设函数f(x)=2sin(2x+π\6)的最大值是M,求及函数的单调递增区间;求M及函数f(x)的单调递增区间;若10个互不相等的正数满足f(Xi)=M,且Xi
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设函数f(x)=2sin(2x+π\6)的最大值是M,求及函数的单调递增区间;求M及函数f(x)的单调递增区间;若10个互不相等的正数满足f(Xi)=M,且Xi
设函数f(x)=2sin(2x+π\6)的最大值是M,求及函数的单调递增区间;
求M及函数f(x)的单调递增区间;若10个互不相等的正数满足f(Xi)=M,且Xi
设函数f(x)=2sin(2x+π\6)的最大值是M,求及函数的单调递增区间;求M及函数f(x)的单调递增区间;若10个互不相等的正数满足f(Xi)=M,且Xi
f(x)max=2=M
当2kπ-π/2≤2x+π\6≤2kπ+π/2
即kπ-π/3≤2x+π\6≤kπ+π/6时f(x)递增
所以f(x)的单调递增区间为[kπ-π/3,kπ+π/6],k∈Z
函数周期为π
因为Xi是10个互不相等的正数,且f(Xi)=M=f(π/6)
所以Xi=π/6,π/6+π,π/6+2π.
所以x1+x2+x3.+x10=(1+2+3+...+9)π+π/6×10=140π/3
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