Question

1.求函数y=sin2x的单调递增区间

Answer 1

y=sin(2x)的单调递增区间为2x∈(2kπ-π/2,2kπ+π/2),k∈Z
即x∈(kπ-π/4,kπ+π/4),k∈Z
y=√3sinx+cosx
=2(√3sinx/2+cosx/2)
=2[sinxcos(π/6)+cosxsin(π/6)]
=2sin(x+π/6)
当x+π/6=2kπ+π/2,即x=2kπ+π/3,k∈Z时,y有最大值2
当x+π/6=2kπ-π/2,即x=2kπ-2π/3,k∈Z时,y有最小值-2
y=2+|cosx|
当x=kπ,k∈Z时,y有最大值3
当x=kπ+π/2,k∈Z时,y有最小值2
y=sin(π/6-x)=-sin(x-π/6)
y=-sin(x-π/6)的单调递减区间为x-π/6∈(2kπ-π/2,2kπ+π/2),k∈Z
即x∈(2kπ-π/3,2kπ+2π/3),k∈Z