Rezolvarea completă a subiectului de Bacalaureat 2016 la Matematică M_tehnologic , 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. Arătați că ( 1 3 − 1 4 ) : 1 12 = 1 \left(\frac13 - \frac14\right) : \frac{1}{12} = 1 ( 3 1 − 4 1 ) : 12 1 = 1
1 3 − 1 4 = 1 12 , 1 12 : 1 12 = 1. \frac13 - \frac14 = \frac{1}{12}, \qquad \frac{1}{12} : \frac{1}{12} = 1. 3 1 − 4 1 = 12 1 , 12 1 : 12 1 = 1.
Barem: 3p + 2p.
2. 4 ( x 1 + x 2 ) − 3 x 1 x 2 = 2 4(x_1 + x_2) - 3x_1x_2 = 2 4 ( x 1 + x 2 ) − 3 x 1 x 2 = 2 x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0
Viète: x 1 + x 2 = 5 x_1 + x_2 = 5 x 1 + x 2 = 5 x 1 x 2 = 6 x_1x_2 = 6 x 1 x 2 = 6 4 ⋅ 5 − 3 ⋅ 6 = 2 4\cdot5 - 3\cdot6 = 2 4 ⋅ 5 − 3 ⋅ 6 = 2 Barem: 2p + 3p.
3. Ecuația x − 1 = 2 \sqrt{x - 1} = 2 x − 1 = 2
x − 1 = 4 ⟺ x = 5 x - 1 = 4 \iff x = 5 x − 1 = 4 ⟺ x = 5 Barem: 3p + 2p.
4. După o ieftinire cu 10%, prețul este 90 lei
p − 0,1 p = 0,9 p = 90 ⟺ p = 100 p - 0{,}1p = 0{,}9p = 90 \iff p = 100 p − 0 , 1 p = 0 , 9 p = 90 ⟺ p = 100
Răspuns: 100 de lei. Barem: 3p + 2p.
5. Lungimea segmentului A B AB A B A ( 5 , 1 ) A(5,1) A ( 5 , 1 ) B ( 3 , 1 ) B(3,1) B ( 3 , 1 )
A B = ( 3 − 5 ) 2 + 0 2 = 2. AB = \sqrt{(3-5)^2 + 0^2} = 2. A B = ( 3 − 5 ) 2 + 0 2 = 2.
Barem: 3p + 2p.
6. Dacă x ∈ ( 0 , π 2 ) x \in \left(0, \frac\pi2\right) x ∈ ( 0 , 2 π ) cos x = 4 5 \cos x = \frac45 cos x = 5 4 sin x = 3 5 \sin x = \frac35 sin x = 5 3
sin 2 x = 1 − 16 25 = 9 25 \sin^2 x = 1 - \frac{16}{25} = \frac{9}{25} sin 2 x = 1 − 25 16 = 25 9 sin x = 3 5 \sin x = \frac35 sin x = 5 3 Barem: 3p + 2p.
Subiectul al II-lea (30 de puncte)
1. A = ( 2 3 3 2 ) A = \begin{pmatrix} 2 & 3\\ 3 & 2\end{pmatrix} A = ( 2 3 3 2 ) B = ( x 1 1 x ) B = \begin{pmatrix} x & 1\\ 1 & x\end{pmatrix} B = ( x 1 1 x )
a) det A = − 5 \det A = -5 det A = − 5
det A = 4 − 9 = − 5. \det A = 4 - 9 = -5. det A = 4 − 9 = − 5.
Barem: 3p + 2p.
b) A ⋅ B = B ⋅ A A\cdot B = B\cdot A A ⋅ B = B ⋅ A
A ⋅ B = ( 2 x + 3 2 + 3 x 3 x + 2 3 + 2 x ) , B ⋅ A = ( 2 x + 3 3 x + 2 2 + 3 x 3 + 2 x ) . A\cdot B = \begin{pmatrix} 2x + 3 & 2 + 3x\\ 3x + 2 & 3 + 2x\end{pmatrix}, \qquad B\cdot A = \begin{pmatrix} 2x + 3 & 3x + 2\\ 2 + 3x & 3 + 2x\end{pmatrix}. A ⋅ B = ( 2 x + 3 3 x + 2 2 + 3 x 3 + 2 x ) , B ⋅ A = ( 2 x + 3 2 + 3 x 3 x + 2 3 + 2 x ) .
