Rezolvarea completă a subiectului de Bacalaureat 2015 la Matematică M_șt-nat , sesiunea de rezervă, varianta 5 . 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. ( 2 − 3 i ) ( 2 + 3 i ) (2 - 3i)(2 + 3i) ( 2 − 3 i ) ( 2 + 3 i )
( 2 − 3 i ) ( 2 + 3 i ) = 4 − 9 i 2 = 4 + 9 = 13. (2 - 3i)(2 + 3i) = 4 - 9i^2 = 4 + 9 = 13. ( 2 − 3 i ) ( 2 + 3 i ) = 4 − 9 i 2 = 4 + 9 = 13.
Răspuns: 13 13 13 Barem: 3p, 2p.
2. f ( f ( 3 ) ) f(f(3)) f ( f ( 3 )) f ( x ) = 2 x − 1 f(x) = 2x - 1 f ( x ) = 2 x − 1
f ( 3 ) = 5 f(3) = 5 f ( 3 ) = 5 f ( 5 ) = 9 f(5) = 9 f ( 5 ) = 9
Răspuns: 9 9 9 Barem: 2p, 3p.
3. Ecuația log 3 ( x 2 + 17 ) = log 3 81 \log_3(x^2 + 17) = \log_3 81 log 3 ( x 2 + 17 ) = log 3 81
x 2 + 17 = 81 ⟺ x 2 = 64 ⟺ x = ± 8 x^2 + 17 = 81 \iff x^2 = 64 \iff x = \pm 8 x 2 + 17 = 81 ⟺ x 2 = 64 ⟺ x = ± 8
Răspuns: x ∈ { − 8 , 8 } x \in \{-8, 8\} x ∈ { − 8 , 8 } Barem: 2p, 3p.
4. Probabilitatea ca un număr de două cifre să fie divizibil cu 5
Cazuri posibile: 90 . Divizibile cu 5: 10 , 15 , … , 95 10, 15, \dots, 95 10 , 15 , … , 95 18 .
p = 18 90 = 1 5 . p = \frac{18}{90} = \frac15. p = 90 18 = 5 1 .
Barem: 1p, 2p, 2p.
5. A ( 1 , a ) A(1, a) A ( 1 , a ) B ( 3 , 2 ) B(3, 2) B ( 3 , 2 ) C ( 2 , 1 ) C(2, 1) C ( 2 , 1 )
m B C = 1 − 2 2 − 3 = 1 m_{BC} = \dfrac{1 - 2}{2 - 3} = 1 m B C = 2 − 3 1 − 2 = 1 m A B = 2 − a 3 − 1 = 2 − a 2 m_{AB} = \dfrac{2 - a}{3 - 1} = \dfrac{2 - a}{2} m A B = 3 − 1 2 − a = 2 2 − a ⟺ m A B = m B C ⟺ 2 − a = 2 ⟺ a = 0 \iff m_{AB} = m_{BC} \iff 2 - a = 2 \iff a = 0 ⟺ m A B = m B C ⟺ 2 − a = 2 ⟺ a = 0
Răspuns: a = 0 a = 0 a = 0 Barem: 2p, 3p.
6. E ( x ) = sin x 3 + cos x 2 E(x) = \sin\frac x3 + \cos\frac x2 E ( x ) = sin 3 x + cos 2 x E ( π 2 ) = 1 + 2 2 E\left(\frac\pi2\right) = \frac{1 + \sqrt2}{2} E ( 2 π ) = 2 1 + 2
E ( π 2 ) = sin π 6 + cos π 4 = 1 2 + 2 2 = 1 + 2 2 . E\left(\frac\pi2\right) = \sin\frac\pi6 + \cos\frac\pi4 = \frac12 + \frac{\sqrt2}{2} = \frac{1 + \sqrt2}{2}. E ( 2 π ) = sin 6 π + cos 4 π = 2 1 + 2 2 = 2 1 + 2 .
Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. A ( a ) = ( 1 2 a 2 a 4 ) A(a) = \begin{pmatrix} 1 & 2a\\ 2a & 4\end{pmatrix} A ( a ) = ( 1 2 a 2 a 4 )
a) A ( 1 ) + A ( − 1 ) = 2 A ( 0 ) A(1) + A(-1) = 2A(0) A ( 1 ) + A ( − 1 ) = 2 A ( 0 )
( 1 2 2 4 ) + ( 1 − 2 − 2 4 ) = ( 2 0 0 8 ) = 2 ( 1 0 0 4 ) = 2 A ( 0 ) . \begin{pmatrix} 1 & 2\\ 2 & 4\end{pmatrix} + \begin{pmatrix} 1 & -2\\ -2 & 4\end{pmatrix} = \begin{pmatrix} 2 & 0\\ 0 & 8\end{pmatrix} = 2\begin{pmatrix} 1 & 0\\ 0 & 4\end{pmatrix} = 2A(0). ( 1 2 2 4 ) + ( 1 − 2 − 2 4 ) = ( 2 0 0 8 ) = 2 ( 1 0 0 4 ) = 2 A ( 0 ) .
Barem: 3p, 2p.
b) a a a det ( A ( a ) ) = 0 \det(A(a)) = 0 det ( A ( a )) = 0
det A ( a ) = 4 − 4 a 2 = 0 ⟺ a ∈ { − 1 , 1 } . \det A(a) = 4 - 4a^2 = 0 \iff a \in \{-1, 1\}. det A ( a ) = 4 − 4 a 2 = 0 ⟺ a ∈ { − 1 , 1 } .
Răspuns: a ∈ { − 1 , 1 } a \in \{-1, 1\} a ∈ { − 1 , 1 } Barem: 3p, 2p.
c) Ecuația A ( 2 ) ⋅ X = A ( 8 ) A(2)\cdot X = A(8) A ( 2 ) ⋅ X = A ( 8 )
A ( 2 ) = ( 1 4 4 4 ) A(2) = \begin{pmatrix} 1 & 4\\ 4 & 4\end{pmatrix} A ( 2 ) = ( 1 4 4 4 ) det A ( 2 ) = − 12 ≠ 0 \det A(2) = -12 \ne 0 det A ( 2 ) = − 12 = 0
( A ( 2 ) ) − 1 = − 1 12 ( 4 − 4 − 4 1 ) = ( − 1 3 1 3 1 3 − 1 12 ) . (A(2))^{-1} = -\frac{1}{12}\begin{pmatrix} 4 & -4\\ -4 & 1\end{pmatrix} = \begin{pmatrix} -\frac13 & \frac13\\ \frac13 & -\frac{1}{12}\end{pmatrix}. ( A ( 2 ) ) − 1 = − 12 1 ( 4 − 4 − 4 1 ) = ( − 3 1 3 1 3 1 − 12 1 ) .
X = ( A ( 2 ) ) − 1 A ( 8 ) = ( − 1 3 1 3 1 3 − 1 12 ) ( 1 16 16 4 ) = ( 5 − 4 − 1 5 ) . X = (A(2))^{-1}A(8) = \begin{pmatrix} -\frac13 & \frac13\\ \frac13 & -\frac{1}{12}\end{pmatrix}\begin{pmatrix} 1 & 16\\ 16 & 4\end{pmatrix} = \begin{pmatrix} 5 & -4\\ -1 & 5\end{pmatrix}. X = ( A ( 2 ) ) − 1 A ( 8 ) = ( − 3 1 3 1 3 1 − 12 1 ) ( 1 16 16 4 ) = ( 5 − 1 − 4 5 ) .
Verificare: A ( 2 ) X = ( 5 − 4 − 4 + 20 20 − 4 − 16 + 20 ) = ( 1 16 16 4 ) = A ( 8 ) A(2)X = \begin{pmatrix} 5 - 4 & -4 + 20\\ 20 - 4 & -16 + 20\end{pmatrix} = \begin{pmatrix} 1 & 16\\ 16 & 4\end{pmatrix} = A(8) A ( 2 ) X = ( 5 − 4 20 − 4 − 4 + 20 − 16 + 20 ) = ( 1 16 16 4 ) = A ( 8 )
Răspuns: X = ( 5 − 4 − 1 5 ) X = \begin{pmatrix} 5 & -4\\ -1 & 5\end{pmatrix} X = ( 5 − 1 − 4 5 ) Barem: 3p, 2p.
