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已知函数F(x)=4sinxsin²(π/4+x/2)+Cos2x%1

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...

(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)函数f(x)=4sinx?sin 2 ( π 4 + x 2 )+cos2x=2sinx[1-cos( π 2 +x )]+cos2x=2sinx+2sin 2 x+cos 2 x-sin 2 x=1+2sinx…(4分)由题意需[- π 2 , 2π 3 ] ?[- π 2ω , π 2ω ] 得ω ∈(0, 3 4 ] …(6分)(2)由题意当 π 6 ≤x≤ 2π 3 时,...

(本小题满分14分)(1) f(x)=4sinx? 1-cos( π 2 +x) 2 +cos2x-1 =2sinx(1+sinx)-2sin 2 x=2sinx.∵ f(ωx)=2sinωx在[- π 2 , 2π 3 ] 是增函数,∴ [- π 2 , 2π 3 ]?[- π 2ω , π 2ω ] ? 2π 3 ≤ π 2ω ,∴ ω∈(0, 3 4 ] (2) [ 1 2 f(x) ] 2...

解: (1) tanx有意义,x≠kπ+ π/2,(k∈Z) 函数定义域为{x|x≠kπ+ π/2,k∈Z} f(x)=4tanxsin(π/2 -x)cos(x- π/3) -√3 =4tanxcosxcos(x-π/3)-√3 =4sinx[cosxcos(π/3)+sinxsin(π/3)] -√3 =4sinx[(1/2)cosx+(√3/2)sinx] -√3 =2sinxcosx+2√3sin²x-√...

(1)f(x)=4sinx?sin2(π4+x2)+2cos2x+1+a=4sinx?1?cos(π2+x)2+2cos2x+1+a=2sinx(1+sinx)+2cos2x+1+a=2sinx+2sin2x+2cos2x+1+a=2sinx+3+a,∵函数f(x)为奇函数,∴a+3=0,即a=-3,f(x)=2sinx,f(2x)=2sin2x≥-3,即sin2x≥-32,∴2kπ+4π3≥...

(1)化简可得f(x)=4sinx?1?cos(π2+x)2+cos2x=2sinx(1+sinx)+1-2sin2x=2sinx+1,∵x∈R,∴sinx∈[-1,1],∴f(x)的值域是[-1,3](2)当x∈[π6,2π3]时,sinx∈[12,1],∴f(x)∈[2,3]由|f(x)-m|<2可得-2<f(x)-m<2,∴f(x)-2<m<f(x...

(1)∵f(x)=2sin2x-23cos2x+1=4sin(2x-π3)+1.又∵π4≤x≤π2,∴π6≤2x-π3≤2π3,即3≤4sin(2x-π3)+1≤5∴f(x)max=5,f(x)min=3(2)∵|f(x)-m|<2,∴m-2<f(x)<m+2又p是q的充分不必要条件∴m?2<3m+2>5,∴3<m<5.∴m的取值范围为(3,5)

前两问基本相同,不在回答,只回答第三问 具体见图片

图象过点(π/4,1/4) f(x)=sinxcosxcosφ+cos²xsinφ+1/2sin(π+φ) =1/2sin2xcosφ+1/2cos2xsinφ+1/2sinφ-1/2sinφ =1/2sin(2x+φ) f(π/4)=1/2sin(π/2+φ)=1/2cosφ=1/4 cosφ=1/2 φ=π/3 2)f(x)=1/2sin(2x+π/3)=1/2sin2[x+π/6] y=g(x)=1/2sin2[x+π...

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