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. The article ______ now will appear in tomorrow’s newspaper.A.writingB.to writeC.being writtenD.write