2. x ∘ y = 2 x y − 6 x − 6 y + 21 x \circ y = 2xy - 6x - 6y + 21 x ∘ y = 2 x y − 6 x − 6 y + 21
a) ( − 3 ) ∘ 3 = 3 (-3)\circ 3 = 3 ( − 3 ) ∘ 3 = 3
( − 3 ) ∘ 3 = − 18 + 18 − 18 + 21 = 3. (-3)\circ 3 = -18 + 18 - 18 + 21 = 3. ( − 3 ) ∘ 3 = − 18 + 18 − 18 + 21 = 3.
Barem: 3p, 2p.
b) x ∘ y = 2 ( x − 3 ) ( y − 3 ) + 3 x \circ y = 2(x - 3)(y - 3) + 3 x ∘ y = 2 ( x − 3 ) ( y − 3 ) + 3
2 ( x y − 3 x − 3 y + 9 ) + 3 = 2 x y − 6 x − 6 y + 21. 2(xy - 3x - 3y + 9) + 3 = 2xy - 6x - 6y + 21. 2 ( x y − 3 x − 3 y + 9 ) + 3 = 2 x y − 6 x − 6 y + 21.
Barem: 2p, 3p.
c) 1 ∘ 2 ∘ 3 ∘ ⋯ ∘ 2015 1\circ\sqrt2\circ\sqrt3\circ\dots\circ\sqrt{2015} 1 ∘ 2 ∘ 3 ∘ ⋯ ∘ 2015
Din b), x ∘ 3 = 3 ∘ y = 3 x \circ 3 = 3 \circ y = 3 x ∘ 3 = 3 ∘ y = 3 3 3 3 9 = 3 \sqrt9 = 3 9 = 3
( 1 ∘ 2 ∘ ⋯ ∘ 8 ) ∘ 3 ∘ ( 10 ∘ ⋯ ∘ 2015 ) = 3. \left(1\circ\sqrt2\circ\dots\circ\sqrt8\right)\circ 3\circ\left(\sqrt{10}\circ\dots\circ\sqrt{2015}\right) = 3. ( 1 ∘ 2 ∘ ⋯ ∘ 8 ) ∘ 3 ∘ ( 10 ∘ ⋯ ∘ 2015 ) = 3.
Răspuns: 3 3 3 Barem: 2p, 3p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = 3 e x + x 2 f(x) = 3e^x + x^2 f ( x ) = 3 e x + x 2
a) lim x → 0 f ( x ) − f ( 0 ) x = 3 \lim\limits_{x\to 0}\dfrac{f(x) - f(0)}{x} = 3 x → 0 lim x f ( x ) − f ( 0 ) = 3
Limita este f ′ ( 0 ) f'(0) f ′ ( 0 ) f ′ ( x ) = 3 e x + 2 x f'(x) = 3e^x + 2x f ′ ( x ) = 3 e x + 2 x f ′ ( 0 ) = 3 f'(0) = 3 f ′ ( 0 ) = 3 Barem: 2p, 3p.
b) Tangenta în punctul de abscisă x = 0 x = 0 x = 0
f ( 0 ) = 3 f(0) = 3 f ( 0 ) = 3 f ′ ( 0 ) = 3 f'(0) = 3 f ′ ( 0 ) = 3 y − 3 = 3 x ⟺ y = 3 x + 3 y - 3 = 3x \iff y = 3x + 3 y − 3 = 3 x ⟺ y = 3 x + 3
Răspuns: y = 3 x + 3 y = 3x + 3 y = 3 x + 3 Barem: 2p, 3p.
c) f f f R \mathbb R R
f ′ ′ ( x ) = 3 e x + 2 > 0 f''(x) = 3e^x + 2 > 0 f ′′ ( x ) = 3 e x + 2 > 0 x x x f f f Barem: 2p, 3p.
