Rezolvarea completă a subiectului de Bacalaureat 2014 la Matematică M_tehnologic , sesiunea specială, varianta 9 . 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 3 ⋅ ( 2 3 − 1 3 ) = 1 3\cdot\left(\frac23 - \frac13\right) = 1 3 ⋅ ( 3 2 − 3 1 ) = 1
2 3 − 1 3 = 1 3 , 3 ⋅ 1 3 = 1. \frac23 - \frac13 = \frac13, \qquad 3\cdot\frac13 = 1. 3 2 − 3 1 = 3 1 , 3 ⋅ 3 1 = 1.
Barem: 2p + 3p.
2. m m m f ( m ) = 1 f(m) = 1 f ( m ) = 1 f ( x ) = x − 4 f(x) = x - 4 f ( x ) = x − 4
m − 4 = 1 ⟺ m = 5 m - 4 = 1 \iff m = 5 m − 4 = 1 ⟺ m = 5 Barem: 3p + 2p.
3. Ecuația 2 x 2 + 1 = 1 \sqrt{2x^2 + 1} = 1 2 x 2 + 1 = 1
2 x 2 + 1 = 1 ⟺ x 2 = 0 ⟺ x = 0 2x^2 + 1 = 1 \iff x^2 = 0 \iff x = 0 2 x 2 + 1 = 1 ⟺ x 2 = 0 ⟺ x = 0 1 = 1 \sqrt1 = 1 1 = 1 Barem: 3p + 2p.
4. Profitul de 100000 lei reprezintă 4% din venituri
4 100 ⋅ x = 100000 ⟺ x = 2 500 000. \frac{4}{100}\cdot x = 100000 \iff x = 2\,500\,000. 100 4 ⋅ x = 100000 ⟺ x = 2 500 000.
Răspuns: veniturile au fost 2 500 000 de lei. Barem: 3p + 2p.
5. Triunghiul A ( 5 , 6 ) A(5,6) A ( 5 , 6 ) B ( 2 , 6 ) B(2,6) B ( 2 , 6 ) C ( 5 , 2 ) C(5,2) C ( 5 , 2 )
A B = 5 − 2 = 3 AB = 5 - 2 = 3 A B = 5 − 2 = 3 A C = 6 − 2 = 4 AC = 6 - 2 = 4 A C = 6 − 2 = 4 B C = 3 2 + 4 2 = 5 BC = \sqrt{3^2 + 4^2} = 5 B C = 3 2 + 4 2 = 5 A B 2 + A C 2 = 9 + 16 = 25 = B C 2 AB^2 + AC^2 = 9 + 16 = 25 = BC^2 A B 2 + A C 2 = 9 + 16 = 25 = B C 2 A A A Barem: 3p + 2p.
6. tg 2 60 ∘ + tg 2 45 ∘ = 4 \operatorname{tg}^2 60^\circ + \operatorname{tg}^2 45^\circ = 4 tg 2 6 0 ∘ + tg 2 4 5 ∘ = 4
( 3 ) 2 + 1 2 = 3 + 1 = 4. (\sqrt3)^2 + 1^2 = 3 + 1 = 4. ( 3 ) 2 + 1 2 = 3 + 1 = 4.
Barem: 2p + 3p.
Subiectul al II-lea (30 de puncte)
1. A = ( 3 1 − 5 − 2 ) A = \begin{pmatrix} 3 & 1\\ -5 & -2\end{pmatrix} A = ( 3 − 5 1 − 2 ) B = ( 2 1 − 5 − 3 ) B = \begin{pmatrix} 2 & 1\\ -5 & -3\end{pmatrix} B = ( 2 − 5 1 − 3 )
a) det A = − 1 \det A = -1 det A = − 1
det A = 3 ⋅ ( − 2 ) − 1 ⋅ ( − 5 ) = − 6 + 5 = − 1. \det A = 3\cdot(-2) - 1\cdot(-5) = -6 + 5 = -1. det A = 3 ⋅ ( − 2 ) − 1 ⋅ ( − 5 ) = − 6 + 5 = − 1.
Barem: 2p + 3p.
b) 2 A ⋅ B − B ⋅ A = I 2 2A\cdot B - B\cdot A = I_2 2 A ⋅ B − B ⋅ A = I 2
A ⋅ B = ( 6 − 5 3 − 3 − 10 + 10 − 5 + 6 ) = I 2 , B ⋅ A = ( 6 − 5 2 − 2 − 15 + 15 − 5 + 6 ) = I 2 . A\cdot B = \begin{pmatrix} 6 - 5 & 3 - 3\\ -10 + 10 & -5 + 6\end{pmatrix} = I_2, \qquad B\cdot A = \begin{pmatrix} 6 - 5 & 2 - 2\\ -15 + 15 & -5 + 6\end{pmatrix} = I_2. A ⋅ B = ( 6 − 5 − 10 + 10 3 − 3 − 5 + 6 ) = I 2 , B ⋅ A = ( 6 − 5 − 15 + 15 2 − 2 − 5 + 6 ) = I 2 .
