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F x sin 2x sinx 1

f'(x)=1/2cos2x*(2x)]+cosx =cos2x+cosx 显然这是偶函数 cos2x+cosx =2cos²x-1+cosx =2(cosx+1/4)²-9/8 -1

f(x)=sin²x+sinxcosx+1 =(1-cos2x)/2+sin2x/2+1 =½(sin2x-cos2x)+3/2 =√2/2sin(2x-π/4)+3/2 ∴最小正周期是π 单调递增区间2x-π/4∈(2kπ-π/2,2kπ+π/2)→x∈(kπ-π/8,kπ+3π/8) 单调递减区间2x-π/4∈(2kπ+π/2,2kπ+3π/2)→x∈(kπ+3π/8,kπ+7π/8)

应该是导数吧?套书上基本求导公式即可,=1-(2/3)cos2x+acosx

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...

(1)定义域:x≠kπ+π/2 k为整数 值域:(-∞,+∞) (2)奇函数 (3)递增:(2kπ-π/2,2kπ+π/2) 递减:(2kπ+π/2,2kπ+3π/2) 答第一问,为何值域是R: 函数在(2kπ-π/2,2kπ+π/2)连续,lim[x-->2kπ-π/2]sinx/√(1-sin^2x)=-∞ lim[x-->2kπ+π/2]sinx/√(...

f(x)=sin2x-2√3sin^2x+√3+1 =sin2x+√3(-2sin^2x+1)+1 =(sin2x+√3cos2x)+1 =(sin2xcos(π/3)+cos2xsin(π/3))*2+1 =2sin(2x+π/3)+1 最小正周期=π -π/2+2kπ

(1)因为f(x)=0,即a=sin2x?sinx=(sinx?12)2?14,a的最大值等于(?1?12)2 ?14=2,a的最小值等于-14,所以,a∈[?14,2].(2)f(x)=-sin2x+sinx+a=?(sinx?12)2+14+a,∴f(x)∈[?2+a,14+a],又∵1≤f(x)≤174恒成立,∴1≤?2+a14+a≤174,∴3≤a≤4....

(1)∵sinx≠0解得x≠kπ(k∈Z),∴函数f(x)的定义域为{x|x≠kπ(k∈Z)}------------------------(2分)∵f(x)=sin2xsinx+2sinx=2cosx+2sinx=22sin(π4+x)---(4分)∴f(x)的最小正周期T=2π1=2π-----------------------------------(6分)(2...

由 f(x)=sinx+cosx知 f'(x)=cosx-sinx(Ⅰ)由f(x)=2f'(x)得3sinx=cosx,有1+sin2xcos2x?sinxcosx=cos2x+2sin2xcos2x?sinxcosx=9sin2x+2sin2x9sin2x?3sinxsinx=116.(Ⅱ)由x∈[0,2π],g(x)=f(x)?f′(x)4+f(x)+f′(x)=2sinx4+2cosx=s...

f(x)=-sin2x+sinx+a=-(sinx-12)2+a+14.由-1≤sinx≤1可以的出函数f(x)的值域为[a-2,a+14],由1≤f(x)≤174得[a-2,a+14]?[1,174].∴a?2≥1a+14≤174?3≤a≤4,故a的范围是3≤a≤4.

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