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Ministerul Educaţiei, Cercetării şi Inovării 
Centrul Naţional pentru Curriculum şi Evaluare în Învăţământul Preuniversitar 
Soluţie 
1. 2(1− x) = x +1+ 4⇒ x = −1 
2. f (0) = −6 , f (x) = 0⇒ x∈{1;−6}⇒ A(0;−6),B(1;0),C(−6;0) 
1 
3. 
sin 
    ∈ −  −  + ∈  
    
x = − , ( ) 1 
2 
x 1 k arcsin 
kπ k 
2 
x 
∈ π π    
Z ,dar x∈[0,2π ]⇒ 7 11 
; 
6 6 
  
. 
4. Numărul cazurilor posibile este 26 . Numărul cazurilor favorabile este 
2 
C6 =15 . 
15 
64 
p = . 
5. 
JJG JJG JJG JJG 
+ + 
r r r 
A B C 
= ⇒ 
3 
r 
G 
JJG G G 
rC = 6i + 6 j 
. 
G G G G GG G G GG GG G G 
6.( )( ) 2 2 2 2 
2u + v ⋅ 2v − u = 4uv − 2u + 2v − uv = 3uv − 2u + 2v 
, 
GG G G 
2 2 1 2 
uv − u + v = ⋅ ⋅ ⋅ − ⋅ + ⋅ 
3 2 2 312 21 22 
2 
G G G G 
, (2u + v)⋅(2v − u) = 9 
. 
BACALAUREAT 2009-MATEMATICĂ - Proba D, tipul subiectului MT1, programa M1

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D mt1 i_061

  • 1. Ministerul Educaţiei, Cercetării şi Inovării Centrul Naţional pentru Curriculum şi Evaluare în Învăţământul Preuniversitar Soluţie 1. 2(1− x) = x +1+ 4⇒ x = −1 2. f (0) = −6 , f (x) = 0⇒ x∈{1;−6}⇒ A(0;−6),B(1;0),C(−6;0) 1 3. sin     ∈ −  −  + ∈      x = − , ( ) 1 2 x 1 k arcsin kπ k 2 x ∈ π π    Z ,dar x∈[0,2π ]⇒ 7 11 ; 6 6   . 4. Numărul cazurilor posibile este 26 . Numărul cazurilor favorabile este 2 C6 =15 . 15 64 p = . 5. JJG JJG JJG JJG + + r r r A B C = ⇒ 3 r G JJG G G rC = 6i + 6 j . G G G G GG G G GG GG G G 6.( )( ) 2 2 2 2 2u + v ⋅ 2v − u = 4uv − 2u + 2v − uv = 3uv − 2u + 2v , GG G G 2 2 1 2 uv − u + v = ⋅ ⋅ ⋅ − ⋅ + ⋅ 3 2 2 312 21 22 2 G G G G , (2u + v)⋅(2v − u) = 9 . BACALAUREAT 2009-MATEMATICĂ - Proba D, tipul subiectului MT1, programa M1