已知定义域为R的函数f(x)对任意实数x、y满足f(x+y)+f(x-y)=2f(x)cosy,且f(0)=0,f(π2)=1.给出下列结论:f(π4)=12;②f(x)为奇函数;③f(x)为周期函数

题目简介

已知定义域为R的函数f(x)对任意实数x、y满足f(x+y)+f(x-y)=2f(x)cosy,且f(0)=0,f(π2)=1.给出下列结论:f(π4)=12;②f(x)为奇函数;③f(x)为周期函数

题目详情

已知定义域为R的函数f(x)对任意实数x、y满足f(x+y)+f(x-y)=2f(x)cosy,且f(0)=0,f(
π
2
)=1
.给出下列结论:f(
π
4
)=
1
2
;②f(x)为奇函数;③f(x)为周期函数;④f(x)在(0,x)内单调递减.其中正确的结论序号是(  )
A.②③B.②④C.①③D.①④
题型:单选题难度:偏易来源:不详

答案

令x=y=class="stub"π
4
,根据f(x+y)+f(x-y)=2f(x)cosy,且f(0)=0,f(class="stub"π
2
)=1

f(class="stub"π
2
)+f(0)=2f(class="stub"π
4
)  •
2
2
f(class="stub"π
4
)=
2
2
故①不对
∵f(x+y)+f(x-y)=2f(x)cosy
令x=0,则
f(y)+f(-y)=f(0)cosy=0
f(-y)=-f(y)
所以f(x)是奇函数   故②对.
令x=class="stub"π
2
,由f(0)=0,f(class="stub"π
2
)=1知④不对
故选A.

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