Rezolvarea completă a subiectului de Bacalaureat 2020 la Matematică M_tehnologic , sesiunea iunie–iulie, varianta 6 . 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 = 6 \sqrt3\left(2\sqrt3 + 1\right) - \sqrt3 = 6 3 ( 2 3 + 1 ) − 3 = 6
2 3 ⋅ 3 + 3 − 3 = 2 ⋅ 3 = 6. ■ 2\sqrt3\cdot\sqrt3 + \sqrt3 - \sqrt3 = 2\cdot 3 = 6.\ \blacksquare 2 3 ⋅ 3 + 3 − 3 = 2 ⋅ 3 = 6. ■
Barem: 3p, 2p.
2. f ( x ) = x 2 − 4 x + 2 f(x) = x^2 - 4x + 2 f ( x ) = x 2 − 4 x + 2 a a a f ( a ) = 2 f(a) = 2 f ( a ) = 2
a 2 − 4 a + 2 = 2 ⟺ a ( a − 4 ) = 0 ⟺ a ∈ { 0 , 4 } a^2 - 4a + 2 = 2 \iff a(a - 4) = 0 \iff a \in \{0, 4\} a 2 − 4 a + 2 = 2 ⟺ a ( a − 4 ) = 0 ⟺ a ∈ { 0 , 4 } Barem: 3p, 2p.
3. Ecuația x − 1 = 3 \sqrt{x - 1} = 3 x − 1 = 3
x − 1 = 9 ⟺ x = 10 x - 1 = 9 \iff x = 10 x − 1 = 9 ⟺ x = 10 9 = 3 \sqrt9 = 3 9 = 3
Răspuns: x = 10 x = 10 x = 10 Barem: 3p, 2p.
4. După o ieftinire cu 10 % 10\% 10%
x − 10 100 x = 180 ⟺ 0,9 x = 180 ⟺ x = 200 x - \frac{10}{100}x = 180 \iff 0{,}9x = 180 \iff x = 200 x − 100 10 x = 180 ⟺ 0 , 9 x = 180 ⟺ x = 200
Răspuns: 200 200 200 Barem: 3p, 2p.
5. Înălțimea din C C C △ A B C \triangle ABC △ A B C A ( 4 , 1 ) A(4,1) A ( 4 , 1 ) B ( − 4 , 1 ) B(-4,1) B ( − 4 , 1 ) C ( 0 , 4 ) C(0,4) C ( 0 , 4 )
A B AB A B y = 1 y = 1 y = 1 C C C 0 0 0 D ( 0 , 1 ) D(0, 1) D ( 0 , 1 ) A B AB A B C A = C B = 5 CA = CB = 5 C A = C B = 5 C D ⊥ A B CD \perp AB C D ⊥ A B
C D = ∣ 4 − 1 ∣ = 3. CD = |4 - 1| = 3. C D = ∣4 − 1∣ = 3.
Răspuns: 3 3 3 Barem: 3p, 2p.
6. Arătați că 3 sin 60 ∘ − 2 cos 45 ∘ = 1 2 \sqrt3\sin 60^\circ - \sqrt2\cos 45^\circ = \frac12 3 sin 6 0 ∘ − 2 cos 4 5 ∘ = 2 1
3 ⋅ 3 2 − 2 ⋅ 2 2 = 3 2 − 1 = 1 2 . ■ \sqrt3\cdot\frac{\sqrt3}{2} - \sqrt2\cdot\frac{\sqrt2}{2} = \frac32 - 1 = \frac12.\ \blacksquare 3 ⋅ 2 3 − 2 ⋅ 2 2 = 2 3 − 1 = 2 1 . ■
Barem: 2p, 3p.
Subiectul al II-lea (30 de puncte)
1. A = ( 1 3 0 2 ) A = \begin{pmatrix}1 & 3\\ 0 & 2\end{pmatrix} A = ( 1 0 3 2 )
a) det A = 1 ⋅ 2 − 3 ⋅ 0 = 2 \det A = 1\cdot 2 - 3\cdot 0 = 2 det A = 1 ⋅ 2 − 3 ⋅ 0 = 2 ■ \blacksquare ■ Barem: 3p, 2p.
b) A ⋅ A = ( 1 9 0 4 ) A\cdot A = \begin{pmatrix}1 & 9\\ 0 & 4\end{pmatrix} A ⋅ A = ( 1 0 9 4 ) 3 A − A 2 = ( 3 9 0 6 ) − ( 1 9 0 4 ) = ( 2 0 0 2 ) = 2 I 2 3A - A^2 = \begin{pmatrix}3 & 9\\ 0 & 6\end{pmatrix} - \begin{pmatrix}1 & 9\\ 0 & 4\end{pmatrix} = \begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix} = 2I_2 3 A − A 2 = ( 3 0 9 6 ) − ( 1 0 9 4 ) = ( 2 0 0 2 ) = 2 I 2 ■ \blacksquare ■ Barem: 2p, 3p.
