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已知函数F(x)=2sin2(π4+x)?3Cos2x?1(x∈R)(1)求F...

解答: (1) f(x)=2sin²(π/4+x)-根号3cos2x =1-cos(π/2+2x)-√3cos2x =sin2x-√3cos2x+1 =2sin(2x-π/3)+1 ∵ x∈[π/4,π/2] ∴ 2x-π/3∈[π/6,2π/3] ∴ sin(2x-π/3)∈[1/2,1] ∴ 2x-π/3=π/6时,f(x)有最小值2 2x-π/3=π/2时,f(x)有最大值3 (2) |f(x...

∵f(x)=4sin^(π/4+x)-2√3cos2x-1 ∴f(x)=2[1-cos(π/2+2x)]-2√3cos2x-1 =2sin2x-2√3cos2x+1 =4sin(2x-π/6)+1 又π/2≤2x≤π,π/3≤2x-π/6≤5/6π ∴f(x)max=5 f(x)min=3 若q|f(x)-m|

(本小题满分14分)解:(1)f(x)=4sinx?1?cos(π2+x)2+cos2x?1=2sinx(1+sinx)-2sin2x=2sinx.∵f(ωx)=2sinωx在[?π2,2π3]是增函数,∴[?π2,2π3]?[?π2ω,π2ω]?2π3≤π2ω,∴ω∈(0,34](2)[12f(x)]2?mf(x)+m2+m?1=sin2x-2msinx+m2+m-1>0因为x∈[π...

(1)f(x)=12(3-cos2x)-32[1-cos(2x-π2)]=32sin2x-12cos2x=sin(2x-π6),令-π2+2kπ≤2x-π6≤π2+2kπ,k∈Z,得到kπ-π6≤x≤kπ+π3,k∈Z,则函数f(x)的单调递增区间[kπ-π6,kπ+π3],k∈Z;(2)由f(B)=1,得到sin(2B-π6)=1,∴2B-π6=π2,即B=...

f(x)=cosxsin(x+π/6)-cos2x-1/4, =cosx(√3/2sinx+1/2cosx)-cos2x-1/4, =√3/2sinxcosx+1/2(cosx)^2-cos2x-1/4, =√3/4sin2x+1/4(1+cos2x)-cos2x-1/4, =√3/4sin2x-3/4cos2x =√3/2(1/2sin2x-√3/2cos2x) =√3/2sin(2x-π/3) 2x-π/3∈[2kπ-π/2,2kπ+π/2]单...

等一下

f(x)=2sin平方(π/4+x)-根号3cos2x,x∈(π/4,π/2) =[1-cos(2x+π/2)]-√3cos2x =1+sin2x-√3cos2x =1+2sin(2x-π/3) f(X)的最大值3和最小值1 x∈(π/4,π/2),y∈(1/2,1) │f(x)-m│<2 f(x)-2

f(x)=4sinxsin²(π/4+x/2)+cos2x-1 =4sinxsin²[(π/2+x)/2]+cos2x-1 =4sinx[1-cos(π/2+x]/2+cos2x-1 =2sinx+2sin²x+1-2sin²x-1 =2sinx ∴f(ωx)=2sinωx x∈[-π/2,2π/3]是增函数 f'(ωx)=2ωcosωx>0 ∵ω>0 ∴cosωx>0 ω·2π/3≤π/2→ω≤3/4...

『根3』+1,-1.(负无穷,『根3』-1)。仅供参考。

解:(1)f(x)=2sin²(π/4+x)-√3cos2x=1-cos(π/2+2x)-√3cos2x=1+sin(2x)-√3cos2x =1+sin(2x-π/3)/2 π/4≤x≤π/2,π/2≤2x≤π,π/6≤2x-π/3≤2π/3,1/4≤sin(2x-π/3)/2≤1/2, 5/4≤f(x)≤3/2。f(x)的最大值和最小值分别为3/2和5/4。 (2)若f(x)≥m,则有...

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