Rezolvarea completă a subiectului de Bacalaureat 2015 la Matematică M_șt-nat , sesiunea august–septembrie, varianta 1 . 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. Al doilea termen al progresiei aritmetice cu a 1 = 1 a_1 = 1 a 1 = 1 r = 2 r = 2 r = 2
a 2 = a 1 + r = 1 + 2 = 3. a_2 = a_1 + r = 1 + 2 = 3. a 2 = a 1 + r = 1 + 2 = 3.
Răspuns: a 2 = 3 a_2 = 3 a 2 = 3 Barem: 3p, 2p.
2. m m m A ( m , 0 ) A(m, 0) A ( m , 0 ) f ( x ) = x + 1 f(x) = x + 1 f ( x ) = x + 1
f ( m ) = 0 ⟺ m + 1 = 0 ⟺ m = − 1 f(m) = 0 \iff m + 1 = 0 \iff m = -1 f ( m ) = 0 ⟺ m + 1 = 0 ⟺ m = − 1
Răspuns: m = − 1 m = -1 m = − 1 Barem: 3p, 2p.
3. Ecuația log 2 ( x 2 + 4 ) = log 2 8 \log_2(x^2 + 4) = \log_2 8 log 2 ( x 2 + 4 ) = log 2 8
x 2 + 4 = 8 ⟺ x 2 = 4 ⟺ x = ± 2 x^2 + 4 = 8 \iff x^2 = 4 \iff x = \pm 2 x 2 + 4 = 8 ⟺ x 2 = 4 ⟺ x = ± 2
Răspuns: x ∈ { − 2 , 2 } x \in \{-2, 2\} x ∈ { − 2 , 2 } Barem: 2p, 3p.
4. Probabilitatea ca un număr din M = { 1 , 2 , … , 8 } M = \{1, 2, \dots, 8\} M = { 1 , 2 , … , 8 }
Cazuri posibile: 8 . Divizibile cu 3: 3 , 6 3, 6 3 , 6 2 .
p = 2 8 = 1 4 . p = \frac28 = \frac14. p = 8 2 = 4 1 .
Barem: 1p, 2p, 2p.
5. a a a u ⃗ = ( a + 1 ) i ⃗ + 4 j ⃗ \vec u = (a + 1)\vec i + 4\vec j u = ( a + 1 ) i + 4 j v ⃗ = i ⃗ + 2 j ⃗ \vec v = \vec i + 2\vec j v = i + 2 j
a + 1 1 = 4 2 ⟺ a + 1 = 2 ⟺ a = 1. \frac{a + 1}{1} = \frac42 \iff a + 1 = 2 \iff a = 1. 1 a + 1 = 2 4 ⟺ a + 1 = 2 ⟺ a = 1.
Răspuns: a = 1 a = 1 a = 1 Barem: 3p, 2p.
6. sin 2 x = 3 2 \sin 2x = \frac{\sqrt3}{2} sin 2 x = 2 3 sin x = 1 2 \sin x = \frac12 sin x = 2 1 x ∈ ( 0 , π 2 ) x \in \left(0, \frac\pi2\right) x ∈ ( 0 , 2 π )
Pe ( 0 , π 2 ) \left(0, \frac\pi2\right) ( 0 , 2 π ) cos x > 0 \cos x > 0 cos x > 0 cos x = 1 − 1 4 = 3 2 \cos x = \sqrt{1 - \frac14} = \frac{\sqrt3}{2} cos x = 1 − 4 1 = 2 3
sin 2 x = 2 sin x cos x = 2 ⋅ 1 2 ⋅ 3 2 = 3 2 . \sin 2x = 2\sin x\cos x = 2\cdot\frac12\cdot\frac{\sqrt3}{2} = \frac{\sqrt3}{2}. sin 2 x = 2 sin x cos x = 2 ⋅ 2 1 ⋅ 2 3 = 2 3 .
Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. A ( a ) = ( a 3 a − 1 2 ) A(a) = \begin{pmatrix} a & 3\\ a - 1 & 2\end{pmatrix} A ( a ) = ( a a − 1 3 2 )
a) A ( 2014 ) + A ( 2016 ) = 2 A ( 2015 ) A(2014) + A(2016) = 2A(2015) A ( 2014 ) + A ( 2016 ) = 2 A ( 2015 )
A ( 2014 ) + A ( 2016 ) = ( 2014 + 2016 6 2013 + 2015 4 ) = ( 4030 6 4028 4 ) = 2 ( 2015 3 2014 2 ) = 2 A ( 2015 ) . A(2014) + A(2016) = \begin{pmatrix} 2014 + 2016 & 6\\ 2013 + 2015 & 4\end{pmatrix} = \begin{pmatrix} 4030 & 6\\ 4028 & 4\end{pmatrix} = 2\begin{pmatrix} 2015 & 3\\ 2014 & 2\end{pmatrix} = 2A(2015). A ( 2014 ) + A ( 2016 ) = ( 2014 + 2016 2013 + 2015 6 4 ) = ( 4030 4028 6 4 ) = 2 ( 2015 2014 3 2 ) = 2 A ( 2015 ) .
Barem: 3p, 2p.
b) a a a det ( A ( a ) ) = 0 \det(A(a)) = 0 det ( A ( a )) = 0
det A ( a ) = 2 a − 3 ( a − 1 ) = 3 − a = 0 ⟺ a = 3. \det A(a) = 2a - 3(a - 1) = 3 - a = 0 \iff a = 3. det A ( a ) = 2 a − 3 ( a − 1 ) = 3 − a = 0 ⟺ a = 3.
Răspuns: a = 3 a = 3 a = 3 Barem: 3p, 2p.
c) Ecuația det ( A ( 2 ) + x A ( 3 ) ) = 0 \det(A(2) + xA(3)) = 0 det ( A ( 2 ) + x A ( 3 )) = 0
A ( 2 ) + x A ( 3 ) = ( 2 + 3 x 3 + 3 x 1 + 2 x 2 + 2 x ) . A(2) + xA(3) = \begin{pmatrix} 2 + 3x & 3 + 3x\\ 1 + 2x & 2 + 2x\end{pmatrix}. A ( 2 ) + x A ( 3 ) = ( 2 + 3 x 1 + 2 x 3 + 3 x 2 + 2 x ) .
det = ( 2 + 3 x ) ( 2 + 2 x ) − ( 3 + 3 x ) ( 1 + 2 x ) = ( 4 + 10 x + 6 x 2 ) − ( 3 + 9 x + 6 x 2 ) = x + 1. \det = (2 + 3x)(2 + 2x) - (3 + 3x)(1 + 2x) = (4 + 10x + 6x^2) - (3 + 9x + 6x^2) = x + 1. det = ( 2 + 3 x ) ( 2 + 2 x ) − ( 3 + 3 x ) ( 1 + 2 x ) = ( 4 + 10 x + 6 x 2 ) − ( 3 + 9 x + 6 x 2 ) = x + 1.
x + 1 = 0 ⟺ x = − 1 x + 1 = 0 \iff x = -1 x + 1 = 0 ⟺ x = − 1
Răspuns: x = − 1 x = -1 x = − 1 Barem: 3p, 2p.
2. x ∗ y = − x y − x − y − 2 x * y = -xy - x - y - 2 x ∗ y = − x y − x − y − 2
a) ( − 1 ) ∗ 1 = − 1 (-1) * 1 = -1 ( − 1 ) ∗ 1 = − 1
( − 1 ) ∗ 1 = 1 + 1 − 1 − 2 = − 1. (-1) * 1 = 1 + 1 - 1 - 2 = -1. ( − 1 ) ∗ 1 = 1 + 1 − 1 − 2 = − 1.
Barem: 3p, 2p.
b) x ∗ y = − ( x + 1 ) ( y + 1 ) − 1 x * y = -(x + 1)(y + 1) - 1 x ∗ y = − ( x + 1 ) ( y + 1 ) − 1
− ( x + 1 ) ( y + 1 ) − 1 = − x y − x − y − 1 − 1 = − x y − x − y − 2 = x ∗ y . -(x+1)(y+1) - 1 = -xy - x - y - 1 - 1 = -xy - x - y - 2 = x * y. − ( x + 1 ) ( y + 1 ) − 1 = − x y − x − y − 1 − 1 = − x y − x − y − 2 = x ∗ y .
