Rezolvarea completă a subiectului de Bacalaureat 2015 la Matematică M_șt-nat , subiectul 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 = 3 + i z_1 = 3 + i z 1 = 3 + i z 2 = 3 − i z_2 = 3 - i z 2 = 3 − i z 1 z 2 z_1z_2 z 1 z 2
z 1 z 2 = ( 3 + i ) ( 3 − i ) = 9 − i 2 = 10 ∈ R . z_1z_2 = (3 + i)(3 - i) = 9 - i^2 = 10 \in \mathbb R. z 1 z 2 = ( 3 + i ) ( 3 − i ) = 9 − i 2 = 10 ∈ R .
Barem: 3p produsul, 2p concluzia.
2. a a a A ( 1 , 1 ) A(1,1) A ( 1 , 1 ) f ( x ) = x + a f(x) = x + a f ( x ) = x + a
f ( 1 ) = 1 ⟺ 1 + a = 1 ⟺ a = 0 f(1) = 1 \iff 1 + a = 1 \iff a = 0 f ( 1 ) = 1 ⟺ 1 + a = 1 ⟺ a = 0 Barem: 3p, 2p.
3. Ecuația x 3 + 2 x − 4 3 = x \sqrt[3]{x^3 + 2x - 4} = x 3 x 3 + 2 x − 4 = x
Ridicăm la cub: x 3 + 2 x − 4 = x 3 ⟺ 2 x = 4 ⟺ x = 2 x^3 + 2x - 4 = x^3 \iff 2x = 4 \iff x = 2 x 3 + 2 x − 4 = x 3 ⟺ 2 x = 4 ⟺ x = 2
Răspuns: x = 2 x = 2 x = 2 Barem: 3p, 2p.
4. Probabilitatea ca un număr din { 1 , … , 80 } \{1, \dots, 80\} { 1 , … , 80 }
Multiplii lui 7: 7 , 14 , … , 77 7, 14, \dots, 77 7 , 14 , … , 77 11 ; posibile: 80 .
p = 11 80 . p = \frac{11}{80}. p = 80 11 .
Barem: 1p, 2p, 2p.
5. O ( 0 , 0 ) O(0,0) O ( 0 , 0 ) A ( 1 , 2 ) A(1,2) A ( 1 , 2 ) B ( 2 , a ) B(2,a) B ( 2 , a )
m O A = 2 m_{OA} = 2 m O A = 2 m O B = a 2 m_{OB} = \dfrac a2 m O B = 2 a ⟺ a 2 = 2 ⟺ a = 4 \iff \dfrac a2 = 2 \iff a = 4 ⟺ 2 a = 2 ⟺ a = 4
Răspuns: a = 4 a = 4 a = 4 Barem: 2p pantele, 3p valoarea.
6. E ( x ) = cos x 2 + sin x E(x) = \cos\dfrac x2 + \sin x E ( x ) = cos 2 x + sin x E ( π 3 ) = 3 E\left(\frac\pi3\right) = \sqrt3 E ( 3 π ) = 3
E ( π 3 ) = cos π 6 + sin π 3 = 3 2 + 3 2 = 3 . E\left(\frac\pi3\right) = \cos\frac\pi6 + \sin\frac\pi3 = \frac{\sqrt3}{2} + \frac{\sqrt3}{2} = \sqrt3. E ( 3 π ) = cos 6 π + sin 3 π = 2 3 + 2 3 = 3 .
Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. A ( x ) = ( 2 x x 2 ) A(x) = \begin{pmatrix}2 & x\\x & 2\end{pmatrix} A ( x ) = ( 2 x x 2 )
a) det ( A ( 0 ) ) = 4 \det(A(0)) = 4 det ( A ( 0 )) = 4
A ( 0 ) = 2 I 2 A(0) = 2I_2 A ( 0 ) = 2 I 2 det = 4 \det = 4 det = 4 Barem: 3p, 2p.
b) a a a A ( 1 ) + A ( 3 ) = a A ( 2 ) A(1) + A(3) = aA(2) A ( 1 ) + A ( 3 ) = a A ( 2 )
A ( 1 ) + A ( 3 ) = ( 4 4 4 4 ) = 2 ( 2 2 2 2 ) = 2 A ( 2 ) . A(1) + A(3) = \begin{pmatrix}4&4\\4&4\end{pmatrix} = 2\begin{pmatrix}2&2\\2&2\end{pmatrix} = 2A(2). A ( 1 ) + A ( 3 ) = ( 4 4 4 4 ) = 2 ( 2 2 2 2 ) = 2 A ( 2 ) .
