Question

高一数学。

Answer 1

解析：
已知f(x)=a*sin(πx+α）+b*cos(πx+β) +1
若f(2010)=-1，那么：
a*sin(2010π+α）+b*cos(2010π+β) +1=-1
即a*sinα +b*cosβ=-2
所以：f(2011)
=a*sin(2011π+α）+b*cos(2011π+β) +1
=a*sin(π+α）+b*cos(π+β) +1
=-a*sinα -b*cosβ +1
=-(a*sinα +b*cosβ)+1
=2+1
=3

Answer 2

答案为1

Answer 3

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Answer 4

答案是3.f(2010)=asin(2010pi+a)+bcos(2010pi+b)+1=asina+bsinb+1=-1,所以asina+bsinb=-2.
f(2011)=asin(2011pi+a)+bcos(2011pi+b)+1=-(asina+bsinb)+1=2+1=3.

Answer 5