已知二次函数f(x)=x2-2x+6,设向量a=(sinx,2),b=(2sinx,),c=(cos2x,1),d=

已知二次函数f(x)=x2-2x+6,设向量a=(sinx,2),b=(2sinx,),c=(cos2x,1),d=(1,2).

(1)分别求a·b和c·d的取值范围;

(2)当x∈[0,π]时,求不等式f(a·b)>f(c·d)的解集.

答案

解:(1)a·b=2sin2x+1≥1   c·d=cos2x+1≥1

(2)∵f(x)图像关于x=1对称,∴f(x)在(1,+∞)内单调递增,

由f(a·b)>f(c·d) a·b>c·d,即2sin2x+1>2cos2x+1

又∵x∈[0,π]  ∴x∈()

故不等式的解集为().

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