定義在R上的函數f(x)滿足f(x)=f(x+2)當x∈[3,5]時,f(x)=2-∣x-4∣,則( )A.f(sinπ/6)<f(cosπ/6)B.f(sin 1)>f(cos 1)C.f(cos2π/3)<(sin2π/3)D.f(cos 2)>f(sin 2)請對各選項作具體分析。謝謝！

熱心網友

解：由已知，f(x)是以2為最小正周期的周期函數，且x∈[3,5]時,f(x)=2-∣x-4∣所以：A.錯f(sinπ/6)=f(0.5)=f(2+0.5)=f(2+2.5)=1.5f(cosπ/6)=f(√3/2)=f[2+(√3/2)]=f[4+(√3/2)]=2-(√3/2)f(cosπ/6)B.錯同理，f(sin1)=f(4+sin1)=2-sin1, f(cos1)=f(4+cos1)=2-cos1因cos1(sin2π/3)D.對f(sin2)=f(4+sin2)=2-︱sin2︱, f(cos2)=f(4+cos2)=2-︱cos2︱因︱sin2︱︱cos2︱所以f(cos2)f(sin2)

熱心網友

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