Răspuns:
2. f: (0,∞) -> R, f(x) = x-2m+2.
Dom. de definitie e (0,∞) => gr. funct. e o semidreapta
Daca Gf ∩ Ox= {A(x,0)} => f(x)=0 <=> x-2m+2=0 <=> x=2m-2 care trb sa fie ≠ (0,∞) <=> 2m≠ (2;+∞) => m≠( 1;+∞). sau m<1
6. α∈ ( π/2, π) ai cos2α = -1/2
| sinα| = √1-2cos2α/2 ( cos2α= 1-2*sin²α => 2*sin²α= 1-2cos2α => sin²α= 1-2cos2α/2 => |sinα| = √1-2 cos2α/2
|sinα| = √1+1/2/2 <=> |sinα|= √(3/4) => sinα= ± √3/ 2
α∈ (π/2, π) => sinα= + √3/2