Cele două matrice coincid pentru orice x x x ■ \blacksquare ■ Barem: 2p + 3p.
c) x x x A ⋅ A − 3 ( A + B ) = I 2 A\cdot A - 3(A + B) = I_2 A ⋅ A − 3 ( A + B ) = I 2
A ⋅ A = ( 13 12 12 13 ) , A + B = ( 2 + x 4 4 2 + x ) , A\cdot A = \begin{pmatrix} 13 & 12\\ 12 & 13\end{pmatrix}, \qquad A + B = \begin{pmatrix} 2 + x & 4\\ 4 & 2 + x\end{pmatrix}, A ⋅ A = ( 13 12 12 13 ) , A + B = ( 2 + x 4 4 2 + x ) ,
A ⋅ A − 3 ( A + B ) = ( 13 − 3 ( 2 + x ) 0 0 13 − 3 ( 2 + x ) ) = ( 7 − 3 x 0 0 7 − 3 x ) . A\cdot A - 3(A + B) = \begin{pmatrix} 13 - 3(2 + x) & 0\\ 0 & 13 - 3(2 + x)\end{pmatrix} = \begin{pmatrix} 7 - 3x & 0\\ 0 & 7 - 3x\end{pmatrix}. A ⋅ A − 3 ( A + B ) = ( 13 − 3 ( 2 + x ) 0 0 13 − 3 ( 2 + x ) ) = ( 7 − 3 x 0 0 7 − 3 x ) .
7 − 3 x = 1 ⟺ x = 2 7 - 3x = 1 \iff x = 2 7 − 3 x = 1 ⟺ x = 2 Răspuns: x = 2 x = 2 x = 2 Barem: 2p + 3p.
2. x ∗ y = 1 3 x y + x + y x * y = \frac13xy + x + y x ∗ y = 3 1 x y + x + y
a) 1 ∗ ( − 3 ) = − 3 1 * (-3) = -3 1 ∗ ( − 3 ) = − 3
1 3 ⋅ 1 ⋅ ( − 3 ) + 1 − 3 = − 1 + 1 − 3 = − 3. \frac13\cdot1\cdot(-3) + 1 - 3 = -1 + 1 - 3 = -3. 3 1 ⋅ 1 ⋅ ( − 3 ) + 1 − 3 = − 1 + 1 − 3 = − 3.
Barem: 3p + 2p.
b) x ∗ y = 1 3 ( x + 3 ) ( y + 3 ) − 3 x * y = \frac13(x+3)(y+3) - 3 x ∗ y = 3 1 ( x + 3 ) ( y + 3 ) − 3
1 3 x y + x + y + 3 − 3 = 1 3 ( x y + 3 x + 3 y + 9 ) − 3 = 1 3 ( x + 3 ) ( y + 3 ) − 3. \frac13xy + x + y + 3 - 3 = \frac13(xy + 3x + 3y + 9) - 3 = \frac13(x+3)(y+3) - 3. 3 1 x y + x + y + 3 − 3 = 3 1 ( x y + 3 x + 3 y + 9 ) − 3 = 3 1 ( x + 3 ) ( y + 3 ) − 3.
Barem: 3p + 2p.
c) x ≠ 0 x \ne 0 x = 0 x ∗ 1 x = − 3 x * \frac1x = -3 x ∗ x 1 = − 3
1 3 ( x + 3 ) ( 1 x + 3 ) − 3 = − 3 ⟺ ( x + 3 ) ( 1 x + 3 ) = 0. \frac13(x + 3)\left(\frac1x + 3\right) - 3 = -3 \iff (x + 3)\left(\frac1x + 3\right) = 0. 3 1 ( x + 3 ) ( x 1 + 3 ) − 3 = − 3 ⟺ ( x + 3 ) ( x 1 + 3 ) = 0.
x + 3 = 0 ⇒ x = − 3 x + 3 = 0 \Rightarrow x = -3 x + 3 = 0 ⇒ x = − 3 1 x + 3 = 0 ⇒ x = − 1 3 \frac1x + 3 = 0 \Rightarrow x = -\frac13 x 1 + 3 = 0 ⇒ x = − 3 1
Răspuns: x ∈ { − 3 , − 1 3 } x \in \left\{-3, -\dfrac13\right\} x ∈ { − 3 , − 3 1 } Barem: 3p + 2p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = x 3 − 3 x f(x) = x^3 - 3x f ( x ) = x 3 − 3 x
a) f ′ ( x ) = 3 ( x − 1 ) ( x + 1 ) f'(x) = 3(x-1)(x+1) f ′ ( x ) = 3 ( x − 1 ) ( x + 1 )
f ′ ( x ) = 3 x 2 − 3 = 3 ( x − 1 ) ( x + 1 ) . f'(x) = 3x^2 - 3 = 3(x-1)(x+1). f ′ ( x ) = 3 x 2 − 3 = 3 ( x − 1 ) ( x + 1 ) .
Barem: 3p + 2p.
b) lim x → 0 f ( x ) + 3 x x = 0 \lim\limits_{x\to0}\dfrac{f(x) + 3x}{x} = 0 x → 0 lim x f ( x ) + 3 x = 0
lim x → 0 x 3 x = lim x → 0 x 2 = 0. \lim_{x\to0}\frac{x^3}{x} = \lim_{x\to0}x^2 = 0. lim x → 0 x x 3 = lim x → 0 x 2 = 0.
