定义一种运算a⊕b=a,a≤bb,a>b,令f(x)=(cos2x+sinx)⊕32,且x∈[0,π2],则函数f(x-π2)的最大值是()A.54B.1C.-1D.-54-数学

题目简介

定义一种运算a⊕b=a,a≤bb,a>b,令f(x)=(cos2x+sinx)⊕32,且x∈[0,π2],则函数f(x-π2)的最大值是()A.54B.1C.-1D.-54-数学

题目详情

定义一种运算a⊕b=
a,a≤b
b,a>b
,令f(x)=(cos2x+sinx)⊕
3
2
,且x∈[0,
π
2
],则函数f(x-
π
2
)的最大值是(  )
A.
5
4
B.1C.-1D.-
5
4
题型:单选题难度:中档来源:滨州一模

答案

∵cos2x+sinx=-sin2x+sinx+1=-(sinx-class="stub"1
2
)2+class="stub"5
4
class="stub"3
2

∴f(x)=(cos2x+sinx)⊗class="stub"3
2
=cos2x+sinx,
∴f(x-class="stub"π
2
)=cos2(x-class="stub"π
2
)+sin(x-class="stub"π
2
)=sin2x-cosx=-(cos2x+cosx+class="stub"1
4
)+1+class="stub"1
4
=-(cosx+class="stub"1
2
)2+class="stub"5
4
class="stub"5
4

故选A.

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