Deci 2 A ⋅ B − B ⋅ A = 2 I 2 − I 2 = I 2 2A\cdot B - B\cdot A = 2I_2 - I_2 = I_2 2 A ⋅ B − B ⋅ A = 2 I 2 − I 2 = I 2 ■ \blacksquare ■ Barem: 2p + 3p.
c) x x x A ⋅ A − x A = I 2 A\cdot A - xA = I_2 A ⋅ A − x A = I 2
A ⋅ A = ( 4 1 − 5 − 1 ) , A ⋅ A − x A = ( 4 − 3 x 1 − x − 5 + 5 x − 1 + 2 x ) . A\cdot A = \begin{pmatrix} 4 & 1\\ -5 & -1\end{pmatrix}, \qquad A\cdot A - xA = \begin{pmatrix} 4 - 3x & 1 - x\\ -5 + 5x & -1 + 2x\end{pmatrix}. A ⋅ A = ( 4 − 5 1 − 1 ) , A ⋅ A − x A = ( 4 − 3 x − 5 + 5 x 1 − x − 1 + 2 x ) .
Egalitatea cu I 2 I_2 I 2 1 − x = 0 1 - x = 0 1 − x = 0 x = 1 x = 1 x = 1 4 − 3 = 1 4 - 3 = 1 4 − 3 = 1 − 5 + 5 = 0 -5 + 5 = 0 − 5 + 5 = 0 − 1 + 2 = 1 -1 + 2 = 1 − 1 + 2 = 1
Răspuns: x = 1 x = 1 x = 1 Barem: 3p + 2p.
2. x ∗ y = 2 ( x + y − 1 ) − x y x * y = 2(x + y - 1) - xy x ∗ y = 2 ( x + y − 1 ) − x y
a) 1 ∗ 2 = 2 1 * 2 = 2 1 ∗ 2 = 2
2 ( 1 + 2 − 1 ) − 2 = 4 − 2 = 2. 2(1 + 2 - 1) - 2 = 4 - 2 = 2. 2 ( 1 + 2 − 1 ) − 2 = 4 − 2 = 2.
Barem: 3p + 2p.
b) x ∗ 2 = 2 ∗ x = 2 x * 2 = 2 * x = 2 x ∗ 2 = 2 ∗ x = 2
x ∗ 2 = 2 ( x + 1 ) − 2 x = 2 , 2 ∗ x = 2 ( 1 + x ) − 2 x = 2. x * 2 = 2(x + 1) - 2x = 2, \qquad 2 * x = 2(1 + x) - 2x = 2. x ∗ 2 = 2 ( x + 1 ) − 2 x = 2 , 2 ∗ x = 2 ( 1 + x ) − 2 x = 2.
Barem: 2p + 3p.
c) x ∗ x = x x * x = x x ∗ x = x
2 ( 2 x − 1 ) − x 2 = x ⟺ − x 2 + 3 x − 2 = 0 ⟺ x 2 − 3 x + 2 = 0 ⟺ ( x − 1 ) ( x − 2 ) = 0. 2(2x - 1) - x^2 = x \iff -x^2 + 3x - 2 = 0 \iff x^2 - 3x + 2 = 0 \iff (x-1)(x-2) = 0. 2 ( 2 x − 1 ) − x 2 = x ⟺ − x 2 + 3 x − 2 = 0 ⟺ x 2 − 3 x + 2 = 0 ⟺ ( x − 1 ) ( x − 2 ) = 0.
Răspuns: x ∈ { 1 , 2 } x \in \{1, 2\} x ∈ { 1 , 2 } Barem: 3p + 2p.
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) lim x → 0 f ( x ) = − 1 \lim\limits_{x\to0} f(x) = -1 x → 0 lim f ( x ) = − 1
f f f lim x → 0 f ( x ) = f ( 0 ) = − 1 ⋅ e 0 = − 1 \lim\limits_{x\to0} f(x) = f(0) = -1\cdot e^0 = -1 x → 0 lim f ( x ) = f ( 0 ) = − 1 ⋅ e 0 = − 1 Barem: 2p + 3p.
b) f ′ ( x ) = e x + f ( x ) f'(x) = e^x + f(x) f ′ ( x ) = e x + f ( x )
f ′ ( x ) = 1 ⋅ e x + ( x − 1 ) e x = e x + f ( x ) . f'(x) = 1\cdot e^x + (x-1)e^x = e^x + f(x). f ′ ( x ) = 1 ⋅ e x + ( x − 1 ) e x = e x + f ( x ) .
