Rezolvarea completă a subiectului de Bacalaureat 2017 la Matematică M_șt-nat , sesiunea iunie–iulie, varianta 2 . 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 treilea termen al progresiei aritmetice cu a 1 = 4 a_1 = 4 a 1 = 4 a 2 = 7 a_2 = 7 a 2 = 7
r = a 2 − a 1 = 3 r = a_2 - a_1 = 3 r = a 2 − a 1 = 3 a 3 = a 2 + r = 10 a_3 = a_2 + r = 10 a 3 = a 2 + r = 10
Răspuns: a 3 = 10 a_3 = 10 a 3 = 10 Barem: 3p rația, 2p a 3 a_3 a 3
2. x 1 , x 2 x_1, x_2 x 1 , x 2 x 2 − 4 x + 1 = 0 x^2 - 4x + 1 = 0 x 2 − 4 x + 1 = 0 4 x 1 x 2 − ( x 1 + x 2 ) = 0 4x_1x_2 - (x_1 + x_2) = 0 4 x 1 x 2 − ( x 1 + x 2 ) = 0
Viète: x 1 + x 2 = 4 x_1 + x_2 = 4 x 1 + x 2 = 4 x 1 x 2 = 1 x_1x_2 = 1 x 1 x 2 = 1 4 ⋅ 1 − 4 = 0 4\cdot 1 - 4 = 0 4 ⋅ 1 − 4 = 0 Barem: 2p, 3p.
3. 2 2 x + 1 = 1 8 2^{2x+1} = \dfrac18 2 2 x + 1 = 8 1
1 8 = 2 − 3 \dfrac18 = 2^{-3} 8 1 = 2 − 3 2 x + 1 = − 3 ⟺ x = − 2 2x + 1 = -3 \iff x = -2 2 x + 1 = − 3 ⟺ x = − 2
Răspuns: x = − 2 x = -2 x = − 2 Barem: 3p, 2p.
4. Probabilitatea ca un număr de două cifre să fie multiplu de 15
90 cazuri posibile; multiplii de 15: 15 , 30 , 45 , 60 , 75 , 90 15, 30, 45, 60, 75, 90 15 , 30 , 45 , 60 , 75 , 90
p = 6 90 = 1 15 . p = \frac{6}{90} = \frac{1}{15}. p = 90 6 = 15 1 .
Răspuns: p = 1 15 p = \dfrac1{15} p = 15 1 Barem: 2p, 2p, 1p.
5. A ( 0 , 1 ) A(0,1) A ( 0 , 1 ) B ( 1 , 1 ) B(1,1) B ( 1 , 1 ) C ( 3 , a ) C(3,a) C ( 3 , a )
m A B = 1 − 1 1 − 0 = 0 m_{AB} = \dfrac{1 - 1}{1 - 0} = 0 m A B = 1 − 0 1 − 1 = 0 m A C = a − 1 3 m_{AC} = \dfrac{a - 1}{3} m A C = 3 a − 1 ⟺ m A B = m A C ⟺ a − 1 = 0 \iff m_{AB} = m_{AC} \iff a - 1 = 0 ⟺ m A B = m A C ⟺ a − 1 = 0
Răspuns: a = 1 a = 1 a = 1 Barem: 2p, 3p.
6. A B = 4 3 AB = 4\sqrt3 A B = 4 3 A C = 4 AC = 4 A C = 4 sin C = 3 2 \sin C = \dfrac{\sqrt3}{2} sin C = 2 3 sin B \sin B sin B
Teorema sinusurilor: A B sin C = A C sin B \dfrac{AB}{\sin C} = \dfrac{AC}{\sin B} sin C A B = sin B A C
sin B = A C ⋅ sin C A B = 4 ⋅ 3 2 4 3 = 1 2 . \sin B = \frac{AC\cdot\sin C}{AB} = \frac{4\cdot\frac{\sqrt3}{2}}{4\sqrt3} = \frac12. sin B = A B A C ⋅ s i n C = 4 3 4 ⋅ 2 3 = 2 1 .
Răspuns: sin B = 1 2 \sin B = \dfrac12 sin B = 2 1 Barem: 3p, 2p.