c) x A − I 2 = ( x − 1 3 x 0 2 x − 1 ) xA - I_2 = \begin{pmatrix}x - 1 & 3x\\ 0 & 2x - 1\end{pmatrix} x A − I 2 = ( x − 1 0 3 x 2 x − 1 )
( x A − I 2 ) 2 = ( ( x − 1 ) 2 9 x 2 − 6 x 0 ( 2 x − 1 ) 2 ) , 5 A − I 2 = ( 4 15 0 9 ) . (xA - I_2)^2 = \begin{pmatrix}(x-1)^2 & 9x^2 - 6x\\ 0 & (2x-1)^2\end{pmatrix},\qquad 5A - I_2 = \begin{pmatrix}4 & 15\\ 0 & 9\end{pmatrix}. ( x A − I 2 ) 2 = ( ( x − 1 ) 2 0 9 x 2 − 6 x ( 2 x − 1 ) 2 ) , 5 A − I 2 = ( 4 0 15 9 ) .
Sistemul: ( x − 1 ) 2 = 4 ⇒ x ∈ { − 1 , 3 } (x - 1)^2 = 4 \Rightarrow x \in \{-1, 3\} ( x − 1 ) 2 = 4 ⇒ x ∈ { − 1 , 3 } ( 2 x − 1 ) 2 = 9 ⇒ x ∈ { − 1 , 2 } (2x - 1)^2 = 9 \Rightarrow x \in \{-1, 2\} ( 2 x − 1 ) 2 = 9 ⇒ x ∈ { − 1 , 2 } 9 x 2 − 6 x = 15 ⇒ x ∈ { − 1 , 5 3 } 9x^2 - 6x = 15 \Rightarrow x \in \{-1, \frac53\} 9 x 2 − 6 x = 15 ⇒ x ∈ { − 1 , 3 5 } x = − 1 x = -1 x = − 1
Răspuns: x = − 1 x = -1 x = − 1 Barem: 3p, 2p.
2. x ∘ y = x 2 + ( x + 1 ) ( y + 1 ) + y 2 x \circ y = x^2 + (x+1)(y+1) + y^2 x ∘ y = x 2 + ( x + 1 ) ( y + 1 ) + y 2
a) 3 ∘ ( − 1 ) = 9 + 4 ⋅ 0 + 1 = 10 3 \circ (-1) = 9 + 4\cdot 0 + 1 = 10 3 ∘ ( − 1 ) = 9 + 4 ⋅ 0 + 1 = 10 ■ \blacksquare ■ Barem: 3p, 2p.
b) x ∘ y = x 2 + ( x + 1 ) ( y + 1 ) + y 2 = y 2 + ( y + 1 ) ( x + 1 ) + x 2 = y ∘ x x \circ y = x^2 + (x+1)(y+1) + y^2 = y^2 + (y+1)(x+1) + x^2 = y \circ x x ∘ y = x 2 + ( x + 1 ) ( y + 1 ) + y 2 = y 2 + ( y + 1 ) ( x + 1 ) + x 2 = y ∘ x ■ \blacksquare ■ Barem: 2p, 3p.
c) x ∘ 1 = x 2 + 2 ( x + 1 ) + 1 = x 2 + 2 x + 1 + 2 = ( x + 1 ) 2 + 2 ≥ 2 x \circ 1 = x^2 + 2(x + 1) + 1 = x^2 + 2x + 1 + 2 = (x + 1)^2 + 2 \ge 2 x ∘ 1 = x 2 + 2 ( x + 1 ) + 1 = x 2 + 2 x + 1 + 2 = ( x + 1 ) 2 + 2 ≥ 2 ■ \blacksquare ■ Barem: 3p, 2p.
Subiectul al III-lea (30 de puncte)
1. f ( x ) = ( x − 1 ) ln x f(x) = (x - 1)\ln x f ( x ) = ( x − 1 ) ln x x > 0 x > 0 x > 0
a) f ′ ( x ) = ln x + ( x − 1 ) ⋅ 1 x = 1 − 1 x + ln x f'(x) = \ln x + (x - 1)\cdot\dfrac1x = 1 - \dfrac1x + \ln x f ′ ( x ) = ln x + ( x − 1 ) ⋅ x 1 = 1 − x 1 + ln x ■ \blacksquare ■ Barem: 3p, 2p.
b) f ( 1 ) = 0 f(1) = 0 f ( 1 ) = 0 f ′ ( 1 ) = 1 − 1 + 0 = 0 f'(1) = 1 - 1 + 0 = 0 f ′ ( 1 ) = 1 − 1 + 0 = 0 y − f ( 1 ) = f ′ ( 1 ) ( x − 1 ) y - f(1) = f'(1)(x - 1) y − f ( 1 ) = f ′ ( 1 ) ( x − 1 ) y = 0 y = 0 y = 0
Răspuns: y = 0 y = 0 y = 0 Barem: 2p, 3p.
c) Pentru x ∈ ( 0 , 1 ] x \in (0, 1] x ∈ ( 0 , 1 ] ln x ≤ 0 \ln x \le 0 ln x ≤ 0 1 x ≥ 1 ⇒ 1 − 1 x ≤ 0 \frac1x \ge 1 \Rightarrow 1 - \frac1x \le 0 x 1 ≥ 1 ⇒ 1 − x 1 ≤ 0 f ′ ( x ) ≤ 0 f'(x) \le 0 f ′ ( x ) ≤ 0 f f f ( 0 , 1 ] (0, 1] ( 0 , 1 ] ■ \blacksquare ■ Barem: 3p, 2p.
