题目

已知在△ABC中,a、b、c分别为角A、B、C的对边,设f(x)=a2x2-(a2-b2)x-4c2.(1)若f(1)=0,且B-C=,求角C;(2)若f(2)=0,求角C的取值范围. 答案:解:(1)由f(1)=0,得a2-a2+b2-4c2=0,∴b=2c.                             又由正弦定理,得b=2RsinB,c=2RsinC,将其代入上式,得sinB=2sinC.             ∵B-C=,∴B=+C,将其代入上式,得sin(+C)=2sinC.                  ∴sincosC+cossinC=2sinC,整理得,sinC=cosC.                           ∴tanC=.                                                                 ∵角C是三角形的内角,∴C=.                                               (2)∵f(2)=0,∴4a2-2a2+2b2-4c2=0,即a2+b2-2c2=0.                            由余弦定理,得cosC=                                             =(当且仅当a=b时取等号).            ∴cosC≥,∠C是锐角.又∵余弦函数在(0,)上递减,∴0<C≤.
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