已知函数f(x)=e x (sinx-cosx),x∈(0,2013π),则函数f(x)的极大值之和为(  ) A.

2025-06-29 04:29:35
推荐回答(1个)
回答1:

∵函数f(x)=e x (sinx-cosx),
∴f′(x)=(e x )′(sinx-cosx)+e x (sinx-cosx)′=2e x sinx,
∵x∈(2kπ,2kπ+π)(k∈Z)时,f′(x)>0,x∈(2kπ+π,2kπ+2π)(k∈Z)时,f′(x)<0,
∴x∈(2kπ,2kπ+π)(k∈Z)时f(x)递增,x∈(2kπ+π,2kπ+2π)(k∈Z)时f(x)递减,
∴当x=2kπ+π(k∈Z)时,f(x)取极大值,
其极大值为f(2kπ+π)=e 2kπ+π [sin(2kπ+π)-cos(2kπ+π)]
=e 2kπ+π ×(0-(-1))
=e 2kπ+π
又x∈(0,2013π),
∴函数f(x)的各极大值之和S=e π +e +e +…+e 2011π
e π (1 -e 2π×1006 )
1 -e
=
e π (1 -e 2012π )
1 -e

故选:B.