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F(x)=Cos^2x%1+sinx+A (1)当F(x)=0有实数解,求A范...

f(x)=-sin²x+sinx+a =-(sinx-1/2)²+a 最大值=a a=1 a>=3 所以 a的取值范围: 3≤a≤17/4

f(x)=a(1+cosx+sinx)+b=2asin(x+π4)+a+b,(1)当a=-1时,由2kπ+π2≤x+π4≤2kπ+32π,得2kπ+π4≤x≤2kπ+54π,∴f(x)的单调增区间为[2kπ+π4,2kπ+54π](k∈Z);(2)∵0≤x≤π,∴π4≤x+π4≤54π,∴-22≤sin(x+π4)≤1,依题意知a≠0,分两种情况考虑:1...

解: (1) f(x)=cos²x-√3sinxcosx+½ =½[1+cos(2x)]-(√3/2)sin(2x)+½ =½cos(2x)-(√3/2)sin(2x)+1 =cos(2x+π/3)+1 最小正周期T=2π/2=π cos(2x+π/3)=1时,f(x)取得最大值f(x)max=1+1=2 cos(2x+π/3)=-1时,f(x)取得最小值f(...

[-2,9/8] 解: f(x) =cos2x+sinx =1-2sin²x+sinx =-2(sinx-1/4)²+9/8 当sinx=1/4时,f(x)取得最大值9/8 当sinx=-1时,f(x)取得最小值-2 ∴ 值域是[-2,9/8]

f(x)=a(cos2x2+12sinx)+b=a2(cosx+sinx)+a2+b=2a2sin(x+π4)+a2+b,(1)当a=2时,f(x)=2sin(x+π4)+b+1,令2kπ+π2≤x+π4≤2kπ+3π2,(k∈Z),解得:2kπ+π4≤x≤2kπ+5π4,(k∈Z),则函数f(x)的单调递减区间为[2kπ+π4,2kπ+5π4](k∈Z)...

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)=1+2cos2x?12cosx+sinx+a2sin(x+π4)=cosx+sinx+a2sin(x+π4)=2sin(x+π4)+a2sin(x+π4)=(2+a2)sin(x+π4).依题意有2+a2=2+3,∴a=±3.故答案为:±3

f(x)=(1+cos2x+8(sinx)^2)/(sin2x) =(1+(1-2(sinx)^2)+8(sinx)^2)/(2sinx*cosx) =(1+3(sinx)^2)/(sinx*cosx) =(4(sinx)^2+(cosx)^2)/(sinx*cosx) =4sinx/cosx + cosx/sinx 设t=sinx/cosx --> cosx/sinx=1/t f(x)=4t+1/t 当00 -->t=sinx/cosx>0 -...

(1)f(x)=a(cosx+1+sinx)+b=2asin(x+π4)+a+b,当a=1时,f(x)=2sin(x+π4)+1+b.∴当2kπ?π2≤x+π4≤2kπ+π2 (k∈Z)时,f(x)是增函数,所以函数f(x)的单调递增区间为[2kπ?3π4,2kπ+π4] (k∈Z);(Ⅱ)由x∈[0,π]得π4≤x+π4≤5π4,∴?22≤sin(x+π4)≤1....

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