Rezolvarea completă a subiectului de Simulare Bacalaureat 2017 la Matematică M_tehnologic (clasa a XII-a). Pentru fiecare cerință găsești enunțul pe scurt, rezolvarea pas cu pas, răspunsul final și cum se punctează conform baremului oficial. Timp de lucru: 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. ( 2 + 3 ) 2 + ( 1 − 2 3 ) 2 = 20 (2 + \sqrt3)^2 + (1 - 2\sqrt3)^2 = 20 ( 2 + 3 ) 2 + ( 1 − 2 3 ) 2 = 20
( 2 + 3 ) 2 = 4 + 4 3 + 3 = 7 + 4 3 , ( 1 − 2 3 ) 2 = 1 − 4 3 + 12 = 13 − 4 3 . (2 + \sqrt3)^2 = 4 + 4\sqrt3 + 3 = 7 + 4\sqrt3, \qquad (1 - 2\sqrt3)^2 = 1 - 4\sqrt3 + 12 = 13 - 4\sqrt3. ( 2 + 3 ) 2 = 4 + 4 3 + 3 = 7 + 4 3 , ( 1 − 2 3 ) 2 = 1 − 4 3 + 12 = 13 − 4 3 .
Suma: 7 + 13 = 20 7 + 13 = 20 7 + 13 = 20 Barem: 2p, 3p.
2. f ( x ) = x 2 − 3 x f(x) = x^2 - 3x f ( x ) = x 2 − 3 x f ( 1 ) ⋅ f ( 2 ) ⋅ f ( 3 ) ⋅ f ( 4 ) f(1)\cdot f(2)\cdot f(3)\cdot f(4) f ( 1 ) ⋅ f ( 2 ) ⋅ f ( 3 ) ⋅ f ( 4 )
f ( 3 ) = 9 − 9 = 0 f(3) = 9 - 9 = 0 f ( 3 ) = 9 − 9 = 0 0 0 0
Răspuns: 0 0 0 Barem: 3p f ( 3 ) = 0 f(3) = 0 f ( 3 ) = 0
3. 8 x = 4 2 x + 1 8^x = 4^{2x+1} 8 x = 4 2 x + 1
2 3 x = 2 4 x + 2 ⟺ 3 x = 4 x + 2 ⟺ x = − 2 2^{3x} = 2^{4x + 2} \iff 3x = 4x + 2 \iff x = -2 2 3 x = 2 4 x + 2 ⟺ 3 x = 4 x + 2 ⟺ x = − 2
Răspuns: x = − 2 x = -2 x = − 2 Barem: 3p, 2p.
4. După o scumpire cu 25 % 25\% 25%
p + 25 100 p = 250 ⟺ 5 4 p = 250 ⟺ p = 200. p + \frac{25}{100}p = 250 \iff \frac54p = 250 \iff p = 200. p + 100 25 p = 250 ⟺ 4 5 p = 250 ⟺ p = 200.
Răspuns: 200 de lei. Barem: 2p, 3p.
5. A ( 1 , 5 ) A(1,5) A ( 1 , 5 ) B ( 1 , 1 ) B(1,1) B ( 1 , 1 ) C ( 5 , 5 ) C(5,5) C ( 5 , 5 ) A B C ABC A B C
A B = 0 + 16 = 4 AB = \sqrt{0 + 16} = 4 A B = 0 + 16 = 4 A C = 16 + 0 = 4 AC = \sqrt{16 + 0} = 4 A C = 16 + 0 = 4 A B = A C AB = AC A B = A C Barem: 2p, 3p.