Barem: 2p, 3p.
c) Ecuația ( x + 2 ) ∗ ( 2 x − 3 ) = 5 (x + 2) * (2x - 3) = 5 ( x + 2 ) ∗ ( 2 x − 3 ) = 5
( x + 2 ) ∗ ( 2 x − 3 ) = − ( x + 3 ) ( 2 x − 2 ) − 1 = 5 ⟺ ( x + 3 ) ( 2 x − 2 ) = − 6 ⟺ 2 x 2 + 4 x − 6 = − 6 ⟺ 2 x ( x + 2 ) = 0. (x + 2) * (2x - 3) = -(x + 3)(2x - 2) - 1 = 5 \iff (x + 3)(2x - 2) = -6 \iff 2x^2 + 4x - 6 = -6 \iff 2x(x + 2) = 0. ( x + 2 ) ∗ ( 2 x − 3 ) = − ( x + 3 ) ( 2 x − 2 ) − 1 = 5 ⟺ ( x + 3 ) ( 2 x − 2 ) = − 6 ⟺ 2 x 2 + 4 x − 6 = − 6 ⟺ 2 x ( x + 2 ) = 0.
Răspuns: x ∈ { − 2 , 0 } x \in \{-2, 0\} x ∈ { − 2 , 0 } Barem: 2p, 3p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x 4 − 8 x 2 + 16 f(x) = x^4 - 8x^2 + 16 f ( x ) = x 4 − 8 x 2 + 16
a) f ′ ( x ) = 4 x ( x − 2 ) ( x + 2 ) f'(x) = 4x(x - 2)(x + 2) f ′ ( x ) = 4 x ( x − 2 ) ( x + 2 )
f ′ ( x ) = 4 x 3 − 16 x = 4 x ( x 2 − 4 ) = 4 x ( x − 2 ) ( x + 2 ) . f'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x - 2)(x + 2). f ′ ( x ) = 4 x 3 − 16 x = 4 x ( x 2 − 4 ) = 4 x ( x − 2 ) ( x + 2 ) .
Barem: 3p, 2p.
b) lim x → + ∞ f ( x ) − x 4 x 2 + 1 \lim\limits_{x\to+\infty}\dfrac{f(x) - x^4}{x^2 + 1} x → + ∞ lim x 2 + 1 f ( x ) − x 4
lim x → + ∞ − 8 x 2 + 16 x 2 + 1 = − 8. \lim_{x\to+\infty}\frac{-8x^2 + 16}{x^2 + 1} = -8. lim x → + ∞ x 2 + 1 − 8 x 2 + 16 = − 8.
Răspuns: − 8 -8 − 8 Barem: 2p, 3p.
c) Punctele graficului în care tangenta este paralelă cu O x Ox O x
Tangentă orizontală ⟺ f ′ ( x ) = 0 ⟺ x ∈ { − 2 , 0 , 2 } \iff f'(x) = 0 \iff x \in \{-2, 0, 2\} ⟺ f ′ ( x ) = 0 ⟺ x ∈ { − 2 , 0 , 2 } f ( ± 2 ) = 16 − 32 + 16 = 0 f(\pm 2) = 16 - 32 + 16 = 0 f ( ± 2 ) = 16 − 32 + 16 = 0 f ( 0 ) = 16 f(0) = 16 f ( 0 ) = 16
Răspuns: ( − 2 , 0 ) (-2, 0) ( − 2 , 0 ) ( 0 , 16 ) (0, 16) ( 0 , 16 ) ( 2 , 0 ) (2, 0) ( 2 , 0 ) Barem: 2p, 3p.
