 1 ≤ 0 ■ \blacksquare ■ Barem: 2p, 3p.
c) 2 I n + 1 + ( n + 1 ) I n = e 2 2I_{n+1} + (n + 1)I_n = e^2 2 I n + 1 + ( n + 1 ) I n = e 2
Prin părți (u = ln n + 1 x u = \ln^{n+1}x u = ln n + 1 x v ′ = x v' = x v ′ = x
I n + 1 = x 2 2 ln n + 1 x ∣ 1 e − ∫ 1 e x 2 2 ⋅ ( n + 1 ) ln n x x d x = e 2 2 − n + 1 2 I n . I_{n+1} = \frac{x^2}{2}\ln^{n+1}x\Big|_1^e - \int_1^e\frac{x^2}{2}\cdot\frac{(n+1)\ln^n x}{x}dx = \frac{e^2}{2} - \frac{n+1}{2}I_n. I n + 1 = 2 x 2 ln n + 1 x 1 e − ∫ 1 e 2 x 2 ⋅ x ( n + 1 ) l n n x d x = 2 e 2 − 2 n + 1 I n .
Înmulțind cu 2: 2 I n + 1 + ( n + 1 ) I n = e 2 2I_{n+1} + (n + 1)I_n = e^2 2 I n + 1 + ( n + 1 ) I n = e 2 ■ \blacksquare ■ Barem: 3p, 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 z 2 = 2 i z^2 = 2i z 2 = 2 i I.3 x = 4 x = 4 x = 4 I.4 p = 1 2 p = \tfrac12 p = 2 1 II.1.a − 1 -1 − 1 II.1.b m ∈ { 0 , 2 } m \in \{0, 2\} m ∈ { 0 , 2 } II.1.c a = − 1 a = -1 a = − 1 II.2.a 3 3 3 II.2.c x ∈ { 2 , 3 } x \in \{2, 3\} x ∈ { 2 , 3 } III.1.b y = 0 y = 0 y = 0 III.2.a e 2 + 1 4 \tfrac{e^2+1}{4} 4 e 2 + 1
Ce merită reținut din varianta 3
Legea 3 − ( x − 3 ) ( y − 3 ) 3 - (x-3)(y-3) 3 − ( x − 3 ) ( y − 3 ) : compunerea repetată dă 3 + ( − 1 ) n − 1 ( x − 3 ) n 3 + (-1)^{n-1}(x-3)^n 3 + ( − 1 ) n − 1 ( x − 3 ) n Raport cu sin \sin sin cos \cos cos : împarte la cos a \cos a cos a tg a \operatorname{tg} a tg a Integrale cu ln n x \ln^n x ln n x : pe [ 1 , e ] [1, e] [ 1 , e ] 0 ≤ ln x ≤ 1 0 \le \ln x \le 1 0 ≤ ln x ≤ 1