定义在R上的奇函数f(x)满足:f(x+1)=f(x-1),且当0≤x≤1时,f(x)=-8x2+8x,则f(-20132)=()A.2B.-1C.-2D.1-高二数学

题目简介

定义在R上的奇函数f(x)满足:f(x+1)=f(x-1),且当0≤x≤1时,f(x)=-8x2+8x,则f(-20132)=()A.2B.-1C.-2D.1-高二数学

题目详情

定义在R上的奇函数f(x)满足:f(x+1)=f(x-1),且当0≤x≤1时,f(x)=-8x2+8x,则f(-
2013
2
)
=(  )
A.2B.-1C.-2D.1
题型:单选题难度:偏易来源:不详

答案

∵f(x+1)=f(x-1),
∴f(x+2)=f(x),
∴f(x)是周期函数,且周期为2,
f(-class="stub"2013
2
)
=f(-class="stub"2013
2
+1006)=f(-class="stub"1
2
),
∵f(x)为奇函数,
∴f(-class="stub"1
2
)=-f(class="stub"1
2
),
∵当0≤x≤1时,f(x)=-8x2+8x,
∴f(class="stub"1
2
)=-8×(class="stub"1
2
)2+8×class="stub"1
2
=2,
f(-class="stub"2013
2
)
=-2.
故选:C.

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