若函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<π/2)的最小值为-2,且它的图象经过点(0,√3)和(5π/6,0)
(1)写出一个满足条件的函数解析式f(x)
(2)若函数f(x)在(0,π/8]上单调递增,求此函数所有可能的解析式
(3)若函数f(x)在[0,2]上恰有一个最大值和最小值,求ω的值.
(1)解析:∵f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<π/2)的最小值为-2,
∴A=|-2|=2==>f(x)=2sin(ωx+φ),
又它的图象经过点(0,√3)和(5π/6,0)
f(0)=2sin(φ)=√3==>φ=π/3或φ=2π/3
∵|φ|<π/2,∴φ=π/3==>f(x)=2sin(ωx+π/3),
∴f(5π/6)=2sin(ω5π/6+π/3)=0
当点(5π/6,0)为半周期点时,ω5π/6+π/3=π==>ω=4/5
当点(5π/6,0)为整周期点时,ω5π/6+π/3=2π==>ω=2
∴满足条件的函数解析式为f(x)=2sin(4/5x+π/3)或f(x)=2sin(2x+π/3)
(2)解析:设函数f(x)在(0,π/8]上单调递增
∵f(x)=2sin(ωx+π/3)
最大的值点ωx+π/3=π/2==>x=π/(6ω)
令π/(6ω)>=π/8==>0<ω<=4/3
∴函数f(x)在(0,π/8]上单调递增,ω取值范围为ω∈(0,4/3]
∵ω=4/5<4/3满足题意,ω=2>4/3不满足题意
综上:满足题意,且在(0,π/8]上单调递增的函数解析式只有f(x)=2sin(4/5x+π/3)
(3) 解析:设函数f(x)在[0,2]上恰有一个最大值和最小值
∵f(x)=2sin(ωx+π/3)
单调递减区间:
2kπ+π/2<=ωx+π/3<=2kπ+3π/2==>2kπ/ω+π/(6ω)<= x<=2kπ/ω+7π/(6ω)
令7π/(6ω)<=2==>ω>=7π/12
2π/ω+π/(6ω)>2==>13π/6*1/ω>2==>ω<13π/12
∴在[0,2]上恰有一个最大值和最小值,ω∈[7π/12,13π/12)
函数f(x)满足题意,且在[0,2]上恰有一个最大值和最小值,ω=2
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