设f0(x)=sin(x),f1(x)=f0'(x),f2(x)=f1'(x),…,fn+1(x)=fn'(x),n∈N,则f2013(x)=()A.sinxB.-sin

题目简介

设f0(x)=sin(x),f1(x)=f0'(x),f2(x)=f1'(x),…,fn+1(x)=fn'(x),n∈N,则f2013(x)=()A.sinxB.-sin

题目详情

设f0(x)=sin(x),f1(x)=f0'(x),f2(x)=f1'(x),…,fn+1(x)=fn'(x),n∈N,则f2013(x)=(  )
A.sinxB.-sinxC.cosxD.-cosx
题型:单选题难度:偏易来源:不详

答案

f0(x)=sinx
f1(x)=f0'(x)=cosx
f2(x)=f1'(x)=-sinx
f3(x)=f2'(x)=-cosx
f4(x)=f3'(x)=sinx

由上面可以看出,以4为周期进行循环
∴f2013(x)=f1(x)=cosx.
故选C.

更多内容推荐