Subiectul al II-lea (30 de puncte)
1. A ( x ) = ( 0 x x 0 ) A(x) = \begin{pmatrix} 0 & x\\ x & 0\end{pmatrix} A ( x ) = ( 0 x x 0 )
a) det ( A ( 1 ) ) = − 1 \det(A(1)) = -1 det ( A ( 1 )) = − 1
det ( 0 1 1 0 ) = 0 − 1 = − 1 \det\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix} = 0 - 1 = -1 det ( 0 1 1 0 ) = 0 − 1 = − 1 Barem: 2p, 3p.
b) A ( x ) A ( y ) = x y I 2 A(x)A(y) = xyI_2 A ( x ) A ( y ) = x y I 2
A ( x ) A ( y ) = ( 0 x x 0 ) ( 0 y y 0 ) = ( x y 0 0 x y ) = x y I 2 . A(x)A(y) = \begin{pmatrix} 0 & x\\ x & 0\end{pmatrix}\begin{pmatrix} 0 & y\\ y & 0\end{pmatrix} = \begin{pmatrix} xy & 0\\ 0 & xy\end{pmatrix} = xyI_2. A ( x ) A ( y ) = ( 0 x x 0 ) ( 0 y y 0 ) = ( x y 0 0 x y ) = x y I 2 .
Barem: 3p, 2p.
c) A ( 3 a ) A ( 3 a + 1 ) A ( 3 a + 2 ) = A ( 27 ) A(3^a)A(3^{a+1})A(3^{a+2}) = A(27) A ( 3 a ) A ( 3 a + 1 ) A ( 3 a + 2 ) = A ( 27 )
Din b): A ( 3 a ) A ( 3 a + 1 ) = 3 2 a + 1 I 2 A(3^a)A(3^{a+1}) = 3^{2a+1}I_2 A ( 3 a ) A ( 3 a + 1 ) = 3 2 a + 1 I 2 A ( 3 a + 2 ) A(3^{a+2}) A ( 3 a + 2 ) A ( 3 2 a + 1 ⋅ 3 a + 2 ) = A ( 3 3 a + 3 ) A(3^{2a+1}\cdot 3^{a+2}) = A(3^{3a+3}) A ( 3 2 a + 1 ⋅ 3 a + 2 ) = A ( 3 3 a + 3 )
3 3 a + 3 = 27 = 3 3 ⟺ 3 a + 3 = 3 ⟺ a = 0. 3^{3a+3} = 27 = 3^3 \iff 3a + 3 = 3 \iff a = 0. 3 3 a + 3 = 27 = 3 3 ⟺ 3 a + 3 = 3 ⟺ a = 0.
Răspuns: a = 0 a = 0 a = 0 Barem: 3p, 2p.
2. f = X 3 + m X 2 + 2 X − 4 f = X^3 + mX^2 + 2X - 4 f = X 3 + m X 2 + 2 X − 4
a) Pentru m = 1 m = 1 m = 1 f ( 1 ) = 0 f(1) = 0 f ( 1 ) = 0
f ( 1 ) = 1 + 1 + 2 − 4 = 0 f(1) = 1 + 1 + 2 - 4 = 0 f ( 1 ) = 1 + 1 + 2 − 4 = 0 Barem: 3p, 2p.
