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已知函数F(x)4sin²(π/4+x)%2√3Cos2x%1,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|

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

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22x-π/3=π/2时,f(x)有最大值3(2)|f(x)-m|m-2且f(x)m-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...

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...

f(x) = 2sin²(π/4+x)+√3cos2x-1 = {1-cos[2(π/4+x)] + √3cos2x - 1 = -cos(π/2+2x) + √3cos2x = -sin2x + √3cos2x = -2(sin2xcosπ/3 - cos2xsinπ/3) = -2sin(2x-π/3) 最小正周期 = 2π/2 = π 2x-π/3∈(2kπ+π/2,2kπ+3π/2),其中k∈Z时单调增...

(1)∵f(x)=2sin2(π4+x)-3cos2x-1=-cos2(x+π4)-3cos2x=sin2x-3cos2x=2sin(2x-π3).则函数的最大值为2,最小值为-2,函数的周期T=2π2=π.(2)∵f(x)=2sin(2x-π3),∴h(x)=f(x+t)=2sin(2x+2t-π3),∵h(x)=f(x+t)的图象关于点...

解: (1)f(x)=cosxsin(x+∏/3)-根号3cos^x+根号3/4 =cosx(sinxcosπ/3+cosxsinπ/3)-√3/2(1+cos2x)+√3/4 =1/2sinxcosx+√3/2cos²x-√3/2(1+cos2x)+√3/4 =1/4sin2x+√3/4(1+cos2x)-√3/2(1+cos2x)+√3/4 =1/4sin2x-√3/4cos2x-√3/2 =1/2(1/2sin2x...

(Ⅰ)f(x)=4cosx1?cos(π2+x)2+3cos2x?2cosx=2cosx(1+sinx)+3cos2x?2cosx=sin2x+3cos2x=2sin(2x+π3).故周期T=2π2=π.(Ⅱ)∵x∈(0,π2),∴π3<2x+π3<4π3,由π3<2x+π3≤π2?0<x≤π12,∴π2≤2x+π3<4π3?π12≤x<π2,f(x)的单调递增区间为x∈(0,...

(1)f(x)=2sin2(π4+x) ?3cos2x ?1=-cos(π2+2x)-3cos2x=sin2x-3cos2x=2sin(2x?π3).由 2kπ-π2≤2x-π3≤2kπ+π2,k∈z,可得 kπ-π12≤x≤kπ+5π12,,k∈z.再由x∈[π4,π2],可得 x∈[π4,5π12],故f(x)的单调递增区间 [π4,5π12].(2)不等式|f(x...

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