6. sin 60 ∘ + tg 45 ∘ = cos 30 ∘ + ctg 45 ∘ \sin 60^\circ + \operatorname{tg} 45^\circ = \cos 30^\circ + \operatorname{ctg} 45^\circ sin 6 0 ∘ + tg 4 5 ∘ = cos 3 0 ∘ + ctg 4 5 ∘
sin 60 ∘ = cos 30 ∘ = 3 2 \sin 60^\circ = \cos 30^\circ = \dfrac{\sqrt3}{2} sin 6 0 ∘ = cos 3 0 ∘ = 2 3 tg 45 ∘ = ctg 45 ∘ = 1 \operatorname{tg}45^\circ = \operatorname{ctg}45^\circ = 1 tg 4 5 ∘ = ctg 4 5 ∘ = 1 3 2 + 1 \dfrac{\sqrt3}{2} + 1 2 3 + 1 Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. A ( x ) = ( x 2 x x ) A(x) = \begin{pmatrix} x & 2\\ x & x\end{pmatrix} A ( x ) = ( x x 2 x )
a) det ( A ( 3 ) ) = 3 \det(A(3)) = 3 det ( A ( 3 )) = 3
3 ⋅ 3 − 3 ⋅ 2 = 9 − 6 = 3 3\cdot 3 - 3\cdot 2 = 9 - 6 = 3 3 ⋅ 3 − 3 ⋅ 2 = 9 − 6 = 3 Barem: 3p, 2p.
b) A ( 2017 + x ) + A ( 2017 − x ) = 2 A ( 2017 ) A(2017 + x) + A(2017 - x) = 2A(2017) A ( 2017 + x ) + A ( 2017 − x ) = 2 A ( 2017 )
( 2017 + x 2 2017 + x 2017 + x ) + ( 2017 − x 2 2017 − x 2017 − x ) = ( 4034 4 4034 4034 ) = 2 A ( 2017 ) . \begin{pmatrix} 2017 + x & 2\\ 2017 + x & 2017 + x\end{pmatrix} + \begin{pmatrix} 2017 - x & 2\\ 2017 - x & 2017 - x\end{pmatrix} = \begin{pmatrix} 4034 & 4\\ 4034 & 4034\end{pmatrix} = 2A(2017). ( 2017 + x 2017 + x 2 2017 + x ) + ( 2017 − x 2017 − x 2 2017 − x ) = ( 4034 4034 4 4034 ) = 2 A ( 2017 ) .
Barem: 3p, 2p.
c) det ( A ( 2 ) + m A ( 1 ) ) = 0 \det(A(2) + mA(1)) = 0 det ( A ( 2 ) + m A ( 1 )) = 0
A ( 2 ) + m A ( 1 ) = ( 2 + m 2 + 2 m 2 + m 2 + m ) , A(2) + mA(1) = \begin{pmatrix} 2 + m & 2 + 2m\\ 2 + m & 2 + m\end{pmatrix}, A ( 2 ) + m A ( 1 ) = ( 2 + m 2 + m 2 + 2 m 2 + m ) ,
det = ( 2 + m ) 2 − ( 2 + m ) ( 2 + 2 m ) = ( 2 + m ) ( 2 + m − 2 − 2 m ) = − m ( m + 2 ) . \det = (2 + m)^2 - (2 + m)(2 + 2m) = (2 + m)(2 + m - 2 - 2m) = -m(m + 2). det = ( 2 + m ) 2 − ( 2 + m ) ( 2 + 2 m ) = ( 2 + m ) ( 2 + m − 2 − 2 m ) = − m ( m + 2 ) .
− m ( m + 2 ) = 0 ⟺ m ∈ { − 2 , 0 } -m(m + 2) = 0 \iff m \in \{-2, 0\} − m ( m + 2 ) = 0 ⟺ m ∈ { − 2 , 0 }
Răspuns: m ∈ { − 2 , 0 } m \in \{-2, 0\} m ∈ { − 2 , 0 } Barem: 3p, 2p.