Răspuns: a = 2 a = 2 a = 2 Barem: 3p, 2p.
c) A ( x ) A ( y ) = 2 A ( x + y ) + x y I 2 A(x)A(y) = 2A(x+y) + xyI_2 A ( x ) A ( y ) = 2 A ( x + y ) + x y I 2
A ( x ) A ( y ) = ( 4 + x y 2 y + 2 x 2 x + 2 y x y + 4 ) , A(x)A(y) = \begin{pmatrix}4 + xy & 2y + 2x\\2x + 2y & xy + 4\end{pmatrix}, A ( x ) A ( y ) = ( 4 + x y 2 x + 2 y 2 y + 2 x x y + 4 ) ,
2 A ( x + y ) + x y I 2 = ( 4 2 ( x + y ) 2 ( x + y ) 4 ) + ( x y 0 0 x y ) = ( 4 + x y 2 ( x + y ) 2 ( x + y ) 4 + x y ) . 2A(x+y) + xyI_2 = \begin{pmatrix}4 & 2(x+y)\\2(x+y) & 4\end{pmatrix} + \begin{pmatrix}xy & 0\\0 & xy\end{pmatrix} = \begin{pmatrix}4 + xy & 2(x+y)\\2(x+y) & 4 + xy\end{pmatrix}. 2 A ( x + y ) + x y I 2 = ( 4 2 ( x + y ) 2 ( x + y ) 4 ) + ( x y 0 0 x y ) = ( 4 + x y 2 ( x + y ) 2 ( x + y ) 4 + x y ) .
Barem: 2p produsul, 3p finalizarea.
2. x ∗ y = 3 x y + 6 x + 6 y + 10 x * y = 3xy + 6x + 6y + 10 x ∗ y = 3 x y + 6 x + 6 y + 10
a) 2 ∗ ( − 2 ) = − 2 2 * (-2) = -2 2 ∗ ( − 2 ) = − 2
3 ⋅ 2 ⋅ ( − 2 ) + 12 − 12 + 10 = − 2. 3\cdot2\cdot(-2) + 12 - 12 + 10 = -2. 3 ⋅ 2 ⋅ ( − 2 ) + 12 − 12 + 10 = − 2.
Barem: 3p, 2p.
b) x ∗ y = 3 ( x + 2 ) ( y + 2 ) − 2 x * y = 3(x+2)(y+2) - 2 x ∗ y = 3 ( x + 2 ) ( y + 2 ) − 2
3 x y + 6 x + 6 y + 12 − 2 = 3 x ( y + 2 ) + 6 ( y + 2 ) − 2 = 3 ( x + 2 ) ( y + 2 ) − 2. 3xy + 6x + 6y + 12 - 2 = 3x(y + 2) + 6(y + 2) - 2 = 3(x+2)(y+2) - 2. 3 x y + 6 x + 6 y + 12 − 2 = 3 x ( y + 2 ) + 6 ( y + 2 ) − 2 = 3 ( x + 2 ) ( y + 2 ) − 2.
Barem: 2p, 3p.
c) x ∗ x ∗ x = x x * x * x = x x ∗ x ∗ x = x
x ∗ x ∗ x = 9 ( x + 2 ) 3 − 2 x * x * x = 9(x+2)^3 - 2 x ∗ x ∗ x = 9 ( x + 2 ) 3 − 2
9 ( x + 2 ) 3 − 2 = x ⟺ 9 ( x + 2 ) 3 = x + 2 ⟺ ( x + 2 ) ( 9 ( x + 2 ) 2 − 1 ) = 0. 9(x+2)^3 - 2 = x \iff 9(x+2)^3 = x + 2 \iff (x + 2)\big(9(x+2)^2 - 1\big) = 0. 9 ( x + 2 ) 3 − 2 = x ⟺ 9 ( x + 2 ) 3 = x + 2 ⟺ ( x + 2 ) ( 9 ( x + 2 ) 2 − 1 ) = 0.
x + 2 = 0 x + 2 = 0 x + 2 = 0 x + 2 = ± 1 3 x + 2 = \pm\frac13 x + 2 = ± 3 1
Răspuns: x ∈ { − 7 3 , − 2 , − 5 3 } x \in \left\{-\frac73, -2, -\frac53\right\} x ∈ { − 3 7 , − 2 , − 3 5 } Barem: 2p expresia, 3p soluțiile.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = ( x + 1 ) e x f(x) = (x + 1)e^x f ( x ) = ( x + 1 ) e x
a) f ′ ( x ) = ( x + 2 ) e x f'(x) = (x + 2)e^x f ′ ( x ) = ( x + 2 ) e x
f ′ ( x ) = e x + ( x + 1 ) e x = ( x + 2 ) e x . f'(x) = e^x + (x + 1)e^x = (x + 2)e^x. f ′ ( x ) = e x + ( x + 1 ) e x = ( x + 2 ) e x .
Barem: 2p, 3p.
b) Tangenta în x = 0 x = 0 x = 0
f ( 0 ) = 1 f(0) = 1 f ( 0 ) = 1 f ′ ( 0 ) = 2 f'(0) = 2 f ′ ( 0 ) = 2 y = 2 x + 1 y = 2x + 1 y = 2 x + 1 Barem: 2p, 3p.
c) f f f [ − 3 , + ∞ ) [-3, +\infty) [ − 3 , + ∞ )
f ′ ′ ( x ) = e x + ( x + 2 ) e x = ( x + 3 ) e x ≥ 0 f''(x) = e^x + (x + 2)e^x = (x + 3)e^x \ge 0 f ′′ ( x ) = e x + ( x + 2 ) e x = ( x + 3 ) e x ≥ 0 x ≥ − 3 x \ge -3 x ≥ − 3 Barem: 2p, 3p.
