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f(x)=根号3sin2x+2cos平方x+m(x€R)若f(x)在区间【0,派/2】上的最小值为3 求f(x)最少正周其和常数 求f(x)单调区间
人气:226 ℃ 时间:2020-05-15 15:36:59
解答
(1)f(x)=√3sin2x+1+cos2x+m=2sin(2x+π/6)+(m+1)
∴最小正周期为2π/2=π
∵x∈[0,π/2]
∴2x+π/6∈[π/6,7π/6]
∴2x+π/6=7π/6时,f(x)最小,即2*(-1/2)+(m+1)=3
∴m=3
(2)单调递增,∴2x+π/6∈[2kπ-π/2,2kπ+π/2]∴x∈[kπ-π/3,kπ+π/6]
单调递减,∴2x+π/6∈[2kπ+π/2,2kπ+3π/2]∴x∈[kπ+π/6,kπ+2π/3]
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