2. Legea x ∗ y = 2 x y + 6 x + 6 y + 15 x * y = 2xy + 6x + 6y + 15 x ∗ y = 2 x y + 6 x + 6 y + 15
a) x ∗ y = 2 ( x + 3 ) ( y + 3 ) − 3 x * y = 2(x + 3)(y + 3) - 3 x ∗ y = 2 ( x + 3 ) ( y + 3 ) − 3
2 ( x + 3 ) ( y + 3 ) − 3 = 2 x y + 6 x + 6 y + 18 − 3 = 2 x y + 6 x + 6 y + 15. 2(x+3)(y+3) - 3 = 2xy + 6x + 6y + 18 - 3 = 2xy + 6x + 6y + 15. 2 ( x + 3 ) ( y + 3 ) − 3 = 2 x y + 6 x + 6 y + 18 − 3 = 2 x y + 6 x + 6 y + 15.
Barem: 2p, 3p.
b) 7 ∗ 98 = 2017 7 * 98 = 2017 7 ∗ 98 = 2017
2 ⋅ 10 ⋅ 101 − 3 = 2020 − 3 = 2017 2\cdot 10\cdot 101 - 3 = 2020 - 3 = 2017 2 ⋅ 10 ⋅ 101 − 3 = 2020 − 3 = 2017 Barem: 3p, 2p.
c) x ∗ ( x + 2 ) = 3 x * (x + 2) = 3 x ∗ ( x + 2 ) = 3
2 ( x + 3 ) ( x + 5 ) − 3 = 3 ⟺ ( x + 3 ) ( x + 5 ) = 3 ⟺ x 2 + 8 x + 12 = 0 ⟺ ( x + 6 ) ( x + 2 ) = 0. 2(x + 3)(x + 5) - 3 = 3 \iff (x + 3)(x + 5) = 3 \iff x^2 + 8x + 12 = 0 \iff (x + 6)(x + 2) = 0. 2 ( x + 3 ) ( x + 5 ) − 3 = 3 ⟺ ( x + 3 ) ( x + 5 ) = 3 ⟺ x 2 + 8 x + 12 = 0 ⟺ ( x + 6 ) ( x + 2 ) = 0.
Răspuns: x ∈ { − 6 , − 2 } x \in \{-6, -2\} x ∈ { − 6 , − 2 } Barem: 3p, 2p.
Subiectul al III-lea (30 de puncte)
1. f : ( 2 , + ∞ ) → R f:(2, +\infty) \to \mathbb R f : ( 2 , + ∞ ) → R f ( x ) = x + 1 + 1 x − 2 f(x) = x + 1 + \dfrac{1}{x - 2} f ( x ) = x + 1 + x − 2 1
a) lim x → 3 f ( x ) − f ( 3 ) x − 3 = 0 \lim\limits_{x\to3}\dfrac{f(x) - f(3)}{x - 3} = 0 x → 3 lim x − 3 f ( x ) − f ( 3 ) = 0
Limita este f ′ ( 3 ) f'(3) f ′ ( 3 ) f ′ ( x ) = 1 − 1 ( x − 2 ) 2 f'(x) = 1 - \dfrac{1}{(x - 2)^2} f ′ ( x ) = 1 − ( x − 2 ) 2 1 f ′ ( 3 ) = 1 − 1 = 0 f'(3) = 1 - 1 = 0 f ′ ( 3 ) = 1 − 1 = 0 Barem: 2p, 3p.
b) Asimptota oblică spre + ∞ +\infty + ∞
m = lim x → + ∞ f ( x ) x = lim x → + ∞ ( 1 + 1 x + 1 x ( x − 2 ) ) = 1 , m = \lim_{x\to+\infty}\frac{f(x)}{x} = \lim_{x\to+\infty}\left(1 + \frac1x + \frac{1}{x(x-2)}\right) = 1, m = lim x → + ∞ x f ( x ) = lim x → + ∞ ( 1 + x 1 + x ( x − 2 ) 1 ) = 1 ,
n = lim x → + ∞ ( f ( x ) − x ) = lim x → + ∞ ( 1 + 1 x − 2 ) = 1. n = \lim_{x\to+\infty}(f(x) - x) = \lim_{x\to+\infty}\left(1 + \frac{1}{x - 2}\right) = 1. n = lim x → + ∞ ( f ( x ) − x ) = lim x → + ∞ ( 1 + x − 2 1 ) = 1.