2. f ( x ) = 1 + x x 2 + 1 − 2 x 2 + 1 f(x) = 1 + \dfrac{x}{x^2 + 1} - \dfrac{2}{x^2 + 1} f ( x ) = 1 + x 2 + 1 x − x 2 + 1 2
a) ( x 2 + 1 ) f ( x ) = x 2 + 1 + x − 2 = x 2 + x − 1 (x^2 + 1)f(x) = x^2 + 1 + x - 2 = x^2 + x - 1 ( x 2 + 1 ) f ( x ) = x 2 + 1 + x − 2 = x 2 + x − 1
∫ 0 1 ( x 2 + x − 1 ) d x = 1 3 + 1 2 − 1 = − 1 6 . ■ \int_0^1 (x^2 + x - 1)\,dx = \frac13 + \frac12 - 1 = -\frac16.\ \blacksquare ∫ 0 1 ( x 2 + x − 1 ) d x = 3 1 + 2 1 − 1 = − 6 1 . ■
Barem: 3p, 2p.
b) F ( x ) = ∫ 0 x f ( t ) d t = [ t + 1 2 ln ( t 2 + 1 ) − 2 a r c t g t ] 0 x = x + 1 2 ln ( x 2 + 1 ) − 2 a r c t g x . F(x) = \int_0^x f(t)\,dt = \left[t + \frac12\ln(t^2 + 1) - 2\,\mathrm{arctg}\,t\right]_0^x = x + \frac12\ln(x^2 + 1) - 2\,\mathrm{arctg}\,x. F ( x ) = ∫ 0 x f ( t ) d t = [ t + 2 1 ln ( t 2 + 1 ) − 2 arctg t ] 0 x = x + 2 1 ln ( x 2 + 1 ) − 2 arctg x .
Barem: 3p, 2p.
c) Pentru x ≠ 0 x \ne 0 x = 0 f ( 1 x ) = 1 + x 1 + x 2 − 2 x 2 1 + x 2 f\left(\frac1x\right) = 1 + \dfrac{x}{1 + x^2} - \dfrac{2x^2}{1 + x^2} f ( x 1 ) = 1 + 1 + x 2 x − 1 + x 2 2 x 2
f ( x ) + f ( 1 x ) = 2 + 2 x x 2 + 1 − 2 + 2 x 2 x 2 + 1 = 2 x x 2 + 1 . f(x) + f\left(\tfrac1x\right) = 2 + \frac{2x}{x^2+1} - \frac{2 + 2x^2}{x^2 + 1} = \frac{2x}{x^2 + 1}. f ( x ) + f ( x 1 ) = 2 + x 2 + 1 2 x − x 2 + 1 2 + 2 x 2 = x 2 + 1 2 x .
∫ 1 2 2 x x 2 + 1 d x = ln ( x 2 + 1 ) ∣ 1 2 = ln 5 − ln 2 = ln 5 2 . ■ \int_1^2\frac{2x}{x^2+1}dx = \ln(x^2 + 1)\Big|_1^2 = \ln 5 - \ln 2 = \ln\frac52.\ \blacksquare ∫ 1 2 x 2 + 1 2 x d x = ln ( x 2 + 1 ) 1 2 = ln 5 − ln 2 = ln 2 5 . ■
Barem: 2p, 3p.
Rezumatul răspunsurilor
Cerință Răspuns I.2 a ∈ { 0 , 4 } a \in \{0, 4\} a ∈ { 0 , 4 } I.3 x = 10 x = 10 x = 10 I.4 200 200 200 I.5 3 3 3 II.1.c x = − 1 x = -1 x = − 1 III.1.b y = 0 y = 0 y = 0 III.2.b F ( x ) = x + 1 2 ln ( x 2 + 1 ) − 2 a r c t g x F(x) = x + \frac12\ln(x^2+1) - 2\,\mathrm{arctg}\,x F ( x ) = x + 2 1 ln ( x 2 + 1 ) − 2 arctg x
Ce merită reținut din varianta 6
Egalitatea de matrice dă un sistem: soluția trebuie să verifice toate ecuațiile.Semnul derivatei se citește din semnul fiecărui termen pe interval.f ( x ) + f ( 1 / x ) f(x) + f(1/x) f ( x ) + f ( 1/ x )