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1.
Ministerul Educaţiei, Cercetării
şi Inovării Centrul Naţional pentru Curriculum şi Evaluare în Învăţământul Preuniversitar Soluţie b 1. 2 b3 = 6 ⋅ 24⇒b3 =12 3 = = ⇒ b1 = 3 . b 2 2 q 2. 3−m2 > 0⇒ m∈(− 3; 3). 3. π = 3 2 π = 3 ⇒ 3 4 3 sin ,sin 3 2 3 2 π = π = − , sin 0,sin 3 3 2 π + 2 π + 3 π + 4 π = 3 . sin sin sin sin 3 3 3 3 2 4. Numărul cazurilor posibile este: 33 . Numărul cazurilor favorabile este = ⇒ p = . JJJG JJJG 1 m = . α = − . BACALAUREAT 2009-MATEMATICĂ - Proba D, tipul subiectului MT1, programa M1 2 3! 6 9 5. 1 3 GP AB = , 1 3 GP = AB⇒ 3 6. cos2α = 2cos2α −1⇒ 7 cos2 9
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