Barem: 2p + 3p.
c) f ( x ) ≥ − 2 f(x) \ge -2 f ( x ) ≥ − 2 x ∈ [ − 1 , + ∞ ) x \in [-1, +\infty) x ∈ [ − 1 , + ∞ )
f ′ ≤ 0 f' \le 0 f ′ ≤ 0 [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ] f ′ ≥ 0 f' \ge 0 f ′ ≥ 0 [ 1 , + ∞ ) [1, +\infty) [ 1 , + ∞ ) [ − 1 , + ∞ ) [-1, +\infty) [ − 1 , + ∞ ) f ( 1 ) = 1 − 3 = − 2 f(1) = 1 - 3 = -2 f ( 1 ) = 1 − 3 = − 2 ■ \blacksquare ■ Barem: 2p + 1p + 1p + 1p.
2. f ( x ) = x 4 + x + 1 f(x) = x^4 + x + 1 f ( x ) = x 4 + x + 1
a) ∫ 0 1 ( f ( x ) − x − 1 ) d x = 1 5 \displaystyle\int_0^1(f(x) - x - 1)\,dx = \frac15 ∫ 0 1 ( f ( x ) − x − 1 ) d x = 5 1
∫ 0 1 x 4 d x = x 5 5 ∣ 0 1 = 1 5 . \int_0^1 x^4\,dx = \frac{x^5}{5}\Big|_0^1 = \frac15. ∫ 0 1 x 4 d x = 5 x 5 0 1 = 5 1 .
Barem: 2p + 3p.
b) ∫ 1 e ( f ( x ) − x 4 − 1 ) ln x d x = e 2 + 1 4 \displaystyle\int_1^e\big(f(x) - x^4 - 1\big)\ln x\,dx = \frac{e^2 + 1}{4} ∫ 1 e ( f ( x ) − x 4 − 1 ) ln x d x = 4 e 2 + 1
f ( x ) − x 4 − 1 = x f(x) - x^4 - 1 = x f ( x ) − x 4 − 1 = x ∫ 1 e x ln x d x \displaystyle\int_1^e x\ln x\,dx ∫ 1 e x ln x d x
x 2 2 ln x ∣ 1 e − ∫ 1 e x 2 2 ⋅ 1 x d x = e 2 2 − x 2 4 ∣ 1 e = e 2 2 − e 2 4 + 1 4 = e 2 + 1 4 . \frac{x^2}{2}\ln x\Big|_1^e - \int_1^e\frac{x^2}{2}\cdot\frac1x\,dx = \frac{e^2}{2} - \frac{x^2}{4}\Big|_1^e = \frac{e^2}{2} - \frac{e^2}{4} + \frac14 = \frac{e^2 + 1}{4}. 2 x 2 ln x 1 e − ∫ 1 e 2 x 2 ⋅ x 1 d x = 2 e 2 − 4 x 2 1 e = 2 e 2 − 4 e 2 + 4 1 = 4 e 2 + 1 .
Barem: 3p + 2p.
c) Aria sub grafic pe [ 0 , 1 ] [0,1] [ 0 , 1 ]
f ( x ) > 0 f(x) > 0 f ( x ) > 0 [ 0 , 1 ] [0,1] [ 0 , 1 ]
A = ∫ 0 1 ( x 4 + x + 1 ) d x = 1 5 + 1 2 + 1 = 17 10 . \mathcal A = \int_0^1(x^4 + x + 1)\,dx = \frac15 + \frac12 + 1 = \frac{17}{10}. A = ∫ 0 1 ( x 4 + x + 1 ) d x = 5 1 + 2 1 + 1 = 10 17 .
Barem: 3p + 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.3 x = 5 x = 5 x = 5 I.4 100 lei I.5 2 2 2 II.1.c x = 2 x = 2 x = 2 II.2.c x ∈ { − 3 , − 1 3 } x \in \{-3, -\frac13\} x ∈ { − 3 , − 3 1 } III.1.b 0 0 0 III.1.c minim f ( 1 ) = − 2 f(1) = -2 f ( 1 ) = − 2 III.2.c 17 10 \frac{17}{10} 10 17
Ce merită reținut din varianta 5
Legea 1 a ( x + a ) ( y + a ) − a \frac1a(x+a)(y+a) - a a 1 ( x + a ) ( y + a ) − a : − a -a − a Matrice simetrice de forma ( p q q p ) \begin{pmatrix}p&q\\q&p\end{pmatrix} ( p q q p ) comută între ele.∫ x ln x d x = x 2 2 ln x − x 2 4 + C \int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C ∫ x ln x d x = 2 x 2 ln x − 4 x 2 + C Minimul pe un interval semi-infinit : compară valoarea în punctul critic cu comportarea la capete.