2. f ( x ) = x 3 + 3 x x 2 + 1 f(x) = \dfrac{x^3 + 3x}{x^2 + 1} f ( x ) = x 2 + 1 x 3 + 3 x
Observație: f ( x ) = x + 2 x x 2 + 1 f(x) = x + \dfrac{2x}{x^2 + 1} f ( x ) = x + x 2 + 1 2 x
a) ∫ − 1 1 ( x 2 + 1 ) f ( x ) d x = 0 \displaystyle\int_{-1}^1(x^2 + 1)f(x)\,dx = 0 ∫ − 1 1 ( x 2 + 1 ) f ( x ) d x = 0
∫ − 1 1 ( x 3 + 3 x ) d x = ( x 4 4 + 3 x 2 2 ) ∣ − 1 1 = ( 1 4 + 3 2 ) − ( 1 4 + 3 2 ) = 0. \int_{-1}^1(x^3 + 3x)\,dx = \left(\frac{x^4}{4} + \frac{3x^2}{2}\right)\Big|_{-1}^1 = \left(\frac14 + \frac32\right) - \left(\frac14 + \frac32\right) = 0. ∫ − 1 1 ( x 3 + 3 x ) d x = ( 4 x 4 + 2 3 x 2 ) − 1 1 = ( 4 1 + 2 3 ) − ( 4 1 + 2 3 ) = 0.
(Sau: integrandul este impar pe un interval simetric.) Barem: 3p, 2p.
b) ∫ 0 1 f ( x ) d x = 1 2 + ln 2 \displaystyle\int_0^1 f(x)\,dx = \frac12 + \ln2 ∫ 0 1 f ( x ) d x = 2 1 + ln 2
∫ 0 1 ( x + 2 x x 2 + 1 ) d x = ( x 2 2 + ln ( x 2 + 1 ) ) ∣ 0 1 = 1 2 + ln 2. \int_0^1\left(x + \frac{2x}{x^2+1}\right)dx = \left(\frac{x^2}{2} + \ln(x^2 + 1)\right)\Big|_0^1 = \frac12 + \ln2. ∫ 0 1 ( x + x 2 + 1 2 x ) d x = ( 2 x 2 + ln ( x 2 + 1 ) ) 0 1 = 2 1 + ln 2.
Barem: 2p descompunerea, 3p calculul.
c) m > 0 m > 0 m > 0 g ( x ) = f ( x ) − x g(x) = f(x) - x g ( x ) = f ( x ) − x ln 2 \ln 2 ln 2
g ( x ) = 2 x x 2 + 1 ≥ 0 g(x) = \dfrac{2x}{x^2 + 1} \ge 0 g ( x ) = x 2 + 1 2 x ≥ 0 [ 0 , m ] [0, m] [ 0 , m ]
A = ∫ 0 m 2 x x 2 + 1 d x = ln ( m 2 + 1 ) = ln 2 ⟺ m 2 = 1 ⟺ m = 1. \mathcal A = \int_0^m\frac{2x}{x^2+1}\,dx = \ln(m^2 + 1) = \ln 2 \iff m^2 = 1 \iff m = 1. A = ∫ 0 m x 2 + 1 2 x d x = ln ( m 2 + 1 ) = ln 2 ⟺ m 2 = 1 ⟺ m = 1.
Răspuns: m = 1 m = 1 m = 1 Barem: 3p aria, 2p valoarea.
Rezumatul răspunsurilor
Cerință Răspuns I.1 z 1 z 2 = 10 z_1z_2 = 10 z 1 z 2 = 10 I.2 a = 0 a = 0 a = 0 I.3 x = 2 x = 2 x = 2 I.4 p = 11 80 p = \frac{11}{80} p = 80 11 I.5 a = 4 a = 4 a = 4 II.1.b a = 2 a = 2 a = 2 II.2.c x ∈ { − 7 3 , − 2 , − 5 3 } x \in \{-\frac73, -2, -\frac53\} x ∈ { − 3 7 , − 2 , − 3 5 } III.1.b y = 2 x + 1 y = 2x + 1 y = 2 x + 1 III.2.c m = 1 m = 1 m = 1
Ce merită reținut din varianta 3
Produsul unui complex cu conjugatul este ∣ z ∣ 2 ∈ R |z|^2 \in \mathbb R ∣ z ∣ 2 ∈ R Legea 3 ( x + 2 ) ( y + 2 ) − 2 3(x+2)(y+2) - 2 3 ( x + 2 ) ( y + 2 ) − 2 : x ∗ x ∗ x = 9 ( x + 2 ) 3 − 2 x * x * x = 9(x+2)^3 - 2 x ∗ x ∗ x = 9 ( x + 2 ) 3 − 2 Funcții impare pe intervale simetrice : integrala este 0 0 0 ∫ 2 x x 2 + 1 = ln ( x 2 + 1 ) \int\frac{2x}{x^2+1} = \ln(x^2 + 1) ∫ x 2 + 1 2 x = ln ( x 2 + 1 ) u ′ / u u'/u u ′ / u