2. f : ( 0 , + ∞ ) → R f:(0, +\infty) \to \mathbb R f : ( 0 , + ∞ ) → R f ( x ) = x + 1 x f(x) = x + \frac1x f ( x ) = x + x 1
a) ∫ 1 3 ( f ( x ) − 1 x ) d x = 4 \displaystyle\int_1^3\left(f(x) - \frac1x\right)dx = 4 ∫ 1 3 ( f ( x ) − x 1 ) d x = 4
∫ 1 3 x d x = x 2 2 ∣ 1 3 = 9 − 1 2 = 4. \int_1^3 x\,dx = \frac{x^2}{2}\Big|_1^3 = \frac{9 - 1}{2} = 4. ∫ 1 3 x d x = 2 x 2 1 3 = 2 9 − 1 = 4.
Barem: 3p, 2p.
b) ∫ 1 2 ( f ( x ) − 1 x ) e x d x = e 2 \displaystyle\int_1^2\left(f(x) - \frac1x\right)e^x\,dx = e^2 ∫ 1 2 ( f ( x ) − x 1 ) e x d x = e 2
Prin părți:
∫ 1 2 x e x d x = x e x ∣ 1 2 − ∫ 1 2 e x d x = 2 e 2 − e − ( e 2 − e ) = e 2 . \int_1^2 xe^x\,dx = xe^x\Big|_1^2 - \int_1^2 e^x\,dx = 2e^2 - e - (e^2 - e) = e^2. ∫ 1 2 x e x d x = x e x 1 2 − ∫ 1 2 e x d x = 2 e 2 − e − ( e 2 − e ) = e 2 .
Barem: 3p, 2p.
c) a > 1 a > 1 a > 1 G f G_f G f O x Ox O x x = 1 x = 1 x = 1 x = a x = a x = a 4 + ln a 4 + \ln a 4 + ln a
f > 0 f > 0 f > 0
A = ∫ 1 a ( x + 1 x ) d x = ( x 2 2 + ln x ) ∣ 1 a = a 2 − 1 2 + ln a . \mathcal A = \int_1^a\left(x + \frac1x\right)dx = \left(\frac{x^2}{2} + \ln x\right)\Big|_1^a = \frac{a^2 - 1}{2} + \ln a. A = ∫ 1 a ( x + x 1 ) d x = ( 2 x 2 + ln x ) 1 a = 2 a 2 − 1 + ln a .
a 2 − 1 2 + ln a = 4 + ln a ⟺ a 2 = 9 ⇒ a = 3 ( a > 1 ) . \frac{a^2 - 1}{2} + \ln a = 4 + \ln a \iff a^2 = 9 \Rightarrow a = 3\ (a > 1). 2 a 2 − 1 + ln a = 4 + ln a ⟺ a 2 = 9 ⇒ a = 3 ( a > 1 ) .
Răspuns: a = 3 a = 3 a = 3 Barem: 3p, 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 13 13 13 I.2 9 9 9 I.3 x ∈ { − 8 , 8 } x \in \{-8, 8\} x ∈ { − 8 , 8 } I.4 p = 1 5 p = \tfrac15 p = 5 1 I.5 a = 0 a = 0 a = 0 II.1.b a ∈ { − 1 , 1 } a \in \{-1, 1\} a ∈ { − 1 , 1 } II.1.c X = ( 5 − 4 − 1 5 ) X = \begin{pmatrix}5&-4\\-1&5\end{pmatrix} X = ( 5 − 1 − 4 5 ) II.2.c 3 3 3 III.1.b y = 3 x + 3 y = 3x + 3 y = 3 x + 3 III.2.c a = 3 a = 3 a = 3
Ce merită reținut din varianta 5
( a − b i ) ( a + b i ) = a 2 + b 2 (a - bi)(a + bi) = a^2 + b^2 ( a − bi ) ( a + bi ) = a 2 + b 2 Coliniaritate prin pante : pantele A B AB A B B C BC B C Element absorbant ascuns : printre k \sqrt k k 9 = 3 \sqrt 9 = 3 9 = 3 ∫ x e x d x = ( x − 1 ) e x + C \int xe^x\,dx = (x - 1)e^x + C ∫ x e x d x = ( x − 1 ) e x + C