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. e
1个回答

∵函数f(x)=e x(sinx-cosx),

∴f′(x)=(e x)′(sinx-cosx)+e x(sinx-cosx)′=2e xsinx,

∵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 2π =

e π (1 -e 2012π )

1 -e 2π .

故选:B.