b) Dacă X + 2 ∣ f X + 2 \mid f X + 2 ∣ f X + 3 X + 3 X + 3 − 1 -1 − 1
f ( − 2 ) = − 8 + 4 m − 4 − 4 = 4 m − 16 = 0 ⇒ m = 4 f(-2) = -8 + 4m - 4 - 4 = 4m - 16 = 0 \Rightarrow m = 4 f ( − 2 ) = − 8 + 4 m − 4 − 4 = 4 m − 16 = 0 ⇒ m = 4 f = X 3 + 4 X 2 + 2 X − 4 f = X^3 + 4X^2 + 2X - 4 f = X 3 + 4 X 2 + 2 X − 4
Restul la X + 3 X + 3 X + 3 f ( − 3 ) = − 27 + 36 − 6 − 4 = − 1 f(-3) = -27 + 36 - 6 - 4 = -1 f ( − 3 ) = − 27 + 36 − 6 − 4 = − 1 ■ \blacksquare ■ Barem: 3p m m m
c) 1 x 1 + 1 x 2 + 1 x 3 + x 1 + x 2 + x 3 = 1 2 \dfrac1{x_1} + \dfrac1{x_2} + \dfrac1{x_3} + x_1 + x_2 + x_3 = \dfrac12 x 1 1 + x 2 1 + x 3 1 + x 1 + x 2 + x 3 = 2 1
Viète: x 1 + x 2 + x 3 = − m x_1 + x_2 + x_3 = -m x 1 + x 2 + x 3 = − m x 1 x 2 + x 2 x 3 + x 3 x 1 = 2 x_1x_2 + x_2x_3 + x_3x_1 = 2 x 1 x 2 + x 2 x 3 + x 3 x 1 = 2 x 1 x 2 x 3 = 4 x_1x_2x_3 = 4 x 1 x 2 x 3 = 4
x 1 x 2 + x 2 x 3 + x 3 x 1 x 1 x 2 x 3 + ( x 1 + x 2 + x 3 ) = 2 4 − m = 1 2 − m . \frac{x_1x_2 + x_2x_3 + x_3x_1}{x_1x_2x_3} + (x_1 + x_2 + x_3) = \frac24 - m = \frac12 - m. x 1 x 2 x 3 x 1 x 2 + x 2 x 3 + x 3 x 1 + ( x 1 + x 2 + x 3 ) = 4 2 − m = 2 1 − m .
1 2 − m = 1 2 ⟺ m = 0 \dfrac12 - m = \dfrac12 \iff m = 0 2 1 − m = 2 1 ⟺ m = 0
Răspuns: m = 0 m = 0 m = 0 Barem: 3p, 2p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x + 2017 e x f(x) = \dfrac{x + 2017}{e^x} f ( x ) = e x x + 2017
a) f ′ ( x ) = − ( x + 2016 ) e x f'(x) = \dfrac{-(x + 2016)}{e^x} f ′ ( x ) = e x − ( x + 2016 )
f ′ ( x ) = e x − ( x + 2017 ) e x e 2 x = 1 − x − 2017 e x = − ( x + 2016 ) e x . f'(x) = \frac{e^x - (x + 2017)e^x}{e^{2x}} = \frac{1 - x - 2017}{e^x} = \frac{-(x + 2016)}{e^x}. f ′ ( x ) = e 2 x e x − ( x + 2017 ) e x = e x 1 − x − 2017 = e x − ( x + 2016 ) .
Barem: 3p, 2p.
b) Tangenta în x = 0 x = 0 x = 0
f ( 0 ) = 2017 f(0) = 2017 f ( 0 ) = 2017 f ′ ( 0 ) = − 2016 f'(0) = -2016 f ′ ( 0 ) = − 2016 y = − 2016 x + 2017 y = -2016x + 2017 y = − 2016 x + 2017 Barem: 2p, 3p.
c) f f f [ − 2015 , + ∞ ) [-2015, +\infty) [ − 2015 , + ∞ )
f ( x ) = − ( x + 2016 ) e − x f(x) = -(x + 2016)e^{-x} f ( x ) = − ( x + 2016 ) e − x f ′ ′ ( x ) = − e − x + ( x + 2016 ) e − x = x + 2015 e x f''(x) = -e^{-x} + (x + 2016)e^{-x} = \dfrac{x + 2015}{e^x} f ′′ ( x ) = − e − x + ( x + 2016 ) e − x = e x x + 2015
Pentru x ≥ − 2015 x \ge -2015 x ≥ − 2015 f ′ ′ ( x ) ≥ 0 f''(x) \ge 0 f ′′ ( x ) ≥ 0 f f f [ − 2015 , + ∞ ) [-2015, +\infty) [ − 2015 , + ∞ ) ■ \blacksquare ■ Barem: 2p, 3p.
