函数y=asin(x+π/6)+b的值域在[-1/2,9/2],求a的值,以及原函数的单调增区间
3个回答

∵数y=asin(x+π/6)+b的值域在[-1/2,9/2]

∴最大值|a|+b=9/2

最小值-|a|+b=-1/2

解得|a|=5/2 b=2

∴a=5/2或a=-5/2

当a=5/2时,y=5/2sin(x+π/6)+2

由2kπ-π/2≤x+π/6≤2kπ+π/2,

得2kπ-2π/3≤x≤2kπ+π/3,k∈Z

∴函数递增区间为

[2kπ-2π/3,2kπ+π/3],k∈Z

当a=-5/2时,y=-5/2sin(x+π/6)+2

由2kπ+π/2≤x+π/6≤2kπ+3π/2,

得2kπ+π/3≤x≤2kπ+4π/3,k∈Z

∴函数递增区间为

[2kπ+π/3,2kπ+4π/3],k∈Z