加冕
幼苗
共回答了22个问题采纳率:95.5% 举报
(1) f(x)=sin(2(x-π/12))
x∈[0,π]时,2(x-π/12)∈[-π/6,11π/6]
单调递减区间是2(x-π/12)∈[π/2,3π/2]
即x-π/12∈[π/4,3π/4]
则x∈[π/3,5π/6]
(2)x∈[-π/12,π/2]时,
2(x-π/12)∈[-π/3,5π/6]
而当2(x-π/12)∈[-π/3,π/3]时,sin(2(x-π/12))∈[sin(-π/3),sin(π/3)]
即sin(2(x-π/12))∈[-√3/2,√3/2] ①
而当2(x-π/12)∈[π/3,5π/6]时,由(1)的讨论可知,此时函数单调递减,即
sin(2(x-π/12))∈[sin(5π/6),sin(π/3)],即sin(2(x-π/12))∈[1/2,√3/2] ②
由①②可知,sin(2(x-π/12))∈[-√3/2,√3/2] ,而f(x)≥a恒成立,则
则a≤-√3/2,即a的取值范围(-∞,-√3/2]
(3)
g(x)=f((x+π/6)/2)=sin(x+π/6-π/6)=sinx
g(x)-1/3=sinx-1/3
在区间[-2π,4π]上,显然满足sinx-1/3=0的零点,
都是满足sinx=1/3的点,而在区间[-2π,2π]上,显然这些点关于y轴对称分布,
则这些零点的和为0
因此只需考察区间(2π,4π]上满足sinx=1/3的点
显然区间[2π,4π]是函数sinx的一个完整周期,共有4个点满足条件,
前两个点关于x=2π+π/2对称分布,后两个点关于x=2π+3π/2对称分布
则这4个点的和是,2(2π+π/2)+2(2π+3π/2)=2(2π+π/2+2π+3π/2)=12π
1年前
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