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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...

f'x = f'(pi/2) cosx -2sin2x ---1 f'(pi/2) = f'(pi/2) cos(pi/2) -2sinpi = f'(pi/2) *0 +2 所以f'(pi/2)=2 带入1)式子就得到f'x =2cosx-2sin2x 然后再把x=pi/4带入就得到了

(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...

(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...

已知函数f(x)=4SinX * Sin^2(π/4+x/2)+Cos2x (1)设W>0为常数,若Y=f(Wx)在区间[-π/2,2π/3]上时增函数,求W的范围!2)设集合A={X| π/6≤X≤2π/3},B={X|f(x)-m<2},若A∪B=B,求实数m的取值范围! 1) f(x)=4sinx*sin^2(π/4+x/2)+cos2x =...

(1)∵f(x)=sin2x+cos2x,∴f(π4)=sinπ2+cosπ2=1(2)f(x)=sin2x+cos2x=2sin(2x+π4),所以函数的最大值为2,此时,2x+π4=2kπ+π2,k∈z,x=kπ+π4,k∈z.

(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)

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-√...

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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