求y=x+|sin2x|单调区间
2个回答

y=x+|sin2x|等价于

y=x+sin2x kπ≤x≤kπ+π/2

y=x-sin2x kπ+π/2≤x≤(k+1)π

对该曲线方程求导,得

y'=1+2cos2x kπ≤x≤kπ+π/2

y'=1-2cos2x kπ+π/2≤x≤(k+1)π

当y'>0时,该函数单调增,即

1+2cos2x>0 kπ≤x≤kπ+π/2

1-2cos2x>0 kπ+π/2≤x≤(k+1)π

即±cos2x>-1/2

在kπ≤x≤kπ+π/2时,当且仅当kπ≤x≤kπ+π/3时,

cos2x>cos2π/3,即cos2x>-1/2

kπ+π/2≤x≤(k+1)π时,当且仅当kπ+π/2≤x≤kπ+5π/6时,

cos2x-1/2,

因此y=x+|sin2x|单调增区间是[kπ,kπ+π/3]U[kπ+π/2,kπ+5π/6],

单调减区间是[kπ+π/3,kπ+π/2]U[kπ+5π/6,kπ+π]

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