2. f : ( 0 , + ∞ ) → R f:(0, +\infty) \to \mathbb R f : ( 0 , + ∞ ) → R f ( x ) = x + 2 x f(x) = \dfrac{x + 2}{x} f ( x ) = x x + 2
a) ∫ 1 2 x f ( x ) d x = 7 2 \displaystyle\int_1^2 xf(x)\,dx = \frac72 ∫ 1 2 x f ( x ) d x = 2 7
∫ 1 2 ( x + 2 ) d x = ( x 2 2 + 2 x ) ∣ 1 2 = 6 − 5 2 = 7 2 . \int_1^2(x + 2)\,dx = \left(\frac{x^2}{2} + 2x\right)\Big|_1^2 = 6 - \frac52 = \frac72. ∫ 1 2 ( x + 2 ) d x = ( 2 x 2 + 2 x ) 1 2 = 6 − 2 5 = 2 7 .
Barem: 3p, 2p.
b) F ( x ) = x + 2 ln x + 2015 F(x) = x + 2\ln x + 2015 F ( x ) = x + 2 ln x + 2015 f f f
F ′ ( x ) = 1 + 2 x = x + 2 x = f ( x ) , x > 0. F'(x) = 1 + \frac2x = \frac{x + 2}{x} = f(x), \quad x > 0. F ′ ( x ) = 1 + x 2 = x x + 2 = f ( x ) , x > 0.
Barem: 3p, 2p.
c) Aria delimitată de graficul lui g ( x ) = ( f ( x ) − 1 ) ln x g(x) = (f(x) - 1)\ln x g ( x ) = ( f ( x ) − 1 ) ln x O x Ox O x x = 1 x = 1 x = 1 x = e x = e x = e
g ( x ) = 2 x ln x ≥ 0 g(x) = \frac2x\ln x \ge 0 g ( x ) = x 2 ln x ≥ 0 [ 1 , e ] [1, e] [ 1 , e ] 2 ln x x = 2 ln x ⋅ ( ln x ) ′ \frac{2\ln x}{x} = 2\ln x\cdot(\ln x)' x 2 l n x = 2 ln x ⋅ ( ln x ) ′
A = ∫ 1 e 2 ln x x d x = ln 2 x ∣ 1 e = 1 − 0 = 1. ■ \mathcal A = \int_1^e\frac{2\ln x}{x}\,dx = \ln^2 x\Big|_1^e = 1 - 0 = 1. \ \blacksquare A = ∫ 1 e x 2 l n x d x = ln 2 x 1 e = 1 − 0 = 1. ■
Barem: 3p, 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 a 2 = 3 a_2 = 3 a 2 = 3 I.2 m = − 1 m = -1 m = − 1 I.3 x ∈ { − 2 , 2 } x \in \{-2, 2\} x ∈ { − 2 , 2 } I.4 p = 1 4 p = \tfrac14 p = 4 1 I.5 a = 1 a = 1 a = 1 II.1.b a = 3 a = 3 a = 3 II.1.c x = − 1 x = -1 x = − 1 II.2.c x ∈ { − 2 , 0 } x \in \{-2, 0\} x ∈ { − 2 , 0 } III.1.b − 8 -8 − 8 III.1.c ( − 2 , 0 ) (-2,0) ( − 2 , 0 ) ( 0 , 16 ) (0,16) ( 0 , 16 ) ( 2 , 0 ) (2,0) ( 2 , 0 )
Ce merită reținut din varianta 1
Tangentă paralelă cu O x Ox O x ⟺ \iff ⟺ Legea − ( x + 1 ) ( y + 1 ) − 1 -(x+1)(y+1) - 1 − ( x + 1 ) ( y + 1 ) − 1 : forma factorizată transformă ecuațiile în ecuații algebrice simple.∫ ln x x d x = ln 2 x 2 \int\frac{\ln x}{x}dx = \frac{\ln^2 x}{2} ∫ x l n x d x = 2 l n 2 x u ⋅ u ′ u\cdot u' u ⋅ u ′ x 4 − 8 x 2 + 16 = ( x 2 − 4 ) 2 x^4 - 8x^2 + 16 = (x^2 - 4)^2 x 4 − 8 x 2 + 16 = ( x 2 − 4 ) 2 ( ± 2 , 0 ) (\pm 2, 0) ( ± 2 , 0 ) ( 0 , 16 ) (0, 16) ( 0 , 16 )