设函数y=f(x)满足对任意的x∈R,f(x)≥0且f2(x+1)+f2(x)=9.已知当x∈[0,1]时,有f(x)=2-|4x-2|,则f(20136)的值为______.-数学

题目简介

设函数y=f(x)满足对任意的x∈R,f(x)≥0且f2(x+1)+f2(x)=9.已知当x∈[0,1]时,有f(x)=2-|4x-2|,则f(20136)的值为______.-数学

题目详情

设函数y=f(x)满足对任意的x∈R,f(x)≥0且f2(x+1)+f2(x)=9.已知当x∈[0,1]时,有f(x)=2-|4x-2|,则f(
2013
6
)
的值为______.
题型:填空题难度:中档来源:盐城二模

答案

∵f2(x+1)+f2(x)=9,即 f2(x+1)=9-f2(x),
∴f2(x+2)=9-f2(x+1),化简可得 f2(x+2)=9-[9-f2(x)]=f2(x).
再由 函数y=f(x)满足对任意的x∈R,f(x)≥0,可得 f(x+2)=f(x),故函数是周期为2的周期函数.
f(class="stub"2013
6
)
=f(336-class="stub"1
2
)=f(-class="stub"1
2
).
又 f2(-class="stub"1
2
)=9-f2(-class="stub"1
2
+1)
=9-f2(class="stub"1
2
),
再由当x∈[0,1]时,有f(x)=2-|4x-2|,可得f(class="stub"1
2
)=2-|4×class="stub"1
2
-2|=2,
故 f2(-class="stub"1
2
)=9-f2(class="stub"1
2
)=9-4=5,故f(-class="stub"1
2
)=
5

f(class="stub"2013
6
)
=f(-class="stub"1
2
)=
5

故答案为
5

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