已知函数f(x)=2√3sinxcosx+2cos²x+m(1)求函数f(x)在[0,π]上得单调递增区间(2)当x∈[0,π/6]时,|f(x)|<4恒成立,求实数m的取值范围
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![已知函数f(x)=2√3sinxcosx+2cos²x+m(1)求函数f(x)在[0,π]上得单调递增区间(2)当x∈[0,π/6]时,|f(x)|<4恒成立,求实数m的取值范围](/uploads/image/z/1615638-30-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D2%E2%88%9A3sinxcosx%2B2cos%26%23178%3Bx%2Bm%EF%BC%881%EF%BC%89%E6%B1%82%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E5%9C%A8%5B0%2C%CF%80%5D%E4%B8%8A%E5%BE%97%E5%8D%95%E8%B0%83%E9%80%92%E5%A2%9E%E5%8C%BA%E9%97%B4%EF%BC%882%EF%BC%89%E5%BD%93x%E2%88%88%5B0%2C%CF%80%2F6%5D%E6%97%B6%2C%7Cf%28x%29%7C%EF%BC%9C4%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知函数f(x)=2√3sinxcosx+2cos²x+m(1)求函数f(x)在[0,π]上得单调递增区间(2)当x∈[0,π/6]时,|f(x)|<4恒成立,求实数m的取值范围
已知函数f(x)=2√3sinxcosx+2cos²x+m
(1)求函数f(x)在[0,π]上得单调递增区间
(2)当x∈[0,π/6]时,|f(x)|<4恒成立,求实数m的取值范围
已知函数f(x)=2√3sinxcosx+2cos²x+m(1)求函数f(x)在[0,π]上得单调递增区间(2)当x∈[0,π/6]时,|f(x)|<4恒成立,求实数m的取值范围
1)f(x)=√3sin2x+cos2x+1+m=2sin(2x+π/6)+m+1
在[0,π], π/6=<2x+π/6<=2π+π/6
递增区间为: π/6=<2x+π/6<=π/2, 及2π-π/2=<2x+π/6<=2π+π/6
即:0=
,f(x)的最大值为f(π/6)=m+3, |m+3|<4, 得: -7
f(x)=1+cos2x+√3sin2x+m
=2sin(2x+π/6)+m+1
(1) 0≤x≤π 0≤2x≤2π π/6≤2x+π/6≤13π/6
单调增区间 【0,π/6】U【2π/3,π】
(2) x∈[0,π/6], 最小值f(0)=m+2
最大值f(π/6)=m+3
...
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f(x)=1+cos2x+√3sin2x+m
=2sin(2x+π/6)+m+1
(1) 0≤x≤π 0≤2x≤2π π/6≤2x+π/6≤13π/6
单调增区间 【0,π/6】U【2π/3,π】
(2) x∈[0,π/6], 最小值f(0)=m+2
最大值f(π/6)=m+3
所以 m+2>-4且m+3<4
所以 m>-6且 m<1
即 -6
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