Barem: 3p + 2p.
c) lim x → 0 f ( x ) + 1 x = 0 \lim\limits_{x\to0}\dfrac{f(x) + 1}{x} = 0 x → 0 lim x f ( x ) + 1 = 0
Cum f ( 0 ) = − 1 f(0) = -1 f ( 0 ) = − 1 lim x → 0 f ( x ) − f ( 0 ) x − 0 = f ′ ( 0 ) = e 0 + f ( 0 ) = 1 − 1 = 0 \lim\limits_{x\to0}\dfrac{f(x) - f(0)}{x - 0} = f'(0) = e^0 + f(0) = 1 - 1 = 0 x → 0 lim x − 0 f ( x ) − f ( 0 ) = f ′ ( 0 ) = e 0 + f ( 0 ) = 1 − 1 = 0 Barem: 2p + 3p.
2. f ( x ) = 3 x 2 + 2 x f(x) = 3x^2 + 2x f ( x ) = 3 x 2 + 2 x
a) ∫ 1 2 3 x 2 d x = 7 \displaystyle\int_1^2 3x^2\,dx = 7 ∫ 1 2 3 x 2 d x = 7
x 3 ∣ 1 2 = 8 − 1 = 7. x^3\Big|_1^2 = 8 - 1 = 7. x 3 1 2 = 8 − 1 = 7.
Barem: 3p + 2p.
b) Primitiva F F F F ( 1 ) = 2014 F(1) = 2014 F ( 1 ) = 2014
F ( x ) = x 3 + x 2 + c F(x) = x^3 + x^2 + c F ( x ) = x 3 + x 2 + c F ( 1 ) = 2 + c = 2014 ⇒ c = 2012 F(1) = 2 + c = 2014 \Rightarrow c = 2012 F ( 1 ) = 2 + c = 2014 ⇒ c = 2012
Răspuns: F ( x ) = x 3 + x 2 + 2012 F(x) = x^3 + x^2 + 2012 F ( x ) = x 3 + x 2 + 2012 Barem: 3p + 2p.
c) n ≥ 2 n \ge 2 n ≥ 2 ∫ 1 n f ( x ) x d x = 13 2 \displaystyle\int_1^n\frac{f(x)}{x}\,dx = \frac{13}{2} ∫ 1 n x f ( x ) d x = 2 13
f ( x ) x = 3 x + 2 \dfrac{f(x)}{x} = 3x + 2 x f ( x ) = 3 x + 2
∫ 1 n ( 3 x + 2 ) d x = ( 3 x 2 2 + 2 x ) ∣ 1 n = 3 n 2 2 + 2 n − 7 2 = 3 n 2 + 4 n − 7 2 . \int_1^n(3x + 2)\,dx = \left(\frac{3x^2}{2} + 2x\right)\Big|_1^n = \frac{3n^2}{2} + 2n - \frac72 = \frac{3n^2 + 4n - 7}{2}. ∫ 1 n ( 3 x + 2 ) d x = ( 2 3 x 2 + 2 x ) 1 n = 2 3 n 2 + 2 n − 2 7 = 2 3 n 2 + 4 n − 7 .
3 n 2 + 4 n − 7 2 = 13 2 ⟺ 3 n 2 + 4 n − 20 = 0 ⟺ ( n − 2 ) ( 3 n + 10 ) = 0 \dfrac{3n^2 + 4n - 7}{2} = \dfrac{13}{2} \iff 3n^2 + 4n - 20 = 0 \iff (n - 2)(3n + 10) = 0 2 3 n 2 + 4 n − 7 = 2 13 ⟺ 3 n 2 + 4 n − 20 = 0 ⟺ ( n − 2 ) ( 3 n + 10 ) = 0 n n n n = 2 n = 2 n = 2
Răspuns: n = 2 n = 2 n = 2 Barem: 3p + 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 m = 5 m = 5 m = 5 I.3 x = 0 x = 0 x = 0 I.4 2 500 000 lei II.1.c x = 1 x = 1 x = 1 II.2.c x ∈ { 1 , 2 } x \in \{1, 2\} x ∈ { 1 , 2 } III.1.c 0 0 0 III.2.b F ( x ) = x 3 + x 2 + 2012 F(x) = x^3 + x^2 + 2012 F ( x ) = x 3 + x 2 + 2012 III.2.c n = 2 n = 2 n = 2
Ce merită reținut din varianta 9
Procente inverse : dacă p % p\% p % x x x A A A x = 100 A p x = \frac{100A}{p} x = p 100 A Legea 2 ( x + y − 1 ) − x y = − ( x − 2 ) ( y − 2 ) + 2 2(x + y - 1) - xy = -(x-2)(y-2) + 2 2 ( x + y − 1 ) − x y = − ( x − 2 ) ( y − 2 ) + 2 : 2 2 2 Dacă A B = I 2 AB = I_2 A B = I 2 , atunci și B A = I 2 BA = I_2 B A = I 2 B = A − 1 B = A^{-1} B = A − 1 Limite f ( x ) − f ( a ) x − a \frac{f(x) - f(a)}{x - a} x − a f ( x ) − f ( a ) = f ′ ( a ) f'(a) f ′ ( a )