Răspuns: y = x + 1 y = x + 1 y = x + 1 Barem: 2p m m m n n n
c) f f f ( 2 , + ∞ ) (2, +\infty) ( 2 , + ∞ )
f ′ ′ ( x ) = 2 ( x − 2 ) 3 > 0 f''(x) = \dfrac{2}{(x - 2)^3} > 0 f ′′ ( x ) = ( x − 2 ) 3 2 > 0 x > 2 x > 2 x > 2 f f f Barem: 3p, 2p.
2. f ( x ) = 1 + ln x f(x) = 1 + \ln x f ( x ) = 1 + ln x F ( x ) = x ln x F(x) = x\ln x F ( x ) = x ln x ( 0 , + ∞ ) (0, +\infty) ( 0 , + ∞ )
a) ∫ 1 e ( f ( x ) − ln x ) d x \displaystyle\int_1^e(f(x) - \ln x)\,dx ∫ 1 e ( f ( x ) − ln x ) d x
∫ 1 e 1 d x = x ∣ 1 e = e − 1. \int_1^e 1\,dx = x\Big|_1^e = e - 1. ∫ 1 e 1 d x = x 1 e = e − 1.
Răspuns: e − 1 e - 1 e − 1 Barem: 3p, 2p.
b) F F F f f f
F ′ ( x ) = ln x + x ⋅ 1 x = ln x + 1 = f ( x ) F'(x) = \ln x + x\cdot\dfrac1x = \ln x + 1 = f(x) F ′ ( x ) = ln x + x ⋅ x 1 = ln x + 1 = f ( x ) Barem: 3p, 2p.
c) ∫ 1 e f ( x ) F ( x ) d x = e 2 2 \displaystyle\int_1^e f(x)F(x)\,dx = \frac{e^2}{2} ∫ 1 e f ( x ) F ( x ) d x = 2 e 2
Cum f = F ′ f = F' f = F ′ ∫ F ′ F d x = F 2 2 \displaystyle\int F'F\,dx = \frac{F^2}{2} ∫ F ′ F d x = 2 F 2
F 2 ( x ) 2 ∣ 1 e = ( e ln e ) 2 − ( 1 ⋅ ln 1 ) 2 2 = e 2 2 . \frac{F^2(x)}{2}\Big|_1^e = \frac{(e\ln e)^2 - (1\cdot\ln 1)^2}{2} = \frac{e^2}{2}. 2 F 2 ( x ) 1 e = 2 ( e l n e ) 2 − ( 1 ⋅ l n 1 ) 2 = 2 e 2 .
Barem: 3p, 2p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 0 0 0 I.3 x = − 2 x = -2 x = − 2 I.4 200 lei II.1.c m ∈ { − 2 , 0 } m \in \{-2, 0\} m ∈ { − 2 , 0 } II.2.b 2017 2017 2017 II.2.c x ∈ { − 6 , − 2 } x \in \{-6, -2\} x ∈ { − 6 , − 2 } III.1.b y = x + 1 y = x + 1 y = x + 1 III.2.a e − 1 e - 1 e − 1 III.2.c e 2 2 \tfrac{e^2}{2} 2 e 2
Ce merită reținut din simulare
Un factor nul anulează produsul : caută întâi un f ( k ) = 0 f(k) = 0 f ( k ) = 0 Asimptota oblică : m = lim f ( x ) x m = \lim\frac{f(x)}{x} m = lim x f ( x ) n = lim ( f ( x ) − m x ) n = \lim(f(x) - mx) n = lim ( f ( x ) − m x ) ∫ F ′ F = F 2 2 \int F'F = \frac{F^2}{2} ∫ F ′ F = 2 F 2