2. f ( x ) = 1 x 2 + 1 f(x) = \dfrac{1}{x^2 + 1} f ( x ) = x 2 + 1 1
a) ∫ 0 1 1 f ( x ) d x = 4 3 \displaystyle\int_0^1 \frac{1}{f(x)}dx = \frac43 ∫ 0 1 f ( x ) 1 d x = 3 4
∫ 0 1 ( x 2 + 1 ) d x = ( x 3 3 + x ) ∣ 0 1 = 1 3 + 1 = 4 3 . \int_0^1(x^2 + 1)\,dx = \left(\frac{x^3}{3} + x\right)\Big|_0^1 = \frac13 + 1 = \frac43. ∫ 0 1 ( x 2 + 1 ) d x = ( 3 x 3 + x ) 0 1 = 3 1 + 1 = 3 4 .
Barem: 3p, 2p.
b) Primitiva F F F F ( 1 ) = π 4 + 1 F(1) = \dfrac\pi4 + 1 F ( 1 ) = 4 π + 1
F ( x ) = arctg x + c F(x) = \operatorname{arctg} x + c F ( x ) = arctg x + c F ( 1 ) = π 4 + c = π 4 + 1 ⇒ c = 1 F(1) = \dfrac\pi4 + c = \dfrac\pi4 + 1 \Rightarrow c = 1 F ( 1 ) = 4 π + c = 4 π + 1 ⇒ c = 1
Răspuns: F ( x ) = arctg x + 1 F(x) = \operatorname{arctg} x + 1 F ( x ) = arctg x + 1 Barem: 2p, 3p.
c) n n n ∫ 0 n x f ( x ) d x = 1 2 ln 5 \displaystyle\int_0^n xf(x)\,dx = \frac12\ln5 ∫ 0 n x f ( x ) d x = 2 1 ln 5
∫ 0 n x x 2 + 1 d x = 1 2 ln ( x 2 + 1 ) ∣ 0 n = 1 2 ln ( n 2 + 1 ) . \int_0^n\frac{x}{x^2 + 1}dx = \frac12\ln(x^2 + 1)\Big|_0^n = \frac12\ln(n^2 + 1). ∫ 0 n x 2 + 1 x d x = 2 1 ln ( x 2 + 1 ) 0 n = 2 1 ln ( n 2 + 1 ) .
ln ( n 2 + 1 ) = ln 5 ⟺ n 2 = 4 ⟺ n = 2 \ln(n^2 + 1) = \ln 5 \iff n^2 = 4 \iff n = 2 ln ( n 2 + 1 ) = ln 5 ⟺ n 2 = 4 ⟺ n = 2 n n n
Răspuns: n = 2 n = 2 n = 2 Barem: 3p, 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.1 a 3 = 10 a_3 = 10 a 3 = 10 I.3 x = − 2 x = -2 x = − 2 I.4 p = 1 15 p = \tfrac1{15} p = 15 1 I.5 a = 1 a = 1 a = 1 I.6 sin B = 1 2 \sin B = \tfrac12 sin B = 2 1 II.1.c a = 0 a = 0 a = 0 II.2.c m = 0 m = 0 m = 0 III.1.b y = − 2016 x + 2017 y = -2016x + 2017 y = − 2016 x + 2017 III.2.b F ( x ) = arctg x + 1 F(x) = \operatorname{arctg}x + 1 F ( x ) = arctg x + 1 III.2.c n = 2 n = 2 n = 2
Ce merită reținut din varianta 2
Restul împărțirii la X − a X - a X − a este f ( a ) f(a) f ( a ) Viète pentru inverse : ∑ 1 x i = x 1 x 2 + x 2 x 3 + x 3 x 1 x 1 x 2 x 3 \sum\frac1{x_i} = \frac{x_1x_2 + x_2x_3 + x_3x_1}{x_1x_2x_3} ∑ x i 1 = x 1 x 2 x 3 x 1 x 2 + x 2 x 3 + x 3 x 1 ∫ x x 2 + 1 d x = 1 2 ln ( x 2 + 1 ) \int\frac{x}{x^2+1}dx = \frac12\ln(x^2+1) ∫ x 2 + 1 x d x = 2 1 ln ( x 2 + 1 )